JEE Main 23 January 2026 Shift 2 — Physics Paper with Answer Key

The afternoon paper of 23 January 2026, the sixth 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 23 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 Block sliding down an incline given a push up — distance before it stops Laws of Motion
27 Same work done isothermally and then adiabatically — the final temperature Thermodynamics
28 Two beads, one on a curved wire and one on a straight wire — which arrives first Work, Energy and Power
29 Two short dipoles 80 cm apart — field at a point equidistant from both Electric Charges and Fields
30 Paratrooper, 2 s of free fall then steady deceleration — height of the aeroplane Motion in a Straight Line
31 Which pair of nuclei are isobars Nuclei
32 Two balancing lengths on a potentiometer — percentage error in the emf ratio Units and Measurements
33 Body at rest explodes into three fragments — speed of the heaviest Centre of Mass
34 Mica sheet slipped into a connected capacitor — its dielectric constant Electrostatic Potential and Capacitance
35 Reflected beam fully polarised — the angle of the refracted beam Wave Optics
36 Speed of light in vacuum against that in a medium of given K and μ Electromagnetic Waves
37 Metal sphere dropped in glycerine — terminal velocity Mechanical Properties of Fluids
38 Prism with a coated back face — incidence needed for total internal reflection Ray Optics
39 Correct truth table for the given gate circuit Semiconductor Electronics
40 Air bubble rising from the bottom of a pool — its volume at the surface Kinetic Theory of Gases
41 Circular loop reshaped into a square in half a second — the emf Electromagnetic Induction
42 Two charges sitting in an external field — the electrostatic energy Electrostatic Potential and Capacitance
43 Loop slid inside a solenoid carrying an alternating current — r.m.s. current Electromagnetic Induction
44 Helium heated under a light piston — how far the piston travels Thermodynamics
45 Current in a triangular loop — field at the centroid Moving Charges and Magnetism
46 numeric Energy released when 47 g of uranium-235 fissions completely Nuclei
47 numeric Powers in a formula for the time to reach terminal velocity Units and Measurements
48 numeric Disc with a piece cut out — moment of inertia of what is left Rotational Motion
49 numeric Two object positions giving images of the same size — the focal length Ray Optics
50 numeric Temperature at which the speed of sound in air doubles Waves

Chapter by chapter

Chapter Questions Marks
Electromagnetic Induction 2 8
Electrostatic Potential and Capacitance 2 8
Nuclei 2 8
Ray Optics 2 8
Thermodynamics 2 8
Units and Measurements 2 8
Centre of Mass 1 4
Electric Charges and Fields 1 4
Electromagnetic Waves 1 4
Kinetic Theory of Gases 1 4
Laws of Motion 1 4
Mechanical Properties of Fluids 1 4
Motion in a Straight Line 1 4
Moving Charges and Magnetism 1 4
Rotational Motion 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-one chapters in twenty-five questions — the thinnest spread of the whole year. Only four chapters got a second question, and seventeen appeared exactly once. Mechanics was split across five different chapters rather than concentrated in two.

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%

The most even split of any January shift, and almost exactly the 2026 average. Nothing was over-weighted and nothing was skipped.

How tough it was

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

The loop sliding inside an alternating-current solenoid was the single hardest question in January: it needed the flux, a derivative, a root-mean-square step and careful powers of ten. The two charges sitting in an external field was the second, because the external potential and the mutual energy both had to be counted and it is easy to drop one.

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  ·  Laws of Motion  ·  on the pattern of Q26

A block is sliding down an incline of slope θ at a steady speed. At t = 0 it is given a push up the slope so that it moves up with speed u. The distance it travels up the incline before its speed falls to zero is ______.

Answer  u2 / 4g sinθ

SolutionSliding down at a steady speed means friction exactly balances gravity on the way down, so μmg cosθ = mg sinθ and therefore μ = tanθ. On the way up both gravity and friction act down the slope, so the deceleration is g sinθ + μg cosθ = g sinθ + tanθ · g cosθ = 2g sinθ. Then S = u2/2a = u2/4g sinθ.

