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JEE Main 31 January 2023 Shift 1 Question Paper with Answers

31 January 2023 · January session · 90 questions

The complete JEE Main 31 January 2023 Shift 1 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 31 January 2023 Shift 1

Q1·PhysicsSingle correct
The maximum potential energy of a block executing simple harmonic motion is 252525 J. AAA is amplitude of oscillation. At A/2A/2A/2, the kinetic energy of the block is :
  1. (A)18.7518.7518.75 J
  2. (B)9.759.759.75 J
  3. (C)37.537.537.5 J
  4. (D)12.512.512.5 J

Correct answer: (A)

Step-by-step solution →
Q2·PhysicsSingle correct
The drift velocity of electrons for a conductor connected in an electrical circuit is VdV_dVd​. The conductor is now replaced by another conductor with same material and same length but double the area of cross section. The applied voltage remains same. The new drift velocity of electrons will be
  1. (A)VdV_dVd​
  2. (B)Vd4\frac{V_d}{4}4Vd​​
  3. (C)2Vd2V_d2Vd​
  4. (D)Vd2\frac{V_d}{2}2Vd​​

Correct answer: (A)

Step-by-step solution →
Q3·PhysicsSingle correct
The initial speed of a projectile fired from ground is uuu. At the highest point during its motion, the speed of projectile is 32u\frac{\sqrt{3}}{2}u23​​u. The time of flight of the projectile is :
  1. (A)2ug\frac{2u}{g}g2u​
  2. (B)u2g\frac{u}{2g}2gu​
  3. (C)3ug\frac{\sqrt{3}u}{g}g3​u​
  4. (D)ug\frac{u}{g}gu​

Correct answer: (D)

Step-by-step solution →
Q4·PhysicsSingle correct
The correct relation between γ=CpCv\gamma=\frac{C_p}{C_v}γ=Cv​Cp​​ and temperature TTT is :
  1. (A)γα T0\gamma \alpha\, T^0γαT0
  2. (B)γα T\gamma \alpha\, TγαT
  3. (C)γα 1T\gamma \alpha\, \frac{1}{\sqrt{T}}γαT​1​
  4. (D)γα 1T\gamma \alpha\, \frac{1}{T}γαT1​

Correct answer: (A)

Step-by-step solution →
Q5·PhysicsSingle correct
The effect of increase in temperature on the number of electrons in conduction band (ne)(n_e)(ne​) and resistance of a semiconductor will be as:
  1. (A)Both nen_ene​ and resistance increase
  2. (B)Both nen_ene​ and resistance decrease
  3. (C)nen_ene​ decreases, resistance increases
  4. (D)nen_ene​ increases, resistance decreases

Correct answer: (D)

Step-by-step solution →
Q6·PhysicsSingle correct
The amplitude of 15sin⁡(1000πt)15\sin(1000\pi t)15sin(1000πt) is modulated by 10sin⁡(4πt)10\sin(4\pi t)10sin(4πt) signal. The amplitude modulated signal contains frequency (ies) of A. 500500500 Hz B. 222 Hz C. 250250250 Hz D. 498498498 Hz E. 502502502 Hz Choose the correct answer from the options given below:
  1. (A)A Only
  2. (B)B Only
  3. (C)A and B Only
  4. (D)A, D and E Only

Correct answer: (D)

Step-by-step solution →
Q7·PhysicsSingle correct
Two polaroids AAA and BBB are placed in such a way that the pass-axis of polaroids are perpendicular to each other. Now, another polaroid CCC is placed between AAA and BBB bisecting angle between them. If intensity of unpolarized light is I0I_0I0​ then intensity of transmitted light after passing through polaroid BBB will be:
  1. (A)I04\frac{I_0}{4}4I0​​
  2. (B)I02\frac{I_0}{2}2I0​​
  3. (C)Zero
  4. (D)I08\frac{I_0}{8}8I0​​

Correct answer: (D)

Step-by-step solution →
Q8·PhysicsSingle correct
As shown in figure, a 707070 kg garden roller is pushed with a force of F⃗=200\vec{F}=200F=200 N at an angle of 30∘30^\circ30∘ with horizontal. The normal reaction on the roller is (Given g=10g=10g=10 m s−2^{-2}−2)
  1. (A)8002800\sqrt{2}8002​ N
  2. (B)2003200\sqrt{3}2003​ N
  3. (C)600600600 N
  4. (D)800800800 N

Correct answer: (D)

Step-by-step solution →
Q9·PhysicsSingle correct
If 100010001000 droplets of water of surface tension 0.070.070.07 N/m, having same radius 111 mm each, combine to from a single drop. In the process the released surface energy is- (Take π=227\pi=\frac{22}{7}π=722​)
  1. (A)8.8×10−58.8 \times 10^{-5}8.8×10−5 J
  2. (B)7.92×10−47.92 \times 10^{-4}7.92×10−4 J
  3. (C)7.92×10−67.92 \times 10^{-6}7.92×10−6 J
  4. (D)9.68×10−49.68 \times 10^{-4}9.68×10−4 J

Correct answer: (B)

Step-by-step solution →
Q10·PhysicsSingle correct
The pressure of a gas changes linearly with volume from AAA to BBB as shown in figure. If no heat is supplied to or extracted from the gas then change in the internal energy of the gas will be
  1. (A)−4.5-4.5−4.5 J
  2. (B)zero
  3. (C)4.54.54.5 J
  4. (D)666 J

Correct answer: (C)

Step-by-step solution →
Q11·PhysicsSingle correct
Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R Assertion A: The beam of electrons show wave nature and exhibit interference and diffraction. Reason R: Davisson Germer Experimentally verified the wave nature of electrons. 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 and R is the correct explanation of A
  2. (B)A is not correct but R is correct
  3. (C)A is correct but R is not correct
  4. (D)Both A and R are correct but R is Not the correct explanation of A

Correct answer: (A)

Step-by-step solution →
Q12·PhysicsSingle correct
A free neutron decays into a proton but a free proton does not decay into neutron. This is because
  1. (A)proton is a charged particle
  2. (B)neutron is an uncharged particle
  3. (C)neutron is a composite particle made of a proton and an electron
  4. (D)neutron has larger rest mass than proton

Correct answer: (D)

Step-by-step solution →
Q13·PhysicsSingle correct
Spherical insulating ball and a spherical metallic ball of same size and mass are dropped from the same height. Choose the correct statement out of the following Assume negligible air friction)
  1. (A)Insulating ball will reach the earth's surface earlier than the metal ball
  2. (B)Metal ball will reach the earth's surface earlier than the insulating ball
  3. (C)Both will reach the earth's surface simultaneously
  4. (D)Time taken by them to reach the earth's surface will be independent of the properties of their materials

Correct answer: (A)

