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JEE Main 29 January 2024 Shift 2 Question Paper with Answers

29 January 2024 · January session · 90 questions

The complete JEE Main 29 January 2024 Shift 2 paper — all 90 questions with the correct answer for each, tagged to the chapter it tests. Free to read, no account needed.

Physics
30
Chemistry
30
Mathematics
30

Physics — JEE Main 29 January 2024 Shift 2

Q1·Physics·Dual Nature of Matter and RadiationSingle correct
Two sources of light emit with a power of 200 W. The ratio of number of photons of visible light emitted by each source having wavelengths 300 nm and 500 nm respectively, will be:
  1. (A)1 : 5
  2. (B)1 : 3
  3. (C)5 : 3
  4. (D)3 : 5

Correct answer: (D)

Step-by-step solution →
Q2·Physics·Electronic DevicesSingle correct
The truth table for the given circuit is:
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (B)

Step-by-step solution →
Q3·Physics·Units and MeasurementsSingle correct
A physical quantity QQQ is found to depend on quantities aaa, bbb, ccc by the relation Q=a4b3c2Q=\dfrac{a^{4}b^{3}}{c^{2}}Q=c2a4b3​. The percentage error in aaa, bbb and ccc are 3%, 4% and 5% respectively. Then, the percentage error in QQQ is:
  1. (A)66%
  2. (B)43%
  3. (C)34%
  4. (D)14%

Correct answer: (C)

Step-by-step solution →
Q4·Physics·Electromagnetic InductionSingle correct
In an a.c. circuit, voltage and current are given by: V=100sin⁡(100 t)V=100\sin(100\,t)V=100sin(100t) V and I=100sin⁡(100 t+π3)I=100\sin\left(100\,t+\dfrac{\pi}{3}\right)I=100sin(100t+3π​) mA respectively. The average power dissipated in one cycle is:
  1. (A)5 W
  2. (B)10 W
  3. (C)2.5 W
  4. (D)25 W

Correct answer: (C)

Step-by-step solution →
Q5·Physics·Kinetic Theory of GasesSingle correct
The temperature of a gas having 2.0×10252.0\times10^{25}2.0×1025 molecules per cubic meter at 1.38 atm (Given, k=1.38×10−23k=1.38\times10^{-23}k=1.38×10−23 JK−1^{-1}−1) is:
  1. (A)500 K
  2. (B)200 K
  3. (C)100 K
  4. (D)300 K

Correct answer: (A)

Step-by-step solution →
Q6·Physics·Laws of MotionSingle correct
A stone of mass 900 g is tied to a string and moved in a vertical circle of radius 1 m making 10 rpm. The tension in the string, when the stone is at the lowest point is (if π2=9.8\pi^{2}=9.8π2=9.8 and g=9.8g=9.8g=9.8 m/s2^{2}2):
  1. (A)97 N
  2. (B)9.8 N
  3. (C)8.82 N
  4. (D)17.8 N

Correct answer: (B)

Step-by-step solution →
Q7·Physics·Work, Energy and PowerSingle correct
The bob of a pendulum was released from a horizontal position. The length of the pendulum is 10 m. If it dissipates 10% of its initial energy against air resistance, the speed with which the bob arrives at the lowest point is: [Use g=10g=10g=10 ms−2^{-2}−2]
  1. (A)656\sqrt{5}65​ ms−1^{-1}−1
  2. (B)565\sqrt{6}56​ ms−1^{-1}−1
  3. (C)555\sqrt{5}55​ ms−1^{-1}−1
  4. (D)252\sqrt{5}25​ ms−1^{-1}−1

Correct answer: (A)

Step-by-step solution →
Q8·Physics·Geometrical OpticsSingle correct
If the distance between object and its two times magnified virtual image produced by a curved mirror is 15 cm, the focal length of the mirror must be:
  1. (A)15 cm
  2. (B)−12-12−12 cm
  3. (C)−10-10−10 cm
  4. (D)103\dfrac{10}{3}310​ cm

Correct answer: (C)

Step-by-step solution →
Q9·Physics·Magnetic Field of CurrentSingle correct
Two particles X and Y having equal charges are being accelerated through the same potential difference. Thereafter they enter normally in a region of uniform magnetic field and describe circular paths of radii R1R_{1}R1​ and R2R_{2}R2​ respectively. The mass ratio of X and Y is:
  1. (A)(R2R1)2\left(\dfrac{R_{2}}{R_{1}}\right)^{2}(R1​R2​​)2
  2. (B)(R1R2)2\left(\dfrac{R_{1}}{R_{2}}\right)^{2}(R2​R1​​)2
  3. (C)(R1R2)\left(\dfrac{R_{1}}{R_{2}}\right)(R2​R1​​)
  4. (D)(R2R1)\left(\dfrac{R_{2}}{R_{1}}\right)(R1​R2​​)

Correct answer: (B)

Step-by-step solution →
Q10·Physics·Wave OpticsSingle correct
In Young's double slit experiment, light from two identical sources are superimposing on a screen. The path difference between the two lights reaching at a point on the screen is 7λ4\dfrac{7\lambda}{4}47λ​. The ratio of intensity of fringe at this point with respect to the maximum intensity of the fringe is:
  1. (A)12\dfrac{1}{2}21​
  2. (B)34\dfrac{3}{4}43​
  3. (C)13\dfrac{1}{3}31​
  4. (D)14\dfrac{1}{4}41​

Correct answer: (A)

Step-by-step solution →
Q11·Physics·Properties of Solids and LiquidsSingle correct
A small liquid drop of radius R is divided into 27 identical liquid drops. If the surface tension is T, then the work done in the process will be:
  1. (A)8πR2T8\pi R^{2}T8πR2T
  2. (B)3πR2T3\pi R^{2}T3πR2T
  3. (C)18πR2T\dfrac{1}{8}\pi R^{2}T81​πR2T
  4. (D)4πR2T4\pi R^{2}T4πR2T

Correct answer: (A)

