JEE Main 21 January 2026 Shift 1 — Physics Paper with Answer Key

The opening shift of JEE Main 2026, the first 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

Date 21 January 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 Satellite moved to a larger orbit — energy that has to be supplied Gravitation
27 Alpha particle fired at a gold foil — closest it can get Atoms
28 Electromagnetic wave — magnetic field from the electric field Electromagnetic Waves
29 A changed Bernoulli equation — dimensions of its two constants Units and Measurements
30 Force that grows with time — velocity and position later Laws of Motion
31 Dielectric slab filling part of a capacitor — new capacitance Electrostatic Potential and Capacitance
32 Sheet over one slit in a double slit set-up — its thickness Wave Optics
33 Light carrying two frequencies — greatest kinetic energy of the electrons Dual Nature of Radiation and Matter
34 Two measured spring constants in parallel — percentage error Units and Measurements
35 Gas geyser heating flowing water — rate at which gas burns Thermal Properties of Matter
36 Charge released on the axis of a vertical solenoid — its acceleration Moving Charges and Magnetism
37 Work done moving a charge between two points Electrostatic Potential and Capacitance
38 Logic circuit — which single gate it behaves as Semiconductor Electronics
39 Rod hanging from two strings — tension the instant one is cut Rotational Motion
40 Aluminium and steel rods joined — length after heating Thermal Properties of Matter
41 Pulses sent along two joined strings — ratio of their travel times Waves
42 Potential energy against distance — ranking the forces Work, Energy and Power
43 Rod sliding in a magnetic field — force needed to keep it moving Electromagnetic Induction
44 Horizontal tube with two standpipes — rate of flow Mechanical Properties of Fluids
45 Loop in a field that varies as a sine — average heat in one period Electromagnetic Induction
46 numeric Two rods joined — moment of inertia about a corner point Rotational Motion
47 numeric Heat generated in a circuit in one minute Current Electricity
48 numeric Beam expander — focal length of the second lens Ray Optics
49 numeric Oxygen heated at constant volume — change in internal energy Kinetic Theory of Gases
50 numeric Compound microscope — the magnification Ray Optics

Chapter by chapter

Chapter Questions Marks
Electromagnetic Induction 2 8
Electrostatic Potential and Capacitance 2 8
Ray Optics 2 8
Rotational Motion 2 8
Thermal Properties of Matter 2 8
Units and Measurements 2 8
Atoms 1 4
Current Electricity 1 4
Dual Nature of Radiation and Matter 1 4
Electromagnetic Waves 1 4
Gravitation 1 4
Kinetic Theory of Gases 1 4
Laws of Motion 1 4
Mechanical Properties of Fluids 1 4
Moving Charges and Magnetism 1 4
Semiconductor Electronics 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 for twenty-five questions.

Where the 100 marks sat

Part of Physics Questions Marks Share
Electrodynamics 7 28 28%
Mechanics 5 20 20%
Properties of Matter and Heat 4 16 16%
Optics and Waves 4 16 16%
Modern Physics and Electronics 3 12 12%
Units, Dimensions and Measurement 2 8 8%

Electrodynamics opened the year where it stayed all through 2026, at 28 marks. Mechanics was a little lighter here than it would be in April, and optics and waves a little heavier.

How tough it was

Level Questions Share Which ones
Straight recall or one step 6 24% 28, 34, 36, 38, 40, 49
Two or three steps 14 56% 26, 27, 29, 30, 31, 32, 33, 35, 37, 41, 42, 44, 48, 50
Long or awkward 5 20% 39, 43, 45, 46, 47

Fewer gift questions than any April shift — only six could be finished on sight. The rod hanging from two strings and the two-rod moment of inertia were the two that ate the most clock. Nothing here was unfair, but very little was free either.

