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JEE Main 29 June 2022 Shift 1 Question Paper with Answers

29 June 2022 · June session · 89 questions

89 of the 90 questions from the JEE Main 29 June 2022 Shift 1 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
30
Chemistry
30
Mathematics
29

Physics — JEE Main 29 June 2022 Shift 1

Q1·PhysicsSingle correct
Two balls A and B are placed at the top of 180 m tall tower. Ball A is released from the top at t = 0 s. Ball B is thrown vertically down with an initial velocity 'u' at t = 2 s. After a certain time, both balls meet 100 m above the ground. Find the value of 'u' in ms−1ms^{-1}ms−1. [use g = 10 ms−2ms^{-2}ms−2] :
  1. (A)10
  2. (B)15
  3. (C)20
  4. (D)30

Correct answer: (D)

Step-by-step solution →
Q2·PhysicsSingle correct
A body of mass M at rest explodes into three pieces, in the ratio of masses 1 : 1 : 2. Two smaller pieces fly off perpendicular to each other with velocities of 30 ms−1ms^{-1}ms−1 and 40 ms−1ms^{-1}ms−1 respectively. The velocity of the third piece will be :
  1. (A)15 ms−1ms^{-1}ms−1
  2. (B)25 ms−1ms^{-1}ms−1
  3. (C)35 ms−1ms^{-1}ms−1
  4. (D)50 ms−1ms^{-1}ms−1

Correct answer: (B)

Step-by-step solution →
Q3·PhysicsSingle correct
The activity of a radioactive material is 2.56×10−32.56 \times 10^{-3}2.56×10−3 Ci. If the half life of the material is 5 days, after how many days the activity will become 2×10−52 \times 10^{-5}2×10−5 Ci?
  1. (A)30 days
  2. (B)35 days
  3. (C)40 days
  4. (D)25 days

Correct answer: (B)

Step-by-step solution →
Q4·PhysicsSingle correct
A spherical shell of 1 kg mass and radius R is rolling with angular speed ω on horizontal plane (as shown in figure). The magnitude of angular momentum of the shell about the origin O is a3R2ω\frac{a}{3}R^2\omega3a​R2ω. The value of a will be :
  1. (A)2
  2. (B)3
  3. (C)5
  4. (D)4

Correct answer: (C)

Step-by-step solution →
Q5·PhysicsSingle correct
A cylinder of fixed capacity of 44.8 litres contains helium gas at standard temperature and pressure. The amount of heat needed to raise the temperature of gas in the cylinder by 20.0°C will be : (Given gas constant R = 8.3 JK−1JK^{-1}JK−1-mol−1mol^{-1}mol−1)
  1. (A)249 J
  2. (B)415 J
  3. (C)498 J
  4. (D)830 J

Correct answer: (C)

Step-by-step solution →
Q6·PhysicsSingle correct
A wire of length L is hanging from a fixed support. The length changes to L1L_1L1​ and L2L_2L2​ when masses 1kg and 2 kg are suspended respectively from its free end. Then the value of L is equal to :
  1. (A)L1L2\sqrt{L_1L_2}L1​L2​​
  2. (B)L1+L22\frac{L_1+L_2}{2}2L1​+L2​​
  3. (C)2L1−L22L_1 - L_22L1​−L2​
  4. (D)3L1−2L23L_1 - 2L_23L1​−2L2​

Correct answer: (C)

Step-by-step solution →
Q7·PhysicsSingle correct
Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R. Assertion A : The photoelectric effect does not take place, if the energy of the incident radiation is less than the work function of a metal. Reason R : Kinetic energy of the photoelectrons is zero, if the energy of the incident radiation is equal to the work function of a metal. 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)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)A is not correct but R is correct

Correct answer: (B)

Step-by-step solution →
Q8·PhysicsSingle correct
A particle of mass 500 gm is moving in a straight line with velocity υ=b x5/2\upsilon = b\,x^{5/2}υ=bx5/2. The work done by the net force during its displacement from x = 0 to x = 4 m is : (Take b = 0.25 m−3/2m^{-3/2}m−3/2 s−1s^{-1}s−1).
  1. (A)2 J
  2. (B)4 J
  3. (C)8 J
  4. (D)16 J

