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JEE Main 28 January 2026 Shift 2 Question Paper with Answers

28 January 2026 · January session · 75 questions

The complete JEE Main 28 January 2026 Shift 2 paper — all 75 questions with the correct answer for each, tagged to the chapter it tests. Free to read, no account needed.

Physics
25
Chemistry
25
Mathematics
25

Physics — JEE Main 28 January 2026 Shift 2

Q1·PhysicsSingle correct
A nucleus has mass number α and radius RαR_{\alpha}Rα​. Another nucleus has mass number β and radius RβR_{\beta}Rβ​. If β = 8α then Rα/RβR_{\alpha}/R_{\beta}Rα​/Rβ​ is :
  1. (A)2
  2. (B)8
  3. (C)1
  4. (D)0.5

Correct answer: (D)

Step-by-step solution →
Q2·PhysicsSingle correct
A plane electromagnetic wave is moving in free space with velocity c = 3×1083 \times 10^{8}3×108 m/s and its electric field is given as E⃗=54sin⁡(kz−ωt)j^\vec{E} = 54\sin(kz - \omega t)\hat{j}E=54sin(kz−ωt)j^​ V / m, where j^\hat{j}j^​ is the unit vector along y-axis. The magnetic field vector B⃗\vec{B}B of the wave is :
  1. (A)−1.8×10−7sin⁡(kz−ωt)i^-1.8 \times 10^{-7}\sin(kz - \omega t)\hat{i}−1.8×10−7sin(kz−ωt)i^ T
  2. (B)1.4×10−7sin⁡(kz−ωt)k^1.4 \times 10^{-7}\sin(kz - \omega t)\hat{k}1.4×10−7sin(kz−ωt)k^ T
  3. (C)1.4×10−7sin⁡(kz−ωt)i^1.4 \times 10^{-7}\sin(kz - \omega t)\hat{i}1.4×10−7sin(kz−ωt)i^ T
  4. (D)+1.8×10−7sin⁡(kz−ωt)i^+1.8 \times 10^{-7}\sin(kz - \omega t)\hat{i}+1.8×10−7sin(kz−ωt)i^ T

Correct answer: (A)

Step-by-step solution →
Q3·PhysicsSingle correct
A biconvex lens is formed by using two thin planoconvex lenses, as shown in the figure. The refractive index and radius of curved surfaces are also mentioned in figure. When an object is placed on the left side of lens at a distance of 30 cm from the biconvex lens, the magnification of the image will be :
  1. (A)–2
  2. (B)+2
  3. (C)+2.5
  4. (D)–2.5

Correct answer: (A)

Step-by-step solution →
Q4·PhysicsSingle correct
The mean free path of a molecule of diameter 5×10−105 \times 10^{-10}5×10−10 m at the temperature 41°C and pressure 1.38×1051.38 \times 10^{5}1.38×105 Pa, is given as ____m. (Given kB=1.38×10−23k_{B} = 1.38 \times 10^{-23}kB​=1.38×10−23 J/K).
  1. (A)22×10−102\sqrt{2} \times 10^{-10}22​×10−10
  2. (B)102×10−810\sqrt{2} \times 10^{-8}102​×10−8
  3. (C)22×10−82\sqrt{2} \times 10^{-8}22​×10−8
  4. (D)2×10−82 \times 10^{-8}2×10−8

Correct answer: (C)

Step-by-step solution →
Q5·PhysicsSingle correct
Two p-n junction diodes D1D_{1}D1​ and D2D_{2}D2​ are connected as shown in figure. A and B are input signals and C is the output. The given circuit will function as a ____.
  1. (A)OR Gate
  2. (B)NOR Gate
  3. (C)NAND Gate
  4. (D)AND Gate

Correct answer: (D)

Step-by-step solution →
Q6·PhysicsSingle correct
A wheatstone bridge is initially at room temperature and all arms of the bridge have same value of resistances (R1=R2=R3=R4R_{1} = R_{2} = R_{3} = R_{4}R1​=R2​=R3​=R4​). When R3R_{3}R3​ resistance is heated to some temperature, its resistance value has gone up by 10%. The potential difference (Va−VbV_{a} - V_{b}Va​−Vb​) (after R3R_{3}R3​ is heated) is _____V.
  1. (A)1.05
  2. (B)0
  3. (C)0.95
  4. (D)2

Correct answer: (C)

Step-by-step solution →
Q7·PhysicsSingle correct
In an experiment, a set of reading are obtained –1.24 mm, 1.25 mm, 1.23 mm, 1.21 mm. The expected least count of the instrument used in recording these readings is _____mm.
  1. (A)0.01
  2. (B)0.001
  3. (C)0.1
  4. (D)0.05

Correct answer: (A)

Step-by-step solution →
Q8·PhysicsSingle correct
A particle starts moving from time t = 0 and its coordinate is given as x(t)=4t3−3tx(t) = 4t^{3} - 3tx(t)=4t3−3t. A. The particle returns to its original position (origin) 0.866 units later B. The particle is 1 unit away from origin at its turning point. C. Acceleration of the particle is non-negative. D. The particle is 0.5 units away from origin at its turning point. E. Particle never turns back as acceleration is non-negative. Choose the correct answer from the options given below :
  1. (A)A,C,D only
  2. (B)A,B,C only
  3. (C)C,E only
  4. (D)A,C only

Correct answer: (B)

Step-by-step solution →
Q9·PhysicsSingle correct
The speed of a longitudinal wave in a metallic bar is 400 m/s. If the density and Young's modulus of the bar material are increased by 0.5% and 1% respectively then the speed of the wave is changed approximately to _____ m/s.
  1. (A)399
  2. (B)398
  3. (C)402
  4. (D)401

Correct answer: (D)

Step-by-step solution →
Q10·PhysicsSingle correct
Identify the correct statements : A. Effective capacitance of a series combination of capacitors is always smaller than the smallest capacitance of the capacitor in the combination. B. When a dielectric medium is placed between the charged plates of a capacitor, displacement of charges cannot occur due to insulation property of dielectric. C. Increasing of area of capacitor plate or decreasing of thickness of dielectric is an alternate method to increase the capacitance. D. For a point charge, concentric spherical shells centered at the location of the charge are equipotential surfaces. Choose the correct answer from the options given below.
  1. (A)A, B and C only
  2. (B)C and D only
  3. (C)A, C and D only
  4. (D)B and D only

Correct answer: (C)

