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JEE Main 28 January 2025 Shift 2 Question Paper with Answers

28 January 2025 · January session · 72 questions

72 of the 75 questions from the JEE Main 28 January 2025 Shift 2 paper, each with its correct answer and tagged to the chapter it tests. Free to read, no account needed.

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

Physics
24
Chemistry
23
Mathematics
25

Physics — JEE Main 28 January 2025 Shift 2

Q1·Physics·Electromagnetic InductionSingle correct
A uniform magnetic field of 0.4 T acts perpendicular to a circular copper disc 20 cm in radius. The disc is having a uniform angular velocity of 10π10\pi10π rad s−1^{-1}−1 about an axis through its centre and perpendicular to the disc. What is the potential difference developed between the axis of the disc and the rim? (π=3.14\pi=3.14π=3.14)
  1. (A)0.0628 V
  2. (B)0.5024 V
  3. (C)0.2512 V
  4. (D)0.1256 V

Correct answer: (C)

Step-by-step solution →
Q2·Physics·Capacitors and DielectricsSingle correct
A parallel plate capacitor of capacitance 1 μ\muμF is charged to a potential difference of 20 V. The distance between plates is 1 μ\muμm. The energy density between plates of capacitor is:
  1. (A)1.8×1031.8\times10^31.8×103 J/m3^33
  2. (B)2×10−42\times10^{-4}2×10−4 J/m3^33
  3. (C)2×1022\times10^22×102 J/m3^33
  4. (D)1.8×1051.8\times10^51.8×105 J/m3^33

Correct answer: (A)

Step-by-step solution →
Q3·Physics·Units and MeasurementsSingle correct
Match List-I (Physical Quantity) with List-II (Dimensional Formula). Choose the correct answer from the options given below:
List-I (Physical Quantity)List-II (Dimensional Formula)
A.Angular ImpulseI.[M0L2T−2][M^0L^2T^{-2}][M0L2T−2]
B.Latent HeatII.[ML2T−3A−1][ML^2T^{-3}A^{-1}][ML2T−3A−1]
C.Electrical resistivityIII.[ML2T−1][ML^2T^{-1}][ML2T−1]
D.Electromotive forceIV.[ML3T−3A−2][ML^3T^{-3}A^{-2}][ML3T−3A−2]
  1. (A)(A)-(III), (B)-(I), (C)-(IV), (D)-(II)
  2. (B)(A)-(I), (B)-(III), (C)-(IV), (D)-(II)
  3. (C)(A)-(III), (B)-(I), (C)-(II), (D)-(IV)
  4. (D)(A)-(II), (B)-(I), (C)-(IV), (D)-(III)

Correct answer: (A)

Step-by-step solution →
Q4·Physics·Kinetic Theory of GasesSingle correct
The ratio of vapour densities of two gases at the same temperature is 425\frac{4}{25}254​, then the ratio of r.m.s. velocities will be:
  1. (A)254\frac{25}{4}425​
  2. (B)25\frac{2}{5}52​
  3. (C)52\frac{5}{2}25​
  4. (D)425\frac{4}{25}254​

Correct answer: (C)

Step-by-step solution →
Q5·Physics·Kinetic Theory of GasesSingle correct
The kinetic energy of translation of the molecules in 50 g of CO2_22​ gas at 17∘17^\circ17∘C is:
  1. (A)3986.3 J
  2. (B)4102.8 J
  3. (C)4205.5 J
  4. (D)3582.7 J

Correct answer: (B)

Step-by-step solution →
Q6·Physics·Geometrical OpticsSingle correct
In a long glass tube, mixture of two liquids A and B with refractive indices 1.3 and 1.4 respectively, forms a convex refractive meniscus towards A. If an object placed at 13 cm from the vertex of the meniscus in A forms an image with a magnification of '-2' then the radius of curvature of meniscus is:
  1. (A)1 cm
  2. (B)13\frac{1}{3}31​ cm
  3. (C)23\frac{2}{3}32​ cm
  4. (D)43\frac{4}{3}34​ cm

Correct answer: (C)

Step-by-step solution →
Q7·Physics·Atoms and NucleiSingle correct
The frequency of revolution of the electron in Bohr's orbit varies with n, the principal quantum number as
  1. (A)1n\frac{1}{n}n1​
  2. (B)1n3\frac{1}{n^3}n31​
  3. (C)1n4\frac{1}{n^4}n41​
  4. (D)1n2\frac{1}{n^2}n21​

Correct answer: (B)

Step-by-step solution →
Q8·Physics·Dual Nature of Matter and RadiationSingle correct
Which of the following phenomena can not be explained by wave theory of light?
  1. (A)Reflection of light
  2. (B)Diffraction of light
  3. (C)Refraction of light
  4. (D)Compton effect

Correct answer: (D)

Step-by-step solution →
Q9·Physics·KinematicsSingle correct
The velocity-time graph of an object moving along a straight line is shown in figure. What is the distance covered by the object between t = 0 to t = 4s?
  1. (A)30 m
  2. (B)10 m
  3. (C)13 m
  4. (D)11 m

Correct answer: (A)

Step-by-step solution →
Q10·Physics·GravitationSingle correct
Earth has mass 8 times and radius 2 times that of a planet. If the escape velocity from the earth is 11.2 km/s, the escape velocity in km/s from the planet will be:
  1. (A)11.2
  2. (B)5.6
  3. (C)2.8
  4. (D)8.4

Correct answer: (B)

