Jarvis OS
PYQ papersPricingSign inGet started
  1. Home
  2. /JEE Main PYQs
  3. /2026
  4. /22 Jan Shift 1

JEE Main 22 January 2026 Shift 1 Question Paper with Answers

22 January 2026 · January session · 72 questions

72 of the 75 questions from the JEE Main 22 January 2026 Shift 1 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
24
Chemistry
25
Mathematics
23

Physics — JEE Main 22 January 2026 Shift 1

Q1·PhysicsSingle correct
A solid sphere of mass 5 kg and radius 10 cm is kept in contact with another solid sphere of mass 10 kg and radius 20 cm. The moment of inertia of this pair of spheres about the tangent passing through the point of contact is__________kg.m2^{2}2.
  1. (A)0.36
  2. (B)0.72
  3. (C)0.18
  4. (D)0.63

Correct answer: (D)

Step-by-step solution →
Q2·PhysicsSingle correct
7.9 MeV α−particle scatters from a target material of atomic number 79. From the given data the estimated diameter of nuclei of the target material is (approximately)_________m. [14π∈o=9×109 Nm2/C2\frac{1}{4\pi \in_{o}} = 9\times 10^{9}\,\mathrm{Nm}^{2}/\mathrm{C}^{2}4π∈o​1​=9×109Nm2/C2 and electron charge =1.6×10−19C=1.6\times 10^{-19}\mathrm{C}=1.6×10−19C]
  1. (A)5.76×10−145.76 \times 10^{-14}5.76×10−14
  2. (B)1.44×10−131.44 \times 10^{-13}1.44×10−13
  3. (C)2.88×10−142.88 \times 10^{-14}2.88×10−14
  4. (D)1.69×10−121.69 \times 10^{-12}1.69×10−12

Correct answer: (A)

Step-by-step solution →
Q3·PhysicsSingle correct
Six point charges are kept 60° apart from each other on the circumference of a circle of radius R as shown in figure. The net electric field at the centre of the circle is _________. (∈o\in_{o}∈o​ is permittivity of free space)
  1. (A)−5Q8π∈oR2(i^+3j^)-\frac{5Q}{8\pi \in_{o} R^{2}}(\hat{i}+\sqrt{3}\hat{j})−8π∈o​R25Q​(i^+3​j^​)
  2. (B)−Q4π∈oR2(3 i^−j^)-\frac{Q}{4\pi \in_{o} R^{2}}\left(\sqrt{3}\,\hat{i}-\hat{j}\right)−4π∈o​R2Q​(3​i^−j^​)
  3. (C)−(5Q8π∈oR2)(i^−3j^)-\left(\frac{5Q}{8\pi \in_{o} R^{2}}\right)\left(\hat{i}-3\hat{j}\right)−(8π∈o​R25Q​)(i^−3j^​)
  4. (D)Q4π∈oR2(3 i^−j^)\frac{Q}{4\pi \in_{o} R^{2}}\left(\sqrt{3}\,\hat{i}-\hat{j}\right)4π∈o​R2Q​(3​i^−j^​)

Correct answer: (B)

Step-by-step solution →
Q4·PhysicsSingle correct
XPQY is a vertical smooth long loop having a total resistance R where PX is parallel to QY and separation between them is lll. A constant magnetic field B perpendicular to the plane of the loop exists in the entire space. A rod CD of length L (L > lll) and mass m is made to slide down from rest under the gravity as shown in figure. The terminal speed acquired by the rod is_________m/s. (g = acceleration due to gravity)
  1. (A)2mgRB2l2\frac{2mgR}{B^{2}l^{2}}B2l22mgR​
  2. (B)8mgRB2l2\frac{8mgR}{B^{2}l^{2}}B2l28mgR​
  3. (C)2mgRB2L2\frac{2mgR}{B^{2}L^{2}}B2L22mgR​
  4. (D)mgRB2l2\frac{mgR}{B^{2}l^{2}}B2l2mgR​

Correct answer: (D)

Step-by-step solution →
Q5·PhysicsSingle correct
The escape velocity from a spherical planet A is 10 km/s. The escape velocity from another planet B whose density and radius are 10% of those of planet A, is_______m/s.
  1. (A)1000
  2. (B)2005200\sqrt{5}2005​
  3. (C)10010100\sqrt{10}10010​
  4. (D)100021000\sqrt{2}10002​

Correct answer: (C)

Step-by-step solution →
Q6·PhysicsSingle correct
A meter bridge with two resistances R1R_{1}R1​ and R2R_{2}R2​ as shown in figure was balanced (null point) at 40 cm from the point P. The null point changed to 50 cm from the point P, when 16 Ω resistance is connected in parallel to R2R_{2}R2​. The values of resistances R1R_{1}R1​ and R2R_{2}R2​ are_______.
  1. (A)R2=16Ω,R1=163ΩR_{2} = 16\Omega, R_{1} = \frac{16}{3}\OmegaR2​=16Ω,R1​=316​Ω
  2. (B)R2=4Ω,R1=43ΩR_{2} = 4\Omega, R_{1} = \frac{4}{3}\OmegaR2​=4Ω,R1​=34​Ω
  3. (C)R2=8Ω,R1=163ΩR_{2} = 8\Omega, R_{1} = \frac{16}{3}\OmegaR2​=8Ω,R1​=316​Ω
  4. (D)R2=12Ω,R1=123ΩR_{2} = 12\Omega, R_{1} = \frac{12}{3}\OmegaR2​=12Ω,R1​=312​Ω

Correct answer: (C)

Step-by-step solution →
Q7·PhysicsSingle correct
A projectile is thrown upward at an angle 60° with the horizontal. The speed of the projectile is 20 m/s when its direction of motion is 45° with the horizontal. The initial speed of the projectile is______m/s.
  1. (A)40240\sqrt{2}402​
  2. (B)40
  3. (C)20320\sqrt{3}203​
  4. (D)20220\sqrt{2}202​

Correct answer: (D)

