JEE Main 28 January 2026 Shift 1 — Physics Paper with Answer Key

The morning paper of 28 January 2026, the ninth 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 28 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 Closed cycle on a p-V diagram — the net work done Thermodynamics
27 Temperature against heat supplied for water from −20 °C to 120 °C Thermal Properties of Matter
28 Drops falling at equal intervals — height of the fourth when the first lands Motion in a Straight Line
29 Ice at −10 °C added to water — how much the water cools Thermal Properties of Matter
30 Two statements on how a wavefront changes at a prism and at a pinhole Wave Optics
31 Two cells giving the same current in series and in parallel — the internal resistance Current Electricity
32 Current rising linearly over one period — its r.m.s. value Alternating Current
33 Two discs joined by a rod — angular acceleration under a given torque Rotational Motion
34 Two wires stretched equally by the same load — ratio of their Young’s moduli Mechanical Properties of Solids
35 Work to bring a third charge to the free corner of a triangle Electrostatic Potential and Capacitance
36 Three parallel current-carrying wires — force on 15 cm of the middle one Moving Charges and Magnetism
37 Nucleus that absorbs protons and neutrons — growth in its surface area Nuclei
38 Potentiometer balance with two different shunts — internal resistance of the cell Current Electricity
39 Current through a silicon diode with a 0.7 V forward drop Semiconductor Electronics
40 Falling body in a medium with resistance proportional to speed — the v-t graph Laws of Motion
41 Block on a rough incline — force needed to bring it down at steady speed Laws of Motion
42 Vernier callipers with a zero error — the corrected length Units and Measurements
43 Biconvex and plano-concave lenses of equal power — ratio of the two radii Ray Optics
44 Refractive index of a medium from the wave equation Electromagnetic Waves
45 Field on the axis of a loop at a distance √3R from its centre Moving Charges and Magnetism
46 numeric de Broglie wavelengths of a deuteron and an alpha particle compared Dual Nature of Radiation and Matter
47 numeric Convex lens dipped in water — change in its focal length Ray Optics
48 numeric Equivalent resistance of the given network between two points Current Electricity
49 numeric First instant at which the potential energy of an oscillator is greatest Oscillations
50 numeric Radius of gyration of a sphere about an axis away from its centre Rotational Motion

Chapter by chapter

Chapter Questions Marks
Current Electricity 3 12
Laws of Motion 2 8
Moving Charges and Magnetism 2 8
Ray Optics 2 8
Rotational Motion 2 8
Thermal Properties of Matter 2 8
Alternating Current 1 4
Dual Nature of Radiation and Matter 1 4
Electromagnetic Waves 1 4
Electrostatic Potential and Capacitance 1 4
Mechanical Properties of Solids 1 4
Motion in a Straight Line 1 4
Nuclei 1 4
Oscillations 1 4
Semiconductor Electronics 1 4
Thermodynamics 1 4
Units and Measurements 1 4
Wave Optics 1 4

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

Eighteen chapters, and Current Electricity again took three questions for 12 marks — two cells, a potentiometer and a resistor network. Together with 24 January Shift 2 these were the two January shifts where circuits paid best.

Where the 100 marks sat

Part of Physics Questions Marks Share
Electrodynamics 8 32 32%
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 1 4 4%

Electrodynamics 32 for the fourth time in January. Four of the ten January shifts put exactly 32 marks there, and none put fewer than 28.

How tough it was

Level Questions Share Which ones
Straight recall or one step 7 28% 30, 32, 35, 44, 45, 46, 49
Two or three steps 12 48% 28, 31, 34, 37, 38, 39, 41, 42, 43, 47, 48, 50
Long or awkward 6 24% 26, 27, 29, 33, 36, 40

The ice-and-water calorimetry was the longest single calculation in January: the ice has to be warmed, then melted, then the melt water warmed, all against the heat the 100 kg of water gives up. Four separate terms, and a sign slip anywhere gives a plausible wrong answer.

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  ·  Thermodynamics  ·  on the pattern of Q26

A gas is taken round a closed rectangular cycle on the p-V diagram: an expansion from V1 to V2 at constant pressure P, a fall to pressure P/2 at constant volume, a compression back to V1 at that pressure, and a rise back to P at constant volume. The net work done in the cycle is ______.

