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JEE Main 25 January 2023 Shift 2 Question Paper with Answers

25 January 2023 · January session · 90 questions

The complete JEE Main 25 January 2023 Shift 2 paper — all 90 questions with the correct answer for each, tagged to the chapter it tests. Free to read, no account needed.

Physics
30
Chemistry
30
Mathematics
30

Physics — JEE Main 25 January 2023 Shift 2

Q1·PhysicsSingle correct
According to law of equipartition of energy the molar specific heat of a diatomic gas at constant volume where the molecule has one additional vibrational mode is:
  1. (A)52R\frac{5}{2}R25​R
  2. (B)92R\frac{9}{2}R29​R
  3. (C)72R\frac{7}{2}R27​R
  4. (D)32R\frac{3}{2}R23​R

Correct answer: (C)

Step-by-step solution →
Q2·PhysicsSingle correct
A wire of length 111 m moving with velocity 888 m/s at right angles to a magnetic field of 222 T. The magnitude of induced emf, between the ends of wire will be
  1. (A)202020 V
  2. (B)888 V
  3. (C)121212 V
  4. (D)161616 V

Correct answer: (D)

Step-by-step solution →
Q3·PhysicsSingle correct
The energy levels of an atom is shown in figure. Which one of these transitions will result in the emission of a photon of wavelength 124.1124.1124.1 nm? Given h=6.62×10−34h = 6.62\times10^{-34}h=6.62×10−34 Js.
  1. (A)D
  2. (B)B
  3. (C)C
  4. (D)A

Correct answer: (A)

Step-by-step solution →
Q4·PhysicsSingle correct
Given below are two statements: Statement I: Stopping potential in photoelectric effect does not depend on the power of the light source. Statement II: For a given metal, the maximum kinetic energy of the photoelectron depends on the wavelength of the incident light. In the light of above statements, choose the most appropriate answer from the options given below
  1. (A)Statement I is correct but statement II is incorrect
  2. (B)Statement I is incorrect but statement II is correct
  3. (C)Both Statement I and Statement II are correct
  4. (D)Both Statement I and Statement II are incorrect

Correct answer: (C)

Step-by-step solution →
Q5·PhysicsSingle correct
The distance travelled by a particle is related to time t as x=4t2x = 4t^2x=4t2. The velocity of the particle at t=5t = 5t=5 s is:
  1. (A)404040 ms−1^{-1}−1
  2. (B)202020 ms−1^{-1}−1
  3. (C)888 ms−1^{-1}−1
  4. (D)252525 ms−1^{-1}−1

Correct answer: (A)

Step-by-step solution →
Q6·PhysicsSingle correct
Match List I (physical quantities) with List II (their dimensional formulae) and choose the correct answer.
List-IList-II
A.Young's Modulus (Y)I.[M L−1 T−1][M\,L^{-1}\,T^{-1}][ML−1T−1]
B.Co-efficient of Viscosity (η\etaη)II.[M L2 T−1][M\,L^2\,T^{-1}][ML2T−1]
C.Planck's Constant (h)IV.[M L2 T−2][M\,L^2\,T^{-2}][ML2T−2]
D.Work Function (φ\varphiφ)III.[M L−1 T−2][M\,L^{-1}\,T^{-2}][ML−1T−2]
  1. (A)A-I, B-II, C-III, D-IV
  2. (B)A-II, B-III, C-IV, D-I
  3. (C)A-I, B-III, C-IV, D-II
  4. (D)A-III, B-I, C-II, D-IV

Correct answer: (D)

Step-by-step solution →
Q7·PhysicsSingle correct
Match List I (layers of the atmosphere / ionosphere) with List II (their approximate heights over Earth's surface) and choose the correct answer.
List-IList-II
A.TroposphereI.Approximate 65−7565 - 7565−75 km over Earth's surface
B.E- Part of StratosphereII.Approximate 300300300 km over Earth's surface
C.F2- Part of ThermosphereIV.Approximate 100100100 km over Earth's surface
D.D- Part of StratosphereIII.Approximate 101010 km over Earth's surface
  1. (A)A-III, B-IV, C-II, D-I
  2. (B)A-III, B-II, C-I, D-IV
  3. (C)A-I, B-IV, C-III, D-II
  4. (D)A-I, B-II, C-IV, D-III

Correct answer: (A)

Step-by-step solution →
Q8·PhysicsSingle correct
The light rays from an object have been reflected towards an observer from a standard flat mirror, the image observed by the observer are: A. Real B. Erect C. Smaller in size then object D. Laterally inverted Choose the most appropriate answer from the options given below:
  1. (A)A, C, and D Only
  2. (B)B and D Only
  3. (C)A and D Only
  4. (D)B and C Only

Correct answer: (B)

Step-by-step solution →
Q9·PhysicsSingle correct
The graph between two temperature scales PPP and QQQ is shown in the figure. Between upper fixed point and lower fixed point there are 150150150 equal divisions of scale P and 100100100 divisions on scale Q. The relationship for conversion between the two scales is given by:
  1. (A)tP100=tQ−180150\frac{t_P}{100} = \frac{t_Q - 180}{150}100tP​​=150tQ​−180​
  2. (B)tQ150=tP−180100\frac{t_Q}{150} = \frac{t_P - 180}{100}150tQ​​=100tP​−180​
  3. (C)tP180−tQ−40100\frac{t_P}{180} - \frac{t_Q - 40}{100}180tP​​−100tQ​−40​
  4. (D)tQ100=tP−30150\frac{t_Q}{100} = \frac{t_P - 30}{150}100tQ​​=150tP​−30​

Correct answer: (D)

Step-by-step solution →
Q10·PhysicsSingle correct
Consider a block kept on an inclined plane (inclined at 45∘45^\circ45∘) as shown in the figure. If the force required to just push it up the incline is 222 times the force required to just prevent it from sliding down, the coefficient of friction between the block and inclined plane (μ\muμ) is equal to:
  1. (A)0.250.250.25
  2. (B)0.500.500.50
  3. (C)0.600.600.60
  4. (D)0.330.330.33

Correct answer: (D)

Step-by-step solution →
Q11·PhysicsSingle correct
Every planet revolves around the sun in an elliptical orbit: A. The force acting on a planet is inversely proportional to square of distance from sun. B. Force acting on planet is inversely proportional to product of the masses of the planet and the sun. C. The Centripetal force acting on the planet is directed away from the sun. D. The square of time period of revolution of planet around sun is directly proportional to cube of semi-major axis of elliptical orbit. Choose the correct answer from the options given below:
  1. (A)B and C only
  2. (B)A and C Only
  3. (C)A and D only
  4. (D)C and D only

Correct answer: (C)

