JEE Main 22 January 2026 Shift 2 — Physics Paper with Answer Key

The afternoon paper of 22 January 2026, the fourth shift 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

Date 22 January 2026
Shift Shift 2 — 3:00 PM to 6:00 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 Pendulum experiment for g — which plot of length against time period is right Units and Measurements
27 Ammeter reading at the instant a switch is closed on a coil circuit Electromagnetic Induction
28 Power line of 2 Ω delivering 1 kW at 250 V — percentage efficiency Current Electricity
29 Photoelectrons with zero kinetic energy — angular frequency of the light Dual Nature of Radiation and Matter
30 Two statements on the period of a satellite skimming the Earth Gravitation
31 Five equal charges at the corners of a pentagon — potential and field at the centre Electrostatic Potential and Capacitance
32 Capillary rise when radius, density and surface tension each fall by 1% Mechanical Properties of Fluids
33 Gap between the longest Paschen and the longest Balmer wavelength Atoms
34 Two statements on splitting the kinetic energy of a many-particle system Centre of Mass
35 Unit of permittivity multiplied into a changing electric field Units and Measurements
36 Open organ pipe — length from the gap between the third and sixth harmonics Waves
37 Where the object must sit in the parallax method for a concave mirror Ray Optics
38 Two statements on the work done by a conservative force Work, Energy and Power
39 Amplitude of the magnetic field in a laser beam of known intensity Electromagnetic Waves
40 Collision frequency in a gas of smaller, heavier molecules at the same T, P and density Kinetic Theory of Gases
41 Two point masses striking the ends of a bar and sticking — ratio of v to ω Rotational Motion
42 Three charged bubbles merging into one — ratio of the potentials Electrostatic Potential and Capacitance
43 Wavelength of the same light on passing from water into a denser medium Ray Optics
44 Which statements about the central maximum in single slit diffraction hold Wave Optics
45 Correct truth table for a given gate combination Semiconductor Electronics
46 numeric Loop spun about its diameter in a field — radius from the emf at 30° Electromagnetic Induction
47 numeric Two masses over a heavy disc pulley — speed after falling 3.6 m Rotational Motion
48 numeric Gas squeezed from 60 cm³ to 20 cm³ as it warms — final pressure Thermodynamics
49 numeric Charged capacitor joined to an uncharged one — charge that ends up on it Electrostatic Potential and Capacitance
50 numeric Mobility of electrons in a wire from the current and the voltage Current Electricity

Chapter by chapter

Chapter Questions Marks
Electrostatic Potential and Capacitance 3 12
Current Electricity 2 8
Electromagnetic Induction 2 8
Ray Optics 2 8
Rotational Motion 2 8
Units and Measurements 2 8
Atoms 1 4
Centre of Mass 1 4
Dual Nature of Radiation and Matter 1 4
Electromagnetic Waves 1 4
Gravitation 1 4
Kinetic Theory of Gases 1 4
Mechanical Properties of Fluids 1 4
Semiconductor Electronics 1 4
Thermodynamics 1 4
Wave Optics 1 4
Waves 1 4
Work, Energy and Power 1 4

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

Twenty chapters in twenty-five questions, the widest spread of any January shift. Five chapters took two questions each and fifteen took exactly one. A student who had left even three chapters untouched would have met a gap here.

Where the 100 marks sat

Part of Physics Questions Marks Share
Electrodynamics 8 32 32%
Mechanics 5 20 20%
Optics and Waves 4 16 16%
Properties of Matter and Heat 3 12 12%
Modern Physics and Electronics 3 12 12%
Units, Dimensions and Measurement 2 8 8%

Electrodynamics at 32 marks again, the same as the 21 January afternoon shift. Heat and properties of matter came down to 12, the lightest it was treated in January.

How tough it was

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

The friendliest January shift on the numbers: eight questions were one line each, and the five numeric questions in Section B were all straightforward. The two Bohr-series wavelengths and the bar struck at both ends were the only real time sinks.

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. NTA never put the January 2026 question papers on its public site — only the final answer key, which is linked further down.

