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JEE Main 23 January 2025 Shift 1 Question Paper with Answers

23 January 2025 · January session · 71 questions

71 of the 75 questions from the JEE Main 23 January 2025 Shift 1 paper, each with its correct answer and tagged to the chapter it tests. Free to read, no account needed.

4 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
21
Chemistry
25
Mathematics
25

Physics — JEE Main 23 January 2025 Shift 1

Q1·Physics·Electromagnetic InductionSingle correct
Regarding self-inductance : A : The self-inductance of the coil depends on its geometry. B : Self-inductance does not depend on the permeability of the medium. C : Self-induced e.m.f. opposes any change in the current in a circuit. D : Self-inductance is electromagnetic analogue of mass in mechanics. E : Work needs to be done against self-induced e.m.f. in establishing the current. Choose the correct answer from the options given below:
  1. (A)A, B, C, D only
  2. (B)A, C, D, E only
  3. (C)A, B, C, E only
  4. (D)B, C, D, E only

Correct answer: (B)

Step-by-step solution →
Q2·Physics·OscillationsSingle correct
A light hollow cube of side length 10 cm and mass 10g, is floating in water. It is pushed down and released to execute simple harmonic oscillations. The time period of oscillations is yπ×10−2y\pi\times10^{-2}yπ×10−2 s, where the value of yyy is (Acceleration due to gravity, g=10g=10g=10 m/s2^22, density of water =103=10^3=103 kg/m3^33)
  1. (A)2
  2. (B)6
  3. (C)4
  4. (D)1

Correct answer: (A)

Step-by-step solution →
Q3·Physics·Properties of Solids and LiquidsSingle correct
Given below are two statements: Statement-I : The hot water flows faster than cold water. Statement-II : Soap water has higher surface tension as compared to fresh water. 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)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 →
Q4·Physics·Dual Nature of Matter and RadiationSingle correct
A sub-atomic particle of mass 10−3010^{-30}10−30 kg is moving with a velocity 2.21×1052.21\times10^52.21×105 m/s. Under the matter wave consideration, the particle will behave closely like _______ (h=6.63×10−34h=6.63\times10^{-34}h=6.63×10−34 J.s)
  1. (A)Infra-red radiation
  2. (B)X-rays
  3. (C)Gamma rays
  4. (D)Visible radiation

Correct answer: (B)

Step-by-step solution →
Q5·Physics·Geometrical OpticsSingle correct
A spherical surface of radius of curvature RRR, separates air from glass (refractive index =1.5=1.5=1.5). The centre of curvature is in the glass medium. A point object ‘O’ placed in air on the optic axis of the surface, so that its real image is formed at ‘I’ inside glass. The line OI intersects the spherical surface at P and PO = PI. The distance PO equals to-
  1. (A)5R
  2. (B)3R
  3. (C)2R
  4. (D)1.5R

Correct answer: (A)

Step-by-step solution →
Q6·Physics·Atoms and NucleiSingle correct
A radioactive nucleus n2n_2n2​ has 3 times the decay constant as compared to the decay constant of another radioactive nucleus n1n_1n1​. If initial number of both nuclei are the same, what is the ratio of number of nuclei of n2n_2n2​ to the number of nuclei of n1n_1n1​, after one half-life of n1n_1n1​?
  1. (A)14\dfrac{1}{4}41​
  2. (B)18\dfrac{1}{8}81​
  3. (C)4
  4. (D)8

Correct answer: (A)

Step-by-step solution →
Q7·Physics·Current ElectricitySingle correct
Identify the valid statements relevant to the given circuit at the instant when the key is closed. A. There will be no current through resistor R. B. There will be maximum current in the connecting wires. C. Potential difference between the capacitor plates A and B is minimum. D. Charge on the capacitor plates is minimum. Choose the correct answer from the options given below :
  1. (A)C, D only
  2. (B)B, C, D only
  3. (C)A, C only
  4. (D)A, B, D only

Correct answer: (B)

Step-by-step solution →
Q8·Physics·Units and MeasurementsSingle correct
The position of a particle moving on x-axis is given by x(t)=Asin⁡t+Bcos⁡2t+Ct2+Dx(t)=A\sin t+B\cos^2 t+Ct^2+Dx(t)=Asint+Bcos2t+Ct2+D, where ttt is time. The dimension of ABCD\dfrac{ABC}{D}DABC​ is-
  1. (A)L
  2. (B)L3T−2L^3T^{-2}L3T−2
  3. (C)L2T−2L^2T^{-2}L2T−2
  4. (D)L2L^2L2

Correct answer: (C)

Step-by-step solution →
Q9·Physics·ThermodynamicsSingle correct
Match List-I (Thermodynamic Process) with List-II (Process Name). Choose the correct answer from the options given below:
List-I (Thermodynamic Process)List-II (Process Name)
A.Pressure varies inversely with volume of an ideal gasI.Adiabatic process
B.Heat absorbed goes partly to increase internal energy and partly to do workII.Isochoric process
C.Heat is neither absorbed nor released by a systemIII.Isothermal process
D.No work is done on or by a gasIV.Isobaric process
  1. (A)A-II, B-IV, C-II, D-III
  2. (B)A-III, B-I, C-IV, D-II
  3. (C)A-I, B-III, C-II, D-IV
  4. (D)A-III, B-IV, C-I, D-II

Correct answer: (D)

Step-by-step solution →
Q10·Physics·Magnetic Field of CurrentSingle correct
Consider a moving coil galvanometer (MCG) : A : The torsional constant in moving coil galvanometer has dimensions [ML2T−2][ML^2T^{-2}][ML2T−2]. B : Increasing the current sensitivity may not necessarily increase the voltage sensitivity. C : If we increase number of turns (N) to its double (2N), then the voltage sensitivity doubles. D : MCG can be converted into an ammeter by introducing a shunt resistance of large value in parallel with galvanometer. E : Current sensitivity of MCG depends inversely on number of turns of coil. Choose the correct answer from the options given below :
  1. (A)A, B only
  2. (B)A, D only
  3. (C)B, D, E only
  4. (D)A, B, E only

Correct answer: (A)