Correct answer: (D)

Step-by-step solution →
Q9·Physics·Magnetic Field of CurrentSingle correct
A charged particle moves along circular path in a uniform magnetic field in a cyclotron. The kinetic energy of the charged particle increases to 4 times its initial value. What will be the ratio of new radius to the original radius of circular path of the charged particle :
  1. (A)1 : 1
  2. (B)1 : 2
  3. (C)2 : 1
  4. (D)1 : 4

Correct answer: (C)

Step-by-step solution →
Q10·PhysicsSingle correct
For a series LCR circuit, I vs ω curve is shown : (a) To the left of ωr\omega_rωr​, the circuit is mainly capacitive. (b) To the left of ωr\omega_rωr​, the circuit is mainly inductive. (c) At ωr\omega_rωr​, impedance of the circuit is equal to the resistance of the circuit. (d) At ωr\omega_rωr​, impedance of the circuit is 0. Choose the most appropriate answer from the options given below :
  1. (A)(a) and (d) only
  2. (B)(b) and (d) only
  3. (C)(a) and (c) only
  4. (D)(b) and (c) only

Correct answer: (C)

Step-by-step solution →
Q11·PhysicsSingle correct
A block of metal weighing 2 kg is resting on a frictionless plane (as shown in figure). It is struck by a jet releasing water at a rate of 1 kgs−1kgs^{-1}kgs−1 and at a speed of 10 ms−1ms^{-1}ms−1. Then, the initial acceleration of the block, in ms−2ms^{-2}ms−2, will be :
  1. (A)3
  2. (B)6
  3. (C)5
  4. (D)4

Correct answer: (C)

Step-by-step solution →
Q12·PhysicsSingle correct
In Vander Waals equation [P+aV2][V−b]=RT\left[P+\frac{a}{V^2}\right][V-b]=RT[P+V2a​][V−b]=RT; P is pressure, V is volume, R is universal gas constant and T is temperature. The ratio of constants ab\frac{a}{b}ba​ is dimensionally equal to :
  1. (A)PV\frac{P}{V}VP​
  2. (B)VP\frac{V}{P}PV​
  3. (C)PV
  4. (D)PV3PV^3PV3

Correct answer: (C)

Step-by-step solution →
Q13·PhysicsSingle correct
Two vectors A⃗\vec{A}A and B⃗\vec{B}B have equal magnitudes. If magnitude of A⃗+B⃗\vec{A}+\vec{B}A+B is equal to two times the magnitude of A⃗−B⃗\vec{A}-\vec{B}A−B, then the angle between A⃗\vec{A}A and B⃗\vec{B}B will be :
  1. (A)sin⁡−1(35)\sin^{-1}\left(\frac{3}{5}\right)sin−1(53​)
  2. (B)sin⁡−1(13)\sin^{-1}\left(\frac{1}{3}\right)sin−1(31​)
  3. (C)cos⁡−1(35)\cos^{-1}\left(\frac{3}{5}\right)cos−1(53​)
  4. (D)cos⁡−1(13)\cos^{-1}\left(\frac{1}{3}\right)cos−1(31​)

Correct answer: (C)

Step-by-step solution →
Q14·PhysicsSingle correct
The escape velocity of a body on a planet 'A' is 12 kms−1kms^{-1}kms−1. The escape velocity of the body on another planet 'B', whose density is four times and radius is half of the planet 'A', is :
  1. (A)12 kms−1kms^{-1}kms−1
  2. (B)24 kms−1kms^{-1}kms−1
  3. (C)36 kms−1kms^{-1}kms−1
  4. (D)6 kms−1kms^{-1}kms−1

Correct answer: (A)