Step-by-step solution →
Q11·PhysicsSingle correct
Number of photons of equal energy emitted per second by a 6 mW laser source operating at 663 nm is _____. (Given : h = 6.63×10−346.63 \times 10^{-34}6.63×10−34 J.s and c = 3×1083 \times 10^{8}3×108 m/s)
  1. (A)5×10165 \times 10^{16}5×1016
  2. (B)5×10155 \times 10^{15}5×1015
  3. (C)10×101510 \times 10^{15}10×1015
  4. (D)2×10162 \times 10^{16}2×1016

Correct answer: (D)

Step-by-step solution →
Q12·PhysicsSingle correct
When the position vector r⃗=xi^+yj^+zk^\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}r=xi^+yj^​+zk^ changes sign as −r⃗-\vec{r}−r, which one of the following vector will not flip under sign change ?
  1. (A)Linear momentum
  2. (B)Velocity
  3. (C)Acceleration
  4. (D)Angular momentum

Correct answer: (D)

Step-by-step solution →
Q13·PhysicsSingle correct
Which one of the following is not a measurable quantity ?
  1. (A)Voltage difference
  2. (B)Resistance
  3. (C)Voltage
  4. (D)Displacement current

Correct answer: (C)

Step-by-step solution →
Q14·Physics·Magnetic Field of CurrentSingle correct
A long cylindrical conductor with large cross section carries an electric current distributed uniformly over its cross-section. Magnetic field due to this current is : A. maximum at either ends of the conductor and minimum at the midpoint B. maximum at the axis of the conductor C. minimum at the surface of the conductor D. minimum at the axis of the conductor E. same at all points in the cross-section of the conductor Choose the correct answer from the options given below :
  1. (A)D Only
  2. (B)A, D Only
  3. (C)B, C Only
  4. (D)E Only

Correct answer: (A)

Step-by-step solution →
Q15·PhysicsSingle correct
A small block of mass m slides down from the top of a frictionless inclined surface, while the inclined plane is moving towards left with constant acceleration a0a_0a0​. The angle between the inclined plane and ground is θ and its base length is L. Assuming that initially the small block is at the top of the inclined plane, the time it takes to reach the lowest point of the inclined plane is _________.
  1. (A)2Lgsin⁡2θ−a0(1+cos⁡2θ)\sqrt{\dfrac{2L}{g\sin 2\theta - a_0(1 + \cos 2\theta)}}gsin2θ−a0​(1+cos2θ)2L​​
  2. (B)4Lgsin⁡2θ−a0(1+cos⁡2θ)\sqrt{\dfrac{4L}{g\sin 2\theta - a_0(1 + \cos 2\theta)}}gsin2θ−a0​(1+cos2θ)4L​​
  3. (C)4Lgcos⁡2θ−a0sin⁡θcos⁡θ\sqrt{\dfrac{4L}{g\cos^2 \theta - a_0 \sin\theta\cos\theta}}gcos2θ−a0​sinθcosθ4L​​
  4. (D)2Lgsin⁡θ−a0cos⁡θ\sqrt{\dfrac{2L}{g\sin \theta - a_0 \cos\theta}}gsinθ−a0​cosθ2L​​

Correct answer: (B)

Step-by-step solution →
Q16·PhysicsSingle correct
Identify the correct statements : A. Electrostatic field lines form closed loops. B. The electric field lines point radially outward when charge is greater than zero. C. The Gauss-Law is valid only for inverse-square force. D. The workdone in moving a charged particle in a static electric field around a closed path is zero. E. The motion of a particle under Coulomb's force must take place in a plane. Choose the correct answer from the options given below :
  1. (A)A, B, D, E Only
  2. (B)A, B, C, D Only
  3. (C)B, C, D, E Only
  4. (D)A, C, E Only

Correct answer: (C)

Step-by-step solution →
Q17·PhysicsSingle correct
As shown in the figure, a spring is kept in a stretched position with some extension by holding the masses 1 kg and 0.2 kg with a separation more than spring natural length and are released. Assuming the horizontal surface to be frictionless, the angular frequency (in SI unit) of the system is :
  1. (A)30
  2. (B)27
  3. (C)20
  4. (D)5

Correct answer: (A)

Step-by-step solution →
Q18·PhysicsSingle correct
For a transparent prism, if the angle of minimum deviation is equal to its refracting angle, the refractive index n of the prism satisfies.
  1. (A)2<n<22\sqrt{2} < n < 2\sqrt{2}2​<n<22​
  2. (B)1<n<21 < n < 21<n<2
  3. (C)n≥2n \geq 2n≥2
  4. (D)2<n<2\sqrt{2} < n < 22​<n<2

Correct answer: (D)

Step-by-step solution →
Q19·PhysicsSingle correct
The time period of a simple harmonic oscillator is T=2πkmT = 2\pi\sqrt{\dfrac{k}{m}}T=2πmk​​ . The measured value of mass (m) of the object is 10 g with an accuracy of 10 mg, and time for 50 oscillations of the spring is found to be 60 s using a watch of 2 s resolution. Percentage error in determination of spring constant(k) is _______%.
  1. (A)3.43
  2. (B)3.35
  3. (C)7.60
  4. (D)6.76

Correct answer: (D)

Step-by-step solution →
Q20·PhysicsSingle correct
Match List-I with List-II. Choose the correct answer from the options given below :
List-IList-II
A.Coefficient of viscosityI.[ML–1^{–1}–1T–2^{–2}–2]
B.Surface tensionII.[ML2^{2}2T–2^{–2}–2]
C.PressureIII.[ML0^{0}0T–2^{–2}–2]
D.Surface energyIV.[ML–1^{–1}–1T–1^{–1}–1]
  1. (A)A-I, B-II, C-IV, D-III
  2. (B)A-IV,B-III,C-I,D-II
  3. (C)A-I, B-III, C-II, D-IV
  4. (D)A-IV, B-I, C-II, D-III

Correct answer: (B)

Step-by-step solution →
Q21·PhysicsNumerical
Two tuning forks A and B are sounded together giving rise to 8 beats in 2 s. When fork A is loaded with wax, the beat frequency is reduced to 4 beats in 2 s. If the original frequency of tuning fork B is 380 Hz, then the original frequency of tuning fork A is _______ Hz.

Correct answer: 384

Step-by-step solution →
Q22·PhysicsNumerical
A thermodynamic system is taken through the cyclic process ABC as shown in the figure. The total work done by the system during the cycle ABC is _______ J.