Q1  ·  Units and Measurement  ·  on the pattern of Q26

In a simple pendulum experiment, g is found by measuring the time period T for several lengths L. Which graph comes out as a straight line through the origin?

Answer  L plotted against T2

SolutionThe period is T = 2π√(L/g), so squaring gives T2 = 4π2L/g, that is L = (g/4π2)T2. So L against T2 is a straight line through the origin, and its slope g/4π2 is what gives g. A plot of L against T is a parabola, and T against L is a square-root curve — neither is a straight line.

Where it goes wrong  Plotting T against L and trying to read a slope off a curve. Square the period first; that is the whole point of the graph.

Q2  ·  Electromagnetic Induction  ·  on the pattern of Q27

A 9 Ω resistor and a 4 mH inductor are joined in series across an 18 V battery through a switch, with an ammeter in the circuit. The ammeter reading at the instant the switch is closed is ______ A.

Answer  zero

SolutionAn inductor opposes any sudden change in the current through it. At the very instant of switching on, the current in the branch must still be what it was a moment before, which was zero, so the ammeter reads zero. The inductor behaves for that instant like a break in the wire. As time passes the current grows as I = (V/R)(1 − e−Rt/L) and finally settles at 18/9 = 2 A, when the inductor behaves like a plain wire.

Where it goes wrong  Giving 2 A, the steady value. That is the reading long afterwards, not at the instant of switching.

Q3  ·  Current Electricity  ·  on the pattern of Q28

A power line of total resistance 2 Ω delivers 1 kW of power at 250 V. The percentage efficiency of the line is ______.

Answer  96.9%

SolutionThe current carried is I = P/V = 1000/250 = 4 A. The line itself wastes I2R = 16 × 2 = 32 W as heat. So the power fed in at the far end is 1000 + 32 = 1032 W, and the efficiency is 1000/1032 = 0.969, that is 96.9%.

Where it goes wrong  Dividing by 1000 instead of 1032. The delivered power is the output; the input is output plus loss.

Q4  ·  Dual Nature of Matter  ·  on the pattern of Q29

Light falls on a metal plate whose work function is 110 × 10−20 J. The photoelectrons come out with zero kinetic energy. The angular frequency of the incident light is ______ rad/s. (h = 6.63 × 10−34 J·s.)

Answer  1.04 × 1016 rad/s

SolutionZero kinetic energy means the photon energy exactly equals the work function, so this is the threshold: hν = φ. Using ω = 2πν the whole thing can be done in one step: ω = 2πφ/h = 6.283 × 1.10 × 10−18/6.63 × 10−34 = 1.04 × 1016 rad/s.

Where it goes wrong  Giving the answer as ν and forgetting the 2π. The question asks for angular frequency, not frequency.

Q5  ·  Gravitation  ·  on the pattern of Q30

Two statements about a satellite circling the Earth just above the surface. Statement I: its period depends on the density of the Earth. Statement II: its period is T = 2π√(Re/g). Which is true?

Answer  Both statements are true

SolutionFor an orbit of radius Re, the gravitational pull supplies the centripetal force: GM/Re2 = ω2Re, and since g = GM/Re2 this gives ω2 = g/Re, that is T = 2π√(Re/g) — Statement II is true. Writing M as (4/3)πRe3ρ turns the same result into T = √(3π/Gρ), in which the radius has vanished and only the density is left — so Statement I is true as well.

Where it goes wrong  Calling Statement I false because the radius seems to matter. For a close orbit it cancels out, and only the density survives.

Q6  ·  Electrostatic Potential  ·  on the pattern of Q31

Five equal positive charges q are placed at the five corners of a regular pentagon of circumradius r. The potential V and the field E at the centre O are ______.

Answer  V = 5q/4πε0r and E = 0

SolutionPotential is a scalar, so the five contributions simply add: V = 5 × q/4πε0r. The field is a vector, and the five charges are arranged with perfect fivefold symmetry about O, so their five equal field vectors are spread 72° apart and add to nothing: E = 0.

