JEE Main 5 April 2026 Shift 1 — Physics Paper with Answer Key

The morning shift of 5 April 2026. Below is the Physics section question by question, the chapter split, where the hundred marks sat, how tough it was, and the official paper and answer key from NTA.

The shift

Date 5 April 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 Vernier callipers — the zero error Units and Measurements
27 Which combination of L, C and R has a given dimension Units and Measurements
28 g at the same distance below and above the surface — the change Gravitation
29 Three masses over a pulley — ratio of two tensions Laws of Motion
30 Wedge pulled sideways with a block sliding on it — time to slide Laws of Motion
31 Young’s modulus when the radius and length are doubled Mechanical Properties of Solids
32 Two statements — change in internal energy, and gamma from degrees of freedom Thermodynamics
33 Five statements on the laws of thermodynamics and on variables Thermodynamics
34 Network of resistors — the three branch currents Current Electricity
35 Electron in an electric field — de Broglie wavelength at a later time Dual Nature of Radiation and Matter
36 Hydrogen transition given by the ratio of orbit radii — wavelength Atoms
37 Displacement current in a capacitor — rate of change of voltage Electromagnetic Waves
38 Grid of cells and resistors — current between two terminals Current Electricity
39 Fringe width when the whole set-up goes into a medium Wave Optics
40 Equilateral prism — minimum deviation from the speed of light in it Ray Optics
41 Photoelectric effect — work function from the stopping potential Dual Nature of Radiation and Matter
42 Microscope objective cut into two plano-convex lenses — tube length Ray Optics
43 Two wires and a pan — largest mass that can be added Mechanical Properties of Solids
44 Resistor and capacitor on a.c. — current at two frequencies Alternating Current
45 Logic circuit driving two LEDs — which inputs light both Semiconductor Electronics
46 numeric Cube under a tangential force — sideways displacement Mechanical Properties of Solids
47 numeric Ball dropped — height where the speed matches g in number Motion in a Straight Line
48 numeric Transverse wave on a string — distance between successive crests Waves
49 numeric Charged particle in a magnetic field — unknown in the acceleration Moving Charges and Magnetism
50 numeric Three equal inductors — ratio of magnetic energies Electromagnetic Induction

Chapter by chapter

Chapter Questions Marks
Mechanical Properties of Solids 3 12
Current Electricity 2 8
Dual Nature of Radiation and Matter 2 8
Laws of Motion 2 8
Ray Optics 2 8
Thermodynamics 2 8
Units and Measurements 2 8
Alternating Current 1 4
Atoms 1 4
Electromagnetic Induction 1 4
Electromagnetic Waves 1 4
Gravitation 1 4
Motion in a Straight Line 1 4
Moving Charges and Magnetism 1 4
Semiconductor Electronics 1 4
Wave Optics 1 4
Waves 1 4

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

Elasticity alone carried 12 marks, the most any one chapter has taken in this session.

Where the 100 marks sat

Part of Physics Questions Marks Share
Electrodynamics 6 24 24%
Properties of Matter and Heat 5 20 20%
Mechanics 4 16 16%
Optics and Waves 4 16 16%
Modern Physics and Electronics 4 16 16%
Units, Dimensions and Measurement 2 8 8%

The flattest spread of the session. No part took more than a quarter of the marks, and elasticity, usually a two-mark corner of the syllabus, was worth twelve. A student who had skipped properties of solids gave away a lot here.

How tough it was

Level Questions Share Which ones
Straight recall or one step 8 32% 26, 27, 31, 37, 39, 41, 46, 48
Two or three steps 12 48% 28, 29, 32, 33, 35, 36, 40, 43, 44, 47, 49, 50
Long or awkward 5 20% 30, 34, 38, 42, 45

Two large circuit diagrams, a wedge with a block sliding on it, a microscope objective cut in half and a logic circuit driving two LEDs were the five that ate the clock. The other twenty were fair, and eight of them could be finished in a minute each.