Answer  (P/2)(V2 − V1), done by the gas

SolutionWork done by a gas is the area under its p-V path, and round a closed cycle it is the area enclosed. The rectangle has height P − P/2 = P/2 and width V2 − V1, so the net work is (P/2)(V2 − V1). The cycle is traced clockwise — expanding at the higher pressure and compressing at the lower — so the work is positive, done by the gas. Nothing is done on the two constant-volume legs.

Where it goes wrong  Adding the areas under all four legs as positive. The expansion counts positive and the compression negative, which is exactly why only the enclosed area is left.

Q2  ·  Thermal Properties of Matter  ·  on the pattern of Q27

Water is heated steadily from −20 °C to 120 °C. Describe how a graph of its temperature against the heat supplied looks.

Answer  a steep rise, a flat step at 0 °C, a shallower rise, a long flat step at 100 °C, then a steep rise again

SolutionFive stages. Ice warms from −20 °C to 0 °C: the slope is 1/(mcice), and since ice has about half the specific heat of water, this part is the steeper of the two solid-liquid rises. At 0 °C the ice melts at constant temperature, so the graph is flat for the latent heat of fusion. Then water warms from 0 °C to 100 °C with a shallower slope. At 100 °C it boils, giving a much longer flat step, because the latent heat of vaporisation is about seven times the heat of fusion. Finally steam warms above 100 °C, with a steep slope again.

Where it goes wrong  Drawing the two flat steps the same length. The boiling step is far longer than the melting one, and that difference is the point of the question.

Q3  ·  Motion in a Straight Line  ·  on the pattern of Q28

Drops fall from a tap at equal intervals of time onto a floor 5 m below. The first drop strikes the floor at the very instant the sixth drop begins to fall. At that instant the fourth drop is ______ m above the ground. (g = 10 m/s2.)

Answer  4.2 m

SolutionThe first drop takes t = √(2 × 5/10) = 1 s to fall 5 m, and in that time five intervals have passed, so each interval is 0.2 s. Counting back, the fourth drop has been falling for two intervals, 0.4 s. It has dropped ½ × 10 × 0.16 = 0.8 m, so it is 5 − 0.8 = 4.2 m above the ground.

Where it goes wrong  Dividing the second by 6 instead of 5. Six drops in the air means five gaps between them.

Q4  ·  Thermal Properties of Matter  ·  on the pattern of Q29

10 kg of ice at −10 °C is added to 100 kg of water at 25 °C, with no heat lost to the surroundings. The temperature of the water falls by ______ °C. (Specific heat of ice 2100 J/kg·°C, of water 4200 J/kg·°C, latent heat of fusion of ice 3.36 × 105 J/kg.)

Answer  10 °C — it settles at 15 °C

SolutionFirst check that all the ice melts. Warming it to 0 °C takes 10 × 2100 × 10 = 2.1 × 105 J and melting it takes 10 × 3.36 × 105 = 3.36 × 106 J, so 3.57 × 106 J in all, while cooling the water right down to 0 °C would release 100 × 4200 × 25 = 1.05 × 107 J. There is plenty, so all the ice melts. Now let T be the final temperature: 100 × 4200 × (25 − T) = 3.57 × 106 + 10 × 4200 × T. That is 1.05 × 107 − 4.2 × 105T = 3.57 × 106 + 4.2 × 104T, so 6.93 × 106 = 4.62 × 105T and T = 15 °C. The water has dropped 10 °C.

Where it goes wrong  Forgetting the melt water. Once the ice has melted, those 10 kg must also be warmed from 0 °C to the final temperature, and that term is easy to leave out.

Q5  ·  Wave Optics  ·  on the pattern of Q30

Two statements. Statement I: a plane wave stays plane after passing through a prism, but may become spherical after passing through a small pinhole. Statement II: the curvature of the spherical wave coming out of a slit increases as the slit is made wider. Which is true?

Answer  Statement I is true, Statement II is false

SolutionA prism simply tilts a plane wavefront — every part of it is delayed by a different but fixed amount, so it stays flat. A pinhole small enough to compare with the wavelength lets through what is effectively a point source, and the wave spreading from it is spherical. So Statement I is right. Statement II has it backwards: the narrower the opening, the more the wave spreads and the greater the curvature. Widening the slit makes the emerging wave flatter, not more curved.

