JEE Main 2025 — 22 January, Shift 1 — Physics Question Paper

The first Physics paper of JEE Main 2025, the morning shift of 22 January and the opening paper 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

Exam JEE Main 2025, Session 1
Date 22 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 Electron fired between charged plates, sideways velocity Electric Charges and Fields
2 Two spheres of the same material radiating, energy ratio Thermal Properties of Matter
3 Work functions of two metals, which one emits Dual Nature of Radiation and Matter
4 Dimensions of magnetic field divided by permeability Units and Measurements
5 de Broglie wavelength in two hydrogen orbits Atoms
6 Which circuit forward biases the diode Semiconductor Electronics
7 Line charge on an edge of a cube, total flux Electric Charges and Fields
8 Resistivity against temperature for a standard resistor Current Electricity
9 Dielectric pushed in with the supply still on Electrostatic Potential and Capacitance
10 Ice at minus ten taken all the way to steam Thermal Properties of Matter
11 Two statements on vernier callipers Units and Measurements
12 Closed and open pipes, harmonics matching Waves
13 Double slit done inside a denser medium Wave Optics
14 Two real cells in parallel, equivalent emf Current Electricity
15 Disc with a circular piece cut out, moment of inertia Rotational Motion
16 Potentiometer with a load on the sliding contact Current Electricity
17 Bob on a string going round, kinetic energy ratio Work, Energy and Power
18 Sphere with a cavity, gravitational pull on a mass Gravitation
19 Lens silvered on one face, object imaged on itself Ray Optics
20 Three lenses formed at a surface, power of the set Ray Optics
21 Projectile, height in the first and the last second Motion in a Plane
22 Car in a convex side mirror, acceleration of the image Ray Optics
23 Three rods of different conductivity, junction temperature Thermal Properties of Matter
24 Two soap bubbles touching, common surface Mechanical Properties of Fluids
25 Two particles, angular momentum of one about the other Rotational Motion

Chapter by chapter

Chapter Questions Marks
Current Electricity 3 12
Ray Optics 3 12
Thermal Properties of Matter 3 12
Electric Charges and Fields 2 8
Rotational Motion 2 8
Units and Measurements 2 8
Atoms 1 4
Dual Nature of Radiation and Matter 1 4
Electrostatic Potential and Capacitance 1 4
Gravitation 1 4
Mechanical Properties of Fluids 1 4
Motion in a Plane 1 4
Semiconductor Electronics 1 4
Wave Optics 1 4
Waves 1 4
Work, Energy and Power 1 4

Where the 100 marks sat

Part of Physics Qs Marks Share Which ones
Mechanics and properties of matter 6 24 24% Q15, 17, 18, 21, 24, 25
Electricity and magnetism 6 24 24% Q1, 7, 8, 9, 14, 16
Optics 4 16 16% Q13, 19, 20, 22
Heat and thermodynamics 3 12 12% Q2, 10, 23
Modern physics and electronics 3 12 12% Q3, 5, 6
Units and measurements 2 8 8% Q4, 11
Waves 1 4 4% Q12

How tough it was

This was the opening shift of JEE Main 2025 and the Physics paper was fair. Sixteen chapters were touched in 25 questions, so nothing was repeated much and nothing was skipped either.

The quick marks sat in Section A — the photoelectric comparison, the diode circuit, the vernier statements and the dimensions question are all one-step if the idea is clear. The slow ones were the disc with a piece cut out, the sphere with a cavity and the lens silvered on one face. Each of those needs a second step that students often forget: the parallel axis shift, the mass of the removed piece, and the fact that light crosses the lens twice.

Section B was friendly. Four of the five numeric questions land on small whole numbers, so an answer that comes out ugly is a sign to check the working.

If you lost marks here, look first at where the second step was dropped, not at the chapter name.

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  ·  Electric Charges and Fields  ·  on the pattern of Q1

A proton enters along the mid line between two parallel plates of length 20 cm with a horizontal speed of 2 × 105 m/s. The uniform field between the plates is 1.0 × 104 V/m and is perpendicular to the motion. Find the vertical component of velocity with which the proton leaves the plates. (mp = 1.6 × 10−27 kg, e = 1.6 × 10−19 C)

Answer  1 × 106 m/s

SolutionThe horizontal speed is not changed by a vertical field, so the time inside is t = L/v = 0.20 / (2 × 105) = 1 × 10−6 s. The vertical acceleration is a = eE/m = (1.6 × 10−19)(1.0 × 104) / (1.6 × 10−27) = 1 × 1012 m/s2. So vy = at = 1012 × 10−6 = 1 × 106 m/s.

