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JEE Main 24 January 2025 Shift 2 Question Paper with Answers

24 January 2025 · January session · 75 questions

The complete JEE Main 24 January 2025 Shift 2 paper — all 75 questions with the correct answer for each, tagged to the chapter it tests. Free to read, no account needed.

Physics
25
Chemistry
25
Mathematics
25

Physics — JEE Main 24 January 2025 Shift 2

Q1·Physics·Wave OpticsSingle correct
Young's double slit interference apparatus is immersed in a liquid of refractive index 1.44. It has slit separation of 1.5 mm. The slits are illuminated by a parallel beam of light whose wavelength in air is 690 nm. The fringe-width recorded behind the plane of slits at a distance of 0.72 m, will be :
  1. (A)0.23 mm
  2. (B)0.33 mm
  3. (C)0.63 mm
  4. (D)0.46 mm

Correct answer: (A)

Step-by-step solution →
Q2·Physics·Electromagnetic WavesSingle correct
Arrange the following in the ascending order of wavelength (λ)(\lambda)(λ): (A) Microwaves (λ1)(\lambda_1)(λ1​) (B) Ultraviolet rays (λ2)(\lambda_2)(λ2​) (C) Infrared rays (λ3)(\lambda_3)(λ3​) (D) X-rays (λ4)(\lambda_4)(λ4​). Choose the most appropriate answer from the options given below :
  1. (A)λ1<λ3<λ2<λ4\lambda_1<\lambda_3<\lambda_2<\lambda_4λ1​<λ3​<λ2​<λ4​
  2. (B)λ3<λ4<λ2<λ1\lambda_3<\lambda_4<\lambda_2<\lambda_1λ3​<λ4​<λ2​<λ1​
  3. (C)λ4<λ2<λ3<λ1\lambda_4<\lambda_2<\lambda_3<\lambda_1λ4​<λ2​<λ3​<λ1​
  4. (D)λ4<λ3<λ1<λ2\lambda_4<\lambda_3<\lambda_1<\lambda_2λ4​<λ3​<λ1​<λ2​

Correct answer: (C)

Step-by-step solution →
Q3·Physics·Magnetic Field of CurrentSingle correct
Given below are two statements. Assertion (A) : An electron in a certain region of uniform magnetic field is moving with constant velocity in a straight line path. Reason (R) : The magnetic field in that region is along the direction of velocity of the electron. In the light of the above statements, choose the correct answer from the options given below :
  1. (A)(A) is false but (R) is true
  2. (B)Both (A) and (R) are true and (R) is the correct explanation of (A)
  3. (C)Both (A) and (R) are true but (R) is NOT the correct explanation of (A)
  4. (D)(A) is true but (R) is false

Correct answer: (B)

Step-by-step solution →
Q4·Physics·Rotational MotionSingle correct
A solid sphere is rolling without slipping on a horizontal plane. The ratio of the linear kinetic energy of the centre of mass of the sphere and rotational kinetic energy is:
  1. (A)25\dfrac{2}{5}52​
  2. (B)52\dfrac{5}{2}25​
  3. (C)34\dfrac{3}{4}43​
  4. (D)43\dfrac{4}{3}34​

Correct answer: (B)

Step-by-step solution →
Q5·Physics·Magnetic Field of CurrentSingle correct
A long straight wire of a circular cross-section with radius "a" carries a steady current I. The current I is uniformly distributed across this cross-section. The plot of magnitude of magnetic field B with distance r from the centre of the wire is given by (choose the correct graph):
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (A)

Step-by-step solution →
Q6·Physics·ThermodynamicsSingle correct
Given below are two statements. Assertion (A) : In an insulated container, a gas is adiabatically shrunk to half of its initial volume. The temperature of the gas decreases. Reason (R) : Free expansion of an ideal gas is an irreversible and an adiabatic process. In the light of the above statement, choose the correct answer from the options given below :
  1. (A)Both (A) and (R) are true and (R) is the correct explanation of (A)
  2. (B)(A) is true but (R) is false
  3. (C)(A) is false but (R) is true
  4. (D)Both (A) and (R) are true but (R) is NOT the correct explanation of (A)

Correct answer: (C)

Step-by-step solution →
Q7·Physics·Electric PotentialSingle correct
In the first configuration (1) as shown in the figure, four identical charges (q0)(q_0)(q0​) are kept at the corners A, B, C and D of a square of side length "a". In the second configuration (2), the same charges are shifted to mid points G, E, F and H, of the sides of the square. If K=14πε0K=\dfrac{1}{4\pi\varepsilon_0}K=4πε0​1​, the difference between the potential energy of configuration (2) and (1) is given by:
  1. (A)Kq02a(42−2)\dfrac{Kq_0^2}{a}\left(4\sqrt2-2\right)aKq02​​(42​−2)
  2. (B)Kq02a(3−2)\dfrac{Kq_0^2}{a}\left(3-\sqrt2\right)aKq02​​(3−2​)
  3. (C)Kq02a(4−22)\dfrac{Kq_0^2}{a}\left(4-2\sqrt2\right)aKq02​​(4−22​)
  4. (D)Kq02a(32−2)\dfrac{Kq_0^2}{a}\left(3\sqrt2-2\right)aKq02​​(32​−2)

Correct answer: (D)

Step-by-step solution →
Q8·Physics·KinematicsSingle correct
The position vector of a moving body at any instant of time is given as r⃗=(5t2i^−5tj^)\vec r=(5t^2\hat i-5t\hat j)r=(5t2i^−5tj^​) m. The magnitude and direction of velocity at t=2t=2t=2 s, is,
  1. (A)5155\sqrt{15}515​ m/s, making an angle of tan⁡−14\tan^{-1}4tan−14 with −-−ve Y axis
  2. (B)5155\sqrt{15}515​ m/s, making an angle of tan⁡−14\tan^{-1}4tan−14 with +++ve X axis
  3. (C)5175\sqrt{17}517​ m/s, making an angle of tan⁡−14\tan^{-1}4tan−14 with −-−ve Y axis
  4. (D)5175\sqrt{17}517​ m/s, making an angle of tan⁡−14\tan^{-1}4tan−14 with +++ve X axis

Correct answer: (C)