Where it goes wrong  Missing the clue in “sliding down at a steady speed”. That one phrase is what fixes μ = tanθ; without it the question cannot be answered at all.

Q2  ·  Thermodynamics  ·  on the pattern of Q27

One mole of an ideal diatomic gas expands from V to 2V isothermally at 27 °C and does W joule of work. The same gas, again starting at 27 °C, now expands adiabatically and does the same amount of work W. Its final temperature is about ______ °C. (loge2 = 0.693.)

Answer  about −56 °C

SolutionThe isothermal work is W = nRT ln2 = 1 × 8.314 × 300 × 0.693 = 1729 J. In an adiabatic expansion no heat enters, so the work comes entirely out of the internal energy: W = nCV(T1 − T2), and for a diatomic gas CV = 5R/2. So 1729 = (5/2)(8.314)(300 − T2), giving 300 − T2 = 83.2 and T2 = 217 K, which is about −56 °C.

Where it goes wrong  Using CP for the adiabatic step. No heat is exchanged, so all the work comes from the internal energy, and that is CV.

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

A bead slides from rest down a smooth wire bent into some curve, from a point A to a point B which is 1.8 m lower. Its speed at B is ______ m/s. (g = 10 m/s2.)

Answer  6 m/s

SolutionThe wire is smooth, so the only force doing work is gravity, and gravity does mgh whatever route is taken. So ½mv2 = mgh gives v = √(2gh) = √(2 × 10 × 1.8) = 6 m/s. The shape of the wire changes how long the bead takes, but not how fast it arrives. The normal force from the wire is always at right angles to the motion, so it never does any work.

Where it goes wrong  Trying to work along the curve. The drop in height is the only thing that matters for the speed.

Q4  ·  Electric Charges and Fields  ·  on the pattern of Q29

A short electric dipole is made of charges +2 μC and −2 μC separated by 1 cm. The electric field at a point on the axis of the dipole, 40 cm from its centre, is ______ N/C. (1/4πε0 = 9 × 109 SI units.)

Answer  5625 N/C, that is 5.625 × 103 N/C

SolutionThe dipole moment is p = q × 2a = 2 × 10−6 × 0.01 = 2 × 10−8 C·m. On the axis of a short dipole, E = 2kp/r3 = 2 × 9 × 109 × 2 × 10−8/(0.4)3 = 360/0.064 = 5625 N/C, pointing along the dipole away from the positive end.

Where it goes wrong  Using kp/r3, the equatorial value, on the axis. The axial field is twice the equatorial one at the same distance.

Q5  ·  Motion in a Straight Line  ·  on the pattern of Q30

A paratrooper jumps from an aeroplane, falls freely for 2 s, then opens the parachute and decelerates steadily at 3 m/s2. At a height of 10 m above the ground he is descending at 5 m/s. The height of the aeroplane was ______ m. (g = 10 m/s2.)

Answer  92.5 m

SolutionTake the two stages one at a time. In 2 s of free fall he reaches 20 m/s and drops 20 m. In the second stage he slows from 20 m/s to 5 m/s at 3 m/s2, covering (400 − 25)/(2 × 3) = 62.5 m. So he has fallen 20 + 62.5 = 82.5 m by the time he is 10 m up, and the aeroplane was at 82.5 + 10 = 92.5 m.

Where it goes wrong  Forgetting the last 10 m. He is not on the ground when the 5 m/s is quoted, so that height has to be added back.

Q6  ·  Nuclei  ·  on the pattern of Q31

Which of these pairs are isobars? (a) 236U and 238U, (b) 198Hg and 197Au, (c) 2H and 3H, (d) 3H and 3He.

Answer  (d) 3H and 3He

SolutionIsobars have the same mass number A but different atomic number Z. Pair (a) is two uranium nuclei with different A, so they are isotopes. Pair (b) has A = 198 and 197, so neither isobars nor isotopes. Pair (c) is deuterium and tritium, both hydrogen, so isotopes again. Pair (d) has A = 3 for both but Z = 1 and Z = 2, so they are isobars.