Step-by-step solution →
Q11·Physics·Electric Field and Coulomb's LawSingle correct
A point particle of charge Q is located at P along the axis of an electric dipole 1 at a distance r as shown in the figure. The point P is also on the equatorial plane of a second electric dipole 2 at a distance r. The dipoles are made of opposite charges q separated by a distance 2a. For the charge particle at P not to experience any net force, which of the following correctly describes the situation?
  1. (A)ar∼20\dfrac{a}{r}\sim 20ra​∼20
  2. (B)ar∼10\dfrac{a}{r}\sim 10ra​∼10
  3. (C)ar∼0.5\dfrac{a}{r}\sim 0.5ra​∼0.5
  4. (D)ar∼3\dfrac{a}{r}\sim 3ra​∼3

Correct answer: (D)

Step-by-step solution →
Q12·Physics·ThermodynamicsSingle correct
A gun fires a lead bullet of temperature 300K into a wooden block. The bullet having melting temperature of 600 K penetrates into the block and melts down. If the total heat required for the process is 625 J, then the mass of the bullet is _______ grams. (Latent heat of fusion of lead =2.5×104=2.5\times10^4=2.5×104 JKg−1^{-1}−1 and specific heat capacity of lead =125=125=125 JKg−1^{-1}−1K−1^{-1}−1)
  1. (A)20
  2. (B)15
  3. (C)10
  4. (D)5

Correct answer: (C)

Step-by-step solution →
Q13·Physics·Geometrical OpticsSingle correct
What is the lateral shift of a ray refracted through a parallel-sided glass slab of thickness ‘h’ in terms of the angle of incidence ‘i’ and angle of refraction ‘r’, if the glass slab is placed in air medium ?
  1. (A)hsin⁡(i−r)tan⁡r\dfrac{h\sin(i-r)}{\tan r}tanrhsin(i−r)​
  2. (B)hcos⁡(i−r)sin⁡r\dfrac{h\cos(i-r)}{\sin r}sinrhcos(i−r)​
  3. (C)h
  4. (D)hsin⁡(i−r)cos⁡r\dfrac{h\sin(i-r)}{\cos r}cosrhsin(i−r)​

Correct answer: (D)

Step-by-step solution →
Q14·Physics·Rotational MotionSingle correct
A solid sphere of mass ‘m’ and radius ‘r’ is allowed to roll without slipping from the highest point of an inclined plane of length ‘L’ and makes an angle 30° with the horizontal. The speed of the particle at the bottom of the plane is v1v_1v1​. If the angle of inclination is increased to 45° while keeping L constant. Then the new speed of the sphere at the bottom of the plane is v2v_2v2​. The ratio of v12:v22v_1^2:v_2^2v12​:v22​ is
  1. (A)1:21:\sqrt21:2​
  2. (B)1:31:31:3
  3. (C)1:21:21:2
  4. (D)1:31:\sqrt31:3​

Correct answer: (A)

Step-by-step solution →
Q15·Physics·Electronic DevicesSingle correct
Refer to the circuit diagram given in the figure, which of the following observation of circuit are correct? A. Total resistance of circuit is 6 Ω. B. Current in Ammeter is 1A. C. Potential across AB is 4 Volts. D. Potential across CD is 4 Volts. E. Total resistance of the circuit is 8Ω. Choose the correct answer from the options given below :
  1. (A)A, B and D only
  2. (B)A, C and D only
  3. (C)B, C and E only
  4. (D)A, B and C only

Correct answer: (A)

Step-by-step solution →
Q16·Physics·Electric Field and Coulomb's LawSingle correct
The electric flux is ϕ=ασ+βλ\phi=\alpha\sigma+\beta\lambdaϕ=ασ+βλ where λ\lambdaλ and σ\sigmaσ are linear and surface charge density, respectively. (αβ)\left(\dfrac{\alpha}{\beta}\right)(βα​) represents
  1. (A)charge
  2. (B)electric field
  3. (C)displacement
  4. (D)area

Correct answer: (C)

Step-by-step solution →
Q17·Physics·Geometrical OpticsSingle correct
Given a thin convex lens (refractive index μ2\mu_2μ2​), kept in a liquid (refractive index μ1\mu_1μ1​, μ1<μ2\mu_1<\mu_2μ1​<μ2​) having radii of curvature ∣R1∣|R_1|∣R1​∣ and ∣R2∣|R_2|∣R2​∣. Its second surface is silver polished. Where should an object be placed on the optic axis so that a real and inverted image is formed at the same place ?
  1. (A)μ1∣R1∣∣R2∣μ2(∣R1∣+∣R2∣)−μ1∣R1∣\dfrac{\mu_1|R_1||R_2|}{\mu_2(|R_1|+|R_2|)-\mu_1|R_1|}μ2​(∣R1​∣+∣R2​∣)−μ1​∣R1​∣μ1​∣R1​∣∣R2​∣​
  2. (B)μ1∣R1∣∣R2∣μ2(∣R1∣+∣R2∣)−μ1∣R2∣\dfrac{\mu_1|R_1||R_2|}{\mu_2(|R_1|+|R_2|)-\mu_1|R_2|}μ2​(∣R1​∣+∣R2​∣)−μ1​∣R2​∣μ1​∣R1​∣∣R2​∣​
  3. (C)μ1∣R1∣∣R2∣μ2(2∣R1∣+∣R2∣)−μ1∣R1∣∣R2∣\dfrac{\mu_1|R_1||R_2|}{\mu_2(2|R_1|+|R_2|)-\mu_1\sqrt{|R_1||R_2|}}μ2​(2∣R1​∣+∣R2​∣)−μ1​∣R1​∣∣R2​∣​μ1​∣R1​∣∣R2​∣​
  4. (D)(μ2+μ1)∣R1∣μ2−μ1\dfrac{(\mu_2+\mu_1)|R_1|}{\mu_2-\mu_1}μ2​−μ1​(μ2​+μ1​)∣R1​∣​

Correct answer: (B)