Where it goes wrong  Assuming an odd number of charges cannot cancel. Symmetry, not parity, is what matters — any regular polygon of equal charges gives zero field at the centre.

Q7  ·  Mechanical Properties of Fluids  ·  on the pattern of Q32

A liquid rises to a height h in a vertical capillary tube. The inner radius of the tube, the density of the liquid and the surface tension of the liquid each decrease by 1%. The height changes by ______ %.

Answer  +1%

SolutionThe rise is h = 2Tcosθ/ρgr. Taking small fractional changes, Δh/h = ΔT/T − Δρ/ρ − Δr/r. Each of the three is −1%, so the sum is (−1) − (−1) − (−1) = +1%. The drop in surface tension would have lowered the column, but the drops in density and radius each raise it, and on balance it rises by 1%.

Where it goes wrong  Adding all three as −1% and answering −3%. Quantities in the denominator come in with the opposite sign.

Q8  ·  Atoms  ·  on the pattern of Q33

The shortest wavelength of the Lyman series is 91 nm. The difference between the longest wavelength of the Paschen series and the longest wavelength of the Balmer series is about ______ nm.

Answer  about 1216 nm

SolutionThe shortest Lyman wavelength is the series limit, ∞ → 1, so 1/91 nm−1 = R, which fixes the Rydberg constant in these units. The longest Balmer wavelength is the smallest jump, 3 → 2: 1/λ = R(1/4 − 1/9) = 5R/36, so λ = 36 × 91/5 = 655 nm. The longest Paschen wavelength is 4 → 3: 1/λ = R(1/9 − 1/16) = 7R/144, so λ = 144 × 91/7 = 1871 nm. The difference is 1871 − 655 = 1216 nm.

Where it goes wrong  Taking the longest wavelength of a series as the limit. The limit is the shortest; the longest comes from the smallest energy gap in the series.

Q9  ·  Centre of Mass  ·  on the pattern of Q34

Two statements about a system of many particles. Statement I: the total kinetic energy is the sum of the kinetic energies of all the particles. Statement II: the total kinetic energy can also be written as the kinetic energy of the centre of mass plus the kinetic energy of the particles measured relative to the centre of mass. Which is true?

Answer  Both statements are true

SolutionStatement I is the definition of the system’s kinetic energy. Statement II is the standard split: writing each particle’s velocity as the centre-of-mass velocity plus a velocity relative to the centre of mass, and squaring, the cross terms add to zero because Σmiui = 0 in the centre-of-mass frame. What is left is ½Mvcm2 + Σ½miui2. Both statements hold, and the second is the more useful one in collision problems.

Where it goes wrong  Calling Statement II false because it looks like double counting. It is the same energy written two ways, and the cross terms vanish exactly.

Q10  ·  Units and Dimensions  ·  on the pattern of Q35

If ε0 is the permittivity of free space and E is an electric field, the SI unit of ε0 multiplied by the rate of change of E with time is ______.

Answer  A/m2

Solutionε0E is the electric displacement, measured in coulomb per square metre. Dividing by time gives coulomb per second per square metre, which is ampere per square metre. That is no accident: ε0dE/dt is exactly the displacement current density that Maxwell added to Ampere’s law, and it has to carry the same unit as any other current density.

Where it goes wrong  Multiplying by time instead of dividing. ε0Et would be C·s/m2, which is not a current density at all.

Q11  ·  Waves  ·  on the pattern of Q36

In an open organ pipe the third and sixth harmonic frequencies differ by 2200 Hz. The length of the pipe is ______ mm. (Speed of sound in air = 330 m/s.)

Answer  225 mm

SolutionAn open pipe has all harmonics, νn = nv/2L. So ν6 − ν3 = 3v/2L = 2200, which gives 2L = 3 × 330/2200 = 0.45 m, so L = 0.225 m = 225 mm.

Where it goes wrong  Using the closed-pipe formula with only odd harmonics. The pipe here is open at both ends, so every integer multiple is present.

Q12  ·  Ray Optics  ·  on the pattern of Q37

In the parallax method for finding the focal length of a concave mirror, where must the object be placed?