Step-by-step solution →
Q15·PhysicsSingle correct
At a certain place the angle of dip is 30° and the horizontal component of earth's magnetic field is 0.5 G. The earth's total magnetic field (in G), at that certain place, is :
  1. (A)13\frac{1}{\sqrt{3}}3​1​
  2. (B)12\frac{1}{2}21​
  3. (C)3\sqrt{3}3​
  4. (D)1

Correct answer: (A)

Step-by-step solution →
Q16·PhysicsSingle correct
A longitudinal wave is represented by x=10sin⁡2π(nt−xλ)x = 10 \sin 2\pi\left(nt - \frac{x}{\lambda}\right)x=10sin2π(nt−λx​) cm. The maximum particle velocity will be four times the wave velocity if the determined value of wavelength is equal to :
  1. (A)2π2\pi2π
  2. (B)5π5\pi5π
  3. (C)π\piπ
  4. (D)5π2\frac{5\pi}{2}25π​

Correct answer: (B)

Step-by-step solution →
Q17·PhysicsSingle correct
A parallel plate capacitor filled with a medium of dielectric constant 10, is connected across a battery and is charged. The dielectric slab is replaced by another slab of dielectric constant 15. Then the energy of capacitor will :
  1. (A)increase by 50%
  2. (B)decrease by 15%
  3. (C)increase by 25%
  4. (D)increase by 33%

Correct answer: (A)

Step-by-step solution →
Q18·PhysicsSingle correct
A positive charge particle of 100 mg is thrown in opposite direction to a uniform electric field of strength 1×1051 \times 10^51×105 NC−1^{-1}−1. If the charge on the particle is 40 μC and the initial velocity is 200 ms−1^{-1}−1, how much distance it will travel before coming to the rest momentarily :
  1. (A)1 m
  2. (B)5 m
  3. (C)10 m
  4. (D)0.5 m

Correct answer: (D)

Step-by-step solution →
Q19·PhysicsSingle correct
Using Young's double slit experiment, a monochromatic light of wavelength 5000 Å produces fringes of fringe width 0.5 mm. If another monochromatic light of wavelength 6000Å is used and the separation between the slits is doubled, then the new fringe width will be :
  1. (A)0.5 mm
  2. (B)1.0 mm
  3. (C)0.6 mm
  4. (D)0.3 mm

Correct answer: (D)

Step-by-step solution →
Q20·PhysicsSingle correct
Only 2% of the optical source frequency is the available channel bandwidth for an optical communicating system operating at 1000 nm. If an audio signal requires a bandwidth of 8 kHz, how many channels can be accommodated for transmission :
  1. (A)375×107375 \times 10^7375×107
  2. (B)75×10775 \times 10^775×107
  3. (C)375×108375 \times 10^8375×108
  4. (D)75×10975 \times 10^975×109

Correct answer: (B)

Step-by-step solution →
Q21·PhysicsNumerical
Two coils require 20 minutes and 60 minutes respectively to produce same amount of heat energy when connected separately to the same source. If they are connected in parallel arrangement to the same source; the time required to produce same amount of heat by the combination of coils, will be__________min.

Correct answer: 15

Step-by-step solution →
Q22·PhysicsNumerical
The intensity of the light from a bulb incident on a surface is 0.22 W/m2^22. The amplitude of the magnetic field in this light-wave is_______×10−9\times 10^{-9}×10−9 T. (Given : Permittivity of vacuum ∈0=8.85×10−12C2N−1m−2\in_0 = 8.85 \times 10^{-12} \mathrm{C}^2\mathrm{N}^{-1}\mathrm{m}^{-2}∈0​=8.85×10−12C2N−1m−2, speed of light in vacuum c = 3×1083 \times 10^83×108 ms−1^{-1}−1)

Correct answer: 43

Step-by-step solution →
Q23·PhysicsNumerical
As per the given figure, two plates A and B of thermal conductivity K and 2 K are joined together to form a compound plate. The thickness of plates are 4.0 cm and 2.5 cm respectively and the area of cross-section is 120 cm2^22 for each plate. The equivalent thermal conductivity of the compound plate is (1+5α)\left(1+\frac{5}{\alpha}\right)(1+α5​) K, then the value of α will be __________.

