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JEE Main 23 January 2025 Shift 2 Question Paper with Answers

23 January 2025 · January session · 75 questions

The complete JEE Main 23 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
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Chemistry
25
Mathematics
25

Physics — JEE Main 23 January 2025 Shift 2

Q1·Physics·KinematicsSingle correct
A ball having kinetic energy KE, is projected at an angle of 60∘60^\circ60∘ from the horizontal. What will be the kinetic energy of ball at the highest point of its flight?
  1. (A)KE8\dfrac{KE}{8}8KE​
  2. (B)KE4\dfrac{KE}{4}4KE​
  3. (C)KE16\dfrac{KE}{16}16KE​
  4. (D)KE2\dfrac{KE}{2}2KE​

Correct answer: (B)

Step-by-step solution →
Q2·Physics·Electric PotentialSingle correct
Two charges 7 μc7\,\mu c7μc and −4 μc-4\,\mu c−4μc are placed at (−7 cm,0,0)(-7\text{ cm}, 0, 0)(−7 cm,0,0) and (7 cm,0,0)(7\text{ cm}, 0, 0)(7 cm,0,0) respectively. Given, ε0=8.85×10−12 C2 N−1 m−2\varepsilon_0=8.85\times10^{-12}\ \mathrm{C^2\,N^{-1}\,m^{-2}}ε0​=8.85×10−12 C2N−1m−2, the electrostatic potential energy of the charge configuration is :
  1. (A)−1.5-1.5−1.5 J
  2. (B)−2.0-2.0−2.0 J
  3. (C)−1.2-1.2−1.2 J
  4. (D)−1.8-1.8−1.8 J

Correct answer: (D)

Step-by-step solution →
Q3·Physics·Geometrical OpticsSingle correct
The refractive index of the material of a glass prism is 3\sqrt33​. The angle of minimum deviation is equal to the angle of the prism. What is the angle of the prism?
  1. (A)50∘50^\circ50∘
  2. (B)60∘60^\circ60∘
  3. (C)58∘58^\circ58∘
  4. (D)48∘48^\circ48∘

Correct answer: (B)

Step-by-step solution →
Q4·Physics·WavesSingle correct
The equation of a transverse wave travelling along a string is y(x,t)=4.0sin⁡[20×10−3x+600t]y(x,t)=4.0\sin[20\times10^{-3}x+600t]y(x,t)=4.0sin[20×10−3x+600t] mm, where xxx is in the mm and ttt is in second. The velocity of the wave is :
  1. (A)+30+30+30 m/s
  2. (B)−60-60−60 m/s
  3. (C)−30-30−30 m/s
  4. (D)+60+60+60 m/s

Correct answer: (C)

Step-by-step solution →
Q5·Physics·Units and MeasurementsSingle correct
The energy of a system is given as E(t)=α3e−βtE(t)=\alpha^3 e^{-\beta t}E(t)=α3e−βt, where ttt is the time and β=0.3 s−1\beta=0.3\ \mathrm{s^{-1}}β=0.3 s−1. The errors in the measurement of α\alphaα and ttt are 1.2%1.2\%1.2% and 1.6%1.6\%1.6%, respectively. At t=5t=5t=5 s, maximum percentage error in the energy is :
  1. (A)4%4\%4%
  2. (B)11.6%11.6\%11.6%
  3. (C)6%6\%6%
  4. (D)8.4%8.4\%8.4%

Correct answer: (C)

Step-by-step solution →
Q6·Physics·Dual Nature of Matter and RadiationSingle correct
In photoelectric effect an em-wave is incident on a metal surface and electrons are ejected from the surface. If the work function of the metal is 2.142.142.14 eV and stopping potential is 222V, what is the wavelength of the em-wave? (Given hc=1242hc=1242hc=1242 eVnm where hhh is the Planck's constant and ccc is the speed of light in vacuum.)
  1. (A)400400400 nm
  2. (B)600600600 nm
  3. (C)200200200 nm
  4. (D)300300300 nm

Correct answer: (D)

Step-by-step solution →
Q7·Physics·Rotational MotionSingle correct
A circular disk of radius R meter and mass M kg is rotating about the axis perpendicular to the disk. An external torque is applied to the disk such that θ(t)=5t2−8t\theta(t)=5t^2-8tθ(t)=5t2−8t, where θ(t)\theta(t)θ(t) is the angular position of the rotating disc as a function of time t. How much power is delivered by the applied torque, when t=2t=2t=2s?
  1. (A)60 MR260\,MR^260MR2
  2. (B)72 MR272\,MR^272MR2
  3. (C)108 MR2108\,MR^2108MR2
  4. (D)8 MR28\,MR^28MR2

Correct answer: (A)

Step-by-step solution →
Q8·Physics·Properties of Solids and LiquidsSingle correct
Water flows in a horizontal pipe whose one end is closed with a valve. The reading of the pressure gauge attached to the pipe is P1P_1P1​. The reading of the pressure gauge falls to P2P_2P2​ when the valve is opened. The speed of water flowing in the pipe is proportional to :
  1. (A)P1−P2\sqrt{P_1-P_2}P1​−P2​​
  2. (B)(P1−P2)2(P_1-P_2)^2(P1​−P2​)2
  3. (C)(P1−P2)4(P_1-P_2)^4(P1​−P2​)4
  4. (D)P1−P2P_1-P_2P1​−P2​

Correct answer: (A)

Step-by-step solution →
Q9·Physics·Units and MeasurementsSingle correct
Match List-I (Electromagnetic Quantity) with List-II (Dimensional Formula). Choose the correct answer from the options given below:
List-I (Electromagnetic Quantity)List-II (Dimensional Formula)
A.Permeability of free spaceI.[M L2 T−2][\mathrm{M\,L^2\,T^{-2}}][ML2T−2]
B.Magnetic fieldII.[M T−2 A−1][\mathrm{M\,T^{-2}\,A^{-1}}][MT−2A−1]
C.Magnetic momentIII.[M L T−2 A−2][\mathrm{M\,L\,T^{-2}\,A^{-2}}][MLT−2A−2]
D.Torsional constantIV.[L2 A][\mathrm{L^2\,A}][L2A]
  1. (A)(A)-(I), (B)-(IV), (C)-(II), (D)-(III)
  2. (B)(A)-(II), (B)-(I), (C)-(III), (D)-(IV)
  3. (C)(A)-(IV), (B)-(III), (C)-(I), (D)-(II)
  4. (D)(A)-(III), (B)-(II), (C)-(IV), (D)-(I)

Correct answer: (D)

Step-by-step solution →
Q10·Physics·GravitationSingle correct
If a satellite orbiting the Earth is 999 times closer to the Earth than the Moon, what is the time period of rotation of the satellite? Given rotational time period of Moon =27=27=27 days and gravitational attraction between the satellite and the moon is neglected.
  1. (A)111 day
  2. (B)818181 days
  3. (C)272727 days
  4. (D)333 days