Step-by-step solution →
Q12·Physics·Work, Energy and PowerSingle correct
A bob of mass 'm' is suspended by a light string of length 'L'. It is imparted a minimum horizontal velocity at the lowest point A such that it just completes half circle reaching the top most position B. The ratio of kinetic energies (K.E.)A(K.E.)B\dfrac{(K.E.)_{A}}{(K.E.)_{B}}(K.E.)B​(K.E.)A​​ is:
  1. (A)3 : 2
  2. (B)5 : 1
  3. (C)2 : 5
  4. (D)1 : 5

Correct answer: (B)

Step-by-step solution →
Q13·Physics·Properties of Solids and LiquidsSingle correct
A wire of length L and radius r is clamped at one end. If its other end is pulled by a force F, its length increases by lll. If the radius of the wire and the applied force both are reduced to half of their original values keeping original length constant, the increase in length will become:
  1. (A)3 times
  2. (B)32\dfrac{3}{2}23​ times
  3. (C)4 times
  4. (D)2 times

Correct answer: (D)

Step-by-step solution →
Q14·Physics·GravitationSingle correct
A planet takes 200 days to complete one revolution around the Sun. If the distance of the planet from Sun is reduced to one fourth of the original distance, how many days will it take to complete one revolution?
  1. (A)25
  2. (B)50
  3. (C)100
  4. (D)20

Correct answer: (A)

Step-by-step solution →
Q15·Physics·Electromagnetic WavesSingle correct
A plane electromagnetic wave of frequency 35 MHz travels in free space along the X-direction. At a particular point (in space and time) E⃗=9.6 j^\vec{E}=9.6\,\hat{j}E=9.6j^​ V/m. The value of magnetic field at this point is:
  1. (A)3.2×10−8 k^3.2\times10^{-8}\,\hat{k}3.2×10−8k^ T
  2. (B)3.2×10−8 i^3.2\times10^{-8}\,\hat{i}3.2×10−8i^ T
  3. (C)9.6 j^9.6\,\hat{j}9.6j^​ T
  4. (D)9.6×10−8 k^9.6\times10^{-8}\,\hat{k}9.6×10−8k^ T

Correct answer: (A)

Step-by-step solution →
Q16·Physics·Current ElectricitySingle correct
In the given circuit, the current in resistance R3R_{3}R3​ is:
  1. (A)1 A
  2. (B)1.5 A
  3. (C)2 A
  4. (D)2.5 A

Correct answer: (A)

Step-by-step solution →
Q17·Physics·KinematicsSingle correct
A particle is moving in a straight line. The variation of position 'x' as a function of time 't' is given as x=(t3−6t2+20t+15)x=(t^{3}-6t^{2}+20t+15)x=(t3−6t2+20t+15) m. The velocity of the body when its acceleration becomes zero is:
  1. (A)4 m/s
  2. (B)8 m/s
  3. (C)10 m/s
  4. (D)6 m/s

Correct answer: (B)

Step-by-step solution →
Q18·Physics·Kinetic Theory of GasesSingle correct
N moles of a polyatomic gas (f=6f=6f=6) must be mixed with two moles of a monoatomic gas so that the mixture behaves as a diatomic gas. The value of N is:
  1. (A)6
  2. (B)3
  3. (C)4
  4. (D)2

Correct answer: (C)

Step-by-step solution →
Q19·Physics·Atoms and NucleiSingle correct
Given below are two statements: Statement I: Most of the mass of the atom and all its positive charge are concentrated in a tiny nucleus and the electrons revolve around it, is Rutherford’s model. Statement II: An atom is a spherical cloud of positive charges with electrons embedded in it, is a special case of Rutherford’s model. In the light of the above statements, choose the most appropriate from the options given below.
  1. (A)Both statement I and statement II are false
  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 true

Correct answer: (C)

Step-by-step solution →
Q20·Physics·Electric Field and Coulomb's LawSingle correct
An electric field is given by (6i^+5j^+3k^)(6\hat{i}+5\hat{j}+3\hat{k})(6i^+5j^​+3k^) N/C. The electric flux through a surface area 30i^30\hat{i}30i^ m2^{2}2 lying in YZ-plane (in SI unit) is:
  1. (A)90
  2. (B)150
  3. (C)180
  4. (D)60

Correct answer: (C)

Step-by-step solution →
Q21·Physics·Properties of Solids and LiquidsNumerical
Two metallic wires P and Q have same volume and are made up of same material. If their area of cross sections are in the ratio 4 : 1 and force F1F_{1}F1​ is applied to P, an extension of Δl\Delta lΔl is produced. The force which is required to produce same extension in Q is F2F_{2}F2​. The value of F1F2\dfrac{F_{1}}{F_{2}}F2​F1​​ is ___.

Correct answer: 16

Step-by-step solution →
Q22·Physics·Electromagnetic InductionNumerical
A horizontal straight wire 5 m long extending from east to west is falling freely at right angle to horizontal component of earth's magnetic field 0.60×10−40.60\times10^{-4}0.60×10−4 Wbm−2^{-2}−2. The instantaneous value of emf induced in the wire when its velocity is 10 ms−1^{-1}−1 is ___ ×10−3\times10^{-3}×10−3 V.

Correct answer: 3

Step-by-step solution →
Q23·Physics·Atoms and NucleiNumerical
Hydrogen atom is bombarded with electrons accelerated through a potential difference of V, which causes excitation of hydrogen atoms. If the experiment is being performed at T=0T=0T=0 K, the minimum potential difference needed to observe any Balmer series lines in the emission spectra will be α10\dfrac{\alpha}{10}10α​ V, where α=\alpha=α= ___.

Correct answer: 121

Step-by-step solution →
Q24·Physics·Magnetic Field of CurrentNumerical
A charge of 4.0 μ\muμC is moving with a velocity of 4.0×1064.0\times10^{6}4.0×106 ms−1^{-1}−1 along the positive y-axis under a magnetic field B⃗\vec{B}B of strength (2k^)(2\hat{k})(2k^) T. The force acting on the charge is xi^x\hat{i}xi^ N. The value of xxx is ___.

