JEE Main 4 April 2026 Shift 1 — Physics Paper with Answer Key

The morning shift of 4 April 2026, the second exam day of JEE Main 2026 Session 2. 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 4 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 Screw gauge — least count from the pitch and the divisions Units and Measurements
27 Bulk modulus of water — percentage drop in volume Mechanical Properties of Solids
28 Rough 45° incline takes longer than a smooth one — coefficient of friction Laws of Motion
29 Light nuclei joining to form a heavier one — energy released Nuclei
30 Sphere cut into a small sphere and a disc — ratio of moments of inertia Rotational Motion
31 Two projectiles at different angles — ratio of their ranges Motion in a Plane
32 Stopping potential against frequency for three metals — which gives the fastest electrons Dual Nature of Radiation and Matter
33 Single slit — distance between two minima on the screen Wave Optics
34 Two wires joined end to end under a load — net extension Mechanical Properties of Solids
35 Two gases mixed — temperature of the mixture Kinetic Theory of Gases
36 Gas on a given pressure–volume relation — difference in temperature of two samples Thermodynamics
37 Raising the range of a voltmeter Current Electricity
38 Two four-bit numbers through a logic circuit — the output Semiconductor Electronics
39 Rod lying on the axis of a concave mirror — length of the image Ray Optics
40 Capacitor plates pulled apart — how the stored energy changes Electrostatic Potential and Capacitance
41 Flat spiral coil of many turns — magnetic moment Moving Charges and Magnetism
42 Capacitor, resistor and two diodes — time constant Semiconductor Electronics
43 Network of cells and resistors — voltage and current between two points Current Electricity
44 Telescope objective — diameter needed for a given resolution Wave Optics
45 Light at the Brewster angle — the reflected wave Wave Optics
46 numeric Two forces acting together — work done over a given displacement Work, Energy and Power
47 numeric Soap bubble — work needed to grow its radius Mechanical Properties of Fluids
48 numeric SHM given as v2 against x — time period Oscillations
49 numeric Force that grows with time — power at a given instant Work, Energy and Power
50 numeric Series LCR with a given power factor — difference of the reactances Alternating Current

Chapter by chapter

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

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

Nineteen chapters for twenty-five questions.

Where the 100 marks sat

Part of Physics Questions Marks Share
Mechanics 6 24 24%
Properties of Matter and Heat 5 20 20%
Electrodynamics 5 20 20%
Optics 4 16 16%
Modern Physics and Electronics 4 16 16%
Units, Dimensions and Measurement 1 4 4%

This is a different shape from 2 April. Optics and modern physics together took 32 marks, well above the 16 to 24 they usually get, and mechanics was the lightest it has been in this session. A student who had left optics and semiconductors for later lost real marks here.

How tough it was

Level Questions Share Which ones
Straight recall or one step 7 28% 26, 27, 31, 32, 37, 47, 49
Two or three steps 11 44% 28, 29, 33, 34, 35, 38, 41, 44, 46, 48, 50
Long or awkward 7 28% 30, 36, 39, 40, 42, 43, 45

Harder than either 2 April shift. Seven heavy questions instead of four. The sphere cut into a sphere and a disc, the gas on an odd pressure–volume law, the rod on the mirror axis, the moving capacitor plates, the diode circuit, the cell network and the Brewster angle wave all needed real setting up. Anyone who spent too long on the first of these ran out of time at the end.

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 4 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 screw gauge the circular scale moves 3 mm along the main scale in 6 complete rotations. The circular scale carries 50 divisions. The least count of the screw gauge is ______ mm.

Answer  1 × 10−2 mm

SolutionPitch is the distance moved in one rotation: 3 mm ÷ 6 = 0.5 mm. Least count = pitch ÷ number of circular scale divisions = 0.5 ÷ 50 = 0.01 mm, that is 1 × 10−2 mm.

Where it goes wrong  Treating 3 mm as the pitch. The pitch is for one rotation, so the total travel must be divided by the number of rotations first.

Q2  ·  Mechanical Properties of Solids  ·  on the pattern of Q27

The pressure on a sample of water is raised by 4.2 × 107 N/m2. The percentage decrease in its volume is ______. (Bulk modulus of water = 2.1 × 109 N/m2)

Answer  2 %

SolutionBulk modulus is B = −ΔP / (ΔV/V), so the fractional change in volume is ΔV/V = ΔP/B = 4.2 × 107 / 2.1 × 109 = 0.02. As a percentage that is 2 %.

Where it goes wrong  Forgetting to turn the fraction into a percentage and answering 0.02.

