JEE Main 6 April 2026 Shift 1 — Physics Paper with Answer Key

The morning shift of 6 April 2026. 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 6 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 Density of a cylinder from three measurements — percentage error Units and Measurements
27 Potential energy given as a formula — dimensions of the constants Units and Measurements
28 Raindrop falling a kilometre — work done by the resistive force Work, Energy and Power
29 Two blocks on an accelerating cube — coefficient of friction Laws of Motion
30 Lift hanging from a wire — largest safe acceleration Mechanical Properties of Solids
31 Spinning sphere brought to rest — torque and number of turns Rotational Motion
32 Incline leading into a vertical loop — release height Work, Energy and Power
33 Three masses at the corners of a triangle — centre of mass Centre of Mass
34 Two joined wires stretching equally — length of the second Mechanical Properties of Solids
35 Two joined vessels, one heated — the final pressure Kinetic Theory of Gases
36 Two points with different path differences — ratio of intensities Wave Optics
37 SHM — time to move from the extreme to a given point Oscillations
38 Half ring of charge — electric field at its centre Electric Charges and Fields
39 LED in a circuit — smallest safe series resistance Semiconductor Electronics
40 Point source of waves — detector moved on the same sphere Electromagnetic Waves
41 Single spherical refracting surface — magnification Ray Optics
42 Magnetic energy inside a small cube at a loop’s centre Moving Charges and Magnetism
43 Inductor with resistance across a battery — two instantaneous voltages Electromagnetic Induction
44 Momenta of the first two photons of the Balmer series Atoms
45 Series LCR circuit — the power factor Alternating Current
46 numeric Heat produced in one resistor of a network Current Electricity
47 numeric Optically active solution between polariser and analyser Wave Optics
48 numeric Lithium bombarded by protons — energy released Nuclei
49 numeric Work done moving a charge through a varying field Electrostatic Potential and Capacitance
50 numeric Gas compressed isothermally — work done on it Thermodynamics

Chapter by chapter

Chapter Questions Marks
Mechanical Properties of Solids 2 8
Units and Measurements 2 8
Wave Optics 2 8
Work, Energy and Power 2 8
Alternating Current 1 4
Atoms 1 4
Centre of Mass 1 4
Current Electricity 1 4
Electric Charges and Fields 1 4
Electromagnetic Induction 1 4
Electromagnetic Waves 1 4
Electrostatic Potential and Capacitance 1 4
Kinetic Theory of Gases 1 4
Laws of Motion 1 4
Moving Charges and Magnetism 1 4
Nuclei 1 4
Oscillations 1 4
Ray Optics 1 4
Rotational Motion 1 4
Semiconductor Electronics 1 4
Thermodynamics 1 4

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

Twenty-one chapters, and only three of them came up twice.

Where the 100 marks sat

Part of Physics Questions Marks Share
Electrodynamics 7 28 28%
Mechanics 6 24 24%
Properties of Matter and Heat 4 16 16%
Optics 3 12 12%
Modern Physics and Electronics 3 12 12%
Units, Dimensions and Measurement 2 8 8%

Exactly the same split as 2 April Shift 2 and 5 April Shift 2. Over six shifts this 28 – 24 – 16 shape has now appeared three times, which makes it the safest thing to plan revision around.

How tough it was

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

The two blocks on the accelerating cube was the hardest single question of the session so far: a pulley, a pseudo force and friction, all in one figure. Anyone who sat on it lost five minutes for four marks.

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 6 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

The density of a uniform cylinder is found by measuring its mass m, length l and diameter d. The measured values are m = 50.0 ± 0.1 g, l = 10.00 ± 0.01 mm and d = 5.00 ± 0.01 mm. The percentage error in the calculated density is ______ %.

Answer  0.7 %

SolutionThe density is ρ = m / (πd2l/4), so ρ depends on m once, on d twice and on l once. The fractional errors add, with the power as a multiplier: Δρ/ρ = Δm/m + 2Δd/d + Δl/l = 0.1/50 + 2(0.01/5) + 0.01/10 = 0.002 + 0.004 + 0.001 = 0.007. As a percentage that is 0.7 %.