Correct answer: 32

Step-by-step solution →
Q25·Physics·OscillationsNumerical
A simple harmonic oscillator has an amplitude A and time period 6π6\pi6π second. Assuming the oscillation starts from its mean position, the time required by it to travel from x=Ax=Ax=A to x=32Ax=\dfrac{\sqrt{3}}{2}Ax=23​​A will be πx\dfrac{\pi}{x}xπ​ s, where x=x=x= ___.

Correct answer: 2

Step-by-step solution →
Q26·Physics·Capacitors and DielectricsNumerical
In the given figure, the charge stored in 6 μ\muμF capacitor, when points A and B are joined by a connecting wire is ___ μ\muμC.

Correct answer: 36

Step-by-step solution →
Q27·Physics·Wave OpticsNumerical
In a single slit diffraction pattern, a light of wavelength 6000 Å is used. The distance between the first and third minima in the diffraction pattern is found to be 3 mm when the screen is placed 50 cm away from slits. The width of the slit is ___ ×10−4\times10^{-4}×10−4 m.

Correct answer: 2

Step-by-step solution →
Q28·Physics·Current ElectricityNumerical
In the given circuit, the current flowing through the resistance 20 Ω\OmegaΩ is 0.3 A, while the ammeter reads 0.9 A. The value of R1R_{1}R1​ is ___ Ω\OmegaΩ.

Correct answer: 30

Step-by-step solution →
Q29·Physics·Laws of MotionNumerical
A particle is moving in a circle of radius 50 cm in such a way that at any instant the normal and tangential components of its acceleration are equal. If its speed at t=0t=0t=0 is 4 m/s, the time taken to complete the first revolution will be 1α[1−e−2π]\dfrac{1}{\alpha}\left[1-e^{-2\pi}\right]α1​[1−e−2π] s, where α=\alpha=α= ___.

Correct answer: 8

Step-by-step solution →
Q30·Physics·Rotational MotionNumerical
A body of mass 5 kg is moving with a uniform speed 323\sqrt{2}32​ ms−1^{-1}−1 in X–Y plane along the line y=x+4y=x+4y=x+4. The angular momentum of the particle about the origin will be ___ kg m2^{2}2s−1^{-1}−1.

Correct answer: 60

Step-by-step solution →

Chemistry — JEE Main 29 January 2024 Shift 2

Q31·Chemistry·Electronic Effects and StabilitySingle correct
The ascending acidity order of the following H atoms is:
  1. (A)C < D < B < A
  2. (B)A < B < C < D
  3. (C)A < B < D < C
  4. (D)D < C < B < A

Correct answer: (A)

Step-by-step solution →
Q32·Chemistry·BiomoleculesSingle correct
Match List I with List II. Choose the correct answer from the options given below:
List I (Bio Polymer)List II (Monomer)
A.StarchI.nucleotide
B.CelluloseII.α\alphaα-glucose
C.Nucleic acidIII.β\betaβ-glucose
D.ProteinIV.α\alphaα-amino acid
  1. (A)A-II, B-I, C-III, D-IV
  2. (B)A-IV, B-II, C-I, D-III
  3. (C)A-I, B-III, C-II, D-IV
  4. (D)A-II, B-III, C-I, D-IV

Correct answer: (D)

Step-by-step solution →
Q33·Chemistry·Aldehydes and KetonesSingle correct
Match List I with List II. Choose the correct answer from the options given below:
List I (Compound)List II (pKa value)
A.EthanolI.10.0
B.PhenolII.15.9
C.m-NitrophenolIII.7.1
D.p-NitrophenolIV.8.3
  1. (A)A-I, B-II, C-III, D-IV
  2. (B)A-IV, B-I, C-II, D-III
  3. (C)A-III, B-IV, C-I, D-II
  4. (D)A-II, B-I, C-IV, D-III

Correct answer: (D)

Step-by-step solution →
Q34·Chemistry·AminesSingle correct
Which of the following reaction is correct?
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (B)

Step-by-step solution →
Q35·Chemistry·IUPAC NomenclatureSingle correct
According to IUPAC system, the compound is named as:
  1. (A)Cyclohex-1-en-2-ol
  2. (B)1-Hydroxyhex-2-ene
  3. (C)Cyclohex-1-en-3-ol
  4. (D)Cyclohex-2-en-1-ol

Correct answer: (D)

Step-by-step solution →
Q36·Chemistry·d- and f-Block ElementsSingle correct
The correct IUPAC name of K2MnO4K_2MnO_4K2​MnO4​ is:
  1. (A)Potassium tetraoxopermanganate (VI)
  2. (B)Potassium tetraoxidomanganate (VI)
  3. (C)Dipotassium tetraoxidomanganate (VII)
  4. (D)Potassium tetraoxidomanganese (VI)

Correct answer: (B)

Step-by-step solution →
Q37·Chemistry·Coordination CompoundsSingle correct
A reagent which gives brilliant red precipitate with Nickel ions in basic medium is:
  1. (A)sodium nitroprusside
  2. (B)neutral FeCl3
  3. (C)meta-dinitrobenzene
  4. (D)dimethyl glyoxime

Correct answer: (D)

Step-by-step solution →
Q38·Chemistry·Alcohols and EthersSingle correct
Phenol treated with chloroform in presence of sodium hydroxide, which further hydrolysed in presence of an acid results:
  1. (A)Salicylic acid
  2. (B)Benzene-1,2-diol
  3. (C)Benzene-1,3-diol
  4. (D)2-Hydroxybenzaldehyde

Correct answer: (D)

Step-by-step solution →
Q39·Chemistry·Atomic StructureSingle correct
Match List I with List II. Choose the correct answer from the options given below:
List I (Spectral Series for Hydrogen)List II (Spectral Region)
A.LymanI.Infrared region
B.BalmerII.UV region
C.PaschenIII.Infrared region
D.PfundIV.Visible region
  1. (A)A-II, B-III, C-I, D-IV
  2. (B)A-I, B-III, C-II, D-IV
  3. (C)A-II, B-IV, C-III, D-I
  4. (D)A-I, B-II, C-III, D-IV