Step-by-step solution →
Q8·PhysicsSingle correct
Given below are two statements: Statement I : Pressure of fluid is exerted only on a solid surface in contact as the fluid-pressure does not exist everywhere in a still fluid. Statement II: Excess potential energy of the molecules on the surface of a liquid, when compared to interior, results in surface tension. In the light of the above statements, choose the correct answer from the options given below.
  1. (A)Statement I is true but Statement II is false
  2. (B)Both Statement I and Statement II are false
  3. (C)Both Statement I and Statement II are true
  4. (D)Statement I is false but Statement II is true

Correct answer: (D)

Step-by-step solution →
Q9·PhysicsSingle correct
Find the correct combination of A, B, C and D inputs which can cause the LED to glow.
  1. (A)0100
  2. (B)0011
  3. (C)1000
  4. (D)1101

Correct answer: (D)

Step-by-step solution →
Q10·PhysicsSingle correct
Net gravitational force at the centre of a square is found to be F1F_{1}F1​ when four particles having mass M, 2M, 3M and 4M are placed at the four corners of the square as shown in figure and it is F2F_{2}F2​ when the positions of 3M and 4M are interchanged. The ratio F1F2\frac{F_{1}}{F_{2}}F2​F1​​ is α5\frac{\alpha}{\sqrt{5}}5​α​. The value of α is _____.
  1. (A)2
  2. (B)3
  3. (C)1
  4. (D)252\sqrt{5}25​

Correct answer: (A)

Step-by-step solution →
Q11·PhysicsSingle correct
The minimum frequency of photon required to break a particle of mass 15.348 amu into 4α particles is ____ kHz. [mass of He nucleus = 4.002 amu, 1 amu = 1.66×10−271.66 \times 10^{-27}1.66×10−27 kg, h = 6.6×10−346.6 \times 10^{-34}6.6×10−34 J.s and c = 3×1083 \times 10^{8}3×108 m/s]
  1. (A)9×10199 \times 10^{19}9×1019
  2. (B)9×10209 \times 10^{20}9×1020
  3. (C)14.94×102014.94 \times 10^{20}14.94×1020
  4. (D)14.94×101914.94 \times 10^{19}14.94×1019

Correct answer: (D)

Step-by-step solution →
Q12·PhysicsSingle correct
A cylindrical tube ABABAB of length lll, closed at both ends contains an ideal gas of 1 mol having molecular weight MMM. The tube is rotated in a horizontal plane with constant angular velocity ω\omegaω about an axis perpendicular to ABABAB and passing through the edge at end A, as shown in the figure. If PAP_{A}PA​ and PBP_{B}PB​ are the pressures at AAA and BBB respectively, then (Consider the temperature is same at all points in the tube)
  1. (A)PB=PAexp⁡(Mω2l2/2RT)P_{B} = P_{A} \exp (M\omega^{2}l^{2} / 2RT)PB​=PA​exp(Mω2l2/2RT)
  2. (B)PB=PAP_{B} = P_{A}PB​=PA​
  3. (C)PB=PAexp⁡(Mω2l2/3RT)P_{B} = P_{A} \exp (M\omega^{2}l^{2} / 3RT)PB​=PA​exp(Mω2l2/3RT)
  4. (D)PB=PAexp⁡(Mω2l2/RT)P_{B} = P_{A} \exp (M\omega^{2}l^{2} / RT)PB​=PA​exp(Mω2l2/RT)

Correct answer: (A)

Step-by-step solution →
Q13·PhysicsSingle correct
Rods x and y of equal dimensions but of different materials are joined as shown in figure. Temperatures of end points AAA and FFF are maintained at 100 °C and 40 °C respectively. Given the thermal conductivity of rod x is three times of that of rod y, the temperature at junction points BBB and EEE are (close to) :
  1. (A)89 °C and 73 °C respectively
  2. (B)80 °C and 60 °C respectively
  3. (C)80 °C and 70 °C respectively
  4. (D)60 °C and 45 °C respectively

Correct answer: (A)

Step-by-step solution →
Q14·PhysicsSingle correct
Consider an equilateral prism (refractive index 2\sqrt{2}2​). A ray of light is incident on its one surface at a certain angle iii. If the emergent ray is found to graze along the other surface then the angle of refraction at the incident surface is close to ____.
  1. (A)15°
  2. (B)20°
  3. (C)40°
  4. (D)30°

Correct answer: (A)

Step-by-step solution →
Q15·PhysicsSingle correct
The volume of an ideal gas increases 8 times and temperature becomes (1/4)th(1/4)^{th}(1/4)th of initial temperature during a reversible change. If there is no exchange of heat in this process (ΔQ = 0) then identify the gas from the following options (Assuming the gases given in the options are ideal gases) :
  1. (A)CO2CO_2CO2​
  2. (B)O2O_2O2​
  3. (C)NH3NH_3NH3​
  4. (D)He

Correct answer: (D)

Step-by-step solution →
Q16·PhysicsSingle correct
Electric field in a region is given by E⃗=Axi^+Byj^\vec{E} = Ax\hat{i} + By\hat{j}E=Axi^+Byj^​, where AAA = 10 V/m2^22 and BBB = 5 V/m2^22. If the electric potential at a point (10, 20) is 500 V, then the electric potential at origin is ______ V.
  1. (A)1000
  2. (B)500
  3. (C)2000
  4. (D)0

Correct answer: (C)

Step-by-step solution →
Q17·PhysicsSingle correct
A simple pendulum has a bob with mass mmm and charge q. The pendulum string has negligible mass. When a uniform and horizontal electric field E⃗\vec{E}E is applied, the tension in the string changes. The final tension in the string, when pendulum attains an equilibrium position is ______. (g : acceleration due to gravity)
  1. (A)mg−qEmg - qEmg−qE
  2. (B)mg+qEmg + qEmg+qE
  3. (C)m2g2+q2E2\sqrt{m^2g^2 + q^2E^2}m2g2+q2E2​
  4. (D)m2g2−q2E2\sqrt{m^2g^2 - q^2E^2}m2g2−q2E2​

Correct answer: (C)

