JEE Main 2025 — 23 January, Shift 1 — Physics Question Paper

The morning shift of 23 January 2025, the third Physics paper of JEE Main 2025 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

Exam JEE Main 2025, Session 1
Date 23 January 2025
Shift Shift 1 — 9:00 AM to 12:00 PM IST
Paper Paper 1 (B.E./B.Tech)
Physics 25 questions — 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
1 RC circuit at the instant the key is closed Current Electricity
2 Magnetic field of a given electromagnetic wave Electromagnetic Waves
3 Two samples, one decaying three times faster Nuclei
4 Floating cube pushed down, time period Oscillations
5 Statements about self-inductance Electromagnetic Induction
6 Velocity-time graph, distance covered Motion in a Straight Line
7 Sphere rolling down two different inclines Rotational Motion
8 Circuit readings, resistance and potentials Current Electricity
9 Spherical surface between air and glass Ray Optics
10 Statements about a moving coil galvanometer Moving Charges and Magnetism
11 Lens in a liquid, back surface silvered Ray Optics
12 Hot water flow and surface tension of soap water Mechanical Properties of Fluids
13 Charge between an axial and an equatorial dipole Electric Charges and Fields
14 Lateral shift through a glass slab Ray Optics
15 Lead bullet melting inside a block Thermal Properties of Matter
16 Matter wave of a sub-atomic particle Dual Nature of Radiation and Matter
17 Matching the four thermodynamic processes Thermodynamics
18 Disc with a hole, centre of mass Centre of Mass
19 Dimensions from a position-time expression Units and Measurements
20 What the ratio of two charge densities stands for Units and Measurements
21 Current when it is changing at a given rate Electromagnetic Induction
22 Two vectors at right angles, find the constant Motion in a Plane
23 Gas compressed suddenly to a quarter of its volume Thermodynamics
24 Work done by a varying force along a path Work, Energy and Power
25 Electrostatic force compared with gravitational force Electric Charges and Fields

Chapter by chapter

Chapter Questions Marks
Ray Optics 3 12
Current Electricity 2 8
Electric Charges and Fields 2 8
Electromagnetic Induction 2 8
Thermodynamics 2 8
Units and Measurements 2 8
Centre of Mass 1 4
Dual Nature of Radiation and Matter 1 4
Electromagnetic Waves 1 4
Mechanical Properties of Fluids 1 4
Motion in a Plane 1 4
Motion in a Straight Line 1 4
Moving Charges and Magnetism 1 4
Nuclei 1 4
Oscillations 1 4
Rotational Motion 1 4
Thermal Properties of Matter 1 4
Work, Energy and Power 1 4

Where the 100 marks sat

Part of Physics Qs Marks Share Which ones
Electricity and magnetism 8 32 32% Q1, 2, 5, 8, 10, 13, 21, 25
Mechanics and properties of matter 6 24 24% Q6, 7, 12, 18, 22, 24
Optics 3 12 12% Q9, 11, 14
Heat and thermodynamics 3 12 12% Q15, 17, 23
Modern physics 2 8 8% Q3, 16
Units and measurements 2 8 8% Q19, 20
Oscillations 1 4 4% Q4

How tough it was

This shift leaned hard on statements. Five questions gave a list of claims and asked which ones hold. That is not hard physics, but it punishes half-learnt definitions, because one wrong claim takes the whole answer down with it.

Electricity and magnetism carried a third of the paper again. Optics gave three questions and all three needed a clean sign convention: the refracting surface, the silvered lens and the slab.

The slow ones were the two dipoles acting on one charge, and the silvered lens sitting in a liquid. Both reward a student who writes the standard formula first and puts the numbers in last.

Section B was fair. Three of the five are a single substitution once the right relation is chosen.

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. The official answer key is linked further down.

Q1  ·  Current Electricity  ·  on the pattern of Q1

An uncharged capacitor C is joined in series with a resistor R and a battery of emf E. The key is closed. At that very instant, what is the current in the circuit and the potential difference across the capacitor?