Step-by-step solution →
Q18·Physics·Electromagnetic WavesSingle correct
The electric field of an electromagnetic wave in free space is E⃗=57cos⁡[7.5×106t−5×10−3(3x+4y)](4i^−3j^)\vec E=57\cos[7.5\times10^6 t-5\times10^{-3}(3x+4y)](4\hat i-3\hat j)E=57cos[7.5×106t−5×10−3(3x+4y)](4i^−3j^​) N/C. The associated magnetic field in Tesla is-
  1. (A)B⃗=573×108cos⁡[7.5×106t−5×10−3(3x+4y)](5k^)\vec B=\dfrac{57}{3\times10^8}\cos[7.5\times10^6 t-5\times10^{-3}(3x+4y)](5\hat k)B=3×10857​cos[7.5×106t−5×10−3(3x+4y)](5k^)
  2. (B)B⃗=573×108cos⁡[7.5×106t−5×10−3(3x+4y)](k^)\vec B=\dfrac{57}{3\times10^8}\cos[7.5\times10^6 t-5\times10^{-3}(3x+4y)](\hat k)B=3×10857​cos[7.5×106t−5×10−3(3x+4y)](k^)
  3. (C)B⃗=−573×108cos⁡[7.5×106t−5×10−3(3x+4y)](5k^)\vec B=-\dfrac{57}{3\times10^8}\cos[7.5\times10^6 t-5\times10^{-3}(3x+4y)](5\hat k)B=−3×10857​cos[7.5×106t−5×10−3(3x+4y)](5k^)
  4. (D)B⃗=−573×108cos⁡[7.5×106t−5×10−3(3x+4y)](k^)\vec B=-\dfrac{57}{3\times10^8}\cos[7.5\times10^6 t-5\times10^{-3}(3x+4y)](\hat k)B=−3×10857​cos[7.5×106t−5×10−3(3x+4y)](k^)

Correct answer: (C)

Step-by-step solution →
Q19·Physics·KinematicsSingle correct
The motion of an airplane is represented by velocity-time graph as shown below. The distance covered by airplane in the first 30.5 second is _______ km.
  1. (A)9
  2. (B)6
  3. (C)3
  4. (D)12

Correct answer: (D)

Step-by-step solution →
Q20·Physics·Rotational MotionSingle correct
Consider a circular disc of radius 20 cm with centre located at the origin. A circular hole of radius 5 cm is cut from this disc in such a way that the edge of the hole touches the edge of the disc. The distance of centre of mass of residual or remaining disc from the origin will be-
  1. (A)2.0 cm
  2. (B)0.5 cm
  3. (C)1.5 cm
  4. (D)1.0 cm

Correct answer: (D)

Step-by-step solution →
Q21·Physics·KinematicsInteger
Two particles are located at equal distance from origin. The position vectors of those are represented by A⃗=2i^+3nj^+2k^\vec A=2\hat i+3n\hat j+2\hat kA=2i^+3nj^​+2k^ and B⃗=2i^−2j^+4pk^\vec B=2\hat i-2\hat j+4p\hat kB=2i^−2j^​+4pk^, respectively. If both the vectors are at right angle to each other, the value of n−1n^{-1}n−1 is _______.

Correct answer: 3

Step-by-step solution →

Chemistry — JEE Main 23 January 2025 Shift 1

Q22·Chemistry·Classification of Elements and Periodicity in PropertiesSingle correct
The element that does not belong to the same period of the remaining elements (modern periodic table) is:
  1. (A)Palladium
  2. (B)Iridium
  3. (C)Osmium
  4. (D)Platinum

Correct answer: (A)

Step-by-step solution →
Q23·Chemistry·Atomic StructureSingle correct
Heat treatment of muscular pain involves radiation of wavelength of about 900 nm. Which spectral line of H atom is suitable for this ? (Given: Rydberg constant RH=105R_H=10^5RH​=105 cm−1^{-1}−1, h=6.6×10−34h=6.6\times10^{-34}h=6.6×10−34 J s, c=3×108c=3\times10^8c=3×108 m/s)
  1. (A)Paschen series, ∞→3
  2. (B)Lyman series, ∞→1
  3. (C)Balmer series, ∞→2
  4. (D)Paschen series, 5→3

Correct answer: (A)

Step-by-step solution →
Q24·Chemistry·p-Block ElementsSingle correct
The incorrect statement among the following is
  1. (A)PH₃ shows lower proton affinity than NH₃.
  2. (B)PF₃ exists but NF₅ does not.
  3. (C)NO₂ can dimerise easily.
  4. (D)SO₂ can act as an oxidizing agent, but not as a reducing agent.

Correct answer: (D)

Step-by-step solution →
Q25·Chemistry·Coordination CompoundsSingle correct
CrCl₃·xNH₃ can exist as a complex. 0.1 molal aqueous solution of this complex shows a depression in freezing point of 0.558°C. Assuming 100% ionisation of this complex and coordination number of Cr is 6, the complex will be (Given Kf=1.86K_f=1.86Kf​=1.86 K kg mol−1^{-1}−1)
  1. (A)[Cr(NH₃)₆]Cl₃
  2. (B)[Cr(NH₃)₄Cl₂]Cl
  3. (C)[Cr(NH₃)₅Cl]Cl₂
  4. (D)[Cr(NH₃)₃Cl₃]

Correct answer: (C)

Step-by-step solution →
Q26·Chemistry·Redox Reactions and ElectrochemistrySingle correct
In the diagram below, the standard electrode potentials are given in volts (over the arrow). FeO42−→+2.0VFe3+→0.8VFe2+→−0.5VFe0FeO_4^{2-}\xrightarrow{+2.0V}Fe^{3+}\xrightarrow{0.8V}Fe^{2+}\xrightarrow{-0.5V}Fe^{0}FeO42−​+2.0V​Fe3+0.8V​Fe2+−0.5V​Fe0. The value of EFeO42−/Fe2+∘E^{\circ}_{FeO_4^{2-}/Fe^{2+}}EFeO42−​/Fe2+∘​ is
  1. (A)1.7 V
  2. (B)1.2 V
  3. (C)2.1 V
  4. (D)1.4 V

Correct answer: (A)

Step-by-step solution →
Q27·Chemistry·Organic Compounds Containing HalogensSingle correct
Match List-I (Name Reaction) with List-II (Product Obtained). Choose the correct answer from the options given below:
List-I (Name Reaction)List-II (Product Obtained)
A.Swarts reactionI.Ethyl benzene
B.Sandmeyer's reactionII.Ethyl iodide
C.Wurtz Fittig reactionIII.Cyanobenzene
D.Finkelstein reactionIV.Ethyl fluoride
  1. (A)A-II, B-III, C-I, D-IV
  2. (B)A-IV, B-I, C-III, D-II
  3. (C)A-IV, B-III, C-I, D-II
  4. (D)A-II, B-I, C-III, D-IV