Where it goes wrong  Treating the plate length as a vertical distance. The 20 cm is travelled horizontally; it only fixes the time.

Q2  ·  Thermal Properties of Matter  ·  on the pattern of Q2

Two spheres of the same material have radii 0.1 m and 0.3 m. The smaller one is at 600 K and the bigger one at 200 K, both in the same surroundings whose effect may be ignored. If the smaller sphere radiates energy E per second, how much does the bigger one radiate?

Answer  E/9

SolutionBy Stefan’s law the power radiated is P = σAT4 = σ(4πr2)T4, so P ∝ r2T4. The ratio is (0.3/0.1)2 × (200/600)4 = 9 × (1/3)4 = 9/81 = 1/9. The bigger sphere radiates E/9.

Where it goes wrong  Using T instead of T4, or forgetting that the area carries r2. Both factors have to be taken together.

Q3  ·  Dual Nature of Radiation and Matter  ·  on the pattern of Q3

Two metals have work functions 2.0 eV and 2.8 eV. Light of wavelength 500 nm falls on both. From which metal are photoelectrons emitted?

Answer  Only from the metal of work function 2.0 eV

SolutionThe photon energy in eV is E = 1240/λ(nm) = 1240/500 = 2.48 eV. Emission needs E greater than the work function. 2.48 > 2.0, so the first metal emits. 2.48 < 2.8, so the second does not.

Where it goes wrong  Comparing wavelengths instead of energies. A longer wavelength means a smaller photon energy, which is the opposite of what students often assume.

Q4  ·  Units and Measurements  ·  on the pattern of Q4

Find the dimensional formula of ε0E2, where E is the electric field.

Answer  [M L−1 T−2]

SolutionThe energy stored per unit volume in an electric field is u = ½ε0E2. So ε0E2 has the dimensions of energy divided by volume: [M L2 T−2] / [L3] = [M L−1 T−2]. That is also the dimension of pressure.

Where it goes wrong  Writing the dimensions of energy. The ½ε0E2 form is energy per unit volume, not energy.

Q5  ·  Atoms  ·  on the pattern of Q5

In a hydrogen atom the de Broglie wavelength of the electron in the n-th Bohr orbit is λn. Find λ3 / λ1.

Answer  3

SolutionBohr’s condition says the orbit holds a whole number of wavelengths: 2πrn = nλn, so λn = 2πrn/n. Since rn ∝ n2, we get λn ∝ n2/n = n. So λ3/λ1 = 3.

Where it goes wrong  Taking λ ∝ r, which gives 9. The orbit of a higher n is longer, but it also holds more wavelengths, and the two effects partly cancel.

Q6  ·  Semiconductor Electronics  ·  on the pattern of Q6

A silicon diode is joined in series with a 1 kΩ resistor across a 5 V battery, with the p side towards the positive terminal. Taking the drop across a conducting silicon diode as 0.7 V, find the current.

Answer  4.3 mA

SolutionWith the p side at the higher potential the diode is forward biased, so it conducts and holds about 0.7 V across itself. The resistor then gets 5 − 0.7 = 4.3 V, and I = 4.3 / 1000 = 4.3 × 10−3 A.

Where it goes wrong  Ignoring the 0.7 V and writing 5 mA. Also check the direction first — reverse the battery and the current is almost zero.

Q7  ·  Electric Charges and Fields  ·  on the pattern of Q7

A point charge q is placed exactly at one corner of a cube. What is the electric flux through the whole cube?

Answer  q / 8ε0

SolutionGauss’s law gives q/ε0 only for a charge fully enclosed. A charge at a corner is shared by eight identical cubes stacked around that point, so each cube takes one eighth of the total flux: q / 8ε0.

Where it goes wrong  Writing q/ε0, or dividing by 6 because a cube has six faces. The sharing is between cubes, not between faces.

Q8  ·  Current Electricity  ·  on the pattern of Q8

A standard resistor in a resistance box is made of manganin, whose temperature coefficient of resistance is about 2 × 10−5 per °C, and not of copper, whose coefficient is about 4 × 10−3 per °C. By what percentage does the manganin resistor change if it warms by 50 °C?

Answer  About 0.1 per cent

SolutionThe fractional change is ΔR/R = αΔT = (2 × 10−5)(50) = 1 × 10−3, which is 0.1 per cent. For copper the same warming would give 4 × 10−3 × 50 = 0.2, that is 20 per cent, which would make the box useless.