Where it goes wrong  Reading “wider slit, more light, more spreading”. More light gets through, but it spreads less; diffraction grows as the opening shrinks.

Q6  ·  Current Electricity  ·  on the pattern of Q31

Two cells of the same emf E and the same internal resistance r are connected to an external 6 Ω resistor, and the current through it comes out the same whether the cells are in series or in parallel. The value of r is ______ Ω.

Answer  6 Ω

SolutionIn series the emf doubles and the internal resistances add: Is = 2E/(6 + 2r). In parallel the emf stays E and the internal resistance halves: Ip = E/(6 + r/2). Setting them equal, 2/(6 + 2r) = 1/(6 + r/2), so 2(6 + r/2) = 6 + 2r, that is 12 + r = 6 + 2r and r = 6 Ω. The two arrangements give the same current exactly when the internal resistance equals the external one.

Where it goes wrong  Doubling the emf in the parallel case too. Cells in parallel share the current, not the voltage.

Q7  ·  Alternating Current  ·  on the pattern of Q32

A current rises linearly with time as i = i0t/T over the period from t = 0 to t = T. Its r.m.s. value over that period is ______.

Answer  i0/√3

SolutionThe r.m.s. value is the square root of the average of i2. Here the average of i2 is (1/T)∫0T i02t2/T2 dt = (i02/T3)(T3/3) = i02/3. Taking the root gives i0/√3. Note that it is not i0/√2 — that factor belongs to a sine wave, and this is a ramp.

Where it goes wrong  Reaching for i0/√2 by habit. The √2 comes from the average of sin2; a different waveform gives a different factor, and here it is √3.

Q8  ·  Rotational Motion  ·  on the pattern of Q33

Two circular discs, each of mass 600 g and radius 10 cm, are joined at their centres by a light rod 30 cm long, and the whole thing turns about the common axis of the discs. A torque of (4/3) × 105 dyne·cm is applied. The angular acceleration is about ______ rad/s2.

Answer  about 2.2 rad/s2

SolutionEach disc turns about its own axis, so each contributes ½MR2 = ½ × 0.6 × (0.1)2 = 3 × 10−3 kg·m2, and the two give I = 6 × 10−3 kg·m2; the light rod lies along the axis and adds nothing. For the torque, 1 dyne·cm = 10−7 N·m, so τ = (4/3) × 105 × 10−7 = 1.33 × 10−2 N·m. Then α = τ/I = 1.33 × 10−2/6 × 10−3 = 2.2 rad/s2.

Where it goes wrong  Treating 1 dyne·cm as 10−5 N·m. The dyne contributes 10−5 and the centimetre another 10−2, so the two together give 10−7.

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

Two wires A and B of different materials, of lengths 6.0 cm and 5.4 cm and cross-sections 3.0 × 10−5 m2 and 4.5 × 10−5 m2, stretch by the same amount under the same load. The ratio of the Young’s modulus of A to that of B is x : 3. The value of x is ______.

Answer  5

SolutionFrom Y = FL/AΔL, with F and ΔL the same for both, Y is proportional to L/A. So YA/YB = (LA/AA) ÷ (LB/AB) = (6.0/3.0) ÷ (5.4/4.5) = 2/1.2 = 5/3. Written as x : 3 that gives x = 5.

Where it goes wrong  Converting the lengths to metres and the areas not at all, or the other way round. Here only the ratio matters, so the centimetres cancel — but the two lengths must be in the same unit as each other.

Q10  ·  Electrostatic Potential  ·  on the pattern of Q35

Charges of 1 nC and 2 nC sit at two corners of an equilateral triangle of side 3 cm. The work done in bringing a 3 nC charge from infinity to the third corner is ______ μJ. (1/4πε0 = 9 × 109 N·m2/C2.)

Answer  2.7 μJ

SolutionThe third corner is 3 cm from each of the other two, so the potential there is k(q1 + q2)/r = 9 × 109 × 3 × 10−9/0.03 = 900 V. The work done in bringing in the third charge from infinity is qV = 3 × 10−9 × 900 = 2.7 × 10−6 J = 2.7 μJ. Potential is a scalar, so the two existing charges simply add.

