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JEE Main 31 January 2023 Shift 2 Question Paper with Answers

31 January 2023 · January session · 89 questions

89 of the 90 questions from the JEE Main 31 January 2023 Shift 2 paper, each with its correct answer and tagged to the chapter it tests. Free to read, no account needed.

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

Physics
29
Chemistry
30
Mathematics
30

Physics — JEE Main 31 January 2023 Shift 2

Q1·PhysicsSingle correct
Given below are two statements: Statement I: In a typical transistor, all three regions emitter, base and collector have same doping level. Statement II: In a transistor, collector is the thickest and base is the thinnest segment. In the light of the 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)Statement I is correct but Statement II is incorrect
  4. (D)Both Statement I and Statement II are incorrect

Correct answer: (B)

Step-by-step solution →
Q2·PhysicsSingle correct
If the two metals AAA and BBB are exposed to radiation of wavelength 350350350 nm. The work functions of metals AAA and BBB are 4.84.84.8 eV and 2.22.22.2 eV. Then choose the correct option.
  1. (A)Both metals A and B will emit photo-electrons
  2. (B)Metal A will not emit photo-electrons
  3. (C)Metal B will not emit photo-electrons
  4. (D)Both metals A and B will not emit photo-electrons

Correct answer: (B)

Step-by-step solution →
Q3·PhysicsSingle correct
Heat energy of 735735735 J is given to a diatomic gas allowing the gas to expand at constant pressure. Each gas molecule rotates around an internal axis but do not oscillate. The increase in the internal energy of the gas will be:
  1. (A)525525525 J
  2. (B)441441441 J
  3. (C)572572572 J
  4. (D)735735735 J

Correct answer: (A)

Step-by-step solution →
Q4·PhysicsSingle correct
Match List I (physical quantities) with List II (dimensional formulae). Choose the correct answer from the options given below:
List IList II
A.Angular momentumI.[ML2T−2][ML^2T^{-2}][ML2T−2]
B.TorqueII.[ML−2T−2][ML^{-2}T^{-2}][ML−2T−2]
C.StressIII.[ML2T−1][ML^2T^{-1}][ML2T−1]
D.Pressure gradientIV.[ML−1T−2][ML^{-1}T^{-2}][ML−1T−2]
  1. (A)A-III, B-I, C-IV, D-II
  2. (B)A-III, B-II, C-IV, D-I
  3. (C)A-IV, B-II, C-I, D-III
  4. (D)A-I, B-IV, C-III, D-II

Correct answer: (A)

Step-by-step solution →
Q5·Physics·Geometrical OpticsSingle correct
A microscope is focused on an object at the bottom of a bucket. If liquid with refractive index 53\dfrac5335​ is poured inside the bucket, then microscope have to be raised by 303030 cm to focus the object again. The height of the liquid in the bucket is:
  1. (A)121212 cm
  2. (B)505050 cm
  3. (C)181818 cm
  4. (D)757575 cm

Correct answer: (D)

Step-by-step solution →
Q6·Physics·Magnetic Field of CurrentSingle correct
The number of turns of the coil of a moving coil galvanometer is increased in order to increase current sensitivity by 50%50\%50%. The percentage change in voltage sensitivity of the galvanometer will be:
  1. (A)0%0\%0%
  2. (B)75%75\%75%
  3. (C)50%50\%50%
  4. (D)100%100\%100%

Correct answer: (A)

Step-by-step solution →
Q7·PhysicsSingle correct
A body is moving with constant speed, in a circle of radius 101010 m. The body completes one revolution in 444 s. At the end of 3rd3^{rd}3rd second, the displacement of body (in m) from its starting point is:
  1. (A)15π15\pi15π
  2. (B)10210\sqrt2102​
  3. (C)303030
  4. (D)5π5\pi5π

Correct answer: (B)

Step-by-step solution →
Q8·PhysicsSingle correct
The H amount of thermal energy is developed by a resistor in 101010 s when a current of 444 A is passed through it. If the current is increased to 161616 A, the thermal energy developed by the resistor in 101010 s will be:
  1. (A)H4\dfrac{H}{4}4H​
  2. (B)16H16H16H
  3. (C)4H4H4H
  4. (D)HHH

Correct answer: (B)

Step-by-step solution →
Q9·Physics·Magnetic Field of CurrentSingle correct
A long conducting wire having a current I flowing through it, is bent into a circular coil of N turns. Then it is bent into a circular coil of n turns. The magnetic field is calculated at the centre of coils in both the cases. The ratio of the magnetic field in first case to that of second case is:
  1. (A)n:Nn:Nn:N
  2. (B)n2:N2n^2:N^2n2:N2
  3. (C)N2:n2N^2:n^2N2:n2
  4. (D)N:nN:nN:n

Correct answer: (C)

Step-by-step solution →
Q10·PhysicsSingle correct
A body weight W, is projected vertically upwards from earth's surface to reach a height above the earth which is equal to nine times the radius of earth. The weight of the body at that height will be:
  1. (A)w100\dfrac{w}{100}100w​
  2. (B)w91\dfrac{w}{91}91w​
  3. (C)w3\dfrac{w}{3}3w​
  4. (D)w9\dfrac{w}{9}9w​

Correct answer: (A)

Step-by-step solution →
Q11·PhysicsSingle correct
The radius of electron's second stationary orbit in Bohr's atom is R. The radius of 3rd orbit will be
  1. (A)R3\dfrac{R}{3}3R​
  2. (B)3R3R3R
  3. (C)2.25R2.25R2.25R
  4. (D)9R9R9R

Correct answer: (C)

Step-by-step solution →
Q12·PhysicsSingle correct
A hypothetical gas expands adiabatically such that its volume changes from 888 litres to 272727 litres. If the ratio of final pressure of the gas to initial pressure of the gas is 1681\dfrac{16}{81}8116​, then the ratio of CpCv\dfrac{C_p}{C_v}Cv​Cp​​ will be:
  1. (A)12\dfrac1221​
  2. (B)45\dfrac4554​
  3. (C)32\dfrac3223​
  4. (D)34\dfrac3443​

Correct answer: (B)