Correct answer: (A)

Step-by-step solution →
Q11·Physics·Electric Field and Coulomb's LawSingle correct
Two point charges −4 μc-4\,\mu c−4μc and 4 μc4\,\mu c4μc, constituting an electric dipole, are placed at (−9,0,0)(-9, 0, 0)(−9,0,0) cm and (9,0,0)(9, 0, 0)(9,0,0) cm in a uniform electric field of strength 104 NC−110^4\ \mathrm{NC^{-1}}104 NC−1. The work done on the dipole in rotating it from the equilibrium through 180∘180^\circ180∘ is :
  1. (A)14.414.414.4 mJ
  2. (B)18.418.418.4 mJ
  3. (C)12.412.412.4 mJ
  4. (D)16.416.416.4 mJ

Correct answer: (A)

Step-by-step solution →
Q12·Physics·Current ElectricitySingle correct
A galvanometer having a coil of resistance 30 Ω30\ \Omega30 Ω need 202020 mA of current for full-scale deflection. If a maximum current of 333 A is to be measured using this galvanometer, the resistance of the shunt to be added to the galvanometer should be 30X Ω\dfrac{30}{X}\ \OmegaX30​ Ω, where XXX is :
  1. (A)447447447
  2. (B)298298298
  3. (C)149149149
  4. (D)596596596

Correct answer: (C)

Step-by-step solution →
Q13·Physics·Wave OpticsSingle correct
The width of one of the two slits in Young's double slit experiment is ddd while that of the other slit is xdxdxd. If the ratio of the maximum to the minimum intensity in the interference pattern on the screen is 9:49:49:4 what is the value of xxx? (Assume that the field strength varies according to the slit width.)
  1. (A)222
  2. (B)333
  3. (C)555
  4. (D)444

Correct answer: (C)

Step-by-step solution →
Q14·Physics·Atoms and NucleiSingle correct
Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : The binding energy per nucleon is found to be practically independent of the atomic number A, for nuclei with mass numbers between 30 and 170. Reason (R) : Nuclear force is long range. 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)(A) is true but (R) is false
  3. (C)Both (A) and (R) are true and (R) is the correct explanation of (A)
  4. (D)Both (A) and (R) are true but (R) is NOT the correct explanation of (A)

Correct answer: (B)

Step-by-step solution →
Q15·Physics·ThermodynamicsSingle correct
Water of mass m gram is slowly heated to increase the temperature from T1T_1T1​ to T2T_2T2​. The change in entropy of the water, given specific heat of water is 1 Jkg−1K−11\ \mathrm{Jkg^{-1}K^{-1}}1 Jkg−1K−1, is :
  1. (A)zero
  2. (B)m(T2−T1)m(T_2-T_1)m(T2​−T1​)
  3. (C)mln⁡ ⁣(T1T2)m\ln\!\left(\dfrac{T_1}{T_2}\right)mln(T2​T1​​)
  4. (D)mln⁡ ⁣(T2T1)m\ln\!\left(\dfrac{T_2}{T_1}\right)mln(T1​T2​​)

Correct answer: (D)

Step-by-step solution →
Q16·Physics·Electronic DevicesSingle correct
What is the current through the battery in the circuit shown below?
  1. (A)1.01.01.0 A
  2. (B)1.51.51.5 A
  3. (C)0.50.50.5 A
  4. (D)0.250.250.25 A

Correct answer: (C)

Step-by-step solution →
Q17·Physics·Electromagnetic WavesSingle correct
A plane electromagnetic wave of frequency 20 MHz travels in free space along the +x+x+x direction. At a particular point in space and time, the electric field vector of the wave is Ey=9.3 V m−1E_y=9.3\ \mathrm{V\,m^{-1}}Ey​=9.3 Vm−1. Then, the magnetic field vector of the wave at that point is :
  1. (A)Bz=9.3×10−8B_z=9.3\times10^{-8}Bz​=9.3×10−8 T
  2. (B)Bz=1.55×10−8B_z=1.55\times10^{-8}Bz​=1.55×10−8 T
  3. (C)Bz=6.2×10−8B_z=6.2\times10^{-8}Bz​=6.2×10−8 T
  4. (D)Bz=3.1×10−8B_z=3.1\times10^{-8}Bz​=3.1×10−8 T

Correct answer: (D)

Step-by-step solution →
Q18·Physics·ThermodynamicsSingle correct
Using the given P-V diagram, the work done by an ideal gas along the path ABCD is :
  1. (A)4 P0V04\,P_0V_04P0​V0​
  2. (B)3 P0V03\,P_0V_03P0​V0​
  3. (C)−4 P0V0-4\,P_0V_0−4P0​V0​
  4. (D)−3 P0V0-3\,P_0V_0−3P0​V0​

Correct answer: (D)

Step-by-step solution →
Q19·Physics·Geometrical OpticsSingle correct
A concave mirror of focal length fff in air is dipped in a liquid of refractive index μ\muμ. Its focal length in the liquid is :
  1. (A)fμ\dfrac{f}{\mu}μf​
  2. (B)fμ−1\dfrac{f}{\mu-1}μ−1f​
  3. (C)μf\mu fμf
  4. (D)fff

Correct answer: (D)

Step-by-step solution →
Q20·Physics·Properties of Solids and LiquidsSingle correct
A massless spring gets elongated by amount x1x_1x1​ under a tension of 5N. Its elongation is x2x_2x2​ under the tension of 7N. For the elongation of (5x1−2x2)(5x_1-2x_2)(5x1​−2x2​), the tension in the spring will be :
  1. (A)151515 N
  2. (B)202020 N
  3. (C)111111 N
  4. (D)393939 N

Correct answer: (C)

Step-by-step solution →
Q21·Physics·Properties of Solids and LiquidsInteger
An air bubble of radius 1.0 mm is observed at a depth of 20 cm below the free surface of a liquid having surface tension 0.095 J/m20.095\ \mathrm{J/m^2}0.095 J/m2 and density 103 kg/m310^3\ \mathrm{kg/m^3}103 kg/m3. The difference between pressure inside the bubble and atmospheric pressure is __________ N/m2\mathrm{N/m^2}N/m2. (Take g=10 m/s2g=10\ \mathrm{m/s^2}g=10 m/s2)

Correct answer: 2190

Step-by-step solution →
Q22·Physics·GravitationInteger
A satellite of mass M2\dfrac{M}{2}2M​ is revolving around earth in a circular orbit at a height of R3\dfrac{R}{3}3R​ from earth surface. The angular momentum of the satellite is MGMRxM\sqrt{\dfrac{GMR}{x}}MxGMR​​. The value of xxx is __________, where MMM and RRR are the mass and radius of earth, respectively. (GGG is the gravitational constant)

Correct answer: 3

Step-by-step solution →
Q23·Physics·Current ElectricityInteger
At steady state the charge on the capacitor, as shown in the circuit below, is __________ μ\muμC.