Step-by-step solution →
Q9·Physics·Rotational MotionSingle correct
A solid sphere and a hollow sphere of the same mass and of same radius are rolled on an inclined plane. Let the time taken to reach the bottom by the solid sphere and the hollow sphere be t1t_1t1​ and t2t_2t2​, respectively, then
  1. (A)t1<t2t_1<t_2t1​<t2​
  2. (B)t1=t2t_1=t_2t1​=t2​
  3. (C)t1=2t2t_1=2t_2t1​=2t2​
  4. (D)t1>t2t_1>t_2t1​>t2​

Correct answer: (A)

Step-by-step solution →
Q10·Physics·ThermodynamicsSingle correct
Which of the following figure represents the relation between Celsius and Fahrenheit temperatures? (Choose the correct C vs F graph.)
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (B)

Step-by-step solution →
Q11·Physics·Magnetic Field of CurrentSingle correct
N equally spaced charges each of value q, are placed on a circle of radius R. The circle rotates about its axis with an angular velocity ω\omegaω as shown in the figure. A bigger Amperian loop B encloses the whole circle where as a smaller Amperian loop A encloses a small segment. The difference between enclosed currents, IA−IBI_A-I_BIA​−IB​, for the given Amperian loops is :
  1. (A)N22πqω\dfrac{N^2}{2\pi}q\omega2πN2​qω
  2. (B)2πNqω\dfrac{2\pi}{N}q\omegaN2π​qω
  3. (C)N2πqω\dfrac{N}{2\pi}q\omega2πN​qω
  4. (D)Nπqω\dfrac{N}{\pi}q\omegaπN​qω

Correct answer: (C)

Step-by-step solution →
Q12·Physics·Dual Nature of Matter and RadiationSingle correct
In photoelectric effect, the stopping potential (V0)(V_0)(V0​) v/s frequency (ν)(\nu)(ν) curve is plotted. (hhh is the Planck's constant and ϕ0\phi_0ϕ0​ is work function of metal) (A) V0V_0V0​ v/s ν\nuν is linear. (B) The slope of V0V_0V0​ v/s ν\nuν curve =ϕ0h=\dfrac{\phi_0}{h}=hϕ0​​. (C) hhh is constant is related to the slope of V0V_0V0​ v/s ν\nuν line. (D) The value of electric charge of electron is not required to determine hhh using the V0V_0V0​ v/s ν\nuν curve. (E) The work function can be estimated without knowing the value of hhh. Choose the correct answer from the options given below :
  1. (A)(A), (B) and (C) only
  2. (B)(C) and (D) only
  3. (C)(A), (C) and (E) only
  4. (D)(D) and (E) only

Correct answer: (C)

Step-by-step solution →
Q13·Physics·ThermodynamicsSingle correct
The magnitude of heat exchanged by a system for the given cyclic process ABCA (as shown in figure, a semicircular loop in the P-V plane with P in kPa and V in cc) is (in SI unit):
  1. (A)10π10\pi10π
  2. (B)5π5\pi5π
  3. (C)zero
  4. (D)40π40\pi40π

Correct answer: (B)

Step-by-step solution →
Q14·Physics·Geometrical OpticsSingle correct
A photograph of a landscape is captured by a drone camera at a height of 18 km. The size of the camera film is 2 cm ×\times× 2 cm and the area of the landscape photographed is 400 km2^22. The focal length of the lens in the drone camera is :
  1. (A)1.8 cm
  2. (B)2.8 cm
  3. (C)2.5 cm
  4. (D)0.9 cm

Correct answer: (A)

Step-by-step solution →
Q15·Physics·Electronic DevicesSingle correct
The output of the circuit (shown in the figure) is low (zero) for : (A) X=0, Y=0 (B) X=0, Y=1 (C) X=1, Y=0 (D) X=1, Y=1. Choose the correct answer from the options given below :
  1. (A)(A), (C) and (D) only
  2. (B)(A), (B) and (C) only
  3. (C)(B), (C) and (D) only
  4. (D)(A), (B) and (D) only

Correct answer: (C)

Step-by-step solution →
Q16·Physics·ThermodynamicsSingle correct
The temperature of a body in air falls from 40∘40^\circ40∘C to 24∘24^\circ24∘C in 4 minutes. The temperature of the air is 16∘16^\circ16∘C. The temperature of the body in the next 4 minutes will be :
  1. (A)143 ∘\dfrac{14}{3}\,^\circ314​∘C
  2. (B)283 ∘\dfrac{28}{3}\,^\circ328​∘C
  3. (C)563 ∘\dfrac{56}{3}\,^\circ356​∘C
  4. (D)423 ∘\dfrac{42}{3}\,^\circ342​∘C

Correct answer: (C)

Step-by-step solution →
Q17·Physics·Units and MeasurementsSingle correct
The energy E and momentum p of a moving body of mass m are related by some equation. Given that c represents the speed of light, identify the correct equation.
  1. (A)E2=pc2+m2c4E^2=pc^2+m^2c^4E2=pc2+m2c4
  2. (B)E2=pc2+m2c2E^2=pc^2+m^2c^2E2=pc2+m2c2
  3. (C)E2=p2c2+m2c2E^2=p^2c^2+m^2c^2E2=p2c2+m2c2
  4. (D)E2=p2c2+m2c4E^2=p^2c^2+m^2c^4E2=p2c2+m2c4

Correct answer: (D)

Step-by-step solution →
Q18·Physics·Electric Field and Coulomb's LawSingle correct
A small uncharged conducting sphere is placed in contact with an identical sphere but having 4×10−84\times10^{-8}4×10−8 C charge and then removed to a distance such that the force of repulsion between them is 9×10−39\times10^{-3}9×10−3 N. The distance between them is (Take 14πε0\dfrac{1}{4\pi\varepsilon_0}4πε0​1​ as 9×1099\times10^99×109 in SI units):
  1. (A)2 cm
  2. (B)3 cm
  3. (C)4 cm
  4. (D)1 cm

Correct answer: (A)

Step-by-step solution →
Q19·Physics·OscillationsSingle correct
A particle oscillates along the x-axis according to the law, x(t)=x0sin⁡2(t2)x(t)=x_0\sin^2\left(\dfrac{t}{2}\right)x(t)=x0​sin2(2t​) where x0=1x_0=1x0​=1 m. The kinetic energy (K) of the particle as a function of x is correctly represented by the graph:
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (A)