Step-by-step solution →
Q14·PhysicsSingle correct
If RRR, XLX_LXL​ and XCX_CXC​ represent resistance, inductive reactance and capacitive reactance. Then which of the following is dimensionless :
  1. (A)RXLXC\frac{R}{X_L X_C}XL​XC​R​
  2. (B)RXLXC\frac{R}{\sqrt{X_L X_C}}XL​XC​​R​
  3. (C)RXLXCR\frac{X_L}{X_C}RXC​XL​​
  4. (D)RXLXCRX_L X_CRXL​XC​

Correct answer: (B)

Step-by-step solution →
Q15·PhysicsSingle correct
100100100 balls each of mass mmm moving with speed vvv simultaneously strike a wall normally and reflected back with same speed, in time ttt sec. The total force exerted by the balls on the wall is
  1. (A)100mvt\frac{100mv}{t}t100mv​
  2. (B)200mvt200mvt200mvt
  3. (C)mv100t\frac{mv}{100t}100tmv​
  4. (D)200mvt\frac{200mv}{t}t200mv​

Correct answer: (D)

Step-by-step solution →
Q16·PhysicsSingle correct
If a source of electromagnetic radiation having power 151515 kW produces 101610^{16}1016 photons per second, the radiation belongs to a part of spectrum is. (Take Planck constant h=6×10−34h = 6 \times 10^{-34}h=6×10−34 Js)
  1. (A)Micro waves
  2. (B)Ultraviolet rays
  3. (C)Gamma rays
  4. (D)Radio waves

Correct answer: (C)

Step-by-step solution →
Q17·PhysicsSingle correct
Which of the following correctly represents the variation of electric potential (VVV) of a charged spherical conductor of radius (RRR) with radial distance (rrr) from the center?
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (A)

Step-by-step solution →
Q18·PhysicsSingle correct
A bar magnet with a magnetic moment 5.0 Am25.0\,\text{Am}^25.0Am2 is placed in parallel position relative to a magnetic field of 0.40.40.4 T. The amount of required work done in turning the magnet from parallel to antiparallel position relative to the field direction is __________.
  1. (A)111 J
  2. (B)444 J
  3. (C)222 J
  4. (D)zero

Correct answer: (B)

Step-by-step solution →
Q19·PhysicsSingle correct
At a certain depth "ddd" below surface of earth, value of acceleration due to gravity becomes four times that of its value at a height 3R3R3R above earth surface. Where RRR is Radius of earth (Take R=6400R = 6400R=6400 km). The depth ddd is equal to
  1. (A)480048004800 km
  2. (B)256025602560 km
  3. (C)640640640 km
  4. (D)526052605260 km

Correct answer: (A)

Step-by-step solution →
Q20·Physics·Magnetism and MatterSingle correct
A rod with circular cross-section area 2 cm22\,\text{cm}^22cm2 and length 404040 cm is wound uniformly with 400400400 turns of an insulated wire. If a current of 0.40.40.4 A flows in the wire windings, the total magnetic flux produced inside windings is 4π×10−64\pi \times 10^{-6}4π×10−6 Wb. The relative permeability of the rod is (Given: Permeability of vacuum μ0=4π×10−7 NA−2\mu_0 = 4\pi \times 10^{-7}\,\text{NA}^{-2}μ0​=4π×10−7NA−2)
  1. (A)516\frac{5}{16}165​
  2. (B)12.512.512.5
  3. (C)125125125
  4. (D)325\frac{32}{5}532​

Correct answer: (A)

Step-by-step solution →
Q21·PhysicsNumerical
In a medium the speed of light wave decreases to 0.20.20.2 times to its speed in free space. The ratio of relative permittivity to the refractive index of the medium is x:1x:1x:1. The value of xxx is (Given speed of light in free space =3×108 m s−1= 3 \times 10^8\,\text{m s}^{-1}=3×108m s−1 and for the given medium μr=1\mu_r = 1μr​=1)

Correct answer: 5

Step-by-step solution →
Q22·PhysicsNumerical
A solid sphere of mass 111 kg rolls without slipping on a plane surface. Its kinetic energy is 7×10−37 \times 10^{-3}7×10−3 J. The speed of the centre of mass of the sphere is __________ cms−1\text{cms}^{-1}cms−1

Correct answer: 10

Step-by-step solution →
Q23·PhysicsNumerical
A lift of mass M=500M = 500M=500 kg is descending with speed of 2 ms−12\,\text{ms}^{-1}2ms−1. Its supporting cable begins to slip thus allowing it to fall with a constant acceleration of 2 ms−22\,\text{ms}^{-2}2ms−2. The kinetic energy of the lift at the end of fall through to a distance of 666 m will be __________ kJ.

Correct answer: 7

Step-by-step solution →
Q24·PhysicsNumerical
In the figure given below, a block of mass M=490M = 490M=490 g placed on a frictionless table is connected with two springs having same spring constant (K=2 N m−1K = 2\,\text{N m}^{-1}K=2N m−1). If the block is horizontally displaced through 'XXX' m then the number of complete oscillations it will make in 14π14\pi14π seconds will be __________.

Correct answer: 20

Step-by-step solution →
Q25·PhysicsNumerical
An inductor of 0.50.50.5 mH, a capacitor of 20 μF20\,\mu\text{F}20μF and resistance of 20 Ω20\,\Omega20Ω are connected in series with a 220220220 V ac source. If the current is in phase with the emf, the amplitude of current of the circuit is x\sqrt{x}x​ A. The value of xxx is-

Correct answer: 242

Step-by-step solution →
Q26·PhysicsNumerical
The speed of a swimmer is 4 km h−14\,\text{km h}^{-1}4km h−1 in still water. If the swimmer makes his strokes normal to the flow of river of width 111 km, he reaches a point 750750750 m down the stream on the opposite bank. The speed of the river water is __________ kmh−1\text{kmh}^{-1}kmh−1.

Correct answer: 3

Step-by-step solution →
Q27·PhysicsNumerical
For hydrogen atom, λ1\lambda_1λ1​ and λ2\lambda_2λ2​ are the wavelengths corresponding to the transitions 1 and 2 respectively as shown in figure. The ratio of λ1\lambda_1λ1​ and λ2\lambda_2λ2​ is x32\frac{x}{32}32x​. The value of xxx is __________.

Correct answer: 27

Step-by-step solution →
Q28·PhysicsNumerical
Two identical cells, when connected either in parallel or in series gives same current in an external resistance 5 Ω5\,\Omega5Ω. The internal resistance of each cell will be __________ Ω\OmegaΩ.