Step-by-step solution →
Q12·Physics·Magnetic Field of CurrentSingle correct
For a moving coil galvanometer, the deflection in the coil is 0.050.050.05 rad when a current of 101010 mA is passed through it. If the torsional constant of suspension wire is 4.0×10−54.0\times10^{-5}4.0×10−5 N m rad−1^{-1}−1, the magnetic field is 0.010.010.01 T and the number of turns in the coil is 200200200, the area of each turn (in cm2^22) is:
  1. (A)1.01.01.0
  2. (B)2.02.02.0
  3. (C)1.51.51.5
  4. (D)0.50.50.5

Correct answer: (A)

Step-by-step solution →
Q13·PhysicsSingle correct
Match List I (fundamental laws of electromagnetism) with List II (their integral form expressions) and choose the correct answer.
List-IList-II
A.Gauss's Law in ElectrostaticsI.∮E⃗⋅dl⃗=−dϕBdt\oint \vec{E}\cdot d\vec{l} = -\frac{d\phi_B}{dt}∮E⋅dl=−dtdϕB​​
B.Faraday's LawII.∮B⃗⋅dA⃗=0\oint \vec{B}\cdot d\vec{A} = 0∮B⋅dA=0
C.Gauss's Law in MagnetismIV.∮E⃗⋅ds⃗=qϵ0\oint \vec{E}\cdot d\vec{s} = \frac{q}{\epsilon_0}∮E⋅ds=ϵ0​q​
D.Ampere-Maxwell LawIII.∮B⃗⋅dl⃗=μ0ic+μ0ϵ0dϕEdt\oint \vec{B}\cdot d\vec{l} = \mu_0 i_c + \mu_0 \epsilon_0 \frac{d\phi_E}{dt}∮B⋅dl=μ0​ic​+μ0​ϵ0​dtdϕE​​
  1. (A)A-IV, B-I, C-II, D-III
  2. (B)A-II, B-III, C-IV, D-I
  3. (C)A-III, B-IV, C-I, D-II
  4. (D)A-I, B-II, C-III, D-IV

Correct answer: (A)

Step-by-step solution →
Q14·PhysicsSingle correct
Two objects are projected with same velocity 'u' however at different angles α\alphaα and β\betaβ with the horizontal. If α+β=90∘\alpha + \beta = 90^\circα+β=90∘, the ratio of horizontal range of the first object to the 2nd object will be:
  1. (A)2:12:12:1
  2. (B)1:21:21:2
  3. (C)1:11:11:1
  4. (D)4:14:14:1

Correct answer: (C)

Step-by-step solution →
Q15·PhysicsSingle correct
A particle executes simple harmonic motion between x=−Ax = -Ax=−A and x=+Ax = +Ax=+A. If time taken by particle to go from x=0x = 0x=0 to A2\frac{A}{2}2A​ is 222 s; then time taken by particle in going from x=A2x = \frac{A}{2}x=2A​ to A is
  1. (A)444 S
  2. (B)1.51.51.5 S
  3. (C)222 S
  4. (D)333 S

Correct answer: (A)

Step-by-step solution →
Q16·PhysicsSingle correct
Match List I with List II: LIST I: (A) Isothermal Process, (B) Adiabatic Process, (C) Isochoric Process, (D) Isobaric Process. LIST II: (I) Work done by the gas decreases internal energy, (II) No change in internal energy, (III) The heat absorbed goes partly to increase internal energy and partly to do work, (IV) No work is done on or by the gas. Choose the correct answer from the options given below:
List-IList-II
A.Isothermal ProcessI.Work done by the gas decreases internal energy
B.Adiabatic ProcessII.No change in internal energy
C.Isochoric ProcessIII.The heat absorbed goes partly to increase internal energy and partly to do work
D.Isobaric ProcessIV.No work is done on or by the gas
  1. (A)A-I, B-II, C-III, D-IV
  2. (B)A-II, B-I, C-IV, D-III
  3. (C)A-II, B-I, C-III, D-IV
  4. (D)A-I, B-II, C-IV, D-III

Correct answer: (B)

Step-by-step solution →
Q17·PhysicsSingle correct
Statement I: When a Si sample is doped with Boron, it becomes P type and when doped by Arsenic it becomes N-type semi conductor such that P-type has excess holes and N-type has excess electrons. Statement II: When such P-type and N-type semi-conductors, are fused to make a junction, a current will automatically flow which can be detected with an externally connected ammeter. In the light of above statements, choose the most appropriate answer from the options given below:
  1. (A)Both Statement I and statement II are correct
  2. (B)Statement I is incorrect but statement II is correct
  3. (C)Both Statement I and Statement II are incorrect
  4. (D)Statement I is correct but statement II is incorrect

Correct answer: (D)

Step-by-step solution →
Q18·PhysicsSingle correct
A point charge of 10 μC10\,\mu C10μC is placed at the origin. At what location on the X-axis should a point charge of 40 μC40\,\mu C40μC be placed so that the net electric field is zero at x=2x = 2x=2 cm on the X-axis?
  1. (A)x=−4x = -4x=−4 cm
  2. (B)x=6x = 6x=6 cm
  3. (C)x=4x = 4x=4 cm
  4. (D)x=8x = 8x=8 cm

Correct answer: (B)

Step-by-step solution →
Q19·PhysicsSingle correct
The resistance of a wire is 5 Ω5\,\Omega5Ω. Its new resistance in ohm if stretched to 5 times of its original length will be:
  1. (A)25
  2. (B)125
  3. (C)5
  4. (D)625

Correct answer: (B)

Step-by-step solution →
Q20·PhysicsSingle correct
A body of mass is taken from earth surface to the height hhh equal to twice the radius of earth (ReR_eRe​), the increase in potential energy will be: ( ggg = acceleration due to gravity on the surface of Earth)
  1. (A)3 mgRe3\,mgR_e3mgRe​
  2. (B)13mgRe\frac{1}{3}mgR_e31​mgRe​
  3. (C)23mgRe\frac{2}{3}mgR_e32​mgRe​
  4. (D)12mgRe\frac{1}{2}mgR_e21​mgRe​

Correct answer: (C)

Step-by-step solution →
Q21·PhysicsNumerical
Two long parallel wires carrying currents 8 A and 15 A in opposite directions are placed at a distance of 7 cm from each other. A point PPP is at equidistant from both the wires such that the lines joining the point PPP to the wires are perpendicular to each other. The magnitude of magnetic field at PPP is ______ ×10−6\times 10^{-6}×10−6 T (Given: 2=1.4\sqrt{2} = 1.42​=1.4)

Correct answer: 68

Step-by-step solution →
Q22·PhysicsNumerical
A spherical drop of liquid splits into 1000 identical spherical drops. If uiu_iui​ is the surface energy of the original drop and ufu_fuf​ is the total surface energy of the resulting drops, then (ignoring evaporation), ufui=(10x)\frac{u_f}{u_i} = \left(\frac{10}{x}\right)ui​uf​​=(x10​). Then value of xxx is ______.