Step-by-step solution →
Q13·PhysicsSingle correct
For a solid rod, the Young's modulus of elasticity is 3.2×1011 Nm−23.2\times10^{11}\,Nm^{-2}3.2×1011Nm−2 and density is 8×1038\times10^38×103 kg m−3^{-3}−3. The velocity of longitudinal wave in the rod will be.
  1. (A)145.75×103 ms−1145.75\times10^3~ms^{-1}145.75×103 ms−1
  2. (B)18.96×103 ms−118.96\times10^3~ms^{-1}18.96×103 ms−1
  3. (C)3.65×103 ms−13.65\times10^3~ms^{-1}3.65×103 ms−1
  4. (D)6.32×103 ms−16.32\times10^3~ms^{-1}6.32×103 ms−1

Correct answer: (D)

Step-by-step solution →
Q14·PhysicsSingle correct
A body of mass 101010 kg is moving with an initial speed of 202020 m/s. The body stops after 555 s due to friction between body and the floor. The value of the coefficient of friction is: (Take acceleration due to gravity g=10 ms−2g=10~ms^{-2}g=10 ms−2)
  1. (A)0.30.30.3
  2. (B)0.50.50.5
  3. (C)0.20.20.2
  4. (D)0.40.40.4

Correct answer: (D)

Step-by-step solution →
Q15·PhysicsSingle correct
Given below are two statements: Statement I: For transmitting a signal, size of antenna (l)(l)(l) should be comparable to wavelength of signal (at least l=λ4l=\dfrac{\lambda}{4}l=4λ​ in dimension). Statement II: In amplitude modulation, amplitude of carrier wave remains constant (unchanged). In the light of the 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)Both Statement I and Statement II are correct
  3. (C)Statement I is incorrect but Statement II is correct
  4. (D)Both Statement I and Statement II are incorrect

Correct answer: (A)

Step-by-step solution →
Q16·Physics·Alternating CurrentsSingle correct
An alternating voltage source V=260sin⁡(628t)V=260\sin(628t)V=260sin(628t) is connected across a pure inductor of 555 mH. Inductive reactance in the circuit is:
  1. (A)0.318 Ω0.318\,\Omega0.318Ω
  2. (B)6.28 Ω6.28\,\Omega6.28Ω
  3. (C)3.14 Ω3.14\,\Omega3.14Ω
  4. (D)0.5 Ω0.5\,\Omega0.5Ω

Correct answer: (C)

Step-by-step solution →
Q17·PhysicsSingle correct
Under the same load, wire A having length 5.05.05.0 m and cross section 2.5×10−5 m22.5\times10^{-5}~m^22.5×10−5 m2 stretches uniformly by the same amount as another wire B of length 6.06.06.0 m and a cross section of 3.0×10−5 m23.0\times10^{-5}~m^23.0×10−5 m2 stretches. The ratio of the Young's modulus of wire A to that of wire B will be:
  1. (A)1:11:11:1
  2. (B)1:101:101:10
  3. (C)1:21:21:2
  4. (D)1:41:41:4

Correct answer: (A)

Step-by-step solution →
Q18·PhysicsSingle correct
Match List I (electromagnetic radiations) with List II (applications). Choose the correct answer from the options given below:
List IList II
A.MicrowavesI.Physiotherapy
B.UV raysII.Treatment of cancer
C.Infra-red lightIII.Lasik eye surgery
D.X-rayIV.Aircraft navigation
  1. (A)A-IV, B-III, C-I, D-II
  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-IV, C-III, D-I

Correct answer: (A)

Step-by-step solution →
Q19·Physics·Electric Field and Coulomb's LawSingle correct
Considering a group of positive charges, which of the following statements is correct?
  1. (A)both the net potential and the net electric field cannot be zero at a point.
  2. (B)net potential of the system at a point can be zero but net electric field can't be zero at that point.
  3. (C)net potential of the system cannot be zero at a point but net electric field can be zero at that point.
  4. (D)Both the net potential and the net field can be zero at a point.

Correct answer: (C)

Step-by-step solution →
Q20·Physics·Alternating CurrentsNumerical
A series LCR circuit consists of R=80 ΩR=80\,\OmegaR=80Ω, XL=100 ΩX_L=100\,\OmegaXL​=100Ω and XC=40 ΩX_C=40\,\OmegaXC​=40Ω. The input voltage is 2500cos⁡(100πt)2500\cos(100\pi t)2500cos(100πt) V. The amplitude of current, in the circuit, is _________ A.

Correct answer: 25

Step-by-step solution →
Q21·PhysicsNumerical
Two bodies are projected from ground with same speeds 40 ms−140~ms^{-1}40 ms−1 at two different angles with respect to horizontal. The bodies were found to have same range. If one of the body was projected at an angle of 60°60°60° with horizontal then sum of the maximum heights, attained by the two projectiles, is _________ m. (Given g=10 ms−2g=10~ms^{-2}g=10 ms−2)

Correct answer: 80

Step-by-step solution →
Q22·PhysicsNumerical
For the given circuit, in the steady state, ∣VB−VD∣=|V_B-V_D|=∣VB​−VD​∣= _________ V.

Correct answer: 1

Step-by-step solution →
Q23·Physics·Capacitors and DielectricsNumerical
Two parallel plate capacitors C1C_1C1​ and C2C_2C2​ each having capacitance of 10 μF10\,\mu F10μF are individually charged by a 100100100 V D.C. source. Capacitor C1C_1C1​ is kept connected to the source and a dielectric slab is inserted between it plates. Capacitor C2C_2C2​ is disconnected from the source and then a dielectric slab is inserted in it. Afterwards the capacitor C1C_1C1​ is also disconnected from the source and the two capacitors are finally connected in parallel combination. The common potential of the combination will be _________ V. (Assuming Dielectric constant =10=10=10)

Correct answer: 55

Step-by-step solution →
Q24·Physics·Wave OpticsNumerical
Two light waves of wavelengths 800800800 and 600600600 nm are used in Young's double slit experiment to obtain interference fringes on a screen placed 777 m away from plane of slits. If the two slits are separated by 0.350.350.35 mm, then shortest distance from the central bright maximum to the point where the bright fringes of the two wavelength coincide will be _________ mm.

Correct answer: 48

Step-by-step solution →
Q25·PhysicsNumerical
A ball is dropped from a height of 202020 m. If the coefficient of restitution for the collision between ball and floor is 0.50.50.5, after hitting the floor, the ball rebounds to a height of _________ m.