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 5 April 2026 Shift 1 paper, with our own numbers. They are not NTA’s wording. For the exact paper, open NTA’s own file, linked further down — that is the only place it should be read from. Every method below was checked against the official options: put NTA’s numbers back in and the answer lands on one of their four choices.

Q1  ·  Units and Measurement  ·  on the pattern of Q26

In a Vernier callipers, when the two jaws touch, the zero of the Vernier scale lies to the right of the zero of the main scale and the 6th Vernier division lines up with a main scale mark. One main scale division is 1 mm and the Vernier scale has 20 divisions. The zero error is ______.

Answer  0.030 cm, a positive zero error

SolutionLeast count = one main scale division ÷ number of Vernier divisions = 1/20 = 0.05 mm. The Vernier zero sitting to the right of the main scale zero is a positive zero error, equal to 6 × 0.05 = 0.30 mm, that is 0.030 cm. Every reading taken with this instrument is 0.030 cm too large, so this has to be subtracted.

Where it goes wrong  Calling it negative. The Vernier zero to the right of the main scale zero is positive; to the left is negative.

Q2  ·  Units and Dimensions  ·  on the pattern of Q27

L, C and R stand for inductance, capacitance and resistance. Which of these has the dimensions of time? (A) RC   (B) L/R   (C) √(LC)

Answer  All three of them

SolutionA charging capacitor decays with the time constant RC, so RC is a time. A current in an inductor decays with the time constant L/R, so L/R is a time. An LC circuit oscillates at ω = 1/√(LC), and since ω has dimensions of T−1, √(LC) is a time. All three are times.

Where it goes wrong  Trying to break every symbol down into M, L, T and A. Remembering the three standard time constants settles this question in seconds.

Q3  ·  Gravitation  ·  on the pattern of Q28

A person moves from a point 32 km below the earth’s surface to a point 32 km above it. The change in g between the two points is about α % of its value at the surface. The value of α is ______. (Radius of the earth = 6400 km)

Answer  0.5

SolutionBelow the surface gd = g(1 − d/R) = g(1 − 32/6400) = 0.995g. Above it, for small heights, gh ≈ g(1 − 2h/R) = g(1 − 64/6400) = 0.990g. The difference is 0.005g, that is 0.5 % of g.

Where it goes wrong  Using the same formula both ways. Going down the fall is d/R; going up it is twice that, 2h/R.

Q4  ·  Laws of Motion  ·  on the pattern of Q29

A light string over a fixed frictionless pulley carries a 5 kg block on one side. On the other side a 4 kg block hangs from the string, and a 2 kg block hangs from the 4 kg block by a second string. T1 is the tension in the string over the pulley and T2 the tension in the lower string. The ratio T1/T2 is ______.

Answer  3

SolutionEvery part of the system has the same magnitude of acceleration a. The string over the pulley has to carry both blocks on its side, so T1 = (4 + 2)(g − a) if that side goes up, while the lower string carries only the 2 kg block, so T2 = 2(g − a). The bracket is the same in both, so it cancels: T1/T2 = 6/2 = 3.

Where it goes wrong  Working out a first. The ratio only needs the masses each string carries; the acceleration cancels, so the numbers never have to be found.

Q5  ·  Laws of Motion  ·  on the pattern of Q30

A block rests on the frictionless 37° incline of a wedge. The wedge is given a constant horizontal acceleration a, pointing into the slope, so that the block stays at rest relative to the wedge. The value of a is ______ m/s2. (tan37° = 3/4, g = 10 m/s2)

Answer  7.5 m/s2

SolutionSit in the wedge’s frame. The block feels its weight mg down, the normal force N at right angles to the slope, and a pseudo force ma pointing backwards. For it to stay put, take components along the slope: ma cosθ = mg sinθ, so a = g tanθ = 10 × 0.75 = 7.5 m/s2. The mass does not matter.

Where it goes wrong  Pointing the pseudo force the same way as the acceleration. In an accelerating frame it always points the opposite way.