Correct answer: 300

Step-by-step solution →
Q23·PhysicsNumerical
An inductor stores 16 J of magnetic field energy and dissipates 32 W of thermal energy due to its resistance when an a.c. current of 2 A (rms) and frequency 50 Hz flows through it. The ratio of inductive reactance to its resistance is _________. (π = 3.14)

Correct answer: 314

Step-by-step solution →
Q24·PhysicsNumerical
A beam of light consisting of wavelengths 650 nm and 550 nm illuminates the Young's double slits with separation of 2 mm such that the interference fringes are formed on a screen, placed at a distance of 1.2 m from the slits. The least distance of a point from the central maximum, where the bright fringes due to both the wavelengths coincide, is _______ ×10−5\times 10^{-5}×10−5 m.

Correct answer: 429

Step-by-step solution →
Q25·PhysicsNumerical
A fly wheel having mass 3 kg and radius 5 m is free to rotate about a horizontal axis. A string having negligible mass is wound around the wheel and the loose end of the string is connected to a 3 kg mass. The mass is kept at rest initially and released. Kinetic energy of the wheel when the mass descends by 3 m is _______ J. (g = 10 m/s²)

Correct answer: 30

Step-by-step solution →

Chemistry — JEE Main 28 January 2026 Shift 2

Q26·ChemistrySingle correct
Identify the correct statements : The presence of −NO2-NO_2−NO2​ group in benzene ring A. activates the ring towards electrophilic substitutions. B. deactivates the ring towards electrophilic substitutions. C. activates the ring towards nucleophilic substitutions. D. deactivates the ring towards nucleophilic substitutions.
  1. (A)B and D Only
  2. (B)C and A Only
  3. (C)A and D Only
  4. (D)B and C Only

Correct answer: (D)

Step-by-step solution →
Q27·ChemistrySingle correct
Given below are two statements : Statement I : The increasing order of boiling point of hydrogen halides is HCl < HBr < HI < HF. Statement II : The increasing order of melting point of hydrogen halides is HCl < HBr < HF < HI. In the light of the above statements, choose the correct answer from the options given below :
  1. (A)Both Statement I and Statement II are true
  2. (B)Statement I is true but Statement II is false
  3. (C)Both Statement I and Statement II are false
  4. (D)Statement I is false but Statement II is true

Correct answer: (A)

Step-by-step solution →
Q28·ChemistrySingle correct
Consider the elements N, P, O, S, Cl and F. The number of valence electrons present in the elements with most and least metallic character from the above list is respectively.
  1. (A)7 and 5
  2. (B)5 and 6
  3. (C)5 and 7
  4. (D)6 and 7

Correct answer: (C)

Step-by-step solution →
Q29·ChemistrySingle correct
Observe the following equilibrium in a 1 L flask. A(g)⇌B(g)A(g) \rightleftharpoons B(g)A(g)⇌B(g) At T(K), the equilibrium concentrations of A and B are 0.5 M and 0.375 M respectively. 0.1 moles of A is added into the flask and heated to T(K) to establish the equilibrium again. The new equilibrium concentrations (in M) of A and B are respectively.
  1. (A)0.367, 0.275
  2. (B)0.53, 0.4
  3. (C)0.742, 0.557
  4. (D)0.557, 0.418

Correct answer: (D)

Step-by-step solution →
Q30·ChemistrySingle correct
The plot of log⁡10K\log_{10}Klog10​K vs 1T\frac{1}{T}T1​ gives a straight line. The intercept and slope respectively are (where K is equilibrium constant).
  1. (A)2.303RΔH∘\frac{2.303R}{\Delta H^\circ}ΔH∘2.303R​ , 2.303RΔS∘\frac{2.303R}{\Delta S^\circ}ΔS∘2.303R​
  2. (B)ΔS∘2.303R\frac{\Delta S^\circ}{2.303R}2.303RΔS∘​ , −ΔH∘2.303R-\frac{\Delta H^\circ}{2.303R}−2.303RΔH∘​
  3. (C)−ΔS∘R2.303-\frac{\Delta S^\circ R}{2.303}−2.303ΔS∘R​ , ΔH∘R2.303\frac{\Delta H^\circ R}{2.303}2.303ΔH∘R​
  4. (D)−ΔH∘2.303R-\frac{\Delta H^\circ}{2.303R}−2.303RΔH∘​ , ΔS∘2.303R\frac{\Delta S^\circ}{2.303R}2.303RΔS∘​

Correct answer: (B)

Step-by-step solution →
Q31·ChemistrySingle correct
The reactions which produce alcohol as the product area : A. CH4+O2→ΔMo2O3CH_4 + O_2 \xrightarrow[\Delta]{Mo_2O_3}CH4​+O2​Mo2​O3​Δ​ B. 2CH3CH3+3O2→Δ(CH3COO)2 Mn2CH_3CH_3 + 3O_2 \xrightarrow[\Delta]{(CH_3COO)_2\,Mn}2CH3​CH3​+3O2​(CH3​COO)2​MnΔ​ C. (CH3)3CH→KMnO4(CH_3)_3CH \xrightarrow{KMnO_4}(CH3​)3​CHKMnO4​​ D. 2CH4+O2→Cu/523 K/100 atm.2CH_4 + O_2 \xrightarrow{Cu/523\,K/100\,atm.}2CH4​+O2​Cu/523K/100atm.​ E. CH3−CH=CH−CH3→KMnO4/H+CH_3-CH=CH-CH_3 \xrightarrow{KMnO_4/H^+}CH3​−CH=CH−CH3​KMnO4​/H+​ Choose the correct answer from the options given below :
  1. (A)A and D Only
  2. (B)A, C and E Only
  3. (C)C and D Only
  4. (D)B, D and E Only

Correct answer: (C)

Step-by-step solution →
Q32·ChemistrySingle correct
Consider the following statements about manganate and permanganate ions. Identify the correct statements : A. The geometry of both manganate and permanganate ions is tetrahedral. B. The oxidation states of Mn in manganate and permanganate are +7 and +6, respectively. C. Oxidation of Mn(II) salt by peroxodisulphate gives manganate ion as the final product. D. Manganate ion is paramagnetic and permanganate ions is diamagnetic. E. Acidified permanganate ion reduces oxalate, nitrite and iodide ions. Choose the correct answer from the options given below:
  1. (A)A, C and D Only
  2. (B)A, B and C Only
  3. (C)A, D and E Only
  4. (D)A and D Only

Correct answer: (D)

Step-by-step solution →
Q33·ChemistrySingle correct
Which of the following reaction is NOT correctly represented ?
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (A)

