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JEE Main 22 January 2026 Shift 2 Question Paper with Answers

22 January 2026 · January session · 72 questions

72 of the 75 questions from the JEE Main 22 January 2026 Shift 2 paper, each with its correct answer and tagged to the chapter it tests. Free to read, no account needed.

3 questions are held back from this paper while we re-check the transcription or the answer key. We would rather show you nothing than show you an answer we are not confident is right.

Physics
23
Chemistry
25
Mathematics
24

Physics — JEE Main 22 January 2026 Shift 2

Q1·PhysicsSingle correct
If ∈, E and t represent the free space permittivity, electric field and time respectively, then the unit of ∈Et\frac{\in E}{t}t∈E​ will be :
  1. (A)Am
  2. (B)Am2^{2}2
  3. (C)A/m2^{2}2
  4. (D)A/m

Correct answer: (C)

Step-by-step solution →
Q2·PhysicsSingle correct
Using a simple pendulum experiment g is determined by measuring its time period T. Which of the following plots represent the correct relation between the pendulum length L and time period T ?
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (B)

Step-by-step solution →
Q3·PhysicsSingle correct
A uniform bar of length 12 cm and mass 20 m lies on a smooth horizontal table. Two point masses m and 2 m are moving in opposite directions with same speed of v and in the same plane as the bar, as shown in figure. These masses strike the bar simultaneously and get stuck to it. After collision the entire system is rotating with angular frequency ω. The ratio of v and ω is :
  1. (A)33
  2. (B)2882\sqrt{88}288​
  3. (C)66
  4. (D)32

Correct answer: (A)

Step-by-step solution →
Q4·PhysicsSingle correct
Three small identical bubbles of water having same charge on each coalesce to form a bigger bubble. Then the ratio of the potentials on one initial bubble and that on the resultant bigger bubble is :
  1. (A)1 : 31/3^{1/3}1/3
  2. (B)1 : 22/3^{2/3}2/3
  3. (C)32/3^{2/3}2/3 : 1
  4. (D)1 : 32/3^{2/3}2/3

Correct answer: (D)

Step-by-step solution →
Q5·PhysicsSingle correct
In parallax method for the determination of focal length of a concave mirror, the object should always be placed :
  1. (A)between the focus (F) and the centre of curvature (C) of the mirror ONLY
  2. (B)at any point beyond the focus (F) of the mirror
  3. (C)beyond the centre of the curvature (C) of the mirror ONLY
  4. (D)between the pole (P) and the focus (F) of the concave mirror ONLY

Correct answer: (B)

Step-by-step solution →
Q6·PhysicsSingle correct
The smallest wavelength of Lyman series is 91 nm. The difference between the largest wavelengths of Paschen and Balmer series is nearly ____ nm.
  1. (A)1875
  2. (B)1550
  3. (C)1217
  4. (D)1784

Correct answer: (C)

Step-by-step solution →
Q7·PhysicsSingle correct
In an open organ pipe ν3\nu_{3}ν3​ and ν6\nu_{6}ν6​ are 3rd3^{\text{rd}}3rd and 6th6^{\text{th}}6th harmonic frequencies, respectively. If ν6−ν3\nu_{6} - \nu_{3}ν6​−ν3​ = 2200 Hz then length of the pipe is ______ mm. (Take velocity of sound in air is 330 m/s.)
  1. (A)275
  2. (B)225
  3. (C)200
  4. (D)250

Correct answer: (B)

Step-by-step solution →
Q8·PhysicsSingle correct
When a part of a straight capillary tube is placed vertically in a liquid, the liquid raises upto certain height h. If the inner radius of the capillary tube, density of the liquid and surface tension of the liquid decrease by 1 % each, then the height of the liquid in the tube will change by _____%.
  1. (A)–1
  2. (B)+3
  3. (C)–3
  4. (D)+1

Correct answer: (D)

Step-by-step solution →
Q9·PhysicsSingle correct
The correct truth table for the given input data of the following logic gate is :
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (B)

Step-by-step solution →
Q10·PhysicsSingle correct
An electric power line having total resistance of 2 Ω, delivers 1 kW of power of 250 V. The percentage efficiency of transmission line is _____ .
  1. (A)96.9
  2. (B)86.5
  3. (C)100
  4. (D)92.5

Correct answer: (A)

Step-by-step solution →
Q11·PhysicsSingle correct
The wavelength of light, while it is passing through water is 540 nm. The refractive index of water is 43\frac{4}{3}34​. The wavelength of the same light when it is passing through a transparent medium having refractive index of 32\frac{3}{2}23​ is _________nm.
  1. (A)380
  2. (B)840
  3. (C)480
  4. (D)540

Correct answer: (C)

Step-by-step solution →
Q12·PhysicsSingle correct
Figure shows the circuit that contains three resistances (9 Ω each) and two inductors (4 mH each). The reading of ammeter at the moment switch K is turned ON, is _________A.
  1. (A)1
  2. (B)zero
  3. (C)3
  4. (D)2

Correct answer: (A)

Step-by-step solution →
Q13·PhysicsSingle correct
Given below are two statements : Statement I : A satellite is moving around earth in the orbit very close to the earth surface. The time period of revolution of satellite depends upon the density of earth. Statement II : The time period of revolution of the satellite is T=2πRegT = 2\pi\sqrt{\dfrac{R_e}{g}}T=2πgRe​​​ (for satellite very close to the earth surface), where ReR_eRe​ radius of earth and ggg acceleration due to gravity. 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 true but Statement II is false
  4. (D)Statement I is false but Statement II is true