Correct answer: 16

Step-by-step solution →
Q24·Physics·Electromagnetic WavesInteger
A time varying potential difference is applied between the plates of a parallel plate capacitor of capacitance 2.5 μ2.5\ \mu2.5 μF. The dielectric constant of the medium between the capacitor plates is 1. It produces an instantaneous displacement current of 0.250.250.25 mA in the intervening space between the capacitor plates, the magnitude of the rate of change of the potential difference will be __________ Vs−1\mathrm{Vs^{-1}}Vs−1.

Correct answer: 100

Step-by-step solution →
Q25·Physics·Alternating CurrentsInteger
In a series LCR circuit, a resistor of 300 Ω300\ \Omega300 Ω, a capacitor of 252525 nF and an inductor of 100100100 mH are used. For maximum current in the circuit, the angular frequency of the ac source is __________ ×104\times10^4×104 radians s−1\mathrm{s^{-1}}s−1.

Correct answer: 2

Step-by-step solution →

Chemistry — JEE Main 23 January 2025 Shift 2

Q26·Chemistry·Chemical ThermodynamicsSingle correct
The effect of temperature on spontaneity of reactions are represented as: (A) ΔH=+, ΔS=−\Delta H=+,\ \Delta S=-ΔH=+, ΔS=−, any T →\to→ Non-spontaneous; (B) ΔH=+, ΔS=+\Delta H=+,\ \Delta S=+ΔH=+, ΔS=+, low T →\to→ spontaneous; (C) ΔH=−, ΔS=−\Delta H=-,\ \Delta S=-ΔH=−, ΔS=−, low T →\to→ Non-spontaneous; (D) ΔH=−, ΔS=+\Delta H=-,\ \Delta S=+ΔH=−, ΔS=+, any T →\to→ spontaneous.
  1. (A)(A), (B) and (D) only
  2. (B)(A) and (D) only
  3. (C)(B) and (C) only
  4. (D)(D) and (C) only

Correct answer: (C)

Step-by-step solution →
Q27·Chemistry·Redox Reactions and ElectrochemistrySingle correct
Standard electrode potentials for a few half cells are mentioned below: ECu2+/Cu∘=0.34E^\circ_{Cu^{2+}/Cu}=0.34ECu2+/Cu∘​=0.34 V, EZn2+/Zn∘=−0.76E^\circ_{Zn^{2+}/Zn}=-0.76EZn2+/Zn∘​=−0.76 V, EAg+/Ag∘=0.80E^\circ_{Ag^+/Ag}=0.80EAg+/Ag∘​=0.80 V, EMg2+/Mg∘=−2.37E^\circ_{Mg^{2+}/Mg}=-2.37EMg2+/Mg∘​=−2.37 V. Which one of the following cells gives the most negative value of ΔG∘\Delta G^\circΔG∘?
  1. (A)Zn ∣ Zn2+(1M) ∣∣ Ag+(1M) ∣ AgZn\,|\,Zn^{2+}(1M)\,||\,Ag^+(1M)\,|\,AgZn∣Zn2+(1M)∣∣Ag+(1M)∣Ag
  2. (B)Zn ∣ Zn2+(1M) ∣∣ Mg2+(1M) ∣ MgZn\,|\,Zn^{2+}(1M)\,||\,Mg^{2+}(1M)\,|\,MgZn∣Zn2+(1M)∣∣Mg2+(1M)∣Mg
  3. (C)Ag ∣ Ag+(1M) ∣∣ Mg2+(1M) ∣ MgAg\,|\,Ag^+(1M)\,||\,Mg^{2+}(1M)\,|\,MgAg∣Ag+(1M)∣∣Mg2+(1M)∣Mg
  4. (D)Cu ∣ Cu2+(1M) ∣∣ Ag+(1M) ∣ AgCu\,|\,Cu^{2+}(1M)\,||\,Ag^+(1M)\,|\,AgCu∣Cu2+(1M)∣∣Ag+(1M)∣Ag

Correct answer: (A)

Step-by-step solution →
Q28·Chemistry·BiomoleculesSingle correct
The α\alphaα-Helix and β\betaβ-Pleated sheet structures of protein are associated with its:
  1. (A)quaternary structure
  2. (B)primary structure
  3. (C)secondary structure
  4. (D)tertiary structure

Correct answer: (C)

Step-by-step solution →
Q29·Chemistry·Aldehydes and KetonesSingle correct
Given below are two statements. Consider the following reaction: R-CHO+H2O⇌KR-CH(OH)2R\text{-CHO}+H_2O\xrightleftharpoons{K}R\text{-CH(OH)}_2R-CHO+H2​OK​R-CH(OH)2​. Statement (I): In the case of formaldehyde (HCHO), K is about 2280, due to small substituents, hydration is faster. Statement (II): In the case of trichloro acetaldehyde (CCl3_33​CHO), K is about 2000 due to −I-I−I effect of −-−Cl. In the light of the above statements, choose the correct answer from the options given below:
  1. (A)Statement I true but Statement II is false
  2. (B)Both Statement I and Statement II are true
  3. (C)Statement I is false but Statement II is true
  4. (D)Both Statement I and Statement II are false

Correct answer: (B)

Step-by-step solution →
Q30·Chemistry·EquilibriumSingle correct
Consider the reaction X2Y(g)⇌X2(g)+12Y2(g)X_2Y(g)\rightleftharpoons X_2(g)+\tfrac12 Y_2(g)X2​Y(g)⇌X2​(g)+21​Y2​(g). The equation representing correct relationship between the degree of dissociation (x)(x)(x) of X2Y(g)X_2Y(g)X2​Y(g) with its equilibrium constant KpK_pKp​ is __________. Assume xxx to be very very small.
  1. (A)x=2Kppx=\sqrt{\dfrac{2K_p}{p}}x=p2Kp​​​
  2. (B)x=2Kp2p3x=\sqrt[3]{\dfrac{2K_p^2}{p}}x=3p2Kp2​​​
  3. (C)x=Kp2px=\sqrt{\dfrac{K_p}{2p}}x=2pKp​​​
  4. (D)x=Kpp3x=\sqrt[3]{\dfrac{K_p}{p}}x=3pKp​​​

Correct answer: (B)