Step-by-step solution →
Q34·ChemistrySingle correct
The wavelength of photon 'A' is 400 nm. The frequency of photon 'B' is 1016s−110^{16}s^{-1}1016s−1. The wave number of photon 'C' is 10410^4104 cm−1cm^{-1}cm−1. The correct order of energy of these photons is :
  1. (A)C > B > A
  2. (B)B > A > C
  3. (C)A > B > C
  4. (D)A > C > B

Correct answer: (B)

Step-by-step solution →
Q35·Chemistry·Electronic Effects and StabilitySingle correct
The cyclic cations having the same number of hyperconjugation are : Choose the correct answer from the options given below :
  1. (A)A and C Only
  2. (B)B and C Only
  3. (C)A and B Only
  4. (D)A, C and D only

Correct answer: (A)

Step-by-step solution →
Q36·ChemistrySingle correct
Structures of four disaccharides are given below. Among the given disaccharides, the non-reducing sugar is :
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (A)

Step-by-step solution →
Q37·ChemistrySingle correct
Match List-I with List-II according to shape. Choose the correct answer from the options given below :
List-IList-II
A.XeO3_{3}3​I.BrF5_{5}5​
B.XeF2_{2}2​II.NH3_{3}3​
C.XeO2_{2}2​F2_{2}2​III.[I3_{3}3​]–^{–}–
D.XeOF4_{4}4​IV.SF4_{4}4​
  1. (A)A-II, B-I, C-III, D-IV
  2. (B)A-II, B-III, C-IV, D-I
  3. (C)A-II, B-III, C-I, D-IV
  4. (D)A-III, B-II, C-IV, D-I

Correct answer: (B)

Step-by-step solution →
Q38·ChemistrySingle correct
A student performed analysis of aliphatic organic compound 'X' which on analysis gave C = 61.01%, H=15.25%, N=23.74%. This compound, on treatment with HNO2/H2OHNO_2/H_2OHNO2​/H2​O produced another compound 'Y' which did not contain any nitrogen atom. However, the compound 'Y' upon controlled oxidation produced another compound 'Z' that responded to iodoform test. The structure of 'X' is:
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (C)

Step-by-step solution →
Q39·ChemistrySingle correct
Consider the following aqueous solutions. I. 2.2 g Glucose in 125 mL of solution. II. 1.9 g Calcium chloride in 250 mL of solution. III. 9.0 g Urea in 500 mL of solution. IV. 20.5 g Aluminium sulphate in 750 mL of solution. The correct increasing order of boiling point of these solutions will be: [Given: Molar mass in g mol−1mol^{-1}mol−1: H=1, C=12, N=14, O=16, Cl=35.5, Ca=40, Al=27 and S=32]
  1. (A)I < II < III < IV
  2. (B)III < I < II < IV
  3. (C)II < III < I < IV
  4. (D)II < III < IV < I

Correct answer: (A)

Step-by-step solution →
Q40·ChemistrySingle correct
The correct order of acidic strength of the major products formed in the given reactions, is : A. PhNH2→(1) NaNO2+HCl (<5∘C)  (2) CuCN  (3) H3O+/Δ[A]PhNH_2 \xrightarrow{(1)\ NaNO_2+HCl\,(<5^\circ C)\ \ (2)\ CuCN\ \ (3)\ H_3O^+/\Delta} [A]PhNH2​(1) NaNO2​+HCl(<5∘C)  (2) CuCN  (3) H3​O+/Δ​[A] B. CH3CH2CHO→Δ[Ag(NH3)2]+, OH−[B]CH_3CH_2CHO \xrightarrow[\Delta]{[Ag(NH_3)_2]^+,\ OH^-} [B]CH3​CH2​CHO[Ag(NH3​)2​]+, OH−Δ​[B] C. CH4+O2→(ii) Na2Cr2O7/H+(i) Mo2O3[C]CH_4 + O_2 \xrightarrow[(ii)\ Na_2Cr_2O_7/H^+]{(i)\ Mo_2O_3} [C]CH4​+O2​(i) Mo2​O3​(ii) Na2​Cr2​O7​/H+​[C] D. PhCH2MgBr+CO2→H3O+Dry ether[D]PhCH_2MgBr + CO_2 \xrightarrow[H_3O^+]{Dry\ ether} [D]PhCH2​MgBr+CO2​Dry etherH3​O+​[D] Choose the correct answer from the options given below :
  1. (A)C > B > A > D
  2. (B)A > D > C > B
  3. (C)A > D > B > C
  4. (D)C > A > D > B

Correct answer: (D)

Step-by-step solution →
Q41·ChemistrySingle correct
Total number of alkali insoluble solid sulphonamides obtained by reaction of given amines with Hinsberg's reagent is ……….. . Aniline, N-Methylaniline, Methanamine, N, N-Dimethylmethanamine, N-Methyl methanamine, Phenylmethanamine, N-propylaniline, N-phenylaniline, N, N-Dimethylaniline, Allyl amine, Isopropyl amine
  1. (A)4
  2. (B)2
  3. (C)8
  4. (D)5

Correct answer: (A)

Step-by-step solution →
Q42·ChemistrySingle correct
Consider the following reactions. Na2B4O7→Δ2X+YNa_2B_4O_7 \xrightarrow{\Delta} 2X + YNa2​B4​O7​Δ​2X+Y CuSO4+Y→Non−Luminous flameZ+SO3CuSO_4 + Y \xrightarrow{Non-Luminous\ flame} Z + SO_3CuSO4​+YNon−Luminous flame​Z+SO3​ 2Z+2X+Carbon→Luminous flame2Q+Na2B4O7+CO2Z+2X+Carbon \xrightarrow{Luminous\ flame} 2Q+Na_2B_4O_7+CO2Z+2X+CarbonLuminous flame​2Q+Na2​B4​O7​+CO The oxidation states of Cu in Z and Q, respectively are :
  1. (A)+2 and +2
  2. (B)+2 and +1
  3. (C)+1 and +2
  4. (D)+1 and +1

Correct answer: (B)

Step-by-step solution →
Q43·ChemistrySingle correct
For the given reaction; CaCO3+2HC1→CaCl2+H2O+CO2CaCO_3 + 2HC1 \rightarrow CaCl_2 + H_2O + CO_2CaCO3​+2HC1→CaCl2​+H2​O+CO2​ If 90 g CaCO3CaCO_3CaCO3​ is added to 300 mL of HCl which contains 38.55% HCl by mass and has density 1.13 g mL−1mL^{-1}mL−1, then which of the following option is correct? Given molar mass of H, Cl, Ca and O are 1, 35.5, 40 and 16 g mol−1mol^{-1}mol−1 respectively.
  1. (A)64.97 g of HCl remains unreacted
  2. (B)32.85 g of CaCO3CaCO_3CaCO3​ remains unreacted
  3. (C)97.30 g of HCl reacted
  4. (D)60.32 g of HCl reamains unreacted