Correct answer: 5

Step-by-step solution →
Q29·Physics·Electric Field and Coulomb's LawNumerical
Expression for an electric field is given by E⃗=4000 x2 i^ Vm\vec{E} = 4000\,x^2\,\hat{i}\,\frac{\text{V}}{\text{m}}E=4000x2i^mV​. The electric flux through the cube of side 202020 cm when placed in electric field (as shown in the figure) is __________ V cm.

Correct answer: 640

Step-by-step solution →
Q30·PhysicsNumerical
A thin rod having a length of 111 m and area of cross-section 3×10−6 m23 \times 10^{-6}\,\text{m}^23×10−6m2 is suspended vertically from one end. The rod is cooled from 210∘210^\circ210∘C to 160∘160^\circ160∘C. After cooling, a mass MMM is attached at the lower end of the rod such that the length of rod again becomes 111 m. Young's modulus and coefficient of linear expansion of the rod are 2×1011 N m−22 \times 10^{11}\,\text{N m}^{-2}2×1011N m−2 and 2×10−5 K−12 \times 10^{-5}\,\text{K}^{-1}2×10−5K−1, respectively. The value of M is __________ kg. (Take g=10 m s−2g = 10\,\text{m s}^{-2}g=10m s−2)

Correct answer: 60

Step-by-step solution →

Chemistry — JEE Main 31 January 2023 Shift 1

Q31·ChemistrySingle correct
Match items of Column I (Mixture of compounds) with Column II (Separation Technique).
Column I (Mixture of compounds)Column II (Separation Technique)
A.H2OH_2OH2​O / CH2Cl2CH_2Cl_2CH2​Cl2​i.Crystallization
B.1-Tetralone / 4-Nitrophenolii.Differential solvent extraction
C.Kerosene / Naphthaleneiii.Column chromatography
D.C6H12O6C_6H_{12}O_6C6​H12​O6​ / NaCliv.Fractional distillation
  1. (A)A-(ii), B-(iii), C-(iv), D-(i)
  2. (B)A-(i), B-(iii), C-(ii), D-(iv)
  3. (C)A-(ii), B-(i), C-(iv), D-(iii)
  4. (D)A-(iii), B-(ii), C-(i), D-(iv)

Correct answer: (A)

Step-by-step solution →
Q32·Chemistry·AminesSingle correct
Consider the following reaction and identify the product B. Ph−NO2→C2H5OHH2/Pd[A]→Pyridine(CH3CO)2O[B]Ph-NO_2 \xrightarrow[C_2H_5OH]{H_2/Pd} [A] \xrightarrow[Pyridine]{(CH_3CO)_2O} [B]Ph−NO2​H2​/PdC2​H5​OH​[A](CH3​CO)2​OPyridine​[B]
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (A)

Step-by-step solution →
Q33·Chemistry·Alcohols and EthersSingle correct
An organic compound 'A' with empirical formula C6H6OC_6H_6OC6​H6​O gives sooty flame on burning. Its reaction with bromine solution in low polarity solvent results in high yield of B. B is
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (C)

Step-by-step solution →
Q34·ChemistrySingle correct
When Cu2+Cu^{2+}Cu2+ ion is treated with KI, a white precipitate, X appears in solution. The solution is titrated with sodium thiosulphate, the compound Y is formed. X and Y respectively are
  1. (A)X = CuI2CuI_2CuI2​, Y = Na2S4O6Na_2S_4O_6Na2​S4​O6​
  2. (B)X = CuI2CuI_2CuI2​, Y = Na2S2O3Na_2S_2O_3Na2​S2​O3​
  3. (C)X = Cu2I2Cu_2I_2Cu2​I2​, Y = Na2S4O5Na_2S_4O_5Na2​S4​O5​
  4. (D)X = Cu2I2Cu_2I_2Cu2​I2​, Y = Na2S4O6Na_2S_4O_6Na2​S4​O6​

Correct answer: (D)

Step-by-step solution →
Q35·ChemistrySingle correct
Choose the correct set of reagents for the following conversion. trans-(Ph−CH=CH−CH3Ph-CH=CH-CH_3Ph−CH=CH−CH3​) →\rightarrow→ cis-(Ph−CH=CH−CH3Ph-CH=CH-CH_3Ph−CH=CH−CH3​)
  1. (A)Br2Br_2Br2​, aq. KOH, NaNH2NaNH_2NaNH2​, Na(LiqNH3NH_3NH3​)
  2. (B)Br2Br_2Br2​, alc. KOH, NaNH2NaNH_2NaNH2​, H2H_2H2​ Lindlar Catalyst
  3. (C)Br2Br_2Br2​, aq. KOH, NaNH2NaNH_2NaNH2​, H2H_2H2​ Lindlar Catalyst
  4. (D)Br2Br_2Br2​, alc. KOH, NaNH2NaNH_2NaNH2​, Na(LiqNH3NH_3NH3​)

Correct answer: (B)

Step-by-step solution →
Q36·ChemistrySingle correct
Consider the following reaction Propanal + Methanal →(i)dil.NaOH,(ii)Δ,(iii)NaCN,(iv)H3O+\xrightarrow{(i) dil.NaOH, (ii) \Delta, (iii) NaCN, (iv) H_3O^+}(i)dil.NaOH,(ii)Δ,(iii)NaCN,(iv)H3​O+​ Product B (C5H8O3C_5H_8O_3C5​H8​O3​) The correct statement for product B is. It is
  1. (A)optically active alcohol and is neutral
  2. (B)racemic mixture and gives a gas with saturated NaHCO3NaHCO_3NaHCO3​ solution
  3. (C)optically active and adds one mole of bromine
  4. (D)racemic mixture and is neutral

Correct answer: (B)

Step-by-step solution →
Q37·ChemistrySingle correct
The methods NOT involved in concentration of ore are A. Liquation B. Leaching C. Electrolysis D. Hydraulic washing E. Froth floatation Choose the correct answer from the options given below:
  1. (A)C, D and E only
  2. (B)B, D and C only
  3. (C)A and C only
  4. (D)B, D and E only

Correct answer: (C)

Step-by-step solution →
Q38·ChemistrySingle correct
A protein 'X' with molecular weight of 70,000u, on hydrolysis gives amino acids. One of these amino acid is
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (D)

Step-by-step solution →
Q39·ChemistrySingle correct
Nd2+Nd^{2+}Nd2+ =
  1. (A)4f34f^34f3
  2. (B)4f46s24f^46s^24f46s2
  3. (C)4f44f^44f4
  4. (D)4f26s24f^26s^24f26s2

Correct answer: (C)