Correct answer: 5

Step-by-step solution →
Q26·PhysicsNumerical
If the binding energy of ground state electron in a hydrogen atom is 13.613.613.6 eV, then, the energy required to remove the electron from the second excited state of Li2+Li^{2+}Li2+ will be x×10−1x\times10^{-1}x×10−1 eV. The value of xxx is _________.

Correct answer: 136

Step-by-step solution →
Q27·PhysicsNumerical
A water heater of power 200020002000 W is used to heat water. The specific heat capacity of water is 420042004200 J kg−1^{-1}−1 K−1^{-1}−1. The efficiency of heater is 70%70\%70%. Time required to heat 222 kg of water from 10°C10°C10°C to 60°C60°C60°C is _________ s. (Assume that the specific heat capacity of water remains constant over the temperature range of the water).

Correct answer: 300

Step-by-step solution →
Q28·PhysicsNumerical
Two discs of same mass and different radii are made of different materials such that their thicknesses are 111 cm and 0.50.50.5 cm respectively. The densities of materials are in the ratio 3:53:53:5. The moment of inertia of these discs respectively about their diameters will be in the ratio of x6\dfrac{x}{6}6x​. The value of xxx is _________.

Correct answer: 5

Step-by-step solution →
Q29·Physics·WavesNumerical
The displacement equations of two interfering waves are given by y1=10sin⁡(ωt+π3)y_1=10\sin\left(\omega t+\dfrac{\pi}{3}\right)y1​=10sin(ωt+3π​) cm, y2=5[sin⁡ωt+3cos⁡ωt]y_2=5[\sin\omega t+\sqrt3\cos\omega t]y2​=5[sinωt+3​cosωt] cm respectively. The amplitude of the resultant wave is _________ cm.

Correct answer: 20

Step-by-step solution →

Chemistry — JEE Main 31 January 2023 Shift 2

Q30·ChemistrySingle correct
Which one of the following statements is incorrect?
  1. (A)van Arkel method is used to purify tungsten.
  2. (B)The malleable iron is prepared from cast iron by oxidising impurities in a reverberatory furnace.
  3. (C)Cast iron is obtained by melting pig iron with scrap iron and coke using hot air blast.
  4. (D)Boron and Indium can be purified by zone refining method.

Correct answer: (A)

Step-by-step solution →
Q31·ChemistrySingle correct
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): The first ionization enthalpy of 3d series elements is more than that of group 2 metals. Reason (R): In 3d series of elements successive filling of d-orbitals takes place. In the light of the above statements, choose the correct answer from the options given below:
  1. (A)Both (A) and (R) are true but (R) is not the correct explanation of (A)
  2. (B)Both (A) and (R) are true and (R) is the correct explanation of (A)
  3. (C)(A) is true but (R) is false
  4. (D)(A) is false but (R) is true

Correct answer: (B)

Step-by-step solution →
Q32·ChemistrySingle correct
Given below are two statements: Statement I: H2O2H_2O_2H2​O2​ is used in the synthesis of Cephalosporin. Statement II: H2O2H_2O_2H2​O2​ is used for the restoration of aerobic conditions to sewage wastes. In the light of the 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)Statement I is correct but Statement II is incorrect
  4. (D)Both Statement I and Statement II are incorrect

Correct answer: (D)

Step-by-step solution →
Q33·ChemistrySingle correct
A hydrocarbon 'X' with formula C6H8C_6H_8C6​H8​ uses two moles H2H_2H2​ on catalytic hydrogenation of its one mole. On ozonolysis, 'X' yields two moles of methane dicarbaldehyde. The hydrocarbon 'X' is:
  1. (A)cyclohexa-1,4-diene
  2. (B)cyclohexa-1,3-diene
  3. (C)1-methylcyclopenta-1,4-diene
  4. (D)hexa-1,3,5-triene

Correct answer: (A)

Step-by-step solution →
Q34·ChemistrySingle correct
Evaluate the following statements for their correctness. A. The elevation in boiling point temperature of water will be same for 0.1M NaCl and 0.1M urea. B. Azeotropic mixtures boil without change in their composition. C. Osmosis always takes place from hypertonic to hypotonic. D. The density of 32% H2SO432\%\,H_2SO_432%H2​SO4​ solution having molarity 4.09M is approximately 1.26 g mL−1^{-1}−1. E. A negatively charged sol is obtained when KI solution is added to silver nitrate solution. Choose the correct answer from the options given below:
  1. (A)A, B and D only
  2. (B)B and D only
  3. (C)B, D and E only
  4. (D)A and C only

Correct answer: (B)

Step-by-step solution →
Q35·ChemistrySingle correct
The Lewis acid character of boron tri halides follows the order:
  1. (A)BI3>BBr3>BCl3>BF3BI_3>BBr_3>BCl_3>BF_3BI3​>BBr3​>BCl3​>BF3​
  2. (B)BBr3>BI3>BCl3>BF3BBr_3>BI_3>BCl_3>BF_3BBr3​>BI3​>BCl3​>BF3​
  3. (C)BCl3>BF3>BBr3>BI3BCl_3>BF_3>BBr_3>BI_3BCl3​>BF3​>BBr3​>BI3​
  4. (D)BF3>BCl3>BBr3>BI3BF_3>BCl_3>BBr_3>BI_3BF3​>BCl3​>BBr3​>BI3​

Correct answer: (A)

Step-by-step solution →
Q36·ChemistrySingle correct
When a hydrocarbon A undergoes complete combustion it requires 11 equivalents of oxygen and produces 4 equivalents of water. What is the molecular formula of A?
  1. (A)C4H8C_4H_8C4​H8​
  2. (B)C11H4C_{11}H_4C11​H4​
  3. (C)C8H8C_8H_8C8​H8​
  4. (D)C11H8C_{11}H_8C11​H8​

Correct answer: (C)

Step-by-step solution →
Q37·ChemistrySingle correct
Arrange the following orbitals in decreasing order of energy. A. n=3, l=0, m=0 B. n=4, l=0, m=0 C. n=3, l=1, m=0 D. n=3, l=2, m=1. The correct option for the order is:
  1. (A)D > B > C > A
  2. (B)D > B > A > C
  3. (C)A > C > B > D
  4. (D)B > D > C > A

Correct answer: (A)