Where it goes wrong  Forgetting the factor of 2 on the diameter. It appears squared in the volume, so its error counts double and here it is the biggest single contribution.

Q2  ·  Units and Dimensions  ·  on the pattern of Q27

The potential energy of a particle changes with the distance x from a fixed origin as V = Ax2/(x3 + B), where A and B are constants. The dimensions of AB are ______.

Answer  [M L6 T−2]

SolutionOnly quantities with the same dimensions can be added, so B must match x3, giving [B] = L3. Now V is an energy, [V] = ML2T−2. Reading the formula, [A]L2/L3 = [A]L−1 = ML2T−2, so [A] = ML3T−2. Multiplying, [AB] = ML3T−2 × L3 = ML6T−2.

Where it goes wrong  Going straight for A. Always read off the added constant first; it is given to you free by the rule that only like quantities add.

Q3  ·  Work, Energy and Power  ·  on the pattern of Q28

A raindrop of mass 2 g starts from rest at a height of 2 km and reaches the ground at 6 m/s. The work done on it by the resistive force of the air is ______ J. (g = 10 m/s2)

Answer  about −39.96 J

SolutionThe work–energy theorem says the total work equals the change in kinetic energy. Gravity does Wg = mgh = 0.002 × 10 × 2000 = 40 J. The kinetic energy gained is ½mv2 = ½ × 0.002 × 36 = 0.036 J. So the resistive work is 0.036 − 40 = −39.96 J. The minus sign says the air took almost all the energy away.

Where it goes wrong  Giving the answer as positive. Resistance always opposes the motion, so its work is negative.

Q4  ·  Laws of Motion  ·  on the pattern of Q29

A block P of mass 2 kg lies on the frictionless top face of a cube, and a block Q of mass 1 kg is pressed against the vertical side face, where the coefficient of friction is μ. They are joined by a light thread running over a frictionless pulley fixed at the top edge. The cube is pushed horizontally with an acceleration of 2g/3 and both blocks stay still relative to it. The smallest value of μ is ______. (g = 10 m/s2)

Answer  0.5

SolutionWork in the cube’s frame, where each block also feels a pseudo force ma backwards. For P on the frictionless top, the only horizontal force left is the tension, so T = mPa = 2 × (2 × 10/3) = 13.33 N. For Q, the side face pushes it with the normal force that supplies its acceleration: N = mQa = 1 × 6.67 = 6.67 N. Vertically, Q has its weight 10 N down and the tension 13.33 N up, so friction must pull down with 3.33 N. For that, μN ≥ 3.33, giving μ ≥ 3.33/6.67 = 0.5.

Where it goes wrong  Assuming friction acts upward. Here the tension is larger than the weight, so friction has to hold Q down, not up.

Q5  ·  Mechanical Properties of Solids  ·  on the pattern of Q30

A lift of mass 1500 kg hangs from a steel wire of radius 5 mm. The greatest stress the wire can take is 3 × 108 N/m2. The largest upward acceleration the lift can have is ______ m/s2. (g = 10 m/s2, π = 3.14)

Answer  about 5.7 m/s2

SolutionThe greatest tension the wire can take is stress × area = 3 × 108 × 3.14 × (0.005)2 = 3 × 108 × 7.85 × 10−5 = 23550 N. For a lift going up with acceleration a, T = m(g + a). So 23550 = 1500(10 + a), giving 10 + a = 15.7 and a = 5.7 m/s2.

Where it goes wrong  Using T = ma and leaving out the weight. The wire has to hold the lift up as well as accelerate it.

Q6  ·  Rotational Motion  ·  on the pattern of Q31

A solid sphere of radius 5 cm and mass 4 kg spins about an axis through its centre at 600 rpm. A constant torque brings it to rest in 20 s. The torque applied and the number of turns it makes before stopping are ______.