Correct answer: 1

Step-by-step solution →
Q23·PhysicsNumerical
A nucleus disintegrates into two smaller parts, which have their velocities in the ratio 3:23:23:2. The ratio of their nuclear sizes will be (x3)13\left(\frac{x}{3}\right)^{\frac{1}{3}}(3x​)31​. The value of 'xxx' is:

Correct answer: 2

Step-by-step solution →
Q24·PhysicsNumerical
A train blowing a whistle of frequency 320 Hz approaches an observer standing on the platform at a speed of 66 m/s. The frequency observed by the observer will be (given speed of sound =330 ms−1= 330\,\text{ms}^{-1}=330ms−1) ______ Hz.

Correct answer: 400

Step-by-step solution →
Q25·PhysicsNumerical
A body of mass 1 kg collides head on elastically with a stationary body of mass 3 kg. After collision, the smaller body reverses its direction of motion and moves with a speed of 2 m/s. The initial speed of the smaller body before collision is ______ ms−1\text{ms}^{-1}ms−1

Correct answer: 4

Step-by-step solution →
Q26·PhysicsNumerical
A series LCR circuit is connected to an AC source of 220 V, 50 Hz. The circuit contains a resistance R=80 ΩR = 80\,\OmegaR=80Ω, an inductor of inductive reactance XL=70 ΩX_L = 70\,\OmegaXL​=70Ω, and a capacitor of capacitive reactance XC=130 ΩX_C = 130\,\OmegaXC​=130Ω. The power factor of circuit is x10\frac{x}{10}10x​. The value of xxx is:

Correct answer: 8

Step-by-step solution →
Q27·PhysicsNumerical
If a solid sphere of mass 5 kg and a disc of mass 4 kg have the same radius. Then the ratio of moment of inertia of the disc about a tangent in its plane to the moment of inertia of the sphere about its tangent will be x7\frac{x}{7}7x​. The value of xxx is ______.

Correct answer: 5

Step-by-step solution →
Q28·PhysicsNumerical
An object is placed on the principal axis of convex lens of focal length 10 cm as shown. A plane mirror is placed on the other side of lens at a distance of 20 cm. The image produced by the plane mirror is 5 cm inside the mirror. The distance of the object from the lens is ______ cm.

Correct answer: 30

Step-by-step solution →
Q29·PhysicsNumerical
A capacitor has capacitance 5 μF5\,\mu F5μF when its parallel plates are separated by air medium of thickness ddd. A slab of material of dielectric constant 1.5 having area equal to that of plates but thickness d2\frac{d}{2}2d​ is inserted between the plates. Capacitance of the capacitor in the presence of slab will be ______ μF\mu FμF.

Correct answer: 6

Step-by-step solution →
Q30·PhysicsNumerical
Two cells are connected between points A and B as shown. Cell 1 has emf of 12 V and internal resistance of 3 Ω3\,\Omega3Ω. Cell 2 has emf of 6 V and internal resistance of 6 Ω6\,\Omega6Ω. An external resistor R of 4 Ω4\,\Omega4Ω is connected across A and B. The current flowing through R will be ______ A.

Correct answer: 1

Step-by-step solution →

Chemistry — JEE Main 25 January 2023 Shift 2

Q31·ChemistrySingle correct
When the hydrogen ion concentration [H+][H^+][H+] changes by a factor of 100010001000, the value of pH of the solution
  1. (A)increases by 2 units
  2. (B)increases by 1000 units
  3. (C)decreases by 2 units
  4. (D)decreases by 3 units

Correct answer: (D)

Step-by-step solution →
Q32·Chemistry·Alcohols and EthersSingle correct
Find out the major product from the following reaction.
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (D)

Step-by-step solution →
Q33·ChemistrySingle correct
Given below are two statements, one is labelled as Assertion A and the other is labelled as Reason R. Assertion A : Carbon forms two important oxides - CO and CO2CO_2CO2​. CO is neutral whereas CO2CO_2CO2​ is acidic in nature. Reason R : CO2CO_2CO2​ can combine with water in a limited way to form carbonic acid, while CO is sparingly soluble in water. In the light of the above statements, choose the most appropriate answer from the options given below
  1. (A)Both A and R are correct but R is NOT the correct explanation of A
  2. (B)A is correct but R is not correct
  3. (C)Both A and R are correct and R is the correct explanation of A
  4. (D)A is not correct but R is correct

Correct answer: (C)

Step-by-step solution →
Q34·ChemistrySingle correct
Given below are two statements, one is labelled as Assertion A and the other is labelled as Reason R. Assertion A : The alkali metals and their salts impart characteristic colour to reducing flame. Reason R : Alkali metals can be detected using flame tests. In the light of the above statements, choose the most appropriate answer from the options given below
  1. (A)A is not correct but R is correct
  2. (B)Both A and R are correct but R is NOT the correct explanation of A
  3. (C)A is correct but R is not correct
  4. (D)Both A and R are correct and R is the correct explanation of A

Correct answer: (A)

Step-by-step solution →
Q35·ChemistrySingle correct
Potassium dichromate acts as a strong oxidizing agent in acidic solution. During this process, the oxidation state change from
  1. (A)+2 to +1
  2. (B)+3 to +1
  3. (C)+6 to +2
  4. (D)+6 to +3

Correct answer: (D)

Step-by-step solution →
Q36·ChemistrySingle correct
Match List I (Name of polymer) with List II (Uses) and choose the correct answer from the options given below.
LIST I (Name of polymer)LIST II (Uses)
A.GlyptalI.Flexible pipes
B.NeopreneII.Synthetic wool
C.AcrilanIII.Paints and Lacquers
D.LDPIV.Gaskets
  1. (A)A-III, B-IV, C-I, D-II
  2. (B)A-III, B-II, C-IV, D-I
  3. (C)A-III, B-I, C-IV, D-II
  4. (D)A-III, B-IV, C-II, D-I

Correct answer: (D)

Step-by-step solution →
Q37·ChemistrySingle correct
Which of the following represents the correct order of metallic character of the given elements ?
  1. (A)Si<Be<Mg<KSi < Be < Mg < KSi<Be<Mg<K
  2. (B)Be<Si<K<MgBe < Si < K < MgBe<Si<K<Mg
  3. (C)Be<Si<Mg<KBe < Si < Mg < KBe<Si<Mg<K
  4. (D)K<Mg<Be<SiK < Mg < Be < SiK<Mg<Be<Si

Correct answer: (A)