Step-by-step solution →
Q20·Physics·Wave OpticsSingle correct
In a Young's double slit experiment, three polarizers are kept as shown in the figure. The transmission axes of P1P_1P1​ and P2P_2P2​ are orthogonal to each other. The polarizer P3P_3P3​ covers both the slits with its transmission axis at 45∘45^\circ45∘ to those of P1P_1P1​ and P2P_2P2​. An unpolarized light of wavelength λ\lambdaλ and intensity I0I_0I0​ is incident on P1P_1P1​ and P2P_2P2​. The intensity at a point after P3P_3P3​ where the path difference between the light waves from s1s_1s1​ and s2s_2s2​ is λ3\dfrac{\lambda}{3}3λ​, is
  1. (A)I02\dfrac{I_0}{2}2I0​​
  2. (B)I04\dfrac{I_0}{4}4I0​​
  3. (C)I0I_0I0​
  4. (D)I03\dfrac{I_0}{3}3I0​​

Correct answer: (B)

Step-by-step solution →
Q21·Physics·Magnetic Field of CurrentInteger
A tightly wound long solenoid carries a current of 1.5 A. An electron is executing uniform circular motion inside the solenoid with a time period of 75 ns. The number of turns per metre in the solenoid is __________. [Take mass of electron =9×10−31=9\times10^{-31}=9×10−31 kg, charge of electron ∣qe∣=1.6×10−19|q_e|=1.6\times10^{-19}∣qe​∣=1.6×10−19 C, μ0=4π×10−7 N/A2\mu_0=4\pi\times10^{-7}\ \mathrm{N/A^2}μ0​=4π×10−7 N/A2, 111 ns =10−9=10^{-9}=10−9 s]

Correct answer: 250

Step-by-step solution →
Q22·Physics·Laws of MotionInteger
A string of length L is fixed at one end and carries a mass of M at the other end. The mass makes (3π)\left(\dfrac{3}{\pi}\right)(π3​) rotations per second about the vertical axis passing through end of the string as shown. The tension in the string is __________ ML.

Correct answer: 36

Step-by-step solution →
Q23·Physics·Dual Nature of Matter and RadiationInteger
The ratio of the power of a light source S1S_1S1​ to the light source S2S_2S2​ is 2. S1S_1S1​ is emitting 2×10152\times10^{15}2×1015 photons per second at 600 nm. If the wavelength of the source S2S_2S2​ is 300 nm, then the number of photons per second emitted by S2S_2S2​ is __________ ×1014\times10^{14}×1014.

Correct answer: 5

Step-by-step solution →
Q24·Physics·Properties of Solids and LiquidsInteger
The increase in pressure required to decrease the volume of a water sample by 0.2% is P×105 Nm−2P\times10^5\ \mathrm{Nm^{-2}}P×105 Nm−2. Bulk modulus of water is 2.15×109 Nm−22.15\times10^9\ \mathrm{Nm^{-2}}2.15×109 Nm−2. The value of P is __________.

Correct answer: 43

Step-by-step solution →
Q25·Physics·GravitationInteger
Acceleration due to gravity on the surface of earth is "g". If the diameter of earth is reduced to one third of its original value and mass remains unchanged, then the acceleration due to gravity on the surface of the earth is __________ g.

Correct answer: 9

Step-by-step solution →

Chemistry — JEE Main 24 January 2025 Shift 2

Q26·Chemistry·Redox Reactions and ElectrochemistrySingle correct
Based on the data given below: ECr2O72−/Cr3+0=1.33E^0_{Cr_2O_7^{2-}/Cr^{3+}}=1.33ECr2​O72−​/Cr3+0​=1.33 V, ECl2/Cl−0=1.36E^0_{Cl_2/Cl^-}=1.36ECl2​/Cl−0​=1.36 V, EMnO4−/Mn2+0=1.51E^0_{MnO_4^-/Mn^{2+}}=1.51EMnO4−​/Mn2+0​=1.51 V, ECr3+/Cr0=−0.74E^0_{Cr^{3+}/Cr}=-0.74ECr3+/Cr0​=−0.74 V. The strongest reducing agent is :
  1. (A)Mn2+Mn^{2+}Mn2+
  2. (B)CrCrCr
  3. (C)MnO4−MnO_4^-MnO4−​
  4. (D)Cl−Cl^-Cl−

Correct answer: (B)

Step-by-step solution →
Q27·Chemistry·Chemical KineticsSingle correct
Given below are two statements: Statement (I): A plot of half-life t1/2t_{1/2}t1/2​ versus initial concentration [R]0[R]_0[R]0​ is a horizontal straight line — is valid for first order reaction. Statement (II): A plot of log⁡[R][R]0\log\dfrac{[R]}{[R]_0}log[R]0​[R]​ versus time is a straight line with slope k2.303\dfrac{k}{2.303}2.303k​ — is valid for first order reaction. In the light of the above statements, choose the correct answer from the options given below :
  1. (A)Both Statement I and Statement II are false
  2. (B)Statement I is false but Statement II is true
  3. (C)Both Statement I and Statement II are true
  4. (D)Statement I is true but Statement II is false

Correct answer: (D)

Step-by-step solution →
Q28·Chemistry·AminesSingle correct
For the reaction in which aniline is converted to o-bromoaniline (2-bromoaniline) as the major product, the correct order of set of reagents for the above conversion is :
  1. (A)Br2 ∣ FeBr3, H2O(Δ), NaOHBr_2\,|\,FeBr_3,\ H_2O(\Delta),\ NaOHBr2​∣FeBr3​, H2​O(Δ), NaOH
  2. (B)H2SO4, Ac2O, Br2, H2O(Δ), NaOHH_2SO_4,\ Ac_2O,\ Br_2,\ H_2O(\Delta),\ NaOHH2​SO4​, Ac2​O, Br2​, H2​O(Δ), NaOH
  3. (C)Ac2O, Br2, H2O(Δ), NaOHAc_2O,\ Br_2,\ H_2O(\Delta),\ NaOHAc2​O, Br2​, H2​O(Δ), NaOH
  4. (D)Ac2O, H2SO4, Br2, NaOHAc_2O,\ H_2SO_4,\ Br_2,\ NaOHAc2​O, H2​SO4​, Br2​, NaOH