Correct answer: (B)

Step-by-step solution →
Q14·PhysicsSingle correct
Given below are two statements : Statement I : An object moves from position r1r_1r1​ to position r2r_2r2​ under a conservative force field F⃗\vec{F}F. The work done by the force is W=−∫r1r2F⃗⋅dr⃗W = -\int_{r_1}^{r_2} \vec{F} \cdot \vec{dr}W=−∫r1​r2​​F⋅dr. Statement II : Any object moving from one location to another location can follow infinite number of paths. Therefore, the amount of work done by the object changes with the path it follows for a conservative force. 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 false but Statement II is true
  3. (C)Statement I is true but Statement II is false
  4. (D)Both Statement I and Statement II are false

Correct answer: (D)

Step-by-step solution →
Q15·PhysicsSingle correct
Five positive charges each having charge qqq are placed at the vertices of a pentagon as shown in the figure. The electric potential (V)(V)(V) and the electric field (E⃗)(\vec{E})(E) at the center O of the pentagon due to these five positive charges are :
  1. (A)V=5q4πε0rV = \dfrac{5q}{4\pi\varepsilon_0 r}V=4πε0​r5q​ and E⃗=0\vec{E} = 0E=0
  2. (B)V=5q4πε0rV = \dfrac{5q}{4\pi\varepsilon_0 r}V=4πε0​r5q​ and E⃗=53q8πε0r2 r^\vec{E} = \dfrac{5\sqrt{3}q}{8\pi\varepsilon_0 r^2}\,\hat{r}E=8πε0​r253​q​r^
  3. (C)V=5q4πε0rV = \dfrac{5q}{4\pi\varepsilon_0 r}V=4πε0​r5q​ and E⃗=5q4πε0r2 r^\vec{E} = \dfrac{5q}{4\pi\varepsilon_0 r^2}\,\hat{r}E=4πε0​r25q​r^
  4. (D)V=0V = 0V=0 and E⃗=0\vec{E} = 0E=0

Correct answer: (A)

Step-by-step solution →
Q16·PhysicsSingle correct
A laser beam has intensity of 4.0×10144.0 \times 10^{14}4.0×1014 W/m2^22. The amplitude of magnetic field associated with beam is __________ T. ( Take ϵ0=8.85×10−12\epsilon_0 = 8.85 \times 10^{-12}ϵ0​=8.85×10−12 C2^22/Nm2^22 and c =3×108= 3 \times 10^{8}=3×108 m/s)
  1. (A)2.0
  2. (B)18.3
  3. (C)5.5
  4. (D)1.83

Correct answer: (D)

Step-by-step solution →
Q17·PhysicsSingle correct
Light is incident on a metallic plate having work function 110×10−20110 \times 10^{-20}110×10−20 J. If the produced photoelectrons have zero kinetic energy then the angular frequency of the incident light is __________ rad/s. (h =6.63×10−34= 6.63 \times 10^{-34}=6.63×10−34 J.s)
  1. (A)1.04×10161.04 \times 10^{16}1.04×1016
  2. (B)1.04×10131.04 \times 10^{13}1.04×1013
  3. (C)1.66×10161.66 \times 10^{16}1.66×1016
  4. (D)1.66×10151.66 \times 10^{15}1.66×1015

Correct answer: (A)

Step-by-step solution →
Q18·PhysicsSingle correct
Given below are two statements : Statement I : For a mechanical system of many particles total kinetic energy is the sum of kinetic energies of all the particles. Statement II : The total kinetic energy can be the sum of kinetic energy of the center of mass w.r.t. to the origin and the kinetic energy of all the particles w.r.t. the center of mass as the reference. 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)Statement I is false but Statement II is true
  4. (D)Both Statement I and Statement II are false

Correct answer: (A)

Step-by-step solution →
Q19·PhysicsNumerical
A conducting circular loop is rotated about its diameter at a constant angular speed of 100 rad/s in a magnetic field of 0.5T perpendicular to the axis of rotation. When the loop is rotated by 30° from the horizontal position, the induced EMF is 15.4 mV. The radius of the loop is _______ mm. (Take π=227)\left(\text{Take } \pi = \dfrac{22}{7}\right)(Take π=722​)

Correct answer: 14

Step-by-step solution →
Q20·PhysicsNumerical
Two masses m and 2m are connected by a light string going over a pulley (disc) of mass 30m with radius r=0.1r = 0.1r=0.1 m. The pulley is mounted in a vertical plane and it is free to rotate about its axis. The 2m mass is released from rest and its speed when it has descended through a height of 3.6 m is ________ m/s. (Assume string does not slip and g =10= 10=10 m/s2^22)

Correct answer: 2

Step-by-step solution →
Q21·PhysicsNumerical
A capacitor PPP with capacitance 10×10−610 \times 10^{-6}10×10−6 F is fully charged with a potential difference of 6.0 V and disconnected from the battery. The charged capacitor PPP is connected across another capacitor QQQ with capacitance 20×10−620 \times 10^{-6}20×10−6 F. The charge on capacitor QQQ when equilibrium is established will be α×10−5\alpha \times 10^{-5}α×10−5 C (assume capacitor QQQ does not have any charge initially), the value of α\alphaα is ________ .

