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JEE Advanced 2026 Paper 1 Question Paper with Answers

48 questions · Physics, Chemistry & Mathematics

The complete JEE Advanced 2026 Paper 1 paper — every question with its correct answer, tagged to the chapter it tests. Free to read, no account needed.

Physics
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Chemistry
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Mathematics
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Physics — JEE Advanced 2026 Paper 1

Q1·PhysicsSingle correct
Consider a large disk of radius RRR and two smaller disks, each of radius r=R/50r = R/50r=R/50, lying on its circumference, as shown in the figure. The smaller disks are initially in contact with each other, with an angular separation Δθ\Delta\thetaΔθ between their centers. They are made to roll without slipping in opposite directions, with constant angular velocities ω\omegaω and 2ω2\omega2ω while the large disk is held stationary. The time τ\tauτ at which the smaller disks are again in contact is: [Use sin⁡(Δθ)=Δθ\sin(\Delta\theta) = \Delta\thetasin(Δθ)=Δθ and ignore gravity.]
  1. (A)τ=51×(2π−451)/ω\tau = 51 \times \left(2\pi - \frac{4}{51}\right)/\omegaτ=51×(2π−514​)/ω
  2. (B)τ=51×(2π−251)/3ω\tau = 51 \times \left(2\pi - \frac{2}{51}\right)/3\omegaτ=51×(2π−512​)/3ω
  3. (C)τ=51×(2π−451)/3ω\tau = 51 \times \left(2\pi - \frac{4}{51}\right)/3\omegaτ=51×(2π−514​)/3ω
  4. (D)τ=51×(2π−251)/ω\tau = 51 \times \left(2\pi - \frac{2}{51}\right)/\omegaτ=51×(2π−512​)/ω

Correct answer: (C)

Step-by-step solution →
Q2·PhysicsSingle correct
Consider a circuit consisting of a capacitor of capacitance CCC and a coil with NNN turns per unit length, cross sectional area SSS and length ddd, where d2≫Sd^2 \gg Sd2≫S. There is another coil of length d/2d/2d/2, cross sectional area S/2S/2S/2 and 2N2N2N turns per unit length completely inside the larger coil, as shown in the figure. The ends of this smaller coil are connected with each other by an insulated conducting wire. The self-inductance of the larger coil is LLL. Neglecting edge effects and all the Ohmic resistances, the resonant frequency of the circuit is:
  1. (A)415 LC\frac{4}{\sqrt{15\,LC}}15LC​4​
  2. (B)65 LC\frac{6}{\sqrt{5\,LC}}5LC​6​
  3. (C)23 LC\frac{2}{\sqrt{3\,LC}}3LC​2​
  4. (D)23 LC\sqrt{\frac{2}{3\,LC}}3LC2​​

Correct answer: (C)

Step-by-step solution →
Q3·PhysicsSingle correct
A solid cylinder of radius RRR rolls without slipping with a center of mass speed v0=gR3v_0 = \sqrt{\frac{gR}{3}}v0​=3gR​​ on a horizontal surface with a vertical edge, as shown in the figure. Here, ggg is the acceleration due to the gravity. At the moment when the cylinder loses contact with the surface due to rotation around the corner, the speed of its center of mass is:
  1. (A)0
  2. (B)5gR7\sqrt{\frac{5gR}{7}}75gR​​
  3. (C)gR15\sqrt{\frac{gR}{15}}15gR​​
  4. (D)3gR7\sqrt{\frac{3gR}{7}}73gR​​

Correct answer: (B)

Step-by-step solution →
Q4·PhysicsSingle correct
A double convex lens made of glass of refractive index 1.5 and radii of curvature of the curved surfaces 20 cm each is immersed in a liquid of refractive index nLn_LnL​. The correct plot showing the variation of the power, in the units of diopter (D)(D)(D), as a function of nLn_LnL​ is:
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (A)

Step-by-step solution →
Q5·PhysicsMultiple correct
Consider a hydrogen atom with vkv_kvk​, rkr_krk​, and KkK_kKk​ denoting the velocity, orbital radius and kinetic energy of the electron in the kthk^{\text{th}}kth orbit, respectively. The electron undergoes a transition from the nthn^{\text{th}}nth orbit, emitting radiation corresponding to the Lyman series. Considering hhh to be the Planck's constant and ϵ0\epsilon_0ϵ0​ the permittivity of the free space, the correct statement(s) is/are:
  1. (A)Magnitude of change in kinetic energy of electron can be expressed as h4π∣nvnrn−v1r1∣\frac{h}{4\pi}\left|\frac{nv_n}{r_n} - \frac{v_1}{r_1}\right|4πh​​rn​nvn​​−r1​v1​​​.
  2. (B)Magnitude of change in de Broglie wavelength of the electron can be expressed as e24ϵ0∣1Kn−1K1∣\frac{e^2}{4\epsilon_0}\left|\frac{1}{K_n} - \frac{1}{K_1}\right|4ϵ0​e2​​Kn​1​−K1​1​​.
  3. (C)Frequency of the radiation emitted can be expressed as e28πϵ0h(1r1−1rn)\frac{e^2}{8\pi\epsilon_0 h}\left(\frac{1}{r_1} - \frac{1}{r_n}\right)8πϵ0​he2​(r1​1​−rn​1​).
  4. (D)Magnitude of change in total energy of the electron can be expressed as h2π∣v1r1−nvnrn∣\frac{h}{2\pi}\left|\frac{v_1}{r_1} - \frac{nv_n}{r_n}\right|2πh​​r1​v1​​−rn​nvn​​​.