Step-by-step solution →
Q40·ChemistrySingle correct
Match List I (molecule/ion) with List II (shape/geometry).
List IList II
A.XeF4XeF_4XeF4​I.See-saw
B.SF4SF_4SF4​II.Square planar
C.NH4+NH_4^+NH4+​III.Bent T-shaped
D.BrF3BrF_3BrF3​IV.Tetrahedral
  1. (A)A-IV, B-III, C-II, D-I
  2. (B)A-IV, B-I, C-II, D-III
  3. (C)A-II, B-I, C-III, D-IV
  4. (D)A-II, B-I, C-IV, D-III

Correct answer: (D)

Step-by-step solution →
Q41·ChemistrySingle correct
Identify X, Y and Z in the following reaction. (Equation not balanced) ClO+NO2→X→H2OY+ZClO + NO_2 \rightarrow X \xrightarrow{H_2O} Y + ZClO+NO2​→XH2​O​Y+Z
  1. (A)X = ClONO2ClONO_2ClONO2​, Y = HOCl, Z = HNO3HNO_3HNO3​
  2. (B)X = ClONO2ClONO_2ClONO2​, Y = HOCl, Z = NO2NO_2NO2​
  3. (C)X = ClNO2ClNO_2ClNO2​, Y = HCl, Z = HNO3HNO_3HNO3​
  4. (D)X = ClNO3ClNO_3ClNO3​, Y = Cl2Cl_2Cl2​, Z = NO2NO_2NO2​

Correct answer: (A)

Step-by-step solution →
Q42·ChemistrySingle correct
The correct increasing order of the ionic radii is
  1. (A)S2−<Cl−<Ca2+<K+S^{2-} < Cl^- < Ca^{2+} < K^+S2−<Cl−<Ca2+<K+
  2. (B)K+<S2−<Ca2+<Cl−K^+ < S^{2-} < Ca^{2+} < Cl^-K+<S2−<Ca2+<Cl−
  3. (C)Ca2+<K+<Cl−<S2−Ca^{2+} < K^+ < Cl^- < S^{2-}Ca2+<K+<Cl−<S2−
  4. (D)Cl−<Ca2+<K+<S2−Cl^- < Ca^{2+} < K^+ < S^{2-}Cl−<Ca2+<K+<S2−

Correct answer: (C)

Step-by-step solution →
Q43·ChemistrySingle correct
Cobalt chloride when dissolved in water forms pink colored complex X which has octahedral geometry. This solution on treating with conc HCl forms deep blue complex, Y which has a Z geometry. X, Y and Z, respectively, are
  1. (A)X = [Co(H2O)6]2+, Y = [CoCl4]2−, Z = Tetrahedral
  2. (B)X = [Co(H2O)6]2+, Y = [CoCl6]3−, Z = Octahedral
  3. (C)X = [Co(H2O)4Cl2]+, Y = [CoCl4]2−, Z = Tetrahedral
  4. (D)X = [Co(H2O)6]3+, Y = [CoCl6]3−, Z = Octahedral

Correct answer: (A)

Step-by-step solution →
Q44·ChemistrySingle correct
H2O2H_2O_2H2​O2​ acts as a reducing agent in
  1. (A)2NaOCl+H2O2→2NaCl+H2O+O22NaOCl + H_2O_2 \rightarrow 2NaCl + H_2O + O_22NaOCl+H2​O2​→2NaCl+H2​O+O2​
  2. (B)Na2S+4H2O2→Na2SO4+4H2ONa_2S + 4H_2O_2 \rightarrow Na_2SO_4 + 4H_2ONa2​S+4H2​O2​→Na2​SO4​+4H2​O
  3. (C)2Fe2++2H++H2O2→2Fe3++2H2O2Fe^{2+} + 2H^+ + H_2O_2 \rightarrow 2Fe^{3+} + 2H_2O2Fe2++2H++H2​O2​→2Fe3++2H2​O
  4. (D)Mn2++2H2O2→MnO2+2H2OMn^{2+} + 2H_2O_2 \rightarrow MnO_2 + 2H_2OMn2++2H2​O2​→MnO2​+2H2​O

Correct answer: (A)

Step-by-step solution →
Q45·ChemistrySingle correct
Adding surfactants in non polar solvent, the micelles structure will look like
  1. (A)a
  2. (B)d
  3. (C)b
  4. (D)c

Correct answer: (A)

Step-by-step solution →
Q46·ChemistrySingle correct
The correct order of melting points of dichlorobenzenes is
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (B)

Step-by-step solution →
Q47·ChemistrySingle correct
The correct order of basicity of oxides of vanadium is
  1. (A)V2O5>V2O4>V2O3V_2O_5 > V_2O_4 > V_2O_3V2​O5​>V2​O4​>V2​O3​
  2. (B)V2O4>V2O3>V2O5V_2O_4 > V_2O_3 > V_2O_5V2​O4​>V2​O3​>V2​O5​
  3. (C)V2O3>V2O5>V2O4V_2O_3 > V_2O_5 > V_2O_4V2​O3​>V2​O5​>V2​O4​
  4. (D)V2O3>V2O4>V2O5V_2O_3 > V_2O_4 > V_2O_5V2​O3​>V2​O4​>V2​O5​

Correct answer: (D)

Step-by-step solution →
Q48·ChemistrySingle correct
Which of the following artificial sweeteners has the highest sweetness value in comparison to cane sugar ?
  1. (A)Sucralose
  2. (B)Aspartame
  3. (C)Alitame
  4. (D)Saccharin

Correct answer: (C)

Step-by-step solution →
Q49·ChemistrySingle correct
Which one of the following statements is correct for electrolysis of brine solution?
  1. (A)Cl2Cl_2Cl2​ is formed at cathode
  2. (B)O2O_2O2​ is formed at cathode
  3. (C)H2H_2H2​ is formed at anode
  4. (D)OH−OH^-OH− is formed at cathode

Correct answer: (D)

Step-by-step solution →
Q50·ChemistrySingle correct
Which transition in the hydrogen spectrum would have the same wavelength as the Balmer type transition from n=4n = 4n=4 to n=2n = 2n=2 of He+He^+He+ spectrum
  1. (A)n=2n = 2n=2 to n=1n = 1n=1
  2. (B)n=1n = 1n=1 to n=2n = 2n=2
  3. (C)n=3n = 3n=3 to n=4n = 4n=4
  4. (D)n=1n = 1n=1 to n=3n = 3n=3

Correct answer: (A)

Step-by-step solution →
Q51·ChemistryNumerical
The oxidation state of phosphorus in hypophosphoric acid is +++ ___ .