Correct answer: 4

Step-by-step solution →
Q22·PhysicsNumerical
A cylindrical conductor of length 2m and area of cross-section 0.2 mm2^22 carries an electric current of 1.6 A when its ends are connected to a 2V battery. Mobility of electrons in the conductor is α×10−3\alpha \times 10^{-3}α×10−3 m2^22/V.s. The value of α\alphaα is : (electron concentration =5×1028= 5 \times 10^{28}=5×1028/m3^33 and electron charge =1.6×10−19= 1.6 \times 10^{-19}=1.6×10−19 C)

Correct answer: 1

Step-by-step solution →
Q23·PhysicsNumerical
An insulated cylinder of volume 60 cm3^33 is filled with a gas at 27°C and 2 atmospheric pressure. Then the gas is compressed making the final volume as 20 cm3^33 while allowing the temperature to rise to 77°C. The final pressure is ________ atmospheric pressure.

Correct answer: 7

Step-by-step solution →

Chemistry — JEE Main 22 January 2026 Shift 2

Q24·ChemistrySingle correct
At T(K), 100 g of 98% H2SO4H_2SO_4H2​SO4​ (w/w) aqueous solution is mixed with 100 g of 49% H2SO4H_2SO_4H2​SO4​ (w/w) aqueous solution. What is the mole fraction of H2SO4H_2SO_4H2​SO4​ in the resultant solution ? (Given : Atomic mass H = 1 u ; S = 32 u; O = 16 u) (Assume that temperature after mixing remains constant)
  1. (A)0.9
  2. (B)0.1
  3. (C)0.337
  4. (D)0.663

Correct answer: (C)

Step-by-step solution →
Q25·ChemistrySingle correct
Consider the following reaction : The product Y formed is :
  1. (A)2-methylhex-2-yne
  2. (B)5-methylhex-2-yne
  3. (C)2-methylhex-3-yne
  4. (D)Isopropylbut-1-yne

Correct answer: (C)

Step-by-step solution →
Q26·ChemistrySingle correct
A+2B→AB2A + 2B \rightarrow AB_2A+2B→AB2​ 36.0 g of 'A' (Molar mass : 60 g mol−1mol^{-1}mol−1) and 56.0 g of 'B' (Molar mass : 80 g mol−1mol^{-1}mol−1) are allowed to react. Which of the following statements are correct ? (A) 'A' is the limting reagent (B) 77.0 g of AB2AB_2AB2​ is formed (C) Molar mass of AB2AB_2AB2​ is 140 g mol−1mol^{-1}mol−1 (D) 15.0 g of A is left unreacted after the completion of reaction. Choose the correct answer from the options given below :
  1. (A)C and D only
  2. (B)A and C only
  3. (C)B and D only
  4. (D)A and B only

Correct answer: (C)

Step-by-step solution →
Q27·ChemistrySingle correct
Given below are two statements : Statement-I L: The first ionization enthalpy of Cr is lower than that of Mn. Statement-II : The second and third ionization enthalpies of Cr are higher than those of Mn. 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)Statement-I is true but Statement-II is false.
  3. (C)Both Statement-I and Statement-II are true.
  4. (D)Statement-I is false but Statement-II is true.

Correct answer: (B)

Step-by-step solution →
Q28·ChemistrySingle correct
The final product [B] is :
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (C)

Step-by-step solution →
Q29·ChemistrySingle correct
When 1 g of compound (X) is subjected to Kjeldahl's method for estimation of nitrogen, 15 mL, 1M H2SO4H_2SO_4H2​SO4​ was neutralized by ammonia evolved. The percentage of nitrogen in compound (X) is :
  1. (A)21
  2. (B)0.42
  3. (C)42
  4. (D)0.21

Correct answer: (C)

Step-by-step solution →
Q30·ChemistrySingle correct
Correct statements regarding Arrhenius equation among the following are : (A) Factor e−Ea/RTe^{-Ea/RT}e−Ea/RT corresponds to fraction of molecules having kinetic energy less than Ea. (B) At a given temperature, lower the Ea, faster is the reaction. (C) Increase in temperature by about 10∘C10^\circ C10∘C doubles the rate of reaction. (D) Plot of log k vs 1T\frac{1}{T}T1​ gives a straight line with slope =−EaR= -\frac{Ea}{R}=−REa​. Choose the correct answer from the options given below :
  1. (A)B and D only
  2. (B)A and B only
  3. (C)A and C only
  4. (D)B and C only

Correct answer: (D)

Step-by-step solution →
Q31·Chemistry·IUPAC NomenclatureSingle correct
The IUPAC name of the following comopound is :
  1. (A)n-propyl-2-bromo-5-methylheptanoate
  2. (B)2-bromo-5-methylhexylpropanoate
  3. (C)2-bromo-5-methylpropanoate
  4. (D)n-propyl-1-bromo-4-methylhexanoate

Correct answer: (A)

Step-by-step solution →
Q32·ChemistrySingle correct
Given below are two statements : Statement-I : Element 'X' and 'Y' are the most and least electronegative elements, respectively among N, As, Sb and P. The nature of the oxides X2O3X_2O_3X2​O3​ and Y2O3Y_2O_3Y2​O3​ is acidic and amphoteric, respectively. Statement-II : BCl3BCl_3BCl3​ is covalent in nature and gets hydrolysed in water. It produces [B(OH)4]−[B(OH)_4]^-[B(OH)4​]− and [B(H2O)6]3+[B(H_2O)_6]^{3+}[B(H2​O)6​]3+ in aqueous medium. 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: (B)