Correct answer: (A), (C)

Step-by-step solution →
Q6·PhysicsMultiple correct
A particle is thrown with a speed vvv from a point OOO at an angle θ\thetaθ with the horizontal plane such that it passes through the point PPP at a height of 1 m and horizontal distance of 5 m from OOO, as shown in the figure. If acceleration due to gravity is ggg ms−2^{-2}−2, then the correct statement(s) is/are:
  1. (A)If θ=45∘\theta = 45^\circθ=45∘, then v=5g2v = \frac{5\sqrt{g}}{2}v=25g​​ ms−1^{-1}−1.
  2. (B)If θ=45∘\theta = 45^\circθ=45∘, the particle reaches its maximum height before it reaches PPP.
  3. (C)If θ=30∘\theta = 30^\circθ=30∘, the particle reaches its maximum height after reaching PPP.
  4. (D)If θ=tan⁡−1(15)\theta = \tan^{-1}\left(\frac{1}{5}\right)θ=tan−1(51​), then v=125gv = 125\sqrt{g}v=125g​ ms−1^{-1}−1.

Correct answer: (A), (B)

Step-by-step solution →
Q7·PhysicsMultiple correct
A quasi-static cycle of a monoatomic ideal gas contains an isothermal process (ab)(ab)(ab), followed by an isochoric process (bc)(bc)(bc) and an adiabatic process (ca)(ca)(ca) as shown in the figure. The volumes of the gas are V1V_1V1​ and V2V_2V2​ at aaa and bbb, respectively. If the cycle has heat input QinQ_{\text{in}}Qin​ and output QoutQ_{\text{out}}Qout​, then the efficiency of the cycle is defined as η=Qin−QoutQin\eta = \frac{Q_{\text{in}} - Q_{\text{out}}}{Q_{\text{in}}}η=Qin​Qin​−Qout​​. The correct statement(s) is/are: [Given: ln⁡2≈0.7\ln 2 \approx 0.7ln2≈0.7]
  1. (A)If V2/V1=8V_2/V_1 = 8V2​/V1​=8, the heat released in the process bcbcbc is smaller than the heat absorbed in the process ababab.
  2. (B)For a given value of V2/V1V_2/V_1V2​/V1​, η\etaη does not depend on the temperature of the isothermal process.
  3. (C)If V2/V1=8V_2/V_1 = 8V2​/V1​=8, then the temperature of the gas at aaa is 4 times the temperature of the gas at ccc.
  4. (D)If V2/V1=8V_2/V_1 = 8V2​/V1​=8, then the pressure of the gas at aaa is 4 times the pressure of the gas at bbb.

Correct answer: (A), (B), (C)

Step-by-step solution →
Q8·PhysicsMultiple correct
The electric field associated with an electromagnetic wave travelling in vacuum is given by E0sin⁡(3y+4z+ωt) i^E_0 \sin(3y + 4z + \omega t)\,\hat{i}E0​sin(3y+4z+ωt)i^, where ω\omegaω is the angular frequency. All quantities are in SI units. The correct statement(s) about this wave is/are: [Given: speed of light in vacuum c=3×108c = 3 \times 10^8c=3×108 ms−1^{-1}−1.]
  1. (A)The wave is travelling in −15(3j^+4k^)-\frac{1}{5}(3\hat{j} + 4\hat{k})−51​(3j^​+4k^) direction.
  2. (B)The magnitude of the wave vector is 0.5 m−1^{-1}−1.
  3. (C)The value of ω\omegaω is 1.5×1091.5 \times 10^91.5×109 rad s−1^{-1}−1.
  4. (D)The magnetic field associated with this wave is given by E0csin⁡(3y+4z+ωt)(4j^−3k^)\frac{E_0}{c}\sin(3y + 4z + \omega t)(4\hat{j} - 3\hat{k})cE0​​sin(3y+4z+ωt)(4j^​−3k^).

Correct answer: (A), (C)

Step-by-step solution →
Q9·PhysicsNumerical
A tank contains two immiscible liquids of densities 6ρ6\rho6ρ and 2ρ2\rho2ρ. The higher density liquid is filled up to a height L/2L/2L/2 from the bottom. A thin rod of density ρ\rhoρ and length LLL is fully immersed and hinged at the bottom so that it can oscillate freely, as shown in the figure. If the rod is slightly disturbed from its equilibrium, the time period of small oscillations is 2πnLg\frac{2\pi}{n}\sqrt{\frac{L}{g}}n2π​gL​​, where ggg is the acceleration due to gravity. The value of nnn is:

Correct answer: 1.73

Step-by-step solution →
Q10·PhysicsNumerical
As shown in the figure, five Carnot engines, each with efficiency η\etaη and same number of cycles per unit time, are operating between six heat reservoirs. The amount of heat released per cycle by one engine is completely absorbed by the next engine. Consider Q0Q_0Q0​ to be the amount of heat absorbed per cycle by the first engine and WWW as the amount of total work done by all the engines per cycle, then the net efficiency of the system is found to be ηnet=WQ0=211243\eta_{\text{net}} = \frac{W}{Q_0} = \frac{211}{243}ηnet​=Q0​W​=243211​. The value of η\etaη is:

Correct answer: 0.33

Step-by-step solution →
Q11·PhysicsNumerical
As shown in the figure, an insulated container is fitted with a thermally conducting but immovable partition (P1)(P_1)(P1​) and a freely movable but thermally insulated piston (P2)(P_2)(P2​). The partition P1P_1P1​ with thermal conductivity KKK, cross sectional area AAA and width xxx divides the container into two sections, S1S_1S1​ and S2S_2S2​, each containing one mole of a monoatomic gas. The piston P2P_2P2​ moves freely such that the gas in S2S_2S2​ is always at the atmospheric pressure. Initially, the difference between the temperatures of S1S_1S1​ and S2S_2S2​ is ΔT0\Delta T_0ΔT0​. The time it takes for the temperature difference to become ΔT02\frac{\Delta T_0}{2}2ΔT0​​ is nxR/KAnxR/KAnxR/KA, where RRR is the universal gas constant. The value of nnn is: [Given: ln⁡2≈0.7\ln 2 \approx 0.7ln2≈0.7]

Correct answer: 0.66

Step-by-step solution →
Q12·PhysicsNumerical
A hollow, right circular cone of base radius RRR and height hhh, with its tip at the origin is rotating about the ZZZ-axis with an angular velocity ω\omegaω, as shown in the figure. The cone carries a total charge QQQ uniformly distributed on its curved surface. The magnitude of magnetic field at a point (0,0,z)(0, 0, z)(0,0,z), where z≫Rz \gg Rz≫R and z≫hz \gg hz≫h, is nμ04πQR2ωz3\frac{n\mu_0}{4\pi}\frac{QR^2\omega}{z^3}4πnμ0​​z3QR2ω​. The value of nnn is:

Correct answer: 0.5

Step-by-step solution →
Q13·PhysicsSingle correct
List-I shows four configurations made of straight and semi-circular narrow tubes containing air. A sound wave of wavelength λ=0.29\lambda = 0.29λ=0.29 m enters these structures at the point SSS and a sound detector is placed at DDD. Between the points SSS and DDD, the sound travels only through the tubes. List-II contains the possible smallest values of lll (refer to the figures) for which the detector DDD records maximum amplitude. Ignore effects of sharp corners. [Given: cos⁡(15∘)=0.97\cos(15^\circ) = 0.97cos(15∘)=0.97] Choose the option that best describes the match between the entries in List-I to those in List-II.
List-IList-II
P.see figure1.1.32 m
Q.see figure2.1.19 m
R.see figure3.0.51 m
S.see figure4.0.29 m
5.0.13 m
  1. (A)P→4, Q→3, R→5, S→1
  2. (B)P→4, Q→3, R→1, S→5
  3. (C)P→3, Q→4, R→1, S→2
  4. (D)P→3, Q→4, R→5, S→2

Correct answer: (D)

Step-by-step solution →
Q14·PhysicsSingle correct
In the List-I, four optical effects are mentioned. The physical phenomena of light which are essential to describe these optical effects are given in List-II. Choose the option which describes the correct match between the entries in List-I to those in List-II.
List-IList-II
P.Colorful sky in north polar region (Aurora Borealis)1.Dispersion and reflection
Q.Partially polarized sun light2.Total internal reflection
R.Rainbow3.Diffraction
S.Dark and bright fringes4.Scattering of light by molecules in the atmosphere
5.Emission of radiation from oxygen and nitrogen atoms excited by charged particles
  1. (A)P→5, Q→4, R→1, S→3
  2. (B)P→4, Q→2, R→1, S→3
  3. (C)P→4, Q→1, R→2, S→3
  4. (D)P→5, Q→4, R→1, S→2

Correct answer: (A)

Step-by-step solution →
Q15·PhysicsSingle correct
List-I contains four conducting loops lying in the XYXYXY plane, as shown in the figures. The loops are rotating about ZZZ axis passing through the point OOO with time period TTT in clockwise direction. The region x>0x > 0x>0 contains a uniform magnetic field BBB in the +z+z+z direction. List-II contains the qualitative variation of the induced current i(t)i(t)i(t) for each of these loops. Choose the option which describes the correct match between the entries in List-I to those in List-II.
  1. (A)P → 5, Q → 4, R → 1, S → 3
  2. (B)P → 3, Q → 2, R → 5, S → 4
  3. (C)P → 3, Q → 2, R → 1, S → 4
  4. (D)P → 5, Q → 1, R → 2, S → 3

Correct answer: (C)

Step-by-step solution →
Q16·PhysicsSingle correct
List-I shows four planar structures made of uniform solid rods each of mass mmm and length lll. In the List-II the possible moment of inertia of these structures about an axis OCO′OCO'OCO′, which lies in the plane of the structures, are given. Choose the option that describes the correct match between the entries in List-I to those in List-II.
List-IList-II
P.see figure1.54ml2\frac{5}{4}ml^245​ml2
Q.see figure2.16ml2\frac{1}{6}ml^261​ml2
R.see figure3.112ml2\frac{1}{12}ml^2121​ml2
S.see figure4.23ml2\frac{2}{3}ml^232​ml2
5.13ml2\frac{1}{3}ml^231​ml2
  1. (A)P→ 5, Q→ 1, R→ 4, S→ 2
  2. (B)P→ 1, Q→ 3, R→ 4, S→ 2
  3. (C)P→ 5, Q→ 3, R→ 2, S→ 1
  4. (D)P→ 5, Q→ 4, R→ 2, S→ 1

Correct answer: (A)

Step-by-step solution →

Chemistry — JEE Advanced 2026 Paper 1

Q17·ChemistrySingle correct
An ideal gas (0.5 mol), initially at 2 bar pressure, is compressed at a constant temperature of 600 K in two steps: first, against a constant external pressure of PPP bar (2<P<82 < P < 82<P<8), and then against constant external pressure of 8 bar. At each step, the compression is stopped only when the pressure of the gas becomes equal to the external pressure. The total work done on the gas in these steps is WWW. Considering all possible values of PPP (2<P<82 < P < 82<P<8) and taking the gas constant as RRR (in J K−1 mol−1\mathrm{J\,K^{-1}\,mol^{-1}}JK−1mol−1), the minimum value of ∣W∣|W|∣W∣ (in J) is
  1. (A)207R207R207R
  2. (B)600R600R600R
  3. (C)630R630R630R
  4. (D)900R900R900R

Correct answer: (B)

Step-by-step solution →
Q18·ChemistrySingle correct
For a reversible reaction R ⇌ P, at constant temperature, both the forward and the backward reactions are first order elementary reactions with rate constants kfk_fkf​ and kbk_bkb​, respectively. At time zero, the concentration of R is [R]0[\mathrm{R}]_0[R]0​ and the concentration of P is zero. At any given time, [R] and [P] are the concentrations of R and P, respectively. If kb=4kfk_b = 4k_fkb​=4kf​, the correct graphical representation of the reaction is
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (C)

Step-by-step solution →
Q19·ChemistrySingle correct
The correct order of dipole moments for the given species is
  1. (A)BF3=NH4+<NF3<NH3\mathrm{BF_3} = \mathrm{NH_4^+} < \mathrm{NF_3} < \mathrm{NH_3}BF3​=NH4+​<NF3​<NH3​
  2. (B)BF3<NH4+<NF3<NH3\mathrm{BF_3} < \mathrm{NH_4^+} < \mathrm{NF_3} < \mathrm{NH_3}BF3​<NH4+​<NF3​<NH3​
  3. (C)NH4+<BF3<NH3<NF3\mathrm{NH_4^+} < \mathrm{BF_3} < \mathrm{NH_3} < \mathrm{NF_3}NH4+​<BF3​<NH3​<NF3​
  4. (D)BF3<NH4+<NH3<NF3\mathrm{BF_3} < \mathrm{NH_4^+} < \mathrm{NH_3} < \mathrm{NF_3}BF3​<NH4+​<NH3​<NF3​