Correct answer: 21

Step-by-step solution →
Q24·PhysicsNumerical
A body is performing simple harmonic with an amplitude of 10 cm. The velocity of the body was tripled by air Jet when it is at 5 cm from its mean position. The new amplitude of vibration is x\sqrt{x}x​ cm. The value of x is___________.

Correct answer: 700

Step-by-step solution →
Q25·PhysicsNumerical
The variation of applied potential and current flowing through a given wire is shown in figure. The length of wire is 31.4 cm. The diameter of wire is measured as 2.4 cm. The resistivity of the given wire is measured as x×10−3x \times 10^{-3}x×10−3 Ω cm. The value of x is ___________. [Take π = 3.14]

Correct answer: 144

Step-by-step solution →
Q26·PhysicsNumerical
300 cal. of heat is given to a heat engine and it rejects 225 cal. of heat. If source temperature is 227°C, then the temperature of sink will be __ ∘^{\circ}∘C.

Correct answer: 102

Step-by-step solution →
Q27·PhysicsNumerical
d1\sqrt{d_1}d1​​ and d2\sqrt{d_2}d2​​ are the impact parameters corresponding to scattering angles 60° and 90° respectively, when an α particle is approaching a gold nucleus. For d1=x d2d_1 = x\, d_2d1​=xd2​, the value of x will be________.

Correct answer: 3

Step-by-step solution →
Q28·PhysicsNumerical
A transistor is used in an amplifier circuit in common emitter mode. If the base current changes by 100 μA, it brings a change of 10 mA in collector current. If the load resistance is 2 kΩ and input resistance is 1 kΩ, the value of power gain is x×104x \times 10^4x×104. The value of x is ________.

Correct answer: 2

Step-by-step solution →
Q29·PhysicsNumerical
A parallel beam of light is allowed to fall on a transparent spherical globe of diameter 30 cm and refractive index 1.5. The distance from the centre of the globe at which the beam of light can converge is _________ mm.

Correct answer: 225

Step-by-step solution →
Q30·PhysicsNumerical
For the network shown below, the value VB−VAV_B - V_AVB​−VA​ is _________V.

Correct answer: 10

Step-by-step solution →

Chemistry — JEE Main 29 June 2022 Shift 1

Q31·ChemistrySingle correct
Production of iron in blast furnace follows the following equation Fe3O4(s)+4CO(g)→3Fe(l)+4CO2(g)Fe_3O_4(s) + 4CO(g) \rightarrow 3Fe(l) + 4CO_2(g)Fe3​O4​(s)+4CO(g)→3Fe(l)+4CO2​(g) when 4.640 kg of Fe3O4Fe_3O_4Fe3​O4​ and 2.520 kg of CO are allowed to react then the amount of iron (in g) produced is : [Given : Molar Atomic mass (g mol−1mol^{-1}mol−1): Fe = 56 Molar Atomic mass (g mol−1mol^{-1}mol−1) : O = 16 Molar Atomic mass (g mol−1mol^{-1}mol−1): = C = 12
  1. (A)1400
  2. (B)2200
  3. (C)3360
  4. (D)4200

Correct answer: (C)

Step-by-step solution →
Q32·ChemistrySingle correct
Which of the following statements are correct ? (A) The electronic configuration of Cr is [Ar] 3d54s13d^5 4s^13d54s1. (B) The magnetic quantum number may have a negative value. (C) In the ground state of an atom, the orbitals are filled in order of their increasing energies. (D) The total number of nodes are given by n – 2 . Choose the most appropriate answer from the options given below :
  1. (A)(A), (C) and (D) only
  2. (B)(A) and (B) only
  3. (C)(A) and (C) only
  4. (D)(A), (B) and (C) only

Correct answer: (D)