Correct answer: (C)

Step-by-step solution →
Q40·Chemistry·p-Block ElementsSingle correct
On passing a gas 'X' through Nessler's reagent, a brown precipitate is obtained. The gas 'X' is:
  1. (A)H2SH_2SH2​S
  2. (B)CO2CO_2CO2​
  3. (C)NH3NH_3NH3​
  4. (D)Cl2Cl_2Cl2​

Correct answer: (C)

Step-by-step solution →
Q41·Chemistry·AminesSingle correct
The product A formed in the following reaction is:
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (C)

Step-by-step solution →
Q42·Chemistry·Aldehydes and KetonesSingle correct
Identify the reagents used for the following conversion:
  1. (A)A = LiAlH4, B = NaOH(aq), C = NH2-NH2/KOH, ethylene glycol
  2. (B)A = LiAlH4, B = NaOH(alc), C = Zn/HCl
  3. (C)A = DIBAL-H, B = NaOH(aq), C = NH2-NH2/KOH, ethylene glycol
  4. (D)A = DIBAL-H, B = NaOH(alc), C = Zn/HCl

Correct answer: (D)

Step-by-step solution →
Q43·Chemistry·d- and f-Block ElementsSingle correct
Which of the following acts as a strong reducing agent? (Atomic number: Ce = 58, Eu = 63, Gd = 64, Lu = 71)
  1. (A)Lu3+Lu^{3+}Lu3+
  2. (B)Gd3+Gd^{3+}Gd3+
  3. (C)Eu2+Eu^{2+}Eu2+
  4. (D)Ce4+Ce^{4+}Ce4+

Correct answer: (C)

Step-by-step solution →
Q44·Chemistry·Purification and Characterisation of Organic CompoundsSingle correct
Chromatographic technique/s based on the principle of differential adsorption is/are: A. Column chromatography B. Thin layer chromatography C. Paper chromatography. Choose the most appropriate answer from the options given below:
  1. (A)B only
  2. (B)A only
  3. (C)A & B only
  4. (D)C only

Correct answer: (C)

Step-by-step solution →
Q45·Chemistry·d- and f-Block ElementsSingle correct
Which of the following statements are correct about Zn, Cd and Hg? A. They exhibit high enthalpy of atomization as the d-subshell is full. B. Zn and Cd do not show variable oxidation state while Hg shows +I and +II. C. Compounds of Zn, Cd and Hg are paramagnetic in nature. D. Zn, Cd and Hg are called soft metals. Choose the most appropriate from the options given below:
  1. (A)B, D only
  2. (B)B, C only
  3. (C)A, D only
  4. (D)C, D only

Correct answer: (A)

Step-by-step solution →
Q46·Chemistry·Classification of Elements and Periodicity in PropertiesSingle correct
The element having the highest first ionization enthalpy is:
  1. (A)Si
  2. (B)Al
  3. (C)N
  4. (D)C

Correct answer: (C)

Step-by-step solution →
Q47·Chemistry·Organic Compounds Containing HalogensSingle correct
Alkyl halide is converted into alkyl isocyanide by reaction with:
  1. (A)NaCN
  2. (B)NH4CN
  3. (C)KCN
  4. (D)AgCN

Correct answer: (D)

Step-by-step solution →
Q48·Chemistry·IsomerismSingle correct
Which one of the following will show geometrical isomerism?
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (C)

Step-by-step solution →
Q49·Chemistry·Classification of Elements and Periodicity in PropertiesSingle correct
Given below are two statements: Statement I: Fluorine has most negative electron gain enthalpy in its group. Statement II: Oxygen has least negative electron gain enthalpy in its group. In the light of the above statements, choose the most appropriate 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: (D)

Step-by-step solution →
Q50·Chemistry·p-Block ElementsSingle correct
Anomalous behaviour of oxygen is due to its:
  1. (A)Large size and high electronegativity
  2. (B)Small size and low electronegativity
  3. (C)Small size and high electronegativity
  4. (D)Large size and low electronegativity

Correct answer: (C)

Step-by-step solution →
Q51·Chemistry·Chemical Bonding and Molecular StructureNumerical
The total number of anti bonding molecular orbitals, formed from 2s and 2p atomic orbitals in a diatomic molecule is ___.

Correct answer: 4

Step-by-step solution →
Q52·Chemistry·p-Block ElementsNumerical
The oxidation number of iron in the compound formed during brown ring test for NO3−NO_3^-NO3−​ ion is ___.

Correct answer: 1

Step-by-step solution →
Q53·Chemistry·EquilibriumNumerical
The following concentrations were observed at 500 K for the formation of NH3NH_3NH3​ from N2N_2N2​ and H2H_2H2​. At equilibrium [N2]=2×10−2[N_2]=2\times10^{-2}[N2​]=2×10−2 M, [H2]=3×10−2[H_2]=3\times10^{-2}[H2​]=3×10−2 M and [NH3]=1.5×10−3[NH_3]=1.5\times10^{-3}[NH3​]=1.5×10−3 M. Equilibrium constant for the reaction is ___ ×10−2\times10^{-2}×10−2.

Correct answer: 417

Step-by-step solution →
Q54·Chemistry·SolutionsNumerical
Molality of 0.8 M H2SO4H_2SO_4H2​SO4​ solution (density 1.06 g cm−3^{-3}−3) is ___ ×10−3\times10^{-3}×10−3 m.

Correct answer: 815

Step-by-step solution →
Q55·Chemistry·Some Basic Concepts in ChemistryNumerical
If 50 mL of 0.5 M oxalic acid is required to neutralise 25 mL of NaOH solution, the amount of NaOH in 50 mL of given NaOH solution is ___ g.

Correct answer: 4

Step-by-step solution →
Q56·Chemistry·Chemical Bonding and Molecular StructureNumerical
The total number of ‘Sigma’ and Pi bonds in 2-formylhex-4-enoic acid is ___.