Q6  ·  Mechanical Properties of Solids  ·  on the pattern of Q31

The Young’s modulus of a steel wire of radius r and length L is Y. If both the radius and the length of the wire are doubled, the value of Y becomes ______.

Answer  It stays exactly the same

SolutionYoung’s modulus is a property of the material, not of the piece of wire. It is the ratio of stress to strain, and both of those are already divided by the size: stress is force per unit area and strain is extension per unit length. Change the shape of the wire and the force needed changes, but Y does not.

Where it goes wrong  Confusing Y with the stiffness of the wire. The stiffness AY/L does change with the shape; the modulus Y does not.

Q7  ·  Thermodynamics  ·  on the pattern of Q32

Statement I: For n moles of an ideal gas the change in internal energy can be written as ΔU = nCv(Tf − Ti) = nR(Tf − Ti)/(γ − 1). Statement II: The ratio γ = Cp/Cv is linked to the degrees of freedom f by γ = 1 + 2/f. Choose the correct option.

Answer  Both statements are true, but Statement II is not the explanation of Statement I

SolutionStatement I is true: ΔU = nCvΔT always, and using Cp − Cv = R with γ = Cp/Cv gives Cv = R/(γ − 1), which is the second form. Statement II is also true, since Cv = fR/2 leads to γ = 1 + 2/f. But II is about counting the degrees of freedom of the molecule, while I rests on Cp − Cv = R. II is a separate fact, not the reason for I.

Where it goes wrong  Marking II as the explanation because both involve γ. Sharing a symbol is not the same as explaining.

Q8  ·  Thermodynamics  ·  on the pattern of Q33

Which of these statements are correct? A. The zeroth law of thermodynamics leads to the idea of temperature. B. The first law of thermodynamics leads to the idea of internal energy. C. In the isothermal expansion of an ideal gas ΔQ = ΔW. D. The product of an intensive and an extensive variable is extensive. E. An extensive variable divided by mass is still extensive.

Answer  A, B, C and D only

SolutionA and B are the standard statements of what those two laws give us. For C, an isothermal change of an ideal gas keeps the temperature fixed, so ΔU = 0 and the first law gives ΔQ = ΔW: true. For D, pressure (intensive) times volume (extensive) is energy, which is extensive: true. E is false, because dividing an extensive variable by mass is exactly how an intensive one is made, as with density and specific heat.

Where it goes wrong  Reading E too fast. Mass is itself extensive, so dividing one extensive quantity by another removes the size dependence.

Q9  ·  Current Electricity  ·  on the pattern of Q34

A 12 V battery of negligible internal resistance is joined to a 2 Ω resistor in series with a parallel pair of 6 Ω and 3 Ω resistors. Find the current drawn from the battery and the current in each of the two parallel resistors.

Answer  3 A from the battery; 1 A in the 6 Ω and 2 A in the 3 Ω

SolutionThe parallel pair is 6 × 3 / (6 + 3) = 2 Ω. The whole circuit is 2 + 2 = 4 Ω, so the battery current is 12/4 = 3 A. All 3 A pass through the series resistor, dropping 6 V, so the remaining 6 V sits across the parallel pair. In the 6 Ω the current is 6/6 = 1 A, and in the 3 Ω it is 6/3 = 2 A. They add back to 3 A, which is the check.

Where it goes wrong  Splitting the current equally between the two branches. The smaller resistance always takes the larger share, in inverse proportion.

Q10  ·  Dual Nature of Radiation  ·  on the pattern of Q35

An electron of mass m moves in a uniform electric field E = −3E0j (E0 > 0) with an initial velocity v0j. Writing λ0 = h/(2mv0), its de Broglie wavelength at time t is ______. (e is the size of the electron’s charge)

Answer  2λ0 / [1 + 3eE0t/(mv0)]

SolutionThe electron carries a negative charge, so the force on it is (−e)(−3E0j) = +3eE0j, along its initial motion. It speeds up with a = 3eE0/m, so v(t) = v0 + 3eE0t/m. Then λ = h/mv = h / [mv0(1 + 3eE0t/mv0)]. Since h/(mv0) = 2λ0, the answer is 2λ0/[1 + 3eE0t/(mv0)].