Where it goes wrong  Swapping isobars and isotopes. Isotopes share Z, the element; isobars share A, the mass number.

Q7  ·  Units and Measurement  ·  on the pattern of Q32

Two cells are compared on a potentiometer and the balancing lengths come out as 200 cm and 100 cm. The least count of the scale is 1 cm. The percentage error in the ratio of the emfs is ______.

Answer  1.5%

SolutionThe ratio of the emfs is the ratio of the balancing lengths, E1/E2 = l1/l2. For a ratio the relative errors add: Δl1/l1 + Δl2/l2 = 1/200 + 1/100 = 0.005 + 0.010 = 0.015, that is 1.5%.

Where it goes wrong  Subtracting the second term because l2 is in the denominator. For the maximum error in a quotient the relative errors are always added.

Q8  ·  Centre of Mass  ·  on the pattern of Q33

A body of mass 14 kg at rest explodes into three fragments whose masses are in the ratio 2 : 2 : 3. The two equal fragments fly off at right angles to each other, each at 18 m/s. The speed of the heaviest fragment is ______ m/s.

Answer  12√2 m/s, about 17 m/s

SolutionThe ratio 2 : 2 : 3 over 14 kg gives masses 4 kg, 4 kg and 6 kg. Each of the two equal fragments carries momentum 4 × 18 = 72 kg·m/s, and since they go off at right angles their combined momentum is 72√2. The body was at rest, so the total momentum must stay zero: the 6 kg piece must carry 72√2 in the opposite direction. So 6v = 72√2 and v = 12√2 m/s.

Where it goes wrong  Adding the two momenta as 144 instead of 72√2. They are perpendicular, so the Pythagorean sum is what the third fragment has to balance.

Q9  ·  Capacitance  ·  on the pattern of Q34

A parallel plate capacitor with a plate separation of 5 mm is charged by a battery. A mica sheet 2 mm thick is now slipped in, with the plates still joined to the battery, and the capacitor is found to draw 25% more charge. The dielectric constant of the mica is ______.

Answer  2

SolutionWith a slab of thickness t and dielectric constant K partly filling a gap d, the capacitance is ε0A/(d − t + t/K). Since the battery stays connected the voltage is fixed, so 25% more charge means 25% more capacitance: 5/(5 − 2 + 2/K) = 1.25. So 3 + 2/K = 4, giving 2/K = 1 and K = 2.

Where it goes wrong  Using C = Kε0A/d, which is for a slab that fills the whole gap. Here 3 mm of the gap is still air.

Q10  ·  Wave Optics  ·  on the pattern of Q35

Unpolarised light falls on a glass plate in air at the particular angle for which the reflected beam is completely linearly polarised. The refracted beam then makes an angle of about ______ with the normal. (Refractive index of glass = 1.52, tan−1(1.52) = 56.7°.)

Answer  about 33.3°

SolutionFull polarisation on reflection happens at the Brewster angle, where tan iB = μ = 1.52, so iB = 56.7°. At that angle the reflected and refracted rays are exactly 90° apart, so the refracted ray makes 90 − 56.7 = 33.3° with the normal. (Checking with Snell’s law: sin56.7°/sin33.3° = 0.836/0.549 = 1.52.)

Where it goes wrong  Giving the Brewster angle itself. The question asks for the refracted ray, which is the complement of it.

Q11  ·  Electromagnetic Waves  ·  on the pattern of Q36

The ratio of the speed of electromagnetic waves in vacuum to their speed in a medium of dielectric constant K = 3 and permeability μ = 2μ0 is ______.

Answer  √6 : 1

SolutionA wave travels at v = 1/√(με). In vacuum that is c = 1/√(μ0ε0); in the medium ε = Kε0 = 3ε0 and μ = 2μ0, so v = c/√(3 × 2) = c/√6. The ratio c : v is therefore √6 : 1, which also means the refractive index of the medium is √6.

Where it goes wrong  Writing n = √K and ignoring the permeability. That shortcut is only valid when the medium is non-magnetic, μ = μ0, which it is not here.

