JEE Main 2025 — 22 January, Shift 2 — Physics Question Paper

The afternoon shift of 22 January 2025, the second 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 22 January 2025
Shift Shift 2 — 3:00 PM to 6:00 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 circling in a magnetic field with Bohr’s condition Moving Charges and Magnetism
2 Body in a vertical circle, kinetic energy at two points Work, Energy and Power
3 Biconvex lens cut into four parts, power of a piece Ray Optics
4 Torque of a force about the origin Rotational Motion
5 Work done is zero, find the missing component Work, Energy and Power
6 Thin film coated on glass, thickness for best transmission Wave Optics
7 Dimensional formula of capacitance Units and Measurements
8 Series LCR at resonance when the resistance doubles Alternating Current
9 Ball falling in glycerine, viscous force at steady speed Mechanical Properties of Fluids
10 Percentage error in the density of a wire Units and Measurements
11 Which thermodynamic variables are extensive Thermodynamics
12 Tube narrowing, speed of water at the exit Mechanical Properties of Fluids
13 Red and blue fringes in a double slit Wave Optics
14 Projectile, loss of kinetic energy up to the top Work, Energy and Power
15 Which gate fits the given truth table Semiconductor Electronics
16 Photoelectric effect when the wavelength is halved Dual Nature of Radiation and Matter
17 Pendulum carried to a heavier, bigger planet Oscillations
18 Ratio of specific heats with and without vibration Kinetic Theory of Gases
19 Short dipole, potential and field off the axis Electric Charges and Fields
20 Loop leaving a field, emf against time Electromagnetic Induction
21 Displacement current through part of a capacitor gap Electromagnetic Waves
22 Proton in crossed fields, then in a circle Moving Charges and Magnetism
23 Liquid filled tube spun about one end, force at the far end Laws of Motion
24 Net current drawn by the given circuit Current Electricity
25 Two parallel wires carrying opposite currents Moving Charges and Magnetism

Chapter by chapter

Chapter Questions Marks
Moving Charges and Magnetism 3 12
Work, Energy and Power 3 12
Mechanical Properties of Fluids 2 8
Units and Measurements 2 8
Wave Optics 2 8
Alternating Current 1 4
Current Electricity 1 4
Dual Nature of Radiation and Matter 1 4
Electric Charges and Fields 1 4
Electromagnetic Induction 1 4
Electromagnetic Waves 1 4
Kinetic Theory of Gases 1 4
Laws of Motion 1 4
Oscillations 1 4
Ray Optics 1 4
Rotational Motion 1 4
Semiconductor Electronics 1 4
Thermodynamics 1 4

Where the 100 marks sat

Part of Physics Qs Marks Share Which ones
Electricity and magnetism 8 32 32% Q1, 8, 19, 20, 21, 22, 24, 25
Mechanics and properties of matter 7 28 28% Q2, 4, 5, 9, 12, 14, 23
Optics 3 12 12% Q3, 6, 13
Heat and thermodynamics 2 8 8% Q11, 18
Modern physics and electronics 2 8 8% Q15, 16
Units and measurements 2 8 8% Q7, 10
Oscillations 1 4 4% Q17

How tough it was

The afternoon paper of the same day went wider than the morning one. Eighteen chapters in 25 questions, and almost a third of the paper sat in electricity and magnetism.

Four questions were pure recall of a definition — extensive and intensive variables, the truth table, the dimensional formula and the fringe comparison. Those are the marks you should never lose.

The slow ones were the cut lens, the dipole off the axis and the spinning tube of liquid. All three punish a student who reaches for a formula before drawing the picture. The spinning tube in particular cannot be done by putting the whole mass at the far end.

Section B was straightforward arithmetic once the right relation was written down.

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  ·  Moving Charges and Magnetism  ·  on the pattern of Q1

An electron moves in a circle in a plane perpendicular to a uniform magnetic field B. If Bohr’s condition mvr = nh/2π is taken to hold, find the radius of the n-th orbit.