Correct answer: (B)

Step-by-step solution →
Q29·Chemistry·Atomic StructureSingle correct
For hydrogen atom, the orbital/s with lowest energy is/are : (A) 4s4s4s (B) 3px3p_x3px​ (C) 3dx2−y23d_{x^2-y^2}3dx2−y2​ (D) 3dz23d_{z^2}3dz2​ (E) 4pz4p_z4pz​. Choose the correct answer from the options given below :
  1. (A)(A) and (E) only
  2. (B)(B) only
  3. (C)(A) only
  4. (D)(B), (C) and (D) only

Correct answer: (D)

Step-by-step solution →
Q30·Chemistry·Some Basic Principles of Organic ChemistrySingle correct
In the given structure CH3-CO-CH=CH-CH2-C≡C-CH=CH-C≡NCH_3\text{-}CO\text{-}CH{=}CH\text{-}CH_2\text{-}C{\equiv}C\text{-}CH{=}CH\text{-}C{\equiv}NCH3​-CO-CH=CH-CH2​-C≡C-CH=CH-C≡N, the number of sp and sp2^22 hybridized carbon atoms present respectively are :
  1. (A)3 and 6
  2. (B)3 and 5
  3. (C)4 and 6
  4. (D)4 and 5

Correct answer: (B)

Step-by-step solution →
Q31·Chemistry·Chemical ThermodynamicsSingle correct
Which of the following mixing of 1M base and 1M acid leads to the largest increase in temperature?
  1. (A)30 mL HCl and 30 mL NaOH
  2. (B)30 mL CH3COOHCH_3COOHCH3​COOH and 30 mL NaOH
  3. (C)50 mL HCl and 20 mL NaOH
  4. (D)45 mL CH3COOHCH_3COOHCH3​COOH and 25 mL NaOH

Correct answer: (A)

Step-by-step solution →
Q32·Chemistry·Chemical Bonding and Molecular StructureSingle correct
Given below are two statements: Statement (I): Experimentally determined oxygen-oxygen bond lengths in O3_33​ are found to be same and the bond length is greater than that of a O=O (double bond) but less than that of a single bond (O–O). Statement (II): The strong lone pair-lone pair repulsion between oxygen atoms is solely responsible for the fact about the bond length in ozone. In the light of the above statements, choose the correct answer from the options given below :
  1. (A)Statement I is true but Statement II is false
  2. (B)Statement I is true and Statement II is true
  3. (C)Both Statement I and Statement II are false
  4. (D)Statement I is false but Statement II is true

Correct answer: (A)

Step-by-step solution →
Q33·Chemistry·Principles of Qualitative AnalysisSingle correct
Find the compound "A" from the following reaction sequences. A→aqua-regiaB→(1) KNO3/NH4OH, (2) AcOHA\xrightarrow{\text{aqua-regia}}B\xrightarrow{\text{(1) }KNO_3/NH_4OH,\ \text{(2) AcOH}}Aaqua-regia​B(1) KNO3​/NH4​OH, (2) AcOH​ yellow ppt (K3[Co(NO2)6]↓K_3[Co(NO_2)_6]\downarrowK3​[Co(NO2​)6​]↓ yellow).
  1. (A)ZnS
  2. (B)CoS
  3. (C)MnS
  4. (D)NiS

Correct answer: (B)

Step-by-step solution →
Q34·Chemistry·EquilibriumSingle correct
For the reaction, H2(g)+I2(g)⇌2HI(g)H_2(g) + I_2(g) \rightleftharpoons 2HI(g)H2​(g)+I2​(g)⇌2HI(g), the attainment of equilibrium is predicted correctly by:
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (B)

Step-by-step solution →
Q35·Chemistry·d- and f-Block ElementsSingle correct
Match List-I (Transition metal ion) with List-II (Spin only magnetic moment B.M.). Choose the correct answer from the options given below:
List-I (Transition metal ion)List-II (Spin only magnetic moment B.M.)
A.Ti3+Ti^{3+}Ti3+I.3.87
B.V2+V^{2+}V2+II.0.00
C.Ni2+Ni^{2+}Ni2+III.1.73
D.Sc3+Sc^{3+}Sc3+IV.2.84
  1. (A)(A)-(III), (B)-(I), (C)-(II), (D)-(IV)
  2. (B)(A)-(III), (B)-(I), (C)-(IV), (D)-(II)
  3. (C)(A)-(IV), (B)-(II), (C)-(III), (D)-(I)
  4. (D)(A)-(II), (B)-(IV), (C)-(I), (D)-(III)

Correct answer: (B)

Step-by-step solution →
Q36·Chemistry·Some Basic Concepts in ChemistrySingle correct
The elemental composition of a compound is 54.2% C, 9.2% H and 36.6% O. If the molar mass of the compound is 132 g mol−1^{-1}−1, the molecular formula of the compound is : [Given : The relative atomic mass of C : H : O = 12 : 1 : 16]
  1. (A)C4H9O3C_4H_9O_3C4​H9​O3​
  2. (B)C6H12O6C_6H_{12}O_6C6​H12​O6​
  3. (C)C6H12O3C_6H_{12}O_3C6​H12​O3​
  4. (D)C4H8O2C_4H_8O_2C4​H8​O2​

Correct answer: (C)

Step-by-step solution →
Q37·Chemistry·Coordination CompoundsSingle correct
When Ethane-1,2-diamine is added progressively to an aqueous solution of Nickel (II) chloride, the sequence of colour change observed will be :
  1. (A)Pale Blue →\to→ Blue →\to→ Green →\to→ Violet
  2. (B)Pale Blue →\to→ Violet →\to→ Violet →\to→ Green
  3. (C)Green →\to→ Pale Blue →\to→ Violet
  4. (D)Violet →\to→ Blue →\to→ Pale Blue →\to→ Green

Correct answer: (C)

Step-by-step solution →
Q38·Chemistry·Coordination CompoundsSingle correct
The conditions and consequence that favours the t2g4 eg0t_{2g}^4\,e_g^0t2g4​eg0​ configuration in a metal complex are :
  1. (A)weak field ligand, high spin complex
  2. (B)strong field ligand, high spin complex
  3. (C)strong field ligand, low spin complex
  4. (D)weak field ligand, low spin complex