Step-by-step solution →
Q31·Chemistry·Principles of Qualitative AnalysisSingle correct
Identify A, B and C in the given below reaction sequence: A→HNO3Pb(NO3)2→H2SO4B→(1) ammonium acetate, (2) acetic acid, (3) K2CrO4CA\xrightarrow{HNO_3}Pb(NO_3)_2\xrightarrow{H_2SO_4}B\xrightarrow{\text{(1) ammonium acetate, (2) acetic acid, (3) }K_2CrO_4}CAHNO3​​Pb(NO3​)2​H2​SO4​​B(1) ammonium acetate, (2) acetic acid, (3) K2​CrO4​​C (Yellow ppt).
  1. (A)PbCl2, PbSO4, PbCrO4PbCl_2,\ PbSO_4,\ PbCrO_4PbCl2​, PbSO4​, PbCrO4​
  2. (B)PbS, PbSO4, PbCrO4PbS,\ PbSO_4,\ PbCrO_4PbS, PbSO4​, PbCrO4​
  3. (C)PbS, PbSO4, Pb(CH3COO)2PbS,\ PbSO_4,\ Pb(CH_3COO)_2PbS, PbSO4​, Pb(CH3​COO)2​
  4. (D)PbCl2, Pb(SO4)2, PbCrO4PbCl_2,\ Pb(SO_4)_2,\ PbCrO_4PbCl2​, Pb(SO4​)2​, PbCrO4​

Correct answer: (B)

Step-by-step solution →
Q32·Chemistry·Alcohols and EthersSingle correct
Given below are two statements: Statement (I): The boiling points of alcohols and phenols increase with increase in the number of C-atoms. Statement (II): The boiling points of alcohols and phenols are higher in comparison to other class of compounds such as ethers, haloalkanes. 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)Statement I is true but Statement II is false
  4. (D)Both Statement I and Statement II are true

Correct answer: (D)

Step-by-step solution →
Q33·Chemistry·SolutionsSingle correct
When a non-volatile solute is added to the solvent, the vapour pressure of the solvent decreases by 10 mm of Hg. The mole fraction of the solute in the solution is 0.2. What would be the mole fraction of the solvent if decrease in vapour pressure is 20 mm of Hg ?
  1. (A)0.6
  2. (B)0.4
  3. (C)0.2
  4. (D)0.8

Correct answer: (A)

Step-by-step solution →
Q34·Chemistry·Atomic StructureSingle correct
Given below are two statements: Statement (I): For a given shell, the total number of allowed orbitals is given by n2n^2n2. Statement (II): For any subshell, the spatial orientation of the orbitals is given by −l-l−l to +l+l+l values including zero. 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 false but Statement II is true
  3. (C)Both Statement I and Statement II are true
  4. (D)Both Statement I and Statement II are false

Correct answer: (C)

Step-by-step solution →
Q35·Chemistry·Organic Compounds Containing HalogensSingle correct
The ascending order of relative rate of solvolysis of following compounds (A), (B), (C) and (D) (shown in the figure) is :
  1. (A)(D) < (A) < (B) < (C)
  2. (B)(C) < (B) < (A) < (D)
  3. (C)(D) < (B) < (A) < (C)
  4. (D)(C) < (D) < (B) < (A)

Correct answer: (C)

Step-by-step solution →
Q36·Chemistry·HydrocarbonsSingle correct
Match List-I (Isomers of C10H14C_{10}H_{14}C10​H14​) with List-II (Ozonolysis product). Choose the correct answer from the options given below:
List-I (Isomers of $C_{10}H_{14}$)List-II (Ozonolysis product)
A.see figureI.see figure
B.see figureII.see figure
C.see figureIII.see figure
D.see figureIV.see figure
  1. (A)(A)-(II), (B)-(III), (C)-(I), (D)-(IV)
  2. (B)(A)-(III), (B)-(IV), (C)-(I), (D)-(II)
  3. (C)(A)-(III), (B)-(II), (C)-(I), (D)-(IV)
  4. (D)(A)-(I), (B)-(IV), (C)-(III), (D)-(II)

Correct answer: (B)

Step-by-step solution →
Q37·Chemistry·Chemical KineticsSingle correct
Which of the following graphs most appropriately represents a zero order reaction?
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (B)

Step-by-step solution →
Q38·Chemistry·Isolation of MetalsSingle correct
Match List-I (Alloy) with List-II (Constituent Metals). Choose the correct answer from the options given below:
List-I (Alloy)List-II (Constituent Metals)
A.BronzeI.Cu, Ni
B.BrassII.Fe, Cr, Ni, C
C.UK silver coinIII.Cu, Zn
D.Stainless SteelIV.Cu, Sn
  1. (A)(A)-(IV), (B)-(II), (C)-(III), (D)-(I)
  2. (B)(A)-(IV), (B)-(III), (C)-(I), (D)-(II)
  3. (C)(A)-(III), (B)-(II), (C)-(IV), (D)-(II)
  4. (D)(A)-(III), (B)-(IV), (C)-(II), (D)-(I)

Correct answer: (B)

Step-by-step solution →
Q39·Chemistry·Coordination CompoundsSingle correct
Identify the coordination complexes in which the central metal ion has d4d^4d4 configuration. (A) [FeO4]2−[FeO_4]^{2-}[FeO4​]2− (B) [Mn(CN)6]3−[Mn(CN)_6]^{3-}[Mn(CN)6​]3− (C) [Fe(CN)6]3−[Fe(CN)_6]^{3-}[Fe(CN)6​]3− (D) Cr2(O–CO–Me)4(H2O)2Cr_2(\text{O–CO–Me})_4(H_2O)_2Cr2​(O–CO–Me)4​(H2​O)2​ (E) [NiF6]2−[NiF_6]^{2-}[NiF6​]2−. Choose the correct answer from the options given below:
  1. (A)(C) and (E) only
  2. (B)(B), (C) and (D) only
  3. (C)(B) and (D) only
  4. (D)(A), (B) and (E) only

Correct answer: (C)

Step-by-step solution →
Q40·Chemistry·p-Block ElementsSingle correct
Given below are the atomic numbers of some group 14 elements. The atomic number of the element with lowest melting point is :
  1. (A)14
  2. (B)6
  3. (C)82
  4. (D)50

Correct answer: (D)

Step-by-step solution →
Q41·Chemistry·EquilibriumSingle correct
pH of water is 7 at 25∘25^\circ25∘C. If water is heated to 80∘80^\circ80∘C, it's pH will :
  1. (A)Decrease
  2. (B)Remains the same
  3. (C)H+H^+H+ concentration increases, OH−OH^-OH− concentration decreases
  4. (D)Increase

Correct answer: (A)

Step-by-step solution →
Q42·Chemistry·Alcohols and EthersSingle correct
Identify the products [A] and [B], respectively in the following reaction: chlorobenzene →(i) NaOH, 623K, 300 atm; (ii) H+[A]→Na2Cr2O7/H2SO4[B]\xrightarrow{\text{(i) NaOH, 623K, 300 atm; (ii) }H^+}[A]\xrightarrow{Na_2Cr_2O_7/H_2SO_4}[B](i) NaOH, 623K, 300 atm; (ii) H+​[A]Na2​Cr2​O7​/H2​SO4​​[B].
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (C)