Answer  r = √(nh / 2πeB)

SolutionIn a magnetic field the radius is r = mv/eB, so mv = eBr. Put this into Bohr’s condition mvr = nh/2π: (eBr)r = nh/2π, so r2 = nh/(2πeB) and r = √(nh / 2πeB).

Where it goes wrong  Using r = nh/(2πmv) and stopping there. The speed is not given, so mv has to be replaced using the magnetic condition first.

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

A body of mass 200 g moves in a vertical circle of radius 1 m. Its speed at the lowest point is 8 m/s. Find the ratio of its kinetic energy at the lowest point to that at the highest point. (g = 10 m/s2)

Answer  8 : 3

SolutionAt the lowest point KE = ½(0.2)(82) = 6.4 J. Climbing to the top raises it by 2R = 2 m, so vtop2 = 64 − 2g(2R) = 64 − 40 = 24 m2/s2 and KEtop = ½(0.2)(24) = 2.4 J. The ratio is 6.4 : 2.4 = 8 : 3.

Where it goes wrong  Using R instead of 2R for the height. The top of the circle is two radii above the bottom.

Q3  ·  Ray Optics  ·  on the pattern of Q3

A thin symmetric biconvex lens of power 5 D is cut into two equal halves by a plane that contains the principal axis. Find the power of each half.

Answer  5 D, unchanged

SolutionPower depends only on the refractive index and the two radii of curvature. A cut along the principal axis leaves both surfaces with the same curvature, so each half still has the same focal length and the same power, 5 D. Only the amount of light gathered falls.

Where it goes wrong  Assuming the power halves. That happens only when the cut is across the axis, which turns the lens into two plano-convex lenses.

Q4  ·  Rotational Motion  ·  on the pattern of Q4

A force F = (î + 2ĵ + 3k̂) N acts on a particle whose position vector is r = (2î + ĵ) m. Find the torque about the origin.

Answer  (3î − 6ĵ + 3k̂) N m

Solutionτ = r × F. Taking the determinant with rows (î, ĵ, k̂), (2, 1, 0) and (1, 2, 3): the î part is 1(3) − 0(2) = 3, the ĵ part is −[2(3) − 0(1)] = −6, and the k̂ part is 2(2) − 1(1) = 3. So τ = 3î − 6ĵ + 3k̂ N m.

Where it goes wrong  Writing F × r. The order matters and reverses every sign.

Q5  ·  Work, Energy and Power  ·  on the pattern of Q5

A force F = (3î + bĵ + 2k̂) N acts on a particle which is displaced by (2î − ĵ + k̂) m. Find b if the work done is zero.

Answer  b = 8

SolutionWork is the dot product: W = (3)(2) + (b)(−1) + (2)(1) = 6 − b + 2 = 8 − b. Setting this to zero gives b = 8.

Where it goes wrong  Dropping the minus sign in the displacement. One wrong sign here flips the answer completely.

Q6  ·  Wave Optics  ·  on the pattern of Q6

A film of refractive index 1.38 is coated on glass of refractive index 1.52 so that light of wavelength 552 nm is reflected as little as possible. Find the smallest thickness of the film.

Answer  100 nm

SolutionLight reflecting off the top of the film goes from a rarer to a denser medium, and so does light reflecting off the film-glass surface, because 1.52 is greater than 1.38. Both reflections flip phase, so the flips cancel and destructive interference needs 2μt = (m + ½)λ. The smallest thickness is t = λ/4μ = 552/(4 × 1.38) = 100 nm.

Where it goes wrong  Using t = λ/2μ. That is the rule when only one of the two reflections flips phase, which depends on the refractive index of the glass underneath.

Q7  ·  Units and Measurements  ·  on the pattern of Q7

Write the dimensional formula of capacitance in terms of M, L, T and charge Q.