Correct answer: 4

Step-by-step solution →
Q52·ChemistryNumerical
The enthalpy change for the conversion of 12Cl2(g)\frac{1}{2}Cl_2(g)21​Cl2​(g) to Cl−(aq)Cl^-(aq)Cl−(aq) is (−)(-)(−) ___ kJ mol−1kJ\,mol^{-1}kJmol−1 (Nearest integer). Given : ΔdisHCl2(g)⊖=240 kJ mol−1\Delta_{dis}H^{\ominus}_{Cl_2(g)} = 240\ kJ\,mol^{-1}Δdis​HCl2​(g)⊖​=240 kJmol−1, ΔegHCl(g)⊖=−350 kJ mol−1\Delta_{eg}H^{\ominus}_{Cl(g)} = -350\ kJ\,mol^{-1}Δeg​HCl(g)⊖​=−350 kJmol−1, ΔhydHCl−(g)⊖=−380 kJ mol−1\Delta_{hyd}H^{\ominus}_{Cl^-(g)} = -380\ kJ\,mol^{-1}Δhyd​HCl−(g)⊖​=−380 kJmol−1

Correct answer: 610

Step-by-step solution →
Q53·ChemistryNumerical
The logarithm of equilibrium constant for the reaction Pd2++4Cl−⇌PdCl42−Pd^{2+} + 4Cl^- \rightleftharpoons PdCl_4^{2-}Pd2++4Cl−⇌PdCl42−​ is ___ (Nearest integer). Given : 2.303RTF=0.06 V\frac{2.303RT}{F} = 0.06\ VF2.303RT​=0.06 V; Pd(aq)2++2e−⇌Pd(s) E⊖=0.83 VPd^{2+}_{(aq)} + 2e^- \rightleftharpoons Pd(s)\ E^{\ominus} = 0.83\ VPd(aq)2+​+2e−⇌Pd(s) E⊖=0.83 V; PdCl42−(aq)+2e−⇌Pd(s)+4Cl−(aq) Eθ=0.65 VPdCl_4^{2-}(aq) + 2e^- \rightleftharpoons Pd(s) + 4Cl^-(aq)\ E^{\theta} = 0.65\ VPdCl42−​(aq)+2e−⇌Pd(s)+4Cl−(aq) Eθ=0.65 V

Correct answer: 6

Step-by-step solution →
Q54·ChemistryNumerical
On complete combustion, 0.4920.4920.492 g of an organic compound gave 0.7920.7920.792 g of CO2CO_2CO2​. The % of carbon in the organic compound is ___ (Nearest integer)

Correct answer: 44

Step-by-step solution →
Q55·ChemistryNumerical
Zinc reacts with hydrochloric acid to give hydrogen and zinc chloride. The volume of hydrogen gas produced at STP from the reaction of 11.511.511.5 g of zinc with excess HCl is ___ L. (Nearest integer) (Given : Molar mass of Zn is 65.465.465.4 g mol−1\,mol^{-1}mol−1 and Molar volume of H2H_2H2​ at STP is 22.722.722.7 L.)

Correct answer: 4

Step-by-step solution →
Q56·ChemistryNumerical
A→BA \rightarrow BA→B. The rate constants of the above reaction at 200200200 K and 300300300 K are 0.03 min−10.03\ min^{-1}0.03 min−1 and 0.05 min−10.05\ min^{-1}0.05 min−1 respectively. The activation energy for the reaction is ___ J (Nearest integer) (Given : ln⁡10=2.3\ln 10 = 2.3ln10=2.3, R=8.3 J K−1 mol−1R = 8.3\ J\,K^{-1}\,mol^{-1}R=8.3 JK−1mol−1, log⁡5=0.70\log 5 = 0.70log5=0.70, log⁡3=0.48\log 3 = 0.48log3=0.48, log⁡2=0.30\log 2 = 0.30log2=0.30)

Correct answer: 2520

Step-by-step solution →
Q57·ChemistryNumerical
For reaction: SO2(g)+12O2(g)⇌SO3(g)SO_2(g) + \frac{1}{2}O_2(g) \rightleftharpoons SO_3(g)SO2​(g)+21​O2​(g)⇌SO3​(g), Kp=2×1012K_p = 2 \times 10^{12}Kp​=2×1012 at 27∘C27^{\circ}C27∘C and 1 atm pressure. The KcK_cKc​ for the same reaction is ___ ×1013\times 10^{13}×1013. (Nearest integer) (Given R=0.082 L atm K−1 mol−1R = 0.082\ L\,atm\,K^{-1}\,mol^{-1}R=0.082 LatmK−1mol−1)

Correct answer: 1

Step-by-step solution →
Q58·ChemistryNumerical
The total pressure of a mixture of non-reacting gases XXX (0.60.60.6 g) and YYY (0.450.450.45 g) in a vessel is 740740740 mm of Hg. The partial pressure of the gas XXX is ___ mm of Hg. (Nearest Integer) (Given : molar mass X=20X = 20X=20 and Y=45Y = 45Y=45 g mol−1\,mol^{-1}mol−1)

Correct answer: 555

Step-by-step solution →
Q59·Chemistry·AminesNumerical
How many of the transformations given below would result in aromatic amines ?

Correct answer: 3

Step-by-step solution →
Q60·ChemistryNumerical
At 27∘C27^{\circ}C27∘C, a solution containing 2.52.52.5 g of solute in 250.0250.0250.0 mL of solution exerts an osmotic pressure of 400400400 Pa. The molar mass of the solute is ___ g mol−1\,mol^{-1}mol−1 (Nearest integer) (Given : R=0.083 L bar K−1 mol−1R = 0.083\ L\,bar\,K^{-1}\,mol^{-1}R=0.083 LbarK−1mol−1)

Correct answer: 62250

Step-by-step solution →

Mathematics — JEE Main 31 January 2023 Shift 1

Q61·MathematicsSingle correct
If the maximum distance of normal to the ellipse x24+y2b2=1, b<2\frac{x^2}{4}+\frac{y^2}{b^2}=1,\ b<24x2​+b2y2​=1, b<2, from the origin is 111, then the eccentricity of the ellipse is :
  1. (A)12\frac{1}{2}21​
  2. (B)34\frac{\sqrt{3}}{4}43​​
  3. (C)32\frac{\sqrt{3}}{2}23​​
  4. (D)12\frac{1}{\sqrt{2}}2​1​

Correct answer: (C)

Step-by-step solution →
Q62·MathematicsSingle correct
Let a differentiable function fff satisfy f(x)+∫3xf(t)t dt=x+1, x≥3f(x)+\int_{3}^{x}\frac{f(t)}{t}\,dt=\sqrt{x+1},\ x\ge 3f(x)+∫3x​tf(t)​dt=x+1​, x≥3. Then 12f(8)12f(8)12f(8) is equal to :
  1. (A)34
  2. (B)1
  3. (C)17
  4. (D)19

Correct answer: (C)