Where it goes wrong  Working out forces and trying to combine them as vectors. The work only needs the potential, which is a single number.

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

Three long straight parallel wires P, Q and R lie in one plane, each carrying 10 A. Q is 2 cm from P and 4 cm from R, with P and R on opposite sides of it. P and R carry current in the same direction and Q in the opposite direction. The force on a 15 cm length of Q is ______ N. (μ0 = 4π × 10−7 T·m/A.)

Answer  7.5 × 10−5 N, away from P

SolutionThe force per metre between two parallel wires is μ0I1I2/2πd = 2 × 10−7 × 100/d. From P, at 2 cm: 2 × 10−5/0.02 = 10−3 N/m. From R, at 4 cm: 5 × 10−4 N/m. Antiparallel currents repel, so Q is pushed away from P and away from R — and those two pushes are in opposite directions. The net is 10−3 − 5 × 10−4 = 5 × 10−4 N/m, directed away from P. Over 0.15 m that is 7.5 × 10−5 N.

Where it goes wrong  Adding the two forces. P and R are on opposite sides, so their pushes on Q oppose each other and the nearer wire wins.

Q12  ·  Nuclei  ·  on the pattern of Q37

A nucleus with mass number 8 is bombarded and absorbs 10 electrons, 10 protons and 9 neutrons. The surface area of the nucleus grows by ______ per cent.

Answer  125% (the new area is 225% of the old)

SolutionOnly nucleons change the size of a nucleus, so the electrons do not count. The mass number goes from 8 to 8 + 10 + 9 = 27. Since R ∝ A1/3, the surface area goes as A2/3: from 82/3 = 4 to 272/3 = 9. So the area becomes 9/4 = 2.25 times as large, that is 225% of the original, which is a growth of 125%.

Where it goes wrong  Counting the 10 electrons into the mass number, or using A1/3 instead of A2/3 for the area.

Q13  ·  Current Electricity  ·  on the pattern of Q38

On a potentiometer, when the cell in the secondary circuit is shunted with 4 Ω the balance length is 120 cm, and when it is shunted with 12 Ω the balance length is 180 cm. The internal resistance of the cell is ______ Ω.

Answer  4 Ω

SolutionWith a shunt R across it, the cell shows a terminal voltage ER/(R + r), and the balance length is proportional to that voltage. So 120 ∝ 4E/(4 + r) and 180 ∝ 12E/(12 + r). Dividing, [4/(4 + r)] ÷ [12/(12 + r)] = 120/180 = 2/3, that is (48 + 4r)/(48 + 12r) = 2/3. Cross-multiplying, 144 + 12r = 96 + 24r, so 12r = 48 and r = 4 Ω.

Where it goes wrong  Using the usual r = R(l1/l2 − 1), which needs one open-circuit length. Here both readings are with shunts, so the two equations have to be divided.

Q14  ·  Semiconductor Electronics  ·  on the pattern of Q39

A silicon diode, which drops 0.7 V when forward biased, is put in series with a 1 kΩ resistor across a 12 V battery. The current through the diode is ______ mA.

Answer  11.3 mA

SolutionIn forward bias the silicon diode behaves as a fixed 0.7 V drop, so the resistor gets the rest: 12 − 0.7 = 11.3 V. The current is then 11.3/1000 = 11.3 × 10−3 A = 11.3 mA, and the same current passes through the diode since they are in series.

Where it goes wrong  Using the whole 12 V across the resistor and answering 12 mA. The 0.7 V is the whole reason the question mentions silicon.

Q15  ·  Laws of Motion  ·  on the pattern of Q40

A particle of mass m falls from rest through a medium whose resistive force is F = −kv, with v the speed and k a constant. Describe the graph of velocity against time.

Answer  it starts at zero with slope g and curves over, flattening out at the terminal speed mg/k

SolutionNewton’s second law gives m dv/dt = mg − kv. At t = 0 the speed is zero, so the resistance is zero and the acceleration is the full g — that is the initial slope. As the speed grows the resistance grows with it and the acceleration falls, so the curve bends over. It levels off when mg = kv, at the terminal speed v = mg/k, which it approaches but never quite reaches. Solving gives v = (mg/k)(1 − e−kt/m), the familiar rising exponential.