Step-by-step solution →
Q38·ChemistrySingle correct
The element playing significant role in neuromuscular function and interneuronal transmission is:
  1. (A)Li
  2. (B)Mg
  3. (C)Be
  4. (D)Ca

Correct answer: (D)

Step-by-step solution →
Q39·ChemistrySingle correct
Given below are two statements: Statement I: Upon heating a borax bead dipped in cupric sulphate in a luminous flame, the colour of the bead becomes green. Statement II: The green colour observed is due to the formation of copper(I) metaborate. In the light of the above statements, choose the most appropriate answer from the options given below:
  1. (A)Both Statement I and Statement II are true
  2. (B)Statement I is true but Statement II is false
  3. (C)Statement I is false but Statement II is true
  4. (D)Both Statement I and Statement II are false

Correct answer: (D)

Step-by-step solution →
Q40·ChemistrySingle correct
Which of the following compounds are not used as disinfectants? A. Chloroxylenol B. Bithional C. Veronal D. Prontosil E. Terpineol. Choose the correct answer from the options given below:
  1. (A)C, D
  2. (B)B, D, E
  3. (C)A, B
  4. (D)A, B, E

Correct answer: (A)

Step-by-step solution →
Q41·ChemistrySingle correct
Incorrect statement for the use of indicators in acid-base titration is:
  1. (A)Methyl orange may be used for a weak acid vs weak base titration.
  2. (B)Phenolphthalein is a suitable indicator for a weak acid vs strong base titration.
  3. (C)Methyl orange is a suitable indicator for a strong acid vs weak base titration.
  4. (D)Phenolphthalein may be used for a strong acid vs strong base titration.

Correct answer: (A)

Step-by-step solution →
Q42·Chemistry·AminesSingle correct
An organic compound [A] (C4H11NC_4H_{11}NC4​H11​N), shows optical activity and gives N2N_2N2​ gas on treatment with HNO2HNO_2HNO2​. The compound [A] reacts with PhSO2ClPhSO_2ClPhSO2​Cl producing a compound which is soluble in KOH.
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (D)

Step-by-step solution →
Q43·ChemistrySingle correct
The normal rain water is slightly acidic and its pH value is 5.6 because of which one of the following:
  1. (A)CO2+H2O→H2CO3CO_2+H_2O\rightarrow H_2CO_3CO2​+H2​O→H2​CO3​
  2. (B)2SO2+O2+2H2O→2H2SO42SO_2+O_2+2H_2O\rightarrow 2H_2SO_42SO2​+O2​+2H2​O→2H2​SO4​
  3. (C)4NO2+O2+2H2O→4HNO34NO_2+O_2+2H_2O\rightarrow 4HNO_34NO2​+O2​+2H2​O→4HNO3​
  4. (D)N2O5+H2O→2HNO3N_2O_5+H_2O\rightarrow 2HNO_3N2​O5​+H2​O→2HNO3​

Correct answer: (A)

Step-by-step solution →
Q44·ChemistrySingle correct
Match List I (adsorption-related terms) with List II. Choose the correct answer from the options given below:
List IList II
A.PhysisorptionI.Single Layer Adsorption
B.ChemisorptionII.20–40 kJ mol−1^{-1}−1
C.N2(g)+3H2(g)→Fe(s)2NH3(g)N_2(g)+3H_2(g)\xrightarrow{Fe(s)}2NH_3(g)N2​(g)+3H2​(g)Fe(s)​2NH3​(g)III.Chromatography
D.Analytical Application of AdsorptionIV.Heterogeneous catalysis
  1. (A)A-II, B-I, C-IV, D-III
  2. (B)A-IV, B-II, C-III, D-I
  3. (C)A-II, B-III, C-I, D-IV
  4. (D)A-III, B-IV, C-I, D-II

Correct answer: (A)

Step-by-step solution →
Q45·Chemistry·Aldehydes and KetonesSingle correct
Cyclohexylamine when treated with nitrous acid yields (P). On treating (P) with PCC results in (Q). When (Q) is heated with dil. NaOH we get (R). The final product (R) is:
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (B)

Step-by-step solution →
Q46·ChemistrySingle correct
In the following halogenated organic compounds the one with maximum number of chlorine atoms in its structure is:
  1. (A)Freon-12
  2. (B)Gammaxene
  3. (C)Chloropicrin
  4. (D)Chloral

Correct answer: (B)

Step-by-step solution →
Q47·ChemistrySingle correct
In Dumas method for the estimation of N2N_2N2​, the sample is heated with copper oxide and the gas evolved is passed over:
  1. (A)Copper oxide
  2. (B)Ni
  3. (C)Pd
  4. (D)Copper gauze

Correct answer: (B)

Step-by-step solution →
Q48·ChemistrySingle correct
Which of the following elements have half-filled f-orbitals in their ground state? (Given atomic number Sm=62; Eu=63; Tb=65; Gd=64; Pm=61) A. Sm B. Eu C. Tb D. Gd E. Pm. Choose the correct answer from the options given below:
  1. (A)A and B only
  2. (B)A and E only
  3. (C)C and D only
  4. (D)B and D only

Correct answer: (D)

Step-by-step solution →
Q49·ChemistrySingle correct
Compound A, C5H10O5C_5H_{10}O_5C5​H10​O5​, gives a tetraacetate with Ac2OAc_2OAc2​O and oxidation of A with Br2−H2OBr_2-H_2OBr2​−H2​O gives an acid, C5H10O6C_5H_{10}O_6C5​H10​O6​. Reduction of A with HI gives isopentane. The possible structure of A is:
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (C)

Step-by-step solution →
Q50·ChemistryNumerical
The rate constant for a first order reaction is 202020 min−1^{-1}−1. The time required for the initial concentration of the reactant to reduce to its 132\dfrac{1}{32}321​ level is _________ ×10−2\times10^{-2}×10−2 min. (Nearest integer) (Given: ln 10 = 2.303, log 2 = 0.3010)