Where it goes wrong  Letting the electron slow down. Two minus signs meet here, so the force ends up along the motion and the wavelength shrinks.

Q11  ·  Atoms  ·  on the pattern of Q36

In a hydrogen atom an electron drops from a higher orbit i to a lower orbit f, with ri : rf = 9 : 1. The wavelength of the photon given out is about ______ nm. (Rydberg constant R = 1.097 × 107 /m)

Answer  about 103 nm

SolutionIn the Bohr model the radius goes as n2, so ri/rf = 9 means ni2/nf2 = 9, that is ni = 3 and nf = 1. Then 1/λ = R(1/nf2 − 1/ni2) = 1.097 × 107 (1 − 1/9) = 9.75 × 106 /m, so λ = 1.03 × 10−7 m = 103 nm. This is the Lyman series, in the ultraviolet.

Where it goes wrong  Taking the ratio of radii as the ratio of n. The radius carries n squared, so the square root has to be taken first.

Q12  ·  Electromagnetic Waves  ·  on the pattern of Q37

A displacement current of 3.0 A is set up in the space between the plates of a 5 μF parallel plate capacitor. The rate of change of the potential difference across the plates is ______ × 105 V/s.

Answer  6.0 × 105 V/s

SolutionThe displacement current between the plates equals the conduction current in the wires, and that is i = C dV/dt. So dV/dt = i/C = 3.0 / (5 × 10−6) = 6.0 × 105 V/s.

Where it goes wrong  Going back to Maxwell’s integral and the changing flux. The one-line result i = C dV/dt does the whole job.

Q13  ·  Current Electricity  ·  on the pattern of Q38

Three identical branches are joined side by side between terminals A and B. Each branch has two 5 V cells in series with a 3 Ω resistor, all the cells facing the same way and of negligible internal resistance. A 4 Ω resistor is connected between A and B. The current through that 4 Ω resistor is ______ A.

Answer  2 A

SolutionEach branch is a 10 V source with 3 Ω in series. Three identical sources in parallel keep the same emf, 10 V, while their resistances combine to 3/3 = 1 Ω. The circuit is then 10 V across 1 + 4 = 5 Ω, so the current is 10/5 = 2 A.

Where it goes wrong  Adding the three branch emfs to get 30 V. Identical sources in parallel share the current; they do not raise the voltage.

Q14  ·  Wave Optics  ·  on the pattern of Q39

In a Young’s double slit experiment the fringe width in air is 3.0 μm. The whole set-up is then placed in a liquid of refractive index 1.25. The new fringe width is ______ μm.

Answer  2.4 μm

SolutionThe fringe width is β = λD/d. Putting the set-up in a liquid shortens the wavelength to λ/n, while D and d are unchanged, so the fringe width becomes β/n = 3.0/1.25 = 2.4 μm. The pattern simply closes up.

Where it goes wrong  Multiplying by n instead of dividing. Light slows in the medium, so the wavelength and the fringes both get smaller.

Q15  ·  Ray Optics  ·  on the pattern of Q40

A ray of light passes through an equilateral prism at minimum deviation, and the angle of minimum deviation is 30°. The speed of light inside the prism material is about ______ × 108 m/s. (c = 3 × 108 m/s)

Answer  about 2.12 × 108 m/s

SolutionAt minimum deviation n = sin[(A + Dm)/2] / sin(A/2). For an equilateral prism A = 60°, so n = sin45°/sin30° = 0.7071/0.5 = 1.414, which is √2. Then the speed inside is v = c/n = 3 × 108/1.414 = 2.12 × 108 m/s.

Where it goes wrong  Using A alone in the top line. Both the prism angle and the deviation go into the numerator, and only the prism angle into the denominator.