Step-by-step solution →
Q43·Chemistry·SolutionsSingle correct
Consider a binary solution of two volatile liquid components 1 and 2. x1x_1x1​ and y1y_1y1​ are the mole fractions of component 1 in liquid and vapour phase, respectively. The slope and intercept of the linear plot of 1x1\dfrac{1}{x_1}x1​1​ vs 1y1\dfrac{1}{y_1}y1​1​ are given respectively as :
  1. (A)P10P20, P20−P10P20\dfrac{P_1^0}{P_2^0},\ \dfrac{P_2^0-P_1^0}{P_2^0}P20​P10​​, P20​P20​−P10​​
  2. (B)P20P10, P10−P20P20\dfrac{P_2^0}{P_1^0},\ \dfrac{P_1^0-P_2^0}{P_2^0}P10​P20​​, P20​P10​−P20​​
  3. (C)P10P20, P10−P20P20\dfrac{P_1^0}{P_2^0},\ \dfrac{P_1^0-P_2^0}{P_2^0}P20​P10​​, P20​P10​−P20​​
  4. (D)P20P10, P20−P10P20\dfrac{P_2^0}{P_1^0},\ \dfrac{P_2^0-P_1^0}{P_2^0}P10​P20​​, P20​P20​−P10​​

Correct answer: (A)

Step-by-step solution →
Q44·Chemistry·Classification of Elements and Periodicity in PropertiesSingle correct
Given below are two statements about X-ray spectra of elements: Statement (I): A plot of ν\sqrt{\nu}ν​ (ν=\nu=ν= frequency of X-rays emitted) vs atomic mass is a straight line. Statement (II): A plot of ν\nuν (ν=\nu=ν= frequency of X-rays emitted) vs atomic number is a straight line. 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 true
  3. (C)Both Statement I and Statement II are false
  4. (D)Statement I is false but Statement II is true

Correct answer: (C)

Step-by-step solution →
Q45·Chemistry·d- and f-Block ElementsSingle correct
Consider the following reactions: K2Cr2O7→KOH, −H2O[A]→H2SO4, −H2O[B]+K2SO4K_2Cr_2O_7\xrightarrow{KOH,\,-H_2O}[A]\xrightarrow{H_2SO_4,\,-H_2O}[B]+K_2SO_4K2​Cr2​O7​KOH,−H2​O​[A]H2​SO4​,−H2​O​[B]+K2​SO4​. The products [A] and [B], respectively are :
  1. (A)K2Cr(OH)4K_2Cr(OH)_4K2​Cr(OH)4​ and Cr2O3Cr_2O_3Cr2​O3​
  2. (B)K2CrO4K_2CrO_4K2​CrO4​ and Cr2O3Cr_2O_3Cr2​O3​
  3. (C)K2CrO4K_2CrO_4K2​CrO4​ and K2Cr2O7K_2Cr_2O_7K2​Cr2​O7​
  4. (D)K2CrO4K_2CrO_4K2​CrO4​ and CrOCrOCrO

Correct answer: (C)

Step-by-step solution →
Q46·Chemistry·Some Basic Concepts in ChemistryInteger
0.01 mole of an organic compound (X) containing 10% hydrogen, on complete combustion produced 0.9 g H2OH_2OH2​O. Molar mass of (X) is __________ g mol−1^{-1}−1.

Correct answer: 100

Step-by-step solution →
Q47·Chemistry·Diazonium Salts and ReactionsInteger
Consider the following sequence of reactions starting from 4-ethoxyaniline: →(i) NaNO2, HCl, 0-5∘C; (ii) HCl (dil)[A]→(i) phenol, NaOH; (ii) HCl (dil)[B] (C14H14N2O2)→(i) NaOH; (ii) H3CCH2Br[C] (C16H18N2O2)\xrightarrow{\text{(i) }NaNO_2,\,HCl,\,0\text{-}5^\circ C;\ \text{(ii) }HCl\,(dil)}[A]\xrightarrow{\text{(i) phenol, }NaOH;\ \text{(ii) }HCl\,(dil)}[B]\ (C_{14}H_{14}N_2O_2)\xrightarrow{\text{(i) }NaOH;\ \text{(ii) }H_3CCH_2Br}[C]\ (C_{16}H_{18}N_2O_2)(i) NaNO2​,HCl,0-5∘C; (ii) HCl(dil)​[A](i) phenol, NaOH; (ii) HCl(dil)​[B] (C14​H14​N2​O2​)(i) NaOH; (ii) H3​CCH2​Br​[C] (C16​H18​N2​O2​). Total number of sp3^33 hybridised carbon atoms in the major product C formed is __________.

Correct answer: 4

Step-by-step solution →
Q48·Chemistry·Some Basic Concepts in ChemistryInteger
When 81.0 g of aluminium is allowed to react with 128.0 g of oxygen gas, the mass of aluminium oxide produced in grams is __________. (Nearest integer) Given: Molar mass of Al =27.0=27.0=27.0 g mol−1^{-1}−1, Molar mass of O =16.0=16.0=16.0 g mol−1^{-1}−1.

Correct answer: 153

Step-by-step solution →
Q49·Chemistry·Chemical ThermodynamicsInteger
The bond dissociation enthalpy of X2X_2X2​, ΔHbond∘\Delta H^\circ_{bond}ΔHbond∘​ calculated from the given data is __________ kJ mol−1^{-1}−1. (Nearest integer) Given: M+X−(s)→M+(g)+X−(g)M^+X^-(s)\to M^+(g)+X^-(g)M+X−(s)→M+(g)+X−(g), ΔHlattice∘=800\Delta H^\circ_{lattice}=800ΔHlattice∘​=800 kJ mol−1^{-1}−1; M(s)→M(g)M(s)\to M(g)M(s)→M(g), ΔHsub∘=100\Delta H^\circ_{sub}=100ΔHsub∘​=100 kJ mol−1^{-1}−1; M(g)→M+(g)+e−(g)M(g)\to M^+(g)+e^-(g)M(g)→M+(g)+e−(g), ΔHIE∘=500\Delta H^\circ_{IE}=500ΔHIE∘​=500 kJ mol−1^{-1}−1; X(g)+e−(g)→X−(g)X(g)+e^-(g)\to X^-(g)X(g)+e−(g)→X−(g), ΔHeg∘=−300\Delta H^\circ_{eg}=-300ΔHeg∘​=−300 kJ mol−1^{-1}−1; M(s)+12X2(g)→M+X−(s)M(s)+\tfrac12 X_2(g)\to M^+X^-(s)M(s)+21​X2​(g)→M+X−(s), ΔHf∘=−400\Delta H^\circ_f=-400ΔHf∘​=−400 kJ mol−1^{-1}−1. [M+X−M^+X^-M+X− is a pure ionic compound and X forms a diatomic molecule X2X_2X2​ in gaseous state.]