Correct answer: 17

Step-by-step solution →
Q51·ChemistryNumerical
Enthalpies of formation of CCl4(g),H2O(g),CO2(g)CCl_4(g), H_2O(g), CO_2(g)CCl4​(g),H2​O(g),CO2​(g) and HCl(g)HCl(g)HCl(g) are −105,−242,−394-105, -242, -394−105,−242,−394 and −92-92−92 kJ mol−1^{-1}−1 respectively. The magnitude of enthalpy of the reaction given below is _________ kJmol−1^{-1}−1. (nearest integer) CCl4(g)+2H2O(g)→CO2(g)+4HCl(g)CCl_4(g)+2H_2O(g)\rightarrow CO_2(g)+4HCl(g)CCl4​(g)+2H2​O(g)→CO2​(g)+4HCl(g)

Correct answer: 173

Step-by-step solution →
Q52·ChemistryNumerical
A sample of a metal oxide has formula M0.83O1.00M_{0.83}O_{1.00}M0.83​O1.00​. The metal M can exist in two oxidation states +2 and +3. In the sample of M0.83O1.00M_{0.83}O_{1.00}M0.83​O1.00​, the percentage of metal ions existing in +2 oxidation state is _________ %. (nearest integer)

Correct answer: 59

Step-by-step solution →
Q53·ChemistryNumerical
The resistivity of a 0.8M solution of an electrolyte is 5×10−3 Ω5\times10^{-3}\,\Omega5×10−3Ω cm. Its molar conductivity is _________ ×104 Ω−1cm2mol−1\times10^4\,\Omega^{-1}cm^2mol^{-1}×104Ω−1cm2mol−1 (Nearest integer)

Correct answer: 25

Step-by-step solution →
Q54·ChemistryNumerical
At 298 K, the solubility of silver chloride in water is 1.434×10−31.434\times10^{-3}1.434×10−3 g L−1^{-1}−1. The value of −log⁡Ksp-\log K_{sp}−logKsp​ for silver chloride is _________ (Given mass of Ag is 107.9 g mol−1^{-1}−1 and mass of Cl is 35.5 g mol−1^{-1}−1)

Correct answer: 10

Step-by-step solution →
Q55·ChemistryNumerical
If the CFSE of [Ti(H2O)6]3+[Ti(H_2O)_6]^{3+}[Ti(H2​O)6​]3+ is −96.0-96.0−96.0 kJ/mol, this complex will absorb maximum at wavelength _________ nm. (nearest integer) Assume Planck's constant (h) =6.4×10−34=6.4\times10^{-34}=6.4×10−34 Js, Speed of light (c) =3.0×108=3.0\times10^8=3.0×108 m/s and Avogadro's Constant (NA)=6×1023(N_A)=6\times10^{23}(NA​)=6×1023/mol

Correct answer: 480

Step-by-step solution →
Q56·ChemistryNumerical
The number of alkali metal(s) from Li, K, Cs, Rb having ionization enthalpy greater than 400 kJ mol−1^{-1}−1 and forming stable super oxide is _________

Correct answer: 2

Step-by-step solution →
Q57·Chemistry·Aldehydes and KetonesNumerical
The number of molecules which gives haloform test among the following molecules is _________

Correct answer: 3

Step-by-step solution →
Q58·ChemistryNumerical
Assume carbon burns according to following equation: 2C(s)+O2(g)→2CO(g)2C_{(s)}+O_{2(g)}\rightarrow 2CO(g)2C(s)​+O2(g)​→2CO(g). When 12 g carbon is burnt in 48 g of oxygen, the volume of carbon monoxide produced is _________ ×10−1\times10^{-1}×10−1 L at STP (nearest integer) [Given: Assume CO as ideal gas, Mass of C is 12 g mol−1^{-1}−1, Mass of O is 16 g mol−1^{-1}−1 and molar volume of an ideal gas at STP is 22.7 L mol−1^{-1}−1]

Correct answer: 227

Step-by-step solution →
Q59·ChemistryNumerical
Amongst the following, the number of species having the linear shape is _________ XeF2,I3+,C2O2,I3−,CO2,SO2,BeCl2XeF_2, I_3^+, C_2O_2, I_3^-, CO_2, SO_2, BeCl_2XeF2​,I3+​,C2​O2​,I3−​,CO2​,SO2​,BeCl2​ and BCl3BCl_3BCl3​

Correct answer: 5

Step-by-step solution →

Mathematics — JEE Main 31 January 2023 Shift 2

Q60·MathematicsSingle correct
The equation e4x+8e3x+13e2x−8ex+1=0, x∈Re^{4x}+8e^{3x}+13e^{2x}-8e^{x}+1=0,\ x\in\mathbb{R}e4x+8e3x+13e2x−8ex+1=0, x∈R has:
  1. (A)four solutions two of which are negative
  2. (B)two solutions and only one of them is negative
  3. (C)two solutions and both are negative
  4. (D)no solution

Correct answer: (C)

Step-by-step solution →
Q61·MathematicsSingle correct
Among the relations S={(a,b):a,b∈R−{0}, 2+ab>0}S=\left\{(a,b):a,b\in\mathbb{R}-\{0\},\,2+\dfrac{a}{b}>0\right\}S={(a,b):a,b∈R−{0},2+ba​>0} and T={(a,b):a,b∈R, a2−b2∈Z}T=\{(a,b):a,b\in\mathbb{R},\,a^2-b^2\in\mathbb{Z}\}T={(a,b):a,b∈R,a2−b2∈Z},
  1. (A)neither SSS nor TTT is transitive
  2. (B)SSS is transitive but TTT is not
  3. (C)TTT is symmetric but SSS is not
  4. (D)both SSS and TTT are symmetric

Correct answer: (C)

Step-by-step solution →
Q62·MathematicsSingle correct
Let α>0\alpha>0α>0. If ∫0αxx+α−xdx=16+20215\displaystyle\int_{0}^{\alpha}\dfrac{x}{\sqrt{x+\alpha}-\sqrt{x}}dx=\dfrac{16+20\sqrt2}{15}∫0α​x+α​−x​x​dx=1516+202​​, then α\alphaα is equal to:
  1. (A)444
  2. (B)222\sqrt222​
  3. (C)2\sqrt22​
  4. (D)222

Correct answer: (D)