Answer  0.004π N·m, about 0.0126 N·m, and 100 turns

SolutionThe moment of inertia is I = (2/5)mR2 = 0.4 × 4 × (0.05)2 = 0.004 kg·m2. The starting rate is 600 rpm = 10 turns per second = 20π rad/s. The angular retardation is α = 20π/20 = π rad/s2, so the torque is τ = Iα = 0.004π N·m. For the turns, the rate falls steadily from 10 per second to zero, so the average is 5 per second, and over 20 s that is 100 turns.

Where it goes wrong  Leaving the speed in rpm. Torque needs rad/s, though for counting the turns the rpm figure is actually the quicker route.

Q7  ·  Work, Energy and Power  ·  on the pattern of Q32

A smooth inclined track ends in a vertical circular loop of radius R. A small body is released from a height h. At the highest point of the loop it presses on the track with a force equal to twice its own weight. If h = αR, the value of α is ______.

Answer  3.5

SolutionAt the top of the loop both the track’s push N and the weight point towards the centre: N + mg = mv2/R. With N = 2mg this gives 3mg = mv2/R, so v2 = 3gR. Now use energy from the release point to the top, which is 2R above the ground: mgh = mg(2R) + ½mv2 = 2mgR + 1.5mgR = 3.5mgR. So h = 3.5R and α = 3.5.

Where it goes wrong  Forgetting that the top of the loop is 2R high, not R. The radius is only half the way up.

Q8  ·  Centre of Mass  ·  on the pattern of Q33

Three masses sit at the corners of a triangle: 1 kg at (0, 0), 2 kg at (4, 0) and 3 kg at (2, 6), all in metres. The centre of mass of the three is at ______.

Answer  (7/3, 3) m, about (2.33, 3)

SolutionTake each coordinate separately, weighting by mass. Total mass is 6 kg. xcm = [1(0) + 2(4) + 3(2)]/6 = 14/6 = 7/3 m. ycm = [1(0) + 2(0) + 3(6)]/6 = 18/6 = 3 m.

Where it goes wrong  Taking the plain average of the three corners. That is the centroid, which only matches the centre of mass when the three masses are equal.

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

Two wires A and B of equal cross-section but different materials are joined end to end. The ratio of the Young’s modulus of A to that of B is 5 : 2. When the joined wire is put under tension, the elongations of A and B turn out to be equal. If A is 3.5 m long, the length of B is ______ m.

Answer  1.4 m

SolutionThe same tension passes through both and they have the same area, so ΔL = FL/AY reduces to ΔL proportional to L/Y. Setting the two elongations equal, LA/YA = LB/YB, so LB = LA(YB/YA) = 3.5 × (2/5) = 1.4 m.

Where it goes wrong  Turning the modulus ratio the wrong way up. The stiffer wire needs to be shorter to stretch by the same amount.

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

Two closed vessels of the same volume are joined by a narrow tube. Both hold air at 90 kPa and 400 K. The first vessel is held at 400 K while the second is raised to 600 K. The final pressure in the vessels is ______ kPa.

Answer  108 kPa

SolutionThe gas is shut in, so the total number of moles does not change, and because the tube is narrow the pressure is the same everywhere. At the start the total is 2 × (90V/R × 400). At the end it is P′V/(R × 400) + P′V/(R × 600). Cancelling V and R, P′(1/400 + 1/600) = 180/400. The bracket is 5/1200, so P′ = 0.45 × 1200/5 = 108 kPa.

Where it goes wrong  Taking the average of the two temperatures. The gas moves from the hot vessel into the cold one, and only the mole count is conserved.

Q11  ·  Wave Optics  ·  on the pattern of Q36

In an interference experiment the path difference between the two waves at a point A on the screen is λ/4, and at another point B it is λ/3. The ratio of the intensities at A and B is ______.