Where it goes wrong  Choosing the material by its resistivity. What a standard resistor needs is a flat resistivity-temperature curve, not a high or low resistivity.

Q9  ·  Electrostatic Potential and Capacitance  ·  on the pattern of Q9

A 20 µF capacitor is charged by a 200 V supply which stays connected. A dielectric of constant K = 3 is then slid in, filling the gap. Find the extra charge drawn and the change in stored energy.

Answer  8 mC drawn in, energy rises by 0.8 J

SolutionThe supply stays on, so V stays at 200 V. Before: C = 20 µF, Q = CV = 4 × 10−3 C, U = ½CV2 = 0.4 J. After: C′ = 60 µF, Q′ = 12 × 10−3 C, U′ = 1.2 J. Extra charge = 12 − 4 = 8 mC and the energy rises by 1.2 − 0.4 = 0.8 J, supplied by the battery.

Where it goes wrong  Holding the charge constant. That is the isolated-capacitor case. Here the battery is still connected, so the voltage is what stays fixed.

Q10  ·  Thermal Properties of Matter  ·  on the pattern of Q10

2 g of ice at −20 °C is turned into steam at 120 °C. Find the heat needed. (cice = 2100, cwater = 4180, csteam = 1920 J kg−1 K−1, Lfusion = 3.35 × 105, Lvaporisation = 2.25 × 106 J kg−1)

Answer  About 6.17 × 103 J

SolutionAdd the five steps for m = 2 × 10−3 kg. Ice −20 to 0: 2100 × 20 = 42000. Melting: 335000. Water 0 to 100: 4180 × 100 = 418000. Boiling: 2250000. Steam 100 to 120: 1920 × 20 = 38400. Total per kg = 3083400 J, so Q = 2 × 10−3 × 3083400 ≈ 6167 J.

Where it goes wrong  Missing one of the two latent heat steps, or starting the steam heating from 0 °C instead of 100 °C.

Q11  ·  Units and Measurements  ·  on the pattern of Q11

A vernier scale has 20 divisions which together cover 19 main scale divisions. One main scale division is 1 mm. Find the vernier constant.

Answer  0.05 mm

SolutionOne vernier division = 19/20 mm = 0.95 mm. The vernier constant is one main scale division minus one vernier division = 1 − 0.95 = 0.05 mm. The same number comes from 1 MSD divided by the number of vernier divisions, but only because 20 vernier divisions match 19 main divisions.

Where it goes wrong  Multiplying one main scale division by the number of vernier divisions. The vernier constant is a difference, and it is always smaller than one main scale division.

Q12  ·  Waves  ·  on the pattern of Q12

A closed organ pipe of length 20 cm and an open organ pipe are filled with the same gas. The 5th harmonic of the closed pipe has the same frequency as the 2nd harmonic of the open pipe. Find the length of the open pipe.

Answer  16 cm

SolutionA closed pipe has only odd harmonics, f = nv/4Lc with n odd, so the 5th harmonic is 5v/(4 × 0.20) = 6.25v. An open pipe has f = nv/2Lo, so its 2nd harmonic is 2v/(2Lo) = v/Lo. Setting them equal, v/Lo = 6.25v, so Lo = 0.16 m = 16 cm.

Where it goes wrong  Using even harmonics in the closed pipe. A closed pipe simply has none.

Q13  ·  Wave Optics  ·  on the pattern of Q13

A double slit experiment is set up in air and the fringe width is β. The whole set-up is then placed in water of refractive index 4/3. What happens to the fringe width?

Answer  It becomes 3β/4, so the fringes come closer

SolutionFringe width is β = λD/d. Inside the medium the speed of light falls while the frequency stays the same, so the wavelength becomes λ/μ. Everything else is unchanged, so the new fringe width is β/μ = 3β/4.

Where it goes wrong  Thinking the frequency changes. Frequency is set by the source and does not change with the medium; the wavelength is what shrinks.

Q14  ·  Current Electricity  ·  on the pattern of Q14

Two cells, one of emf 2 V with internal resistance 1 Ω and the other of emf 6 V with internal resistance 2 Ω, are joined in parallel. Find the equivalent emf and the equivalent internal resistance.

Answer  10/3 V ≈ 3.33 V and 2/3 Ω ≈ 0.67 Ω

SolutionFor cells in parallel, Eeq = (E1/r1 + E2/r2) / (1/r1 + 1/r2) = (2/1 + 6/2) / (1 + 0.5) = 5/1.5 = 10/3 V. The internal resistances are simply in parallel: req = (1 × 2)/(1 + 2) = 2/3 Ω.