Step-by-step solution →
Q63·MathematicsSingle correct
The complex number z=i−1cos⁡π3+isin⁡π3z=\dfrac{i-1}{\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}}z=cos3π​+isin3π​i−1​ is equal to:
  1. (A)2i(cos⁡5π12−isin⁡5π12)\sqrt2 i\left(\cos\frac{5\pi}{12}-i\sin\frac{5\pi}{12}\right)2​i(cos125π​−isin125π​)
  2. (B)2(cos⁡π12+isin⁡π12)\sqrt2\left(\cos\frac{\pi}{12}+i\sin\frac{\pi}{12}\right)2​(cos12π​+isin12π​)
  3. (C)2(cos⁡5π12+isin⁡5π12)\sqrt2\left(\cos\frac{5\pi}{12}+i\sin\frac{5\pi}{12}\right)2​(cos125π​+isin125π​)
  4. (D)cos⁡π12−isin⁡π12\cos\frac{\pi}{12}-i\sin\frac{\pi}{12}cos12π​−isin12π​

Correct answer: (C)

Step-by-step solution →
Q64·MathematicsSingle correct
Let y=y(x)y=y(x)y=y(x) be the solution of the differential equation (3y2−5x2)y dx+2x(x2−y2)dy=0(3y^2-5x^2)y\,dx+2x(x^2-y^2)dy=0(3y2−5x2)ydx+2x(x2−y2)dy=0 such that y(1)=1y(1)=1y(1)=1. Then ∣(y(2))3−12y(2)∣\left|(y(2))^3-12y(2)\right|​(y(2))3−12y(2)​ is equal to:
  1. (A)16216\sqrt2162​
  2. (B)32232\sqrt2322​
  3. (C)323232
  4. (D)646464

Correct answer: (B)

Step-by-step solution →
Q65·Mathematics·Limits and ContinuitySingle correct
lim⁡x→∞(3x+1+3x−1)6+(3x+1−3x−1)6(x+x2−1)6+(x−x2−1)6x3\displaystyle\lim_{x\to\infty}\dfrac{(\sqrt{3x+1}+\sqrt{3x-1})^6+(\sqrt{3x+1}-\sqrt{3x-1})^6}{(x+\sqrt{x^2-1})^6+(x-\sqrt{x^2-1})^6}x^3x→∞lim​(x+x2−1​)6+(x−x2−1​)6(3x+1​+3x−1​)6+(3x+1​−3x−1​)6​x3
  1. (A)does not exist
  2. (B)is equal to 272727
  3. (C)is equal to 272\dfrac{27}{2}227​
  4. (D)is equal to 999

Correct answer: (B)

Step-by-step solution →
Q66·MathematicsSingle correct
The foot of perpendicular from the origin OOO to a plane PPP which meets the co-ordinate axes at the points A,B,CA,B,CA,B,C is (2,a,4), a∈N(2,a,4),\ a\in\mathbb{N}(2,a,4), a∈N. If the volume of the tetrahedron OABCOABCOABC is 144144144 unit3^33, then which of the following points is NOT on PPP?
  1. (A)(0,6,3)(0,6,3)(0,6,3)
  2. (B)(0,4,4)(0,4,4)(0,4,4)
  3. (C)(2,2,4)(2,2,4)(2,2,4)
  4. (D)(3,0,4)(3,0,4)(3,0,4)

Correct answer: (D)

Step-by-step solution →
Q67·MathematicsSingle correct
Let (a,b)⊂(0,2π)(a,b)\subset(0,2\pi)(a,b)⊂(0,2π) be the largest interval for which sin⁡−1(sin⁡θ)−cos⁡−1(sin⁡θ)>0, θ∈(0,2π)\sin^{-1}(\sin\theta)-\cos^{-1}(\sin\theta)>0,\ \theta\in(0,2\pi)sin−1(sinθ)−cos−1(sinθ)>0, θ∈(0,2π), holds. If αx2+βx+sin⁡−1(x2−6x+10)+cos⁡−1(x2−6x+10)=0\alpha x^2+\beta x+\sin^{-1}(x^2-6x+10)+\cos^{-1}(x^2-6x+10)=0αx2+βx+sin−1(x2−6x+10)+cos−1(x2−6x+10)=0 and α−β=b−a\alpha-\beta=b-aα−β=b−a, then α\alphaα is equal to:
  1. (A)π16\dfrac{\pi}{16}16π​
  2. (B)π48\dfrac{\pi}{48}48π​
  3. (C)π12\dfrac{\pi}{12}12π​
  4. (D)π8\dfrac{\pi}{8}8π​

Correct answer: (C)

Step-by-step solution →
Q68·MathematicsSingle correct
Let the mean and standard deviation of marks of class A of 100 students be respectively 40 and α (>0)\alpha\,(>0)α(>0), and the mean and standard deviation of marks of class B of nnn students be respectively 55 and 30−α30-\alpha30−α. If the mean and variance of the marks of the combined class of 100+n100+n100+n students are respectively 50 and 350, then the sum of variances of classes AAA and BBB is:
  1. (A)650650650
  2. (B)450450450
  3. (C)900900900
  4. (D)500500500

Correct answer: (D)

Step-by-step solution →
Q69·MathematicsSingle correct
The absolute minimum value, of the function f(x)=∣x2−x+1∣+[x2−x+1]f(x)=|x^2-x+1|+[x^2-x+1]f(x)=∣x2−x+1∣+[x2−x+1], where [t][t][t] denotes the greatest integer function, in the interval [−1,2][-1,2][−1,2], is:
  1. (A)14\dfrac1441​
  2. (B)32\dfrac3223​
  3. (C)54\dfrac5445​
  4. (D)34\dfrac3443​

Correct answer: (D)

Step-by-step solution →
Q70·MathematicsSingle correct
Let H be the hyperbola, whose foci are (1±2,0)(1\pm\sqrt2,0)(1±2​,0) and eccentricity is 2\sqrt22​. Then the length of its latus rectum is
  1. (A)32\dfrac3223​
  2. (B)222
  3. (C)333
  4. (D)52\dfrac5225​

Correct answer: (B)

Step-by-step solution →
Q71·MathematicsSingle correct
Let a1,a2,a3,…a_1,a_2,a_3,\ldotsa1​,a2​,a3​,… be an A.P. If a7=3a_7=3a7​=3, the product a1a4a_1a_4a1​a4​ is minimum and the sum of its first nnn terms is zero, then n!−4an(n+2)n!-4a_{n(n+2)}n!−4an(n+2)​ is equal to:
  1. (A)999
  2. (B)334\dfrac{33}{4}433​
  3. (C)3814\dfrac{381}{4}4381​
  4. (D)242424