Step-by-step solution →
Q38·ChemistrySingle correct
Match List I with List II (industrial processes/catalysts and their applications) and choose the correct answer from the options given below.
LIST ILIST II
A.Cobalt catalystI.(H2+Cl2)(H_2 + Cl_2)(H2​+Cl2​) production
B.SyngasII.Water gas production
C.Nickel catalystIII.Coal gasification
D.Brine solutionIV.Methanol production
  1. (A)A-IV, B-I, C-II, D-III
  2. (B)A-IV, B-III, C-II, D-I
  3. (C)A-II, B-III, C-IV, D-I
  4. (D)A-IV, B-III, C-I, D-II

Correct answer: (B)

Step-by-step solution →
Q39·ChemistrySingle correct
Match List I (Amines) with List II (pKbpK_bpKb​ values) and choose the correct answer from the options given below.
LIST I (Amines)LIST II ($pK_b$)
A.AnilineI.3.25
B.EthanamineII.3.00
C.N-EthylethanamineIII.9.38
D.N,N-DiethylethanamineIV.3.29
  1. (A)A-III, B-IV, C-II, D-I
  2. (B)A-III, B-II, C-I, D-IV
  3. (C)A-I, B-IV, C-II, D-III
  4. (D)A-III, B-II, C-IV, D-I

Correct answer: (A)

Step-by-step solution →
Q40·Chemistry·IsomerismSingle correct
Match List I (Isomeric pairs) with List II (Type of isomers) and choose the correct answer from the options given below.
LIST I (Isomeric pairs)LIST II (Type of isomers)
A.Propanamine and N-MethylethanamineI.Metamers
B.Hexan-2-one and Hexan-3-oneII.Positional isomers
C.Ethanamide and HydroxyethanimineIII.Functional isomers
D.o-nitrophenol and p-nitrophenolIV.Tautomers
  1. (A)A-II, B-III, C-I, D-IV
  2. (B)A-III, B-I, C-IV, D-II
  3. (C)A-III, B-IV, C-I, D-II
  4. (D)A-IV, B-III, C-I, D-II

Correct answer: (B)

Step-by-step solution →
Q41·ChemistrySingle correct
What is the mass ratio of ethylene glycol (C2H6O2C_2H_6O_2C2​H6​O2​, molar mass =62= 62=62 g/mol) required for making 500500500 g of 0.250.250.25 molal aqueous solution and 250250250 mL of 0.250.250.25 molal aqueous solution?
  1. (A)1 : 1
  2. (B)2 : 1
  3. (C)1 : 2
  4. (D)3 : 1

Correct answer: (B)

Step-by-step solution →
Q42·ChemistrySingle correct
Match List I (Coordination entity) with List II (Wavelength of light absorbed in nm) and choose the correct answer from the options given below.
LIST I (Coordination entity)LIST II (Wavelength of light absorbed in nm)
A.[CoCl(NH3)5]2+[CoCl(NH_3)_5]^{2+}[CoCl(NH3​)5​]2+I.310
B.[Co(NH3)6]3+[Co(NH_3)_6]^{3+}[Co(NH3​)6​]3+II.475
C.[Co(CN)6]3−[Co(CN)_6]^{3-}[Co(CN)6​]3−III.535
D.[Cu(H2O)4]2+[Cu(H_2O)_4]^{2+}[Cu(H2​O)4​]2+IV.600
  1. (A)A-III, B-I, C-II, D-IV
  2. (B)A-IV, B-I, C-III, D-II
  3. (C)A-III, B-II, C-I, D-IV
  4. (D)A-II, B-III, C-IV, D-I

Correct answer: (C)

Step-by-step solution →
Q43·ChemistrySingle correct
Given below are two statements, one is labelled as Assertion A and the other is labelled as Reason R. Assertion A : Butylated hydroxy anisole when added to butter increases its shelf life. Reason R : Butylated hydroxy anisole is more reactive towards oxygen than food. In the light of the above statements, choose the most appropriate answer from the options given below
  1. (A)A is correct but R is not correct
  2. (B)A is not correct but R is correct
  3. (C)Both A and R are correct and R is the correct explanation of A
  4. (D)Both A and R are correct but R is NOT the correct explanation of A

Correct answer: (C)

Step-by-step solution →
Q44·Chemistry·IsomerismSingle correct
The isomeric deuterated bromide with molecular formula C4H8DBrC_4H_8DBrC4​H8​DBr having two chiral carbon atoms is
  1. (A)2-Bromo-2-deuterobutane
  2. (B)2-Bromo-1-deuterobutane
  3. (C)2-Bromo-1-deutero-2-methylpropane
  4. (D)2-Bromo-3-deuterobutane

Correct answer: (D)

Step-by-step solution →
Q45·ChemistrySingle correct
A chloride salt solution acidified with dil. HNO3HNO_3HNO3​ gives a curdy white precipitate, [A], on addition of AgNO3AgNO_3AgNO3​. [A] on treatment with NH4OHNH_4OHNH4​OH gives a clear solution, B. A and B are respectively
  1. (A)AgClAgClAgCl & (NH4)[Ag(OH)2](NH_4)[Ag(OH)_2](NH4​)[Ag(OH)2​]
  2. (B)AgClAgClAgCl & [Ag(NH3)2]Cl[Ag(NH_3)_2]Cl[Ag(NH3​)2​]Cl
  3. (C)H[AgCl3]H[AgCl_3]H[AgCl3​] & (NH4)[Ag(OH)2](NH_4)[Ag(OH)_2](NH4​)[Ag(OH)2​]
  4. (D)H[AgCl3]H[AgCl_3]H[AgCl3​] & [Ag(NH3)2]Cl[Ag(NH_3)_2]Cl[Ag(NH3​)2​]Cl

Correct answer: (B)

Step-by-step solution →
Q46·ChemistrySingle correct
Statement I : Dipole moment is a vector quantity and by convention it is depicted by a small arrow with tail on the negative centre and head pointing towards the positive centre. Statement II : The crossed arrow of the dipole moment symbolizes the direction of the shift of charges in the molecules. In the light of the above statements, choose the most appropriate answer from the options given below:
  1. (A)Statement I is incorrect but Statement II is correct
  2. (B)Statement I is correct but Statement II is incorrect
  3. (C)Both Statement I and Statement II are incorrect
  4. (D)Both Statement I and Statement II are correct

Correct answer: (B)

Step-by-step solution →
Q47·ChemistrySingle correct
Identify the major product 'A' in the reaction shown in the figure.
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (D)

Step-by-step solution →
Q48·ChemistrySingle correct
A. Ammonium salts produce haze in atmosphere. B. Ozone gets produced when atmospheric oxygen reacts with chlorine radicals. C. Polychlorinated biphenyls act as cleansing solvents. D. 'Blue baby' syndrome occurs due to the presence of excess of sulphate ions in water. Choose the correct answer from the options given below:
  1. (A)A and D only
  2. (B)A, B and C only
  3. (C)A and C only
  4. (D)B and C only