Step-by-step solution →
Q63·MathematicsSingle correct
For all z∈Cz\in Cz∈C on the curve C1:∣z∣=4C_1:|z|=4C1​:∣z∣=4, let the locus of the point z+1zz+\frac{1}{z}z+z1​ be the curve C2C_2C2​. Then :
  1. (A)the curve C1C_1C1​ lies inside C2C_2C2​
  2. (B)the curve C2C_2C2​ lies inside C1C_1C1​
  3. (C)the curves C1C_1C1​ and C2C_2C2​ intersect at 4 points
  4. (D)the curves C1C_1C1​ and C2C_2C2​ intersect at 2 points

Correct answer: (C)

Step-by-step solution →
Q64·MathematicsSingle correct
y=f(x)=sin⁡3(π3(cos⁡(π32(−4x3+5x2+1)32)))y=f(x)=\sin^{3}\left(\frac{\pi}{3}\left(\cos\left(\frac{\pi}{3\sqrt{2}}(-4x^{3}+5x^{2}+1)^{\frac{3}{2}}\right)\right)\right)y=f(x)=sin3(3π​(cos(32​π​(−4x3+5x2+1)23​))). Then, at x=1x=1x=1,
  1. (A)2 y′−3π2y=0\sqrt{2}\,y'-3\pi^{2}y=02​y′−3π2y=0
  2. (B)y′+3π2y=0y'+3\pi^{2}y=0y′+3π2y=0
  3. (C)2y′+3π2y=02y'+3\pi^{2}y=02y′+3π2y=0
  4. (D)2y′+3π2y=02y'+\sqrt{3}\pi^{2}y=02y′+3​π2y=0

Correct answer: (C)

Step-by-step solution →
Q65·MathematicsSingle correct
A wire of length 20 m is to be cut into two pieces. A piece of length l1l_1l1​ is bent to make a square of area A1A_1A1​ and the other piece of length l2l_2l2​ is made into a circle of area A2A_2A2​. If 2A1+3A22A_1+3A_22A1​+3A2​ is minimum then (πl1):l2(\pi l_1):l_2(πl1​):l2​ is equal to :
  1. (A)1:61:61:6
  2. (B)6:16:16:1
  3. (C)3:13:13:1
  4. (D)4:14:14:1

Correct answer: (B)

Step-by-step solution →
Q66·MathematicsSingle correct
Let a circle C1C_1C1​ be obtained on rolling the circle x2+y2−4x−6y+11=0x^2+y^2-4x-6y+11=0x2+y2−4x−6y+11=0 upwards 4 units on the tangent TTT to it at the point (3,2)(3,2)(3,2). Let C2C_2C2​ be the image of C1C_1C1​ in TTT. Let AAA and BBB be the centers of circles C1C_1C1​ and C2C_2C2​ respectively, and MMM and NNN be respectively the feet of perpendiculars drawn from AAA and BBB on the xxx-axis. Then the area of the trapezium AMNB is :
  1. (A)4(1+2)4(1+\sqrt{2})4(1+2​)
  2. (B)3+223+2\sqrt{2}3+22​
  3. (C)2(1+2)2(1+\sqrt{2})2(1+2​)
  4. (D)2(2+2)2(2+\sqrt{2})2(2+2​)

Correct answer: (A)

Step-by-step solution →
Q67·MathematicsSingle correct
A bag contains 6 balls. Two balls are drawn from it at random and both are found to be black. The probability that the bag contains at least 5 black balls is
  1. (A)37\frac{3}{7}73​
  2. (B)57\frac{5}{7}75​
  3. (C)56\frac{5}{6}65​
  4. (D)27\frac{2}{7}72​

Correct answer: (B)

Step-by-step solution →
Q68·MathematicsSingle correct
Let y=f(x)y=f(x)y=f(x) represent a parabola with focus (−12,0)\left(-\frac{1}{2},0\right)(−21​,0) and directrix y=−12y=-\frac{1}{2}y=−21​. Then S={x∈R:tan⁡−1(f(x))+sin⁡−1(f(x)+1)=π2}S=\left\{x\in\mathbb{R}:\tan^{-1}(\sqrt{f(x)})+\sin^{-1}(\sqrt{f(x)+1})=\frac{\pi}{2}\right\}S={x∈R:tan−1(f(x)​)+sin−1(f(x)+1​)=2π​}:
  1. (A)contains exactly two elements
  2. (B)contains exactly one element
  3. (C)is an empty set
  4. (D)is an infinite set

Correct answer: (A)

Step-by-step solution →
Q69·MathematicsSingle correct
Let a⃗=2i^+j^+k^\vec{a}=2\hat{i}+\hat{j}+\hat{k}a=2i^+j^​+k^, and b⃗\vec{b}b and c⃗\vec{c}c be two nonzero vectors such that ∣a⃗+b⃗+c⃗∣=∣a⃗+b⃗−c⃗∣|\vec{a}+\vec{b}+\vec{c}|=|\vec{a}+\vec{b}-\vec{c}|∣a+b+c∣=∣a+b−c∣ and b⃗⋅c⃗=0\vec{b}\cdot\vec{c}=0b⋅c=0. Consider the following two statements: (A) ∣a⃗+λc⃗∣≥∣a⃗∣|\vec{a}+\lambda\vec{c}|\ge|\vec{a}|∣a+λc∣≥∣a∣ for all λ∈R\lambda\in\mathbb{R}λ∈R. (B) a⃗\vec{a}a and c⃗\vec{c}c are always parallel. Then.
  1. (A)both (A) and (B) are correct
  2. (B)only (A) is correct
  3. (C)neither (A) nor (B) is correct
  4. (D)only (B) is correct

Correct answer: (B)

Step-by-step solution →
Q70·MathematicsSingle correct
The value of ∫π/3π/2(2+3sin⁡x)sin⁡x(1+cos⁡x) dx\int_{\pi/3}^{\pi/2}\frac{(2+3\sin x)}{\sin x(1+\cos x)}\,dx∫π/3π/2​sinx(1+cosx)(2+3sinx)​dx is equal to
  1. (A)103−3−log⁡e3\frac{10}{3}-\sqrt{3}-\log_{e}\sqrt{3}310​−3​−loge​3​
  2. (B)72−3−log⁡e3\frac{7}{2}-\sqrt{3}-\log_{e}\sqrt{3}27​−3​−loge​3​
  3. (C)−2+33+log⁡e3-2+3\sqrt{3}+\log_{e}\sqrt{3}−2+33​+loge​3​
  4. (D)103−3+log⁡e3\frac{10}{3}-\sqrt{3}+\log_{e}\sqrt{3}310​−3​+loge​3​

Correct answer: (D)