Answer  Current is E/R, the largest it will ever be; the capacitor voltage is zero

SolutionAn uncharged capacitor holds no charge, so there is no potential difference across it at t = 0. The whole emf therefore sits across R and the current is E/R. As charge builds up the capacitor voltage rises and the current falls, reaching zero when the capacitor is full.

Where it goes wrong  Thinking a capacitor blocks current from the start. It blocks only when it is fully charged; at the first instant it behaves like a plain wire.

Q2  ·  Electromagnetic Waves  ·  on the pattern of Q2

An electromagnetic wave travels through free space along the +z direction. Its electric field points along +x with amplitude 60 N/C. Find the amplitude and direction of the magnetic field. (c = 3 × 108 m/s)

Answer  2 × 10−7 T, along +y

SolutionIn an electromagnetic wave the two fields are tied by E0 = cB0, so B0 = 60/(3 × 108) = 2 × 10−7 T. The direction follows from E × B pointing the way the wave travels. With E along +x and travel along +z, B must be along +y, since x̂ × ŷ = ẑ.

Where it goes wrong  Multiplying by c instead of dividing, or taking B along the direction of travel. B is always perpendicular to both E and the direction of travel.

Q3  ·  Nuclei  ·  on the pattern of Q3

Two samples start with the same number of nuclei. The second decays with three times the decay constant of the first. After one half-life of the first sample, find the ratio of the number left in the second to that in the first.

Answer  1 : 4

SolutionAfter a time t the count is N0e−λt. Take t equal to the first sample’s half-life, so e−λt = 1/2 and N1 = N0/2. For the second, e−3λt = (e−λt)3 = 1/8, so N2 = N0/8. The ratio is (1/8)/(1/2) = 1/4.

Where it goes wrong  Answering 1/8. That is the fraction left in the second sample, not the ratio between the two.

Q4  ·  Oscillations  ·  on the pattern of Q4

A hollow cube of side 10 cm and mass 100 g floats in water. It is pushed down a little and released. Find the time period of the oscillation. (g = 10 m/s2, density of water = 1000 kg/m3)

Answer  About 0.20 s

SolutionPush it down by x and the extra buoyant force is ρAgx, a restoring force proportional to x. So the motion is simple harmonic with k = ρAg = 1000 × 0.01 × 10 = 100 N/m. Then T = 2π√(m/k) = 2π√(0.1/100) = 2π(0.0316) ≈ 0.20 s.

Where it goes wrong  Using the volume of the cube in place of the base area. Only the face area matters, because that is what sets how much extra water is pushed aside.

Q5  ·  Electromagnetic Induction  ·  on the pattern of Q5

A coil with an air core has self-inductance L. The core is replaced by a material of relative permeability 500, nothing else changing. Find the new self-inductance.

Answer  500 L

SolutionSelf-inductance depends on the shape and size of the coil and on the medium inside it, through L ∝ μr. Filling the core with a material of relative permeability 500 makes the inductance 500 times larger.

Where it goes wrong  Believing inductance depends on geometry alone. Geometry and the core material both count, which is exactly why iron cores are used.

Q6  ·  Motion in a Straight Line  ·  on the pattern of Q6

A vehicle starts from rest, speeds up uniformly to 40 m/s in 10 s, holds that speed for 20 s, then slows uniformly to rest in another 10 s. Find the total distance covered.

Answer  1.2 km

SolutionUse the area under the velocity-time graph. Speeding up: ½(10)(40) = 200 m. Steady: 40 × 20 = 800 m. Slowing: ½(10)(40) = 200 m. Total = 1200 m = 1.2 km.

Where it goes wrong  Multiplying the top speed by the total time. That counts the two sloping parts at full speed, which they never reach for most of their duration.

Q7  ·  Rotational Motion  ·  on the pattern of Q7

A solid sphere rolls without slipping from the top of an incline of fixed length L, first when the incline is at 30° and then at 60°. Find the ratio v12 : v22 of the squared speeds at the bottom.

Answer  1 : √3

SolutionFor a solid sphere rolling without slipping, energy conservation gives v2 = (10/7)gh, and the drop is h = L sinθ. With L fixed, v2 ∝ sinθ. So the ratio is sin30° : sin60° = (1/2) : (√3/2) = 1 : √3.