Correct answer: (D)

Step-by-step solution →
Q72·MathematicsSingle correct
If a point P(α,β,γ)P(\alpha,\beta,\gamma)P(α,β,γ) satisfying (α β γ)(2108938848)=(0 0 0)(\alpha\ \beta\ \gamma)\begin{pmatrix}2&10&8\\9&3&8\\8&4&8\end{pmatrix}=(0\ 0\ 0)(α β γ)​298​1034​888​​=(0 0 0) lies on the plane 2x+4y+3z=52x+4y+3z=52x+4y+3z=5, then 6α+9β+7γ6\alpha+9\beta+7\gamma6α+9β+7γ is equal to:
  1. (A)−1-1−1
  2. (B)115\dfrac{11}{5}511​
  3. (C)54\dfrac5445​
  4. (D)111111

Correct answer: (D)

Step-by-step solution →
Q73·MathematicsSingle correct
Let a⃗=i^+2j^+3k^, b⃗=i^−j^+2k^\vec a=\hat i+2\hat j+3\hat k,\ \vec b=\hat i-\hat j+2\hat ka=i^+2j^​+3k^, b=i^−j^​+2k^ and c⃗=5i^−3j^+3k^\vec c=5\hat i-3\hat j+3\hat kc=5i^−3j^​+3k^ be three vectors. If r⃗\vec rr is a vector such that r⃗×b⃗=c⃗×b⃗\vec r\times\vec b=\vec c\times\vec br×b=c×b and r⃗⋅a⃗=0\vec r\cdot\vec a=0r⋅a=0, then 25∣r⃗∣225|\vec r|^225∣r∣2 is equal to
  1. (A)560560560
  2. (B)449449449
  3. (C)339339339
  4. (D)336336336

Correct answer: (C)

Step-by-step solution →
Q74·MathematicsSingle correct
Let the plane P:8x+α1y+α2z+12=0P:8x+\alpha_1 y+\alpha_2 z+12=0P:8x+α1​y+α2​z+12=0 be parallel to the line L:x+22=y−33=z+45L:\dfrac{x+2}{2}=\dfrac{y-3}{3}=\dfrac{z+4}{5}L:2x+2​=3y−3​=5z+4​. If the intercept of PPP on the yyy-axis is 1, then the distance between PPP and LLL is:
  1. (A)72\sqrt{\dfrac72}27​​
  2. (B)27\sqrt{\dfrac27}72​​
  3. (C)614\dfrac{6}{\sqrt{14}}14​6​
  4. (D)14\sqrt{14}14​

Correct answer: (D)

Step-by-step solution →
Q75·MathematicsSingle correct
Let PPP be the plane, passing through the point (1,−1,−5)(1,-1,-5)(1,−1,−5) and perpendicular to the line joining the points (4,1,−3)(4,1,-3)(4,1,−3) and (2,4,3)(2,4,3)(2,4,3). Then the distance of PPP from the point (3,−2,2)(3,-2,2)(3,−2,2) is
  1. (A)555
  2. (B)444
  3. (C)777
  4. (D)666

Correct answer: (A)

Step-by-step solution →
Q76·MathematicsSingle correct
The number of values of r∈{p,q,∼p,∼q}r\in\{p,q,\sim p,\sim q\}r∈{p,q,∼p,∼q} for which ((p∧q)⇒(r∨q))∧((p∧r)⇒q)((p\wedge q)\Rightarrow(r\vee q))\wedge((p\wedge r)\Rightarrow q)((p∧q)⇒(r∨q))∧((p∧r)⇒q) is a tautology, is:
  1. (A)333
  2. (B)444
  3. (C)111
  4. (D)222

Correct answer: (D)

Step-by-step solution →
Q77·MathematicsSingle correct
The set of all values of a2a^2a2 for which the line x+y=0x+y=0x+y=0 bisects two distinct chords drawn from a point P(1+a2,1−a2)P\left(\dfrac{1+a}{2},\dfrac{1-a}{2}\right)P(21+a​,21−a​) on the circle 2x2+2y2−(1+a)x−(1−a)y=02x^2+2y^2-(1+a)x-(1-a)y=02x2+2y2−(1+a)x−(1−a)y=0, is equal to:
  1. (A)(0,4](0,4](0,4]
  2. (B)(4,∞)(4,\infty)(4,∞)
  3. (C)(2,12](2,12](2,12]
  4. (D)(8,∞)(8,\infty)(8,∞)

Correct answer: (D)

Step-by-step solution →
Q78·MathematicsSingle correct
If ϕ(x)=1x∫π4x(42sin⁡t−3ϕ′(t))dt, x>0\phi(x)=\dfrac{1}{\sqrt{x}}\displaystyle\int_{\frac{\pi}{4}}^{x}\left(4\sqrt2\sin t-3\phi'(t)\right)dt,\ x>0ϕ(x)=x​1​∫4π​x​(42​sint−3ϕ′(t))dt, x>0, then ϕ′(π4)\phi'\left(\dfrac{\pi}{4}\right)ϕ′(4π​) is equal to:
  1. (A)86+π\dfrac{8}{6+\sqrt\pi}6+π​8​
  2. (B)46+π\dfrac{4}{6+\sqrt\pi}6+π​4​
  3. (C)8π\dfrac{8}{\sqrt\pi}π​8​
  4. (D)46−π\dfrac{4}{6-\sqrt\pi}6−π​4​

Correct answer: (A)

Step-by-step solution →
Q79·MathematicsSingle correct
Let f:R−{2,6}→Rf:\mathbb{R}-\{2,6\}\to\mathbb{R}f:R−{2,6}→R be real valued function defined as f(x)=x2+2x+1x2−8x+12f(x)=\dfrac{x^2+2x+1}{x^2-8x+12}f(x)=x2−8x+12x2+2x+1​. Then range of fff is
  1. (A)(−∞,−214]∪(0,∞)\left(-\infty,-\dfrac{21}{4}\right]\cup(0,\infty)(−∞,−421​]∪(0,∞)
  2. (B)(−∞,−214]∪[1,∞)\left(-\infty,-\dfrac{21}{4}\right]\cup[1,\infty)(−∞,−421​]∪[1,∞)
  3. (C)(−∞,−214]∪[214,∞)\left(-\infty,-\dfrac{21}{4}\right]\cup\left[\dfrac{21}{4},\infty\right)(−∞,−421​]∪[421​,∞)
  4. (D)(−∞,−214]∪[0,∞)\left(-\infty,-\dfrac{21}{4}\right]\cup[0,\infty)(−∞,−421​]∪[0,∞)