Correct answer: (C)

Step-by-step solution →
Q39·Chemistry·Electronic Effects and StabilitySingle correct
Identify correct statement/s : (A) −OCH3-OCH_3−OCH3​ and −NHCOCH3-NHCOCH_3−NHCOCH3​ are activating group (B) −CN-CN−CN and −OH-OH−OH are meta directing group (C) −CN-CN−CN and −SO3H-SO_3H−SO3​H are meta directing group (D) Activating groups act as ortho- and para- directing groups (E) Halides are activating groups. Choose the correct answer from the options given below :
  1. (A)(A), (C) and (D) only
  2. (B)(A), (B) and (D) only
  3. (C)(A) only
  4. (D)(A) and (C) only

Correct answer: (A)

Step-by-step solution →
Q40·Chemistry·Classification of Elements and Periodicity in PropertiesSingle correct
Given below are two statements: Statement (I): The first ionization energy of Pb is greater than that of Sn. Statement (II): The first ionization energy of Ge is greater than that of Si. In the light of the above statements, choose the correct answer from the options given below :
  1. (A)Statement I is true but Statement II is false
  2. (B)Both Statement I and Statement II are false
  3. (C)Statement I is false but Statement II is true
  4. (D)Both Statement I and Statement II are true

Correct answer: (A)

Step-by-step solution →
Q41·Chemistry·Chemical ThermodynamicsSingle correct
S(g)+32O2(g)→SO3(g)+2xS(g)+\dfrac{3}{2}O_2(g)\to SO_3(g)+2xS(g)+23​O2​(g)→SO3​(g)+2x kcal; SO2(g)+12O2(g)→SO3(g)+ySO_2(g)+\dfrac{1}{2}O_2(g)\to SO_3(g)+ySO2​(g)+21​O2​(g)→SO3​(g)+y kcal. The heat of formation of SO2(g)SO_2(g)SO2​(g) is given by :
  1. (A)2xy\dfrac{2x}{y}y2x​ kcal
  2. (B)y−2xy-2xy−2x kcal
  3. (C)2x+y2x+y2x+y kcal
  4. (D)x+yx+yx+y kcal

Correct answer: (B)

Step-by-step solution →
Q42·Chemistry·Aldehydes and KetonesSingle correct
Match List-I (Aldehyde Synthesis) with List-II (Name Reaction). Choose the correct answer from the options given below:
List-I (Aldehyde Synthesis)List-II (Name Reaction)
A.RCN→(i) SnCl2,3HCl (ii) H2ORCHORCN\xrightarrow{\text{(i) }SnCl_2,3HCl\ \text{(ii) }H_2O}RCHORCN(i) SnCl2​,3HCl (ii) H2​O​RCHOI.Etard reaction
B.C6H5COCl→H2, Pd-BaSO4C6H5CHOC_6H_5COCl\xrightarrow{H_2,\ Pd\text{-}BaSO_4}C_6H_5CHOC6​H5​COClH2​, Pd-BaSO4​​C6​H5​CHOII.Gattermann-Koch reaction
C.C6H5CH3→(i) CrO2Cl2,CS2 (ii) H2OC6H5CHOC_6H_5CH_3\xrightarrow{\text{(i) }CrO_2Cl_2,CS_2\ \text{(ii) }H_2O}C_6H_5CHOC6​H5​CH3​(i) CrO2​Cl2​,CS2​ (ii) H2​O​C6​H5​CHOIII.Rosenmund reduction
D.C6H6→(i) CO,HCl (ii) anhydrous AlCl3/CuClC6H5CHOC_6H_6\xrightarrow{\text{(i) }CO,HCl\ \text{(ii) anhydrous }AlCl_3/CuCl}C_6H_5CHOC6​H6​(i) CO,HCl (ii) anhydrous AlCl3​/CuCl​C6​H5​CHOIV.Stephen reaction
  1. (A)(A)-(IV), (B)-(III), (C)-(I), (D)-(II)
  2. (B)(A)-(III), (B)-(IV), (C)-(II), (D)-(I)
  3. (C)(A)-(I), (B)-(III), (C)-(II), (D)-(IV)
  4. (D)(A)-(III), (B)-(IV), (C)-(I), (D)-(II)

Correct answer: (A)

Step-by-step solution →
Q43·Chemistry·Organic Compounds Containing HalogensSingle correct
3-Bromo-5-iodobenzyl chloride is treated with AgCN. The structure of the major product formed is:
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (B)

Step-by-step solution →
Q44·Chemistry·BiomoleculesSingle correct
Match List-I (Nucleobase) with List-II (Structure). Choose the correct answer from the options given below:
List-I (Nucleobase)List-II (Structure)
A.AdenineI.see figure
B.CytosineII.see figure
C.ThymineIII.see figure
D.UracilIV.see figure
  1. (A)(A)-(III), (B)-(IV), (C)-(II), (D)-(I)
  2. (B)(A)-(III), (B)-(II), (C)-(IV), (D)-(I)
  3. (C)(A)-(IV), (B)-(III), (C)-(II), (D)-(I)
  4. (D)(A)-(III), (B)-(IV), (C)-(I), (D)-(II)

Correct answer: (A)

Step-by-step solution →
Q45·Chemistry·Classification of Elements and Periodicity in PropertiesSingle correct
The successive 5 ionisation energies of an element are 800, 2427, 3658, 25024 and 32824 kJ/mol, respectively. By using the above values predict the group in which the above element is present :
  1. (A)Group 2
  2. (B)Group 13
  3. (C)Group 4
  4. (D)Group 14

Correct answer: (B)

Step-by-step solution →
Q46·Chemistry·SolutionsInteger
The observed and normal molar masses of compound MX2MX_2MX2​ are 65.6 and 164 respectively. The percent degree of ionisation of MX2MX_2MX2​ is __________ %. (Nearest integer)

Correct answer: 75

Step-by-step solution →
Q47·Chemistry·IsomerismInteger
The possible number of stereoisomers for 5-phenylpent-4-en-2-ol is __________.