Q12  ·  Mechanical Properties of Fluids  ·  on the pattern of Q37

A small metal sphere of diameter 2 mm and density 10.5 g/cm3 is dropped into glycerine of viscosity 10 poise and density 1.5 g/cm3. Its terminal velocity is ______ cm/s. (g = 10 m/s2.)

Answer  2 cm/s

SolutionConvert everything first: r = 1 mm = 10−3 m, ρsphere = 10500 kg/m3, ρliquid = 1500 kg/m3, and 10 poise = 1 Pa·s. Then v = 2r2(ρ − σ)g/9η = 2 × 10−6 × 9000 × 10/(9 × 1) = 0.18/9 = 0.02 m/s = 2 cm/s.

Where it goes wrong  Leaving the viscosity in poise. One poise is 0.1 Pa·s, so 10 poise is exactly 1 in SI — convenient here, but it still has to be checked.

Q13  ·  Ray Optics  ·  on the pattern of Q38

A prism of angle 75° is made of material of refractive index √3, and its back face is coated with a thin film of refractive index 1.5. For total internal reflection to happen at that back face, the angle of incidence at the front face must be less than about ______.

Answer  about 27°

SolutionAt the coated face the light tries to go from glass (√3) into film (1.5), so the critical angle there is given by sin C = 1.5/√3 = 0.866, that is C = 60°. Total internal reflection needs r2 ≥ 60°. Since r1 + r2 = A = 75°, this means r1 ≤ 15°. At the front face sin i = √3 sin r1 ≤ √3 × 0.2588 = 0.448, so i ≤ about 26.6°. Any steeper incidence lets the light out instead.

Where it goes wrong  Using 1/√3 for sin C, as though the film were air. The rarer medium at that face is the film, so its index goes on top.

Q14  ·  Semiconductor Electronics  ·  on the pattern of Q39

Inputs A and B are fed to a NOR gate, and the output of the NOR gate is fed to a NOT gate, giving Y. Write the truth table for Y.

Answer  Y = A OR B — Y is 0 only when A = B = 0

SolutionA NOR gate is an OR followed by an inversion, so putting another inversion after it undoes the first: the pair behaves as a plain OR gate. Row by row: (0,0) gives NOR = 1, NOT = 0; (0,1) gives NOR = 0, NOT = 1; (1,0) the same, Y = 1; (1,1) gives NOR = 0, NOT = 1. So Y is 1 whenever at least one input is 1.

Where it goes wrong  Assuming two negations somewhere in a circuit always cancel. They do here because they sit one straight after the other; a NOT on each input instead would give something quite different.

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

An air bubble of volume 2.9 cm3 rises from the bottom of a swimming pool 5 m deep, where the water is at 17 °C, to the surface, where it is at 27 °C. The volume of the bubble at the surface is ______ cm3. (g = 10 m/s2, density of water = 103 kg/m3, atmospheric pressure = 105 Pa.)

Answer  4.5 cm3

SolutionAt the bottom the pressure is atmospheric plus the water column: 105 + 1000 × 10 × 5 = 1.5 × 105 Pa. At the surface it is 105 Pa. The air in the bubble is a fixed amount of gas, so P1V1/T1 = P2V2/T2 with T in kelvin: (1.5 × 105)(2.9)/290 = (105)V2/300. That gives V2 = 4.5 cm3.

Where it goes wrong  Using only the water pressure at the bottom and leaving out the atmosphere above it. Both press on the bubble.

Q16  ·  Electromagnetic Induction  ·  on the pattern of Q41

A circular loop of radius 7 cm lies in a uniform field of 0.2 T at right angles to its plane. The loop is reshaped into a square in 0.5 s. The emf induced is ______ mV. (π = 22/7.)

Answer  1.32 mV

SolutionThe wire keeps its length, so the circumference 2πr = 2 × (22/7) × 7 = 44 cm becomes the perimeter of the square, giving a side of 11 cm. The circle’s area is πr2 = (22/7)(49) = 154 cm2, the square’s is 121 cm2, so the area shrinks by 33 cm2 = 33 × 10−4 m2. The flux change is 0.2 × 33 × 10−4 = 6.6 × 10−4 Wb, and dividing by 0.5 s gives 1.32 × 10−3 V = 1.32 mV.