Step-by-step solution →
Q33·ChemistrySingle correct
Match List-I with List-II. Choose the correct answer from the options given below :
Reaction of glucose withProduct formed
A.HydroxylamineI.Glouconic acid
B.Br2Br_2Br2​ waterII.Glucose pentacetate
C.Excess acetic anhydrideIII.Saccharic acid
D.Concentrated HNO3HNO_3HNO3​IV.Glucoxime
  1. (A)A-I, B-III, C-IV, D-II
  2. (B)A-IV, B-I, C-II, D-III
  3. (C)A-III, B-I, C-IV, D-II
  4. (D)A-IV, B-III, C-II, D-I

Correct answer: (B)

Step-by-step solution →
Q34·ChemistrySingle correct
Among H2SH_2SH2​S, H2OH_2OH2​O, NF3NF_3NF3​, NH3NH_3NH3​ and CHCl3CHCl_3CHCl3​, identify the molecule (X) with lowest dipole moment value. The number of lone pairs of electrons present on the central atom of the molecule (X) is :
  1. (A)2
  2. (B)0
  3. (C)1
  4. (D)3

Correct answer: (C)

Step-by-step solution →
Q35·ChemistrySingle correct
Given below are two statements : Statement-I : C < O < N < F is the correct order in terms of first ionization enthalpy values. Statement-II : S > Se > Te > Po > O is the correct order in terms of the magnitude of electron gain enthalpy values. In the light of the above statements, choose the correct answer from the options given below :
  1. (A)Statement-I is false but Statement-II is true
  2. (B)Both Statement-I and Statement-II are true.
  3. (C)Both Statement-I and Statement-II are false.
  4. (D)Statement-I is true but Statement-II is false.

Correct answer: (B)

Step-by-step solution →
Q36·ChemistrySingle correct
Which of the following mixture gives a buffer solution with pH = 9.25 ? Given : pKbpK_bpKb​ (NH4OHNH_4OHNH4​OH) = 4.75
  1. (A)0.2M NH4OHNH_4OHNH4​OH (0.4 L) + 0.1M HCl (1L)
  2. (B)0.2M NH4OHNH_4OHNH4​OH (0.5 L) + 0.1M HCl (0.5 L)
  3. (C)0.5M NH4OHNH_4OHNH4​OH (0.2 L) + 0.2M HCl (0.5 L)
  4. (D)0.4M NH4OHNH_4OHNH4​OH (1 L) + 0.1M HCl (1L)

Correct answer: (B)

Step-by-step solution →
Q37·ChemistrySingle correct
The energy of first (lowest) Balmer line of H atom is x J. The energy (in J) of second Balmer line of H atom is :
  1. (A)x2x^{2}x2
  2. (B)x1.35\frac{x}{1.35}1.35x​
  3. (C)2x
  4. (D)1.35 x

Correct answer: (D)

Step-by-step solution →
Q38·ChemistrySingle correct
Identify the correct statements : A. Hydrated salts can be used as primary standard. B. Primary standard should not undergo any reaction with air. C. Reactions of primary standard with another substance should be instantaneous and stoichiometric. D. Primary standard should not be soluble in water. E. Primary standard should have low relative molar mass. Choose the correct answer from the options given below :
  1. (A)A, B, C and E only
  2. (B)A, B, and C only
  3. (C)A, B and E only
  4. (D)D and E only

Correct answer: (B)

Step-by-step solution →
Q39·ChemistrySingle correct
[Ni(PPh3)2Cl2][Ni(PPh_{3})_{2}Cl_{2}][Ni(PPh3​)2​Cl2​] is a paramagnetic complex. Identify the INCORRECT statements about this complex. A. The complex exhibits geometrical isomerism. B. The complex is white in colour. C. The calculated spin-only magnetic moment of the complex is 2.84 BM. D. The calculated CFSE (Crystal Field Stabilization Energy) of Ni in this complex is −0.8Δ0-0.8\Delta_{0}−0.8Δ0​. E. The geometrical arrangement of ligands in this complex is similar to that in Ni(CO)4Ni(CO)_{4}Ni(CO)4​. Choose the correct answer from the options given below :
  1. (A)A and B only
  2. (B)A, B and D only
  3. (C)C and D only
  4. (D)C, D and E only

Correct answer: (B)

Step-by-step solution →
Q40·ChemistrySingle correct
Consider the following reduction processes : Al3++3e−→Al(s)Al^{3+} + 3e^{-} \rightarrow Al(s)Al3++3e−→Al(s), Eo=−1.66E^{o} = -1.66Eo=−1.66 V Fe3++e−→Fe2+Fe^{3+} + e^{-} \rightarrow Fe^{2+}Fe3++e−→Fe2+, Eo=+0.77E^{o} = +0.77Eo=+0.77 V Co3++e−→Co2+Co^{3+} + e^{-} \rightarrow Co^{2+}Co3++e−→Co2+, Eo=+1.81E^{o} = +1.81Eo=+1.81 V Cr3++3e−→Cr(s)Cr^{3+} + 3e^{-} \rightarrow Cr(s)Cr3++3e−→Cr(s), Eo=−0.74E^{o} = -0.74Eo=−0.74 V The tendency to act as reducing agent decreases in the order :
  1. (A)Al>Cr>Fe2+>Co2+Al > Cr > Fe^{2+} > Co^{2+}Al>Cr>Fe2+>Co2+
  2. (B)Al>Fe2+>Cr>Co2+Al > Fe^{2+} > Cr > Co^{2+}Al>Fe2+>Cr>Co2+
  3. (C)Al>Cr>Co2+>Fe2+Al > Cr > Co^{2+} > Fe^{2+}Al>Cr>Co2+>Fe2+
  4. (D)Cr>Fe2+>Al>Co2+Cr > Fe^{2+} > Al > Co^{2+}Cr>Fe2+>Al>Co2+