Correct answer: 200

Step-by-step solution →
Q50·Chemistry·HydrocarbonsInteger
A compound 'X' absorbs 2 moles of hydrogen and 'X' upon oxidation with KMnO4/H+KMnO_4/H^+KMnO4​/H+ gives CH3COCH3CH_3COCH_3CH3​COCH3​, CH3COOHCH_3COOHCH3​COOH and CH3COCH2CH2COOHCH_3COCH_2CH_2COOHCH3​COCH2​CH2​COOH. The total number of σ\sigmaσ bonds present in the compound 'X' is __________.

Correct answer: 27

Step-by-step solution →

Mathematics — JEE Main 23 January 2025 Shift 2

Q51·Mathematics·Binomial Theorem and Its Simple ApplicationsSingle correct
If in the expansion of (1+x)p(1−x)q(1+x)^p(1-x)^q(1+x)p(1−x)q, the coefficients of xxx and x2x^2x2 are 111 and −2-2−2, respectively, then p2+q2p^2+q^2p2+q2 is equal to :
  1. (A)8
  2. (B)18
  3. (C)13
  4. (D)20

Correct answer: (C)

Step-by-step solution →
Q52·Mathematics·Sets, Relations and FunctionsSingle correct
Let A={(x,y)∈R×R:∣x+y∣≥3}A=\{(x,y)\in\mathbb{R}\times\mathbb{R}:|x+y|\ge 3\}A={(x,y)∈R×R:∣x+y∣≥3} and B={(x,y)∈R×R:∣x∣+∣y∣≤3}B=\{(x,y)\in\mathbb{R}\times\mathbb{R}:|x|+|y|\le 3\}B={(x,y)∈R×R:∣x∣+∣y∣≤3}. If C={(x,y)∈A∩B:x=0 or y=0}C=\{(x,y)\in A\cap B: x=0 \text{ or } y=0\}C={(x,y)∈A∩B:x=0 or y=0}, then ∑(x,y)∈C∣x+y∣\displaystyle\sum_{(x,y)\in C}|x+y|(x,y)∈C∑​∣x+y∣ is :
  1. (A)15
  2. (B)18
  3. (C)24
  4. (D)12

Correct answer: (D)

Step-by-step solution →
Q53·Mathematics·Matrices and DeterminantsSingle correct
The system of equations x+y+z=6, x+2y+5z=9, x+5y+λz=μ,x+y+z=6,\ x+2y+5z=9,\ x+5y+\lambda z=\mu,x+y+z=6, x+2y+5z=9, x+5y+λz=μ, has no solution if :
  1. (A)λ=17, μ≠18\lambda=17,\ \mu\neq 18λ=17, μ=18
  2. (B)λ≠17, μ=18\lambda\neq 17,\ \mu=18λ=17, μ=18
  3. (C)λ=15, μ≠18\lambda=15,\ \mu\neq 18λ=15, μ=18
  4. (D)λ=17, μ=18\lambda=17,\ \mu=18λ=17, μ=18

Correct answer: (A)

Step-by-step solution →
Q54·Mathematics·Indefinite IntegrationSingle correct
Let ∫x3sin⁡x dx=g(x)+C\displaystyle\int x^3\sin x\,dx=g(x)+C∫x3sinxdx=g(x)+C, where CCC is the constant of integration. If 8[g ⁣(π2)+g′ ⁣(π2)]=απ3+βπ2+γ, α,β,γ∈Z8\left[g\!\left(\tfrac{\pi}{2}\right)+g'\!\left(\tfrac{\pi}{2}\right)\right]=\alpha\pi^3+\beta\pi^2+\gamma,\ \alpha,\beta,\gamma\in\mathbb{Z}8[g(2π​)+g′(2π​)]=απ3+βπ2+γ, α,β,γ∈Z, then α+β−γ\alpha+\beta-\gammaα+β−γ equals :
  1. (A)55
  2. (B)47
  3. (C)48
  4. (D)62

Correct answer: (A)

Step-by-step solution →
Q55·Mathematics·Straight LinesSingle correct
A rod of length eight units moves such that its ends AAA and BBB always lie on the lines x−y+2=0x-y+2=0x−y+2=0 and y+2=0y+2=0y+2=0, respectively. If the locus of the point PPP, that divides the rod ABABAB internally in the ratio 2:12:12:1 is 9(x2+αy2+βxy+γx+28y)−76=09(x^2+\alpha y^2+\beta xy+\gamma x+28y)-76=09(x2+αy2+βxy+γx+28y)−76=0, then α−β−γ\alpha-\beta-\gammaα−β−γ is equal to :
  1. (A)24
  2. (B)23
  3. (C)21
  4. (D)22

Correct answer: (B)

Step-by-step solution →
Q56·Mathematics·Three Dimensional GeometrySingle correct
The distance of the line x−22=y−63=z−34\dfrac{x-2}{2}=\dfrac{y-6}{3}=\dfrac{z-3}{4}2x−2​=3y−6​=4z−3​ from the point (1,4,0)(1,4,0)(1,4,0) along the line x1=y−22=z+33\dfrac{x}{1}=\dfrac{y-2}{2}=\dfrac{z+3}{3}1x​=2y−2​=3z+3​ is :
  1. (A)17\sqrt{17}17​
  2. (B)14\sqrt{14}14​
  3. (C)15\sqrt{15}15​
  4. (D)13\sqrt{13}13​

Correct answer: (B)

Step-by-step solution →
Q57·Mathematics·Vector AlgebraSingle correct
Let the point AAA divide the line segment joining the points P(−1,−1,2)P(-1,-1,2)P(−1,−1,2) and Q(5,5,10)Q(5,5,10)Q(5,5,10) internally in the ratio r:1r:1r:1 (r>0)(r>0)(r>0). If OOO is the origin and (OQ→⋅OA→−15 ∣OP→×OA→∣2)=10\left(\overrightarrow{OQ}\cdot\overrightarrow{OA}-\tfrac{1}{5}\,|\overrightarrow{OP}\times\overrightarrow{OA}|^2\right)=10(OQ​⋅OA−51​∣OP×OA∣2)=10, then the value of rrr is :
  1. (A)14
  2. (B)3
  3. (C)7\sqrt{7}7​
  4. (D)7

Correct answer: (D)

Step-by-step solution →
Q58·Mathematics·Area Under CurvesSingle correct
If the area of the region {(x,y):−1≤x≤1, 0≤y≤a+e∣x∣−e−x, a>0}\{(x,y):-1\le x\le 1,\ 0\le y\le a+e^{|x|}-e^{-x},\ a>0\}{(x,y):−1≤x≤1, 0≤y≤a+e∣x∣−e−x, a>0} is e2+8e+1e\dfrac{e^2+8e+1}{e}ee2+8e+1​, then the value of aaa is :
  1. (A)7
  2. (B)6
  3. (C)8
  4. (D)5