Correct answer: (A)

Step-by-step solution →
Q20·ChemistrySingle correct
Considering LiBH4\mathrm{LiBH_4}LiBH4​ reduces an ester group to the corresponding alcohol and does not reduce a carboxylic acid group, the correct statement about the major products P, Q, R and S is
  1. (A)P & Q are identical, and R & S are diastereomers.
  2. (B)P & Q are diastereomers, and R & S are identical.
  3. (C)P & Q are diastereomers, and R & S are diastereomers.
  4. (D)P & Q are identical, and R & S are identical.

Correct answer: (C)

Step-by-step solution →
Q21·ChemistryMultiple correct
The 2s and the 2p orbital energies of hydrogen atom are E2s(H)E_{2s}(\mathrm{H})E2s​(H) and E2p(H)E_{2p}(\mathrm{H})E2p​(H), respectively. The 2s and the 2p orbital energies of lithium atom are E2s(Li)E_{2s}(\mathrm{Li})E2s​(Li) and E2p(Li)E_{2p}(\mathrm{Li})E2p​(Li), respectively. The correct option(s) about the orbital energies is(are)
  1. (A)E2s(Li)<E2p(Li)E_{2s}(\mathrm{Li}) < E_{2p}(\mathrm{Li})E2s​(Li)<E2p​(Li)
  2. (B)E2s(H)=E2p(H)E_{2s}(\mathrm{H}) = E_{2p}(\mathrm{H})E2s​(H)=E2p​(H)
  3. (C)E2p(H)<E2s(Li)E_{2p}(\mathrm{H}) < E_{2s}(\mathrm{Li})E2p​(H)<E2s​(Li)
  4. (D)E2s(H)>E2s(Li)E_{2s}(\mathrm{H}) > E_{2s}(\mathrm{Li})E2s​(H)>E2s​(Li)

Correct answer: (A), (B), (D)

Step-by-step solution →
Q22·ChemistryMultiple correct
Correct statement(s) about the compounds X, Y and Z is(are)
  1. (A)X is used for sterilizing drinking water.
  2. (B)Y has a planar structure.
  3. (C)Z is used in the enrichment of 235U^{235}\mathrm{U}235U.
  4. (D)Y is a stronger Lewis base than ammonia.

Correct answer: (A), (C)

Step-by-step solution →
Q23·ChemistryMultiple correct
Reaction of PtF6\mathrm{PtF_6}PtF6​ with oxygen (O2\mathrm{O_2}O2​) gas results in the formation of an ionic compound, X+Y−\mathrm{X^+Y^-}X+Y−. Correct statement(s) is(are)
  1. (A)The bond order of X+\mathrm{X^+}X+ is 1.5.
  2. (B)Valence ddd-orbitals of the metal ion in X+Y−\mathrm{X^+Y^-}X+Y− has 5 electrons.
  3. (C)PtF6\mathrm{PtF_6}PtF6​ acts as an oxidant in this reaction.
  4. (D)PtF6\mathrm{PtF_6}PtF6​ acts as a fluorinating agent in this reaction.

Correct answer: (B), (C)

Step-by-step solution →
Q24·ChemistryMultiple correct
In the following reaction sequence, Q, R, S and T are the major products. The correct statement(s) about Q, R, S and T is(are)
  1. (A)S on warming with ammoniacal AgNO3\mathrm{AgNO_3}AgNO3​ results in the formation of silver mirror.
  2. (B)Q on treatment with Cl2\mathrm{Cl_2}Cl2​(excess)/UV gives gammaxane.
  3. (C)T is a heterocyclic compound.
  4. (D)R on acid catalyzed intramolecular cyclization followed by treatment with Zn-Hg/HCl gives 9,10-dihydroxyanthracene.

Correct answer: (A), (B), (C)

Step-by-step solution →
Q25·ChemistryNumerical
Two cylinders, both fitted with frictionless pistons, are filled with mixtures of He and Ar gases. In the first cylinder, the masses of He and Ar are m1m_1m1​ and m2m_2m2​, respectively. In the second cylinder, the masses of He and Ar are m2m_2m2​ and m1m_1m1​, respectively. The molar mass of Ar is 10 times the molar mass of He. The external pressure applied by the piston on the first cylinder needs to be 5 times that on the second cylinder so that the volume of the gas mixtures in both the cylinders are equal at the same temperature. Assuming He and Ar behave like ideal gases, the value of (m1/m2)(m_1/m_2)(m1​/m2​) is ____.

Correct answer: 9.8

Step-by-step solution →
Q26·ChemistryNumerical
The total number of all possible isomers for the square planar complex with formula K[M(NCS)(NO2)(gly)]\mathrm{K[M(NCS)(NO_2)(gly)]}K[M(NCS)(NO2​)(gly)] is ____. (MMM = metal ion and gly = NH2CH2COO−\mathrm{NH_2CH_2COO^-}NH2​CH2​COO−)

Correct answer: 8

Step-by-step solution →
Q27·ChemistryNumerical
The sum of total number of carbonyl groups (>C=O>\mathrm{C}=\mathrm{O}>C=O) present in the major products X and Y in the following reactions is ____.

Correct answer: 4

Step-by-step solution →
Q28·ChemistryNumerical
Treatment of buta-1,3-diyne with NaNH2\mathrm{NaNH_2}NaNH2​ (2 equivalents), followed by reaction with excess of trans-CH3-CH=CH-CH2-Br\mathit{trans}\text{-}\mathrm{CH_3}\text{-}\mathrm{CH}=\mathrm{CH}\text{-}\mathrm{CH_2}\text{-}\mathrm{Br}trans-CH3​-CH=CH-CH2​-Br gives X as the major product. The maximum number of carbon atoms that are collinear (in a straight line) in X is ____.