Correct answer: 4

Step-by-step solution →
Q48·Chemistry·Chemical KineticsInteger
Consider a complex reaction taking place in three steps with rate constants k1k_1k1​, k2k_2k2​ and k3k_3k3​ respectively. The overall rate constant k is given by the expression k=k1k3k2k=\sqrt{\dfrac{k_1k_3}{k_2}}k=k2​k1​k3​​​. If the activation energies of the three steps are 60, 30 and 10 kJ mol−1^{-1}−1 respectively, then the overall energy of activation in kJ mol−1^{-1}−1 is __________. (Nearest integer)

Correct answer: 20

Step-by-step solution →
Q49·Chemistry·Some Basic Principles of Organic ChemistryInteger
The hydrocarbon (X) with molar mass 80 g mol−1^{-1}−1 and 90% carbon has __________ degree of unsaturation.

Correct answer: 3

Step-by-step solution →
Q50·Chemistry·Purification and Characterisation of Organic CompoundsInteger
In Carius method of estimation of halogen, 0.25 g of an organic compound gave 0.15 g of silver bromide (AgBr). The percentage of Bromine in the organic compound is __________ ×10−1\times10^{-1}×10−1 %. (Nearest integer) (Given : Molar mass of Ag = 108 and Br is 80 g mol−1^{-1}−1)

Correct answer: 255

Step-by-step solution →

Mathematics — JEE Main 24 January 2025 Shift 2

Q51·Mathematics·EllipseSingle correct
The equation of the chord, of the ellipse x225+y216=1\dfrac{x^2}{25}+\dfrac{y^2}{16}=125x2​+16y2​=1, whose mid-point is (3,1)(3,1)(3,1) is :
  1. (A)48x+25y=16948x+25y=16948x+25y=169
  2. (B)4x+122y=1344x+122y=1344x+122y=134
  3. (C)25x+101y=17625x+101y=17625x+101y=176
  4. (D)5x+16y=315x+16y=315x+16y=31

Correct answer: (A)

Step-by-step solution →
Q52·Mathematics·Sets, Relations and FunctionsSingle correct
The function f:(−∞,∞)→(−∞,1)f:(-\infty,\infty)\to(-\infty,1)f:(−∞,∞)→(−∞,1), defined by f(x)=2x−2−x2x+2−xf(x)=\dfrac{2^x-2^{-x}}{2^x+2^{-x}}f(x)=2x+2−x2x−2−x​ is :
  1. (A)One-one but not onto
  2. (B)Onto but not one-one
  3. (C)Both one-one and onto
  4. (D)Neither one-one nor onto

Correct answer: (A)

Step-by-step solution →
Q53·Mathematics·Inverse Trigonometric FunctionsSingle correct
If α>β>γ>0\alpha>\beta>\gamma>0α>β>γ>0, then the expression cot⁡−1 ⁣{β+1+β2α−β}+cot⁡−1 ⁣{γ+1+γ2β−γ}+cot⁡−1 ⁣{α+1+α2γ−α}\cot^{-1}\!\left\{\beta+\dfrac{1+\beta^2}{\alpha-\beta}\right\}+\cot^{-1}\!\left\{\gamma+\dfrac{1+\gamma^2}{\beta-\gamma}\right\}+\cot^{-1}\!\left\{\alpha+\dfrac{1+\alpha^2}{\gamma-\alpha}\right\}cot−1{β+α−β1+β2​}+cot−1{γ+β−γ1+γ2​}+cot−1{α+γ−α1+α2​} is equal to:
  1. (A)π2−(α+β+γ)\dfrac{\pi}{2}-(\alpha+\beta+\gamma)2π​−(α+β+γ)
  2. (B)3π3\pi3π
  3. (C)000
  4. (D)π\piπ

Correct answer: (D)

Step-by-step solution →
Q54·Mathematics·Differential EquationsSingle correct
Let f:(0,∞)→Rf:(0,\infty)\to\mathbb{R}f:(0,∞)→R be a function differentiable at all points of its domain and satisfies the condition x2f′(x)=2xf(x)+3x^2 f'(x)=2x f(x)+3x2f′(x)=2xf(x)+3, with f(1)=4f(1)=4f(1)=4. Then 2f(2)2f(2)2f(2) is equal to:
  1. (A)29
  2. (B)19
  3. (C)39
  4. (D)23

Correct answer: (C)

Step-by-step solution →
Q55·Mathematics·Sets, Relations and FunctionsSingle correct
Let A={x∈(0,π)−{π2}:log⁡(2/π)∣sin⁡x∣+log⁡(2/π)∣cos⁡x∣=2}A=\left\{x\in(0,\pi)-\left\{\tfrac{\pi}{2}\right\}:\log_{(2/\pi)}|\sin x|+\log_{(2/\pi)}|\cos x|=2\right\}A={x∈(0,π)−{2π​}:log(2/π)​∣sinx∣+log(2/π)​∣cosx∣=2} and B={x≥0:x(x−4)−3∣x−2∣+6=0}B=\left\{x\ge 0:\sqrt{x}(\sqrt{x}-4)-3|\sqrt{x}-2|+6=0\right\}B={x≥0:x​(x​−4)−3∣x​−2∣+6=0}. Then n(A∪B)n(A\cup B)n(A∪B) is equal to:
  1. (A)4
  2. (B)2
  3. (C)8
  4. (D)6

Correct answer: (C)

Step-by-step solution →
Q56·Mathematics·Vector AlgebraSingle correct
Let the position vectors of three vertices of a triangle be 4p⃗+q⃗−3r⃗4\vec p+\vec q-3\vec r4p​+q​−3r, −5p⃗+q⃗+2r⃗-5\vec p+\vec q+2\vec r−5p​+q​+2r and 2p⃗−q⃗+2r⃗2\vec p-\vec q+2\vec r2p​−q​+2r. If the position vectors of the orthocenter and the circumcenter of the triangle are p⃗+q⃗+r⃗4\dfrac{\vec p+\vec q+\vec r}{4}4p​+q​+r​ and αp⃗+βq⃗+γr⃗\alpha\vec p+\beta\vec q+\gamma\vec rαp​+βq​+γr respectively, then α+2β+5γ\alpha+2\beta+5\gammaα+2β+5γ is equal to:
  1. (A)3
  2. (B)1
  3. (C)6
  4. (D)4