Q16  ·  Dual Nature of Radiation  ·  on the pattern of Q41

Light of wavelength 400 nm falls on a metal and the stopping potential is 0.5 V. The work function of the metal is α × 10−19 J. The value of α is about ______. (h = 6.62 × 10−34 J·s, e = 1.6 × 10−19 C, c = 3 × 108 m/s)

Answer  about 4.17

SolutionThe photon energy is E = hc/λ = (6.62 × 10−34 × 3 × 108) / (4 × 10−7) = 4.97 × 10−19 J. The maximum kinetic energy is eV0 = 1.6 × 10−19 × 0.5 = 0.8 × 10−19 J. Einstein’s equation gives φ = E − eV0 = 4.97 − 0.80 = 4.17, in units of 10−19 J.

Where it goes wrong  Subtracting 0.5 instead of eV0. The stopping potential is in volts, so it has to be multiplied by the electron charge first.

Q17  ·  Ray Optics  ·  on the pattern of Q42

A symmetric biconvex lens of focal length 20 cm is cut into two identical plano-convex lenses by a cut right across it, perpendicular to the principal axis. The focal length of each plano-convex lens is ______ cm.

Answer  40 cm

SolutionFor the original lens, 1/f = (n − 1)(1/R + 1/R) = 2(n − 1)/R. After the cut each half has one curved face and one flat face, and a flat face has an infinite radius, so 1/f′ = (n − 1)(1/R) = half of 1/f. That makes f′ = 2f = 40 cm.

Where it goes wrong  Cutting the focal length in half. Taking away one curved surface takes away half the bending power, so the focal length doubles.

Q18  ·  Mechanical Properties of Solids  ·  on the pattern of Q43

Two steel wires hang one below the other, both with breaking stress 12 × 108 N/m2. The upper wire has cross-section 0.006 cm2 and the lower 0.003 cm2. A 30 kg block hangs from the upper wire, a 10 kg block hangs below it from the lower wire, and a light pan hangs from the bottom. The largest mass that can be put in the pan without breaking either wire is ______ kg. (g = 10 m/s2)

Answer  26 kg

SolutionTake each wire in turn. The upper wire can hold F = stress × area = 12 × 108 × 6 × 10−7 = 720 N, that is 72 kg. It already carries 30 + 10 = 40 kg, so it allows 32 kg more. The lower wire can hold 12 × 108 × 3 × 10−7 = 360 N, that is 36 kg, and it already carries 10 kg, so it allows 26 kg more. The weaker limit decides: 26 kg.

Where it goes wrong  Checking only the upper wire because it carries more. The lower one is thinner, and here it is the one that gives way first.

Q19  ·  Alternating Current  ·  on the pattern of Q44

An a.c. source of angular frequency ω is connected across a resistor R and a capacitor C in series, and the current is I. When the frequency is lowered to ω/3 with the voltage unchanged, the current falls to I/2. The ratio of the resistance to the reactance, R/XC, at the original frequency is ______.

Answer  √(5/3)

SolutionWrite X for XC at ω. Since XC = 1/ωC, lowering the frequency to ω/3 raises the reactance to 3X. The impedances are Z1 = √(R2 + X2) and Z2 = √(R2 + 9X2). The current halves, so Z2 = 2Z1, and squaring gives R2 + 9X2 = 4R2 + 4X2. That leaves 5X2 = 3R2, so R/X = √(5/3).

Where it goes wrong  Making the reactance smaller when the frequency drops. For a capacitor it is the other way round: lower frequency means larger reactance.

Q20  ·  Semiconductor Electronics  ·  on the pattern of Q45

In a logic circuit, inputs A and B feed an OR gate. LED-1 is driven by the output of that OR gate. That same output and input C feed an AND gate, whose output drives LED-2. Which input combination makes both LEDs glow?

Answer  A = 0, B = 1, C = 1 (any case with at least one of A and B equal to 1, and C = 1)

SolutionLED-1 glows when the OR output is 1, which needs A or B to be 1. LED-2 glows when (A + B)·C is 1, which needs the OR output to be 1 and C to be 1 as well. So both glow whenever at least one of A and B is 1 and C is 1. Of the usual choices, A = 0, B = 1, C = 1 fits; A = 1, B = 0, C = 0 fails because C is 0.