Step-by-step solution →
Q33·ChemistrySingle correct
Arrange the following in the decreasing order of their covalent character : (A) LiCl (B) NaCl (C) KCl (D) CsCl Question: Choose the most appropriate answer from the options given below :
  1. (A)(A) > (C) > (B) > (D)
  2. (B)(B) > (A) > (C) > (D)
  3. (C)(A) > (B) > (C) > (D)
  4. (D)(A) > (B) > (D) > (C)

Correct answer: (C)

Step-by-step solution →
Q34·ChemistrySingle correct
The solubility of AgCl will be maximum in which of the following ?
  1. (A)0.01 M KCl
  2. (B)0.01 M HCl
  3. (C)0.01 M AgNO3AgNO_3AgNO3​
  4. (D)Deionised water

Correct answer: (D)

Step-by-step solution →
Q35·ChemistrySingle correct
Which of the following is a correct statement ?
  1. (A)Brownian motion destabilises sols.
  2. (B)Any amount of dispersed phase can be added to emulsion without destabilising it.
  3. (C)Mixing two oppositely charged sols in equal amount neutralises charges and stabilises colloids.
  4. (D)Presence of equal and similar charges on colloidal particles provides stability to the colloidal solution.

Correct answer: (D)

Step-by-step solution →
Q36·ChemistrySingle correct
The electronic configuration of Pt (atomic number 78) is:
  1. (A)[Xe] 4f145d96s14f^{14} 5d^9 6s^14f145d96s1
  2. (B)[Kr] 4f145d104f^{14} 5d^{10}4f145d10
  3. (C)[Xe] 4f145d104f^{14} 5d^{10}4f145d10
  4. (D)[Xe] 4f145d86s24f^{14} 5d^8 6s^24f145d86s2

Correct answer: (A)

Step-by-step solution →
Q37·ChemistrySingle correct
In isolation of which one of the following metals from their ores, the use of cyanide salt is not commonly involved ?
  1. (A)Zinc
  2. (B)Gold
  3. (C)Silver
  4. (D)Copper

Correct answer: (D)

Step-by-step solution →
Q38·ChemistrySingle correct
Which one of the following reactions indicates the reducing ability of hydrogen peroxide in basic medium ?
  1. (A)HOCl+H2O2→H3O++Cl−+O2HOCl + H_2O_2 \rightarrow H_3O^+ + Cl^- + O_2HOCl+H2​O2​→H3​O++Cl−+O2​
  2. (B)PbS+4H2O2→PbSO4+4H2OPbS + 4H_2O_2 \rightarrow PbSO_4 + 4H_2OPbS+4H2​O2​→PbSO4​+4H2​O
  3. (C)2MnO4−+3H2O2→2MnO2+3O2+2H2O+2OH−2MnO_4^- + 3H_2O_2 \rightarrow 2MnO_2 + 3O_2 + 2H_2O + 2OH^-2MnO4−​+3H2​O2​→2MnO2​+3O2​+2H2​O+2OH−
  4. (D)Mn2++H2O2→Mn4++2OH−Mn^{2+} + H_2O_2 \rightarrow Mn^{4+} + 2OH^-Mn2++H2​O2​→Mn4++2OH−

Correct answer: (C)

Step-by-step solution →
Q39·ChemistrySingle correct
Match the List-I with List- II. Choose the most appropriate answer from the options given below:
List-I (Metal)List-II (Emitted light wavelength (nm))
A.LiI.670.8
B.NaII.589.2
C.RbIII.780.0
D.CsIV.455.5
  1. (A)(A)-(I), (B)-(II), (C)-(III), (D)-(IV)
  2. (B)(A)-(III), (B)-(II), (C)-(I), (D)-(IV)
  3. (C)(A)-(III), (B)-( I), (C)-(II), (D)-(IV)
  4. (D)(A)-(IV), (B)-(II), (C)-(I), (D)-(III)

Correct answer: (A)