Correct answer: (C)

Step-by-step solution →
Q28·Chemistry·BiomoleculesSingle correct
Given below are two statements: Statement I: Fructose does not contain an aldehydic group but still reduces Tollen's reagent. Statement II: In the presence of base, fructose undergoes rearrangement to give glucose. In the light of the above statements, choose the correct answer 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 →
Q29·Chemistry·Some Basic Concepts in ChemistrySingle correct
2.8×10−32.8\times10^{-3}2.8×10−3 mol of CO₂ is left after removing 102110^{21}1021 molecules from its ‘x’ mg sample. The mass of CO₂ taken initially is (Given : NA=6.02×1023N_A=6.02\times10^{23}NA​=6.02×1023 mol−1^{-1}−1)
  1. (A)196.2 mg
  2. (B)98.3 mg
  3. (C)150.4 mg
  4. (D)48.2 mg

Correct answer: (A)

Step-by-step solution →
Q30·Chemistry·Chemical ThermodynamicsSingle correct
Ice at −5°C is heated to become vapor with temperature of 110°C at atmospheric pressure. The entropy change associated with this process can be obtained from :
  1. (A)∫268383Cp dT+ΔHmelting273+ΔHboiling373\int_{268}^{383}C_p\,dT+\dfrac{\Delta H_{melting}}{273}+\dfrac{\Delta H_{boiling}}{373}∫268383​Cp​dT+273ΔHmelting​​+373ΔHboiling​​
  2. (B)∫268273Cp,mTdT+ΔHm,fusionTf+∫273373Cp,mTdT+ΔHm,vaporisationTb+∫373383Cp,mTdT\int_{268}^{273}\dfrac{C_{p,m}}{T}dT+\dfrac{\Delta H_{m,fusion}}{T_f}+\int_{273}^{373}\dfrac{C_{p,m}}{T}dT+\dfrac{\Delta H_{m,vaporisation}}{T_b}+\int_{373}^{383}\dfrac{C_{p,m}}{T}dT∫268273​TCp,m​​dT+Tf​ΔHm,fusion​​+∫273373​TCp,m​​dT+Tb​ΔHm,vaporisation​​+∫373383​TCp,m​​dT
  3. (C)∫268383Cp dT+qrevT\int_{268}^{383}C_p\,dT+\dfrac{q_{rev}}{T}∫268383​Cp​dT+Tqrev​​
  4. (D)∫268273Cp,m dT+ΔHm,fusionTf+ΔHm,vaporisationTb+∫273373Cp,m dT+∫373383Cp,m dT\int_{268}^{273}C_{p,m}\,dT+\dfrac{\Delta H_{m,fusion}}{T_f}+\dfrac{\Delta H_{m,vaporisation}}{T_b}+\int_{273}^{373}C_{p,m}\,dT+\int_{373}^{383}C_{p,m}\,dT∫268273​Cp,m​dT+Tf​ΔHm,fusion​​+Tb​ΔHm,vaporisation​​+∫273373​Cp,m​dT+∫373383​Cp,m​dT

Correct answer: (B)

Step-by-step solution →
Q31·Chemistry·Coordination CompoundsSingle correct
The d-electronic configuration of an octahedral Co(II) complex having magnetic moment of 3.95 BM is :
  1. (A)t2g6eg1t_{2g}^{6}e_{g}^{1}t2g6​eg1​
  2. (B)t2g5eg0t_{2g}^{5}e_{g}^{0}t2g5​eg0​
  3. (C)t2g5eg2t_{2g}^{5}e_{g}^{2}t2g5​eg2​
  4. (D)e4t23e^{4}t_{2}^{3}e4t23​

Correct answer: (C)

Step-by-step solution →
Q32·Chemistry·Coordination CompoundsSingle correct
The complex that shows Facial – Meridional isomerism is
  1. (A)[Co(NH₃)₃Cl₃]
  2. (B)[Co(NH₃)₄Cl₂]⁺
  3. (C)[Co(en)₃]³⁺
  4. (D)[Co(en)₂Cl₂]⁺

Correct answer: (A)

Step-by-step solution →
Q33·Chemistry·Aldehydes and KetonesSingle correct
The major product of the following reaction is : CH3CH2CHO→refluxexcess HCHO, alkaliCH_3CH_2CHO\xrightarrow[\text{reflux}]{\text{excess HCHO, alkali}}CH3​CH2​CHOexcess HCHO, alkalireflux​ ?
  1. (A)CH3−CH2−CH2−OHCH_3-CH_2-CH_2-OHCH3​−CH2​−CH2​−OH
  2. (B)CH3−CH(CH2OH)−CHOCH_3-CH(CH_2OH)-CHOCH3​−CH(CH2​OH)−CHO
  3. (C)CH3−C(CH2OH)2−CH2OHCH_3-C(CH_2OH)_2-CH_2OHCH3​−C(CH2​OH)2​−CH2​OH
  4. (D)CH3−C(=CH2)−CHOCH_3-C(=CH_2)-CHOCH3​−C(=CH2​)−CHO

Correct answer: (C)

Step-by-step solution →
Q34·Chemistry·Electronic Effects and StabilitySingle correct
The correct stability order of the following species/molecules is :
  1. (A)q > r > p
  2. (B)r > q > p
  3. (C)q > p > r
  4. (D)p > q > r

Correct answer: (A)

Step-by-step solution →
Q35·Chemistry·HydrocarbonsSingle correct
Propane molecule on chlorination under photochemical condition gives two di-chloro products, "x" and "y". Amongst "x" and "y", "x" is an optically active molecule. How many tri-chloro products (consider only structural isomers) will be obtained from "x" when it is further treated with chlorine under the photochemical condition?
  1. (A)4
  2. (B)2
  3. (C)5
  4. (D)3

Correct answer: (D)

Step-by-step solution →
Q36·Chemistry·Alcohols and EthersSingle correct
What amount of bromine will be required to convert 2 g of phenol into 2, 4, 6-tribromophenol ? (Given molar mass in g mol−1^{-1}−1 of C, H, O, Br are 12, 1, 16, 80 respectively)
  1. (A)10.22 g
  2. (B)6.0 g
  3. (C)4.0 g
  4. (D)20.44 g