Where it goes wrong  Drawing a straight line, or a curve that starts flat. The slope is largest at the very start, when there is no resistance yet, and shrinks from there.

Q16  ·  Laws of Motion  ·  on the pattern of Q41

A block of mass 5 kg is on a plane inclined at 30° to the horizontal, with a coefficient of friction 2/√3 between them. The force that must be applied so the block slides down at a steady speed is ______ N. (g = 10 m/s2.)

Answer  25 N, applied down the slope

SolutionGravity pulls the block down the slope with mg sin30° = 5 × 10 × 0.5 = 25 N. The normal force is mg cos30° = 50 × √3/2 = 25√3 N, so friction can be as much as μN = (2/√3)(25√3) = 50 N, and as the block slides down it acts up the slope at that value. For steady speed the forces must balance: 25 + F = 50, so F = 25 N, pushing down the slope. Left alone the block would not move at all, since 25 N of gravity cannot overcome 50 N of friction.

Where it goes wrong  Assuming the push must be up the slope to control the block. With this much friction the block needs help to get moving at all, so the force goes downhill.

Q17  ·  Units and Measurement  ·  on the pattern of Q42

With the jaws of a vernier calliper closed, the zero of the vernier scale lies to the right of the zero of the main scale and the 4th vernier mark coincides with a main scale mark. Measuring a cylinder, the main scale reads 15 divisions and the 5th vernier division coincides with a main scale mark. The length of the cylinder is ______ mm. (Least count = 0.1 mm, 1 main scale division = 1 mm.)

Answer  15.1 mm

SolutionThe vernier zero sitting to the right of the main zero is a positive zero error, equal to 4 × 0.1 = +0.4 mm. The observed reading is 15 + 5 × 0.1 = 15.5 mm. A positive error means the instrument reads high, so it is subtracted: 15.5 − 0.4 = 15.1 mm.

Where it goes wrong  Adding the 0.4 mm. “To the right” means the jaws already show a reading with nothing between them, so that much must come off.

Q18  ·  Ray Optics  ·  on the pattern of Q43

A biconvex lens of refractive index 1.5 and a plano-concave lens of refractive index 1.7 have powers of equal magnitude. The curved face of the plano-concave lens has exactly the same radius as the back surface of the biconvex lens. The ratio of the front to the back radius of the biconvex lens is ______.

Answer  5 : 2

SolutionCall the two radii of the biconvex lens R1 and R2 in magnitude. Its power is (1.5 − 1)(1/R1 + 1/R2) = 0.5(1/R1 + 1/R2). The plano-concave lens has one flat face and one curved face of radius R2, so the magnitude of its power is (1.7 − 1)/R2 = 0.7/R2. Setting them equal: 0.5/R1 + 0.5/R2 = 0.7/R2, so 0.5/R1 = 0.2/R2 and R1/R2 = 0.5/0.2 = 5/2.

Where it goes wrong  Writing the biconvex power as (n − 1)(1/R1 − 1/R2) with both radii positive. For a biconvex lens the two centres of curvature are on opposite sides, so the two terms add.

Q19  ·  Electromagnetic Waves  ·  on the pattern of Q44

The electric field of an electromagnetic wave in a medium is E(x, t) = 25 sin(2.0 × 1015t − 107x), in SI units. The refractive index of the medium is ______.

Answer  1.5

SolutionComparing with sin(ωt − kx), ω = 2.0 × 1015 rad/s and k = 107 m−1. The wave speed is v = ω/k = 2.0 × 108 m/s, and the refractive index is n = c/v = 3 × 108/2 × 108 = 1.5. The amplitude 25 plays no part.

Where it goes wrong  Treating 2.0 × 1015 as a frequency rather than an angular frequency. It does not matter here, since the 2π cancels in ω/k, but it does if the wavelength is asked for.

Q20  ·  Moving Charges and Magnetism  ·  on the pattern of Q45

The magnetic field at the centre of a circular current-carrying loop of radius R is 16 μT. The field at a point on the axis a distance √3R from the centre is ______ μT.