Where it goes wrong  Expecting the equivalent emf to be smaller than both. It always lies between the two emfs. Only the internal resistance comes out smaller than either.

Q15  ·  Rotational Motion  ·  on the pattern of Q15

A uniform disc of mass M and radius R has a circular piece of radius R/2 removed, the centre of the hole being at a distance R/2 from the centre of the disc. Find the moment of inertia of what is left, about the central axis perpendicular to the disc.

Answer  13MR2/32

SolutionMass goes as area, so the removed piece has mass M/4. Its moment of inertia about the central axis is ½(M/4)(R/2)2 + (M/4)(R/2)2 = MR2/32 + MR2/16 = 3MR2/32. The full disc gives ½MR2 = 16MR2/32. Subtracting, the remainder is 13MR2/32.

Where it goes wrong  Forgetting the parallel axis shift for the removed piece. Its own centre is not on the axis you are taking moments about.

Q16  ·  Current Electricity  ·  on the pattern of Q16

A potentiometer wire of total resistance 10 Ω is connected across a 5 V battery of negligible internal resistance. The sliding contact sits at the midpoint and a 10 Ω resistor is joined between the contact and one end of the wire. Find the current drawn from the battery.

Answer  0.6 A

SolutionThe slider splits the wire into two halves of 5 Ω each. The 10 Ω resistor sits across one half, giving (5 × 10)/15 = 10/3 Ω. That is in series with the other half: R = 5 + 10/3 = 25/3 Ω. So I = 5 / (25/3) = 0.6 A.

Where it goes wrong  Putting the 10 Ω across the whole wire. The slider divides the wire, and only the part it covers is in parallel with the load.

Q17  ·  Work, Energy and Power  ·  on the pattern of Q17

A bob on a light string of length L is given a horizontal speed at the lowest point which is just enough to complete the vertical circle. Find the ratio of its kinetic energy at the lowest point to that at the highest point.

Answer  5

SolutionTo just complete the circle, at the top gravity alone supplies the centripetal force: mg = mvtop2/L, so vtop2 = gL. Energy conservation from top to bottom gives vbottom2 = vtop2 + 4gL = 5gL. The kinetic energies are ½m(5gL) and ½m(gL), so the ratio is 5.

Where it goes wrong  Taking the speed at the top as zero. On a string the bob must still be moving at the top, otherwise the string goes slack before it gets there.

Q18  ·  Gravitation  ·  on the pattern of Q18

A point mass m sits at a distance 2R from the centre of a uniform solid sphere of mass M and radius R, and feels a force F1. A sphere of radius R/2 is now removed from the big sphere, the cavity centred at R/2 from the centre on the line joining the centre to m. The force becomes F2. Find F1 : F2.

Answer  9 : 7

SolutionF1 = GMm/(2R)2 = GMm/4R2. The removed sphere has one eighth the volume, so its mass is M/8, and its centre is 2R − R/2 = 1.5R from m. Its pull would have been G(M/8)m/(1.5R)2 = GMm/18R2. So F2 = GMm(1/4 − 1/18)/R2 = 7GMm/36R2, and F1 : F2 = (9/36) : (7/36) = 9 : 7.

Where it goes wrong  Using M/2 for the removed piece because the radius is halved. Mass goes with the cube of the radius, so it is M/8.

Q19  ·  Ray Optics  ·  on the pattern of Q19

A thin equiconvex lens of glass (μ = 1.5) has each face of radius 20 cm. One face is silvered. At what distance from the lens must an object be placed on the axis so that its image forms on the object itself?

Answer  10 cm

SolutionA silvered lens behaves as a mirror of power P = 2Plens + Pmirror. For the lens, 1/f = (μ − 1)(2/R) = 0.5 × 2/20, so flens = 20 cm. For the silvered surface, fmirror = R/2 = 10 cm. So P = 2/20 + 1/10 = 0.2 per cm, giving an equivalent mirror of focal length 5 cm. The image falls on the object when the object is at the centre of curvature, that is at 2 × 5 = 10 cm.

Where it goes wrong  Using the lens focal length alone. The light passes through the lens, reflects, and passes through it again, so the lens power is counted twice.

Q20  ·  Ray Optics  ·  on the pattern of Q20

A thin lens is made of glass of refractive index 1.5, with the first surface of radius +20 cm and the second of radius −20 cm. Find its power in dioptres.