Where it goes wrong  Taking the square’s side as the diameter or the radius. The wire’s length is what is conserved, so the circumference becomes the perimeter.

Q17  ·  Electrostatic Potential  ·  on the pattern of Q42

Charges of 7 μC and −2 μC are placed at (−9, 0, 0) cm and (9, 0, 0) cm in an external field E = (A/r2)r̂, where A = 9 × 105 N·C−1m2. Taking the potential at infinity to be zero, the electrostatic energy of the arrangement is ______ J.

Answer  49.3 J

SolutionThree pieces have to be added. The external field corresponds to a potential V = A/r, which at r = 0.09 m is 9 × 105/0.09 = 107 V. The two charges sitting in it contribute q1V + q2V = (7 × 10−6)(107) + (−2 × 10−6)(107) = 70 − 20 = 50 J. Then the two charges also attract each other, 18 cm apart: kq1q2/r = 9 × 109 × (7 × 10−6)(−2 × 10−6)/0.18 = −0.7 J. Total: 50 − 0.7 = 49.3 J.

Where it goes wrong  Counting only the mutual energy, or only the energy in the external field. Both belong in the total, and here the external part is by far the bigger of the two.

Q18  ·  Electromagnetic Induction  ·  on the pattern of Q43

A long solenoid of 500 turns per metre carries I = 10 sin(ωt) A with ω = 1000 rad/s. A circular loop of radius 1 cm and resistance 10 Ω lies coaxially well inside it. The r.m.s. current in the loop is α/√2 microampere. The value of α is about ______.

Answer  about 197

SolutionInside the solenoid B = μ0nI, so the flux through the loop is Φ = μ0nIπa2 = 4π × 10−7 × 500 × π × 10−4 × 10 sinωt = 1.974 × 10−6 sinωt Wb. The emf is its rate of change, with peak value 1.974 × 10−6 × 1000 = 1.974 × 10−3 V, so the peak current is 1.974 × 10−4 A = 197.4 μA. Dividing by √2 gives the r.m.s. value, so α ≈ 197. The field inside a long solenoid is uniform, so sliding the loop along contributes nothing extra.

Where it goes wrong  Using the solenoid’s own radius for the flux area. The flux that matters is the flux through the small loop, so it is the loop’s area that goes in.

Q19  ·  Thermodynamics  ·  on the pattern of Q44

One mole of helium is held in a cylinder of cross-section 20 cm2 under a light, frictionless, movable piston, with atmospheric pressure 105 Pa outside. The gas is heated slowly with 100 J of heat. The piston moves out by ______ cm.

Answer  20 cm

SolutionA light free piston keeps the pressure constant at 105 Pa. For a monatomic gas the heat splits in a fixed ratio: Q = nCPΔT = (5/2)nRΔT while the work is W = nRΔT, so W = (2/5)Q = 40 J. The other 60 J goes into the internal energy. Then W = P·A·x, so x = 40/(105 × 20 × 10−4) = 40/200 = 0.2 m = 20 cm.

Where it goes wrong  Putting all 100 J into the work. At constant pressure only two fifths of the heat becomes work for a monatomic gas; the rest warms the gas.

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

A current of 2 A flows round a loop in the shape of an equilateral triangle of side 4√3 cm. The magnetic field at its centroid is α × 10−5 T. The value of α is ______. (μ0 = 4π × 10−7 SI units.)

Answer  3√3, about 5.2

SolutionThe centroid is a/2√3 = 4√3/(2√3) = 2 cm from each side, so d = 0.02 m. Each side is a finite wire, and from the foot of the perpendicular each half of it subtends an angle whose tangent is (a/2)/d = 2√3/2 = √3, that is 60°. For a finite wire, B = (μ0I/4πd)(sinθ1 + sinθ2) = (10−7 × 2/0.02)(2 × √3/2) = √3 × 10−5 T. All three sides give fields in the same direction, so the total is 3√3 × 10−5 T and α = 3√3.