Correct answer: (A)

Step-by-step solution →
Q41·ChemistrySingle correct
The compound A, C8H8O2C_{8}H_{8}O_{2}C8​H8​O2​ reacts with acetophenone to form a single product via cross-Aldol condensation. The compound A on reaction with conc. NaOH forms a substituted benzyl alcohol as
  1. (A)2-hydroxy acetophenone
  2. (B)4-methyoxy benzaldehyde
  3. (C)4-hydroxy benzylaldehyde
  4. (D)4-methyl benzoic acid

Correct answer: (B)

Step-by-step solution →
Q42·ChemistrySingle correct
3, 3-Dimenthyl-2-butanol cannot be prepared by : Choose the correct answer from the options given below :
  1. (A)B only
  2. (B)B and E only
  3. (C)B and C only
  4. (D)B, C and E only

Correct answer: (B)

Step-by-step solution →
Q43·ChemistrySingle correct
The dibromo compound [P] (molecular formula : C9H10Br2C_{9}H_{10}Br_{2}C9​H10​Br2​) when heated with excess sodamide followed by treatment with dilute HCl gives [Q]. On warming [Q] with mercuric sulphate and dilute sulphuric acid yield [R] which gives positive Iodoform test but negative Tollen's test. The compound [P] is :
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (C)

Step-by-step solution →
Q44·ChemistryNumerical
Consider the following electrochemical cell : Pt ∣ O2(g) (1bar) ∣ HCl (aq) ∣∣ M2+(aq, 1.0 M) ∣ M(s)Pt\ |\ O_{2}(g)\ (1bar)\ |\ HCl\ (aq)\ ||\ M^{2+}(aq,\ 1.0\ M)\ |\ M(s)Pt ∣ O2​(g) (1bar) ∣ HCl (aq) ∣∣ M2+(aq, 1.0 M) ∣ M(s) The pH above which, oxygen gas would start to evolve at anode is ________ (nearest integer). [Given : EM2+/M0=0.994VE^{0}_{M^{2+}/M} = 0.994VEM2+/M0​=0.994V, EO2/H2O0=1.23VE^{0}_{O_{2}/H_{2}O} = 1.23VEO2​/H2​O0​=1.23V standard reduction potential and RTF(2.303)=0.059\frac{RT}{F}(2.303) = 0.059FRT​(2.303)=0.059 V at the given condition]

Correct answer: 4

Step-by-step solution →
Q45·ChemistryNumerical
If the enthalpy of sublimation of Li is 155 kJ mol−1mol^{-1}mol−1, enthalpy of dissociation of F2F_{2}F2​ is 150 kJ mol−1mol^{-1}mol−1, ionization enthalpy of Li is 520 kJ mol−1mol^{-1}mol−1, electron gain enthalpy of F is −313-313−313 kJ mol−1mol^{-1}mol−1, standard enthalpy of formation of LiF is −594-594−594 kJ mol−1mol^{-1}mol−1. The magnitude of lattice enthalpy of LiF is _________ kJ mol−1mol^{-1}mol−1 (Nearest integer).

Correct answer: 1031

Step-by-step solution →
Q46·ChemistryNumerical
Among the following oxides of 3d elements, the number of mixed oxides are_________. Ti2O3Ti_{2}O_{3}Ti2​O3​, V2O4V_{2}O_{4}V2​O4​, Cr2O3Cr_{2}O_{3}Cr2​O3​, Mn3O4Mn_{3}O_{4}Mn3​O4​, Fe3O4Fe_{3}O_{4}Fe3​O4​, Fe2O3Fe_{2}O_{3}Fe2​O3​, Co3O4Co_{3}O_{4}Co3​O4​

Correct answer: 3

Step-by-step solution →
Q47·ChemistryNumerical
The mass of benzanilide obtained from the benzoylation reaction of 5.8 g of aniline, if yield of product is 82%, is__________g (nearest integer). (Given molar mass in g mol−1mol^{-1}mol−1 H:1, C:12, N:14, O:16)

Correct answer: 10

Step-by-step solution →
Q48·ChemistryNumerical
Consider A →k1\xrightarrow{k_{1}}k1​​ B and C →k2\xrightarrow{k_{2}}k2​​ D are two reactions. If the rate constant (k1)(k_{1})(k1​) of the A →\rightarrow→ B reaction can be expressed by the following equation log⁡10k=14.34−1.5×104T/K\log_{10} k = 14.34 - \frac{1.5 \times 10^{4}}{T/K}log10​k=14.34−T/K1.5×104​ and activation energy of C →\rightarrow→ D reaction (Ea2)(Ea_{2})(Ea2​) is 15\frac{1}{5}51​ th of the A →\rightarrow→ B reaction (Ea1)(Ea_{1})(Ea1​), then the value of (Ea2)(Ea_{2})(Ea2​) is ______ kJ mol−1mol^{-1}mol−1. (Nearest Integer)

Correct answer: 57

Step-by-step solution →

Mathematics — JEE Main 22 January 2026 Shift 2

Q49·MathematicsSingle correct
Let n be the number obtained on rolling a fair die. If the probability that the system x−ny+z=6x - ny + z = 6x−ny+z=6 x+(n−2)y+(n+1)z=8x + (n-2)y + (n+1)z = 8x+(n−2)y+(n+1)z=8 (n−1)y+z=1(n-1)y + z = 1(n−1)y+z=1 Has a unique solution is k6\frac{k}{6}6k​, then the sum of k and all possible values of n is :
  1. (A)21
  2. (B)24
  3. (C)20
  4. (D)22