Correct answer: (D)

Step-by-step solution →
Q59·Mathematics·Limits and ContinuitySingle correct
A spherical chocolate ball has a layer of ice-cream of uniform thickness around it. When the thickness of the ice-cream layer is 111 cm, the ice-cream melts at the rate of 81 cm3/min81\,\text{cm}^3/\text{min}81cm3/min and the thickness of the ice-cream layer decreases at the rate of 14π\dfrac{1}{4\pi}4π1​ cm/min. The surface area (in cm2\text{cm}^2cm2) of the chocolate ball (without the ice-cream layer) is :
  1. (A)225π225\pi225π
  2. (B)128π128\pi128π
  3. (C)196π196\pi196π
  4. (D)256π256\pi256π

Correct answer: (D)

Step-by-step solution →
Q60·Mathematics·Statistics and ProbabilitySingle correct
A board has 16 squares as shown in the figure : Out of these 16 squares, two squares are chosen at random. The probability that they have no side in common is :
  1. (A)45\dfrac{4}{5}54​
  2. (B)710\dfrac{7}{10}107​
  3. (C)35\dfrac{3}{5}53​
  4. (D)2330\dfrac{23}{30}3023​

Correct answer: (A)

Step-by-step solution →
Q61·Mathematics·Differential EquationsSingle correct
Let x=x(y)x=x(y)x=x(y) be the solution of the differential equation y=(x−ydxdy)sin⁡ ⁣(xy), y>0y=\left(x-y\dfrac{dx}{dy}\right)\sin\!\left(\dfrac{x}{y}\right),\ y>0y=(x−ydydx​)sin(yx​), y>0 and x(1)=π2x(1)=\dfrac{\pi}{2}x(1)=2π​. Then cos⁡(x(2))\cos\big(x(2)\big)cos(x(2)) is equal to :
  1. (A)1−2(log⁡e2)21-2(\log_e 2)^21−2(loge​2)2
  2. (B)2(log⁡e2)2−12(\log_e 2)^2-12(loge​2)2−1
  3. (C)2(log⁡e2)−12(\log_e 2)-12(loge​2)−1
  4. (D)1−2(log⁡e2)1-2(\log_e 2)1−2(loge​2)

Correct answer: (B)

Step-by-step solution →
Q62·Mathematics·Trigonometric FunctionsSingle correct
Let the range of the function f(x)=6+16cos⁡x⋅cos⁡ ⁣(π3−x)⋅cos⁡ ⁣(π3+x)⋅sin⁡3x⋅cos⁡6x, x∈Rf(x)=6+16\cos x\cdot\cos\!\left(\dfrac{\pi}{3}-x\right)\cdot\cos\!\left(\dfrac{\pi}{3}+x\right)\cdot\sin 3x\cdot\cos 6x,\ x\in\mathbb{R}f(x)=6+16cosx⋅cos(3π​−x)⋅cos(3π​+x)⋅sin3x⋅cos6x, x∈R be [α,β][\alpha,\beta][α,β]. Then the distance of the point (α,β)(\alpha,\beta)(α,β) from the line 3x+4y+12=03x+4y+12=03x+4y+12=0 is :
  1. (A)11
  2. (B)8
  3. (C)10
  4. (D)9

Correct answer: (A)

Step-by-step solution →
Q63·Mathematics·CirclesSingle correct
Let the shortest distance from (a,0), a>0,(a,0),\ a>0,(a,0), a>0, to the parabola y2=4xy^2=4xy2=4x be 444. Then the equation of the circle passing through the point (a,0)(a,0)(a,0) and the focus of the parabola, and having its centre on the axis of the parabola is :
  1. (A)x2+y2−6x+5=0x^2+y^2-6x+5=0x2+y2−6x+5=0
  2. (B)x2+y2−4x+3=0x^2+y^2-4x+3=0x2+y2−4x+3=0
  3. (C)x2+y2−10x+9=0x^2+y^2-10x+9=0x2+y2−10x+9=0
  4. (D)x2+y2−8x+7=0x^2+y^2-8x+7=0x2+y2−8x+7=0

Correct answer: (A)

Step-by-step solution →
Q64·Mathematics·Sets, Relations and FunctionsSingle correct
Let X=R×RX=\mathbb{R}\times\mathbb{R}X=R×R. Define a relation RRR on XXX as: (a1,b1) R (a2,b2)⇔b1=b2(a_1,b_1)\,R\,(a_2,b_2)\Leftrightarrow b_1=b_2(a1​,b1​)R(a2​,b2​)⇔b1​=b2​. Statement-I: RRR is an equivalence relation. Statement-II: For some (a,b)∈X(a,b)\in X(a,b)∈X, the set S={(x,y)∈X:(x,y) R (a,b)}S=\{(x,y)\in X:(x,y)\,R\,(a,b)\}S={(x,y)∈X:(x,y)R(a,b)} represents a line parallel to y=xy=xy=x. In the light of the above statements, choose the correct answer from the options given below :
  1. (A)Both Statement-I and Statement-II are false.
  2. (B)Statement-I is true but Statement-II is false.
  3. (C)Both Statement-I and Statement-II are true.
  4. (D)Statement-I is false but Statement-II is true.

Correct answer: (B)

Step-by-step solution →
Q65·Mathematics·EllipseSingle correct
The length of the chord of the ellipse x24+y22=1\dfrac{x^2}{4}+\dfrac{y^2}{2}=14x2​+2y2​=1, whose mid-point is (1,12)\left(1,\dfrac{1}{2}\right)(1,21​), is :
  1. (A)2315\dfrac{2}{3}\sqrt{15}32​15​
  2. (B)5315\dfrac{5}{3}\sqrt{15}35​15​
  3. (C)1315\dfrac{1}{3}\sqrt{15}31​15​
  4. (D)15\sqrt{15}15​

Correct answer: (A)

Step-by-step solution →
Q66·Mathematics·Matrices and DeterminantsSingle correct
Let A=[aij]A=[a_{ij}]A=[aij​] be a 3×33\times 33×3 matrix such that A[010]=[001], A[413]=[110]A\begin{bmatrix}0\\1\\0\end{bmatrix}=\begin{bmatrix}0\\0\\1\end{bmatrix},\ A\begin{bmatrix}4\\1\\3\end{bmatrix}=\begin{bmatrix}1\\1\\0\end{bmatrix}A​010​​=​001​​, A​413​​=​110​​ and A[212]=[100]A\begin{bmatrix}2\\1\\2\end{bmatrix}=\begin{bmatrix}1\\0\\0\end{bmatrix}A​212​​=​100​​, then a23a_{23}a23​ equals :
  1. (A)−1-1−1
  2. (B)000
  3. (C)222
  4. (D)111

Correct answer: (A)