Answer  Anywhere beyond the focus F of the mirror

SolutionThe parallax method works by moving a pin until it coincides with the image and the two stop shifting against each other when the eye moves. That needs a real image to coincide with, and a concave mirror gives a real image only when the object is further from the mirror than the focus. Anywhere beyond F will do — between F and C, at C, or beyond C — so the condition is simply that the object is outside the focus. Inside the focus the image is virtual and behind the mirror, and no pin can be placed on it.

Where it goes wrong  Insisting on the object being beyond C. That gives a real image too, but it is not the only position that does.

Q13  ·  Work, Energy and Power  ·  on the pattern of Q38

Two statements. Statement I: for an object moving from r1 to r2 in a conservative force field F, the work done by the force is W = −∫F·dr. Statement II: since an object can take infinitely many paths between two points, the work done by a conservative force depends on the path. Which is true?

Answer  Both statements are false

SolutionThe work done by the force is W = +∫F·dr. The minus sign belongs to the change in potential energy, ΔU = −∫F·dr, so Statement I has the sign on the wrong quantity. Statement II contradicts the very definition of conservative: the work between two points is the same along every path, which is why a potential energy can be defined at all. Both are false.

Where it goes wrong  Letting the familiar minus sign pass unchecked. It is correct for ΔU and wrong for W, and the two are easy to swap.

Q14  ·  Electromagnetic Waves  ·  on the pattern of Q39

A laser beam has an intensity of 4.0 × 1014 W/m2. The amplitude of the magnetic field in the beam is ______ T. (ε0 = 8.85 × 10−12 C2/N·m2, c = 3 × 108 m/s.)

Answer  about 1.83 T

SolutionThe intensity of an electromagnetic wave can be written in terms of the magnetic amplitude as I = cB02/2μ0. Rearranging, B0 = √(2μ0I/c) = √(2 × 4π × 10−7 × 4.0 × 1014/3 × 108) = √(3.35) = 1.83 T.

Where it goes wrong  Working out E0 first and then dividing by c, which is fine but invites a slip. Going straight through the magnetic form is one step.

Q15  ·  Kinetic Theory of Gases  ·  on the pattern of Q40

Two boxes hold ideal gases A and B at the same temperature, the same pressure and the same number density. A molecule of A is half the size of one of B, and four times as heavy. If the collision frequency in B is 32 × 1018 per second, the collision frequency in A is ______ per second.

Answer  4 × 1018 per second

SolutionThe collision frequency goes as √2 πd2n vrms, so with n the same it depends on d2 and on vrms, which itself goes as 1/√m at fixed temperature. Halving the diameter divides d2 by 4; making the molecule four times heavier divides the speed by 2. Together the frequency falls by a factor of 8: 32 × 1018/8 = 4 × 1018 per second.

Where it goes wrong  Changing the number density because the molecules are smaller. The question fixes the number density, so only size and mass matter.

Q16  ·  Rotational Motion  ·  on the pattern of Q41

A uniform bar of length 12 cm and mass 20m lies on a smooth horizontal table. Two point masses, each of mass m, move in opposite directions with the same speed v in the plane of the table, at right angles to the bar. They strike the two ends of the bar at the same instant and stick to it. The whole system then turns with angular velocity ω. The ratio v/ω is ______.

Answer  0.26 m

SolutionThe two incoming momenta are equal and opposite, so the total linear momentum is zero and the centre of mass stays where it was, at the middle of the bar. Angular momentum about that point is conserved: each mass arrives with mv(L/2), and because they come from opposite sides and move oppositely, the two turn the system the same way, giving mvL in all. After the collision the moment of inertia is that of the bar, 20mL2/12 = 5mL2/3, plus the two stuck masses at L/2, which add 2m(L/2)2 = mL2/2. Total I = (13/6)mL2. So ω = mvL/[(13/6)mL2] = 6v/13L, and v/ω = 13L/6 = 13 × 0.12/6 = 0.26 m.

Where it goes wrong  Treating the two angular momenta as cancelling. Opposite velocities arriving at opposite ends spin the bar the same way, so they add.