Correct answer: (C)

Step-by-step solution →
Q49·ChemistrySingle correct
Given below are two statements: Statement I : In froth floatation method a rotating paddle agitates the mixture to drive air out of it. Statement II : Iron pyrites are generally avoided for extraction of iron due to environmental reasons. 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 false
  3. (C)Statement I is true but Statement II is false
  4. (D)Both Statement I and Statement II are true

Correct answer: (A)

Step-by-step solution →
Q50·ChemistrySingle correct
Which one among the following metals is the weakest reducing agent?
  1. (A)Li
  2. (B)K
  3. (C)Rb
  4. (D)Na

Correct answer: (D)

Step-by-step solution →
Q51·ChemistryNumerical
Total number of moles of AgCl precipitated on addition of excess of AgNO3AgNO_3AgNO3​ to one mole each of the following complexes [Co(NH3)4Cl2]Cl[Co(NH_3)_4Cl_2]Cl[Co(NH3​)4​Cl2​]Cl, [Ni(H2O)6]Cl2[Ni(H_2O)_6]Cl_2[Ni(H2​O)6​]Cl2​, [Pt(NH3)2Cl2][Pt(NH_3)_2Cl_2][Pt(NH3​)2​Cl2​] and [Pd(NH3)4]Cl2[Pd(NH_3)_4]Cl_2[Pd(NH3​)4​]Cl2​ is

Correct answer: 5

Step-by-step solution →
Q52·ChemistryNumerical
The number of incorrect statement/s from the following is/are A. Water vapours are adsorbed by anhydrous calcium chloride. B. There is a decrease in surface energy during adsorption. C. As the adsorption proceeds, ΔH\Delta HΔH becomes more and more negative. D. Adsorption is accompanied by decrease in entropy of the system.

Correct answer: 2

Step-by-step solution →
Q53·ChemistryNumerical
Number of hydrogen atoms per molecule of a hydrocarbon A having 85.8% carbon is (Given: Molar mass of A =84= 84=84 g mol−1^{-1}−1)

Correct answer: 12

Step-by-step solution →
Q54·ChemistryNumerical
The number of given orbitals which have electron density along the axis is Px,Py,Pz,dxy,dyz,dxz,dz2,dx2−y2P_x, P_y, P_z, d_{xy}, d_{yz}, d_{xz}, d_{z^2}, d_{x^2-y^2}Px​,Py​,Pz​,dxy​,dyz​,dxz​,dz2​,dx2−y2​

Correct answer: 5

Step-by-step solution →
Q55·ChemistryNumerical
28.0 L of CO2CO_2CO2​ is produced on complete combustion of 16.8 L gaseous mixture of ethene and methane at 25∘C25^\circ C25∘C and 1 atm. Heat evolved during the combustion process is _____ kJ. Given : ΔHc(CH4)=−900\Delta H_c(CH_4) = -900ΔHc​(CH4​)=−900 kJ mol−1^{-1}−1 ΔHc(C2H4)=−1400\Delta H_c(C_2H_4) = -1400ΔHc​(C2​H4​)=−1400 kJ mol−1^{-1}−1

Correct answer: 847

Step-by-step solution →
Q56·ChemistryNumerical
Pt(s)∣H2(g)(1 bar)∣∣H+(aq)(1 M) ∣∣ M3+(aq),M+(aq)∣Pt(s)Pt(s)|H_2(g)(1\,bar)||H^+(aq)(1\,M)\,||\,M^{3+}(aq), M^+(aq)|Pt(s)Pt(s)∣H2​(g)(1bar)∣∣H+(aq)(1M)∣∣M3+(aq),M+(aq)∣Pt(s) The EcellE_{cell}Ecell​ for the given cell is 0.1115 V at 298 K when [M+(aq)][M3+(aq)]=10a\frac{[M^+(aq)]}{[M^{3+}(aq)]} = 10^a[M3+(aq)][M+(aq)]​=10a. The value of aaa is Given : Eθ M3+/M+=0.2E^\theta\,M^{3+}/M^+ = 0.2EθM3+/M+=0.2 V 2.303RTF=0.059\frac{2.303RT}{F} = 0.059F2.303RT​=0.059 V

Correct answer: 3

Step-by-step solution →
Q57·ChemistryNumerical
The number of pairs of the solutions having the same value of the osmotic pressure from the following is (Assume 100% ionization) A. 0.500 M C2H5OHC_2H_5OHC2​H5​OH (aq) and 0.25 M KBr (aq) B. 0.100 M K4[Fe(CN)6]K_4[Fe(CN)_6]K4​[Fe(CN)6​] (aq) and 0.100 M FeSO4(NH4)2SO4FeSO_4(NH_4)_2SO_4FeSO4​(NH4​)2​SO4​ (aq) C. 0.05 M K4[Fe(CN)6]K_4[Fe(CN)_6]K4​[Fe(CN)6​] (aq) and 0.25 M NaCl (aq) D. 0.15 M NaCl(aq) and 0.1 M BaCl2BaCl_2BaCl2​(aq) E. 0.02 M KCl⋅MgCl2⋅6H2OKCl\cdot MgCl_2\cdot 6H_2OKCl⋅MgCl2​⋅6H2​O(aq) and 0.05 M KCl(aq)

Correct answer: 4

Step-by-step solution →
Q58·ChemistryNumerical
A first order reaction has the rate constant, =4.6×10−3= 4.6 \times 10^{-3}=4.6×10−3 s−1^{-1}−1. The number of correct statement/s from the following is/are Given: log⁡3=0.48\log 3 = 0.48log3=0.48 A. Reaction completes in 1000 s. B. The reaction has a half-life of 500 s. C. The time required for 10% completion is 25 times the time required for 90% completion. D. The degree of dissociation is equal to (1−e−kt)(1 - e^{-kt})(1−e−kt). E. The rate and the rate constant have the same unit.

Correct answer: 1

Step-by-step solution →
Q59·ChemistryNumerical
Based on the given figure, the number of correct statement/s is/are _____ A. Surface tension is the outcome of equal attractive and repulsive forces acting on the liquid molecule in bulk. B. Surface tension is due to uneven forces acting on the molecules present on the surface. C. The molecule in the bulk can never come to the liquid surface. D. The molecules on the surface are responsible for vapours pressure if system is a closed system.