Correct answer: (D)

Step-by-step solution →
Q80·MathematicsNumerical
Let A=[aij], aij∈Z∩[0,4], 1≤i,j≤2A=[a_{ij}],\ a_{ij}\in\mathbb{Z}\cap[0,4],\ 1\le i,j\le2A=[aij​], aij​∈Z∩[0,4], 1≤i,j≤2. The number of matrices AAA such that the sum of all entries is a prime number p∈(2,13)p\in(2,13)p∈(2,13) is

Correct answer: 204

Step-by-step solution →
Q81·MathematicsNumerical
Let AAA be a n×nn\times nn×n matrix such that ∣A∣=2|A|=2∣A∣=2. If the determinant of the matrix Adj(2⋅Adj(2A−1))\mathrm{Adj}\left(2\cdot\mathrm{Adj}(2A^{-1})\right)Adj(2⋅Adj(2A−1)) is 2842^{84}284, then nnn is equal to

Correct answer: 84

Step-by-step solution →
Q82·MathematicsNumerical
If the constant term in the binomial expansion of (x5/22−4xl)9\left(\dfrac{x^{5/2}}{2}-\dfrac{4}{x^{l}}\right)^{9}(2x5/2​−xl4​)9 is −84-84−84 and the coefficient of x−3lx^{-3l}x−3l is 2αβ2^{\alpha}\beta2αβ, where β<0\beta<0β<0 is an odd number, then ∣αl−β∣|\alpha l-\beta|∣αl−β∣ is equal to

Correct answer: 98

Step-by-step solution →
Q83·MathematicsNumerical
Let SSS be the set of all a∈Na\in\mathbb{N}a∈N such that the area of the triangle formed by the tangent at the point P(b,c), b,c∈NP(b,c),\ b,c\in\mathbb{N}P(b,c), b,c∈N, on the parabola y2=2axy^2=2axy2=2ax and the lines x=b, y=0x=b,\ y=0x=b, y=0 is 161616 unit2^22, then ∑a∈Sa\displaystyle\sum_{a\in S}aa∈S∑​a is equal to

Correct answer: 146

Step-by-step solution →
Q84·MathematicsNumerical
Let the area of the region {(x,y):∣2x−1∣≤y≤∣x2−x∣, 0≤x≤1}\{(x,y):|2x-1|\le y\le|x^2-x|,\ 0\le x\le1\}{(x,y):∣2x−1∣≤y≤∣x2−x∣, 0≤x≤1} be AAA. Then (6A+11)2(6A+11)^2(6A+11)2 is equal to

Correct answer: 125

Step-by-step solution →
Q85·MathematicsNumerical
The coefficient of x−6x^{-6}x−6, in the expansion of (4x5+52x2)9\left(\dfrac{4x}{5}+\dfrac{5}{2x^2}\right)^{9}(54x​+2x25​)9, is

Correct answer: 5040

Step-by-step solution →
Q86·MathematicsNumerical
Let AAA be the event that the absolute difference between two randomly chosen real numbers in the sample space [0,60][0,60][0,60] is less than or equal to aaa. If P(A)=1136P(A)=\dfrac{11}{36}P(A)=3611​, then aaa is equal to

Correct answer: 10

Step-by-step solution →
Q87·MathematicsNumerical
If 2n+1Pn−1:2n−1Pn=11:21^{2n+1}P_{n-1}:{}^{2n-1}P_{n}=11:212n+1Pn−1​:2n−1Pn​=11:21, then n2+n+15n^2+n+15n2+n+15 is equal to:

Correct answer: 45

Step-by-step solution →
Q88·MathematicsNumerical
Let a⃗,b⃗,c⃗\vec a,\vec b,\vec ca,b,c be three vectors such that ∣a⃗∣=31, 4∣b⃗∣=∣c⃗∣=2|\vec a|=\sqrt{31},\ 4|\vec b|=|\vec c|=2∣a∣=31​, 4∣b∣=∣c∣=2 and 2(a⃗×b⃗)=3(c⃗×a⃗)2(\vec a\times\vec b)=3(\vec c\times\vec a)2(a×b)=3(c×a). If the angle between b⃗\vec bb and c⃗\vec cc is 2π3\dfrac{2\pi}{3}32π​, then (a⃗×c⃗a⃗⋅b⃗)2\left(\dfrac{\vec a\times\vec c}{\vec a\cdot\vec b}\right)^2(a⋅ba×c​)2 is equal to

Correct answer: 3

Step-by-step solution →
Q89·MathematicsNumerical
The sum 12−2⋅32+3⋅52−4⋅72+5⋅92−⋯+15⋅2921^2-2\cdot3^2+3\cdot5^2-4\cdot7^2+5\cdot9^2-\cdots+15\cdot29^212−2⋅32+3⋅52−4⋅72+5⋅92−⋯+15⋅292 is

Correct answer: 6952

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
  • d- and f-Block Elements 126/186
  • Thermodynamics 154/186
  • Aldehydes and Ketones 135/186
  • Equilibrium 163/186
  • Solutions 158/186
  • Laws of Motion 130/186
  • Chemical Thermodynamics 165/186
  • Units and Measurements 149/186
  • 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
  • Wave Optics 130/186
  • Area Under Curves 139/186
  • Kinetic Theory of Gases 135/186
  • Purification and Characterisation of Organic Compounds 116/186
  • Alternating Currents 108/186
  • Organic Compounds Containing Halogens 109/186
  • Waves 109/186
  • Capacitors and Dielectrics 115/186
  • Atoms 112/186
  • Parabola 101/186
  • Statistics 118/186
  • Isolation of Metals 106/186
  • s-Block Elements 88/186
  • Surface Chemistry 98/186
  • Inverse Trigonometric Functions 93/186
  • Environmental Chemistry 83/186
  • Hydrogen 81/186
  • Hyperbola 77/186
  • Solid State 63/186
  • Chemistry in Everyday Life 60/186
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