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  ·  Gravitation  ·  on the pattern of Q26

A satellite of mass 200 kg is in a circular orbit of radius 2RE. The energy that must be supplied to move it into a circular orbit of radius 4RE is α × 106 J. The value of α is ______. (RE = 6 × 106 m, g = 10 m/s2)

Answer  1500

SolutionFor a circular orbit of radius r the total energy is E = −GMm/2r, and GM = gRE2. At r = 2RE, E1 = −gREm/4 = −(10)(6 × 106)(200)/4 = −3 × 109 J. At r = 4RE, E2 = −gREm/8 = −1.5 × 109 J. The energy to be supplied is E2 − E1 = 1.5 × 109 J = 1500 × 106 J.

Where it goes wrong  Using only the potential energy. The satellite also slows down as it climbs, so the kinetic part must be counted, and that is exactly what the −GMm/2r form already does.

Q2  ·  Atoms  ·  on the pattern of Q27

An alpha particle of energy 5.0 MeV is fired at a thin gold foil. The closest it can get to a gold nucleus is ______ m. (Atomic number of gold = 79, 1/4πε0 = 9 × 109 SI units, e = 1.6 × 10−19 C)

Answer  about 4.55 × 10−14 m

SolutionAt the closest point the alpha particle has stopped, so all its kinetic energy has turned into electrostatic potential energy: E = (1/4πε0)(2e)(Ze)/r. First convert the energy: 5.0 MeV = 5 × 106 × 1.6 × 10−19 = 8 × 10−13 J. Then r = 9 × 109 × 2 × 79 × (1.6 × 10−19)2 / (8 × 10−13) = 4.55 × 10−14 m.

Where it goes wrong  Using e instead of 2e for the alpha particle. It is a helium nucleus and carries twice the electron’s charge.

Q3  ·  Electromagnetic Waves  ·  on the pattern of Q28

The electric field in a plane electromagnetic wave is Ey = 60 sin[0.5 × 103 x − 1.5 × 1011 t] V/m. The magnetic field that goes with it is ______ T.

Answer  Bz = 2 × 10−7 sin[0.5 × 103 x − 1.5 × 1011 t] T

SolutionThe amplitude follows from B0 = E0/c = 60/(3 × 108) = 2 × 10−7 T. Check the speed from the wave itself: ω/k = 1.5 × 1011/0.5 × 103 = 3 × 108 m/s, which agrees. The form (kx − ωt) means the wave travels along +x, and with E along y the field B must be along z, since j × k = i. The two are in step, so the sine stays exactly the same.

Where it goes wrong  Flipping a sign inside the bracket. The electric and magnetic fields of a plane wave rise and fall together; only the direction and the size differ.

Q4  ·  Units and Dimensions  ·  on the pattern of Q29

Consider a changed Bernoulli equation: [P + A/(Bt)2] + ρg(h + Bt) + ½ρv2 = constant, where t has the dimensions of time. The dimensions of A and B respectively are ______.

Answer  [MLT−2] and [M0LT−1]

SolutionOnly like quantities can be added. Bt sits beside h, so [Bt] = L and [B] = LT−1, which is a velocity. Next, A/(Bt)2 sits beside P, so [A] = [P][(Bt)2] = (ML−1T−2)(L2) = MLT−2, which is a force.

Where it goes wrong  Forgetting to square Bt. The extra L2 is what turns A from a pressure-like quantity into a force.

Q5  ·  Laws of Motion  ·  on the pattern of Q30

A 2 kg mass starts from rest at the origin at t = 0 and moves under the force F = (6t2i − 4tj) N, with t in seconds. Its velocity and position at t = 2 s are ______.

Answer  v = 8i − 4j m/s and r = 4i − 2.67j m

SolutionThe acceleration is a = F/m = (3t2i − 2tj) m/s2. Starting from rest, integrate once: v = (t3i − t2j), which at t = 2 is 8i − 4j m/s. Starting from the origin, integrate again: r = (t4/4 i − t3/3 j), which at t = 2 is 4i − (8/3)j m.

Where it goes wrong  Using v = at and s = ½at2 with the value of a at t = 2. Those formulas need a constant acceleration, and here it keeps changing.