Correct answer: 2

Step-by-step solution →
Q60·ChemistryNumerical
Number of compounds giving (i) red colouration with ceric ammonium nitrate and also (ii) positive iodoform test from the following is

Correct answer: 3

Step-by-step solution →

Mathematics — JEE Main 25 January 2023 Shift 2

Q61·MathematicsSingle correct
Let Δ,∇∈{∧,∨}\Delta, \nabla \in \{\wedge, \vee\}Δ,∇∈{∧,∨} be such that (p→q) Δ (p ∇ q)(p \to q)\,\Delta\,(p\,\nabla\,q)(p→q)Δ(p∇q) is a tautology. Then
  1. (A)Δ=∨, ∇=∨\Delta = \vee,\ \nabla = \veeΔ=∨, ∇=∨
  2. (B)Δ=∨, ∇=∧\Delta = \vee,\ \nabla = \wedgeΔ=∨, ∇=∧
  3. (C)Δ=∧, ∇=∨\Delta = \wedge,\ \nabla = \veeΔ=∧, ∇=∨
  4. (D)Δ=∧, ∇=∧\Delta = \wedge,\ \nabla = \wedgeΔ=∧, ∇=∧

Correct answer: (A)

Step-by-step solution →
Q62·MathematicsSingle correct
If the four points, whose position vectors are 3i^−4j^+2k^3\hat{i} - 4\hat{j} + 2\hat{k}3i^−4j^​+2k^, i^+2j^−k^\hat{i} + 2\hat{j} - \hat{k}i^+2j^​−k^, −2i^−j^+3k^-2\hat{i} - \hat{j} + 3\hat{k}−2i^−j^​+3k^ and 5i^−2αj^+4k^5\hat{i} - 2\alpha\hat{j} + 4\hat{k}5i^−2αj^​+4k^ are coplanar, then α\alphaα is equal to
  1. (A)7317\dfrac{73}{17}1773​
  2. (B)10717\dfrac{107}{17}17107​
  3. (C)−7317\dfrac{-73}{17}17−73​
  4. (D)−10717\dfrac{-107}{17}17−107​

Correct answer: (A)

Step-by-step solution →
Q63·MathematicsSingle correct
The foot of perpendicular of the point (2,0,5)(2,0,5)(2,0,5) on the line x+12=y−15=z+1−1\dfrac{x+1}{2} = \dfrac{y-1}{5} = \dfrac{z+1}{-1}2x+1​=5y−1​=−1z+1​ is (α,β,γ)(\alpha, \beta, \gamma)(α,β,γ). Then, which of the following is NOT correct?
  1. (A)βγ=−5\dfrac{\beta}{\gamma} = -5γβ​=−5
  2. (B)γα=58\dfrac{\gamma}{\alpha} = \dfrac{5}{8}αγ​=85​
  3. (C)αβ=−8\dfrac{\alpha}{\beta} = -8βα​=−8
  4. (D)αβγ=415\dfrac{\alpha\beta}{\gamma} = \dfrac{4}{15}γαβ​=154​

Correct answer: (A)

Step-by-step solution →
Q64·MathematicsSingle correct
The equations of two sides of a variable triangle are x=0x = 0x=0 and y=3y = 3y=3, and its third side is a tangent to parabola y2=6xy^2 = 6xy2=6x. The locus of its circumcentre is:
  1. (A)4y2−18y−3x−18=04y^2 - 18y - 3x - 18 = 04y2−18y−3x−18=0
  2. (B)4y2−18y−3x+18=04y^2 - 18y - 3x + 18 = 04y2−18y−3x+18=0
  3. (C)4y2−18y+3x+18=04y^2 - 18y + 3x + 18 = 04y2−18y+3x+18=0
  4. (D)4y2+18y+3x+18=04y^2 + 18y + 3x + 18 = 04y2+18y+3x+18=0

Correct answer: (C)

Step-by-step solution →
Q65·MathematicsSingle correct
Let f(x)=2xn+λf(x) = 2x^n + \lambdaf(x)=2xn+λ, λ∈R\lambda \in \mathbb{R}λ∈R, n∈Nn \in \mathbb{N}n∈N, and f(4)=133f(4) = 133f(4)=133, f(5)=255f(5) = 255f(5)=255. Then the sum of all the positive integer divisors of (f(3)−f(2))(f(3) - f(2))(f(3)−f(2)) is
  1. (A)60
  2. (B)59
  3. (C)61
  4. (D)58

Correct answer: (A)

Step-by-step solution →
Q66·MathematicsSingle correct
∑k=0651−kC3\displaystyle\sum_{k=0}^{6} {}^{51-k}C_3k=0∑6​51−kC3​ is equal to
  1. (A)51C4−45C4{}^{51}C_4 - {}^{45}C_451C4​−45C4​
  2. (B)52C3−45C3{}^{52}C_3 - {}^{45}C_352C3​−45C3​
  3. (C)52C4−45C4{}^{52}C_4 - {}^{45}C_452C4​−45C4​
  4. (D)51C3−45C3{}^{51}C_3 - {}^{45}C_351C3​−45C3​

Correct answer: (C)

Step-by-step solution →
Q67·MathematicsSingle correct
Let the function f(x)=2x3+(2p−7)x2+3(2p−9)x−6f(x) = 2x^3 + (2p - 7)x^2 + 3(2p - 9)x - 6f(x)=2x3+(2p−7)x2+3(2p−9)x−6 have a maxima for some value of x<0x < 0x<0 and a minima for some value of x>0x > 0x>0. Then, the set of all values of ppp is
  1. (A)(0,92)\left(0, \dfrac{9}{2}\right)(0,29​)
  2. (B)(−∞,92)\left(-\infty, \dfrac{9}{2}\right)(−∞,29​)
  3. (C)(−92,92)\left(-\dfrac{9}{2}, \dfrac{9}{2}\right)(−29​,29​)
  4. (D)(92,∞)\left(\dfrac{9}{2}, \infty\right)(29​,∞)

Correct answer: (B)

Step-by-step solution →
Q68·MathematicsSingle correct
Let A=[110310−310110]A = \begin{bmatrix} \dfrac{1}{\sqrt{10}} & \dfrac{3}{\sqrt{10}} \\ \dfrac{-3}{\sqrt{10}} & \dfrac{1}{\sqrt{10}} \end{bmatrix}A=​10​1​10​−3​​10​3​10​1​​​ and B=[1−i01]B = \begin{bmatrix} 1 & -i \\ 0 & 1 \end{bmatrix}B=[10​−i1​], where i=−1i = \sqrt{-1}i=−1​. If M=ATBAM = A^{T}BAM=ATBA, then the inverse of the matrix AM2023ATAM^{2023}A^{T}AM2023AT is
  1. (A)[10−2023i1]\begin{bmatrix} 1 & 0 \\ -2023i & 1 \end{bmatrix}[1−2023i​01​]
  2. (B)[1−2023i01]\begin{bmatrix} 1 & -2023i \\ 0 & 1 \end{bmatrix}[10​−2023i1​]
  3. (C)[102023i1]\begin{bmatrix} 1 & 0 \\ 2023i & 1 \end{bmatrix}[12023i​01​]
  4. (D)[12023i01]\begin{bmatrix} 1 & 2023i \\ 0 & 1 \end{bmatrix}[10​2023i1​]