Step-by-step solution →
Q71·MathematicsSingle correct
Let the shortest distance between the lines L:x−5−2=y−λ0=z+λ1, λ≥0L:\frac{x-5}{-2}=\frac{y-\lambda}{0}=\frac{z+\lambda}{1},\ \lambda\ge 0L:−2x−5​=0y−λ​=1z+λ​, λ≥0 and L1:x+1=y−1=4−zL_1:x+1=y-1=4-zL1​:x+1=y−1=4−z be 262\sqrt{6}26​. If (α,β,γ)(\alpha,\beta,\gamma)(α,β,γ) lies on LLL, then which of the following is NOT possible ?
  1. (A)α−2γ=19\alpha-2\gamma=19α−2γ=19
  2. (B)2α+γ=72\alpha+\gamma=72α+γ=7
  3. (C)2α−γ=92\alpha-\gamma=92α−γ=9
  4. (D)α+2γ=24\alpha+2\gamma=24α+2γ=24

Correct answer: (D)

Step-by-step solution →
Q72·MathematicsSingle correct
For the system of linear equations x+y+z=6x+y+z=6x+y+z=6, αx+βy+7z=3\alpha x+\beta y+7z=3αx+βy+7z=3, x+2y+3z=14x+2y+3z=14x+2y+3z=14 which of the following is NOT true ?
  1. (A)If α=β\alpha=\betaα=β and α≠7\alpha\neq 7α=7, then the system has a unique solution
  2. (B)If α=β=7\alpha=\beta=7α=β=7, then the system has no solution
  3. (C)For every point (α,β)≠(7,7)(\alpha,\beta)\neq(7,7)(α,β)=(7,7) on the line x−2y+7=0x-2y+7=0x−2y+7=0, the system has infinitely many solutions
  4. (D)There is a unique point (α,β)(\alpha,\beta)(α,β) on the line x+2y+18=0x+2y+18=0x+2y+18=0 for which the system has infinitely many solutions

Correct answer: (C)

Step-by-step solution →
Q73·MathematicsSingle correct
If the domain of the function f(x)=[x]1+x2f(x)=\frac{[x]}{1+x^{2}}f(x)=1+x2[x]​, where [x][x][x] is greatest integer ≤x\le x≤x, is [2,6)[2,6)[2,6), then its range is
  1. (A)(526,25]\left(\frac{5}{26},\frac{2}{5}\right](265​,52​]
  2. (B)(537,25]−{929,27109,1889,953}\left(\frac{5}{37},\frac{2}{5}\right]-\left\{\frac{9}{29},\frac{27}{109},\frac{18}{89},\frac{9}{53}\right\}(375​,52​]−{299​,10927​,8918​,539​}
  3. (C)(537,25]\left(\frac{5}{37},\frac{2}{5}\right](375​,52​]
  4. (D)(526,25]−{929,27109,1889,953}\left(\frac{5}{26},\frac{2}{5}\right]-\left\{\frac{9}{29},\frac{27}{109},\frac{18}{89},\frac{9}{53}\right\}(265​,52​]−{299​,10927​,8918​,539​}

Correct answer: (C)

Step-by-step solution →
Q74·MathematicsSingle correct
Let R be a relation on N×NN\times NN×N defined by (a,b)R(c,d)(a,b)R(c,d)(a,b)R(c,d) if and only if ad(b−c)=bc(a−d)ad(b-c)=bc(a-d)ad(b−c)=bc(a−d). Then R is
  1. (A)transitive but neither reflexive nor symmetric
  2. (B)symmetric but neither reflexive nor transitive
  3. (C)symmetric and transitive but not reflexive
  4. (D)reflexive and symmetric but not transitive

Correct answer: (B)

Step-by-step solution →
Q75·MathematicsSingle correct
(S1) (p⇒q)∨(p∧(∼q))(p\Rightarrow q)\vee(p\wedge(\sim q))(p⇒q)∨(p∧(∼q)) is a tautology. (S2) ((∼p)⇒(∼q))∧((∼p)∨q)((\sim p)\Rightarrow(\sim q))\wedge((\sim p)\vee q)((∼p)⇒(∼q))∧((∼p)∨q) is a contradiction. Then
  1. (A)both (S1) and (S2) are correct
  2. (B)only (S1) is correct
  3. (C)only (S2) is correct
  4. (D)both (S1) and (S2) are wrong

Correct answer: (B)

Step-by-step solution →
Q76·MathematicsSingle correct
If the sum and product of four positive consecutive terms of a G.P., are 126 and 1296, respectively, then the sum of common ratios of all such GPs is
  1. (A)777
  2. (B)333
  3. (C)92\frac{9}{2}29​
  4. (D)141414

Correct answer: (A)

Step-by-step solution →
Q77·MathematicsSingle correct
Let A=(10004−1012−3)A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 12 & -3 \end{pmatrix}A=​100​0412​0−1−3​​. Then the sum of the diagonal elements of the matrix (A+I)11(A + I)^{11}(A+I)11 is equal to
  1. (A)614461446144
  2. (B)205020502050
  3. (C)409740974097
  4. (D)409440944094

Correct answer: (C)

Step-by-step solution →
Q78·MathematicsSingle correct
The number of real roots of the equation x2−4x+3+x2−9=4x2−14x+6\sqrt{x^2 - 4x + 3} + \sqrt{x^2 - 9} = \sqrt{4x^2 - 14x + 6}x2−4x+3​+x2−9​=4x2−14x+6​, is :
  1. (A)333
  2. (B)111
  3. (C)222
  4. (D)000

Correct answer: (B)

Step-by-step solution →
Q79·MathematicsSingle correct
If sin⁡−1α17+cos⁡−145−tan⁡−17736=0, 0<α<13\sin^{-1}\frac{\alpha}{17} + \cos^{-1}\frac{4}{5} - \tan^{-1}\frac{77}{36} = 0,\ 0 < \alpha < 13sin−117α​+cos−154​−tan−13677​=0, 0<α<13, then sin⁡−1(sin⁡α)+cos⁡−1(cos⁡α)\sin^{-1}(\sin \alpha) + \cos^{-1}(\cos \alpha)sin−1(sinα)+cos−1(cosα) is equal to
  1. (A)161616
  2. (B)000
  3. (C)π\piπ
  4. (D)16−5π16 - 5\pi16−5π

Correct answer: (C)

Step-by-step solution →
Q80·MathematicsSingle correct
Let α∈(0,1)\alpha \in (0,1)α∈(0,1) and β=log⁡e(1−α)\beta = \log_e(1 - \alpha)β=loge​(1−α). Let Pn(x)=x+x22+x33+⋯+xnn, x∈(0,1)P_n(x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \cdots + \frac{x^n}{n},\ x \in (0,1)Pn​(x)=x+2x2​+3x3​+⋯+nxn​, x∈(0,1). Then the integral ∫0αt501−t dt\int_0^{\alpha} \frac{t^{50}}{1 - t}\, dt∫0α​1−tt50​dt is equal to
  1. (A)β+P50(α)\beta + P_{50}(\alpha)β+P50​(α)
  2. (B)P50(α)−βP_{50}(\alpha) - \betaP50​(α)−β
  3. (C)β−P50(α)\beta - P_{50}(\alpha)β−P50​(α)
  4. (D)−(β+P50(α))-(\beta + P_{50}(\alpha))−(β+P50​(α))