Correct answer: (A)

Step-by-step solution →
Q57·Mathematics·Limits and ContinuitySingle correct
Let [x][x][x] denote the greatest integer function, and let mmm and nnn respectively be the numbers of the points, where the function f(x)=[x]+∣x−2∣f(x)=[x]+|x-2|f(x)=[x]+∣x−2∣, −2<x<3-2<x<3−2<x<3, is not continuous and not differentiable. Then m+nm+nm+n is equal to:
  1. (A)6
  2. (B)9
  3. (C)8
  4. (D)7

Correct answer: (C)

Step-by-step solution →
Q58·Mathematics·Straight LinesSingle correct
Let the points (112,α)\left(\dfrac{11}{2},\alpha\right)(211​,α) lie on or inside the triangle with sides x+y=11x+y=11x+y=11, x+2y=16x+2y=16x+2y=16 and 2x+3y=292x+3y=292x+3y=29. Then the product of the smallest and the largest values of α\alphaα is equal to:
  1. (A)22
  2. (B)44
  3. (C)33
  4. (D)55

Correct answer: (C)

Step-by-step solution →
Q59·Mathematics·Sequence and SeriesSingle correct
In an arithmetic progression, if S40=1030S_{40}=1030S40​=1030 and S12=57S_{12}=57S12​=57, then S30−S10S_{30}-S_{10}S30​−S10​ is equal to:
  1. (A)510
  2. (B)515
  3. (C)525
  4. (D)505

Correct answer: (B)

Step-by-step solution →
Q60·Mathematics·Sequence and SeriesSingle correct
If 7=5+17(5+α)+172(5+2α)+173(5+3α)+…∞7=5+\dfrac{1}{7}(5+\alpha)+\dfrac{1}{7^2}(5+2\alpha)+\dfrac{1}{7^3}(5+3\alpha)+\ldots\infty7=5+71​(5+α)+721​(5+2α)+731​(5+3α)+…∞, then the value of α\alphaα is:
  1. (A)1
  2. (B)67\dfrac{6}{7}76​
  3. (C)6
  4. (D)17\dfrac{1}{7}71​

Correct answer: (C)

Step-by-step solution →
Q61·Mathematics·Matrices and DeterminantsSingle correct
If the system of equations x+2y−3z=2, 2x+λy+5z=5, 14x+3y+μz=33x+2y-3z=2,\ 2x+\lambda y+5z=5,\ 14x+3y+\mu z=33x+2y−3z=2, 2x+λy+5z=5, 14x+3y+μz=33 has infinitely many solutions, then λ+μ\lambda+\muλ+μ is equal to:
  1. (A)13
  2. (B)10
  3. (C)11
  4. (D)12

Correct answer: (D)

Step-by-step solution →
Q62·Mathematics·Application of DerivativesSingle correct
Let (2,3)(2,3)(2,3) be the largest open interval in which the function f(x)=2log⁡e(x−2)−x2+ax+1f(x)=2\log_e(x-2)-x^2+ax+1f(x)=2loge​(x−2)−x2+ax+1 is strictly increasing and (b,c)(b,c)(b,c) be the largest open interval, in which the function g(x)=(x−1)3(x+2−a)2g(x)=(x-1)^3(x+2-a)^2g(x)=(x−1)3(x+2−a)2 is strictly decreasing, then 100(a+b−c)100(a+b-c)100(a+b−c) is equal to:
  1. (A)280
  2. (B)360
  3. (C)420
  4. (D)160

Correct answer: (B)

Step-by-step solution →
Q63·Mathematics·Binomial Theorem and Its Simple ApplicationsSingle correct
Suppose A and B are the coefficients of 30th30^{\text{th}}30th and 12th12^{\text{th}}12th terms respectively in the binomial expansion of (1+x)2n−1(1+x)^{2n-1}(1+x)2n−1. If 2A=5B2A=5B2A=5B, then nnn is equal to:
  1. (A)22
  2. (B)21
  3. (C)20
  4. (D)19

Correct answer: (B)

Step-by-step solution →
Q64·Mathematics·Vector AlgebraSingle correct
Let a⃗=3i^−j^+2k^\vec a=3\hat i-\hat j+2\hat ka=3i^−j^​+2k^, b⃗=a⃗×(i^−3k^)\vec b=\vec a\times(\hat i-3\hat k)b=a×(i^−3k^) and c⃗=b⃗×k^\vec c=\vec b\times\hat kc=b×k^. Then the projection of c⃗−2j^\vec c-2\hat jc−2j^​ on a⃗\vec aa is:
  1. (A)373\sqrt737​
  2. (B)14\sqrt{14}14​
  3. (C)2142\sqrt{14}214​
  4. (D)272\sqrt727​

Correct answer: (C)

Step-by-step solution →
Q65·Mathematics·Matrices and DeterminantsSingle correct
For some a,ba,ba,b, let f(x)=∣a+sin⁡xx1ba1+sin⁡xxba1b+sin⁡xx∣f(x)=\begin{vmatrix} a+\dfrac{\sin x}{x} & 1 & b \\ a & 1+\dfrac{\sin x}{x} & b \\ a & 1 & b+\dfrac{\sin x}{x}\end{vmatrix}f(x)=​a+xsinx​aa​11+xsinx​1​bbb+xsinx​​​, x≠0x\neq 0x=0, lim⁡x→0f(x)=λ+μa+νb\displaystyle\lim_{x\to 0}f(x)=\lambda+\mu a+\nu bx→0lim​f(x)=λ+μa+νb. Then (λ+μ+ν)2(\lambda+\mu+\nu)^2(λ+μ+ν)2 is equal to:
  1. (A)25
  2. (B)9
  3. (C)36
  4. (D)16

Correct answer: (D)

Step-by-step solution →
Q66·Mathematics·Permutations and CombinationsSingle correct
Group A consists of 7 boys and 3 girls, while group B consists of 6 boys and 5 girls. The number of ways, 4 boys and 4 girls can be invited for a picnic if 5 of them must be from group A and the remaining 3 from group B, is equal to:
  1. (A)8575
  2. (B)9100
  3. (C)8925
  4. (D)8750

Correct answer: (C)