Q6  ·  Electrostatic Potential and Capacitance  ·  on the pattern of Q31

A parallel plate capacitor has capacitance C with vacuum between the plates. A slab of thickness one quarter of the plate separation, with relative permittivity K, is slid in. The new capacitance is ______.

Answer  4KC/(3K + 1)

SolutionWith a slab inside, the gap behaves as two capacitors in series: the empty part of thickness 3d/4 and the slab of thickness d/4. The result is C′ = ε0A / [(d − t) + t/K] with t = d/4, so the bracket is 3d/4 + d/4K = d(3K + 1)/4K. Since C = ε0A/d, this gives C′ = 4KC/(3K + 1). As a check, K = 1 gives C′ = C, as it must.

Where it goes wrong  Multiplying the whole capacitance by K. That is only right when the slab fills the entire gap.

Q7  ·  Wave Optics  ·  on the pattern of Q32

In a double slit experiment the slits are 0.2 cm apart and the screen is 80 cm away. One slit is covered with a transparent sheet of thickness t and refractive index 1.5, and the central fringe moves by 0.3 cm. The value of t is ______ cm.

Answer  1.5 × 10−3 cm

SolutionCovering one slit adds an extra path (n − 1)t on that side, and the pattern slides until that is made up. The shift on the screen is (n − 1)tD/d. Putting the numbers in: 0.3 = (0.5)t(80)/0.2 = 200t, so t = 1.5 × 10−3 cm.

Where it goes wrong  Writing nt instead of (n − 1)t. The sheet replaces an equal thickness of air, so only the difference counts.

Q8  ·  Dual Nature of Radiation  ·  on the pattern of Q33

Light described by E = 50[sin(2 × 1015 t) + sin(8 × 1015 t)] (SI units) falls on a metal of work function 2.0 eV. The greatest kinetic energy of the ejected photoelectrons is about ______ eV. (h = 6.6 × 10−34 J·s, e = 1.6 × 10−19 C)

Answer  about 3.25 eV

SolutionThe beam carries two frequencies, and the fastest electron comes from the higher one. Take ω = 8 × 1015 rad/s, so ν = ω/2π = 1.27 × 1015 Hz and the photon energy is hν = 8.4 × 10−19 J = 5.25 eV. Einstein’s equation then gives KEmax = 5.25 − 2.0 = 3.25 eV.

Where it goes wrong  Reading 8 × 1015 as the frequency. It is the angular frequency, so it has to be divided by 2π before h is used.

Q9  ·  Units and Measurement  ·  on the pattern of Q34

Two spring constants are measured as k1 = (20 ± 0.3) N/m and k2 = (30 ± 0.2) N/m. The springs are then joined in parallel. The percentage error in the equivalent spring constant is ______ %.

Answer  1 %

SolutionSprings in parallel simply add their stiffness, so k = k1 + k2 = 50 N/m. For a sum the absolute errors add: Δk = 0.3 + 0.2 = 0.5 N/m. The percentage error is (0.5/50) × 100 = 1 %.

Where it goes wrong  Adding the percentage errors instead of the absolute ones. Percentages add for products and quotients; for a sum it is the absolute errors that add.

Q10  ·  Thermal Properties of Matter  ·  on the pattern of Q35

A gas geyser heats water flowing at 6.0 litre per minute from 25 °C to 75 °C. The rate at which the gas burns is ______ g/s. (Heat of combustion of the gas = 2.5 × 104 J/g, specific heat capacity of water = 4200 J/kg·°C)

Answer  0.84 g/s

SolutionSix litres a minute is 6 kg per 60 s, that is 0.1 kg/s. The heat needed each second is m c ΔT = 0.1 × 4200 × 50 = 21000 J. Each gram of gas gives 2.5 × 104 J, so the gas burns at 21000/25000 = 0.84 g/s.

Where it goes wrong  Leaving the flow in litres per minute. Everything else is per second, so the rate has to be converted first.