Step-by-step solution →
Q18·PhysicsSingle correct
Match the LIST-I with LIST-II Choose the correct answer from the options given below:
List-IList-II
A.Spring constantI.ML2T−2K−1ML^2T^{-2}K^{-1}ML2T−2K−1
B.Thermal conductivityII.ML0T−2ML^0T^{-2}ML0T−2
C.Boltzmann constantIII.ML2T−3A−2ML^2T^{-3}A^{-2}ML2T−3A−2
D.Inductive reactanceIV.MLT−3K−1MLT^{-3}K^{-1}MLT−3K−1
  1. (A)A-II, B-I, C-IV, D-III
  2. (B)A-I, B-IV, C-II, D-III
  3. (C)A-III, B-II, C-IV, D-I
  4. (D)A-II, B-IV, C-I, D-III

Correct answer: (D)

Step-by-step solution →
Q19·PhysicsSingle correct
Three identical coils C1C_1C1​, C2C_2C2​ and C3C_3C3​ are closely placed such that they share a common axis. C2C_2C2​ is exactly midway. C1C_1C1​ carries current III in anti-clockwise direction while C3C_3C3​ carries current III in clockwise direction. An induced current flows through C2C_2C2​ will be in clockwise direction when
  1. (A)C1C_1C1​ and C3C_3C3​ move with equal speeds away from C2C_2C2​
  2. (B)C1C_1C1​ moves towards C2C_2C2​ and C3C_3C3​ moves away from C2C_2C2​
  3. (C)C1C_1C1​ moves away from C2C_2C2​ and C3C_3C3​ moves towards C2C_2C2​
  4. (D)C1C_1C1​ and C3C_3C3​ move with equal speeds towards C2C_2C2​

Correct answer: (B)

Step-by-step solution →
Q20·PhysicsNumerical
Two loudspeakers (L1L_1L1​ and L2L_2L2​) are placed with a separation of 10 m, as shown in figure. Both speakers are fed with an audio input signal of same frequency with constant volume. A voice recorder, initially at point AAA, at equidistance to both loud speakers, is moved by 25 m along the line ABABAB while monitoring the audio signal. The measured signal was found to undergo 10 cycles of minima and maxima during the movement. The frequency of the input signal is ______ Hz (Speed of sound in air is 324 m/s and 5\sqrt{5}5​ = 2.23)

Correct answer: 600

Step-by-step solution →
Q21·PhysicsNumerical
The electric field of a plane electromagnetic wave, travelling in an unknown non-magnetic medium is given by, Ey=20sin⁡(3×106x−4.5×1014t)E_y = 20 \sin(3 \times 10^{6}x - 4.5 \times 10^{14}t)Ey​=20sin(3×106x−4.5×1014t) V/m (where x, t and other values have S.I. units). The dielectric constant of the medium is ____. (speed of light in free space is 3×1083 \times 10^{8}3×108 m/s)

Correct answer: 4

Step-by-step solution →
Q22·PhysicsNumerical
A parallel beam of light travelling in air (refractive index 1.0) is incident on a convex spherical glass surface of radius of curvature 50 cm. Refractive index of glass is 1.5. The rays converge to a point at a distance x cm from the centre of the curvature of the spherical surface. The value of x is ____ cm.

Correct answer: 100

Step-by-step solution →
Q23·PhysicsNumerical
A circular disc has radius R1R_1R1​ and thickness T1T_1T1​. Another circular disc made of the same material has radius R2R_2R2​ and thickness T2T_2T2​. If the moment of inertia of both discs are same and R1R2=2\frac{R_1}{R_2} = 2R2​R1​​=2 then T1T2=1α\frac{T_1}{T_2} = \frac{1}{\alpha}T2​T1​​=α1​. The value of α is ____.

Correct answer: 16

Step-by-step solution →
Q24·PhysicsNumerical
Inductance of a coil with 10410^4104 turns is 10 mH and it is connected to a dc source of 10 V with internal resistance of 10Ω. The energy density in the inductor when the current reaches (1e)\left(\frac{1}{e}\right)(e1​) of its maximum value is απ×1e2\alpha\pi \times \frac{1}{e^2}απ×e21​ J/m3^33. The value of α is ____. (μ0=4π×10−7\mu_0 = 4\pi \times 10^{-7}μ0​=4π×10−7 Tm/A).

Correct answer: 20

Step-by-step solution →

Chemistry — JEE Main 22 January 2026 Shift 1

Q25·ChemistrySingle correct
Consider the transition metal ions Mn3+^{3+}3+, Cr3+^{3+}3+, Fe3+^{3+}3+ and Co3+^{3+}3+ and all form low spin octahedral complexes. The correct decreasing order of unpaired electrons in their respective d-orbitals of the complexes is
  1. (A)Cr3+^{3+}3+ > Fe3+^{3+}3+ > Co3+^{3+}3+ > Mn3+^{3+}3+
  2. (B)Mn3+^{3+}3+ > Fe3+^{3+}3+ > Co3+^{3+}3+ > Cr3+^{3+}3+
  3. (C)Fe3+^{3+}3+ > Co3+^{3+}3+ > Mn3+^{3+}3+ > Cr3+^{3+}3+
  4. (D)Cr3+^{3+}3+ > Mn3+^{3+}3+ > Fe3+^{3+}3+ > Co3+^{3+}3+

Correct answer: (D)

Step-by-step solution →
Q26·ChemistrySingle correct
The formal changes on the atoms marked as (1) to (4) in the Lewis representation of HnO3_33​ molecule respectively are
  1. (A)+1, 0, 0, –1
  2. (B)0, –1, 0, +1
  3. (C)0, +1, 0, –1
  4. (D)0, 0, –1, +1

Correct answer: (C)

Step-by-step solution →
Q27·Chemistry·Alcohols and EthersSingle correct
Given below are two statements : Statement I : Phenol on treatment with CHCl3_33​/aq. KOH under refluxing condition, followed by acidification produces p-hydroxy benzladehyde as the major product and o-hydroxy benzaldehyde as the minor product. Statement II : The mixture of p-hydroxybenzaldehyde and o-hydroxybenzaldehyde can be easily separated through steam distillation. 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: (D)