Where it goes wrong  Taking the drop as L instead of L sinθ. The length of the slope is given, not the height.

Q8  ·  Current Electricity  ·  on the pattern of Q8

A 12 V battery of negligible internal resistance feeds a 4 Ω resistor in series with a parallel pair of 6 Ω and 12 Ω. Find the total resistance and the current drawn.

Answer  8 Ω and 1.5 A

SolutionThe parallel pair is (6 × 12)/(6 + 12) = 4 Ω. Adding the series 4 Ω gives 8 Ω. The current is 12/8 = 1.5 A.

Where it goes wrong  Adding all three resistances. Work the parallel part out first, then put it in series.

Q9  ·  Ray Optics  ·  on the pattern of Q9

A spherical surface of radius R separates air from glass of refractive index 1.5, with the centre of curvature inside the glass. A point object in air forms a real image inside the glass at the same distance from the surface as the object. Find that distance.

Answer  5R

SolutionUse μ2/v − μ1/u = (μ2 − μ1)/R with the object at u = −d and the image at v = +d, taking R positive. Then 1.5/d + 1/d = 0.5/R, so 2.5/d = 0.5/R and d = 5R.

Where it goes wrong  Dropping the minus sign on u. The object sits on the incoming side, so its distance is negative in the Cartesian convention.

Q10  ·  Moving Charges and Magnetism  ·  on the pattern of Q10

The number of turns in a moving coil galvanometer is doubled, with the same wire, same area and same spring. What happens to its current sensitivity and to its voltage sensitivity?

Answer  Current sensitivity doubles; voltage sensitivity stays the same

SolutionCurrent sensitivity is NAB/k, so doubling N doubles it. But doubling the turns with the same wire also doubles the coil resistance. Voltage sensitivity is current sensitivity divided by resistance, so the two doublings cancel and it is unchanged.

Where it goes wrong  Assuming both sensitivities double. The extra wire brings extra resistance with it, and that is what holds the voltage sensitivity back.

Q11  ·  Ray Optics  ·  on the pattern of Q11

A thin convex lens of focal length 20 cm is silvered on its back surface, whose radius of curvature is 30 cm. Find the focal length of the mirror it now behaves as.

Answer  6 cm, behaving as a concave mirror

SolutionA silvered lens acts as a mirror of power P = 2Plens + Pmirror. Here Plens = 1/20 and Pmirror = 2/R = 2/30. So P = 2(1/20) + 2/30 = 0.1 + 0.0667 = 1/6 per cm, giving an equivalent focal length of 6 cm.

Where it goes wrong  Counting the lens power once. Light goes through the lens, reflects off the silver, and comes back through the lens, so it is counted twice.

Q12  ·  Mechanical Properties of Fluids  ·  on the pattern of Q12

Two claims. One: hot water flows faster than cold water. Two: soap water has a higher surface tension than fresh water. Which are true?

Answer  The first is true, the second is false

SolutionViscosity of a liquid falls as temperature rises, so hot water offers less internal friction and flows faster. Soap is a surface-active agent: it lowers the surface tension of water, it does not raise it.

Where it goes wrong  Thinking soap raises surface tension because it makes bubbles. It is the opposite — bubbles form easily precisely because the surface tension has been lowered.

Q13  ·  Electric Charges and Fields  ·  on the pattern of Q13

A charge Q sits on the axis of one dipole at distance r, and at the same distance r on the equatorial line of a second, identical dipole. Compare the two forces on Q.

Answer  The axial dipole pulls or pushes twice as hard, a ratio of 2 : 1

SolutionThe field of a short dipole is 2kp/r3 along its axis and kp/r3 on its perpendicular bisector. At equal distances with equal dipole moments, the axial field is exactly twice the equatorial one, so the forces are in the ratio 2 : 1. For them to balance, the dipole moments would have to be in the ratio 1 : 2.

Where it goes wrong  Treating the two positions as equivalent. The axial point always feels twice the field of the equatorial point at the same distance.