Q3  ·  Laws of Motion  ·  on the pattern of Q28

A block takes 25 % more time to slide from the top to the bottom of a rough inclined plane than it takes on an identical smooth plane. Both planes are at 45° to the horizontal. The coefficient of kinetic friction between the block and the rough plane is ______.

Answer  0.36

SolutionOn the smooth plane a1 = g sin45°. On the rough one a2 = g sin45° − μg cos45°, and since the angle is 45° the sine and cosine are equal, so a2 = a1(1 − μ). The same distance from rest gives t proportional to 1/√a, so t2/t1 = √(a1/a2) = 1.25. Squaring, a1/a2 = 1.5625, so 1 − μ = 1/1.5625 = 0.64 and μ = 0.36.

Where it goes wrong  Taking the ratio of times as the ratio of accelerations. Time goes as one over the square root of acceleration, so the ratio must be squared.

Q4  ·  Nuclei  ·  on the pattern of Q29

Two nuclei of mass number 4 combine with one nucleus of mass number 6 to give a nucleus of mass number 14. The binding energies per nucleon for mass numbers 4, 6 and 14 are 7.0 MeV, 5.5 MeV and 7.5 MeV. The energy released in the process is ______ MeV.

Answer  16 MeV

SolutionTotal binding energy before = 2 × (4 × 7.0) + (6 × 5.5) = 56 + 33 = 89 MeV. After = 14 × 7.5 = 105 MeV. The products are more tightly bound, and the extra binding energy comes out as released energy: 105 − 89 = 16 MeV.

Where it goes wrong  Using the binding energies per nucleon directly instead of multiplying each by its own mass number first.

Q5  ·  Rotational Motion  ·  on the pattern of Q30

A solid sphere of mass M and radius R is cut into two unequal pieces. The smaller piece, of mass M/64, is made into a solid sphere of radius r, and the larger piece is made into a circular disc of radius 2R. If I1 is the moment of inertia of the small sphere about a diameter and I2 that of the disc about its own diameter, then I2/I1 = ______.

Answer  2520

SolutionThe density does not change, so volume goes with mass: (4/3)πr3 = (1/64)(4/3)πR3, which gives r = R/4. Then I1 = (2/5)(M/64)(R/4)2 = MR2/2560. The disc has mass 63M/64 and radius 2R, and about its own diameter I2 = (1/4)ma2 = (1/4)(63M/64)(4R2) = 63MR2/64. Dividing, I2/I1 = (63/64) × 2560 = 2520.

Where it goes wrong  Using (1/2)ma2 for the disc. That is about the axis through its centre and perpendicular to it; about a diameter it is half of that.

Q6  ·  Motion in a Plane  ·  on the pattern of Q31

Two projectiles are thrown with the same speed at 15° and 45° to the horizontal. The ratio of their ranges is 1 : x. The value of x is ______.

Answer  2

SolutionThe range of a projectile is R = u2sin2θ/g, and the speed is the same for both, so the ranges go as sin2θ. For 15°, sin30° = 0.5. For 45°, sin90° = 1. The ratio is 0.5 : 1, which is 1 : 2, so x = 2.

Where it goes wrong  Using sinθ instead of sin2θ. The range formula carries twice the angle.

Q7  ·  Dual Nature of Radiation  ·  on the pattern of Q32

Stopping potential is plotted against the frequency of the incident light for three metals X1, X2 and X3. The three lines are parallel and meet the frequency axis at 1.0, 1.5 and 2.0 × 1014 Hz respectively. For the same incident frequency, which metal gives out electrons with the greatest kinetic energy?

Answer  X1

SolutionThe point where a line cuts the frequency axis is the threshold frequency, and the work function is φ = hν0. The smallest threshold belongs to X1, so X1 has the smallest work function. Einstein’s equation gives KE = hν − φ, so for the same ν the smallest φ leaves the most kinetic energy. The answer is X1.

Where it goes wrong  Looking at the slope. All three lines have the same slope h/e; only the intercept separates the metals.

Q8  ·  Wave Optics  ·  on the pattern of Q33

A slit of width a is lit by light of wavelength λ. The screen is at a distance D from the slit. The distance on the screen between the 1st and the 4th minima on the same side of the centre is ______.

Answer  3λD/a

SolutionFor single slit diffraction the minima sit where a sinθ = nλ. For small angles sinθ = y/D, so yn = nλD/a. The 4th minimum is at 4λD/a and the 1st at λD/a, so the gap between them is 3λD/a.