Step-by-step solution →
Q28·ChemistrySingle correct
The energy required by electrons, present in the first Bohr orbit of hydrogen atom to be excited to second Bohr orbit is ________ J mol−1^{-1}−1. Given : RH_HH​ = 2.18 × 10−11^{-11}−11 ergs.
  1. (A)1.635 × 10−18^{-18}−18
  2. (B)9.835 × 105^{5}5
  3. (C)9.835 × 1012^{12}12
  4. (D)1.635 × 10−11^{-11}−11

Correct answer: (B)

Step-by-step solution →
Q29·ChemistrySingle correct
A 'p'-block element (E) and hydrogen form a binary cation (EHx_xx​)+^{+}+, while EH3_33​ on treatment with K2_22​HgI4_44​ in alkaline medium gives a precipitate of basic mercury(II)amido-iodine. Given below are first ionisation enthalpy values (kJ mol−1^{-1}−1) for first element each from group 13, 14, 15 and 16. Identify the correct first ionisation enthalpy value for element E.
  1. (A)1312
  2. (B)1086
  3. (C)1402
  4. (D)801

Correct answer: (C)

Step-by-step solution →
Q30·ChemistrySingle correct
In the reaction, 2Al(s) + 6HCl(aq) → 2Al3+^{3+}3+(aq) + 6Cl−^{-}−(aq) + 3H2_22​(g)
  1. (A)11.2 L H2_22​(g) at STP is produced for every mole of HCl consumed.
  2. (B)67.2 L H2_22​(g) at STP is produced for every mole of Al that reacts.
  3. (C)12 L HCl(aq) is consumed for every 6L H2_22​(g) produced.
  4. (D)33.6 L H2_22​(g) is produced regardless of temperature and pressure for every mole of Al that reacts.

Correct answer: (A)

Step-by-step solution →
Q31·ChemistrySingle correct
Consider a solution of CO2_22​(g) dissolved in water in a closed container. Which one of the following plots correctly represents variation of log (partial pressure of CO2_22​ in vapour phase above water) [y-axis] with log (mole fraction of CO2_22​ in water) [x-axis] at 25°C ?
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (C)

Step-by-step solution →
Q32·ChemistrySingle correct
A first row transition metla (M) does not liberate H2_22​ gas from dilute HCl. 1 mol of aqueous solution of MSO4_44​ is treated with excess of aqueous KCN and then H2_22​S(g) is passed through the solution. The amount of MS (metal sulphide) formed from the above reaction is _______ mol.
  1. (A)2
  2. (B)1
  3. (C)3
  4. (D)0

Correct answer: (D)

Step-by-step solution →
Q33·ChemistrySingle correct
The correct order of reactivity of CH3_33​Br in methanol with the following nucleophiles is F−^{-}−, I−^{-}−, C2_22​H5_55​O−^{-}− and C6_66​H5_55​O−^{-}−
  1. (A)I−^{-}− > C6_66​H5_55​O−^{-}− > F−^{-}− > C2_22​H5_55​O−^{-}−
  2. (B)I−^{-}− > C2_22​H5_55​O−^{-}− > C6_66​H5_55​O−^{-}− > F−^{-}−
  3. (C)I−^{-}− > C2_22​H5_55​O−^{-}− > F−^{-}− > C6_66​H5_55​O−^{-}−
  4. (D)I−^{-}− > F−^{-}− > C6_66​H5_55​O−^{-}− > C2_22​H5_55​O−^{-}−

Correct answer: (B)

Step-by-step solution →
Q34·ChemistrySingle correct
Match the LIST-I (Thermodynamic Process) with LIST-II (Magnitude in kJ) Choose the correct answer from the option given below :
List-I (Thermodynamic Process)List-II (Magnitude in kJ)
A.Work done in reversible, isothermal expansion of 2 mol of ideal gas from 2 dm3^33 to 20 dm3^33 at 300 K.I.4
B.Work done in irreversible isothermal expansion of 1 mol ideal gas from 1 m3^33 to 3 m3^33 at 300 K against A constant pressure of 3kPa.II.11.5
C.Change in internal energy for adiabatic expansion of a 1 mol ideal gas with change of temperature = 320 K and C‾V=32R\overline{C}_V = \frac{3}{2}RCV​=23​R.III.6
D.Change in enthalpy at constant pressure of 1 mole ideal gas with change of temperature = 337 K and C‾P=52R\overline{C}_P = \frac{5}{2}RCP​=25​R.IV.7
  1. (A)A-III, B-II, C-IV, D-I
  2. (B)A-II, B-III, C-I, D-IV
  3. (C)A-I, B-II, C-III, D-IV
  4. (D)A-II, B-I, C-III, D-IV

Correct answer: (B)

Step-by-step solution →
Q35·ChemistrySingle correct
Given below are two statements : Statement I : Benzene is nitrated to give nitrobenzene, which on further treatment with CH3_33​COCl / AlCl3_33​ will give Statement II : NO2_22​ group is a m-directing, and deactivating group. In the light of the above statements, choose the most appropriate answer from the options given below.
  1. (A)Statement I is correct but Statement II is incorrect.
  2. (B)Both Statement I and Statement II are correct.
  3. (C)Statement I is incorrect but Statement II is correct.
  4. (D)Both Statement I and Statement II is are incorrect.

Correct answer: (C)

Step-by-step solution →
Q36·ChemistrySingle correct
A → products (First order reaction). Three sets of experiment were performed for a reaction under similar experimental conditions. Run 1 ⇒ 100 mL of 10 M solution of reactant A Run 2 ⇒ 200 mL of 10 M solution of reactant A Run 3 ⇒ 100 mL of 10 M solution of reactant A+ 100 mL of H2_22​O added. The correct variation of rate of reaction is
  1. (A)Run 1 = Run 2 = Run 3
  2. (B)Run 3 < Run 1 = Run 2
  3. (C)Run 3 < Run 1 < Run 2
  4. (D)Run 1 < Run 2 < Run 3

Correct answer: (B)