Q11  ·  Moving Charges and Magnetism  ·  on the pattern of Q36

A current-carrying solenoid is held vertically. A particle of mass m with charge Q is released from rest on the axis of the solenoid. Taking g as the acceleration due to gravity, the acceleration a of the particle satisfies ______.

Answer  a = g

SolutionInside a solenoid the magnetic field points along the axis. The particle is released from rest on that same axis, so it starts falling straight down the axis and its velocity is parallel to the field at every moment. The magnetic force qv × B is then zero throughout. Nothing but gravity acts, so a = g.

Where it goes wrong  Assuming a magnetic field must slow the charge down. A magnetic force needs the velocity to cross the field; parallel to it, there is no force at all.

Q12  ·  Electrostatic Potential  ·  on the pattern of Q37

A point charge of 2 × 10−8 C sits at the origin. A charge of 3 μC is carried from the point A(6, 6, 3) m to the point B(2, 2, 1) m. The work that has to be done by an outside agent is ______ J. (1/4πε0 = 9 × 109 SI units)

Answer  1.2 × 10−4 J

SolutionThe potential depends only on the distance from the origin. rA = √(36 + 36 + 9) = 9 m and rB = √(4 + 4 + 1) = 3 m. The work done by the field is q(VA − VB) = (3 × 10−6)(9 × 109)(2 × 10−8)(1/9 − 1/3) = −1.2 × 10−4 J. Both charges are positive and B is nearer, so the outside agent has to push against the repulsion and does +1.2 × 10−4 J.

Where it goes wrong  Using the straight-line distance from A to B. Only the two distances from the source charge matter, because the field is conservative.

Q13  ·  Semiconductor Electronics  ·  on the pattern of Q38

Two inputs A and B each pass through a NOT gate, and the two outputs feed an OR gate. The whole circuit works as a ______ gate.

Answer  NAND gate

SolutionAfter the NOT gates the signals are A′ and B′, and the OR gate gives A′ + B′. By De Morgan’s law that is the same as (A·B)′, which is a NAND gate. Checking the only case that matters: with A = B = 1 both inverters give 0, the OR gives 0, and NAND is indeed 0 there and 1 everywhere else.

Where it goes wrong  Answering NOR. NOR is (A + B)′; this circuit is the other De Morgan pair.

Q14  ·  Rotational Motion  ·  on the pattern of Q39

A uniform rod of mass m and length l hangs horizontally from two identical light inextensible strings, one tied at each end. One string is cut. The tension in the other string immediately afterwards is ______.

Answer  mg/4

SolutionThe instant the first string is cut the rod begins to turn about the point where the other string meets it. Taking torques about that point, mg(l/2) = Iα with I = ml2/3, so α = 3g/2l. The centre of mass is l/2 from the pivot, so it starts down with acm = α(l/2) = 3g/4. Now use Newton’s second law on the whole rod: mg − T = m(3g/4), giving T = mg/4.

Where it goes wrong  Answering mg/2, which is the tension before the cut. The moment the rod is free to turn, the centre of mass starts accelerating and the tension drops.

Q15  ·  Thermal Properties of Matter  ·  on the pattern of Q40

An aluminium rod and a steel rod of equal length and cross-section are joined end to end to make a total length of 100 cm at 20 °C. The coefficients of linear expansion are 24 × 10−6 /°C for aluminium and 12 × 10−6 /°C for steel. The length of the composite rod at 120 °C is ______ cm.

Answer  100.18 cm

SolutionEach rod is 50 cm long and the temperature rises by 100 °C. The aluminium part grows by 50 × 24 × 10−6 × 100 = 0.12 cm and the steel part by 50 × 12 × 10−6 × 100 = 0.06 cm. Adding, the rod grows 0.18 cm, so its new length is 100.18 cm.

Where it goes wrong  Using 100 cm as the length of each piece. The two together make 100 cm, so each starts at 50.

Q16  ·  Waves  ·  on the pattern of Q41

Two strings A and B, of linear densities 1 × 10−4 kg/m and 4 × 10−4 kg/m and lengths 3.0 m and 2.0 m, are joined end to end. The free ends are tied to rigid supports and the tension throughout is 400 N. Identical pulses are sent in from the two far ends at the same instant and take times t1 and t2 to reach the joint. The ratio t1/t2 is ______.