Answer  2 : 1

SolutionThe intensity is I = Imaxcos2(φ/2), and the phase is φ = 2πΔ/λ, so I = Imaxcos2(πΔ/λ). At A, Δ = λ/4, so cos245° = 1/2. At B, Δ = λ/3, so cos260° = 1/4. The ratio is (1/2)/(1/4) = 2.

Where it goes wrong  Putting the whole phase inside the cosine instead of half of it. The half is what links intensity to path difference for two sources.

Q12  ·  Oscillations  ·  on the pattern of Q37

A particle performs simple harmonic motion of amplitude A and time period 6 s. The time it takes to move from x = A to x = A/2 is ______ s.

Answer  1 s

SolutionStart the clock at the extreme, so x = A cosωt with ω = 2π/T = 2π/6 = π/3 rad/s. Setting x = A/2 gives cosωt = ½, so ωt = π/3 and t = (π/3)/(π/3) = 1 s, which is one sixth of the period.

Where it goes wrong  Starting from x = A with a sine. From the extreme the motion is a cosine; the sine starts from the centre.

Q13  ·  Electric Charges and Fields  ·  on the pattern of Q38

A thin half ring of radius 20 cm carries a charge Q spread evenly along it. The electric field at the centre of the half ring is 90 V/m. The value of Q is about ______ nC. (1/4πε0 = 9 × 109 N·m2/C2, π = 3.14)

Answer  about 0.63 nC

SolutionFor a half ring the components along the diameter cancel in pairs and only the perpendicular ones survive, which gives E = (1/4πε0)(2λ/R) at the centre, where λ = Q/πR. Putting the second into the first, E = (1/4πε0)(2Q/πR2) = 9 × 109 × 2Q/(3.14 × 0.04) = 1.43 × 1011Q. Setting that to 90 gives Q = 6.3 × 10−10 C, that is about 0.63 nC.

Where it goes wrong  Expecting zero, as for a full ring. Half a ring has nothing to cancel the perpendicular components, so the field does not vanish.

Q14  ·  Semiconductor Electronics  ·  on the pattern of Q39

An LED is rated at a maximum power of 3 mW and drops 2 V across itself when lit. It is connected through a series resistor RS to a 5 V supply. The smallest safe value of RS is ______ kΩ.

Answer  2 kΩ

SolutionThe largest current the LED can take is I = P/V = 3 × 10−3/2 = 1.5 mA. The LED holds 2 V, so the resistor takes the remaining 5 − 2 = 3 V. The smallest safe resistance is RS = 3/(1.5 × 10−3) = 2000 Ω = 2 kΩ. Anything smaller lets too much current through.

Where it goes wrong  Dividing the full 5 V by the current. The LED keeps its own 2 V whatever happens, so only the rest falls across the resistor.

Q15  ·  Electromagnetic Waves  ·  on the pattern of Q40

A point source emits electromagnetic waves in free space. A detector at a distance L measures an intensity I0. The detector is moved to another point on the same spherical surface, so that the angle between the two positions, measured at the source, is 60°. The intensity measured now is ______.

Answer  I0, exactly the same

SolutionA point source sends out energy equally in every direction, so the intensity depends only on the distance: I = P/4πr2. Both positions lie on the same sphere of radius L, so r has not changed and neither has I. The 60° is there only to tempt you into using it.

Where it goes wrong  Applying Malus’s law and multiplying by cos260°. That is for polarisers, and nothing here is polarising anything.

Q16  ·  Ray Optics  ·  on the pattern of Q41

A single spherical refracting surface of radius R separates air from a medium of refractive index 1.5. A point object is placed in the air at a distance 3R from the surface, on the side from which the surface bulges away. The magnification of the image is ______.

Answer  −2, that is inverted and twice the size

SolutionFor one refracting surface, n2/v − n1/u = (n2 − n1)/R. With n1 = 1, n2 = 1.5 and u = −3R: 1.5/v + 1/3R = 0.5/R. So 1.5/v = 0.5/R − 1/3R = 1/6R, giving v = 9R. The magnification for a single surface is m = n1v/n2u = (1 × 9R)/(1.5 × (−3R)) = −2.