Where it goes wrong  Using the double slit condition a sinθ = (n + ½)λ. For a single slit the whole number multiples give the minima, not the maxima.

Q9  ·  Mechanical Properties of Solids  ·  on the pattern of Q34

A wire A of length 0.5 m and Young’s modulus 1 × 1011 N/m2 is joined end to end to a wire B of the same radius, whose length and Young’s modulus are both twice those of A. The pair hangs from a rigid support and a 1 kg load is fixed to the free end. The net increase in the length of the combination is about ______ mm. (radius of both wires 0.2 mm, g = 10 m/s2, the mass of the wires is neglected)

Answer  about 0.8 mm

SolutionThe same force passes through both wires and they have the same radius, so the stress is the same in both: F/A = 10 / (π × (2 × 10−4)2) = 7.96 × 107 N/m2. For a wire, ΔL = (stress × L)/Y. For A, ΔL = 7.96 × 107 × 0.5 / 1011 = 0.40 mm. For B both L and Y double, so the two changes cancel and ΔL is again 0.40 mm. The total is about 0.80 mm.

Where it goes wrong  Assuming the longer wire must stretch more. Here its larger Young’s modulus cancels its larger length exactly.

Q10  ·  Kinetic Theory of Gases  ·  on the pattern of Q35

One gas has n1 moles at temperature T1, volume V1 and pressure P1. A second gas has n2 moles at T2, V2 and P2. They are mixed, and the mixture has pressure P and volume V. The temperature of the mixture is ______.

Answer  T = T1T2PV / (T2P1V1 + T1P2V2)

SolutionNothing is created or destroyed, so the moles add: n = n1 + n2. Writing each with PV = nRT gives PV/RT = P1V1/RT1 + P2V2/RT2. Cancelling R, PV/T = P1V1/T1 + P2V2/T2. Rearranging for T gives T = T1T2PV / (T2P1V1 + T1P2V2).

Where it goes wrong  Taking the plain average (T1 + T2)/2. That is only right when the two samples have equal numbers of moles.

Q11  ·  Thermodynamics  ·  on the pattern of Q36

An ideal gas follows P = P0[1 + (V0/V)2]−1, where P0 and V0 are constants. Two samples A and B, of 3 moles each, are at volumes V0 and 2V0 on this curve. The difference in their temperatures, TB − TA, is ______. (R = gas constant)

Answer  11P0V0 / 30R

SolutionFrom PV = nRT, T = PV/nR = P0V / {nR[1 + (V0/V)2]}. For A, V = V0 and the bracket is 2, so TA = P0V0/(3R × 2) = P0V0/6R. For B, V = 2V0 and the bracket is 1 + ¼ = 5/4, so TB = 2P0V0 × 4 / (3R × 5) = 8P0V0/15R. Subtracting, 8/15 − 1/6 = 16/30 − 5/30 = 11/30, so the difference is 11P0V0/30R.

Where it goes wrong  Putting V0/V the wrong way up in the bracket. For sample B it is ½, so its square is ¼, not 4.

Q12  ·  Current Electricity  ·  on the pattern of Q37

A voltmeter of internal resistance x Ω reads up to 25 V. To make it read up to 100 V, what change is needed?

Answer  Connect a resistor of 3x Ω in series with the voltmeter

SolutionA voltmeter is made to read higher by putting a large resistance in series, so that most of the extra voltage falls across it. The rule is R = x(Vnew/Vold − 1). Here Vnew/Vold = 100/25 = 4, so R = x(4 − 1) = 3x Ω, in series.

Where it goes wrong  Putting the resistor in parallel. Parallel is for an ammeter, where a shunt carries the extra current; a voltmeter needs a series multiplier.

Q13  ·  Semiconductor Electronics  ·  on the pattern of Q38

Two four-bit binary numbers A = 1100 and B = 1010 are fed bit by bit into an OR gate, and the output of that gate is passed through a NOT gate. The final output Y is ______.

Answer  Y = 0001

SolutionWork bit by bit. A OR B: 1 or 1 = 1, 1 or 0 = 1, 0 or 1 = 1, 0 or 0 = 0, which gives 1110. Passing that through a NOT gate flips every bit: 0001.

Where it goes wrong  Adding the two numbers as binary arithmetic. A logic gate works on each bit on its own, with no carry.

Q14  ·  Ray Optics  ·  on the pattern of Q39

A rod of length 15 cm lies along the principal axis of a concave mirror of focal length 10 cm, with its nearer end 20 cm from the mirror. The length of the image is ______ cm.