Answer  2 μT

SolutionOn the axis, B = μ0IR2/2(R2 + x2)3/2, and at the centre this becomes μ0I/2R. Dividing one by the other gives the ratio R3/(R2 + x2)3/2. With x = √3R the bracket is 4R2, whose 3/2 power is 8R3, so the ratio is 1/8. Hence B = 16/8 = 2 μT.

Where it goes wrong  Using an inverse square instead of the 3/2 power. The axial formula falls off faster than that near the loop, and the √3 is chosen so that the bracket comes out exactly 4R2.

Q21  ·  Dual Nature of Matter  ·  numeric  ·  on the pattern of Q46

The ratio of the de Broglie wavelength of a deuteron of kinetic energy E to that of an alpha particle of kinetic energy 2E is n : 1. The value of n is ______. (Take the proton and neutron masses as equal.)

Answer  2

Solutionλ = h/√(2mK). A deuteron is two nucleons, so m = 2mp and 2mK = 2(2mp)(E) = 4mpE. An alpha particle is four nucleons, so 2mK = 2(4mp)(2E) = 16mpE. The wavelengths go as the inverse square roots, so λd/λα = √(16/4) = 2. Hence n = 2.

Where it goes wrong  Forgetting that the alpha particle has both four times the mass and twice the energy. Each of those shrinks its wavelength, and together they give the factor of two.

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

A convex lens of refractive index 1.5 has focal length f = 18 cm in air. It is immersed in water of refractive index 4/3. The difference between its focal lengths in water and in air is α × f. The value of α is ______.

Answer  3

SolutionThe lens maker’s formula gives 1/f = (nlens/nmedium − 1)K, where K depends only on the two radii. In air the bracket is 1.5 − 1 = 0.5; in water it is 1.5/(4/3) − 1 = 1.125 − 1 = 0.125. So the focal length is 0.5/0.125 = 4 times longer in water: fwater = 72 cm. The difference is 72 − 18 = 54 cm, which is 3 × 18, so α = 3.

Where it goes wrong  Using 1.5 − 4/3 for the bracket. It is the ratio of the indices that goes in, not their difference.

Q23  ·  Current Electricity  ·  numeric  ·  on the pattern of Q48

Resistors of 1 Ω, 2 Ω and 3 Ω are all joined in parallel between the points A and B. The equivalent resistance is x/11 Ω. The value of x is ______.

Answer  6

SolutionFor resistors in parallel the reciprocals add: 1/R = 1/1 + 1/2 + 1/3 = (6 + 3 + 2)/6 = 11/6. So R = 6/11 Ω, and x = 6. Notice the result is smaller than the smallest of the three, as a parallel combination always must be.

Where it goes wrong  Forgetting to invert at the end and giving 11/6. A quick sanity check settles it: the answer has to be below 1 Ω.

Q24  ·  Oscillations  ·  numeric  ·  on the pattern of Q49

A particle in simple harmonic motion of period T has displacement x(t) = A sinωt. Its potential energy first reaches its greatest value at t = T/2β. The value of β is ______.

Answer  2

SolutionThe potential energy is ½kx2, so it is greatest where the displacement is greatest, that is where sinωt = 1. That first happens at ωt = π/2, and since ω = 2π/T this gives t = T/4. Setting T/2β = T/4 gives β = 2.

Where it goes wrong  Answering t = T/2, where the particle passes through the centre again. That is where the potential energy is least, not greatest.

Q25  ·  Rotational Motion  ·  numeric  ·  on the pattern of Q50

A solid sphere of radius 10 cm turns about an axis 15 cm from its centre. Its radius of gyration about that axis is n cm. The value of n is about ______.

Answer  √265, about 16.3 cm

SolutionAbout its own centre a solid sphere has I = (2/5)MR2, and the parallel axis theorem adds Md2: I = M[(2/5)(100) + 225] = M(40 + 225) = 265M, in cm2. The radius of gyration is defined by I = Mk2, so k = √265 = 16.3 cm.

Where it goes wrong  Adding 10 and 15 to get 25 cm, or using (2/5)R2 alone. The radius of gyration is a root-mean-square distance, so the two contributions add as squares.

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 other January 2026 dates were 21, 22, 23 and 24 January, two shifts on each.

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