Correct answer: (D)

Step-by-step solution →
Q50·MathematicsSingle correct
If the mean deviation about the median of the numbers k, 2k, 3k, ...., 1000k is 500, then k2k^{2}k2 is equal to :
  1. (A)16
  2. (B)4
  3. (C)1
  4. (D)9

Correct answer: (B)

Step-by-step solution →
Q51·MathematicsSingle correct
The number of elements in the relation R={(x,y):4x2+y2<52, x, y∈Z}R = \{(x,y): 4x^{2} + y^{2} < 52,\ x,\ y \in Z\}R={(x,y):4x2+y2<52, x, y∈Z} is
  1. (A)77
  2. (B)89
  3. (C)67
  4. (D)86

Correct answer: (A)

Step-by-step solution →
Q52·MathematicsSingle correct
Let S={z∈C:4z2+zˉ=0}S = \left\{ z \in \mathbb{C} : 4z^{2} + \bar{z} = 0 \right\}S={z∈C:4z2+zˉ=0}. Then ∑z∈S∣z∣2\sum\limits_{z \in S} |z|^{2}z∈S∑​∣z∣2 is equal to :
  1. (A)316\frac{3}{16}163​
  2. (B)764\frac{7}{64}647​
  3. (C)116\frac{1}{16}161​
  4. (D)564\frac{5}{64}645​

Correct answer: (A)

Step-by-step solution →
Q53·Mathematics·Limits and ContinuitySingle correct
If lim⁡x→0e(a−1)x+2cos⁡bx+(c−2)e−xxcos⁡x−log⁡e(1+x)=2\lim\limits_{x \to 0} \frac{e^{(a-1)x} + 2\cos bx + (c-2)e^{-x}}{x \cos x - \log_{e}(1+x)} = 2x→0lim​xcosx−loge​(1+x)e(a−1)x+2cosbx+(c−2)e−x​=2, then a2+b2+c2a^{2} + b^{2} + c^{2}a2+b2+c2 is equal to :
  1. (A)5
  2. (B)3
  3. (C)7
  4. (D)9

Correct answer: (C)

Step-by-step solution →
Q54·MathematicsSingle correct
Among the statements (S1) : If A(5, -1) and B(-2, 3) are two vertices of a triangle, whose orthocentre is (0, 0), then its third vertex is (-4, -7) and (S2) : If positive numbers 2a, b, c are three consecutive terms of an A.P., then the lines ax+by+c=0ax + by + c = 0ax+by+c=0 are concurrent at (2, -2),
  1. (A)Only (S1) is correct
  2. (B)Only (S2) is correct
  3. (C)Both are incorrect
  4. (D)Both are correct

Correct answer: (D)

Step-by-step solution →
Q55·MathematicsSingle correct
Let a⃗=2i^−j^+k^\vec{a} = 2\hat{i} - \hat{j} + \hat{k}a=2i^−j^​+k^ and b⃗=λj^+2k^\vec{b} = \lambda\hat{j} + 2\hat{k}b=λj^​+2k^, λ∈Z\lambda \in Zλ∈Z be two vectors, Let c⃗=a⃗×b⃗\vec{c} = \vec{a} \times \vec{b}c=a×b and d⃗\vec{d}d be a vector of magnitude 2 in yz-plane. If ∣c⃗∣=53|\vec{c}| = \sqrt{53}∣c∣=53​, then the maximum possible value of (c⃗⋅d⃗)2\left(\vec{c} \cdot \vec{d}\right)^{2}(c⋅d)2 is equal to :
  1. (A)26
  2. (B)104
  3. (C)208
  4. (D)52

Correct answer: (C)

Step-by-step solution →
Q56·MathematicsSingle correct
If X=[xyz]X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}X=​xyz​​ is a solution of the system of equations AX=BAX = BAX=B, where adj A=[422−5051−23]A = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}A=​4−51​20−2​253​​ and B=[402]B = \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix}B=​402​​, then ∣x+y+z∣|x + y + z|∣x+y+z∣ is equal to :
  1. (A)3
  2. (B)32\frac{3}{2}23​
  3. (C)1
  4. (D)2

Correct answer: (D)

Step-by-step solution →
Q57·MathematicsSingle correct
Let L be the line x+12=y+13=z+36\frac{x+1}{2} = \frac{y+1}{3} = \frac{z+3}{6}2x+1​=3y+1​=6z+3​ and let S be the set of all points (a, b, c) on L, whose distance from the line x+12=y+13=z−90\frac{x+1}{2} = \frac{y+1}{3} = \frac{z-9}{0}2x+1​=3y+1​=0z−9​ along the line L is 7. Then ∑(a,b,c)∈S(a+b+c)\sum\limits_{(a,b,c) \in S} (a + b + c)(a,b,c)∈S∑​(a+b+c) is equal to :
  1. (A)34
  2. (B)28
  3. (C)40
  4. (D)6

Correct answer: (A)

Step-by-step solution →
Q58·MathematicsSingle correct
Let P (10, 215)\left(10,\ 2\sqrt{15}\right)(10, 215​) be a point on the hyperbola x2a2−y2b2=1\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1a2x2​−b2y2​=1, whose foci are S and S'. If the length of its latus rectum is 8, then the square of the area of ΔPSS′\Delta PSS'ΔPSS′ is equal to :
  1. (A)4200
  2. (B)900
  3. (C)1462
  4. (D)2700

Correct answer: (D)