Correct answer: 6

Step-by-step solution →
Q29·ChemistrySingle correct
List-I contains various physical/chemical processes, and List-II contains combinations of changes in enthalpy (ΔH\Delta HΔH) and entropy (ΔS\Delta SΔS). Match each entry in List-I to the appropriate entry in List-II, and choose the correct option.
List-IList-II
P.Physisorption1.ΔH>0\Delta H > 0ΔH>0 and ΔS>0\Delta S > 0ΔS>0
Q.Diamond ⟶ Graphite2.ΔH<0\Delta H < 0ΔH<0 and ΔS<0\Delta S < 0ΔS<0
R.Denaturation of protein3.ΔH<0\Delta H < 0ΔH<0 and ΔS=0\Delta S = 0ΔS=0
S.Propene ⟶ Cyclopropane4.ΔH>0\Delta H > 0ΔH>0 and ΔS<0\Delta S < 0ΔS<0
5.ΔH<0\Delta H < 0ΔH<0 and ΔS>0\Delta S > 0ΔS>0
  1. (A)P → 2; Q → 3; R → 5; S → 4
  2. (B)P → 4; Q → 3; R → 5; S → 1
  3. (C)P → 2; Q → 5; R → 1; S → 4
  4. (D)P → 2; Q → 5; R → 1; S → 3

Correct answer: (C)

Step-by-step solution →
Q30·ChemistrySingle correct
Consider the following species: SOCl2\mathrm{SOCl_2}SOCl2​, XeOF4\mathrm{XeOF_4}XeOF4​, ClF3\mathrm{ClF_3}ClF3​, ClF5\mathrm{ClF_5}ClF5​, XeF5+\mathrm{XeF_5^+}XeF5+​, SO32−\mathrm{SO_3^{2-}}SO32−​, XeF3+\mathrm{XeF_3^+}XeF3+​, SF4\mathrm{SF_4}SF4​ List-I contains different molecular shapes and List-II contains total number of species with the same molecular shapes from the given species. Match each entry in List-I with the appropriate entry in List-II and choose the correct option.
List-IList-II
P.See-saw1.one
Q.T-Shaped2.two
R.Trigonal Planar3.three
S.Square Pyramidal4.four
5.zero
  1. (A)P → 1; Q → 2; R → 5; S → 3
  2. (B)P → 5; Q → 4; R → 2; S → 3
  3. (C)P → 3; Q → 2; R → 1; S → 4
  4. (D)P → 1; Q → 3; R → 5; S → 4

Correct answer: (A)

Step-by-step solution →
Q31·ChemistrySingle correct
The List-II contains products obtained from the reaction of compounds in List-I with O3/Zn\mathrm{O_3/Zn}O3​/Zn-H2O\mathrm{H_2O}H2​O followed by cyclization (via more stable enolate) in the presence of aqueous NaOH. Match each entry in List-I with appropriate entry in List-II and choose the correct option.
  1. (A)P ⟶ 2; Q ⟶ 4; R ⟶ 1; S ⟶ 3
  2. (B)P ⟶ 3; Q ⟶ 4; R ⟶ 5; S ⟶ 2
  3. (C)P ⟶ 2; Q ⟶ 1; R ⟶ 5; S ⟶ 3
  4. (D)P ⟶ 3; Q ⟶ 5; R ⟶ 4; S ⟶ 2

Correct answer: (C)

Step-by-step solution →
Q32·ChemistrySingle correct
Match the major products obtained in the reactions given in List-I with the corresponding structures in List-II and choose the correct option.
  1. (A)P → 2; Q → 1; R → 5; S → 4
  2. (B)P → 1; Q → 2; R → 4; S → 5
  3. (C)P → 1; Q → 2; R → 3; S → 4
  4. (D)P → 2; Q → 1; R → 3; S → 5

Correct answer: (B)

Step-by-step solution →

Mathematics — JEE Advanced 2026 Paper 1

Q33·MathematicsSingle correct
Consider the function f:(0,∞)→(−∞,∞)f : (0, \infty) \to (-\infty, \infty)f:(0,∞)→(−∞,∞) given by f(x)=x log⁡e(x)−x+1f(x) = \sqrt{x}\, \log_e(x) - x + 1f(x)=x​loge​(x)−x+1. Then which one of the following statements is TRUE ?
  1. (A)The derivative of the function fff is decreasing in the interval (0,1)(0, 1)(0,1)
  2. (B)The function fff has a local maximum at some point a∈(0,∞)a \in (0, \infty)a∈(0,∞)
  3. (C)The function fff has a local minimum at some point b∈(0,∞)b \in (0, \infty)b∈(0,∞)
  4. (D)The function fff has NEITHER a point of local maximum NOR a point of local minimum in the interval (0,∞)(0, \infty)(0,∞)

Correct answer: (D)

Step-by-step solution →
Q34·MathematicsSingle correct
Let PPP be the point on the parabola y=x2y = x^2y=x2 such that the slope of the tangent to the parabola at the point PPP is 4. Let QQQ be the point in the first quadrant lying on the circle x2+y2=2x^2 + y^2 = 2x2+y2=2 such that the slope of the tangent to the circle at the point QQQ is −1-1−1. Let RRR be the point in the first quadrant lying on the ellipse x2+4y2=8x^2 + 4y^2 = 8x2+4y2=8 such that the slope of the tangent to the ellipse at the point RRR is −12-\frac{1}{2}−21​. Then the radius of the circle passing through the points P,QP, QP,Q and RRR is
  1. (A)10\sqrt{10}10​
  2. (B)5\sqrt{5}5​
  3. (C)52\sqrt{\frac{5}{2}}25​​
  4. (D)252\sqrt{5}25​

Correct answer: (C)