Answer  3 : 4

SolutionA pulse on a stretched string travels at v = √(T/μ). For A, vA = √(400/10−4) = 2000 m/s; for B, vB = √(400/4 × 10−4) = 1000 m/s. The times are t1 = 3/2000 = 1.5 ms and t2 = 2/1000 = 2.0 ms, so the ratio is 1.5 : 2 = 3 : 4.

Where it goes wrong  Assuming the heavier string carries the pulse faster. A larger linear density sits in the denominator, so it slows the pulse down.

Q17  ·  Work, Energy and Power  ·  on the pattern of Q42

The potential energy of a particle changes along x as a straight line in four regions: over AB it rises 4 J in 2 m, over BC it stays flat for 3 m, over CD it falls 6 J in 1 m, and over DE it rises 3 J in 3 m. Ranking the magnitudes of the force on the particle from largest to smallest gives ______.

Answer  FCD > FAB > FDE > FBC

SolutionThe force is F = −dV/dx, so its size is simply the steepness of the graph. CD gives 6/1 = 6 N, AB gives 4/2 = 2 N, DE gives 3/3 = 1 N, and BC is flat, so 0. Putting them in order: CD, AB, DE, BC.

Where it goes wrong  Ranking by the height of the graph instead of its slope. A large potential energy with a flat graph means no force at all.

Q18  ·  Electromagnetic Induction  ·  on the pattern of Q43

A 1 m long metal rod slides on rails and completes a circuit whose plane is perpendicular to a magnetic field of 0.20 T. The total resistance of the circuit is 4 Ω. The force needed to keep the rod moving to the right at a steady 2.0 m/s is ______ N.

Answer  0.02 N

SolutionThe moving rod generates an emf of Blv = 0.20 × 1 × 2.0 = 0.4 V, so the current is I = 0.4/4 = 0.1 A. That current in the field feels a backward force F = BIl = 0.20 × 0.1 × 1 = 0.02 N. Because the speed is steady, the applied force exactly matches it: 0.02 N.

Where it goes wrong  Expecting zero force because the speed is constant. Constant speed means no net force, which is exactly why the push must equal the magnetic drag.

Q19  ·  Mechanical Properties of Fluids  ·  on the pattern of Q44

Water flows through a horizontal tube. Two vertical standpipes show a height difference of 4 cm between points A and B, where the cross-sections are 8 cm2 and 4 cm2. The rate of flow is about ______ cm3/s. (g = 10 m/s2)

Answer  about 413 cm3/s

SolutionContinuity gives vB = 2vA, since B is half as wide. The standpipes measure the pressure difference, ρgh, and Bernoulli at the same height gives ρgh = ½ρ(vB2 − vA2) = ½ρ(3vA2). So gh = 1.5vA2, that is 10 × 0.04 = 1.5vA2, giving vA = 0.516 m/s = 51.6 cm/s. The rate of flow is AAvA = 8 × 51.6 ≈ 413 cm3/s.

Where it goes wrong  Mixing metres and centimetres halfway through. Work the speed out in SI, then convert once at the end.

Q20  ·  Electromagnetic Induction  ·  on the pattern of Q45

A conducting circular loop of area 2.0 m2 lies perpendicular to a magnetic field B = sin(50t) tesla. The resistance of the loop is 50 Ω. The average thermal energy dissipated in the loop over one full period is ______ J.

Answer  4π J, about 12.6 J

SolutionThe flux is Φ = BA = 2 sin(50t), so the emf is −dΦ/dt = −100 cos(50t) V, with a peak of 100 V. For a sinusoid the average power is (peak)2/2R = 10000/100 = 100 W. One period lasts T = 2π/50 = 0.126 s, so the energy in one period is 100 × 2π/50 = 4π ≈ 12.6 J.