Step-by-step solution →
Q37·Chemistry·Aldehydes and KetonesSingle correct
Match the LIST-I (Reagents) with LIST-II (Name of Reaction involving carbonyl compound) Choose the correct answer from the options given below
List-I (Reagents)List-II (Name of Reaction involving carbonyl compound)
A.NH2_22​– NH2_22​, KOHI.Tollen's Test
B.Ag(NH3_33​)2_22​OHII.Clemmensen Reduction
C.Aq. CuSO4_44​, Sodium Potassium tartarate, KOHIII.Wolff-Kishner Reduction
D.Zn – Hg, HClIV.Fehling's Test
  1. (A)A-III, B-I, C-IV, D-II
  2. (B)A-II, B-I, C-IV, D-III
  3. (C)A-IV, B-III, C-II, D-I
  4. (D)A-III, B-IV, C-I, D-II

Correct answer: (A)

Step-by-step solution →
Q38·ChemistrySingle correct
Given below are two statements : Statement I : The halogen that makes longest bond with hydrogen in HX, has the smallest covalent radius in its group. Statement II : A group 15 element's hydride EH3_33​ has the lowest boiling point among corresponding hydrides of other group 15 elements. The maximum covalency of that element E is 4. 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)Both Statement I and Statement II are false.
  4. (D)Statement I is true but Statement II is false.

Correct answer: (C)

Step-by-step solution →
Q39·ChemistrySingle correct
Given below are two statements: Statement I : Sucrose is dextrorotatory. However sucrose upon hydrolysis gives a solution having mixture of products. This solution shows laevorotation. Statement II : Hydrolysis of sucrose gives glucose and fructose. Since the laevorotation of glucose is more than the dextrorotation of fructose the resulting solution becomes laevorotatory. 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 false.
  3. (C)Both Statement I and Statement II are true.
  4. (D)Statement I is true but Statement II is false.

Correct answer: (D)

Step-by-step solution →
Q40·Chemistry·Diazonium Salts and ReactionsSingle correct
'A' is a neutral organic compound (M. F : C8_88​H9_99​ON ). On treatment with aqueous Br2_22​/HO(−)^{(-)}(−), 'A' forms a compound 'B' which is soluble in dilute acid. 'B' on treatment with aqueous NaNO2_22​/HCl(0–5°C) produces a compound 'C' which on treatment with CuCN/NaCN produces 'D' Hydrolysis of 'D' produces 'E' which is also obtainable from the hydrolysis of 'A'. 'E' on treatment with acidified KMnO4_44​ produces 'F'. 'F' contains two different types of hydrogen atoms. The structure of 'A' is
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (C)

Step-by-step solution →
Q41·ChemistrySingle correct
The correct order of the rate of reaction of the following reactants with nucleophile by SN_NN​1 mechanism is : (Given : Structure I and II are rigid)
  1. (A)IV < III < II < I
  2. (B)III < I < II < IV
  3. (C)II < I< III < IV
  4. (D)I < II < III < IV

Correct answer: (C)

Step-by-step solution →
Q42·ChemistrySingle correct
Two p-block elements X and Y form fluroides of the type EF3_33​. The fluoride compound XF3_33​ is a Lewis acid and YF3_33​ is a Lewis base. The hybridization of the central atoms of XF3_33​ and YF3_33​ respectively are
  1. (A)Both sp3^33
  2. (B)sp2^22 and sp3^33
  3. (C)sp3^33 and sp2^22
  4. (D)Both sp2^22

Correct answer: (B)

Step-by-step solution →
Q43·ChemistrySingle correct
As compared with chlorocyclohexane, which of the following statements correctly apply to chlorobenzene ? A. The magnitude of negative charge is more on chlorine atoms B. The C – Cl bond has partial double bond character C. C – Cl bond is less polar D. C – Cl bond is longer due to repulsion between delocalised electrons of the aromatic ring and lone pairs of electrons of chlorine. E. The C–Cl bond is formed using sp2^22 hybridised orbital of carbon. Choose the correct answer from the options given below :
  1. (A)A, C and E only
  2. (B)B, C and D only
  3. (C)A, D and E only
  4. (D)B, C and E only

Correct answer: (D)

Step-by-step solution →
Q44·ChemistrySingle correct
Given below are two statements: Statement I : The Henry's law constant KH_HH​ is constant with respect to variations in solution's concentration over the range for which the solutions is ideally dilute. Statement II : KH_HH​ does not differ for the same solute in different solvents. In the light of the above statements, choose the correct answer from the options.
  1. (A)Statement I is false but Statement II is true.
  2. (B)Statement I is true but Statement II is false.
  3. (C)Both Statement I and Statement II are true.
  4. (D)Both Statement I and Statement II are false.

Correct answer: (B)

Step-by-step solution →
Q45·ChemistryNumerical
The cycloalkane (X) on bromination consumes one mole of bromine per mole of (X) and gives the product (Y) in which C:Br ratio is 3 : 1. The percentage of bromine in the product (Y) is ________ %. (Nearest integer) (Given : Molar mass in g mol−1^{-1}−1 H : 1, C : 12, O : 16, Br : 80 )

Correct answer: 66

Step-by-step solution →
Q46·Chemistry·Redox Reactions and ElectrochemistryNumerical
Consider the following electrochemical cell at 298K Pt∣\left|\right.∣HSnO2−_2^-2−​(aq)∣\left|\right.∣Sn(OH)62−_6^{2-}62−​(aq)∣\left|\right.∣Bi2_22​O3_33​(s)∣\left|\right.∣Bi(s). If the reaction quotient at a given time is 106^66, then the cell EMF (Ecell_{cell}cell​) is ________ × 10−1^{-1}−1V (Nearest integer). Given the standard half-cell reduction potential as EBi2O3/Bi,OH−0=−0.44E^{0}_{Bi_2O_3/Bi,OH^-} = -0.44EBi2​O3​/Bi,OH−0​=−0.44 V and ESn(OH)62−/HSnO2−,OH0=−0.90E^{0}_{Sn(OH)_6^{2-}/HSnO_2^-,OH} = -0.90ESn(OH)62−​/HSnO2−​,OH0​=−0.90 V

Correct answer: 4

Step-by-step solution →
Q47·ChemistryNumerical
The temperature at which the rate constants of the given below two gaseous reactions become equal is ________ K. (Nearest integer). X ⟶ Y k1=106e−30000Tk_1 = 10^6 e^{\frac{-30000}{T}}k1​=106eT−30000​ P ⟶ Q k2=104e−24000Tk_2 = 10^4 e^{\frac{-24000}{T}}k2​=104eT−24000​ Given : ln 10 = 2.303