Answer  6 cm

SolutionFind the image of each end separately with 1/v + 1/u = 1/f, taking f = −10 cm. Nearer end at u = −20: 1/v = −1/10 + 1/20 = −1/20, so v = −20 cm. Far end at u = −35: 1/v = −1/10 + 1/35 = −5/70, so v = −14 cm. The two image points are 20 cm and 14 cm from the mirror, so the image is 6 cm long.

Where it goes wrong  Multiplying the rod’s length by a single magnification. The magnification is different at the two ends, so each end has to be traced on its own.

Q15  ·  Electrostatic Potential and Capacitance  ·  on the pattern of Q40

A parallel plate air capacitor is charged and then disconnected from the battery. Its plates are then pulled apart at a steady speed v. If x is the separation at any instant, the rate of change of the stored electrostatic energy is proportional to xα. The value of α is ______.

Answer  0

SolutionOnce the battery is removed the charge Q is fixed. The energy is U = Q2/2C and C = ε0A/x, so U = Q2x / 2ε0A. Differentiating, dU/dt = (Q2/2ε0A)(dx/dt) = Q2v / 2ε0A. There is no x left in it, so the rate is constant and α = 0.

Where it goes wrong  Keeping the battery connected in your head. At constant voltage U = ε0AV2/2x and the rate goes as x−2; at constant charge it does not depend on x at all.

Q16  ·  Moving Charges and Magnetism  ·  on the pattern of Q41

An insulated wire is wound into a flat spiral coil of N = 100 turns. The innermost turn has radius 2 cm and the outermost 4 cm. A current of 30 mA flows through it. The magnetic moment of the coil is α × 10−2 A·m2. The value of α is about ______.

Answer  about 0.88

SolutionEach turn has a different radius, so take the mean of the squared radii across the spiral. For radii running evenly from r1 to r2 the mean of r2 is (r12 + r1r2 + r22)/3. So m = NIπ(r12 + r1r2 + r22)/3 = 100 × 0.03 × π × (4 + 8 + 16) × 10−4/3 = 3π × 28 × 10−4/3 = 8.8 × 10−3 A·m2, which is 0.88 × 10−2.

Where it goes wrong  Using the plain average radius (r1 + r2)/2 and squaring it. The area depends on r2, so the average has to be taken of r2, not of r.

Q17  ·  Semiconductor Electronics  ·  on the pattern of Q42

A capacitor of 30 μF, a resistor of 200 Ω and two identical diodes are joined in one loop, with the two diodes in parallel with each other and that pair in series with the resistor and the capacitor. The forward resistance of a diode is 20 Ω and its reverse resistance is very large. The time constant of the circuit is α × 10−3 s. The value of α is ______.

Answer  6.6

SolutionThe two diodes face opposite ways, so for a given direction of current only one is forward biased and conducts; the other is as good as an open wire. The conducting path is therefore 20 Ω in series with 200 Ω, that is 220 Ω. The time constant is τ = RC = 220 × 30 × 10−6 = 6.6 × 10−3 s, so α = 6.6.

Where it goes wrong  Putting the two diode resistances in parallel to get 10 Ω. Only one diode conducts at a time, so there is nothing to put in parallel.

Q18  ·  Current Electricity  ·  on the pattern of Q43

Three identical cells of emf 18 V and negligible internal resistance are joined in parallel. The combination is connected through a 1 Ω resistor to a parallel pair of 3 Ω and 6 Ω resistors. The voltage across the parallel pair and the current drawn from the cells are ______.

Answer  12 V and 6 A

SolutionIdentical cells in parallel give the same emf as one of them, 18 V. The parallel pair is 3 × 6 / (3 + 6) = 2 Ω. The total resistance is 1 + 2 = 3 Ω, so the current from the cells is 18/3 = 6 A. The voltage across the parallel pair is 6 × 2 = 12 V, with the other 6 V dropped across the 1 Ω resistor.

Where it goes wrong  Adding the emfs of the three cells to get 54 V. Cells in parallel share the current, they do not add their voltages.

Q19  ·  Wave Optics  ·  on the pattern of Q44

A telescope with an objective of diameter D is used to look at a distant star emitting light of wavelength 500 nm, at a resolution of 2.5 × 10−7 radian. The value of D is ______ cm.

Answer  244 cm

SolutionThe limit of resolution of a telescope is θ = 1.22λ/D. Rearranging, D = 1.22λ/θ = 1.22 × 500 × 10−9 / 2.5 × 10−7 = 1.22 × 2 = 2.44 m = 244 cm.

Where it goes wrong  Dropping the factor 1.22. That factor is what separates the circular aperture from a slit.