Step-by-step solution →
Q35·MathematicsSingle correct
Which one of the following matrices can be obtained by performing elementary row transformations on the 3×33 \times 33×3 identity matrix ?
  1. (A)[111111111]\begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}​111​111​111​​
  2. (B)[111234121]\begin{bmatrix} 1 & 1 & 1 \\ 2 & 3 & 4 \\ 1 & 2 & 1 \end{bmatrix}​121​132​141​​
  3. (C)[111234258]\begin{bmatrix} 1 & 1 & 1 \\ 2 & 3 & 4 \\ 2 & 5 & 8 \end{bmatrix}​122​135​148​​
  4. (D)[111−112023]\begin{bmatrix} 1 & 1 & 1 \\ -1 & 1 & 2 \\ 0 & 2 & 3 \end{bmatrix}​1−10​112​123​​

Correct answer: (B)

Step-by-step solution →
Q36·MathematicsSingle correct
Considering only the principal values of the inverse trigonometric functions, the value of cot⁡−1(cot⁡(−11))+10sin⁡(2cos⁡−1(12))+10sin⁡(2tan⁡−1(2))\cot^{-1}(\cot(-11)) + 10 \sin\left( 2 \cos^{-1}\left( \frac{1}{\sqrt{2}} \right) \right) + 10 \sin(2 \tan^{-1}(2))cot−1(cot(−11))+10sin(2cos−1(2​1​))+10sin(2tan−1(2)) is
  1. (A)3π+73\pi + 73π+7
  2. (B)777
  3. (C)4π+74\pi + 74π+7
  4. (D)3π−53\pi - 53π−5

Correct answer: (C)

Step-by-step solution →
Q37·MathematicsMultiple correct
Suppose that Box I contains 6 red balls and 9 green balls, and Box II contains 8 red balls and 12 green balls. All the balls of Box I and Box II are mixed together and a ball is chosen at random from them. Let E1E_1E1​ be the event that the ball chosen belonged to Box I and let E2E_2E2​ be the event that the ball chosen belonged to Box II. Let F1F_1F1​ be the event that the ball chosen is red and let F2F_2F2​ be the event that the ball chosen is green. Then which of the following statements is (are) TRUE ?
  1. (A)The events E1E_1E1​ and F1F_1F1​ are independent
  2. (B)The events E2E_2E2​ and F2F_2F2​ are dependent
  3. (C)The conditional probability P(F1∣E1)P(F_1 \mid E_1)P(F1​∣E1​) is equal to the conditional probability P(F1∣E2)P(F_1 \mid E_2)P(F1​∣E2​)
  4. (D)The conditional probability P(F1∣E1)P(F_1 \mid E_1)P(F1​∣E1​) is greater than the conditional probability P(F2∣E2)P(F_2 \mid E_2)P(F2​∣E2​)

Correct answer: (A), (C)

Step-by-step solution →
Q38·MathematicsMultiple correct
Let PPP be the plane such that it contains the straight line x−12=y−33=z+21\frac{x-1}{2} = \frac{y-3}{3} = \frac{z+2}{1}2x−1​=3y−3​=1z+2​ and is perpendicular to the plane x+2y+3z=4x + 2y + 3z = 4x+2y+3z=4. Let P1P_1P1​ be the plane which passes through the point (4,2,2)(4, 2, 2)(4,2,2) and is parallel to PPP. Then which of the following statements is (are) TRUE ?
  1. (A)The equation of the plane PPP is 7x−5y+z=−107x - 5y + z = -107x−5y+z=−10
  2. (B)The distance between the planes PPP and P1P_1P1​ is 30
  3. (C)The distance of the plane PPP from the origin is 232\sqrt{3}23​
  4. (D)The acute angle between the plane PPP and the plane 2x+2y+z=32x + 2y + z = 32x+2y+z=3 is cos⁡−1(133)\cos^{-1}\left( \frac{1}{3\sqrt{3}} \right)cos−1(33​1​)

Correct answer: (A), (D)

Step-by-step solution →
Q39·MathematicsMultiple correct
Let R\mathbb{R}R denote the set of all real numbers. Let f:R→Rf : \mathbb{R} \to \mathbb{R}f:R→R be an arbitrary function and let g:R→Rg : \mathbb{R} \to \mathbb{R}g:R→R be the function defined by g(x)=xf(x)g(x) = x f(x)g(x)=xf(x), for all x∈Rx \in \mathbb{R}x∈R. Then which of the following statements is (are) TRUE ?
  1. (A)The function ggg is always continuous at x=0x = 0x=0
  2. (B)If fff is continuous at x=0x = 0x=0, then ggg is differentiable at x=0x = 0x=0
  3. (C)If ggg is differentiable at x=0x = 0x=0, then fff is continuous at x=0x = 0x=0
  4. (D)If ggg is differentiable at x=0x = 0x=0, then lim⁡x→0f(x)\lim_{x \to 0} f(x)limx→0​f(x) exists

Correct answer: (B), (D)

Step-by-step solution →
Q40·MathematicsMultiple correct
Consider the matrix M=[2−110]M = \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix}M=[21​−10​]. Let p,q,r,s,a,b,cp, q, r, s, a, b, cp,q,r,s,a,b,c and ddd be integers such that M26=[pqrs]M^{26} = \begin{bmatrix} p & q \\ r & s \end{bmatrix}M26=[pr​qs​] and ∑k=126Mk=[abcd]\sum_{k=1}^{26} M^k = \begin{bmatrix} a & b \\ c & d \end{bmatrix}∑k=126​Mk=[ac​bd​]. Then which of the following statements is (are) TRUE ?
  1. (A)There exists a 2×22 \times 22×2 invertible matrix NNN with real entries such that MN=N[1101]MN = N \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}MN=N[10​11​]
  2. (B)The value of aaa is 378
  3. (C)For any two given integers mmm and nnn, there exist unique integers xxx and yyy such that px+qy=mpx + qy = mpx+qy=m and rx+sy=nrx + sy = nrx+sy=n
  4. (D)For each positive real number ttt, the system of linear equations (a+t)x+by=1(a + t)x + by = 1(a+t)x+by=1 and cx+(d+t)y=−1cx + (d + t)y = -1cx+(d+t)y=−1 has a unique solution

Correct answer: (A), (C), (D)

Step-by-step solution →
Q41·MathematicsInteger
Let S={1,2,3,…,10}S = \{1, 2, 3, \ldots, 10\}S={1,2,3,…,10}. Consider the set X={R:R is an equivalence relation on the set S such that R has exactly 42 elements}X = \{R : R \text{ is an equivalence relation on the set } S \text{ such that } R \text{ has exactly 42 elements}\}X={R:R is an equivalence relation on the set S such that R has exactly 42 elements}. Then the number of elements in XXX is ______.