Correct answer: 22

Step-by-step solution →
Q57·Chemistry·Chemical KineticsNumerical
The half-life of radioisotope bromine-82 is 36 hours. The fraction which remains after one day is ___ ×10−2\times10^{-2}×10−2. (Given antilog 0.2006 = 1.587)

Correct answer: 63

Step-by-step solution →
Q58·Chemistry·Chemical ThermodynamicsNumerical
Standard enthalpy of vapourisation for CCl4CCl_4CCl4​ is 30.5 kJ mol−1^{-1}−1. Heat required for vapourisation of 284 g of CCl4CCl_4CCl4​ at constant temperature is ___ kJ. (Given molar mass in g mol−1^{-1}−1: C = 12, Cl = 35.5)

Correct answer: 56

Step-by-step solution →
Q59·Chemistry·Redox Reactions and ElectrochemistryNumerical
A constant current was passed through a solution of AuCl4−AuCl_4^-AuCl4−​ ion between gold electrodes. After a period of 10.0 minutes, the increase in mass of cathode was 1.314 g. The total charge passed through the solution is ___ ×10−2\times10^{-2}×10−2 F. (Given atomic mass of Au = 197)

Correct answer: 2

Step-by-step solution →
Q60·Chemistry·Chemical Bonding and Molecular StructureNumerical
The total number of molecules with zero dipole moment among CH4CH_4CH4​, BF3BF_3BF3​, H2OH_2OH2​O, HFHFHF, NH3NH_3NH3​, CO2CO_2CO2​ and SO2SO_2SO2​ is ___.

Correct answer: 3

Step-by-step solution →

Mathematics — JEE Main 29 January 2024 Shift 2

Q61·Mathematics·Matrices and DeterminantsSingle correct
Let A=[2126211332]A=\begin{bmatrix}2 & 1 & 2\\ 6 & 2 & 11\\ 3 & 3 & 2\end{bmatrix}A=​263​123​2112​​ and P=[120502715]P=\begin{bmatrix}1 & 2 & 0\\ 5 & 0 & 2\\ 7 & 1 & 5\end{bmatrix}P=​157​201​025​​. The sum of the prime factors of ∣P−1AP−2I∣|P^{-1}AP-2I|∣P−1AP−2I∣ is equal to:
  1. (A)26
  2. (B)27
  3. (C)66
  4. (D)23

Correct answer: (A)

Step-by-step solution →
Q62·Mathematics·Permutations and CombinationsSingle correct
Number of ways of arranging 8 identical books into 4 identical shelves where any number of shelves may remain empty is equal to:
  1. (A)18
  2. (B)16
  3. (C)12
  4. (D)15

Correct answer: (D)

Step-by-step solution →
Q63·Mathematics·Three Dimensional GeometrySingle correct
Let P(3,2,3)P(3,2,3)P(3,2,3), Q(4,6,2)Q(4,6,2)Q(4,6,2) and R(7,3,2)R(7,3,2)R(7,3,2) be the vertices of △PQR\triangle PQR△PQR. Then, the angle ∠QPR\angle QPR∠QPR is:
  1. (A)π6\dfrac{\pi}{6}6π​
  2. (B)cos⁡−1(718)\cos^{-1}\left(\dfrac{7}{18}\right)cos−1(187​)
  3. (C)cos⁡−1(118)\cos^{-1}\left(\dfrac{1}{18}\right)cos−1(181​)
  4. (D)π3\dfrac{\pi}{3}3π​

Correct answer: (D)

Step-by-step solution →
Q64·Mathematics·Statistics and ProbabilitySingle correct
If the mean and variance of five observations are 245\dfrac{24}{5}524​ and 19425\dfrac{194}{25}25194​ respectively and the mean of first four observations is 72\dfrac{7}{2}27​, then the variance of the first four observations is equal to:
  1. (A)45\dfrac{4}{5}54​
  2. (B)7712\dfrac{77}{12}1277​
  3. (C)54\dfrac{5}{4}45​
  4. (D)1054\dfrac{105}{4}4105​

Correct answer: (C)

Step-by-step solution →
Q65·Mathematics·Limits and ContinuitySingle correct
The function f(x)=2x+3(x)23, x∈Rf(x)=2x+3(x)^{\frac{2}{3}},\ x\in\mathbb{R}f(x)=2x+3(x)32​, x∈R, has:
  1. (A)exactly one point of local minima and no point of local maxima
  2. (B)exactly one point of local maxima and no point of local minima
  3. (C)exactly one point of local maxima and exactly one point of local minima
  4. (D)exactly two points of local maxima and exactly one point of local minima

Correct answer: (C)

Step-by-step solution →
Q66·Mathematics·Complex NumbersSingle correct
Let rrr and θ\thetaθ respectively be the modulus and amplitude of the complex number z=2−i(2tan⁡5π8)z=2-i\left(2\tan\dfrac{5\pi}{8}\right)z=2−i(2tan85π​), then (r,θ)(r,\theta)(r,θ) is equal to:
  1. (A)(2sec⁡3π8,3π8)\left(2\sec\dfrac{3\pi}{8},\dfrac{3\pi}{8}\right)(2sec83π​,83π​)
  2. (B)(2sec⁡3π8,5π8)\left(2\sec\dfrac{3\pi}{8},\dfrac{5\pi}{8}\right)(2sec83π​,85π​)
  3. (C)(2sec⁡5π8,3π8)\left(2\sec\dfrac{5\pi}{8},\dfrac{3\pi}{8}\right)(2sec85π​,83π​)
  4. (D)(2sec⁡11π8,11π8)\left(2\sec\dfrac{11\pi}{8},\dfrac{11\pi}{8}\right)(2sec811π​,811π​)

Correct answer: (A)