Q17  ·  Electrostatic Potential  ·  on the pattern of Q42

Three identical water bubbles, each carrying the same charge, merge into one bigger bubble. The ratio of the potential on one of the original bubbles to the potential on the bigger one is ______.

Answer  1 : 32/3

SolutionWater is conserved, so the volumes add: 3 × (4/3)πr3 = (4/3)πR3, giving R = 31/3r. Charge is conserved too, so the big bubble carries 3q. The potential of a charged sphere is kQ/radius, so the small one has kq/r and the big one k(3q)/(31/3r) = 32/3kq/r. The ratio is therefore 1 : 32/3, about 1 : 2.08.

Where it goes wrong  Adding the radii instead of the volumes. The radius grows only as the cube root of the number merged.

Q18  ·  Ray Optics  ·  on the pattern of Q43

Light passing through water has a wavelength of 540 nm. The refractive index of water is 4/3. The wavelength of the same light while passing through a transparent medium of refractive index 3/2 is ______ nm.

Answer  480 nm

SolutionThe frequency never changes when light enters a new medium; the wavelength does, as λ = λvacuum/μ. From the water figure, λvacuum = 540 × 4/3 = 720 nm. In the new medium, λ = 720/(3/2) = 480 nm.

Where it goes wrong  Scaling 540 nm by 4/3 ÷ 3/2 in one go without care over which way round it goes. Find the vacuum wavelength first; it is the fixed reference.

Q19  ·  Wave Optics  ·  on the pattern of Q44

For single slit diffraction, which of these are true? A. The width of the central maximum grows when the wavelength is increased at fixed slit width. B. It grows when the wavelength is decreased at fixed slit width. C. It grows when the slit is made narrower at fixed wavelength. D. It grows when the slit is made wider at fixed wavelength.

Answer  A and C only

SolutionThe first minimum sits where a sinθ = λ, so the angular half-width of the central maximum is λ/a, and on a screen a distance D away the full width is 2λD/a. The wavelength is on top, so a longer wavelength widens it — A is true and B is false. The slit width is underneath, so a narrower slit widens it — C is true and D is false.

Where it goes wrong  Carrying over the intuition from a double slit, where a smaller separation widens the fringes. The direction is the same here, but the quantity is the slit width itself, not a separation.

Q20  ·  Semiconductor Electronics  ·  on the pattern of Q45

Inputs A and B are fed to an OR gate and also to an AND gate. The two outputs are then fed to a second AND gate, whose output is Y. Write the truth table for Y.

Answer  Y = A AND B — Y is 1 only when A = B = 1

SolutionTake the four cases. A = 0, B = 0: OR gives 0, AND gives 0, so Y = 0. A = 0, B = 1: OR gives 1, AND gives 0, so Y = 0. A = 1, B = 0: same, Y = 0. A = 1, B = 1: OR gives 1, AND gives 1, so Y = 1. The whole arrangement reduces to a plain AND gate, because (A + B)·(A·B) = A·B — whenever A·B is 1, A + B must be 1 too.

Where it goes wrong  Writing out the table row by row without noticing the arrangement simplifies. Spotting that one input already implies the other saves all four rows.

Q21  ·  Electromagnetic Induction  ·  numeric  ·  on the pattern of Q46

A conducting circular loop is spun about its own diameter at a steady 100 rad/s in a magnetic field of 0.5 T at right angles to the axis of rotation. When the loop has turned 30° from the horizontal position, the induced emf is 15.4 mV. The radius of the loop is ______ mm. (π = 22/7.)

Answer  14 mm

SolutionFor a loop spun in a field the emf is BAωsinθ, where θ is measured from the position of zero emf. At 30°, sinθ = ½, so 15.4 × 10−3 = 0.5 × πr2 × 100 × ½ = 25πr2. With π = 22/7 that is 78.57r2, so r2 = 1.96 × 10−4 and r = 0.014 m = 14 mm.

Where it goes wrong  Using cos30° because 30° sounds like an angle from the field. Here it is measured from the position where the emf is zero, so it is the sine that enters.