Q14  ·  Ray Optics  ·  on the pattern of Q14

A ray enters a parallel sided glass slab of thickness h at an angle of incidence i and refracts at an angle r. Write the lateral shift of the emerging ray.

Answer  h sin(i − r) / cos r

SolutionThe ray travels a path length h/cos r inside the slab. The emerging ray is parallel to the original, and the perpendicular gap between them is that path length times sin(i − r). So the shift is h sin(i − r)/cos r.

Where it goes wrong  Writing h tan(i − r). That treats the slab thickness as the path length, but the ray travels a longer, slanted path inside.

Q15  ·  Thermal Properties of Matter  ·  on the pattern of Q15

A lead bullet at 300 K strikes a wooden block and melts completely. Lead melts at 600 K, its specific heat is 125 J kg−1 K−1 and its latent heat of fusion is 2.5 × 104 J kg−1. If the total heat needed is 1250 J, find the mass of the bullet.

Answer  20 g

SolutionPer kilogram the heat needed is the warming plus the melting: 125 × (600 − 300) + 2.5 × 104 = 37500 + 25000 = 62500 J. So m = 1250/62500 = 0.02 kg = 20 g.

Where it goes wrong  Leaving out the latent heat. Melting takes heat at a fixed temperature, and here it is nearly half the total.

Q16  ·  Dual Nature of Radiation and Matter  ·  on the pattern of Q16

A particle of mass 10−30 kg moves at 6.63 × 105 m/s. Find its de Broglie wavelength and say which part of the spectrum it matches. (h = 6.63 × 10−34 J s)

Answer  1 nm, which is the X-ray range

Solutionλ = h/mv = (6.63 × 10−34)/(10−30 × 6.63 × 105) = (6.63 × 10−34)/(6.63 × 10−25) = 10−9 m = 1 nm. Wavelengths around a nanometre sit in the X-ray band.

Where it goes wrong  Using energy instead of momentum. The de Broglie relation needs mv, not ½mv2.

Q17  ·  Thermodynamics  ·  on the pattern of Q17

Match each description with its process: (a) temperature held fixed, (b) no heat enters or leaves, (c) no work done, (d) pressure held fixed.

Answer  (a) isothermal, (b) adiabatic, (c) isochoric, (d) isobaric

SolutionIsothermal means constant temperature, so for an ideal gas the internal energy does not change and all the heat goes into work. Adiabatic means no heat crosses the boundary, so any work is paid for out of internal energy. Isochoric means constant volume, and with no volume change no work is done. Isobaric means constant pressure, where the heat splits between internal energy and work.

Where it goes wrong  Mixing up adiabatic and isochoric. One has no heat flow, the other has no work. They are different things.

Q18  ·  Centre of Mass  ·  on the pattern of Q18

A uniform disc of radius 30 cm has a hole of radius 10 cm cut out, placed so that the edge of the hole just touches the edge of the disc. How far from the centre of the original disc is the centre of mass of what remains?

Answer  2.5 cm, on the side away from the hole

SolutionMass goes with area, so the hole carries (10/30)2 = 1/9 of the disc’s mass. Its centre is at 30 − 10 = 20 cm from the centre. Treating the hole as negative mass, x = −[(M/9)(20)]/[M − M/9] = −(20/9)/(8/9) = −2.5 cm.

Where it goes wrong  Using the ratio of radii, 1/3, for the mass of the hole. Area goes with the square of the radius, so it is 1/9.

Q19  ·  Units and Measurements  ·  on the pattern of Q19

The position of a particle is x(t) = A sinωt + B cosωt + Ct + D. Find the dimensions of ABC/D.

Answer  [L2 T−1]

SolutionEvery term added together must have the dimension of length. Sine and cosine are pure numbers, so A, B and D are all lengths, and Ct is a length, which makes C a length per time. So ABC/D = (L × L × LT−1)/L = L2T−1.

Where it goes wrong  Giving ωt a dimension. The argument of a sine must be a pure number, which is what fixes ω as one over time.

Q20  ·  Units and Measurements  ·  on the pattern of Q20

λ is a linear charge density and σ is a surface charge density. What does λ/σ stand for dimensionally?