Step-by-step solution →
Q11·Physics·OscillationsSingle correct
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): Knowing initial position x0x_0x0​ and initial momentum p0p_0p0​ is enough to determine the position and momentum at any time t for a simple harmonic motion with a given angular frequency ω\omegaω. Reason (R): The amplitude and phase can be expressed in terms of x0x_0x0​ and p0p_0p0​. In the light of the above statements, choose the correct answer from the options given below:
  1. (A)Both (A) and (R) are true and (R) is NOT the correct explanation of (A)
  2. (B)(A) is false but (R) is true
  3. (C)(A) is true but (R) is false
  4. (D)Both (A) and (R) are true and (R) is the correct explanation of (A)

Correct answer: (D)

Step-by-step solution →
Q12·Physics·Geometrical OpticsSingle correct
A concave mirror produces an image of an object such that the distance between the object and image is 20 cm. If the magnification of the image is '-3', then the magnitude of the radius of curvature of the mirror is:
  1. (A)3.75 cm
  2. (B)30 cm
  3. (C)7.5 cm
  4. (D)15 cm

Correct answer: (D)

Step-by-step solution →
Q13·Physics·Work, Energy and PowerSingle correct
A body of mass 4 kg is placed on a plane at a point P having coordinate (3, 4) m. Under the action of force F⃗=(2i^+3j^)\vec{F}=(2\hat{i}+3\hat{j})F=(2i^+3j^​) N, it moves to a new point Q having coordinates (6, 10) m in 4 sec. The average power and instantaneous power at the end of 4 sec are in the ratio of:
  1. (A)13 : 6
  2. (B)6 : 13
  3. (C)1 : 2
  4. (D)4 : 3

Correct answer: (B)

Step-by-step solution →
Q14·Physics·Electronic DevicesSingle correct
In the circuit shown here, assuming threshold voltage of diode is negligibly small, then voltage VABV_{AB}VAB​ is correctly represented by:
  1. (A)VABV_{AB}VAB​ would be zero at all times
  2. (B)a full sine wave (both half cycles appear)
  3. (C)positive half-wave humps only
  4. (D)negative half-wave humps only

Correct answer: (D)

Step-by-step solution →
Q15·Physics·Magnetic Field of CurrentSingle correct
An infinite wire has a circular bend of radius a, and carrying a current I as shown in figure. The magnitude of magnetic field at the origin O of the arc is given by:
  1. (A)μ0I4πa[π2+1]\frac{\mu_0 I}{4\pi a}\left[\frac{\pi}{2}+1\right]4πaμ0​I​[2π​+1]
  2. (B)μ0I4πa[3π2+1]\frac{\mu_0 I}{4\pi a}\left[\frac{3\pi}{2}+1\right]4πaμ0​I​[23π​+1]
  3. (C)μ0I2πa[π2+2]\frac{\mu_0 I}{2\pi a}\left[\frac{\pi}{2}+2\right]2πaμ0​I​[2π​+2]
  4. (D)μ0I4πa[3π2+2]\frac{\mu_0 I}{4\pi a}\left[\frac{3\pi}{2}+2\right]4πaμ0​I​[23π​+2]

Correct answer: (B)

Step-by-step solution →
Q16·Physics·Rotational MotionSingle correct
A uniform rod of mass 250 g having length 100 cm is balanced on a sharp edge at 40 cm mark. A mass of 400 g is suspended at 10 cm mark. To maintain the balance of the rod, the mass to be suspended at 90 cm mark, is:
  1. (A)300 g
  2. (B)190 g
  3. (C)200 g
  4. (D)290 g

Correct answer: (B)

Step-by-step solution →
Q17·Physics·Properties of Solids and LiquidsSingle correct
A 400 g solid cube having an edge of length 10 cm floats in water. How much volume of the cube is outside the water? (Given: density of water = 1000 kg m−3^{-3}−3)
  1. (A)1400 cm3^33
  2. (B)4000 cm3^33
  3. (C)400 cm3^33
  4. (D)600 cm3^33

Correct answer: (D)

Step-by-step solution →
Q18·Physics·Electromagnetic WavesSingle correct
The magnetic field of an E.M. wave is given by B⃗=(32i^+12j^)30sin⁡[ω(t−zc)]\vec{B}=\left(\frac{\sqrt{3}}{2}\hat{i}+\frac{1}{2}\hat{j}\right)30\sin\left[\omega\left(t-\frac{z}{c}\right)\right]B=(23​​i^+21​j^​)30sin[ω(t−cz​)] (S.I. Units). The corresponding electric field in S.I. units is:
  1. (A)E⃗=(12i^−32j^)30csin⁡[ω(t−zc)]\vec{E}=\left(\frac{1}{2}\hat{i}-\frac{\sqrt{3}}{2}\hat{j}\right)30c\sin\left[\omega\left(t-\frac{z}{c}\right)\right]E=(21​i^−23​​j^​)30csin[ω(t−cz​)]
  2. (B)E⃗=(34i^+14j^)30ccos⁡[ω(t−zc)]\vec{E}=\left(\frac{3}{4}\hat{i}+\frac{1}{4}\hat{j}\right)30c\cos\left[\omega\left(t-\frac{z}{c}\right)\right]E=(43​i^+41​j^​)30ccos[ω(t−cz​)]
  3. (C)E⃗=(12i^+32j^)30csin⁡[ω(t−zc)]\vec{E}=\left(\frac{1}{2}\hat{i}+\frac{\sqrt{3}}{2}\hat{j}\right)30c\sin\left[\omega\left(t-\frac{z}{c}\right)\right]E=(21​i^+23​​j^​)30csin[ω(t−cz​)]
  4. (D)E⃗=(32i^−12j^)30csin⁡[ω(t+zc)]\vec{E}=\left(\frac{\sqrt{3}}{2}\hat{i}-\frac{1}{2}\hat{j}\right)30c\sin\left[\omega\left(t+\frac{z}{c}\right)\right]E=(23​​i^−21​j^​)30csin[ω(t+cz​)]