Q22  ·  Rotational Motion  ·  numeric  ·  on the pattern of Q47

Masses m and 2m are joined by a light string passing over a pulley which is a disc of mass 30m and radius 0.1 m, mounted in a vertical plane and free to turn about its axis. The 2m mass is released from rest. Its speed after it has fallen 3.6 m is ______ m/s. (The string does not slip; g = 10 m/s2.)

Answer  2 m/s

SolutionFor masses over a pulley of moment of inertia I, a = (2m − m)g/(2m + m + I/r2). The disc gives I = ½(30m)r2, so I/r2 = 15m and the bracket is 3m + 15m = 18m. Hence a = mg/18m = 10/18 = 5/9 m/s2. Then v2 = 2as = 2 × (5/9) × 3.6 = 4, so v = 2 m/s.

Where it goes wrong  Adding the pulley’s full mass 30m to the bracket. It is I/r2 that goes in, which for a disc is half the mass.

Q23  ·  Thermodynamics  ·  numeric  ·  on the pattern of Q48

A cylinder of volume 60 cm3 holds a gas at 27 °C and 2 atmospheres. The gas is squeezed to a final volume of 20 cm3 while its temperature is allowed to rise to 77 °C. The final pressure is ______ atmospheres.

Answer  7 atmospheres

SolutionThe amount of gas does not change, so P1V1/T1 = P2V2/T2 with the temperatures in kelvin: 300 K and 350 K. So 2 × 60/300 = P2 × 20/350, that is 0.4 = P2/17.5, giving P2 = 7 atmospheres.

Where it goes wrong  Putting 27 and 77 straight into the ratio. The gas equation needs absolute temperature, so add 273 to both.

Q24  ·  Capacitance  ·  numeric  ·  on the pattern of Q49

A capacitor P of 10 μF is charged fully to 6.0 V and disconnected from the battery. It is then connected across an uncharged capacitor Q of 20 μF. The charge that ends up on Q is α × 10−5 C. The value of α is ______.

Answer  4

SolutionThe charge on P before the connection is 10 × 6 = 60 μC, and nothing can leave the pair, so 60 μC is shared. Once settled, the two have the same voltage, so they behave like one capacitor of 30 μF: V = 60/30 = 2 V. The charge on Q is then 20 × 2 = 40 μC = 4 × 10−5 C, so α = 4. The other 20 μC stays on P.

Where it goes wrong  Assuming the charge splits in half. It splits in proportion to the capacitances, because the voltage is what the two end up sharing.

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

A cylindrical conductor 2 m long with a cross-section of 0.2 mm2 carries 1.6 A when its ends are joined to a 2 V battery. The mobility of the electrons is α × 10−3 m2/V·s. The value of α is ______. (Electron concentration = 5 × 1028/m3, e = 1.6 × 10−19 C.)

Answer  1

SolutionThe drift speed comes from I = nAevd: vd = 1.6/(5 × 1028 × 0.2 × 10−6 × 1.6 × 10−19) = 1.6/1600 = 10−3 m/s. The field inside is E = V/L = 2/2 = 1 V/m. Mobility is drift speed per unit field, so μ = 10−3/1 = 10−3 m2/V·s, giving α = 1.

Where it goes wrong  Leaving the area in mm2. 0.2 mm2 is 0.2 × 10−6 m2, not 0.2 × 10−3.

The official answer key

NTA did not publish the Session 1 question papers. They were shown only inside each candidate’s own login during the answer key challenge window in February 2026, and that window is closed. The one official file for this shift is the final answer key, which covers all ten January shifts. Shift 1 of the same day, 22 January 2026, is on its own page, and the other January dates are 21, 23, 24 and 28 January.

Official final answer key →

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

Sit a paper in the same shape, free

25 questions. 45 minutes. Out of 100. Marked the way JEE Main marks it, with a hint and the answer key for every question you miss. No payment, no card, and no account needed to start.

Start the free paper →

Reading a past paper tells you what was asked. Sitting one on the clock tells you what you can do. Only the second one moves your score.

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