Answer  A length

SolutionLinear charge density is charge per unit length and surface charge density is charge per unit area. Dividing one by the other, the charge cancels and what is left is area divided by length, which is a length.

Where it goes wrong  Calling it an area. Work the units through slowly: (C/m) ÷ (C/m2) = m.

Q21  ·  Electromagnetic Induction  ·  on the pattern of Q21

A coil of inductance 3 H and resistance 12 Ω is connected to a 60 V source. At an instant the current is changing at 8 A/s. Find the current at that instant.

Answer  3 A

SolutionGoing round the loop, the source emf is shared by the resistor and the back emf of the inductor: E = IR + L(dI/dt). So 60 = 12I + 3(8) = 12I + 24, which gives 12I = 36 and I = 3 A.

Where it goes wrong  Writing I = E/R = 5 A. That is only the final steady current, after the current has stopped changing.

Q22  ·  Motion in a Plane  ·  on the pattern of Q22

The vectors A = 2î + 3nĵ + 2k̂ and B = 2î − 2ĵ + 4k̂ are at right angles to each other. Find n.

Answer  n = 2

SolutionTwo vectors are perpendicular when their dot product is zero. A · B = (2)(2) + (3n)(−2) + (2)(4) = 4 − 6n + 8 = 12 − 6n. Setting this to zero gives n = 2.

Where it goes wrong  Setting the magnitudes equal instead. Equal length and perpendicular are two different conditions.

Q23  ·  Thermodynamics  ·  on the pattern of Q23

An ideal gas at 0 °C is compressed suddenly to one fourth of its volume. For this gas the ratio of specific heats is 3/2. Find the final temperature.

Answer  546 K, a rise of 273 K

SolutionA sudden compression is adiabatic, so TVγ−1 stays constant. Here γ − 1 = 1/2, so T2 = T1(V1/V2)1/2 = 273 × 41/2 = 273 × 2 = 546 K.

Where it goes wrong  Using PV = constant. That is the isothermal law and it applies to a slow squeeze, not a sudden one.

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

A force F = (2xî + 3ĵ) N acts on a particle which moves from (0, 0) to (2 m, 4 m). Find the work done.

Answer  16 J

SolutionWork is the integral of F · dr. The x part gives ∫2x dx from 0 to 2 = [x2] = 4 J. The y part gives ∫3 dy from 0 to 4 = 12 J. Adding, the work done is 16 J.

Where it goes wrong  Multiplying force by distance. The x component changes along the path, so it must be integrated, not multiplied.

Q25  ·  Electric Charges and Fields  ·  on the pattern of Q25

Two particles carry charges 1.6 × 10−19 C and 3.2 × 10−19 C, with masses 1 × 10−27 kg and 4 × 10−27 kg. Find the ratio of the electrostatic force between them to the gravitational force. (k = 9 × 109, G = 6.67 × 10−11 SI units)

Answer  About 1.7 × 1036

SolutionBoth forces fall off as 1/r2, so the separation cancels out of the ratio. Fe/Fg = kq1q2/(Gm1m2). The top is 9 × 109 × 5.12 × 10−38 = 4.61 × 10−28. The bottom is 6.67 × 10−11 × 4 × 10−54 = 2.67 × 10−64. The ratio is about 1.7 × 1036.

Where it goes wrong  Looking for the separation in the question. It is not needed, because it cancels, and that is the point the question is making.

The official paper and answer key

NTA publishes the question papers and the final answer key for each session on its own site. Always check against that file, not a coaching copy: jeemain.nta.nic.in.

Everything above is written by AcceleratePhysics. We do not reprint NTA’s question wording or its answer key on this page.

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25 questions  ·  45 minutes  ·  out of 100  ·  +4 / −1 / 0  ·  score card at the end

AcceleratePhysics

Study resources

JEE Main Physics

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

NEET(UG) Physics

Syllabus, unit by unit
Chapter-wise weightage
Formula sheet
Chapterwise questions with answer key
Timed test series: Sanjeevani 2027 · Sanjeevani 2028 · see all

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

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