Step-by-step solution →
Q67·Mathematics·Trigonometric FunctionsSingle correct
The sum of the solutions x∈Rx\in\mathbb{R}x∈R of the equation 3cos⁡2x+cos⁡32xcos⁡6x−sin⁡6x=x3−x2+6\dfrac{3\cos 2x+\cos^3 2x}{\cos^6 x-\sin^6 x}=x^3-x^2+6cos6x−sin6x3cos2x+cos32x​=x3−x2+6 is:
  1. (A)0
  2. (B)1
  3. (C)−1-1−1
  4. (D)3

Correct answer: (C)

Step-by-step solution →
Q68·Mathematics·Vector AlgebraSingle correct
Let OA→=a⃗, OB→=12a⃗+4b⃗\overrightarrow{OA}=\vec a,\ \overrightarrow{OB}=12\vec a+4\vec bOA=a, OB=12a+4b and OC→=b⃗\overrightarrow{OC}=\vec bOC=b, where OOO is the origin. If SSS is the parallelogram with adjacent sides OAOAOA and OCOCOC, then area of the quadrilateral OABCarea of S\dfrac{\text{area of the quadrilateral }OABC}{\text{area of }S}area of Sarea of the quadrilateral OABC​ is equal to:
  1. (A)6
  2. (B)10
  3. (C)7
  4. (D)8

Correct answer: (D)

Step-by-step solution →
Q69·Mathematics·Sequence and SeriesSingle correct
If log⁡ea\log_e aloge​a, log⁡eb\log_e bloge​b, log⁡ec\log_e cloge​c are in an A.P. and log⁡ea−log⁡e2b\log_e a-\log_e 2bloge​a−loge​2b, log⁡e2b−log⁡e3c\log_e 2b-\log_e 3cloge​2b−loge​3c, log⁡e3c−log⁡ea\log_e 3c-\log_e aloge​3c−loge​a are also in an A.P., then a:b:ca:b:ca:b:c is equal to:
  1. (A)9:6:49:6:49:6:4
  2. (B)16:4:116:4:116:4:1
  3. (C)25:10:425:10:425:10:4
  4. (D)6:3:26:3:26:3:2

Correct answer: (A)

Step-by-step solution →
Q70·Mathematics·Indefinite IntegrationSingle correct
If ∫sin⁡32x+cos⁡32xsin⁡3x cos⁡3x sin⁡(x−θ) dx=Acos⁡θtan⁡x−sin⁡θ+Bcos⁡θ−sin⁡θcot⁡x+C\displaystyle\int\dfrac{\sin^{\frac{3}{2}}x+\cos^{\frac{3}{2}}x}{\sqrt{\sin^3 x\,\cos^3 x\,\sin(x-\theta)}}\,dx=A\sqrt{\cos\theta\tan x-\sin\theta}+B\sqrt{\cos\theta-\sin\theta\cot x}+C∫sin3xcos3xsin(x−θ)​sin23​x+cos23​x​dx=Acosθtanx−sinθ​+Bcosθ−sinθcotx​+C, where CCC is the integration constant, then ABABAB is equal to:
  1. (A)4 cosec(2θ)4\,\mathrm{cosec}(2\theta)4cosec(2θ)
  2. (B)4sec⁡θ4\sec\theta4secθ
  3. (C)2sec⁡θ2\sec\theta2secθ
  4. (D)8 cosec(2θ)8\,\mathrm{cosec}(2\theta)8cosec(2θ)

Correct answer: (D)

Step-by-step solution →
Q71·Mathematics·Straight LinesSingle correct
The distance of the point (2,3)(2,3)(2,3) from the line 2x−3y+28=02x-3y+28=02x−3y+28=0, measured parallel to the line 3x−y+1=0\sqrt{3}x-y+1=03​x−y+1=0, is equal to:
  1. (A)424\sqrt{2}42​
  2. (B)636\sqrt{3}63​
  3. (C)3+423+4\sqrt{2}3+42​
  4. (D)4+634+6\sqrt{3}4+63​

Correct answer: (D)

Step-by-step solution →
Q72·Mathematics·Differential EquationsSingle correct
If sin⁡(yx)=log⁡e∣x∣+α2\sin\left(\dfrac{y}{x}\right)=\log_e|x|+\dfrac{\alpha}{2}sin(xy​)=loge​∣x∣+2α​ is the solution of the differential equation xcos⁡(yx)dydx=ycos⁡(yx)+xx\cos\left(\dfrac{y}{x}\right)\dfrac{dy}{dx}=y\cos\left(\dfrac{y}{x}\right)+xxcos(xy​)dxdy​=ycos(xy​)+x and y(1)=π3y(1)=\dfrac{\pi}{3}y(1)=3π​, then α2\alpha^2α2 is equal to:
  1. (A)3
  2. (B)12
  3. (C)4
  4. (D)9

Correct answer: (A)

Step-by-step solution →
Q73·Mathematics·Sequence and SeriesSingle correct
If each term of a geometric progression a1,a2,a3,…a_1,a_2,a_3,\ldotsa1​,a2​,a3​,… with a1=18a_1=\dfrac{1}{8}a1​=81​ and a2≠a1a_2\neq a_1a2​=a1​, is the arithmetic mean of the next two terms and Sn=a1+a2+…+anS_n=a_1+a_2+\ldots+a_nSn​=a1​+a2​+…+an​, then S20−S18S_{20}-S_{18}S20​−S18​ is equal to:
  1. (A)2152^{15}215
  2. (B)−218-2^{18}−218
  3. (C)2182^{18}218
  4. (D)−215-2^{15}−215

Correct answer: (D)

Step-by-step solution →
Q74·Mathematics·Straight LinesSingle correct
Let AAA be the point of intersection of the lines 3x+2y=143x+2y=143x+2y=14, 5x−y=65x-y=65x−y=6 and BBB be the point of intersection of the lines 4x+3y=84x+3y=84x+3y=8, 6x+y=56x+y=56x+y=5. The distance of the point P(5,−2)P(5,-2)P(5,−2) from the line ABABAB is:
  1. (A)132\dfrac{13}{2}213​
  2. (B)8
  3. (C)52\dfrac{5}{2}25​
  4. (D)6

Correct answer: (D)