Correct answer: (D)

Step-by-step solution →
Q69·MathematicsSingle correct
Let a⃗=−i^−j^+k^\vec{a} = -\hat{i} - \hat{j} + \hat{k}a=−i^−j^​+k^, a⃗⋅b⃗=1\vec{a} \cdot \vec{b} = 1a⋅b=1 and a⃗×b⃗=i^−j^\vec{a} \times \vec{b} = \hat{i} - \hat{j}a×b=i^−j^​. Then a⃗−6b⃗\vec{a} - 6\vec{b}a−6b is equal to
  1. (A)3(i^−j^+k^)3(\hat{i} - \hat{j} + \hat{k})3(i^−j^​+k^)
  2. (B)(i^+j^−k^)(\hat{i} + \hat{j} - \hat{k})(i^+j^​−k^)
  3. (C)3(i^+j^+k^)3(\hat{i} + \hat{j} + \hat{k})3(i^+j^​+k^)
  4. (D)3(i^−j^−k^)3(\hat{i} - \hat{j} - \hat{k})3(i^−j^​−k^)

Correct answer: (C)

Step-by-step solution →
Q70·MathematicsSingle correct
The integral 16∫12dxx3(x2+2)216\displaystyle\int_{1}^{2} \dfrac{dx}{x^3 (x^2 + 2)^2}16∫12​x3(x2+2)2dx​ is equal to
  1. (A)1112−log⁡e4\dfrac{11}{12} - \log_e 41211​−loge​4
  2. (B)116−log⁡e4\dfrac{11}{6} - \log_e 4611​−loge​4
  3. (C)116+log⁡e4\dfrac{11}{6} + \log_e 4611​+loge​4
  4. (D)1112+log⁡e4\dfrac{11}{12} + \log_e 41211​+loge​4

Correct answer: (B)

Step-by-step solution →
Q71·MathematicsSingle correct
Let TTT and CCC respectively be the transverse and conjugate axes of the hyperbola 16x2−y2+64x+4y+44=016x^2 - y^2 + 64x + 4y + 44 = 016x2−y2+64x+4y+44=0. Then the area of the region above the parabola x2=y+4x^2 = y + 4x2=y+4, below the transverse axis TTT and on the right of the conjugate axis CCC is:
  1. (A)46+2834\sqrt{6} + \dfrac{28}{3}46​+328​
  2. (B)46−4434\sqrt{6} - \dfrac{44}{3}46​−344​
  3. (C)46+4434\sqrt{6} + \dfrac{44}{3}46​+344​
  4. (D)46−2834\sqrt{6} - \dfrac{28}{3}46​−328​

Correct answer: (A)

Step-by-step solution →
Q72·MathematicsSingle correct
Let NNN be the sum of the numbers appeared when two fair dice are rolled and let the probability that N−2N - 2N−2, 3N\sqrt{3N}3N​, N+2N + 2N+2 are in geometric progression be k48\dfrac{k}{48}48k​. Then the value of kkk is
  1. (A)8
  2. (B)16
  3. (C)2
  4. (D)4

Correct answer: (D)

Step-by-step solution →
Q73·Mathematics·Limits and ContinuitySingle correct
If the function f(x)={(1+∣cos⁡x∣)λ∣cos⁡x∣,0<x<π2μ,x=π2cot⁡6xecot⁡4x,π2<x<πf(x) = \begin{cases} (1 + |\cos x|)^{\frac{\lambda}{|\cos x|}}, & 0 < x < \dfrac{\pi}{2} \\ \mu, & x = \dfrac{\pi}{2} \\ \dfrac{\cot 6x}{e^{\cot 4x}}, & \dfrac{\pi}{2} < x < \pi \end{cases}f(x)=⎩⎨⎧​(1+∣cosx∣)∣cosx∣λ​,μ,ecot4xcot6x​,​0<x<2π​x=2π​2π​<x<π​ is continuous at x=π2x = \dfrac{\pi}{2}x=2π​, then 9λ+6log⁡eμ+μ6−e6λ9\lambda + 6\log_e \mu + \mu^6 - e^{6\lambda}9λ+6loge​μ+μ6−e6λ is equal to
  1. (A)10
  2. (B)2e4+82e^4 + 82e4+8
  3. (C)11
  4. (D)8

Correct answer: (A)

Step-by-step solution →
Q74·MathematicsSingle correct
The number of functions f:{1,2,3,4}→{a∈Z:∣a∣≤8}f : \{1,2,3,4\} \to \{a \in \mathbb{Z} : |a| \le 8\}f:{1,2,3,4}→{a∈Z:∣a∣≤8} satisfying f(n)+1nf(n+1)=1f(n) + \dfrac{1}{n}f(n+1) = 1f(n)+n1​f(n+1)=1, ∀n∈{1,2,3}\forall n \in \{1,2,3\}∀n∈{1,2,3} is
  1. (A)1
  2. (B)4
  3. (C)2
  4. (D)3

Correct answer: (C)

Step-by-step solution →
Q75·MathematicsSingle correct
Let y=y(t)y = y(t)y=y(t) be a solution of the differential equation dydt+αy=γe−βt\dfrac{dy}{dt} + \alpha y = \gamma e^{-\beta t}dtdy​+αy=γe−βt where α>0\alpha > 0α>0, β>0\beta > 0β>0 and γ>0\gamma > 0γ>0. Then lim⁡t→∞y(t)\displaystyle\lim_{t \to \infty} y(t)t→∞lim​y(t)
  1. (A)is −1-1−1
  2. (B)is 1
  3. (C)does not exist
  4. (D)is 0

Correct answer: (D)

Step-by-step solution →
Q76·MathematicsSingle correct
Let zzz be a complex number such that ∣z−2iz+i∣=2\left|\dfrac{z-2i}{z+i}\right|=2​z+iz−2i​​=2, z≠−iz\neq -iz=−i. Then zzz lies on the circle of radius 222 and centre
  1. (A)(2,0)(2,0)(2,0)
  2. (B)(0,2)(0,2)(0,2)
  3. (C)(0,−2)(0,-2)(0,−2)
  4. (D)(0,0)(0,0)(0,0)

Correct answer: (C)

Step-by-step solution →
Q77·MathematicsSingle correct
Let A,B,CA, B, CA,B,C be 3×33\times 33×3 matrices such that AAA is symmetric and BBB and CCC are skew-symmetric. Consider the statements (S1) A13B26−B26A13A^{13}B^{26}-B^{26}A^{13}A13B26−B26A13 is symmetric (S2) A26C13−C13A26A^{26}C^{13}-C^{13}A^{26}A26C13−C13A26 is symmetric Then,
  1. (A)Only S2 is true
  2. (B)Both S1 and S2 are false
  3. (C)Only S1 is true
  4. (D)Both S1 and S2 are true