Correct answer: (A)

Step-by-step solution →
Q37·Chemistry·d- and f-Block ElementsSingle correct
The correct set of ions (aqueous solution) with same colour from the following is :
  1. (A)V²⁺, Cr³⁺, Mn³⁺
  2. (B)Zn²⁺, V³⁺, Fe³⁺
  3. (C)Ti⁴⁺, V⁴⁺, Mn²⁺
  4. (D)Sc³⁺, Ti³⁺, Cr²⁺

Correct answer: (A)

Step-by-step solution →
Q38·Chemistry·Purification and Characterisation of Organic CompoundsSingle correct
Given below are two statements : Statement I : In Lassaigne's test, the covalent organic molecules are transformed into ionic compounds. Statement II : The sodium fusion extract of an organic compound having N and S gives prussian blue colour with FeSO₄ and Na₄[Fe(CN)₆]. 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)Both Statement I and Statement II are false
  3. (C)Statement I is false but Statement II is true
  4. (D)Statement I is true but Statement II is false

Correct answer: (D)

Step-by-step solution →
Q39·Chemistry·EquilibriumSingle correct
Which of the following happens when NH₄OH is added gradually to the solution containing 1M A³⁺ and 1M B³⁺ ions ? Given : Ksp[A(OH)3]=9×10−10K_{sp}[A(OH)_3]=9\times10^{-10}Ksp​[A(OH)3​]=9×10−10 and Ksp[B(OH)3]=27×10−18K_{sp}[B(OH)_3]=27\times10^{-18}Ksp​[B(OH)3​]=27×10−18 at 298 K.
  1. (A)B(OH)₃ will precipitate before A(OH)₃
  2. (B)A(OH)₃ and B(OH)₃ will precipitate together
  3. (C)A(OH)₃ will precipitate before B(OH)₃
  4. (D)Both A(OH)₃ and B(OH)₃ do not show precipitation with NH₄OH

Correct answer: (A)

Step-by-step solution →
Q40·Chemistry·Chemical Bonding and Molecular StructureSingle correct
Match List-I (Octet Behaviour) with List-II (Example Molecules). Choose the correct answer from the options given below:
List-I (Octet Behaviour)List-II (Example Molecules)
A.Molecules obeying octet ruleI.NO, NO₂
B.Molecules with incomplete octetII.BCl₃, AlCl₃
C.Molecules with incomplete octet with odd electronIII.H₂SO₄, PCl₅
D.Molecules with expanded octetIV.CCl₄, CO₂
  1. (A)A-IV, B-II, C-I, D-III
  2. (B)A-III, B-II, C-I, D-IV
  3. (C)A-IV, B-I, C-III, D-II
  4. (D)A-II, B-IV, C-III, D-I

Correct answer: (A)

Step-by-step solution →
Q41·Chemistry·AminesSingle correct
Which among the following react with Hinsberg's reagent? (A) C6H5−NH2C_6H_5{-}NH_2C6​H5​−NH2​ (2) C6H5−N(CH3)2C_6H_5{-}N(CH_3)_2C6​H5​−N(CH3​)2​ (C) CH3−NH2CH_3{-}NH_2CH3​−NH2​ (4) N(CH3)3N(CH_3)_3N(CH3​)3​ (E) C6H5−NH−C6H5C_6H_5{-}NH{-}C_6H_5C6​H5​−NH−C6​H5​. Choose the correct answer from the options given below :
  1. (A)B and D only
  2. (B)C and D only
  3. (C)A, B and E only
  4. (D)A, C and E only

Correct answer: (D)

Step-by-step solution →
Q42·Chemistry·EquilibriumInteger
If 1 mM solution of ethylamine produces pH = 9, then the ionization constant (KbK_bKb​) of ethylamine is 10−x10^{-x}10−x. The value of x is _______ (nearest integer). [The degree of ionization of ethylamine can be neglected with respect to unity.]

Correct answer: 7

Step-by-step solution →
Q43·Chemistry·Purification and Characterisation of Organic CompoundsInteger
During "S" estimation, 160 mg of an organic compound gives 466 mg of barium sulphate. The percentage of Sulphur in the given compound is _______ %. (Given molar mass in g mol−1^{-1}−1 of Ba : 137, S : 32, O : 16)

Correct answer: 40

Step-by-step solution →
Q44·Chemistry·Diazonium Salts and ReactionsInteger
Consider the following sequence of reactions to produce major product (A). The starting compound (shown in the figure) is treated successively with: i) Br₂, Fe; ii) Sn, HCl; iii) NaNO₂, HCl, 273 K; iv) H₃PO₂, H₂O. Molar mass of product (A) is _______ g mol−1^{-1}−1. (Given molar mass in g mol−1^{-1}−1 of C : 12, H : 1, O : 16, Br : 80, N : 14, P : 31)

Correct answer: 171

Step-by-step solution →
Q45·Chemistry·Chemical KineticsInteger
For the thermal decomposition of N₂O₅(g) at constant volume, the following table can be formed, for the reaction 2N2O5(g)→2N2O4(g)+O2(g)2N_2O_5(g)\rightarrow 2N_2O_4(g)+O_2(g)2N2​O5​(g)→2N2​O4​(g)+O2​(g). (Table — S.No. 1: Time = —, Total pressure = 0.6 atm; S.No. 2: Time = 100 s, Total pressure = x). x=x=x= _______ ×10−3\times10^{-3}×10−3 atm [nearest integer]. (Given: Rate constant for the reaction is 4.606×10−24.606\times10^{-2}4.606×10−2 s−1^{-1}−1.)