Correct answer: (A)

Step-by-step solution →
Q19·Physics·Laws of MotionSingle correct
A balloon and its content having mass M is moving up with an acceleration 'a'. The mass that must be released from the content so that the balloon starts moving up with an acceleration '3a' will be: (Take 'g' as acceleration due to gravity)
  1. (A)3Ma2a−g\frac{3Ma}{2a-g}2a−g3Ma​
  2. (B)3Ma2a+g\frac{3Ma}{2a+g}2a+g3Ma​
  3. (C)2Ma3a+g\frac{2Ma}{3a+g}3a+g2Ma​
  4. (D)2Ma3a−g\frac{2Ma}{3a-g}3a−g2Ma​

Correct answer: (C)

Step-by-step solution →
Q20·Physics·Electromagnetic InductionInteger
A conducting bar moves on two conducting rails as shown in the figure. A constant magnetic field B exists into the page. The bar starts to move from the vertex at time t = 0 with a constant velocity. If the induced EMF is E∝tnE\propto t^nE∝tn, then value of n is ______.

Correct answer: 1

Step-by-step solution →
Q21·Physics·Electric Field and Coulomb's LawInteger
An electric dipole of dipole moment 6×10−66\times10^{-6}6×10−6 Cm is placed in uniform electric field of magnitude 10610^6106 V/m. Initially, the dipole moment is parallel to electric field. The work that needs to be done on the dipole to make its dipole moment opposite to the field, will be ______ J.

Correct answer: 12

Step-by-step solution →
Q22·Physics·Properties of Solids and LiquidsInteger
The volume contraction of a solid copper cube of edge length 10 cm, when subjected to a hydraulic pressure of 7×1067\times10^67×106 Pa, would be ______ mm3^33. (Given bulk modulus of copper = 1.4×10111.4\times10^{11}1.4×1011 Nm−2^{-2}−2)

Correct answer: 50

Step-by-step solution →
Q23·Physics·Current ElectricityInteger
The value of current I in the electrical circuit as given below, when potential at A is equal to the potential at B, will be ______ A.

Correct answer: 2

Step-by-step solution →
Q24·Physics·Geometrical OpticsInteger
A thin transparent film with refractive index 1.4, is held on circular ring of radius 1.8 cm. The fluid in the film evaporates such that transmission through the film at wavelength 560 nm goes to a minimum every 12 seconds. Assuming that the film is flat on its two sides, the rate of evaporation is ______ π×10−13\pi\times10^{-13}π×10−13 m3^33/s.

Correct answer: 54

Step-by-step solution →

Chemistry — JEE Main 28 January 2025 Shift 2

Q25·Chemistry·Chemical KineticsSingle correct
Consider the elementary reaction A(g) + B(g) →\to→ C(g) + D(g). If the volume of reaction mixture is suddenly reduced to 13\frac{1}{3}31​ of its initial volume, the reaction rate will become 'x' times of the original reaction rate. The value of x is:
  1. (A)19\frac{1}{9}91​
  2. (B)9
  3. (C)13\frac{1}{3}31​
  4. (D)3

Correct answer: (B)

Step-by-step solution →
Q26·Chemistry·d- and f-Block ElementsSingle correct
The amphoteric oxide among V2_22​O3_33​, V2_22​O4_44​ and V2_22​O5_55​ upon reaction with alkali leads to formation of an oxide anion. The oxidation state of V in the oxide anion is:
  1. (A)+3
  2. (B)+7
  3. (C)+5
  4. (D)+4

Correct answer: (C)

Step-by-step solution →
Q27·Chemistry·BiomoleculesSingle correct
Match List-I (Saccharides) with List-II (Glycosidic-linkages found). Choose the correct answer from the options given below:
List-I (Saccharides)List-II (Glycosidic-linkages found)
A.SucroseI.α 1−4\alpha\,1-4α1−4
B.MaltoseII.α 1−4\alpha\,1-4α1−4 and α 1−6\alpha\,1-6α1−6
C.LactoseIII.α 1−β 2\alpha\,1-\beta\,2α1−β2
D.AmylopectinIV.β 1−4\beta\,1-4β1−4
  1. (A)(A)-(III), (B)-(I), (C)-(IV), (D)-(II)
  2. (B)(A)-(IV), (B)-(II), (C)-(I), (D)-(III)
  3. (C)(A)-(II), (B)-(IV), (C)-(I), (D)-(III)
  4. (D)(A)-(I), (B)-(II), (C)-(III), (D)-(IV)

Correct answer: (A)

Step-by-step solution →
Q28·Chemistry·HydrocarbonsSingle correct
Identify product [A], [B] and [C] in the following reaction sequence: CH3_33​-C≡\equiv≡CH →Pd/C, H2\xrightarrow{\text{Pd/C, H}_2}Pd/C, H2​​ [A] →(i) O3 (ii) Zn, H2O\xrightarrow{\text{(i) O}_3\text{ (ii) Zn, H}_2\text{O}}(i) O3​ (ii) Zn, H2​O​ [B] + [C]
  1. (A)[A]: CH3−CH=CH2CH_3-CH=CH_2CH3​−CH=CH2​, [B]: CH3CHOCH_3CHOCH3​CHO, [C]: HCHOHCHOHCHO
  2. (B)[A]: CH2=CH2CH_2=CH_2CH2​=CH2​, [B]: CH3COCH3CH_3COCH_3CH3​COCH3​, [C]: HCHOHCHOHCHO
  3. (C)[A]: CH3−CH=CH2CH_3-CH=CH_2CH3​−CH=CH2​, [B]: CH3CHOCH_3CHOCH3​CHO, [C]: CH3CH2OHCH_3CH_2OHCH3​CH2​OH
  4. (D)[A]: CH3CH2CH3CH_3CH_2CH_3CH3​CH2​CH3​, [B]: CH3CHOCH_3CHOCH3​CHO, [C]: HCHOHCHOHCHO