Correct answer: (A)

Step-by-step solution →
Q44·ChemistrySingle correct
The correct increasing order of spin-only magnetic moment values of the complex ions [MnBr4]2−[MnBr_4]^{2-}[MnBr4​]2− (A), [Cu(H2O)6]2+[Cu(H_2O)_6]^{2+}[Cu(H2​O)6​]2+ (B), [Ni(CN4)]2−[Ni(CN_4)]^{2-}[Ni(CN4​)]2− (C) and [Ni(H2O)6]2+[Ni(H_2O)_6]^{2+}[Ni(H2​O)6​]2+ (D) is:
  1. (A)A = B < C < D
  2. (B)A = B < D < C
  3. (C)C = D < B < A
  4. (D)C < B < D < A

Correct answer: (D)

Step-by-step solution →
Q45·ChemistrySingle correct
A student has been given 0.314 g of an organic compound and asked to estimate Sulphur. During the experiment, the student has obtained 0.4813 g of barium sulphate. The percentage of sulphur present in the compound is _________ . (Given Molar mass in g mol−1mol^{-1}mol−1 S:32, BaSO4BaSO_4BaSO4​ : 233)
  1. (A)42.10%
  2. (B)63.15%
  3. (C)21.05%
  4. (D)48.24%

Correct answer: (C)

Step-by-step solution →
Q46·ChemistryNumerical
Two positively charged particles m1m_1m1​ and m2m_2m2​ have been accelerated across the same potential difference of 200 keV as shown below. [Given mass of m1m_1m1​ = 1 amu and m2m_2m2​ = 4 amu] The deBroglie wavelength of m1m_1m1​ will be x times of m2m_2m2​. The value of x is ________. (nearest integer)

Correct answer: 2

Step-by-step solution →
Q47·ChemistryNumerical
A→BA \rightarrow BA→B (first reaction) C→DC \rightarrow DC→D (second reaction) Consider the above two first-order reactions. The rate constant for first reaction at 500 K is double of the same at 300 K. At 500 K, 50% of the reaction becomes complete in 2 hour. The activation energy of the second reaction is half of that of first reaction. If the rate constant at 500 K of the second reaction becomes double of the rate constant of first reaction at the same temperature; then rate constant for the second reaction at 300 K is ________ ×10−1\times 10^{-1}×10−1 hour−1hour^{-1}hour−1 (nearest integer).

Correct answer: 5

Step-by-step solution →
Q48·ChemistryNumerical
For strong electrolyte Λm\Lambda_mΛm​ increases slowly with dilution and can be represented by the equation Λm=Λm∘−Ac1/2\Lambda_m = \Lambda^\circ_m - Ac^{1/2}Λm​=Λm∘​−Ac1/2 Molar conductivity values of the solutions of strong electrolyte AB at 18°C are given below : c [mol L−1L^{-1}L−1]: 0.04 | 0.09 | 0.16 | 0.25 Λm\Lambda_mΛm​ [S cm2cm^2cm2 mol−1mol^{-1}mol−1]: 96.1 | 95.7 | 95.3 | 94.9 The value of constant A based on the above data [in S cm2cm^2cm2 mol−1mol^{-1}mol−1/(mol/L)1/2(mol/L)^{1/2}(mol/L)1/2] unit is ________.

Correct answer: 4

Step-by-step solution →
Q49·ChemistryNumerical
A volume of x mL of 5 M NaHCO3NaHCO_3NaHCO3​ solution was mixed with 10 mL of 2 M H2CO3H_2CO_3H2​CO3​ solution to make an electrolytic buffer. If the same buffer was used in the following electrochemical cell to record a cell potential of 235.3 mV, then the value of x = __________ mL (nearest integer). Sn(s)∣Sn(OH)62−Sn(s) | Sn(OH)_6^{2-}Sn(s)∣Sn(OH)62−​ (0.5 M) ∣HSnO2−| HSnO_2^-∣HSnO2−​ (0.05 M) ∣OH−∣Bi2O3(s)∣Bi(s)| OH^- | Bi_2O_3(s) | Bi(s)∣OH−∣Bi2​O3​(s)∣Bi(s) Consider upto one place of decimal for intermediate calculations [Given : EHSnO2−∣Sn(OH)62−∘=−0.9 VE^\circ_{HSnO_2^-|Sn(OH)_6^{2-}} = -0.9\,VEHSnO2−​∣Sn(OH)62−​∘​=−0.9V EBi2O3∣Bi∘=−0.44VE^\circ_{Bi_2O_3|Bi} = -0.44VEBi2​O3​∣Bi∘​=−0.44V pKa(H2CO3)=6.11pKa_{(H_2CO_3)} = 6.11pKa(H2​CO3​)​=6.11 2.303RTF=0.059V\frac{2.303RT}{F} = 0.059VF2.303RT​=0.059V Antilog⁡(1.29)=19.5Anti\log(1.29) = 19.5Antilog(1.29)=19.5 ]

Correct answer: 78

Step-by-step solution →
Q50·ChemistryNumerical
The number of isoelectronic species among Sc3+Sc^{3+}Sc3+, Cr2+Cr^{2+}Cr2+, Mn3+Mn^{3+}Mn3+, Co3+Co^{3+}Co3+ and Fe3+Fe^{3+}Fe3+ is 'n'. If 'n' moles of AgCl is formed during the reaction of complex with formula CoCl3(en)2NH3CoCl_3(en)_2NH_3CoCl3​(en)2​NH3​ with excess of AgNO3AgNO_3AgNO3​ solution, then the number of electrons present in the t2gt_{2g}t2g​ orbital of the complex is ________.

Correct answer: 6

Step-by-step solution →

Mathematics — JEE Main 28 January 2026 Shift 2

Q51·MathematicsSingle correct
Given below two statements : Statement I : 2513+2013+813+31325^{13} + 20^{13} + 8^{13} + 3^{13}2513+2013+813+313 is divisible by 7. Statement II : The integral part of (7+43)25\left(7+4\sqrt{3}\right)^{25}(7+43​)25 is an odd number. In the light of the above statements, choose the correct answer from the options given below :
  1. (A)Both Statement I and Statement II are false.
  2. (B)Both Statement I and Statement II are true.
  3. (C)Statement I is false but Statement II is true.
  4. (D)Statement I is true but Statement II is false.