Step-by-step solution →
Q75·Mathematics·Inverse Trigonometric FunctionsSingle correct
Let x=mnx=\dfrac{m}{n}x=nm​ (mmm, nnn are co-prime natural numbers) be a solution of the equation cos⁡(2sin⁡−1x)=19\cos\left(2\sin^{-1}x\right)=\dfrac{1}{9}cos(2sin−1x)=91​ and let α,β (α>β)\alpha,\beta\,(\alpha>\beta)α,β(α>β) be the roots of the equation mx2−nx−m+n=0mx^2-nx-m+n=0mx2−nx−m+n=0. Then the point (α,β)(\alpha,\beta)(α,β) lies on the line:
  1. (A)3x+2y=23x+2y=23x+2y=2
  2. (B)5x−8y=−95x-8y=-95x−8y=−9
  3. (C)3x−2y=−23x-2y=-23x−2y=−2
  4. (D)5x+8y=95x+8y=95x+8y=9

Correct answer: (D)

Step-by-step solution →
Q76·Mathematics·Limits and ContinuitySingle correct
The function f(x)=xx2−6x−16, x∈R−{−2,8}f(x)=\dfrac{x}{x^2-6x-16},\ x\in\mathbb{R}-\{-2,8\}f(x)=x2−6x−16x​, x∈R−{−2,8}:
  1. (A)increasing in (−2,8)(-2,8)(−2,8) and decreasing in (−∞,−2)∪(8,∞)(-\infty,-2)\cup(8,\infty)(−∞,−2)∪(8,∞)
  2. (B)decreasing in (−∞,−2)∪(−2,8)(-\infty,-2)\cup(-2,8)(−∞,−2)∪(−2,8) and increasing in (8,∞)(8,\infty)(8,∞)
  3. (C)decreasing in (−∞,−2)∪(8,∞)(-\infty,-2)\cup(8,\infty)(−∞,−2)∪(8,∞) and increasing in (−2,8)(-2,8)(−2,8)
  4. (D)decreasing in R−{−2,8}\mathbb{R}-\{-2,8\}R−{−2,8}

Correct answer: (D)

Step-by-step solution →
Q77·Mathematics·Limits and ContinuitySingle correct
Let y=log⁡e(1−x21+x2), −1<x<1y=\log_e\left(\dfrac{1-x^2}{1+x^2}\right),\ -1<x<1y=loge​(1+x21−x2​), −1<x<1. Then at x=12x=\dfrac{1}{2}x=21​, the value of 225(y′−y′′)225(y'-y'')225(y′−y′′) is equal to:
  1. (A)732
  2. (B)746
  3. (C)742
  4. (D)736

Correct answer: (D)

Step-by-step solution →
Q78·Mathematics·Sets, Relations and FunctionsSingle correct
If RRR is the smallest equivalence relation on the set {1,2,3,4}\{1,2,3,4\}{1,2,3,4} such that {(1,2),(1,3)}⊂R\{(1,2),(1,3)\}\subset R{(1,2),(1,3)}⊂R, then the number of elements in RRR is:
  1. (A)10
  2. (B)12
  3. (C)8
  4. (D)15

Correct answer: (A)

Step-by-step solution →
Q79·Mathematics·Statistics and ProbabilitySingle correct
An integer is chosen at random from the integers 1,2,3,…,501,2,3,\ldots,501,2,3,…,50. The probability that the chosen integer is a multiple of atleast one of 4,64,64,6 and 777 is:
  1. (A)825\dfrac{8}{25}258​
  2. (B)2150\dfrac{21}{50}5021​
  3. (C)950\dfrac{9}{50}509​
  4. (D)1450\dfrac{14}{50}5014​

Correct answer: (B)

Step-by-step solution →
Q80·Mathematics·Vector AlgebraSingle correct
Let a unit vector u^=xi^+yj^+zk^\hat u=x\hat i+y\hat j+z\hat ku^=xi^+yj^​+zk^ make angles π2,π3\dfrac{\pi}{2},\dfrac{\pi}{3}2π​,3π​ and 2π3\dfrac{2\pi}{3}32π​ with the vectors 12i^+12j^\dfrac{1}{\sqrt2}\hat i+\dfrac{1}{\sqrt2}\hat j2​1​i^+2​1​j^​, 12j^+12k^\dfrac{1}{\sqrt2}\hat j+\dfrac{1}{\sqrt2}\hat k2​1​j^​+2​1​k^ and 12i^+12k^\dfrac{1}{\sqrt2}\hat i+\dfrac{1}{\sqrt2}\hat k2​1​i^+2​1​k^ respectively. If v⃗=12i^+12j^+12k^\vec v=\dfrac{1}{\sqrt2}\hat i+\dfrac{1}{\sqrt2}\hat j+\dfrac{1}{\sqrt2}\hat kv=2​1​i^+2​1​j^​+2​1​k^, then ∣u^−v⃗∣2|\hat u-\vec v|^2∣u^−v∣2 is equal to:
  1. (A)112\dfrac{11}{2}211​
  2. (B)52\dfrac{5}{2}25​
  3. (C)9
  4. (D)7

Correct answer: (B)

Step-by-step solution →
Q81·Mathematics·Quadratic EquationsNumerical
Let α,β\alpha,\betaα,β be the roots of the equation x2−6 x+3=0x^2-\sqrt6\,x+3=0x2−6​x+3=0 such that Im(α)>Im(β)\text{Im}(\alpha)>\text{Im}(\beta)Im(α)>Im(β). Let a,ba,ba,b be integers not divisible by 3 and nnn be a natural number such that α99β+α98=3n(a+ib), i=−1\dfrac{\alpha^{99}}{\beta}+\alpha^{98}=3^n(a+ib),\ i=\sqrt{-1}βα99​+α98=3n(a+ib), i=−1​. Then n+a+bn+a+bn+a+b is equal to ___.