Correct answer: (A)

Step-by-step solution →
Q29·Chemistry·EquilibriumSingle correct
Arrange the following in increasing order of solubility product: Ca(OH)2_22​, AgBr, PbS, HgS.
  1. (A)PbS < HgS < Ca(OH)2_22​ < AgBr
  2. (B)HgS < PbS < AgBr < Ca(OH)2_22​
  3. (C)Ca(OH)2_22​ < AgBr < HgS < PbS
  4. (D)HgS < AgBr < PbS < Ca(OH)2_22​

Correct answer: (B)

Step-by-step solution →
Q30·Chemistry·Purification and Characterisation of Organic CompoundsSingle correct
The purification method based on the following physical transformation is: Solid →Heat\xrightarrow{\text{Heat}}Heat​ Vapour →Cool\xrightarrow{\text{Cool}}Cool​ Solid.
  1. (A)Sublimation
  2. (B)Distillation
  3. (C)Crystallization
  4. (D)Extraction

Correct answer: (A)

Step-by-step solution →
Q31·Chemistry·BiomoleculesSingle correct
Identify correct conversion during acidic hydrolysis from the following: (A) starch gives galactose. (B) cane sugar gives equal amount of glucose and fructose. (C) milk sugar gives glucose and galactose. (D) amylopectin gives glucose and fructose. (E) amylose gives only glucose. Choose the correct answer from the options given below:
  1. (A)(C), (D) and (E) only
  2. (B)(A), (B) and (C) only
  3. (C)(B), (C) and (E) only
  4. (D)(B), (C) and (D) only

Correct answer: (C)

Step-by-step solution →
Q32·Chemistry·Chemical ThermodynamicsSingle correct
An ideal gas undergoes a cyclic transformation starting from the point A and coming back to the same point by tracing the path A →\to→ B →\to→ C →\to→ D →\to→ A as shown in the three cases above. Choose the correct option regarding ΔU\Delta UΔU.
  1. (A)ΔU\Delta UΔU(Case-III) > ΔU\Delta UΔU(Case-II) > ΔU\Delta UΔU(Case-I)
  2. (B)ΔU\Delta UΔU(Case-I) > ΔU\Delta UΔU(Case-II) > ΔU\Delta UΔU(Case-III)
  3. (C)ΔU\Delta UΔU(Case-I) > ΔU\Delta UΔU(Case-III) > ΔU\Delta UΔU(Case-II)
  4. (D)ΔU\Delta UΔU(Case-I) = ΔU\Delta UΔU(Case-II) = ΔU\Delta UΔU(Case-III)

Correct answer: (D)

Step-by-step solution →
Q33·Chemistry·Organic Compounds Containing HalogensSingle correct
The product B formed in the following reaction sequence is (4-methylstyrene, shown in figure) →HCl\xrightarrow{\text{HCl}}HCl​ (A)(Major) →AgCN\xrightarrow{\text{AgCN}}AgCN​ (B)(Major). Choose the correct structure of B from the options shown in the figure:
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (D)

Step-by-step solution →
Q34·Chemistry·SolutionsSingle correct
Concentrated nitric acid is labelled as 75% by mass. The volume in mL of the solution which contains 30 g of nitric acid is: (Given: Density of nitric acid solution is 1.25 g mL−1^{-1}−1)
  1. (A)45
  2. (B)55
  3. (C)32
  4. (D)40

Correct answer: (C)

Step-by-step solution →
Q35·Chemistry·Coordination CompoundsSingle correct
Match List-I (Complex) with List-II (Hybridisation of central metal ion). Choose the correct answer from the options given below:
List-I (Complex)List-II (Hybridisation of central metal ion)
A.[CoF6_66​]3−^{3-}3−I.d2^22sp3^33
B.[NiCl4_44​]2−^{2-}2−II.sp3^33
C.[Co(NH3_33​)6_66​]3+^{3+}3+III.sp3^33d2^22
D.[Ni(CN)4_44​]2−^{2-}2−IV.dsp2^22
  1. (A)(A)-(I), (B)-(IV), (C)-(III), (D)-(II)
  2. (B)(A)-(III), (B)-(II), (C)-(I), (D)-(IV)
  3. (C)(A)-(I), (B)-(II), (C)-(III), (D)-(IV)
  4. (D)(A)-(III), (B)-(IV), (C)-(I), (D)-(II)

Correct answer: (B)

Step-by-step solution →
Q36·Chemistry·HydrocarbonsSingle correct
The total number of compounds from below (shown in the figure) when treated with hot KMnO4_44​ giving benzoic acid is:
  1. (A)3
  2. (B)4
  3. (C)6
  4. (D)5

Correct answer: (D)

Step-by-step solution →
Q37·Chemistry·Organic Compounds Containing HalogensSingle correct
The major product of the following reaction (dibromide shown in the figure) →KOH(EtOH),Δ\xrightarrow{\text{KOH(EtOH)}, \Delta}KOH(EtOH),Δ​ Major product, is:
  1. (A)6-Phenylhepta-2,4-diene
  2. (B)2-Phenylhepta-2,5-diene
  3. (C)6-Phenylhepta-3,5-diene
  4. (D)2-Phenylhepta-2,4-diene