Answer  [M−1 L−2 T2 Q2]

SolutionCapacitance is C = Q/V, and potential is work per unit charge, V = [M L2 T−2]/[Q]. So C = Q ÷ (M L2 T−2 Q−1) = [M−1 L−2 T2 Q2].

Where it goes wrong  Forgetting that potential already carries a charge in its denominator, so the charge ends up squared.

Q8  ·  Alternating Current  ·  on the pattern of Q8

In a series LCR circuit driven at resonance the current amplitude is I0. The resistance is now made three times larger, with L, C and the source unchanged. Find the new current amplitude at resonance.

Answer  I0/3

SolutionAt resonance the inductive and capacitive reactances are equal and cancel, so the impedance is just R. The current amplitude is E0/R, which is inversely proportional to R. Three times the resistance gives one third the current.

Where it goes wrong  Trying to bring L and C into the answer. At resonance they cancel exactly, which is the whole point of resonance.

Q9  ·  Mechanical Properties of Fluids  ·  on the pattern of Q9

A solid ball of mass M is dropped into a tall jar of liquid whose density is one third that of the ball. After it reaches a steady speed, find the viscous force on the ball.

Answer  2Mg/3

SolutionAt the steady speed the ball is in equilibrium: weight = buoyancy + viscous force. The buoyant force equals the weight of liquid displaced, which is (1/3) of the ball’s weight because the volume is the same and the density is one third. So the viscous force is Mg − Mg/3 = 2Mg/3.

Where it goes wrong  Writing Mg for the viscous force. Buoyancy is already carrying part of the weight.

Q10  ·  Units and Measurements  ·  on the pattern of Q10

The density of a wire is found from mass (0.80 ± 0.004) g, radius (0.40 ± 0.01) cm and length (20.00 ± 0.10) cm. Find the maximum percentage error in the density.

Answer  6 per cent

SolutionDensity is ρ = m/(πr2L), so the fractional errors add as Δρ/ρ = Δm/m + 2Δr/r + ΔL/L. In percentages that is 0.5 + 2(2.5) + 0.5 = 6 per cent.

Where it goes wrong  Leaving out the factor of 2 on the radius. The radius is squared in the formula, so its error counts twice.

Q11  ·  Thermodynamics  ·  on the pattern of Q11

Of volume, density, internal energy and temperature, which are extensive and which are intensive?

Answer  Volume and internal energy are extensive; density and temperature are intensive

SolutionAn extensive quantity doubles when you take twice as much of the substance. Volume and internal energy do. Temperature does not, and density does not either, because it is a ratio of two extensive quantities, mass and volume, so the doubling cancels.

Where it goes wrong  Calling density extensive because mass is. Any ratio of two extensive quantities comes out intensive.

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

Water enters the wide part of a pipe of radius 3 cm at 2 m/s and leaves through a narrow part of radius 1 cm. Find the speed at the exit.

Answer  18 m/s

SolutionFor an incompressible liquid A1v1 = A2v2, and area goes as r2. So v2 = v1(r1/r2)2 = 2 × (3/1)2 = 18 m/s.

Where it goes wrong  Using the ratio of radii without squaring, which gives 6 m/s.

Q13  ·  Wave Optics  ·  on the pattern of Q13

The same double slit set-up is lit first by blue light of 450 nm and then by red light of 650 nm. Which gives wider fringes, and by what factor?

Answer  Red, wider by 13/9, about 1.44 times

SolutionFringe width is β = λD/d. With D and d fixed, β is directly proportional to the wavelength. Red has the longer wavelength, so its fringes are wider, by 650/450 = 13/9 ≈ 1.44.

Where it goes wrong  Saying red fringes are closer. A bigger wavelength spreads the pattern out, it does not squeeze it.

Q14  ·  Work, Energy and Power  ·  on the pattern of Q14

A ball of mass 200 g is projected at 30 m/s at 60° to the horizontal. Find the loss in its kinetic energy by the time it reaches the highest point.