Correct answer: (B)

Step-by-step solution →
Q52·MathematicsSingle correct
The sum of the coefficients of x499x^{499}x499 and x500x^{500}x500 in (1+x)1000+x(1+x)999+x2(1+x)998+……+x1000(1 + x)^{1000} + x(1 + x)^{999} + x^{2}(1 + x)^{998} + \ldots\ldots + x^{1000}(1+x)1000+x(1+x)999+x2(1+x)998+……+x1000 is
  1. (A)1001C501^{1001}C_{501}1001C501​
  2. (B)1002C500^{1002}C_{500}1002C500​
  3. (C)1002C501^{1002}C_{501}1002C501​
  4. (D)1000C501^{1000}C_{501}1000C501​

Correct answer: (B)

Step-by-step solution →
Q53·MathematicsSingle correct
Let A be the focus of the parabola y2=8xy^{2} = 8xy2=8x. Let the line y=mx+cy = mx + cy=mx+c intersect the parabola at two distinct points B and C. If the centroid of the triangle ABC is (73,43)\left(\dfrac{7}{3}, \dfrac{4}{3}\right)(37​,34​), then (BC)2(BC)^{2}(BC)2 is equal to :
  1. (A)41
  2. (B)80
  3. (C)89
  4. (D)32

Correct answer: (B)

Step-by-step solution →
Q54·MathematicsSingle correct
The probability distribution of a random variable X is given below : X4k307k327k347k367k387k407k6kP(X)2151152151511521515115\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline X & 4k & \frac{30}{7}k & \frac{32}{7}k & \frac{34}{7}k & \frac{36}{7}k & \frac{38}{7}k & \frac{40}{7}k & 6k \\ \hline P(X) & \frac{2}{15} & \frac{1}{15} & \frac{2}{15} & \frac{1}{5} & \frac{1}{15} & \frac{2}{15} & \frac{1}{5} & \frac{1}{15} \\ \hline \end{array}XP(X)​4k152​​730​k151​​732​k152​​734​k51​​736​k151​​738​k152​​740​k51​​6k151​​​ If E(X)=26315E(X) = \dfrac{263}{15}E(X)=15263​, then P(X<20)P(X < 20)P(X<20) is equal to :
  1. (A)35\dfrac{3}{5}53​
  2. (B)815\dfrac{8}{15}158​
  3. (C)1115\dfrac{11}{15}1511​
  4. (D)1415\dfrac{14}{15}1514​

Correct answer: (C)

Step-by-step solution →
Q55·Mathematics·Limits and ContinuitySingle correct
Let f(x)=lim⁡θ→0(cos⁡πx−x(2θ)sin⁡(x−1)1+x(2θ)(x−1)),x∈Rf(x) = \lim\limits_{\theta \to 0}\left(\dfrac{\cos \pi x - x^{\left(\frac{2}{\theta}\right)}\sin(x-1)}{1 + x^{\left(\frac{2}{\theta}\right)}(x-1)}\right), x \in Rf(x)=θ→0lim​(1+x(θ2​)(x−1)cosπx−x(θ2​)sin(x−1)​),x∈R. Consider the following two statements : (I) f(x) is discontinous at x = 1. (II) f(x) is continous at x = - 1. Then,
  1. (A)Neither (I) nor (II) is True
  2. (B)Both (I) and (II) are True
  3. (C)Only (II) is True
  4. (D)Only (I) is True

Correct answer: (A)

Step-by-step solution →
Q56·MathematicsSingle correct
Considering the principal values of inverse trigonometric functions, the value of the expression tan⁡(2sin⁡−1(213)−2cos⁡−1(310))\tan\left(2\sin^{-1}\left(\dfrac{2}{\sqrt{13}}\right) - 2\cos^{-1}\left(\dfrac{3}{\sqrt{10}}\right)\right)tan(2sin−1(13​2​)−2cos−1(10​3​)) is equal to :
  1. (A)−3356-\dfrac{33}{56}−5633​
  2. (B)3356\dfrac{33}{56}5633​
  3. (C)1663\dfrac{16}{63}6316​
  4. (D)−1663-\dfrac{16}{63}−6316​

Correct answer: (B)

Step-by-step solution →
Q57·MathematicsSingle correct
Let the arithmetic mean of 1a\dfrac{1}{a}a1​ and 1b\dfrac{1}{b}b1​ be 516\dfrac{5}{16}165​, a > 2. If α is such that a, 4, α, b are in A.P., then the equation αx2−ax+2(α−2b)=0\alpha x^{2} - ax + 2(\alpha - 2b) = 0αx2−ax+2(α−2b)=0 has :
  1. (A)One root in (1,4) and another in (–2,0)
  2. (B)One root in (0,2) and another in (– 4, –2)
  3. (C)Complex roots of magnitude less than 2
  4. (D)Both roots in the interval (–2, 0)

Correct answer: (A)

Step-by-step solution →
Q58·MathematicsSingle correct
Given below are two statements : Statement I : The function f : R → R defined by f(x)=x1+∣x∣f(x) = \dfrac{x}{1+|x|}f(x)=1+∣x∣x​ is one-one. Statement II : The function f : R → R defined by f(x)=x2+4x−30x2−8x+18f(x) = \dfrac{x^{2}+4x-30}{x^{2}-8x+18}f(x)=x2−8x+18x2+4x−30​ is many-one. In the light of the above statements, choose the correct answer from the options given below :
  1. (A)Both Statement I and Statement II are false.
  2. (B)Both Statement I and Statement II are true.
  3. (C)Statement I is false but Statement II is true .
  4. (D)Statement I is true but Statement II is false.