Correct answer: (D)

Step-by-step solution →
Q38·Chemistry·Classification of Elements and Periodicity in PropertiesSingle correct
Given below are two statements: Statement (I): According to the Law of Octaves, the elements were arranged in the increasing order of their atomic number. Statement (II): Meyer observed a periodically repeated pattern upon plotting physical properties of certain elements against their respective atomic numbers. In the light of the above statements, choose the correct answer from the options given below:
  1. (A)Statement I is false but Statement II is true
  2. (B)Both Statement I and Statement II are true
  3. (C)Statement I is true but Statement II is false
  4. (D)Both Statement I and Statement II are false

Correct answer: (D)

Step-by-step solution →
Q39·Chemistry·Chemical KineticsSingle correct
For bacterial growth in a cell culture, growth law is very similar to the law of radioactive decay. Which of the following graphs is most suitable to represent bacterial colony growth?
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (D)

Step-by-step solution →
Q40·Chemistry·Atomic StructureSingle correct
Which of the following is/are not correct with respect to energy of atomic orbitals of hydrogen atom? (A) 1s<2p<3d<4s1s<2p<3d<4s1s<2p<3d<4s (B) 1s<2s=2p<3p1s<2s=2p<3p1s<2s=2p<3p (C) 1s<2s<2p<3p1s<2s<2p<3p1s<2s<2p<3p (D) 1s<2s<4s<3d1s<2s<4s<3d1s<2s<4s<3d. Choose the correct answer from the options given below:
  1. (A)(B) and (D) only
  2. (B)(A) and (C) only
  3. (C)(C) and (D) only
  4. (D)(A) and (B) only

Correct answer: (C)

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Q41·Chemistry·AminesSingle correct
Identify correct statements: (A) Primary amines do not give diazonium salts when reacted with NaNO2_22​ in acidic condition. (B) Aliphatic and aromatic primary amines on heating with CHCl3_33​ and ethanolic KOH form carbylamines. (C) Secondary and tertiary amines also give carbylamine reaction. (D) Benzenesulfonyl chloride is known as Hinsberg's reagent. (E) Tertiary amines reacts with benzenesulfonyl chloride easily. Choose the correct answer from the options given below:
  1. (A)(B) and (D) only
  2. (B)(A) and (B) only
  3. (C)(D) and (E) only
  4. (D)(B) and (C) only

Correct answer: (A)

Step-by-step solution →
Q42·Chemistry·IsomerismSingle correct
Given below are two statements: Statement (I): the two compounds shown in the figure are isomeric compounds. Statement (II): the two amines shown in the figure are functional group isomers. In the light of the above statements, choose the correct answer from the options given below:
  1. (A)Both Statement I and Statement II are false
  2. (B)Both Statement I and Statement II are true
  3. (C)Statement I is true but Statement II is false
  4. (D)Statement I is false but Statement II is true

Correct answer: (B)

Step-by-step solution →
Q43·Chemistry·d- and f-Block ElementsInteger
The spin only magnetic moment (μ\muμ) value (B.M.) of the compound with strongest oxidising power among Mn2_22​O3_33​, TiO and VO is ______ B.M. (Nearest integer).

Correct answer: 5

Step-by-step solution →
Q44·Chemistry·Chemical ThermodynamicsInteger
Consider the following data: Heat of formation of CO2_22​(g) = −393.5-393.5−393.5 kJ mol−1^{-1}−1, Heat of formation of H2_22​O(ℓ\ellℓ) = −286.0-286.0−286.0 kJ mol−1^{-1}−1, Heat of combustion of benzene = −3267.0-3267.0−3267.0 kJ mol−1^{-1}−1. The heat of formation of benzene is ______ kJ mol−1^{-1}−1. (Nearest integer)

Correct answer: 48

Step-by-step solution →
Q45·Chemistry·Redox Reactions and ElectrochemistryInteger
Electrolysis of 600 mL aqueous solution of NaCl for 5 min changes the pH of the solution to 12. The current in Amperes used for the given electrolysis is ______. (Nearest integer)

Correct answer: 2

Step-by-step solution →
Q46·Chemistry·p-Block ElementsInteger
A group 15 element forms dπ\piπ-dπ\piπ bond with transition metals. It also forms hydride, which is a strongest base among the hydrides of other group members that form dπ\piπ-dπ\piπ bond. The atomic number of this element is ______.

Correct answer: 15

Step-by-step solution →
Q47·Chemistry·Chemical Bonding and Molecular StructureInteger
Total number of molecules/species from following which will be paramagnetic is ______. O2_22​, O2+_2^+2+​, O2−_2^-2−​, NO, NO2_22​, CO, K2_22​[NiCl4_44​], [Co(NH3_33​)6_66​]Cl3_33​, K2_22​[Ni(CN)4_44​]

Correct answer: 6

Step-by-step solution →

Mathematics — JEE Main 28 January 2025 Shift 2

Q48·Mathematics·Statistics and ProbabilitySingle correct
Bag B1B_1B1​ contains 6 white and 4 blue balls, Bag B2B_2B2​ contains 4 white and 6 blue balls, and Bag B3B_3B3​ contains 5 white and 5 blue balls. One of the bags is selected at random and a ball is drawn from it. If the ball is white, then the probability, that the ball is drawn from Bag B2B_2B2​, is:
  1. (A)13\frac{1}{3}31​
  2. (B)415\frac{4}{15}154​
  3. (C)23\frac{2}{3}32​
  4. (D)25\frac{2}{5}52​