Step-by-step solution →
Q67·Mathematics·Complex NumbersSingle correct
The number of complex numbers zzz, satisfying ∣z∣=1|z|=1∣z∣=1 and ∣zzˉ+zˉz∣=1\left|\dfrac{z}{\bar z}+\dfrac{\bar z}{z}\right|=1​zˉz​+zzˉ​​=1, is :
  1. (A)6
  2. (B)4
  3. (C)10
  4. (D)8

Correct answer: (D)

Step-by-step solution →
Q68·Mathematics·Three Dimensional GeometrySingle correct
If the square of the shortest distance between the lines x−21=y−12=z+3−3\dfrac{x-2}{1}=\dfrac{y-1}{2}=\dfrac{z+3}{-3}1x−2​=2y−1​=−3z+3​ and x+12=y+34=z+5−5\dfrac{x+1}{2}=\dfrac{y+3}{4}=\dfrac{z+5}{-5}2x+1​=4y+3​=−5z+5​ is mn\dfrac{m}{n}nm​, where m,nm,nm,n are coprime numbers, then m+nm+nm+n is equal to :
  1. (A)6
  2. (B)9
  3. (C)21
  4. (D)14

Correct answer: (B)

Step-by-step solution →
Q69·Mathematics·Definite IntegrationSingle correct
If I=∫0π/2sin⁡2xsin⁡4x+cos⁡4x dxI=\displaystyle\int_0^{\pi/2}\dfrac{\sin^2 x}{\sin^4 x+\cos^4 x}\,dxI=∫0π/2​sin4x+cos4xsin2x​dx, then ∫0π/2xsin⁡xcos⁡xsin⁡4x+cos⁡4x dx\displaystyle\int_0^{\pi/2}\dfrac{x\sin x\cos x}{\sin^4 x+\cos^4 x}\,dx∫0π/2​sin4x+cos4xxsinxcosx​dx equals :
  1. (A)π216\dfrac{\pi^2}{16}16π2​
  2. (B)π24\dfrac{\pi^2}{4}4π2​
  3. (C)π28\dfrac{\pi^2}{8}8π2​
  4. (D)π212\dfrac{\pi^2}{12}12π2​

Correct answer: (A)

Step-by-step solution →
Q70·Mathematics·Limits and ContinuitySingle correct
lim⁡x→∞(2x2−3x+5)(3x−1)x/2(3x2+5x+4)(3x+2)x\displaystyle\lim_{x\to\infty}\dfrac{\left(2x^2-3x+5\right)\left(3x-1\right)^{x/2}}{\left(3x^2+5x+4\right)\sqrt{\left(3x+2\right)^x}}x→∞lim​(3x2+5x+4)(3x+2)x​(2x2−3x+5)(3x−1)x/2​ is equal to :
  1. (A)23e\dfrac{2}{\sqrt{3e}}3e​2​
  2. (B)2e3\dfrac{2e}{\sqrt{3}}3​2e​
  3. (C)2e3\dfrac{2e}{3}32e​
  4. (D)23e\dfrac{2}{3\sqrt{e}}3e​2​

Correct answer: (D)

Step-by-step solution →
Q71·Mathematics·Permutations and CombinationsInteger
The number of ways, 555 boys and 444 girls can sit in a row so that either all the boys sit together or no two boys sit together, is __________.

Correct answer: 17280

Step-by-step solution →
Q72·Mathematics·Quadratic EquationsInteger
Let α,β\alpha,\betaα,β be the roots of the equation x2−ax−b=0x^2-ax-b=0x2−ax−b=0 with Im⁡(α)<Im⁡(β)\operatorname{Im}(\alpha)<\operatorname{Im}(\beta)Im(α)<Im(β). Let Pn=αn−βnP_n=\alpha^n-\beta^nPn​=αn−βn. If P3=−57 i, P4=−37 i, P5=117 iP_3=-5\sqrt7\,i,\ P_4=-3\sqrt7\,i,\ P_5=11\sqrt7\,iP3​=−57​i, P4​=−37​i, P5​=117​i and P6=457 iP_6=45\sqrt7\,iP6​=457​i, then ∣α4+β4∣|\alpha^4+\beta^4|∣α4+β4∣ is equal to __________.

Correct answer: 31

Step-by-step solution →
Q73·Mathematics·CirclesInteger
The focus of the parabola y2=4x+16y^2=4x+16y2=4x+16 is the centre of the circle CCC of radius 555. If the values of λ\lambdaλ, for which CCC passes through the point of intersection of the lines 3x−y=03x-y=03x−y=0 and x+λy=4x+\lambda y=4x+λy=4, are λ1\lambda_1λ1​ and λ2, λ1<λ2\lambda_2,\ \lambda_1<\lambda_2λ2​, λ1​<λ2​, then 12λ1+29λ212\lambda_1+29\lambda_212λ1​+29λ2​ is equal to __________.

Correct answer: 15

Step-by-step solution →
Q74·Mathematics·Statistics and ProbabilityInteger
The variance of the numbers 8, 21, 34, 47, …, 3208,\ 21,\ 34,\ 47,\ \ldots,\ 3208, 21, 34, 47, …, 320, is __________.

Correct answer: 8788

Step-by-step solution →
Q75·Mathematics·Sequence and SeriesInteger
The roots of the quadratic equation 3x2−px+q=03x^2-px+q=03x2−px+q=0 are 10th10^{\text{th}}10th and 11th11^{\text{th}}11th terms of an arithmetic progression with common difference 32\dfrac{3}{2}23​. If the sum of the first 111111 terms of this arithmetic progression is 888888, then q−2pq-2pq−2p is equal to __________.

Correct answer: 474

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
  • Binomial Theorem and Its Simple Applications 158/186
  • Electromagnetic Waves 144/186
  • 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
  • Chemical Thermodynamics 165/186
  • Units and Measurements 149/186
  • Hydrocarbons 126/186
  • Complex Numbers 165/186
  • Trigonometric Functions 144/186
  • Biomolecules 162/186
  • Chemical Kinetics 169/186
  • Gravitation 152/186
  • Electric Field and Coulomb's Law 133/186
  • Atomic Structure 161/186
  • Electronic Devices 145/186
  • Circles 142/186
  • Dual Nature of Matter and Radiation 155/186
  • Quadratic Equations 148/186
  • Some Basic Concepts in Chemistry 129/186
  • Wave Optics 130/186
  • Area Under Curves 139/186
  • Alcohols and Ethers 106/186
  • Alternating Currents 108/186
  • Nuclei 116/186
  • Organic Compounds Containing Halogens 109/186
  • Straight Lines 114/186
  • Waves 109/186
  • Classification of Elements and Periodicity in Properties 113/186
  • Statistics 118/186
  • Ellipse 103/186
  • Isolation of Metals 106/186
  • Electric Potential 63/186
  • Indefinite Integration 66/186
  • Principles of Qualitative Analysis 58/186
  • Diazonium Salts and Reactions 53/186
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