Q20  ·  Wave Optics  ·  on the pattern of Q45

Unpolarised light falls on the flat surface of a transparent medium of refractive index √3 kept in air, at the Brewster angle. The angle of incidence, the angle between the reflected and refracted rays, and the state of the reflected light are ______.

Answer  60°; the reflected and refracted rays are 90° apart; the reflected light is fully plane polarised, with its vibrations perpendicular to the plane of incidence

SolutionBrewster’s law says tanθB = n, so tanθB = √3 and θB = 60°. At this angle the reflected and refracted rays are always at right angles to each other. The reflected beam has only the component whose electric field is perpendicular to the plane of incidence, so it is completely plane polarised.

Where it goes wrong  Saying the refracted ray is polarised too. The refracted ray is only partly polarised; it is the reflected one that is fully polarised at this angle.

Q21  ·  Work, Energy and Power  ·  numeric  ·  on the pattern of Q46

A 1 kg block is acted on at the same time by two forces (3i + 2j + k) N and (i − 2j + 3k) N, and moves 10 m along the direction (3i + 4k)/5. The work done in this process is ______ J.

Answer  56 J

SolutionAdd the forces first: (3 + 1)i + (2 − 2)j + (1 + 3)k = 4i + 4k N. The displacement is 10 m along a unit vector, which is 10 × (3i + 4k)/5 = 6i + 8k m. Work is the dot product: W = 4(6) + 0(0) + 4(8) = 24 + 32 = 56 J.

Where it goes wrong  Finding the work of each force separately and then adding the magnitudes. Adding the vectors first is quicker and avoids sign slips.

Q22  ·  Mechanical Properties of Fluids  ·  numeric  ·  on the pattern of Q47

The surface tension of a soap solution is 5 × 10−2 N/m. The work needed to increase the radius of a soap bubble from 1 cm to 2 cm is α × 10−6 J. The value of α is ______. (π = 22/7)

Answer  377

SolutionA bubble has two surfaces, so W = 2T × 4π(r22 − r12) = 8πT(r22 − r12). Here r22 − r12 = 4 × 10−4 − 1 × 10−4 = 3 × 10−4 m2. So W = 8 × (22/7) × 0.05 × 3 × 10−4 = 3.77 × 10−4 J = 377 × 10−6 J, giving α = 377.

Where it goes wrong  Using 4πT instead of 8πT. A drop has one surface, a bubble has two.

Q23  ·  Oscillations  ·  numeric  ·  on the pattern of Q48

The velocity of a particle in simple harmonic motion along the x-axis obeys v2 = 36 − 9x2, where x is the displacement. The time period of the motion is ______ s.

Answer  2π/3 s, about 2.09 s

SolutionFor SHM, v2 = ω2(A2 − x2). Write the given relation in the same shape: v2 = 9(4 − x2). Comparing, ω2 = 9 so ω = 3 rad/s, and A = 2 m. The time period is T = 2π/ω = 2π/3 s.

Where it goes wrong  Reading ω straight off the constant 36. The coefficient of x2 is what gives ω2; the constant gives ω2A2.

Q24  ·  Work, Energy and Power  ·  numeric  ·  on the pattern of Q49

A body of mass 3 kg starts from rest under a time dependent force F = (3t i + 6t2 j) N, where i and j are unit vectors along x and y. The power produced by the force at t = 2 s is ______ W.

Answer  140 W

SolutionAcceleration is a = F/m = (t i + 2t2 j) m/s2. Starting from rest, integrate: v = (t2/2) i + (2t3/3) j. At t = 2 s, v = 2i + (16/3)j and F = 6i + 24j. Power is the dot product: P = 6(2) + 24(16/3) = 12 + 128 = 140 W.

Where it goes wrong  Using v = at with the value of a at t = 2 s. The acceleration is changing, so the velocity has to be found by integrating.

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

An inductor, a capacitor and a 60 Ω resistor are joined in series across an a.c. supply. The power factor of the circuit is 0.6. The difference between the inductive and the capacitive reactance is ______ Ω.

Answer  80 Ω

SolutionThe power factor is cosφ = R/Z, so 0.6 = 60/Z and Z = 100 Ω. For a series LCR circuit Z2 = R2 + (XL − XC)2, so (XL − XC)2 = 1002 − 602 = 6400 and the difference is 80 Ω.

Where it goes wrong  Subtracting R from Z to get 40 Ω. The three quantities form a right triangle, so they add as squares, not directly.

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 →

Other shifts of this session: 2 April Shift 1 · 2 April Shift 2.

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

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