Correct answer: 1303

Step-by-step solution →
Q48·ChemistryNumerical
Sodium fusion extract of an organic compound (Y) with CHCl3_33​ and chlorine water gives violet color to the CHCl3_33​ layer. 0.15g of (Y) gave 0.12 g of the silver halide precipitate in Carius method. Percentage of halogen in the compound (Y) is __________ . (Nearest integer). (Given : molar mass g mol−1^{-1}−1 C : 12, H : 1, Cl : 35.5, Br : 80, I : 127)

Correct answer: 43

Step-by-step solution →
Q49·ChemistryNumerical
Dissociation of a gas A2_22​ takes place according to the following chemical reactions. At equilibrium, the total pressure is 1 bar at 300K. A2_22​(g) ⇌ 2A(g) The standard Gibbs energy of formation of the involved substances has been provided below: Substance | ΔGf∘\Delta G_f^{\circ}ΔGf∘​ / kJ mol−1^{-1}−1 A2_22​ | −100.00 A | −50.832 The degree of dissociation of A2_22​(g) is given by (x × 10−2^{-2}−2)1/2^{1/2}1/2 where x = __________. (Nearest integer). [Given : R = 8 J mol−1^{-1}−1 K−1^{-1}−1, log 2 = 0.3010, log 3 = 0.48]

Correct answer: 33

Step-by-step solution →

Mathematics — JEE Main 22 January 2026 Shift 1

Q50·MathematicsSingle correct
Let AB→=2i^+4j^−5k\overrightarrow{AB} = 2\hat{i} + 4\hat{j} - 5kAB=2i^+4j^​−5k and AD→=i^+2j^+λk\overrightarrow{AD} = \hat{i} + 2\hat{j} + \lambda kAD=i^+2j^​+λk, λ∈R\lambda \in \mathbb{R}λ∈R. Let the projection of the vector v⃗=i^+j^+k^\vec{v} = \hat{i} + \hat{j} + \hat{k}v=i^+j^​+k^ on the diagonal AC→\overrightarrow{AC}AC of the parallelogram ABCD be of length one unit. If α, β, where α > β, be the roots of the equation λ2x2−6λx+5=0\lambda^2 x^2 - 6\lambda x + 5 = 0λ2x2−6λx+5=0, then 2α − β is equal to
  1. (A)1
  2. (B)4
  3. (C)3
  4. (D)6

Correct answer: (C)

Step-by-step solution →
Q51·MathematicsSingle correct
Let the relation R on the set M = {1, 2, 3,.......16} be given by R={(x,y):4y=5x−3, x,y∈M}R = \{(x,y) : 4y = 5x - 3,\ x, y \in M\}R={(x,y):4y=5x−3, x,y∈M}. Then the minimum number of elements required to be added in R, in order to make the relation symmetric, is equal to
  1. (A)1
  2. (B)2
  3. (C)4
  4. (D)3

Correct answer: (B)

Step-by-step solution →
Q52·MathematicsSingle correct
Let the line x = −1 divide the area of the region {(x,y):1+x2≤y≤3−x}\left\{(x,y) : 1 + x^2 \le y \le 3 - x\right\}{(x,y):1+x2≤y≤3−x} in the ratio m : n, gcd (m, n) = 1. Then m + n is equal to
  1. (A)25
  2. (B)28
  3. (C)26
  4. (D)27

Correct answer: (D)

Step-by-step solution →
Q53·MathematicsSingle correct
Two distinct numbers a and b are selected at random from 1, 2, 3,......, 50. The probability, that their product ab is divisible by 3, is
  1. (A)5611225\frac{561}{1225}1225561​
  2. (B)6641225\frac{664}{1225}1225664​
  3. (C)2721225\frac{272}{1225}1225272​
  4. (D)825\frac{8}{25}258​

Correct answer: (B)

Step-by-step solution →
Q54·MathematicsSingle correct
Let f(x)=x2025−x2000f(x) = x^{2025} - x^{2000}f(x)=x2025−x2000, x∈[0,1]x \in [0, 1]x∈[0,1] and the minimum value of the function f(x) in the interval [0, 1] be (80)80(n)−81(80)^{80} (n)^{-81}(80)80(n)−81. Then n is equal to
  1. (A)−81
  2. (B)−40
  3. (C)−41
  4. (D)−80

Correct answer: (A)

Step-by-step solution →
Q55·MathematicsSingle correct
Let P(α, β, γ) be the point on the line x−12=y+1−3=z\frac{x-1}{2} = \frac{y+1}{-3} = z2x−1​=−3y+1​=z at a distance 4144\sqrt{14}414​ from the point (1, −1, 0) and nearer to the origin. Then the shortest distance, between the lines x−α1=y−β2=z−γ3\frac{x-\alpha}{1} = \frac{y-\beta}{2} = \frac{z-\gamma}{3}1x−α​=2y−β​=3z−γ​ and x+52=y−101=z−31\frac{x+5}{2} = \frac{y-10}{1} = \frac{z-3}{1}2x+5​=1y−10​=1z−3​ , is equal to
  1. (A)7547\sqrt{\frac{5}{4}}745​​
  2. (B)4754\sqrt{\frac{7}{5}}457​​
  3. (C)4574\sqrt{\frac{5}{7}}475​​
  4. (D)2742\sqrt{\frac{7}{4}}247​​

Correct answer: (B)

Step-by-step solution →
Q56·MathematicsSingle correct
If a random variable x has the probability distribution x : 0, 1, 2, 3, 4, 5, 6, 7 p(x) : 0, 2k, k, 3k, 2k22k^22k2, 2k, k2+kk^2 + kk2+k, 7k27k^27k2 then P(3<x≤6)P(3 < x \le 6)P(3<x≤6) is equal to
  1. (A)0.34
  2. (B)0.22
  3. (C)0.64
  4. (D)0.33

Correct answer: (D)