Step-by-step solution →
Q67·Mathematics·Area Under CurvesSingle correct
The area of the region enclosed by the curves y=exy=e^xy=ex, y=∣ex−1∣y=|e^x-1|y=∣ex−1∣ and y-axis is:
  1. (A)1+log⁡e21+\log_e 21+loge​2
  2. (B)2log⁡e22\log_e 22loge​2
  3. (C)2log⁡e2−12\log_e 2-12loge​2−1
  4. (D)1−log⁡e21-\log_e 21−loge​2

Correct answer: (D)

Step-by-step solution →
Q68·Mathematics·Quadratic EquationsSingle correct
The number of real solution(s) of the equation x2+3x+2=min⁡{∣x−3∣,∣x+2∣}x^2+3x+2=\min\{|x-3|,|x+2|\}x2+3x+2=min{∣x−3∣,∣x+2∣} is :
  1. (A)2
  2. (B)0
  3. (C)3
  4. (D)1

Correct answer: (A)

Step-by-step solution →
Q69·Mathematics·Statistics and ProbabilitySingle correct
Let A=[aij]A=[a_{ij}]A=[aij​] be a square matrix of order 2 with entries 0 or 1. Let E be the event that A is an invertible matrix. Then the probability P(E)P(E)P(E) is :
  1. (A)58\dfrac{5}{8}85​
  2. (B)316\dfrac{3}{16}163​
  3. (C)18\dfrac{1}{8}81​
  4. (D)38\dfrac{3}{8}83​

Correct answer: (D)

Step-by-step solution →
Q70·Mathematics·ParabolaSingle correct
If the equation of the parabola with vertex V(32,3)V\left(\dfrac{3}{2},3\right)V(23​,3) and the directrix x+2y=0x+2y=0x+2y=0 is αx2+βy2−γxy−30x−60y+225=0\alpha x^2+\beta y^2-\gamma xy-30x-60y+225=0αx2+βy2−γxy−30x−60y+225=0, then α+β+γ\alpha+\beta+\gammaα+β+γ is equal to:
  1. (A)6
  2. (B)8
  3. (C)7
  4. (D)9

Correct answer: (D)

Step-by-step solution →
Q71·Mathematics·Permutations and CombinationsInteger
Number of functions f:{1,2,…,100}→{0,1}f:\{1,2,\ldots,100\}\to\{0,1\}f:{1,2,…,100}→{0,1}, that assign 1 to exactly one of the positive integers less than or equal to 98, is equal to __________.

Correct answer: 392

Step-by-step solution →
Q72·Mathematics·Three Dimensional GeometryInteger
Let P be the image of the point Q(7,−2,5)Q(7,-2,5)Q(7,−2,5) in the line L:x−12=y+13=z4L:\dfrac{x-1}{2}=\dfrac{y+1}{3}=\dfrac{z}{4}L:2x−1​=3y+1​=4z​ and R(5,p,q)R(5,p,q)R(5,p,q) be a point on L. Then the square of the area of △PQR\triangle PQR△PQR is __________.

Correct answer: 957

Step-by-step solution →
Q73·Mathematics·Differential EquationsInteger
Let y=y(x)y=y(x)y=y(x) be the solution of the differential equation 2cos⁡xdydx=sin⁡2x−4ysin⁡x2\cos x\dfrac{dy}{dx}=\sin 2x-4y\sin x2cosxdxdy​=sin2x−4ysinx, x∈(0,π2)x\in\left(0,\dfrac{\pi}{2}\right)x∈(0,2π​). If y(π3)=0y\left(\dfrac{\pi}{3}\right)=0y(3π​)=0, then y(π4)+y′(π4)y\left(\dfrac{\pi}{4}\right)+y'\left(\dfrac{\pi}{4}\right)y(4π​)+y′(4π​) is equal to __________.

Correct answer: 1

Step-by-step solution →
Q74·Mathematics·HyperbolaInteger
Let H1:x2a2−y2b2=1H_1:\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1H1​:a2x2​−b2y2​=1 and H2:−x2A2+y2B2=1H_2:-\dfrac{x^2}{A^2}+\dfrac{y^2}{B^2}=1H2​:−A2x2​+B2y2​=1 be two hyperbolas having length of latus rectums 15215\sqrt2152​ and 12512\sqrt5125​ respectively. Let their eccentricities be e1=52e_1=\sqrt{\dfrac{5}{2}}e1​=25​​ and e2e_2e2​ respectively. If the product of the lengths of their transverse axes is 10010100\sqrt{10}10010​, then 25e2225e_2^225e22​ is equal to __________.

Correct answer: 55

Step-by-step solution →
Q75·Mathematics·Indefinite IntegrationInteger
If ∫2x2+5x+9x2+x+1 dx=xx2+x+1+αx2+x+1+βlog⁡e∣x+12+x2+x+1∣+C\displaystyle\int\dfrac{2x^2+5x+9}{\sqrt{x^2+x+1}}\,dx=x\sqrt{x^2+x+1}+\alpha\sqrt{x^2+x+1}+\beta\log_e\left|x+\dfrac12+\sqrt{x^2+x+1}\right|+C∫x2+x+1​2x2+5x+9​dx=xx2+x+1​+αx2+x+1​+βloge​​x+21​+x2+x+1​​+C, where C is the constant of integration, then α+2β\alpha+2\betaα+2β is equal to __________.

Correct answer: 16

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
  • Sequence and Series 164/186
  • 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
  • Application of Derivatives 139/186
  • Limits and Continuity 149/186
  • d- and f-Block Elements 126/186
  • Thermodynamics 154/186
  • 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
  • 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
  • Some Basic Concepts in Chemistry 129/186
  • Wave Optics 130/186
  • Area Under Curves 139/186
  • Purification and Characterisation of Organic Compounds 116/186
  • Oscillations 117/186
  • Organic Compounds Containing Halogens 109/186
  • Straight Lines 114/186
  • Classification of Elements and Periodicity in Properties 113/186
  • Parabola 101/186
  • Ellipse 103/186
  • Inverse Trigonometric Functions 93/186
  • Electronic Effects and Stability 74/186
  • Hyperbola 77/186
  • Electric Potential 63/186
  • Indefinite Integration 66/186
  • Principles of Qualitative Analysis 58/186
  • Isomerism 51/186
  • IUPAC Nomenclature 37/186
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