Correct answer: (D)

Step-by-step solution →
Q81·MathematicsNumerical
Let α>0\alpha > 0α>0, be the smallest number such that the expansion of (x23+2x3)30\left(x^{\frac{2}{3}} + \frac{2}{x^3}\right)^{30}(x32​+x32​)30 has a term βx−α, β∈N\beta x^{-\alpha},\ \beta \in \mathbb{N}βx−α, β∈N. Then α\alphaα is equal to

Correct answer: 2

Step-by-step solution →
Q82·MathematicsNumerical
Let for x∈Rx \in \mathbb{R}x∈R, f(x)=x+∣x∣2f(x) = \frac{x + |x|}{2}f(x)=2x+∣x∣​ and g(x)={x,x<0x2,x≥0g(x) = \begin{cases} x, & x < 0 \\ x^2, & x \ge 0 \end{cases}g(x)={x,x2,​x<0x≥0​. Then area bounded by the curve y=(f∘g)(x)y = (f \circ g)(x)y=(f∘g)(x) and the lines y=0, 2y−x=15y = 0,\ 2y - x = 15y=0, 2y−x=15 is equal to

Correct answer: 72

Step-by-step solution →
Q83·MathematicsNumerical
Number of 4-digit numbers that are less than or equal to 2800 and either divisible by 3 or by 11, is equal to

Correct answer: 710

Step-by-step solution →
Q84·MathematicsNumerical
If the variance of the frequency distribution with values xi=2,3,4,5,6,7,8x_i = 2, 3, 4, 5, 6, 7, 8xi​=2,3,4,5,6,7,8 and corresponding frequencies fi=3,6,16,α,9,5,6f_i = 3, 6, 16, \alpha, 9, 5, 6fi​=3,6,16,α,9,5,6 is 3, then α\alphaα is equal to

Correct answer: 5

Step-by-step solution →
Q85·MathematicsNumerical
Let θ\thetaθ be the angle between the planes P1:r⃗⋅(i^+j^+2k^)=9P_1: \vec{r} \cdot (\hat{i} + \hat{j} + 2\hat{k}) = 9P1​:r⋅(i^+j^​+2k^)=9 and P2:r⃗⋅(2i^−j^+k^)=15P_2: \vec{r} \cdot (2\hat{i} - \hat{j} + \hat{k}) = 15P2​:r⋅(2i^−j^​+k^)=15. Let LLL be the line that meets P2P_2P2​ at the point (4,−2,5)(4, -2, 5)(4,−2,5) and makes an angle θ\thetaθ with the normal of P2P_2P2​. If α\alphaα is the angle between LLL and P2P_2P2​, then (tan⁡2θ)(cot⁡2α)(\tan^2 \theta)(\cot^2 \alpha)(tan2θ)(cot2α) is equal to

Correct answer: 9

Step-by-step solution →
Q86·MathematicsNumerical
Let 5 digit numbers be constructed using the digits 0,2,3,4,7,90, 2, 3, 4, 7, 90,2,3,4,7,9 with repetition allowed, and are arranged in ascending order with serial numbers. Then the serial number of the number 42923 is

Correct answer: 2997

Step-by-step solution →
Q87·MathematicsNumerical
Let a⃗\vec{a}a and b⃗\vec{b}b be two vectors such that ∣a⃗∣=14|\vec{a}| = \sqrt{14}∣a∣=14​, ∣b⃗∣=6|\vec{b}| = \sqrt{6}∣b∣=6​ and ∣a⃗×b⃗∣=48|\vec{a} \times \vec{b}| = \sqrt{48}∣a×b∣=48​. Then (a⃗⋅b⃗)2(\vec{a} \cdot \vec{b})^2(a⋅b)2 is equal to

Correct answer: 36

Step-by-step solution →
Q88·MathematicsNumerical
Let the line L:x−12=y+1−1=z−31L: \frac{x-1}{2} = \frac{y+1}{-1} = \frac{z-3}{1}L:2x−1​=−1y+1​=1z−3​ intersect the plane 2x+y+3z=162x + y + 3z = 162x+y+3z=16 at the point PPP. Let the point QQQ be the foot of perpendicular from the point R(1,−1,−3)R(1, -1, -3)R(1,−1,−3) on the line LLL. If α\alphaα is the area of triangle PQRPQRPQR, then α2\alpha^2α2 is equal to

Correct answer: 180

Step-by-step solution →
Q89·MathematicsNumerical
Let a1,a2,…,ana_1, a_2, \ldots, a_na1​,a2​,…,an​ be in A.P. If a5=2a7a_5 = 2a_7a5​=2a7​ and a11=18a_{11} = 18a11​=18, then 12(1a10+a11+1a11+a12+⋯+1a17+a18)12\left(\frac{1}{\sqrt{a_{10}} + \sqrt{a_{11}}} + \frac{1}{\sqrt{a_{11}} + \sqrt{a_{12}}} + \cdots + \frac{1}{\sqrt{a_{17}} + \sqrt{a_{18}}}\right)12(a10​​+a11​​1​+a11​​+a12​​1​+⋯+a17​​+a18​​1​) is equal to

Correct answer: 8

Step-by-step solution →
Q90·MathematicsNumerical
The remainder on dividing 5995^{99}599 by 11 is :

Correct answer: 9

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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
  • Chemical Bonding and Molecular Structure 151/186
  • Application of Derivatives 139/186
  • 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
  • Hydrocarbons 126/186
  • Complex Numbers 165/186
  • 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
  • Amines 133/186
  • Quadratic Equations 148/186
  • Some Basic Concepts in Chemistry 129/186
  • Electromagnetic Induction 120/186
  • Wave Optics 130/186
  • Kinetic Theory of Gases 135/186
  • Alcohols and Ethers 106/186
  • Purification and Characterisation of Organic Compounds 116/186
  • Oscillations 117/186
  • Alternating Currents 108/186
  • Nuclei 116/186
  • Organic Compounds Containing Halogens 109/186
  • Atoms 112/186
  • Statistics 118/186
  • Ellipse 103/186
  • Isolation of Metals 106/186
  • Surface Chemistry 98/186
  • Inverse Trigonometric Functions 93/186
  • Hydrogen 81/186
  • Electric Potential 63/186
  • Chemistry in Everyday Life 60/186
  • States of Matter: Gases and Liquids 52/186
  • Magnetism and Matter 50/186
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