Step-by-step solution →
Q57·MathematicsSingle correct
The number of distinct real solutions of the equation x∣x+4∣+3∣x+2∣+10=0x|x + 4| + 3|x + 2| + 10 = 0x∣x+4∣+3∣x+2∣+10=0 is
  1. (A)3
  2. (B)1
  3. (C)0
  4. (D)2

Correct answer: (B)

Step-by-step solution →
Q58·MathematicsSingle correct
Let f:[1,∞)→Rf : [1, \infty) \rightarrow \mathbb{R}f:[1,∞)→R be a differentiable function, If 6∫1xf(t) dt=3xf(x)+x3−46\int_1^x f(t)\,dt = 3xf(x) + x^3 - 46∫1x​f(t)dt=3xf(x)+x3−4 for all x≥1x \ge 1x≥1, then the value of f(2) − f(3) is
  1. (A)−4
  2. (B)−3
  3. (C)4
  4. (D)3

Correct answer: (D)

Step-by-step solution →
Q59·MathematicsSingle correct
If the line αx + 2y = 1, where α ∈ ℝ, does not meet the hyperbola x2−9y2=9x^2 - 9y^2 = 9x2−9y2=9, then a possible value of α is :
  1. (A)0.6
  2. (B)0.8
  3. (C)0.5
  4. (D)0.7

Correct answer: (B)

Step-by-step solution →
Q60·MathematicsSingle correct
If the image of the point P (1, 2 , a) in the line x−63=y−72=7−z2\frac{x-6}{3} = \frac{y-7}{2} = \frac{7-z}{2}3x−6​=2y−7​=27−z​ is Q(5, b, c), then a2+b2+c2a^2 + b^2 + c^2a2+b2+c2 is equal to
  1. (A)293
  2. (B)264
  3. (C)298
  4. (D)283

Correct answer: (C)

Step-by-step solution →
Q61·MathematicsSingle correct
Let the set of all values of r, for which the circles (x+1)2+(y+4)2=r2(x + 1)^2 + (y + 4)^2 = r^2(x+1)2+(y+4)2=r2 and x2+y2−4x−2y−4=0x^2 + y^2 - 4x - 2y - 4 = 0x2+y2−4x−2y−4=0 intersect at two distinct points be the interval (α, β). Then αβ is equal to
  1. (A)25
  2. (B)20
  3. (C)21
  4. (D)24

Correct answer: (A)

Step-by-step solution →
Q62·MathematicsSingle correct
If A=[2335]A = \begin{bmatrix} 2 & 3 \\ 3 & 5 \end{bmatrix}A=[23​35​], then the determinant of the matrix (A2025−3A2024+A2023)(A^{2025} - 3A^{2024} + A^{2023})(A2025−3A2024+A2023) is
  1. (A)28
  2. (B)12
  3. (C)24
  4. (D)16

Correct answer: (D)

Step-by-step solution →
Q63·MathematicsSingle correct
The value of ∫−π2π2(1[x]+4)dx\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\frac{1}{[x]+4}\right)dx∫−2π​2π​​([x]+41​)dx, where [∙][\bullet][∙] denotes the greatest integer function, is
  1. (A)160(21π−1)\frac{1}{60}(21\pi - 1)601​(21π−1)
  2. (B)160(π−7)\frac{1}{60}(\pi - 7)601​(π−7)
  3. (C)760(3π−1)\frac{7}{60}(3\pi - 1)607​(3π−1)
  4. (D)760(π−3)\frac{7}{60}(\pi - 3)607​(π−3)

Correct answer: (C)

Step-by-step solution →
Q64·MathematicsSingle correct
If the chord joining the points P1(x1,y1)P_{1}(x_{1}, y_{1})P1​(x1​,y1​) and P2(x2,y2)P_{2}(x_{2}, y_{2})P2​(x2​,y2​) on the parabola y2=12xy^{2} = 12xy2=12x subtends a right angle at the vertex of the parabola, then x1x2−y1y2x_{1}x_{2} - y_{1}y_{2}x1​x2​−y1​y2​ is equal to
  1. (A)288
  2. (B)280
  3. (C)284
  4. (D)292

Correct answer: (A)

Step-by-step solution →
Q65·MathematicsSingle correct
The number of solutions of tan⁡−14x+tan⁡−16x=π6\tan^{-1} 4x + \tan^{-1}6x = \frac{\pi}{6}tan−14x+tan−16x=6π​, where −126<x<126-\frac{1}{2\sqrt{6}} < x < \frac{1}{2\sqrt{6}}−26​1​<x<26​1​ is equal to
  1. (A)3
  2. (B)0
  3. (C)1
  4. (D)2

Correct answer: (C)

Step-by-step solution →
Q66·MathematicsSingle correct
Let the solution curve of the differential equation xdy−ydx=x2+y2 dxxdy - ydx = \sqrt{x^{2} + y^{2}}\,dxxdy−ydx=x2+y2​dx, x>0x > 0x>0, y(1)=0y(1) = 0y(1)=0, be y=y(x)y = y(x)y=y(x). Then y(3)y(3)y(3) is equal to
  1. (A)4
  2. (B)6
  3. (C)1
  4. (D)2

Correct answer: (A)

Step-by-step solution →
Q67·MathematicsSingle correct
If the sum of the first four terms of an A.P. is 6 and the sum of its first six terms is 4, then the sum of its first twelve terms is
  1. (A)–20
  2. (B)–24
  3. (C)–26
  4. (D)–22

Correct answer: (D)

Step-by-step solution →
Q68·MathematicsNumerical
Let α=−1+i32\alpha = \frac{-1+i\sqrt{3}}{2}α=2−1+i3​​ and β=−1−i32\beta = \frac{-1-i\sqrt{3}}{2}β=2−1−i3​​, i=−1i = \sqrt{-1}i=−1​. If (7−7α+9β)20+(9+7α−7β)20+(−7+9α+7β)20+(14+7α+7β)20=m10(7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = m^{10}(7−7α+9β)20+(9+7α−7β)20+(−7+9α+7β)20+(14+7α+7β)20=m10, then m is _________.