Correct answer: 2520

Step-by-step solution →
Q42·MathematicsInteger
Consider the function f:(−π2,π2)→(−∞,∞)f : \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \to (-\infty, \infty)f:(−2π​,2π​)→(−∞,∞) defined by f(x)=(∣x∣+∣x−1∣)sin⁡x+[xsin⁡x]f(x) = (|x| + |x - 1|) \sin x + [x \sin x]f(x)=(∣x∣+∣x−1∣)sinx+[xsinx], where [xsin⁡x][x \sin x][xsinx] is the greatest integer less than or equal to xsin⁡xx \sin xxsinx. Let α\alphaα be the total number of points in the interval (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)(−2π​,2π​) at which fff is NOT continuous, and let β\betaβ be the total number of points in the interval (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)(−2π​,2π​) at which fff is NOT differentiable. Then the value of α+β\alpha + \betaα+β is ______.

Correct answer: 5

Step-by-step solution →
Q43·MathematicsInteger
The number of ways to distribute 10 identical red pens and 14 identical blue pens among four persons such that each person gets 6 pens, is ______.

Correct answer: 206

Step-by-step solution →
Q44·MathematicsInteger
Let α=(1−2cos⁡(π11))(1−2cos⁡(3π11))(1−2cos⁡(9π11))(1−2cos⁡(27π11))(1−2cos⁡(81π11))\alpha = \left(1 - 2\cos\left(\frac{\pi}{11}\right)\right) \left(1 - 2\cos\left(\frac{3\pi}{11}\right)\right) \left(1 - 2\cos\left(\frac{9\pi}{11}\right)\right) \left(1 - 2\cos\left(\frac{27\pi}{11}\right)\right) \left(1 - 2\cos\left(\frac{81\pi}{11}\right)\right)α=(1−2cos(11π​))(1−2cos(113π​))(1−2cos(119π​))(1−2cos(1127π​))(1−2cos(1181π​)). Then the value of 5−α25 - \alpha^25−α2 is ______.

Correct answer: 4

Step-by-step solution →
Q45·MathematicsSingle correct
Match each entry in List-I to the correct entry in List-II and choose the correct option.
List-IList-II
P.If α\alphaα and β\betaβ are the distinct roots of the equation x2+x+1=0x^2 + x + 1 = 0x2+x+1=0, then the quadratic equation with roots 1(α+1)2026\frac{1}{(\alpha + 1)^{2026}}(α+1)20261​ and 1(β+1)2026\frac{1}{(\beta + 1)^{2026}}(β+1)20261​ is1.x2+x+1=0x^2 + x + 1 = 0x2+x+1=0
Q.If α\alphaα and β\betaβ are the distinct roots of the equation x2+x+1=0x^2 + x + 1 = 0x2+x+1=0, then the quadratic equation with roots 1(α+1)2027\frac{1}{(\alpha + 1)^{2027}}(α+1)20271​ and 1(β+1)2027\frac{1}{(\beta + 1)^{2027}}(β+1)20271​ is2.x2−x+1=0x^2 - x + 1 = 0x2−x+1=0
R.If γ\gammaγ and δ\deltaδ are the distinct roots of the equation x2−x+1=0x^2 - x + 1 = 0x2−x+1=0, then the value of 1(γ−1)2026+1(δ−1)2026\frac{1}{(\gamma - 1)^{2026}} + \frac{1}{(\delta - 1)^{2026}}(γ−1)20261​+(δ−1)20261​ is3.x2+x−1=0x^2 + x - 1 = 0x2+x−1=0
S.If ppp and rrr are the distinct roots of the equation x2+x−1=0x^2 + x - 1 = 0x2+x−1=0, then the value of 1(p+1)3+1(r+1)3\frac{1}{(p + 1)^3} + \frac{1}{(r + 1)^3}(p+1)31​+(r+1)31​ is4.−1-1−1
5.−4-4−4
  1. (A)(P) → (1), (Q) → (2), (R) → (5), (S) → (4)
  2. (B)(P) → (3), (Q) → (1), (R) → (4), (S) → (5)
  3. (C)(P) → (1), (Q) → (2), (R) → (4), (S) → (5)
  4. (D)(P) → (2), (Q) → (3), (R) → (5), (S) → (4)

Correct answer: (C)

Step-by-step solution →
Q46·MathematicsSingle correct
Match each entry in List-I to the correct entry in List-II and choose the correct option.
List-IList-II
P.The number of elements in the set {x∈[−π,π]:sin⁡6x+cos⁡4x=1}\{x \in [-\pi, \pi] : \sin^6 x + \cos^4 x = 1\}{x∈[−π,π]:sin6x+cos4x=1}1.is 1
Q.The number of elements in the set {x∈[−π2,π2]:sin⁡2x+cos⁡6x=1}\left\{x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] : \sin^2 x + \cos^6 x = 1\right\}{x∈[−2π​,2π​]:sin2x+cos6x=1}2.is 2
R.The number of elements in the set {x∈[−π,π]:cos⁡2(x2)−sin⁡2x=12}\left\{x \in [-\pi, \pi] : \cos^2\left(\frac{x}{2}\right) - \sin^2 x = \frac{1}{2}\right\}{x∈[−π,π]:cos2(2x​)−sin2x=21​}3.is 3
S.The number of elements in the set {x∈[−2π,2π]:6sin⁡2(x2)−cos⁡3x=3}\left\{x \in [-2\pi, 2\pi] : 6\sin^2\left(\frac{x}{2}\right) - \cos 3x = 3\right\}{x∈[−2π,2π]:6sin2(2x​)−cos3x=3}4.is 4
5.is 5
  1. (A)(P) → (2), (Q) → (5), (R) → (3), (S) → (4)
  2. (B)(P) → (5), (Q) → (3), (R) → (2), (S) → (4)
  3. (C)(P) → (5), (Q) → (4), (R) → (1), (S) → (3)
  4. (D)(P) → (4), (Q) → (3), (R) → (2), (S) → (5)