Where it goes wrong  Checking only the final output. Each LED has its own condition, and both have to hold at once.

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

A cube of side 4 cm is made of a material of modulus of rigidity 2 × 105 N/m2. A tangential force of 16 N is applied to its top face while the bottom face is held fixed. The sideways displacement of the top face is ______ mm.

Answer  2 mm

SolutionThe shearing stress is F/A = 16 / (0.04)2 = 16/1.6 × 10−3 = 104 N/m2. The shearing strain is stress ÷ modulus of rigidity = 104 / 2 × 105 = 0.05. The strain is the displacement divided by the height, so the displacement is 0.05 × 0.04 m = 2 × 10−3 m = 2 mm.

Where it goes wrong  Using the volume or the side length in place of the face area. The shearing stress uses the area of the face the force acts on.

Q22  ·  Motion in a Straight Line  ·  numeric  ·  on the pattern of Q47

A ball is dropped from rest from a height of 20 m above the ground. The height above the ground at which the number measuring its speed in m/s equals the number measuring g in m/s2 is ______ m. (g = 10 m/s2, neglect air resistance)

Answer  15 m

SolutionThe condition is that the speed is 10 m/s. Starting from rest, v2 = 2g×(distance fallen), so 100 = 20 × distance and the ball has fallen 5 m. It started at 20 m, so it is 20 − 5 = 15 m above the ground.

Where it goes wrong  Giving the distance fallen rather than the height above the ground. The question asks where the ball is, not how far it has come.

Q23  ·  Waves  ·  numeric  ·  on the pattern of Q48

A transverse wave on a string is y = 4 sin(50t + 0.025x + π/3), where x and y are in cm and t in seconds. The least distance between two successive crests is about ______ cm. (π = 3.14)

Answer  about 251 cm

SolutionSuccessive crests are one wavelength apart. Comparing with y = A sin(ωt + kx + φ), the wave number is k = 0.025 cm−1. Then λ = 2π/k = 6.28/0.025 = 251.2 cm, so about 251 cm.

Where it goes wrong  Using the coefficient of t. That one gives the angular frequency and so the time period; the wavelength comes from the coefficient of x.

Q24  ·  Moving Charges and Magnetism  ·  numeric  ·  on the pattern of Q49

A charged particle moving in a uniform magnetic field B = (2i + 3j) T has an acceleration (6i − xj) m/s2. The value of x is ______.

Answer  4

SolutionThe magnetic force is qv × B, which is always at right angles to B. The acceleration is along that force, so it must also be perpendicular to B, and their dot product is zero: (6)(2) + (−x)(3) = 0. That gives 12 = 3x, so x = 4.

Where it goes wrong  Reaching for the velocity. It is never given and never needed; the right angle between the acceleration and the field settles it in one line.

Q25  ·  Electromagnetic Induction  ·  numeric  ·  on the pattern of Q50

Three inductors of the same inductance L are connected so that L1 is in series with a parallel pair L2 and L3. A steady current I flows into the circuit. If Ui is the magnetic energy stored in the whole circuit and Uf that stored in L2 alone, the ratio Ui/Uf is ______. (Ignore mutual inductance)

Answer  6

SolutionAll of I passes through L1, then splits equally between L2 and L3 because they are identical, so each carries I/2. The energy in an inductor is ½Li2. For L1 that is ½LI2, and for each of L2 and L3 it is ½L(I/2)2 = LI2/8. The total is ½LI2 + LI2/8 + LI2/8 = (3/4)LI2. Dividing by LI2/8 gives 6.

Where it goes wrong  Sending the full current I through L2. It only gets half, and since the energy goes as the square of the current, that is a factor of four.

The official paper and answer key

Both are NTA’s own files, opened straight from the government store. Nothing here is a copy.

Official question paper →Official answer key →

Nearby shifts: 4 April Shift 1 · 4 April Shift 2.

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