Correct answer: 49

Step-by-step solution →
Q69·MathematicsNumerical
Let A be a 3×33 \times 33×3 matrix such that A+AT=OA + A^{T} = OA+AT=O. If A[1−10]=[332]A\begin{bmatrix}1\\-1\\0\end{bmatrix} = \begin{bmatrix}3\\3\\2\end{bmatrix}A​1−10​​=​332​​, A2[1−10]=[−319−24]A^{2}\begin{bmatrix}1\\-1\\0\end{bmatrix} = \begin{bmatrix}-3\\19\\-24\end{bmatrix}A2​1−10​​=​−319−24​​ and det⁡(adj(2adj(A+I)))=(2)α.(3)β.(11)γ\det(\text{adj}(2\text{adj}(A + I))) = (2)^{\alpha}.(3)^{\beta}.(11)^{\gamma}det(adj(2adj(A+I)))=(2)α.(3)β.(11)γ, α\alphaα, β\betaβ, γ\gammaγ are non-negative integers, then α+β+γ\alpha + \beta + \gammaα+β+γ is equal to ______

Correct answer: 18

Step-by-step solution →
Q70·MathematicsNumerical
If ∫(sin⁡x)−112(cos⁡x)−52dx=−p1q1(cot⁡x)92−p2q2(cot⁡x)52−p3q3(cot⁡x)12+p4q4(cot⁡x)−32+C\int (\sin x)^{\frac{-11}{2}} (\cos x)^{\frac{-5}{2}} dx = -\frac{p_{1}}{q_{1}}(\cot x)^{\frac{9}{2}} - \frac{p_{2}}{q_{2}}(\cot x)^{\frac{5}{2}} - \frac{p_{3}}{q_{3}}(\cot x)^{\frac{1}{2}} + \frac{p_{4}}{q_{4}}(\cot x)^{\frac{-3}{2}} + C∫(sinx)2−11​(cosx)2−5​dx=−q1​p1​​(cotx)29​−q2​p2​​(cotx)25​−q3​p3​​(cotx)21​+q4​p4​​(cotx)2−3​+C, where pip_{i}pi​ and qiq_{i}qi​ are positive integers with gcd⁡(pi,qi)=1\gcd(p_{i}, q_{i}) = 1gcd(pi​,qi​)=1 for i=1,2,3,4i = 1, 2, 3, 4i=1,2,3,4 and C is the constant of integration, then 15p1p2p3p4q1q2q3q4\frac{15p_{1}p_{2}p_{3}p_{4}}{q_{1}q_{2}q_{3}q_{4}}q1​q2​q3​q4​15p1​p2​p3​p4​​ is equal to _______.

Correct answer: 16

Step-by-step solution →
Q71·MathematicsNumerical
If cos⁡248∘−sin⁡212∘sin⁡224∘−sin⁡26∘=α+β52\frac{\cos^{2} 48^{\circ} - \sin^{2} 12^{\circ}}{\sin^{2} 24^{\circ} - \sin^{2} 6^{\circ}} = \frac{\alpha + \beta\sqrt{5}}{2}sin224∘−sin26∘cos248∘−sin212∘​=2α+β5​​, where α\alphaα, β∈N\beta \in \mathbb{N}β∈N, then α+β\alpha + \betaα+β is equal to _______.

Correct answer: 4

Step-by-step solution →
Q72·MathematicsNumerical
Let ABC be a triangle. Consider four points p1p_{1}p1​, p2p_{2}p2​, p3p_{3}p3​, p4p_{4}p4​ on the side AB, five points p5p_{5}p5​, p6p_{6}p6​, p7p_{7}p7​, p8p_{8}p8​, p9p_{9}p9​ on the side BC and four points p10p_{10}p10​, p11p_{11}p11​, p12p_{12}p12​, p13p_{13}p13​ on the side AC. None of these points is a vertex of the triangle ABC. Then the total number of pentagons, that can be formed by taking all the vertices from the points p1p_{1}p1​, p2p_{2}p2​, .... p13p_{13}p13​, is _________.

Correct answer: 660

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
  • Geometrical Optics 172/186
  • Kinematics 156/186
  • Vector Algebra 173/186
  • Differential Equations 167/186
  • Probability 176/186
  • Permutations and Combinations 162/186
  • Electromagnetic Waves 144/186
  • Chemical Bonding and Molecular Structure 151/186
  • Application of Derivatives 139/186
  • Thermodynamics 154/186
  • Aldehydes and Ketones 135/186
  • Equilibrium 163/186
  • Solutions 158/186
  • Chemical Thermodynamics 165/186
  • Units and Measurements 149/186
  • 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
  • Circles 142/186
  • Quadratic Equations 148/186
  • Some Basic Concepts in Chemistry 129/186
  • Electromagnetic Induction 120/186
  • Area Under Curves 139/186
  • Kinetic Theory of Gases 135/186
  • Alcohols and Ethers 106/186
  • Purification and Characterisation of Organic Compounds 116/186
  • Nuclei 116/186
  • Organic Compounds Containing Halogens 109/186
  • Waves 109/186
  • Atoms 112/186
  • Classification of Elements and Periodicity in Properties 113/186
  • Parabola 101/186
  • Inverse Trigonometric Functions 93/186
  • Hyperbola 77/186
  • Electric Potential 63/186
  • Indefinite Integration 66/186
  • Diazonium Salts and Reactions 53/186
← 21 Jan Shift 2 2026All papers22 Jan Shift 2 2026 →

Attempt this paper under exam timing.

Take the 22 January 2026 Shift 1 paper as a timed mock and Jarvis marks it, then tells you which errors were conceptual gaps, which were silly mistakes, and which pattern you have now repeated. Step-by-step solutions for every question included.

Attempt this paper freeSee pricing

Free plan, no time limit · No credit card needed

Jarvis OS

AI-powered JEE preparation.

Question papers

JEE Main PYQsJEE Advanced PYQsPhysics PYQsChemistry PYQsMaths PYQs

Product

TourPricingSign inCreate account

Legal

Privacy PolicyTerms of ServiceRefund & CancellationShipping & DeliveryContact Us

Operated by

Venyou Craft Private Limited

info@jarvisos.net

© 2026 Jarvis OS