Where it goes wrong  Using the peak emf for the power. The average of a squared sine is half, which is where the factor of 2 in 2R comes from.

Q21  ·  Rotational Motion  ·  numeric  ·  on the pattern of Q46

Two identical thin rods, each of mass M and length L, are joined into a T shape: the second rod is attached by its own centre to one end of the first and lies at right angles to it. The moment of inertia of the pair about an axis through the free end P of the first rod, perpendicular to the plane of the two rods, is (x/12)ML2. The value of x is ______.

Answer  17

SolutionTake the rods one at a time. The first rod turns about its own end, so I1 = ML2/3 = 4ML2/12. The second rod has its centre a distance L from P, and about its own centre, perpendicular to the plane, it is ML2/12; the parallel axis theorem adds ML2, giving 13ML2/12. Adding the two, I = 17ML2/12, so x = 17.

Where it goes wrong  Treating the second rod as a point mass at L, which gives 12/12 and misses the ML2/12 of its own spread.

Q22  ·  Current Electricity  ·  numeric  ·  on the pattern of Q47

A 9 V battery with an internal resistance of 1 Ω is connected across a 3 Ω and a 6 Ω resistor joined in parallel. The heat generated in those two resistors in one minute is ______ J.

Answer  1080 J

SolutionThe parallel pair is 3 × 6/(3 + 6) = 2 Ω, so the circuit is 1 + 2 = 3 Ω and the current from the battery is 9/3 = 3 A. The power in the external pair is I2R = 9 × 2 = 18 W, and over 60 s that is 18 × 60 = 1080 J. The internal resistance takes another 9 W, which the question does not ask for.

Where it goes wrong  Dropping the internal resistance when working out the current. It is only left out of the heat, not out of the circuit.

Q23  ·  Ray Optics  ·  numeric  ·  on the pattern of Q48

A collimated beam of light 3 mm across travels along the x-axis. It is to be widened into a collimated beam 15 mm across using two convex lenses. If the first lens has a focal length of 30 mm, the focal length of the second lens is ______ mm.

Answer  150 mm

SolutionIn a beam expander the two lenses share a focus, so the separation is f1 + f2 and the beam widens in the ratio of the focal lengths: d2/d1 = f2/f1. Here the beam grows 15/3 = 5 times, so f2 = 5 × 30 = 150 mm, and the lenses stand 180 mm apart.

Where it goes wrong  Trying to use the lens formula. Both the incoming and outgoing beams are parallel, so only the ratio of focal lengths matters.

Q24  ·  Kinetic Theory of Gases  ·  numeric  ·  on the pattern of Q49

8 mole of oxygen is heated at constant volume from 20 °C to 45 °C. The change in the internal energy of the gas is ______ calorie. (Molar specific heat at constant pressure CP = 7 cal/mol·°C, R = 2 cal/mol·°C)

Answer  1000 calorie

SolutionMayer’s relation gives CV = CP − R = 7 − 2 = 5 cal/mol·°C. The change in internal energy is ΔU = nCVΔT = 8 × 5 × 25 = 1000 calorie. Since the volume is fixed, no work is done and all of this comes from the heat supplied.

Where it goes wrong  Using CP because the value is the one given. ΔU always uses CV, and CP is given only so that CV can be found.

Q25  ·  Ray Optics  ·  numeric  ·  on the pattern of Q50

In a compound microscope the objective has focal length 2 cm, the eyepiece has focal length 5 cm and the tube length is 25 cm. The magnification for normal adjustment is ______. (Least distance of distinct vision = 25 cm)

Answer  62.5

SolutionFor normal adjustment the final image is at infinity, and the magnification is the product of the two stages: m = (L/fo) × (D/fe). Putting the numbers in, m = (25/2) × (25/5) = 12.5 × 5 = 62.5.

Where it goes wrong  Using (1 + D/fe) for the eyepiece. That form is for the image held at the near point, not at infinity.

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. The January 2026 shifts were held on 21, 22, 23, 24 and 28 January, two shifts a day.

Official final answer key →

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

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

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