Where it goes wrong  Using m = v/u as for a lens. A single refracting surface carries the two refractive indices in its magnification formula.

Q17  ·  Moving Charges and Magnetism  ·  on the pattern of Q42

A small cube of side 2 mm sits at the centre of a circular loop of radius 20 cm carrying a current of 4 A. The magnetic energy stored inside the cube is α × 10−14 J. The value of α is about ______. (μ0 = 4π × 10−7 T·m/A, π = 3.14)

Answer  about 50.3

SolutionThe field at the centre of a loop is B = μ0I/2R = (4π × 10−7 × 4)/(2 × 0.2) = 1.26 × 10−5 T. Because the cube is tiny, the field is as good as uniform across it. The energy per unit volume is u = B2/2μ0 = (1.26 × 10−5)2/(2 × 1.26 × 10−6) = 6.28 × 10−5 J/m3. The volume is (2 × 10−3)3 = 8 × 10−9 m3, so U = 5.03 × 10−13 J, that is 50.3 × 10−14 J.

Where it goes wrong  Using the side of the cube instead of its volume. The energy density is per cubic metre, so the side has to be cubed.

Q18  ·  Electromagnetic Induction  ·  on the pattern of Q43

An inductor of 20 mH with a resistance of 50 Ω is connected through a switch to a battery of emf 2.0 V. After the switch is closed, the ratio of the instantaneous voltage across the inductance when the current is 10 mA to that when it is 20 mA is ______.

Answer  3 : 2

SolutionGoing round the loop, E = iR + L(di/dt), so the voltage across the inductance itself is L(di/dt) = E − iR. At i = 10 mA that is 2 − (0.01 × 50) = 1.5 V. At i = 20 mA it is 2 − (0.02 × 50) = 1.0 V. The ratio is 1.5 : 1.0 = 3 : 2. Note that the inductance itself never enters the answer.

Where it goes wrong  Reaching for the exponential growth formula. The current is given at each instant, so the loop equation settles it in one line.

Q19  ·  Atoms  ·  on the pattern of Q44

The ratio of the momenta of the photons of the 1st and the 2nd lines of the Balmer series of hydrogen is α : β. The possible values of α and β are ______.

Answer  20 and 27

SolutionThe Balmer series ends at n = 2, so the first line is 3 to 2 and the second is 4 to 2. Using 1/λ = R(1/4 − 1/n2): for 3 to 2, 1/4 − 1/9 = 5/36; for 4 to 2, 1/4 − 1/16 = 3/16. A photon’s momentum is h/λ, so the momenta are in the same ratio as those numbers: (5/36) : (3/16) = 80 : 108 = 20 : 27.

Where it goes wrong  Starting the Balmer series at 2 to 1. The series is named for where the electron lands, and Balmer lands on n = 2.

Q20  ·  Alternating Current  ·  on the pattern of Q45

A series LCR circuit is driven at a frequency where the resistance is 60 Ω, the inductive reactance is 30 Ω and the capacitive reactance is 110 Ω. The power factor of the circuit is ______.

Answer  0.6

SolutionThe net reactance is XL − XC = 30 − 110 = −80 Ω; the minus sign only says the circuit is capacitive. The impedance is Z = √(602 + 802) = √10000 = 100 Ω. The power factor is cosφ = R/Z = 60/100 = 0.6.

Where it goes wrong  Adding the two reactances instead of subtracting. They act in opposite directions, so one is taken off the other.

Q21  ·  Current Electricity  ·  numeric  ·  on the pattern of Q46

A 12 V battery of negligible internal resistance drives current through a 2 Ω resistor in series with a parallel pair of 6 Ω and 3 Ω resistors. The heat produced in the 6 Ω resistor in 50 s is ______ J.