Step-by-step solution →
Q59·MathematicsSingle correct
The area of the region A={(x, y):4x2+y2≤8 and y2≤4x}A = \{(x,\ y) : 4x^{2} + y^{2} \le 8 \text{ and } y^{2} \le 4x\}A={(x, y):4x2+y2≤8 and y2≤4x} is :
  1. (A)π2+2\frac{\pi}{2} + 22π​+2
  2. (B)π+23\pi + \frac{2}{3}π+32​
  3. (C)π+4\pi + 4π+4
  4. (D)π2+13\frac{\pi}{2} + \frac{1}{3}2π​+31​

Correct answer: (B)

Step-by-step solution →
Q60·MathematicsSingle correct
Let α\alphaα, β\betaβ be the roots of the quadratic equation 12x2−20x+3λ=012x^{2} - 20x + 3\lambda = 012x2−20x+3λ=0, λ∈Z\lambda \in \mathbb{Z}λ∈Z. If 12≤∣β−α∣≤32\frac{1}{2} \le |\beta - \alpha| \le \frac{3}{2}21​≤∣β−α∣≤23​, then the sum of all possible values of λ\lambdaλ is :
  1. (A)6
  2. (B)1
  3. (C)3
  4. (D)4

Correct answer: (C)

Step-by-step solution →
Q61·MathematicsSingle correct
Let the domain of the function f(x)=log⁡3log⁡5(7−log⁡2(x2−10x+85))+sin⁡−1(∣3x−717−x∣)f(x) = \log_3 \log_5 \left(7 - \log_2 (x^2 - 10x + 85)\right) + \sin^{-1}\left(\left|\frac{3x - 7}{17 - x}\right|\right)f(x)=log3​log5​(7−log2​(x2−10x+85))+sin−1(​17−x3x−7​​) be (α, β]. Then α + β is equal to :
  1. (A)10
  2. (B)12
  3. (C)9
  4. (D)8

Correct answer: (C)

Step-by-step solution →
Q62·Mathematics·DifferentiabilitySingle correct
Let [•] denote the greatest integer function, and let f(x)=min⁡{2x,x2}f(x) = \min\left\{\sqrt{2}x, x^2\right\}f(x)=min{2​x,x2}. Let S = {x ∈ (–2, 2) : the function g(x)=∣x∣[x2]g(x) = |x|[x^2]g(x)=∣x∣[x2] is discontinuous at x}. Then ∑x∈Sf(x)\sum_{x \in S} f(x)∑x∈S​f(x) equals :
  1. (A)2−22 - \sqrt{2}2−2​
  2. (B)26−322\sqrt{6} - 3\sqrt{2}26​−32​
  3. (C)1−21 - \sqrt{2}1−2​
  4. (D)6−22\sqrt{6} - 2\sqrt{2}6​−22​

Correct answer: (C)

Step-by-step solution →
Q63·MathematicsSingle correct
Let S and S' be the foci of the ellipse x225+y29=1\frac{x^2}{25} + \frac{y^2}{9} = 125x2​+9y2​=1 and P (α, β) be a point on the ellipse in the first quadrant. If (SP)2+(S′P)2−SP∙S′P=37(SP)^2 + (S'P)^2 - SP \bullet S'P = 37(SP)2+(S′P)2−SP∙S′P=37, then α2+β2α^2 + β^2α2+β2 is equal to :
  1. (A)15
  2. (B)11
  3. (C)17
  4. (D)13

Correct answer: (D)

Step-by-step solution →
Q64·MathematicsSingle correct
Let the locus of the mid-point of the chord through the origin O of the parabola y2=4xy^2 = 4xy2=4x be the curve S. Let P be any point on S. Then the locus of the point, which internally divides OP in the ratio 3 : 1, is :
  1. (A)3y2=2x3y^2 = 2x3y2=2x
  2. (B)2y2=3x2y^2 = 3x2y2=3x
  3. (C)3x2=2y3x^2 = 2y3x2=2y
  4. (D)2x2=3y2x^2 = 3y2x2=3y

Correct answer: (B)

Step-by-step solution →
Q65·MathematicsSingle correct
Let f(x)=[x]2−[x+3]−3f(x) = [x]^2 - [x + 3] - 3f(x)=[x]2−[x+3]−3, x∈Rx \in \mathbb{R}x∈R where [•] is the greatest integer function. Then
  1. (A)f(x)>0f(x) > 0f(x)>0 only for x∈[4,∞)x \in [4, \infty)x∈[4,∞)
  2. (B)f(x)<0f(x) < 0f(x)<0 only for x∈[−1,3)x \in [-1, 3)x∈[−1,3)
  3. (C)∫02f(x)dx=−6\int_0^2 f(x)dx = -6∫02​f(x)dx=−6
  4. (D)f(x)=0f(x) = 0f(x)=0 for finitely many values of x.