Correct answer: (B)

Step-by-step solution →
Q59·MathematicsSingle correct
An ellipse has its center at (1,-2), one focus at (3,-2) and one vertex at (5, - 2). Then the length of its latus rectum is :
  1. (A)163\dfrac{16}{\sqrt{3}}3​16​
  2. (B)6
  3. (C)434\sqrt{3}43​
  4. (D)636\sqrt{3}63​

Correct answer: (B)

Step-by-step solution →
Q60·MathematicsSingle correct
Let the ellipse E:x2144+y2169=1E : \dfrac{x^{2}}{144} + \dfrac{y^{2}}{169} = 1E:144x2​+169y2​=1 and the hyperbola H:x216−y2λ2=−1H : \dfrac{x^{2}}{16} - \dfrac{y^{2}}{\lambda^{2}} = -1H:16x2​−λ2y2​=−1 have the same foci. If e and L respectively denote the eccentricity and the length of the latus rectum of H, then the value of 24(e + L) is :
  1. (A)296
  2. (B)126
  3. (C)148
  4. (D)67

Correct answer: (A)

Step-by-step solution →
Q61·MathematicsSingle correct
Let P1:y=4x2P_{1} : y = 4x^{2}P1​:y=4x2 and P2:y=x2+27P_{2} : y = x^{2} + 27P2​:y=x2+27 be two parabolas. If the area of the bounded region enclosed between P1P_{1}P1​ and P2P_{2}P2​ is six times the area of the bounded region enclosed between the line y=αxy = \alpha xy=αx, α > 0 and P1P_{1}P1​, then α is equal to :
  1. (A)8
  2. (B)15
  3. (C)12
  4. (D)6

Correct answer: (C)

Step-by-step solution →
Q62·MathematicsSingle correct
Let the circle x2+y2=4x^{2} + y^{2} = 4x2+y2=4 intersect x-axis at the points A(a, 0), a > 0 and B(b,0). Let P(2 cosα, 2 sinα), 0<α<π20 < \alpha < \dfrac{\pi}{2}0<α<2π​ and Q(2 cosβ, 2 sinβ) be two points such that (α−β)=π2(\alpha - \beta) = \dfrac{\pi}{2}(α−β)=2π​. Then the point of intersection of AQ and BP lies on :
  1. (A)x2+y2−4y−4=0x^{2} + y^{2} - 4y - 4 = 0x2+y2−4y−4=0
  2. (B)x2+y2−4x−4=0x^{2} + y^{2} - 4x - 4 = 0x2+y2−4x−4=0
  3. (C)x2+y2−4x−4y=0x^{2} + y^{2} - 4x - 4y = 0x2+y2−4x−4y=0
  4. (D)x2+y2−4x−4y−4=0x^{2} + y^{2} - 4x - 4y - 4 = 0x2+y2−4x−4y−4=0

Correct answer: (A)

Step-by-step solution →
Q63·MathematicsSingle correct
Let [·] denote the greatest integer function. Then ∫−π2π2(12(3+[x])3+[sin⁡x]+[cos⁡x])dx\displaystyle\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\dfrac{12(3+[x])}{3+[\sin x]+[\cos x]}\right) dx−2π​∫2π​​(3+[sinx]+[cosx]12(3+[x])​)dx is equal to:
  1. (A)15π+415\pi + 415π+4
  2. (B)11π+211\pi + 211π+2
  3. (C)13π+113\pi + 113π+1
  4. (D)12π+512\pi + 512π+5

Correct answer: (B)

Step-by-step solution →
Q64·MathematicsSingle correct
Let y = y(x) be the solution of the differential equation xdydx−y=x2cot⁡x,x∈(0,π)x\frac{dy}{dx} - y = x^2 \cot x, x \in (0, \pi)xdxdy​−y=x2cotx,x∈(0,π). If y(π2)=π2y\left(\frac{\pi}{2}\right) = \frac{\pi}{2}y(2π​)=2π​, then 6y(π6)−8y(π4)6y\left(\frac{\pi}{6}\right) - 8y\left(\frac{\pi}{4}\right)6y(6π​)−8y(4π​) is equal to :
  1. (A)3π3\pi3π
  2. (B)- 3π3\pi3π
  3. (C)- π\piπ
  4. (D)π\piπ

Correct answer: (C)

Step-by-step solution →
Q65·MathematicsSingle correct
The sum of all the elements in the range of f(x)=Sgn(sinx)+Sgn(cosx)+Sgn(tanx)+Sgn(cotx), x≠nπ2x \neq \frac{n\pi}{2}x=2nπ​, n ∈\in∈ Z\mathbf{Z}Z, where Sgn(t) = {1,ift>0−1ift<0\begin{cases} 1, & \text{if} \quad t > 0 \\ -1 & \text{if} \quad t < 0 \end{cases}{1,−1​ift>0ift<0​, is
  1. (A)4
  2. (B)2
  3. (C)–2
  4. (D)0

Correct answer: (B)

Step-by-step solution →
Q66·MathematicsSingle correct
Let Q(a, b, c) be the image of the point P(3, 2, 1) in the line x−11=y2=z−11\frac{x-1}{1} = \frac{y}{2} = \frac{z-1}{1}1x−1​=2y​=1z−1​. Then the distance of Q from the line x−93=y−92=z−5−2\frac{x-9}{3} = \frac{y-9}{2} = \frac{z-5}{-2}3x−9​=2y−9​=−2z−5​ is
  1. (A)6
  2. (B)8
  3. (C)7
  4. (D)5

Correct answer: (C)

Step-by-step solution →
Q67·MathematicsSingle correct
Let P be a point in the plane of the vector AB→=3i^+j^−k^\overrightarrow{AB} = 3\hat{i} + \hat{j} - \hat{k}AB=3i^+j^​−k^ and AC→=i^−j^+3k^\overrightarrow{AC} = \hat{i} - \hat{j} + 3\hat{k}AC=i^−j^​+3k^ such that P is equidistant from the lines AB and AC. If ∣AP→∣=52\left|\overrightarrow{AP}\right| = \frac{\sqrt{5}}{2}​AP​=25​​, then the area of the triangle ABP is :
  1. (A)2
  2. (B)32\frac{3}{2}23​
  3. (C)304\frac{\sqrt{30}}{4}430​​
  4. (D)264\frac{\sqrt{26}}{4}426​​

Correct answer: (C)

Step-by-step solution →
Q68·MathematicsSingle correct
Let A={z∈C:∣z−2∣≤4}A = \left\{z \in \mathbb{C} : |z - 2| \leq 4\right\}A={z∈C:∣z−2∣≤4} and B={z∈C:∣z−2∣+∣z+2∣=5}B = \left\{z \in \mathbb{C} : |z - 2| + |z + 2| = 5\right\}B={z∈C:∣z−2∣+∣z+2∣=5}. Then the max {∣z1−z2∣:z1∈A and z2∈B}\left\{|z_1 - z_2| : z_1 \in A \text{ and } z_2 \in B\right\}{∣z1​−z2​∣:z1​∈A and z2​∈B} is
  1. (A)152\frac{15}{2}215​
  2. (B)8
  3. (C)172\frac{17}{2}217​
  4. (D)9