Correct answer: 897

Step-by-step solution →
Q46·Chemistry·Chemical ThermodynamicsInteger
The standard enthalpy and standard entropy of decomposition of N₂O₄ to NO₂ are 55.0 kJ/mol and 175.0 J/K/mol respectively. The standard free energy change for this reaction at 25°C in J mol−1^{-1}−1 is _______ (Nearest integer)

Correct answer: 2850

Step-by-step solution →

Mathematics — JEE Main 23 January 2025 Shift 1

Q47·Mathematics·Definite IntegrationSingle correct
The value of ∫e2e41x(e((log⁡ex)2+1)−1e((log⁡ex)2+1)−1+e((6−log⁡ex)2+1)−1)dx\displaystyle\int_{e^2}^{e^4}\frac{1}{x}\left(\frac{e^{\left((\log_e x)^2+1\right)^{-1}}}{e^{\left((\log_e x)^2+1\right)^{-1}}+e^{\left((6-\log_e x)^2+1\right)^{-1}}}\right)dx∫e2e4​x1​(e((loge​x)2+1)−1+e((6−loge​x)2+1)−1e((loge​x)2+1)−1​)dx is
  1. (A)log⁡e2\log_e 2loge​2
  2. (B)222
  3. (C)111
  4. (D)e2e^2e2

Correct answer: (C)

Step-by-step solution →
Q48·Mathematics·Indefinite IntegrationSingle correct
Let I(x)=∫dx(x−11)11/13(x+15)15/13\displaystyle I(x)=\int\frac{dx}{(x-11)^{11/13}(x+15)^{15/13}}I(x)=∫(x−11)11/13(x+15)15/13dx​. If I(37)−I(24)=14(1b1/13−1c1/13)I(37)-I(24)=\frac{1}{4}\left(\frac{1}{b^{1/13}}-\frac{1}{c^{1/13}}\right)I(37)−I(24)=41​(b1/131​−c1/131​), b,c∈Nb,c\in\mathbb{N}b,c∈N, then 3(b+c)3(b+c)3(b+c) is equal to
  1. (A)404040
  2. (B)393939
  3. (C)222222
  4. (D)262626

Correct answer: (B)

Step-by-step solution →
Q49·Mathematics·Limits and ContinuitySingle correct
If the function f(x)={2x{sin⁡(k1+1)x+sin⁡(k2−1)x},x<04,x=02xlog⁡e(2+k1x2+k2x),x>0f(x)=\begin{cases}\dfrac{2}{x}\{\sin(k_1+1)x+\sin(k_2-1)x\}, & x<0\\[2mm] 4, & x=0\\[2mm] \dfrac{2}{x}\log_e\left(\dfrac{2+k_1x}{2+k_2x}\right), & x>0\end{cases}f(x)=⎩⎨⎧​x2​{sin(k1​+1)x+sin(k2​−1)x},4,x2​loge​(2+k2​x2+k1​x​),​x<0x=0x>0​ is continuous at x=0x=0x=0, then k12+k22k_1^2+k_2^2k12​+k22​ is equal to
  1. (A)888
  2. (B)202020
  3. (C)555
  4. (D)101010

Correct answer: (D)

Step-by-step solution →
Q50·Mathematics·ParabolaSingle correct
If the line 3x−2y+12=03x-2y+12=03x−2y+12=0 intersects the parabola 4y=3x24y=3x^24y=3x2 at the points AAA and BBB, then at the vertex of the parabola, the line segment ABABAB subtends an angle equal to
  1. (A)tan⁡−1(119)\tan^{-1}\left(\dfrac{11}{9}\right)tan−1(911​)
  2. (B)π2−tan⁡−1(32)\dfrac{\pi}{2}-\tan^{-1}\left(\dfrac{3}{2}\right)2π​−tan−1(23​)
  3. (C)tan⁡−1(45)\tan^{-1}\left(\dfrac{4}{5}\right)tan−1(54​)
  4. (D)tan⁡−1(97)\tan^{-1}\left(\dfrac{9}{7}\right)tan−1(79​)

Correct answer: (D)

Step-by-step solution →
Q51·Mathematics·Differential EquationsSingle correct
Let a curve y=f(x)y=f(x)y=f(x) pass through the points (0,5)(0,5)(0,5) and (log⁡e2,k)(\log_e 2, k)(loge​2,k). If the curve satisfies the differential equation 2(3+y)e2xdx−(7+e2x)dy=02(3+y)e^{2x}dx-(7+e^{2x})dy=02(3+y)e2xdx−(7+e2x)dy=0, then kkk is equal to
  1. (A)161616
  2. (B)888
  3. (C)323232
  4. (D)444

Correct answer: (B)

Step-by-step solution →
Q52·Mathematics·Sets, Relations and FunctionsSingle correct
Let f(x)=log⁡exf(x)=\log_e xf(x)=loge​x and g(x)=x4−2x3+3x2−2x+22x2−2x+1g(x)=\dfrac{x^4-2x^3+3x^2-2x+2}{2x^2-2x+1}g(x)=2x2−2x+1x4−2x3+3x2−2x+2​. Then the domain of f∘gf\circ gf∘g is
  1. (A)R\mathbb{R}R
  2. (B)(0,∞)(0,\infty)(0,∞)
  3. (C)[0,∞)[0,\infty)[0,∞)
  4. (D)[1,∞)[1,\infty)[1,∞)

Correct answer: (A)

Step-by-step solution →
Q53·Mathematics·Vector AlgebraSingle correct
Let the arc ACACAC of a circle subtend a right angle at the centre OOO. If the point BBB on the arc ACACAC divides the arc ACACAC such that length of arc ABlength of arc BC=15\dfrac{\text{length of arc }AB}{\text{length of arc }BC}=\dfrac{1}{5}length of arc BClength of arc AB​=51​, and OC→=α OA→+β OB→\overrightarrow{OC}=\alpha\,\overrightarrow{OA}+\beta\,\overrightarrow{OB}OC=αOA+βOB, then α+2(3−1)β\alpha+\sqrt2(\sqrt3-1)\betaα+2​(3​−1)β is equal to
  1. (A)2−32-\sqrt32−3​
  2. (B)232\sqrt323​
  3. (C)535\sqrt353​
  4. (D)2+32+\sqrt32+3​

Correct answer: (A)

Step-by-step solution →
Q54·Mathematics·Sequence and SeriesSingle correct
If the first term of an A.P. is 3 and the sum of its first four terms is equal to one-fifth of the sum of the next four terms, then the sum of the first 20 terms is equal to
  1. (A)−1200-1200−1200
  2. (B)−1080-1080−1080
  3. (C)−1020-1020−1020
  4. (D)−120-120−120

Correct answer: (B)