Correct answer: (B)

Step-by-step solution →
Q49·Mathematics·Vector AlgebraSingle correct
Let A, B, C be three points in xy-plane, whose position vector are given by 3i^+j^\sqrt{3}\hat{i}+\hat{j}3​i^+j^​, i^+3j^\hat{i}+\sqrt{3}\hat{j}i^+3​j^​ and ai^+(1−a)j^a\hat{i}+(1-a)\hat{j}ai^+(1−a)j^​ respectively with respect to the origin O. If the distance of the point C from the point bisecting the angle between the vectors OA⃗\vec{OA}OA and OB⃗\vec{OB}OB is 92\frac{9}{\sqrt{2}}2​9​, then the sum of all the possible values of a is:
  1. (A)1
  2. (B)92\frac{9}{2}29​
  3. (C)0
  4. (D)2

Correct answer: (A)

Step-by-step solution →
Q50·Mathematics·Vector AlgebraSingle correct
If the components of a⃗=αi^+βj^+γk^\vec{a}=\alpha\hat{i}+\beta\hat{j}+\gamma\hat{k}a=αi^+βj^​+γk^ along and perpendicular to b⃗=3i^+j^−k^\vec{b}=3\hat{i}+\hat{j}-\hat{k}b=3i^+j^​−k^ respectively, are 1611(3i^+j^−k^)\frac{16}{11}(3\hat{i}+\hat{j}-\hat{k})1116​(3i^+j^​−k^) and 111(−4i^−5j^−17k^)\frac{1}{11}(-4\hat{i}-5\hat{j}-17\hat{k})111​(−4i^−5j^​−17k^), then α2+β2+γ2\alpha^2+\beta^2+\gamma^2α2+β2+γ2 is equal to:
  1. (A)23
  2. (B)18
  3. (C)16
  4. (D)26

Correct answer: (D)

Step-by-step solution →
Q51·Mathematics·Quadratic EquationsSingle correct
If α+iβ\alpha+i\betaα+iβ and γ+iδ\gamma+i\deltaγ+iδ are the roots of x2−(3−2i)x−(2i−2)=0x^2-(3-2i)x-(2i-2)=0x2−(3−2i)x−(2i−2)=0, i=−1i=\sqrt{-1}i=−1​, then αγ+βδ\alpha\gamma+\beta\deltaαγ+βδ is equal to:
  1. (A)6
  2. (B)2
  3. (C)-2
  4. (D)-6

Correct answer: (B)

Step-by-step solution →
Q52·Mathematics·EllipseSingle correct
If the midpoint of a chord of the ellipse x29+y24=1\frac{x^2}{9}+\frac{y^2}{4}=19x2​+4y2​=1 is (2,43)\left(\sqrt{2},\frac{4}{3}\right)(2​,34​), and the length of the chord is 2α3\frac{2\sqrt{\alpha}}{3}32α​​, then α\alphaα is:
  1. (A)18
  2. (B)22
  3. (C)26
  4. (D)20

Correct answer: (B)

Step-by-step solution →
Q53·Mathematics·Permutations and CombinationsSingle correct
Let S be the set of all the words that can be formed by arranging all the letters of the word GARDEN. From the set S, one word is selected at random. The probability that the selected word will NOT have vowels in alphabetical order is:
  1. (A)14\frac{1}{4}41​
  2. (B)23\frac{2}{3}32​
  3. (C)13\frac{1}{3}31​
  4. (D)12\frac{1}{2}21​

Correct answer: (D)

Step-by-step solution →
Q54·Mathematics·Definite IntegrationSingle correct
Let f be a real valued continuous function defined on the positive real axis such that g(x)=∫0xt f(t)dtg(x)=\int_0^x t\,f(t)dtg(x)=∫0x​tf(t)dt. If g(x3)=x6+x7g(x^3)=x^6+x^7g(x3)=x6+x7, then value of ∑r=115f(r3)\sum_{r=1}^{15} f(r^3)∑r=115​f(r3) is:
  1. (A)320
  2. (B)340
  3. (C)270
  4. (D)310

Correct answer: (D)

Step-by-step solution →
Q55·Mathematics·Three Dimensional GeometrySingle correct
The square of the distance of the point (157,327,7)\left(\frac{15}{7},\frac{32}{7},7\right)(715​,732​,7) from the line x+13=y+35=z+57\frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7}3x+1​=5y+3​=7z+5​ in the direction of the vector i^+4j^+7k^\hat{i}+4\hat{j}+7\hat{k}i^+4j^​+7k^ is:
  1. (A)54
  2. (B)41
  3. (C)66
  4. (D)44

Correct answer: (C)

Step-by-step solution →
Q56·Mathematics·Area Under CurvesSingle correct
The area of the region bounded by the curves x(1+y2)=1x(1+y^2)=1x(1+y2)=1 and y2=2xy^2=2xy2=2x is:
  1. (A)2(π2−13)2\left(\frac{\pi}{2}-\frac{1}{3}\right)2(2π​−31​)
  2. (B)π4−13\frac{\pi}{4}-\frac{1}{3}4π​−31​
  3. (C)π2−13\frac{\pi}{2}-\frac{1}{3}2π​−31​
  4. (D)12(π2−13)\frac{1}{2}\left(\frac{\pi}{2}-\frac{1}{3}\right)21​(2π​−31​)

Correct answer: (C)