Answer  +5 D

SolutionThe lens maker formula gives 1/f = (μ − 1)(1/R1 − 1/R2) = 0.5 × (1/20 − 1/(−20)) = 0.5 × (2/20) = 1/20 per cm. So f = 20 cm = 0.2 m and P = 1/f = +5 D.

Where it goes wrong  Dropping the minus sign on R2. With both radii taken positive the two terms cancel and the lens looks as if it has no power at all.

Q21  ·  Motion in a Plane  ·  on the pattern of Q21

A particle is projected at 30° to the horizontal with a speed of 40 m/s. The height gained in the first second is h0 and the height gained in the last second before it reaches the top is h1. Find h0 : h1. (g = 10 m/s2)

Answer  3 : 1

SolutionThe vertical component is uy = 40 sin30° = 20 m/s, so the particle reaches the top at t = 2 s. Height at 1 s = 20(1) − 5(1) = 15 m, so h0 = 15 m. Height at 2 s = 20(2) − 5(4) = 20 m, so the last second gives h1 = 20 − 15 = 5 m. The ratio is 3 : 1.

Where it goes wrong  Taking the last second as the one after the top. The question stops at the highest point, so the window is from 1 s to 2 s.

Q22  ·  Ray Optics  ·  on the pattern of Q22

The convex side mirror of a parked car has radius of curvature 2 m. A car behind approaches at a steady 20 m/s. When it is 10 m away, how fast is its image moving?

Answer  20/121 m/s, about 0.17 m/s

SolutionFor a convex mirror f = R/2 = +1 m, and with the object at u = −10 m the magnification is m = f/(f − u) = 1/(1 + 10) = 1/11. For motion along the axis the image speed is m2 times the object speed, so it is (1/121)(20) = 20/121 ≈ 0.17 m/s.

Where it goes wrong  Using m instead of m2. Along the axis the image speed carries the square of the magnification, which is why a car behind looks far slower than it is.

Q23  ·  Thermal Properties of Matter  ·  on the pattern of Q23

Two rods of the same length and the same cross section, with thermal conductivities 2k and 3k, are joined end to end. The free end of the first is at 100 °C and the free end of the second at 0 °C. Find the steady temperature of the junction.

Answer  40 °C

SolutionIn the steady state the same heat current flows through both rods. With equal lengths and areas, θ = (k1θ1 + k2θ2) / (k1 + k2) = (2k × 100 + 3k × 0) / (5k) = 40 °C.

Where it goes wrong  Taking the plain average, 50 °C. The junction always sits nearer the end of the better conductor, because that rod needs a smaller temperature drop to carry the same heat.

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

Two soap bubbles of radii 3 cm and 6 cm are in contact with each other. Find the radius of curvature of the common surface between them.

Answer  6 cm

SolutionThe excess pressure inside a soap bubble is 4T/r, so the smaller bubble has the higher pressure. The common surface bulges into the bigger bubble and feels the difference: 4T/r = 4T/r1 − 4T/r2, so r = r1r2/(r2 − r1) = (3 × 6)/(6 − 3) = 6 cm.

Where it goes wrong  Adding the radii, or using 2T/r. A soap bubble has two surfaces, so the excess pressure is 4T/r, not 2T/r.

Q25  ·  Rotational Motion  ·  on the pattern of Q25

A particle of mass 2 kg moves with a constant velocity (3î + 4ĵ) m/s. At one instant its position vector is 5ĵ m. Find the magnitude of its angular momentum about the origin.

Answer  30 kg m2 s−1

SolutionAngular momentum is L = m(r × v). Here r × v = (5ĵ) × (3î + 4ĵ) = 5 × 3 (ĵ × î) = −15k̂, whose magnitude is 15. So |L| = 2 × 15 = 30 kg m2 s−1. Since no force acts, this value stays the same at every instant.

Where it goes wrong  Multiplying mass, speed and distance without the cross product. Only the part of the position vector perpendicular to the velocity counts.

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 your answers against that file, not against any coaching copy: jeemain.nta.nic.in.

Everything above — the chapter list, the marks split and the 25 solved questions — is written by AcceleratePhysics. We do not reprint NTA’s question wording or its answer key on this page.

Sit a paper in the same shape, free

Reading a solved paper feels easy.
Sitting one against the clock tells you the truth.

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
Chapter-wise weightage
Formula sheet
Chapterwise questions with answer key
Previous years papers, date and shift
Timed test series: Dhurandhar 2027 · Dhurandhar 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

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

A question a day on Telegram: NEET(UG) · JEE Main

For the Love of Physics
Scroll to Top
FollowFacebook·YouTube·Instagram