Correct answer: (B)

Step-by-step solution →
Q66·MathematicsSingle correct
Let fff and g be functions satisfying f(x+y)=f(x) f(y)f(x+y) = f(x)\, f(y)f(x+y)=f(x)f(y), f(1)=7f(1) = 7f(1)=7 and g(x+y)=g(xy)g(x+y) = g(xy)g(x+y)=g(xy), g(1)=1g(1) = 1g(1)=1, for all x,y∈Nx, y \in \mathbb{N}x,y∈N. ∑x=1n(f(x)g(x))=19607\sum_{x=1}^{n}\left(\frac{f(x)}{g(x)}\right) = 19607∑x=1n​(g(x)f(x)​)=19607, then n is equal to :
  1. (A)7
  2. (B)5
  3. (C)6
  4. (D)4

Correct answer: (B)

Step-by-step solution →
Q67·MathematicsSingle correct
Let CrC_rCr​ denote the coefficient of xrx^rxr in the binomial expansion of (1+x)n(1 + x)^n(1+x)n, n∈Nn \in \mathbb{N}n∈N, 0≤r≤n0 \le r \le n0≤r≤n. If Pn=C0−C1+223C2−234C3+.....+(−2)nn+1CnP_n = C_0 - C_1 + \frac{2^2}{3}C_2 - \frac{2^3}{4}C_3 + ..... + \frac{(-2)^n}{n+1}C_nPn​=C0​−C1​+322​C2​−423​C3​+.....+n+1(−2)n​Cn​, then the value of ∑n=1251P2n\sum_{n=1}^{25} \frac{1}{P_{2n}}∑n=125​P2n​1​ equals.
  1. (A)580
  2. (B)525
  3. (C)650
  4. (D)675

Correct answer: (D)

Step-by-step solution →
Q68·MathematicsNumerical
Let a vector a⃗=2i^−j^+λk^\vec{a} = \sqrt{2}\hat{i} - \hat{j} + \lambda\hat{k}a=2​i^−j^​+λk^, λ>0\lambda > 0λ>0, make an obtuse angle with the vector b⃗=−λ2i^+42j^+42k^\vec{b} = -\lambda^2\hat{i} + 4\sqrt{2}\hat{j} + 4\sqrt{2}\hat{k}b=−λ2i^+42​j^​+42​k^ and an angle θ, π6<θ<π2\frac{\pi}{6} < \theta < \frac{\pi}{2}6π​<θ<2π​, with the positive z-axis. If the set of all possible values of λ is (α, β) – {γ}, then α + β + γ is equal to ______.

Correct answer: 5

Step-by-step solution →
Q69·MathematicsNumerical
Let [•] be the greatest integer function. If α=∫064(x1/3−[x1/3])dx\alpha = \int_0^{64}\left(x^{1/3} - \left[x^{1/3}\right]\right)dxα=∫064​(x1/3−[x1/3])dx, then 1π∫0απ(sin⁡2θsin⁡6θ+cos⁡6θ)dθ\frac{1}{\pi}\int_0^{\alpha\pi}\left(\frac{\sin^2\theta}{\sin^6\theta + \cos^6\theta}\right)d\thetaπ1​∫0απ​(sin6θ+cos6θsin2θ​)dθ is equal to ____.

Correct answer: 36

Step-by-step solution →
Q70·MathematicsNumerical
Let cos⁡(α+β)=−110\cos(\alpha + \beta) = -\frac{1}{10}cos(α+β)=−101​ and sin⁡(α−β)=38\sin(\alpha - \beta) = \frac{3}{8}sin(α−β)=83​, where 0<α<π30 < \alpha < \frac{\pi}{3}0<α<3π​ and 0<β<π40 < \beta < \frac{\pi}{4}0<β<4π​. If tan⁡2α=3(1−r5)11(s+5)\tan 2\alpha = \frac{3(1 - r\sqrt{5})}{\sqrt{11}(s + \sqrt{5})}tan2α=11​(s+5​)3(1−r5​)​, r, s ∈ N\mathbb{N}N, then r + s is equal to______ .

Correct answer: 20

Step-by-step solution →
Q71·MathematicsNumerical
Suppose a, b, c are in A.P. and a2a^2a2, 2b22b^22b2, c2c^2c2 are in G.P. If a<b<ca < b < ca<b<c and a+b+c=1a + b + c = 1a+b+c=1, then 9(a2+b2+c2)9(a^2 + b^2 + c^2)9(a2+b2+c2) is equal to______ .

Correct answer: 9

Step-by-step solution →
Q72·MathematicsNumerical
Let S be the set of the first 11 natural numbers. Then the number of elements in A = {B ⊆ S : n(B) ≥ 2 and the product of all elements of B is even} is ______ .

Correct answer: 1979

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.

  • Properties of Solids and Liquids 172/186
  • 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
  • Vector Algebra 173/186
  • Permutations and Combinations 162/186
  • Binomial Theorem and Its Simple Applications 158/186
  • Electromagnetic Waves 144/186
  • Chemical Bonding and Molecular Structure 151/186
  • Limits and Continuity 149/186
  • d- and f-Block Elements 126/186
  • Aldehydes and Ketones 135/186
  • Equilibrium 163/186
  • Chemical Thermodynamics 165/186
  • Hydrocarbons 126/186
  • Complex Numbers 165/186
  • Trigonometric Functions 144/186
  • Biomolecules 162/186
  • Chemical Kinetics 169/186
  • Gravitation 152/186
  • Electric Field and Coulomb's Law 133/186
  • Atomic Structure 161/186
  • Electronic Devices 145/186
  • Dual Nature of Matter and Radiation 155/186
  • Amines 133/186
  • Quadratic Equations 148/186
  • Work, Energy and Power 132/186
  • Some Basic Concepts in Chemistry 129/186
  • Electromagnetic Induction 120/186
  • 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
  • Straight Lines 114/186
  • Waves 109/186
  • Capacitors and Dielectrics 115/186
  • Atoms 112/186
  • Classification of Elements and Periodicity in Properties 113/186
  • Parabola 101/186
  • Statistics 118/186
  • Ellipse 103/186
  • Differentiability 91/186
  • Hyperbola 77/186
  • Experimental Skills 68/186
  • Electric Potential 63/186
  • IUPAC Nomenclature 37/186
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