Step-by-step solution →
Q57·Mathematics·Matrices and DeterminantsSingle correct
Let A=[12−201]A=\begin{bmatrix}\frac{1}{\sqrt{2}}&-2\\0&1\end{bmatrix}A=[2​1​0​−21​] and P=[cos⁡θ−sin⁡θsin⁡θcos⁡θ]P=\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}P=[cosθsinθ​−sinθcosθ​], θ>0\theta>0θ>0. If B=PAPTB=PAP^TB=PAPT, C=PTB10PC=P^TB^{10}PC=PTB10P and the sum of the diagonal elements of C is mn\frac{m}{n}nm​, where gcd⁡(m,n)=1\gcd(m,n)=1gcd(m,n)=1, then m+nm+nm+n is:
  1. (A)65
  2. (B)127
  3. (C)258
  4. (D)2049

Correct answer: (A)

Step-by-step solution →
Q58·Mathematics·Indefinite IntegrationSingle correct
If f(x)=∫1x1/4(1+x1/4)dxf(x)=\int\frac{1}{x^{1/4}(1+x^{1/4})}dxf(x)=∫x1/4(1+x1/4)1​dx, f(0)=−6f(0)=-6f(0)=−6, then f(1)f(1)f(1) is equal to:
  1. (A)log⁡e2+2\log_e 2+2loge​2+2
  2. (B)4(log⁡e2−2)4(\log_e 2-2)4(loge​2−2)
  3. (C)2−log⁡e22-\log_e 22−loge​2
  4. (D)4(log⁡e2+2)4(\log_e 2+2)4(loge​2+2)

Correct answer: (B)

Step-by-step solution →
Q59·Mathematics·Definite IntegrationSingle correct
Let f:R→Rf:R\to Rf:R→R be a twice differentiable function such that f(2)=1f(2)=1f(2)=1. If F(x)=x f(x)F(x)=x\,f(x)F(x)=xf(x) for all x∈Rx\in Rx∈R, ∫02x F′(x)dx=6\int_0^2 x\,F'(x)dx=6∫02​xF′(x)dx=6 and ∫02x2F′′(x)dx=40\int_0^2 x^2 F''(x)dx=40∫02​x2F′′(x)dx=40, then F′(2)+∫02F(x)dxF'(2)+\int_0^2 F(x)dxF′(2)+∫02​F(x)dx is equal to:
  1. (A)11
  2. (B)15
  3. (C)9
  4. (D)13

Correct answer: (A)

Step-by-step solution →
Q60·Mathematics·Sequence and SeriesSingle correct
For positive integers n, if 4an=(n2+5n+6)4a_n=(n^2+5n+6)4an​=(n2+5n+6) and Sn=∑k=1n1akS_n=\sum_{k=1}^{n}\frac{1}{a_k}Sn​=∑k=1n​ak​1​, then the value of 507 S2025507\,S_{2025}507S2025​ is:
  1. (A)540
  2. (B)1350
  3. (C)675
  4. (D)135

Correct answer: (C)

Step-by-step solution →
Q61·Mathematics·Sets, Relations and FunctionsSingle correct
Let f:[0,3]→Af:[0,3]\to Af:[0,3]→A be defined by f(x)=2x3−15x2+36x+7f(x)=2x^3-15x^2+36x+7f(x)=2x3−15x2+36x+7 and g:[0,∞)→Bg:[0,\infty)\to Bg:[0,∞)→B be defined by g(x)=x2025x2025+1g(x)=\frac{x^{2025}}{x^{2025}+1}g(x)=x2025+1x2025​. If both the functions are onto and S={x∈Z:x∈AS=\{x\in Z:x\in AS={x∈Z:x∈A or x∈B}x\in B\}x∈B}, then n(S)n(S)n(S) is equal to:
  1. (A)30
  2. (B)36
  3. (C)29
  4. (D)31

Correct answer: (A)

Step-by-step solution →
Q62·Mathematics·Inverse Trigonometric FunctionsSingle correct
Let [x][x][x] denote the greatest integer less than or equal to x. Then domain of f(x)=sec⁡−1(2[x]+1)f(x)=\sec^{-1}(2[x]+1)f(x)=sec−1(2[x]+1) is:
  1. (A)(−∞,−1]∪[0,∞)(-\infty,-1]\cup[0,\infty)(−∞,−1]∪[0,∞)
  2. (B)(−∞,∞)(-\infty,\infty)(−∞,∞)
  3. (C)(−∞,−1]∪[1,∞)(-\infty,-1]\cup[1,\infty)(−∞,−1]∪[1,∞)
  4. (D)(−∞,∞)−{0}(-\infty,\infty)-\{0\}(−∞,∞)−{0}

Correct answer: (B)

Step-by-step solution →
Q63·Mathematics·Trigonometric FunctionsSingle correct
If ∑r=113{1sin⁡(π4+(r−1)π6)sin⁡(π4+rπ6)}=a3+b\sum_{r=1}^{13}\left\{\frac{1}{\sin\left(\frac{\pi}{4}+(r-1)\frac{\pi}{6}\right)\sin\left(\frac{\pi}{4}+r\frac{\pi}{6}\right)}\right\}=a\sqrt{3}+b∑r=113​{sin(4π​+(r−1)6π​)sin(4π​+r6π​)1​}=a3​+b, a,b∈Za,b\in Za,b∈Z, then a2+b2a^2+b^2a2+b2 is equal to:
  1. (A)10
  2. (B)2
  3. (C)8
  4. (D)4

Correct answer: (C)