Answer  67.5 J

SolutionAt launch KE = ½(0.2)(302) = 90 J. At the top only the horizontal component survives: v = 30 cos60° = 15 m/s, so KE = ½(0.2)(152) = 22.5 J. The loss is 90 − 22.5 = 67.5 J.

Where it goes wrong  Taking the kinetic energy at the top as zero. The ball still moves horizontally there; only the vertical part dies.

Q15  ·  Semiconductor Electronics  ·  on the pattern of Q15

A gate gives an output of 1 only when both of its inputs are 0. Name the gate.

Answer  NOR gate

SolutionOR gives 1 whenever any input is 1, so its output is 0 only for inputs 0 and 0. A NOR gate is an OR followed by a NOT, so it gives 1 exactly in that one case and 0 in every other.

Where it goes wrong  Choosing NAND. NAND gives 0 only when both inputs are 1, which is the mirror image of this.

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

Light of wavelength λ falling on a metal of work function 2 eV ejects electrons of maximum kinetic energy 3 eV. Find the maximum kinetic energy when light of wavelength λ/2 is used.

Answer  8 eV

SolutionFrom Einstein’s equation the photon energy is hc/λ = KE + φ = 3 + 2 = 5 eV. Halving the wavelength doubles the photon energy to 10 eV. The work function does not change, so the new maximum kinetic energy is 10 − 2 = 8 eV.

Where it goes wrong  Doubling the kinetic energy to 6 eV. It is the photon energy that doubles, and the work function is subtracted only once.

Q17  ·  Oscillations  ·  on the pattern of Q17

A simple pendulum is taken to a planet whose mass is 9 times and radius 3 times that of the Earth. What happens to its time period?

Answer  It stays the same

SolutionSurface gravity is g = GM/R2. With mass 9 times and radius 3 times, g becomes 9/32 = 1 times, that is unchanged. Since T = 2π√(L/g) and both L and g are unchanged, the period is the same.

Where it goes wrong  Bringing the mass of the bob into it. The bob mass cancels out of the period formula altogether.

Q18  ·  Kinetic Theory of Gases  ·  on the pattern of Q18

For a rigid diatomic gas the ratio of specific heats is γ1. For a diatomic gas that also has a vibrational mode active, it is γ2. Compare them.

Answer  γ2 is smaller than γ1

SolutionA rigid diatomic molecule has 5 degrees of freedom, so Cv = 5R/2 and γ1 = 7/5 = 1.40. One active vibrational mode adds 2 more, giving Cv = 7R/2 and γ2 = 9/7 ≈ 1.29. So γ2 < γ1.

Where it goes wrong  Assuming more ways to store energy means a bigger γ. Extra modes raise Cv, and γ = 1 + R/Cv falls towards 1.

Q19  ·  Electric Charges and Fields  ·  on the pattern of Q19

A short dipole sits at the origin with its moment along the x axis. At a point on the x axis at distance r the potential is V0 and the field is E0. Find the potential and the field at a point on the y axis at distance 2r.

Answer  Potential zero, field E0/16

SolutionOn the axis V0 = kp/r2 and E0 = 2kp/r3, so kp/r3 = E0/2. Every point on the perpendicular bisector is equally far from both charges, so the potential there is exactly zero. The equatorial field at distance 2r is kp/(2r)3 = kp/8r3 = (E0/2)/8 = E0/16.

Where it goes wrong  Halving V0 instead of writing zero. The equatorial potential of a dipole is zero at every distance, not just far away.

Q20  ·  Electromagnetic Induction  ·  on the pattern of Q20

A rectangular loop is pulled out of a uniform magnetic field at a constant speed. Describe how the induced emf behaves with time while the loop is partly inside the field.

Answer  It stays constant while the loop is leaving, then drops to zero

SolutionOnly the part of the loop still inside the field has flux through it. At a constant speed the area inside shrinks at a constant rate, so dΦ/dt is constant and the emf is a flat line. The moment the loop is fully out there is no flux change left and the emf falls to zero.