Correct answer: (C)

Step-by-step solution →
Q69·MathematicsSingle correct
Let f(x)=∫dxx(23)+2x(12)f(x) = \int \frac{dx}{x^{\left(\frac{2}{3}\right)} + 2x^{\left(\frac{1}{2}\right)}}f(x)=∫x(32​)+2x(21​)dx​ be such that f(0)=−26+24log⁡e(2)f(0) = -26 + 24 \log_e(2)f(0)=−26+24loge​(2). If f(1)=a+blog⁡e(3)f(1) = a + b \log_e(3)f(1)=a+bloge​(3), where a, b ∈\in∈ Z\mathbf{Z}Z, then a + b is equal to:
  1. (A)–18
  2. (B)–5
  3. (C)–11
  4. (D)–26

Correct answer: (C)

Step-by-step solution →
Q70·MathematicsSingle correct
6326+10.1325+10.2324+10.22323+...+10.2243\frac{6}{3^{26}} + \frac{10.1}{3^{25}} + \frac{10.2}{3^{24}} + \frac{10.2^2}{3^{23}} + ...+ \frac{10.2^{24}}{3}3266​+32510.1​+32410.2​+32310.22​+...+310.224​ is equal to
  1. (A)2252^{25}225
  2. (B)2262^{26}226
  3. (C)3253^{25}325
  4. (D)3263^{26}326

Correct answer: (B)

Step-by-step solution →
Q71·MathematicsNumerical
If ∑r=125(rr4+r2+1)=pq\sum_{r=1}^{25}\left(\frac{r}{r^4 + r^2 + 1}\right) = \frac{p}{q}∑r=125​(r4+r2+1r​)=qp​, where p and q are positive integers such that gcd (p, q) = 1, then p + q is equal to ______.

Correct answer: 976

Step-by-step solution →
Q72·MathematicsNumerical
Three persons enter in a lift at the ground floor. The lift will go upto 10th10^{th}10th floor. The number of ways, in which the three persons can exit the lift at three different floors, if the lift does not stop at first, second and third floors, is equal to _______ .

Correct answer: 210

Step-by-step solution →
Q73·MathematicsNumerical
Let f be a differentiable function satisfying f(x)=1−2x+∫0xe(x−t) f(t) dtf(x) = 1 - 2x + \int_0^x e^{(x-t)}\, f(t)\, dtf(x)=1−2x+∫0x​e(x−t)f(t)dt, x ∈\in∈ R\mathbf{R}R and let g(x)=∫0x(f(t)+2)15 (t−4)6 (t+12)17 dtg(x) = \int_0^x (f(t) + 2)^{15}\, (t - 4)^6\, (t + 12)^{17}\, dtg(x)=∫0x​(f(t)+2)15(t−4)6(t+12)17dt, x ∈\in∈ R\mathbf{R}R. If p and q are respectively the points of local minima and local maxima of g, then the value of ∣p+q∣|p + q|∣p+q∣ is equal to _________ .

Correct answer: 9

Step-by-step solution →
Q74·MathematicsNumerical
If the distance of the point P(43, α\alphaα, β\betaβ), β<0\beta < 0β<0, from the line r⃗=4i^−k^+μ(2i^+3k^)\vec{r} = 4\hat{i} - \hat{k} + \mu(2\hat{i} + 3\hat{k})r=4i^−k^+μ(2i^+3k^), μ∈R\mu \in \mathbf{R}μ∈R along a line with direction ratios 3, –1, 0 is 131013\sqrt{10}1310​, then α2+β2\alpha^2 + \beta^2α2+β2 is equal to ________.

Correct answer: 170

Step-by-step solution →
Q75·MathematicsNumerical
Let A = [3−41−1]\begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix}[31​−4−1​] and B be two matrices such that A100=100B+IA^{100} = 100B + IA100=100B+I. Then the sum of all the elements of B100B^{100}B100 is ______.

Correct answer: 0

Step-by-step solution →

Chapters tested in this paper

Open any chapter to practise every past-year question on that topic across all papers, and see how often it has actually appeared.

  • Three Dimensional Geometry 176/186
  • Matrices and Determinants 180/186
  • Coordination Compounds 176/186
  • Sets, Relations and Functions 165/186
  • Current Electricity 160/186
  • Sequence and Series 164/186
  • p-Block Elements 164/186
  • Definite Integration 168/186
  • Rotational Motion 172/186
  • Redox Reactions and Electrochemistry 177/186
  • Geometrical Optics 172/186
  • Kinematics 156/186
  • Vector Algebra 173/186
  • Differential Equations 167/186
  • Probability 176/186
  • Permutations and Combinations 162/186
  • Magnetic Field of Current 147/186
  • Binomial Theorem and Its Simple Applications 158/186
  • Electromagnetic Waves 144/186
  • Application of Derivatives 139/186
  • Limits and Continuity 149/186
  • d- and f-Block Elements 126/186
  • Thermodynamics 154/186
  • Equilibrium 163/186
  • Solutions 158/186
  • Laws of Motion 130/186
  • Chemical Thermodynamics 165/186
  • Units and Measurements 149/186
  • Hydrocarbons 126/186
  • Complex Numbers 165/186
  • Biomolecules 162/186
  • Chemical Kinetics 169/186
  • Electric Field and Coulomb's Law 133/186
  • Atomic Structure 161/186
  • Electronic Devices 145/186
  • Circles 142/186
  • Dual Nature of Matter and Radiation 155/186
  • Amines 133/186
  • Quadratic Equations 148/186
  • Some Basic Concepts in Chemistry 129/186
  • Wave Optics 130/186
  • Area Under Curves 139/186
  • Kinetic Theory of Gases 135/186
  • Alcohols and Ethers 106/186
  • Purification and Characterisation of Organic Compounds 116/186
  • Oscillations 117/186
  • Alternating Currents 108/186
  • Nuclei 116/186
  • Organic Compounds Containing Halogens 109/186
  • Waves 109/186
  • Capacitors and Dielectrics 115/186
  • Classification of Elements and Periodicity in Properties 113/186
  • Parabola 101/186
  • Ellipse 103/186
  • Inverse Trigonometric Functions 93/186
  • Electronic Effects and Stability 74/186
  • Hyperbola 77/186
  • Experimental Skills 68/186
  • Electric Potential 63/186
  • Indefinite Integration 66/186
  • Carboxylic Acids and Derivatives 54/186
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