Correct answer: (B)

Step-by-step solution →
Q47·MathematicsSingle correct
For real numbers α,β,γ,δ\alpha, \beta, \gamma, \deltaα,β,γ,δ and μ\muμ, consider the matrix M=[α12−1213β13γδμ]M = \begin{bmatrix} \alpha & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{3}} & \beta & \frac{1}{\sqrt{3}} \\ \gamma & \delta & \mu \end{bmatrix}M=​α3​1​γ​2​1​βδ​−2​1​3​1​μ​​. Suppose that MMT=IMM^T = IMMT=I, where MTM^TMT is the transpose of the matrix MMM, and III is the 3×33 \times 33×3 identity matrix. Let u⃗=α i^+13 j^+γ k^,v⃗=12 i^+β j^+δ k^\vec{u} = \alpha\,\hat{i} + \frac{1}{\sqrt{3}}\,\hat{j} + \gamma\,\hat{k}, \vec{v} = \frac{1}{\sqrt{2}}\,\hat{i} + \beta\,\hat{j} + \delta\,\hat{k}u=αi^+3​1​j^​+γk^,v=2​1​i^+βj^​+δk^ and w⃗=−12 i^+13 j^+μ k^\vec{w} = -\frac{1}{\sqrt{2}}\,\hat{i} + \frac{1}{\sqrt{3}}\,\hat{j} + \mu\,\hat{k}w=−2​1​i^+3​1​j^​+μk^. Match each entry in List-I to the correct entry in List-II and choose the correct option.
List-IList-II
P.The value of γ2+δ2\gamma^2 + \delta^2γ2+δ2 is1.0
Q.If xu⃗+yv⃗+zw⃗=j^x\vec{u} + y\vec{v} + z\vec{w} = \hat{j}xu+yv+zw=j^​ for some real numbers x,yx, yx,y and zzz, then the value of xxx is2.1
R.The value of ∣u⃗⋅(v⃗×w⃗)∣|\vec{u} \cdot (\vec{v} \times \vec{w})|∣u⋅(v×w)∣ is3.12\frac{1}{\sqrt{2}}2​1​
S.The value of ∣u⃗×(v⃗×w⃗)∣|\vec{u} \times (\vec{v} \times \vec{w})|∣u×(v×w)∣ is4.13\frac{1}{\sqrt{3}}3​1​
5.56\frac{5}{6}65​
  1. (A)(P) → (5), (Q) → (4), (R) → (2), (S) → (1)
  2. (B)(P) → (4), (Q) → (5), (R) → (1), (S) → (2)
  3. (C)(P) → (5), (Q) → (3), (R) → (2), (S) → (1)
  4. (D)(P) → (5), (Q) → (4), (R) → (1), (S) → (2)

Correct answer: (A)

Step-by-step solution →
Q48·MathematicsSingle correct
Match each entry in List-I to the correct entry in List-II and choose the correct option.
List-IList-II
P.The circle with centre (1,2)(1, 2)(1,2) and touching the straight line 3x+4y=13x + 4y = 13x+4y=1, passes through1.the point (1,1)(1, 1)(1,1)
Q.The common tangent to the circle x2+y2=2x^2 + y^2 = 2x2+y2=2 and the parabola y2=8xy^2 = 8xy2=8x with positive slope, passes through2.the point (7,9)(7, 9)(7,9)
R.Let MMM be the end point of the latus rectum of the ellipse 3x2+4y2=483x^2 + 4y^2 = 483x2+4y2=48 such that MMM lies in the first quadrant. Then the normal to the ellipse drawn at MMM passes through3.the point (3,2)(3, 2)(3,2)
S.Let HHH be the hyperbola whose centre is at the origin, one of the foci is at (5,0)(5, 0)(5,0), and one directrix is 5x+16=05x + 16 = 05x+16=0. Then HHH passes through4.the point (2,5)(2, 5)(2,5)
5.the point (8,33)(8, 3\sqrt{3})(8,33​)
  1. (A)(P) → (3), (Q) → (4), (R) → (1), (S) → (2)
  2. (B)(P) → (3), (Q) → (2), (R) → (1), (S) → (5)
  3. (C)(P) → (3), (Q) → (2), (R) → (4), (S) → (5)
  4. (D)(P) → (4), (Q) → (1), (R) → (2), (S) → (3)

Correct answer: (B)

Step-by-step solution →

Chapters tested in this paper

  • Three Dimensional Geometry 176/186
  • Matrices and Determinants 180/186
  • Coordination Compounds 176/186
  • Sets, Relations and Functions 165/186
  • p-Block Elements 164/186
  • Rotational Motion 172/186
  • Geometrical Optics 172/186
  • Kinematics 156/186
  • Vector Algebra 173/186
  • Probability 176/186
  • Permutations and Combinations 162/186
  • Magnetic Field of Current 147/186
  • Electromagnetic Waves 144/186
  • Chemical Bonding and Molecular Structure 151/186
  • Application of Derivatives 139/186
  • Thermodynamics 154/186
  • Aldehydes and Ketones 135/186
  • Chemical Thermodynamics 165/186
  • Hydrocarbons 126/186
  • Trigonometric Functions 144/186
  • Chemical Kinetics 169/186
  • Atomic Structure 161/186
  • Circles 142/186
  • Amines 133/186
  • Quadratic Equations 148/186
  • Electromagnetic Induction 120/186
  • Wave Optics 130/186
  • Oscillations 117/186
  • Waves 109/186
  • Atoms 112/186
  • Differentiability 91/186
  • Inverse Trigonometric Functions 93/186
  • Carboxylic Acids and Derivatives 54/186
  • States of Matter: Gases and Liquids 52/186
  • Isomerism 51/186
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