Answer  300 J

SolutionThe parallel pair is 6 × 3/(6 + 3) = 2 Ω, so the whole circuit is 2 + 2 = 4 Ω and the battery current is 12/4 = 3 A. The voltage across the parallel pair is 3 × 2 = 6 V, so the current in the 6 Ω resistor is 1 A. Its power is I2R = 1 × 6 = 6 W, and in 50 s the heat is 6 × 50 = 300 J.

Where it goes wrong  Using the full 3 A in the 6 Ω resistor. The current splits at the junction, and the larger resistance takes the smaller share.

Q22  ·  Wave Optics  ·  numeric  ·  on the pattern of Q47

Unpolarised light of intensity I0 passes through a polariser, then through an optically active solution, and finally reaches an analyser whose axis is parallel to the polariser’s. The light leaving the analyser has intensity I0/4. The angle through which the solution turns the plane of polarisation is ______ degrees.

Answer  45°

SolutionUnpolarised light loses half at the first polariser, leaving I0/2. The solution turns the plane by θ without changing the intensity. At the analyser, whose axis is still at the original angle, Malus’s law gives I = (I0/2)cos2θ. Setting that to I0/4, cos2θ = ½ and θ = 45°.

Where it goes wrong  Forgetting the first halving. The solution does not reduce the intensity at all; only the polariser and the analyser do.

Q23  ·  Nuclei  ·  numeric  ·  on the pattern of Q48

A 7Li nucleus is struck by a proton and turns into two 4He nuclei. The energy released in one such reaction is about ______ MeV. [mass of 7Li = 7.0183 u, mass of 4He = 4.004 u, mass of proton = 1.008 u, 1 u = 931 MeV/c2]

Answer  about 17.0 MeV

SolutionThe reaction is 7Li + p → 2 4He. The mass before is 7.0183 + 1.008 = 8.0263 u and the mass after is 2 × 4.004 = 8.008 u. The mass lost is 0.0183 u, and turning that into energy gives 0.0183 × 931 = 17.0 MeV. Check the bookkeeping first: 7 + 1 = 8 nucleons, and 3 + 1 = 4 protons, which matches two helium nuclei.

Where it goes wrong  Using one helium nucleus instead of two. Balancing the nucleons before anything else prevents it.

Q24  ·  Electrostatic Potential  ·  numeric  ·  on the pattern of Q49

A charge of 2 C moves from the point (0, −1, −2) to the point (3, 2, 1) in an electric field E = 2xi + 3y2j + 4k N/C. The work done in moving the charge is ______ J.

Answer  60 J

SolutionThe work is W = q∫E·dr, and since each component of E depends only on its own coordinate the three integrals separate. ∫2x dx from 0 to 3 is x2 = 9. ∫3y2dy from −1 to 2 is y3 = 8 − (−1) = 9. ∫4 dz from −2 to 1 is 4 × 3 = 12. Adding, 9 + 9 + 12 = 30, and multiplying by the charge, W = 2 × 30 = 60 J.

Where it goes wrong  Worrying about the path taken. The field is conservative, so only the two end points matter.

Q25  ·  Thermodynamics  ·  numeric  ·  on the pattern of Q50

A gas is compressed isothermally to one quarter of its initial volume of 4 litre by applying the needed pressure. The bulk modulus of the gas is 2 × 105 N/m2. The magnitude of the work done on the gas is about ______ J.

Answer  about 1.11 × 103 J

SolutionFor an ideal gas held at constant temperature the bulk modulus is simply equal to the pressure, so the starting pressure is P0 = 2 × 105 N/m2. The work in an isothermal change is W = P0V0 ln(V0/V) = 2 × 105 × 4 × 10−3 × ln4 = 800 × 1.386 = 1109 J, done on the gas.

Where it goes wrong  Using W = PΔV. The pressure is climbing all through the compression, so the logarithm is needed.

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 →

Nearby shifts: 5 April Shift 1 · 5 April Shift 2.

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