Step-by-step solution →
Q55·Mathematics·Three Dimensional GeometrySingle correct
Let PPP be the foot of the perpendicular from the point Q(10,−3,−1)Q(10,-3,-1)Q(10,−3,−1) on the line x−37=y−2−1=z+1−2\dfrac{x-3}{7}=\dfrac{y-2}{-1}=\dfrac{z+1}{-2}7x−3​=−1y−2​=−2z+1​. Then the area of the right angled triangle PQRPQRPQR, where RRR is the point (3,−2,1)(3,-2,1)(3,−2,1), is
  1. (A)9159\sqrt{15}915​
  2. (B)30\sqrt{30}30​
  3. (C)8158\sqrt{15}815​
  4. (D)3303\sqrt{30}330​

Correct answer: (D)

Step-by-step solution →
Q56·Mathematics·Complex NumbersSingle correct
Let ∣zˉ−i2zˉ+i∣=13\left|\dfrac{\bar z-i}{2\bar z+i}\right|=\dfrac{1}{3}​2zˉ+izˉ−i​​=31​, z∈Cz\in\mathbb{C}z∈C, be the equation of a circle with center at CCC. If the area of the triangle, whose vertices are at the points (0,0)(0,0)(0,0), CCC and (α,0)(\alpha,0)(α,0) is 11 square units, then α2\alpha^2α2 equals
  1. (A)100100100
  2. (B)505050
  3. (C)12125\dfrac{121}{25}25121​
  4. (D)8125\dfrac{81}{25}2581​

Correct answer: (A)

Step-by-step solution →
Q57·Mathematics·Sets, Relations and FunctionsSingle correct
Let R={(1,2),(2,3),(3,3)}R=\{(1,2),(2,3),(3,3)\}R={(1,2),(2,3),(3,3)} be a relation defined on the set {1,2,3,4}\{1,2,3,4\}{1,2,3,4}. Then the minimum number of elements, needed to be added in RRR so that RRR becomes an equivalence relation, is
  1. (A)101010
  2. (B)888
  3. (C)999
  4. (D)777

Correct answer: (D)

Step-by-step solution →
Q58·Mathematics·Permutations and CombinationsSingle correct
The number of words, which can be formed using all the letters of the word "DAUGHTER", so that all the vowels never come together, is
  1. (A)340003400034000
  2. (B)370003700037000
  3. (C)360003600036000
  4. (D)350003500035000

Correct answer: (C)

Step-by-step solution →
Q59·Mathematics·Straight LinesSingle correct
Let the area of a △PQR\triangle PQR△PQR with vertices P(5,4)P(5,4)P(5,4), Q(−2,4)Q(-2,4)Q(−2,4) and R(a,b)R(a,b)R(a,b) be 35 square units. If its orthocenter and centroid are O(2,145)O\left(2,\dfrac{14}{5}\right)O(2,514​) and C(c,d)C(c,d)C(c,d) respectively, then c+2dc+2dc+2d is equal to
  1. (A)73\dfrac{7}{3}37​
  2. (B)333
  3. (C)222
  4. (D)83\dfrac{8}{3}38​

Correct answer: (B)

Step-by-step solution →
Q60·Mathematics·Inverse Trigonometric FunctionsSingle correct
If π2≤x≤3π4\dfrac{\pi}{2}\le x\le\dfrac{3\pi}{4}2π​≤x≤43π​, then cos⁡−1(1213cos⁡x+513sin⁡x)\cos^{-1}\left(\dfrac{12}{13}\cos x+\dfrac{5}{13}\sin x\right)cos−1(1312​cosx+135​sinx) is equal to
  1. (A)x−tan⁡−143x-\tan^{-1}\dfrac{4}{3}x−tan−134​
  2. (B)x−tan⁡−1512x-\tan^{-1}\dfrac{5}{12}x−tan−1125​
  3. (C)x+tan⁡−145x+\tan^{-1}\dfrac{4}{5}x+tan−154​
  4. (D)x+tan⁡−1512x+\tan^{-1}\dfrac{5}{12}x+tan−1125​

Correct answer: (B)

Step-by-step solution →
Q61·Mathematics·Trigonometric FunctionsSingle correct
The value of (sin⁡70∘)(cot⁡10∘cot⁡70∘−1)(\sin 70^\circ)(\cot 10^\circ\cot 70^\circ-1)(sin70∘)(cot10∘cot70∘−1) is
  1. (A)111
  2. (B)000
  3. (C)32\dfrac{3}{2}23​
  4. (D)23\dfrac{2}{3}32​

Correct answer: (A)

Step-by-step solution →
Q62·Mathematics·Statistics and ProbabilitySingle correct
Marks obtained by all the students of class 12 are presented in a frequency distribution with classes of equal width. Let the median of this grouped data be 14 with median class interval 12-18 and median class frequency 12. If the number of students whose marks are less than 12 is 18, then the total number of students is
  1. (A)484848
  2. (B)444444
  3. (C)404040
  4. (D)525252

Correct answer: (B)

Step-by-step solution →
Q63·Mathematics·Vector AlgebraSingle correct
Let the position vectors of the vertices AAA, BBB and CCC of a tetrahedron ABCDABCDABCD be i^+2j^+k^\hat i+2\hat j+\hat ki^+2j^​+k^, i^+3j^−2k^\hat i+3\hat j-2\hat ki^+3j^​−2k^ and 2i^+j^−k^2\hat i+\hat j-\hat k2i^+j^​−k^ respectively. The altitude from the vertex DDD to the opposite face ABCABCABC meets the median line segment through AAA of the triangle ABCABCABC at the point EEE. If the length of ADADAD is 1103\dfrac{\sqrt{110}}{3}3110​​ and the volume of the tetrahedron is 80562\dfrac{\sqrt{805}}{6\sqrt2}62​805​​, then the position vector of EEE is
  1. (A)12(i^+4j^+7k^)\dfrac{1}{2}(\hat i+4\hat j+7\hat k)21​(i^+4j^​+7k^)
  2. (B)112(7i^+4j^+3k^)\dfrac{1}{12}(7\hat i+4\hat j+3\hat k)121​(7i^+4j^​+3k^)
  3. (C)16(12i^+12j^+k^)\dfrac{1}{6}(12\hat i+12\hat j+\hat k)61​(12i^+12j^​+k^)
  4. (D)16(7i^+12j^+k^)\dfrac{1}{6}(7\hat i+12\hat j+\hat k)61​(7i^+12j^​+k^)