Where it goes wrong  Expecting the emf to tail off slowly. The rate of change of area is fixed by the speed, not by how much is left inside.

Q21  ·  Electromagnetic Waves  ·  on the pattern of Q21

A parallel plate capacitor of plate area 20 cm2 is being charged by a current of 4 A. Find the displacement current through a flat surface of area 5 cm2 drawn inside the gap, parallel to the plates.

Answer  1 A

SolutionWhile the capacitor charges, the changing electric field between the plates is spread evenly over the plate area, so the displacement current density is uniform. The part passing through 5 cm2 is therefore 4 × (5/20) = 1 A.

Where it goes wrong  Taking the whole 4 A through the smaller surface. Only the share of the area counts.

Q22  ·  Moving Charges and Magnetism  ·  on the pattern of Q22

A proton moves undeflected at 3 × 105 m/s through crossed electric and magnetic fields. When the electric field is switched off, it moves in a circle of radius 3 cm. Find the electric field. (m = 1.6 × 10−27 kg, e = 1.6 × 10−19 C)

Answer  3 × 104 N/C

SolutionWith only the magnetic field left, r = mv/eB gives B = mv/er = (1.6 × 10−27 × 3 × 105)/(1.6 × 10−19 × 0.03) = 0.1 T. Going undeflected means the electric force balanced the magnetic force, eE = evB, so E = vB = 3 × 105 × 0.1 = 3 × 104 N/C.

Where it goes wrong  Starting with the electric field. The radius gives B first, and only then does E = vB follow.

Q23  ·  Laws of Motion  ·  on the pattern of Q23

A tube of length 2 m, closed at both ends and filled with a liquid of total mass 3M, is spun in a horizontal plane about one of its ends with angular velocity ω. Find the force the liquid exerts on the far end.

Answer  3Mω2

SolutionTake a slice of thickness dx at distance x. Its mass is (3M/L)dx and it needs a centripetal force (3M/L)ω2x dx. Adding from 0 to L gives F = (3M/L)ω2(L2/2) = 3Mω2L/2. With L = 2 m, F = 3Mω2.

Where it goes wrong  Putting the whole mass at the far end, which doubles the answer. The liquid near the axis needs much less force, so the average works out at L/2.

Q24  ·  Current Electricity  ·  on the pattern of Q24

A 12 V battery of negligible internal resistance is connected across a 6 Ω and a 12 Ω resistor joined in parallel. Find the current drawn from the battery.

Answer  3 A

SolutionThe parallel combination is (6 × 12)/(6 + 12) = 4 Ω. The current drawn is I = 12/4 = 3 A. As a check, the branches carry 12/6 = 2 A and 12/12 = 1 A, which add to 3 A.

Where it goes wrong  Adding the resistances as if in series. In parallel the total is always smaller than the smaller one.

Q25  ·  Moving Charges and Magnetism  ·  on the pattern of Q25

Two long parallel wires 10 cm apart carry 6 A and 4 A in opposite directions. Find the magnetic field at a point on the line joining them, 5 cm beyond the 4 A wire. (μ0 = 4π × 10−7 SI units)

Answer  8 × 10−6 T

SolutionThe point is 5 cm from the 4 A wire and 15 cm from the 6 A wire. Using B = μ0I/2πd: from the 4 A wire, B = (2 × 10−7 × 4)/0.05 = 1.6 × 10−5 T; from the 6 A wire, B = (2 × 10−7 × 6)/0.15 = 0.8 × 10−5 T. The currents are opposite, so outside the pair the two fields point opposite ways and subtract: 1.6 − 0.8 = 0.8 × 10−5 = 8 × 10−6 T.

Where it goes wrong  Adding the two magnitudes. For opposite currents the fields add only between the wires, and subtract outside them.

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

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