Where it goes wrong  Using the infinite-wire formula μ0I/2πd. Each side is short, so the sines of the two end angles have to be put in.

Q21  ·  Nuclei  ·  numeric  ·  on the pattern of Q46

The average energy released per fission of a uranium-235 nucleus is 190 MeV. When every atom in 47 g of pure uranium-235 undergoes fission, the energy released is α × 1023 MeV. The value of α is ______. (Avogadro number = 6 × 1023 per mole.)

Answer  228

Solution47 g of uranium-235 is 47/235 = 0.2 mole, so it holds 0.2 × 6 × 1023 = 1.2 × 1023 atoms. Each gives 190 MeV, so the total is 1.2 × 1023 × 190 = 228 × 1023 MeV, and α = 228.

Where it goes wrong  Treating 47 g as one mole. The molar mass is 235 g, so 47 g is a fifth of a mole.

Q22  ·  Units and Dimensions  ·  numeric  ·  on the pattern of Q47

A ball of radius r and density ρ, dropped through a viscous liquid of density σ and viscosity η, reaches its terminal velocity after a time t given by t = Aρarbηcσd, where A is a dimensionless constant and a, b, c, d are integers. The value of (b + c)/(a + d) is ______.

Answer  1

SolutionWrite out the dimensions: [ρ] = [σ] = ML−3, [r] = L, [η] = ML−1T−1. Matching T on both sides gives −c = 1, so c = −1. Matching M gives a + c + d = 0, so a + d = 1. Matching L gives −3a + b − c − 3d = 0, that is b = 3(a + d) + c = 3 − 1 = 2. So (b + c)/(a + d) = (2 − 1)/1 = 1. Note that a and d are not separately fixed, because ρ and σ have the same dimensions — but the question only asks for their sum.

Where it goes wrong  Giving up because a and d cannot be found one by one. The combination asked for is exactly the one that is determined.

Q23  ·  Rotational Motion  ·  numeric  ·  on the pattern of Q48

A uniform circular disc of mass M and radius r has a concentric circular hole of radius r/2 cut out of it. The moment of inertia of what is left, about the axis of the disc, is (x/32)Mr2. The value of x is ______.

Answer  15

SolutionTreat the cut as a subtraction. The whole disc gives ½Mr2. The piece removed is a disc of radius r/2, and since the material is uniform its mass is M × (r/2)2/r2 = M/4; its own moment of inertia about the same axis is ½(M/4)(r/2)2 = Mr2/32. So what is left has ½Mr2 − Mr2/32 = 16Mr2/32 − Mr2/32 = 15Mr2/32, and x = 15.

Where it goes wrong  Using M for the mass of the removed piece as well. The hole is a quarter of the area, so it carries a quarter of the mass.

Q24  ·  Ray Optics  ·  numeric  ·  on the pattern of Q49

A thin lens forms images of the same size when the object is placed at 8 cm and at 24 cm from it. The focal length of the lens is ______ cm.

Answer  16 cm

SolutionEqual image size at two different object distances happens when one position is inside the focus and the other outside it, and the result is simply f = (u1 + u2)/2 = (8 + 24)/2 = 16 cm. Checking: with f = 16 and u = −8, v = −16 and the magnification is +2 (virtual, erect); with u = −24, v = +48 and the magnification is −2 (real, inverted). Same size, opposite kind.

Where it goes wrong  Assuming both images must be real. One of the two is virtual, which is exactly why the sizes can match.

Q25  ·  Waves  ·  numeric  ·  on the pattern of Q50

The speed of sound in air doubles when the temperature is raised from 0 °C to α °C. The value of α is ______.

Answer  819

SolutionIn a gas the speed of sound goes as √T, with T in kelvin. Doubling the speed therefore needs four times the absolute temperature: 4 × 273 = 1092 K. In celsius that is 1092 − 273 = 819 °C.

Where it goes wrong  Doubling 0 °C, which gives nothing, or working in celsius at all. The square-root law needs absolute temperature.

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, 23 January 2026, is on its own page, and the other January dates are 21, 22, 24 and 28 January.

Official final answer key →

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

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