Correct answer: (A)

Step-by-step solution →
Q78·MathematicsSingle correct
The number of numbers, strictly between 500050005000 and 100001000010000 that can be formed using the digits 1,3,5,7,91,3,5,7,91,3,5,7,9 without repetition, is
  1. (A)121212
  2. (B)120120120
  3. (C)727272
  4. (D)666

Correct answer: (C)

Step-by-step solution →
Q79·MathematicsSingle correct
Let f:R→Rf:\mathbb{R}\to\mathbb{R}f:R→R be a function defined by f(x)=log⁡m{2(sin⁡x−cos⁡x)+m−2}f(x)=\log_{\sqrt{m}}\{\sqrt{2}(\sin x-\cos x)+m-2\}f(x)=logm​​{2​(sinx−cosx)+m−2}, for some mmm, such that the range of fff is [0,2][0,2][0,2]. Then the value of mmm is
  1. (A)555
  2. (B)444
  3. (C)333
  4. (D)222

Correct answer: (A)

Step-by-step solution →
Q80·MathematicsSingle correct
The shortest distance between the lines x+1=2y=−12zx+1=2y=-12zx+1=2y=−12z and x=y+2=6z−6x=y+2=6z-6x=y+2=6z−6 is
  1. (A)32\dfrac{3}{2}23​
  2. (B)222
  3. (C)52\dfrac{5}{2}25​
  4. (D)333

Correct answer: (B)

Step-by-step solution →
Q81·MathematicsNumerical
25%25\%25% of the population are smokers. A smoker has 272727 times more chances to develop lung cancer than a non smoker. A person is diagnosed with lung cancer and the probability that this person is a smoker is k10\dfrac{k}{10}10k​. Then the value of kkk is.

Correct answer: 9

Step-by-step solution →
Q82·MathematicsNumerical
The remainder when (2023)2023(2023)^{2023}(2023)2023 is divided by 353535 is

Correct answer: 7

Step-by-step solution →
Q83·MathematicsNumerical
Let a∈Ra\in\mathbb{R}a∈R and let α,β\alpha,\betaα,β be the roots of the equation x2+6014x+a=0x^{2}+60^{\frac{1}{4}}x+a=0x2+6041​x+a=0. If α4+β4=−30\alpha^{4}+\beta^{4}=-30α4+β4=−30, then the product of all possible values of aaa is

Correct answer: 45

Step-by-step solution →
Q84·MathematicsNumerical
For two positive numbers a,ba,ba,b such that a,ba,ba,b and 118\dfrac{1}{18}181​ are in a geometric progression, while 1a,10\dfrac{1}{a},10a1​,10 and 1b\dfrac{1}{b}b1​ are in an arithmetic progression, then 16a+b16a+b16a+b is equal to

Correct answer: 3

Step-by-step solution →
Q85·MathematicsNumerical
If mmm and nnn respectively are the numbers of positive and negative values of θ\thetaθ in the interval [−π,π][-\pi,\pi][−π,π] that satisfy the equation cos⁡2θcos⁡θ2=cos⁡3θcos⁡9θ2\cos 2\theta\cos\dfrac{\theta}{2}=\cos 3\theta\cos\dfrac{9\theta}{2}cos2θcos2θ​=cos3θcos29θ​, then mnmnmn is equal to

Correct answer: 25

Step-by-step solution →
Q86·MathematicsNumerical
If the shortest distance between the line joining the points (1,2,3)(1,2,3)(1,2,3) and (2,3,4)(2,3,4)(2,3,4), and the line x−12=y+1−1=z−20\dfrac{x-1}{2}=\dfrac{y+1}{-1}=\dfrac{z-2}{0}2x−1​=−1y+1​=0z−2​ is aaa, then 28a228a^{2}28a2 is equal to

Correct answer: 18

Step-by-step solution →
Q87·MathematicsNumerical
Points P(−3,2)P(-3,2)P(−3,2), Q(9,10)Q(9,10)Q(9,10) and R(a,4)R(a,4)R(a,4) lie on a circle CCC with PRPRPR as its diameter. The tangents to CCC at the points QQQ and RRR intersect at the point SSS. If SSS lies on the line 2x−ky=12x-ky=12x−ky=1, then kkk is equal to

Correct answer: 3

Step-by-step solution →
Q88·MathematicsNumerical
Suppose Anil's mother wants to give 555 whole fruits to Anil from a basket of 777 red apples, 555 white apples and 888 oranges. If in the selected 555 fruits, at least 222 oranges, at least one red apple and at least one white apple must be given, then the number of ways Anil's mother can offer 555 fruits to Anil is

Correct answer: 6860

Step-by-step solution →
Q89·MathematicsNumerical
If ∫1/33∣log⁡ex∣ dx=mnlog⁡e(n2e)\displaystyle\int_{1/3}^{3}|\log_{e}x|\,dx=\dfrac{m}{n}\log_{e}\left(\dfrac{n^{2}}{e}\right)∫1/33​∣loge​x∣dx=nm​loge​(en2​), where mmm and nnn are coprime natural numbers, then m2+n2−5m^{2}+n^{2}-5m2+n2−5 is equal to

Correct answer: 20

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Q90·MathematicsNumerical
A triangle is formed by the X-axis, the Y-axis and the line 3x+4y=603x+4y=603x+4y=60. Then the number of points P(a,b)P(a,b)P(a,b) which lie strictly inside the triangle, where aaa is an integer and bbb is a multiple of aaa, is

Correct answer: 31

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
  • Application of Derivatives 139/186
  • Limits and Continuity 149/186
  • Thermodynamics 154/186
  • Aldehydes and Ketones 135/186
  • Equilibrium 163/186
  • Solutions 158/186
  • Laws of Motion 130/186
  • Units and Measurements 149/186
  • Complex Numbers 165/186
  • Trigonometric Functions 144/186
  • 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
  • Amines 133/186
  • Quadratic Equations 148/186
  • Work, Energy and Power 132/186
  • Some Basic Concepts in Chemistry 129/186
  • Electromagnetic Induction 120/186
  • Kinetic Theory of Gases 135/186
  • Alcohols and Ethers 106/186
  • Oscillations 117/186
  • Alternating Currents 108/186
  • Nuclei 116/186
  • Straight Lines 114/186
  • Waves 109/186
  • Capacitors and Dielectrics 115/186
  • Atoms 112/186
  • Isolation of Metals 106/186
  • s-Block Elements 88/186
  • Surface Chemistry 98/186
  • Environmental Chemistry 83/186
  • Hyperbola 77/186
  • Polymers 64/186
  • Principles of Qualitative Analysis 58/186
  • Chemistry in Everyday Life 60/186
  • States of Matter: Gases and Liquids 52/186
  • Isomerism 51/186
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