Correct answer: 49

Step-by-step solution →
Q82·Mathematics·Matrices and DeterminantsNumerical
Let for any three distinct consecutive terms a,b,ca,b,ca,b,c of an A.P, the lines ax+by+c=0ax+by+c=0ax+by+c=0 be concurrent at the point PPP and Q (α,β)Q\,(\alpha,\beta)Q(α,β) be a point such that the system of equations x+y+z=6x+y+z=6x+y+z=6, 2x+5y+αz=β2x+5y+\alpha z=\beta2x+5y+αz=β and x+2y+3z=4x+2y+3z=4x+2y+3z=4, has infinitely many solutions. Then (PQ)2(PQ)^2(PQ)2 is equal to ___.

Correct answer: 113

Step-by-step solution →
Q83·Mathematics·ParabolaNumerical
Let P(α,β)P(\alpha,\beta)P(α,β) be a point on the parabola y2=4xy^2=4xy2=4x. If PPP also lies on the chord of the parabola x2=8yx^2=8yx2=8y whose mid point is (1,54)\left(1,\dfrac{5}{4}\right)(1,45​). Then (α−28)(β−8)(\alpha-28)(\beta-8)(α−28)(β−8) is equal to ___.

Correct answer: 192

Step-by-step solution →
Q84·Mathematics·Definite IntegrationNumerical
If ∫π/6π/31−sin⁡2x dx=α+β2+γ3\displaystyle\int_{\pi/6}^{\pi/3}\sqrt{1-\sin 2x}\,dx=\alpha+\beta\sqrt2+\gamma\sqrt3∫π/6π/3​1−sin2x​dx=α+β2​+γ3​, where α,β\alpha,\betaα,β and γ\gammaγ are rational numbers, then 3α+4β−γ3\alpha+4\beta-\gamma3α+4β−γ is equal to ___.

Correct answer: 6

Step-by-step solution →
Q85·Mathematics·Area Under CurvesNumerical
Let the area of the region {(x,y):0≤x≤3, 0≤y≤min⁡{x2+2, 2x+2}}\{(x,y):0\le x\le 3,\ 0\le y\le\min\{x^2+2,\,2x+2\}\}{(x,y):0≤x≤3, 0≤y≤min{x2+2,2x+2}} be AAA. Then 12A12A12A is equal to ___.

Correct answer: 164

Step-by-step solution →
Q86·Mathematics·Three Dimensional GeometryNumerical
Let OOO be the origin, and MMM and NNN be the points on the lines x−54=y−41=z−53\dfrac{x-5}{4}=\dfrac{y-4}{1}=\dfrac{z-5}{3}4x−5​=1y−4​=3z−5​ and x+812=y+25=z+119\dfrac{x+8}{12}=\dfrac{y+2}{5}=\dfrac{z+11}{9}12x+8​=5y+2​=9z+11​ respectively such that MNMNMN is the shortest distance between the given lines. Then OM→⋅ON→\overrightarrow{OM}\cdot\overrightarrow{ON}OM⋅ON is equal to ___.

Correct answer: 9

Step-by-step solution →
Q87·Mathematics·Limits and ContinuityNumerical
Let f(x)=lim⁡r→x{2r2[(f(r))2−f(x)f(r)]r2−x2−r3ef(r)r}f(x)=\sqrt{\displaystyle\lim_{r\to x}\left\{\dfrac{2r^2[(f(r))^2-f(x)f(r)]}{r^2-x^2}-r^3 e^{\frac{f(r)}{r}}\right\}}f(x)=r→xlim​{r2−x22r2[(f(r))2−f(x)f(r)]​−r3erf(r)​}​ be differentiable in (−∞,0)∪(0,∞)(-\infty,0)\cup(0,\infty)(−∞,0)∪(0,∞) and f(1)=1f(1)=1f(1)=1. Then the value of eaeaea, such that f(a)=0f(a)=0f(a)=0, is equal to ___.

Correct answer: 2

Step-by-step solution →
Q88·Mathematics·Binomial Theorem and Its Simple ApplicationsNumerical
Remainder when 64323264^{32^{32}}643232 is divided by 999 is equal to ___.

Correct answer: 1

Step-by-step solution →
Q89·Mathematics·Sets, Relations and FunctionsNumerical
Let the set C={(x,y)∣x2−2y=2023, x,y∈N}C=\{(x,y)\mid x^2-2^y=2023,\ x,y\in\mathbb{N}\}C={(x,y)∣x2−2y=2023, x,y∈N}. Then ∑(x,y)∈C(x+y)\displaystyle\sum_{(x,y)\in C}(x+y)(x,y)∈C∑​(x+y) is equal to ___.

Correct answer: 46

Step-by-step solution →
Q90·Mathematics·Limits and ContinuityNumerical
Let the slope of the line 45x+5y+3=045x+5y+3=045x+5y+3=0 be 27r1+9r2227r_1+\dfrac{9r_2}{2}27r1​+29r2​​ for some r1,r2∈Rr_1,r_2\in\mathbb{R}r1​,r2​∈R. Then lim⁡x→3(∫3x8t23r2x2−r2x2−r1x3−3x dt)\displaystyle\lim_{x\to 3}\left(\int_3^x\dfrac{8t^2}{\dfrac{3r_2 x}{2}-r_2 x^2-r_1 x^3-3x}\,dt\right)x→3lim​​∫3x​23r2​x​−r2​x2−r1​x3−3x8t2​dt​ is equal to ___.

Correct answer: 12

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
  • Magnetic Field of Current 147/186
  • 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
  • Solutions 158/186
  • Laws of Motion 130/186
  • Chemical Thermodynamics 165/186
  • Units and Measurements 149/186
  • 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
  • Oscillations 117/186
  • Organic Compounds Containing Halogens 109/186
  • Straight Lines 114/186
  • Capacitors and Dielectrics 115/186
  • Atoms 112/186
  • Classification of Elements and Periodicity in Properties 113/186
  • Parabola 101/186
  • Statistics 118/186
  • Inverse Trigonometric Functions 93/186
  • Electronic Effects and Stability 74/186
  • Indefinite Integration 66/186
  • Isomerism 51/186
  • IUPAC Nomenclature 37/186
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