Step-by-step solution →
Q64·Mathematics·Straight LinesSingle correct
Two equal sides of an isosceles triangle are along −x+2y=4-x+2y=4−x+2y=4 and x+y=4x+y=4x+y=4. If m is the slope of its third side, then the sum, of all possible distinct values of m, is:
  1. (A)-6
  2. (B)12
  3. (C)6
  4. (D)−210-2\sqrt{10}−210​

Correct answer: (C)

Step-by-step solution →
Q65·Mathematics·Binomial Theorem and Its Simple ApplicationsSingle correct
Let the coefficients of three consecutive terms TrT_rTr​, Tr+1T_{r+1}Tr+1​ and Tr+2T_{r+2}Tr+2​ in the binomial expansion of (a+b)12(a+b)^{12}(a+b)12 be in a G.P. and let p be the number of all possible values of r. Let q be the sum of all rational terms in the binomial expansion of (34+43)12\left(\sqrt[4]{3}+\sqrt[3]{4}\right)^{12}(43​+34​)12. Then p+q is equal to:
  1. (A)283
  2. (B)295
  3. (C)287
  4. (D)299

Correct answer: (A)

Step-by-step solution →
Q66·Mathematics·HyperbolaSingle correct
If A and B are the points of intersection of the circle x2+y2−8x=0x^2+y^2-8x=0x2+y2−8x=0 and the hyperbola x29−y24=1\frac{x^2}{9}-\frac{y^2}{4}=19x2​−4y2​=1 and a point P moves on the line 2x−3y+4=02x-3y+4=02x−3y+4=0, then the centroid of △PAB\triangle PAB△PAB lies on the line:
  1. (A)4x−9y=124x-9y=124x−9y=12
  2. (B)x+9y=36x+9y=36x+9y=36
  3. (C)9x−9y=329x-9y=329x−9y=32
  4. (D)6x−9y=206x-9y=206x−9y=20

Correct answer: (D)

Step-by-step solution →
Q67·Mathematics·Sets, Relations and FunctionsSingle correct
Let f:R−{0}→(−∞,1)f:R-\{0\}\to(-\infty,1)f:R−{0}→(−∞,1) be a polynomial of degree 2, satisfying f(x) f(1x)=f(x)+f(1x)f(x)\,f\left(\frac{1}{x}\right)=f(x)+f\left(\frac{1}{x}\right)f(x)f(x1​)=f(x)+f(x1​). If f(K)=−2Kf(K)=-2Kf(K)=−2K, then the sum of squares of all possible values of K is:
  1. (A)1
  2. (B)6
  3. (C)7
  4. (D)9

Correct answer: (B)

Step-by-step solution →
Q68·Mathematics·Permutations and CombinationsInteger
The number of natural numbers, between 212 and 999, such that the sum of their digits is 15, is ______.

Correct answer: 64

Step-by-step solution →
Q69·Mathematics·Limits and ContinuityInteger
Let f(x)=lim⁡n→∞∑r=0n(tan⁡(x/2r+1)+tan⁡3(x/2r+1)1−tan⁡2(x/2r+1))f(x)=\lim_{n\to\infty}\sum_{r=0}^{n}\left(\frac{\tan(x/2^{r+1})+\tan^3(x/2^{r+1})}{1-\tan^2(x/2^{r+1})}\right)f(x)=limn→∞​∑r=0n​(1−tan2(x/2r+1)tan(x/2r+1)+tan3(x/2r+1)​). Then lim⁡x→0ex−ef(x)x−f(x)\lim_{x\to0}\frac{e^x-e^{f(x)}}{x-f(x)}limx→0​x−f(x)ex−ef(x)​ is equal to ______.

Correct answer: 1

Step-by-step solution →
Q70·Mathematics·Sequence and SeriesInteger
The interior angles of a polygon with n sides, are in an A.P. with common difference 6∘6^\circ6∘. If the largest interior angle of the polygon is 219∘219^\circ219∘, then n is equal to ______.

Correct answer: 20

Step-by-step solution →
Q71·Mathematics·ParabolaInteger
Let A and B be the two points of intersection of the line y+5=0y+5=0y+5=0 and the mirror image of the parabola y2=4xy^2=4xy2=4x with respect to the line x+y+4=0x+y+4=0x+y+4=0. If d denotes the distance between A and B, and a denotes the area of △SAB\triangle SAB△SAB, where S is the focus of the parabola y2=4xy^2=4xy2=4x, then the value of (a+d)(a+d)(a+d) is ______.

Correct answer: 14

Step-by-step solution →
Q72·Mathematics·Differential EquationsInteger
If y=y(x)y=y(x)y=y(x) is the solution of the differential equation 4−x2dydx=((sin⁡−1x2)2−y)sin⁡−1x2\sqrt{4-x^2}\frac{dy}{dx}=\left(\left(\sin^{-1}\frac{x}{2}\right)^2-y\right)\sin^{-1}\frac{x}{2}4−x2​dxdy​=((sin−12x​)2−y)sin−12x​, −2≤x≤2-2\le x\le2−2≤x≤2, y(2)=π2−84y(2)=\frac{\pi^2-8}{4}y(2)=4π2−8​, then y2(0)y^2(0)y2(0) is equal to ______.

Correct answer: 4

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
  • Equilibrium 163/186
  • Solutions 158/186
  • Laws of Motion 130/186
  • Chemical Thermodynamics 165/186
  • Units and Measurements 149/186
  • Hydrocarbons 126/186
  • 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
  • Electromagnetic Induction 120/186
  • Area Under Curves 139/186
  • Kinetic Theory of Gases 135/186
  • 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
  • Ellipse 103/186
  • Inverse Trigonometric Functions 93/186
  • Hyperbola 77/186
  • Indefinite Integration 66/186
  • Isomerism 51/186
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