Correct answer: (D)

Step-by-step solution →
Q64·Mathematics·Matrices and DeterminantsSingle correct
If AAA, BBB and (adj⁡(A−1)+adj⁡(B−1))(\operatorname{adj}(A^{-1})+\operatorname{adj}(B^{-1}))(adj(A−1)+adj(B−1)) are non-singular matrices of same order, then the inverse of A(adj⁡(A−1)+adj⁡(B−1))−1BA\big(\operatorname{adj}(A^{-1})+\operatorname{adj}(B^{-1})\big)^{-1}BA(adj(A−1)+adj(B−1))−1B, is equal to
  1. (A)AB−1+A−1BAB^{-1}+A^{-1}BAB−1+A−1B
  2. (B)adj⁡(B−1)+adj⁡(A−1)\operatorname{adj}(B^{-1})+\operatorname{adj}(A^{-1})adj(B−1)+adj(A−1)
  3. (C)1∣AB∣(adj⁡(B)+adj⁡(A))\dfrac{1}{|AB|}\big(\operatorname{adj}(B)+\operatorname{adj}(A)\big)∣AB∣1​(adj(B)+adj(A))
  4. (D)AB−1∣A∣+BA−1∣B∣\dfrac{AB^{-1}}{|A|}+\dfrac{BA^{-1}}{|B|}∣A∣AB−1​+∣B∣BA−1​

Correct answer: (C)

Step-by-step solution →
Q65·Mathematics·Matrices and DeterminantsSingle correct
If the system of equations (λ−1)x+(λ−4)y+λz=5(\lambda-1)x+(\lambda-4)y+\lambda z=5(λ−1)x+(λ−4)y+λz=5 λx+(λ−1)y+(λ−4)z=7\lambda x+(\lambda-1)y+(\lambda-4)z=7λx+(λ−1)y+(λ−4)z=7 (λ+1)x+(λ+2)y−(λ+2)z=9(\lambda+1)x+(\lambda+2)y-(\lambda+2)z=9(λ+1)x+(λ+2)y−(λ+2)z=9 has infinitely many solutions, then λ2+λ\lambda^2+\lambdaλ2+λ is equal to
  1. (A)101010
  2. (B)121212
  3. (C)666
  4. (D)202020

Correct answer: (B)

Step-by-step solution →
Q66·Mathematics·Statistics and ProbabilitySingle correct
One die has two faces marked 1, two faces marked 2, one face marked 3 and one face marked 4. Another die has one face marked 1, two faces marked 2, two faces marked 3 and one face marked 4. The probability of getting the sum of numbers to be 4 or 5, when both the dice are thrown together, is
  1. (A)12\dfrac{1}{2}21​
  2. (B)35\dfrac{3}{5}53​
  3. (C)23\dfrac{2}{3}32​
  4. (D)49\dfrac{4}{9}94​

Correct answer: (A)

Step-by-step solution →
Q67·Mathematics·Indefinite IntegrationInteger
If the area of the larger portion bounded between the curves x2+y2=25x^2+y^2=25x2+y2=25 and y=∣x−1∣y=|x-1|y=∣x−1∣ is 14(bπ+c)\dfrac{1}{4}(b\pi+c)41​(bπ+c), b,c∈Nb,c\in\mathbb{N}b,c∈N, then b+cb+cb+c is equal to _______

Correct answer: 77

Step-by-step solution →
Q68·Mathematics·Binomial Theorem and Its Simple ApplicationsInteger
The sum of all rational terms in the expansion of (1+21/3+31/2)6(1+2^{1/3}+3^{1/2})^6(1+21/3+31/2)6 is equal to _______

Correct answer: 612

Step-by-step solution →
Q69·Mathematics·CirclesInteger
Let the circle CCC touch the line x−y+1=0x-y+1=0x−y+1=0, have the centre on the positive x-axis, and cut off a chord of length 413\dfrac{4}{\sqrt{13}}13​4​ along the line −3x+2y=1-3x+2y=1−3x+2y=1. Let HHH be the hyperbola x2α2−y2β2=1\dfrac{x^2}{\alpha^2}-\dfrac{y^2}{\beta^2}=1α2x2​−β2y2​=1, whose one of the foci is the centre of CCC and the length of the transverse axis is the diameter of CCC. Then 2α2+3β22\alpha^2+3\beta^22α2+3β2 is equal to _______

Correct answer: 19

Step-by-step solution →
Q70·Mathematics·Limits and ContinuityInteger
If the set of all values of aaa, for which the equation 5x3−15x−a=05x^3-15x-a=05x3−15x−a=0 has three distinct real roots, is the interval (α,β)(\alpha,\beta)(α,β), then β−2α\beta-2\alphaβ−2α is equal to _______

Correct answer: 30

Step-by-step solution →
Q71·Mathematics·Quadratic EquationsInteger
If the equation a(b−c)x2+b(c−a)x+c(a−b)=0a(b-c)x^2+b(c-a)x+c(a-b)=0a(b−c)x2+b(c−a)x+c(a−b)=0 has equal roots, where a+c=15a+c=15a+c=15 and b=365b=\dfrac{36}{5}b=536​, then a2+c2a^2+c^2a2+c2 is equal to _______

Correct answer: 117

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
  • Thermodynamics 154/186
  • Aldehydes and Ketones 135/186
  • Equilibrium 163/186
  • 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
  • Electric Field and Coulomb's Law 133/186
  • Atomic Structure 161/186
  • Electronic Devices 145/186
  • Circles 142/186
  • Dual Nature of Matter and Radiation 155/186
  • Amines 133/186
  • Quadratic Equations 148/186
  • Some Basic Concepts in Chemistry 129/186
  • Electromagnetic Induction 120/186
  • Alcohols and Ethers 106/186
  • Purification and Characterisation of Organic Compounds 116/186
  • Oscillations 117/186
  • Nuclei 116/186
  • Organic Compounds Containing Halogens 109/186
  • Straight Lines 114/186
  • 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
  • Diazonium Salts and Reactions 53/186
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