Step-by-step solution →
Q40·ChemistrySingle correct
Match the List-I with List- II. Choose the most appropriate answer from the option given below:
List-I (Metal)List-II Application
A.CsI.High temperature thermometer
B.GaII.Water repellent sprays
C.BIII.Photoelectric cells
D.SiIV.Bullet proof vest
  1. (A)(A)-(III), (B)-(I), (C)-(IV), (D)-(II)
  2. (B)(A)-(IV), (B)-(III), (C)-(II), (D)-(I)
  3. (C)(A)-(II), (B)-(III), (C)-(IV), (D)-(I)
  4. (D)(A)-(I), (B)-(IV), (C)-(II), (D)-(III)

Correct answer: (A)

Step-by-step solution →
Q41·ChemistrySingle correct
The oxoacid of phosphorus that is easily obtained from a reaction of alkali and white phosphorus and has two P-H bonds, is :
  1. (A)Phosphonic acid
  2. (B)Phosphinic acid
  3. (C)Pyrophosphorus acid
  4. (D)Hypophosphoric acid

Correct answer: (B)

Step-by-step solution →
Q42·ChemistrySingle correct
The acid that is believed to be mainly responsible for the damage of Taj Mahal is
  1. (A)Sulfuric acid
  2. (B)Hydrofluoric acid
  3. (C)Phosphoric acid
  4. (D)Hydrochloric acid

Correct answer: (A)

Step-by-step solution →
Q43·ChemistrySingle correct
Two isomers 'A' and 'B' with molecular formula C4H8C_4H_8C4​H8​ give different products on oxidation with KMnO4KMnO_4KMnO4​ in acidic medium. Isomer 'A' on reaction with KMnO4/H+KMnO_4/H^+KMnO4​/H+ results in effervescence of a gas and gives ketone. The compound 'A' is
  1. (A)But-1-ene
  2. (B)cis-But-2-ene
  3. (C)trans-But-2ene
  4. (D)2-methyl propene

Correct answer: (D)

Step-by-step solution →
Q44·ChemistrySingle correct
In the given conversion the compound A is:
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (B)

Step-by-step solution →
Q45·ChemistrySingle correct
Given below are two statements : Statement I : The esterification of carboxylic acid with an alcohol is a nucleophilic acyl substitution. Statement II : Electron withdrawing groups in the carboxylic acid will increase the rate of esterification reaction. Choose the most appropriate option :
  1. (A)Both Statement I and Statement II are correct.
  2. (B)Both Statement I and Statement II are incorrect.
  3. (C)Statement I is correct but Statement II is incorrect.
  4. (D)Statement I is incorrect but Statement II is correct.

Correct answer: (A)

Step-by-step solution →
Q46·ChemistrySingle correct
Consider the above reaction, the product A and product B respectively are
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (C)

Step-by-step solution →
Q47·ChemistrySingle correct
The polymer, which can be stretched and retains its original status on releasing the force is
  1. (A)Bakelite
  2. (B)Nylon 6,6
  3. (C)Buna-N
  4. (D)Terylene

Correct answer: (C)

Step-by-step solution →
Q48·ChemistrySingle correct
Sugar moiety in DNA and RNA molecules respectively are
  1. (A)β-D-2-deoxyribose, β -D-deoxyribose
  2. (B)β-D-2-deoxyribose, β -D-ribose
  3. (C)β-D-ribose, β -D-2-deoxyribose
  4. (D)β-D-deoxyribose, β -D-2-deoxyribose

Correct answer: (B)

Step-by-step solution →
Q49·ChemistrySingle correct
Which of the following compound does not contain sulphur atom ?
  1. (A)Cimetidine
  2. (B)Ranitidine
  3. (C)Histamine
  4. (D)Saccharin

Correct answer: (C)

Step-by-step solution →
Q50·ChemistrySingle correct
Given below are two statements. Statement I : Phenols are weakly acidic. Statement II : Therefore they are freely soluble in NaOH solution and are weaker acids than alcohols and water. Choose the most appropriate option:
  1. (A)Both Statement I and Statement II are correct.
  2. (B)Both Statement I and Statement II are incorrect.
  3. (C)Statement I is correct but Statement II is incorrect.
  4. (D)Statement I is incorrect but Statement II is correct.

Correct answer: (C)

Step-by-step solution →
Q51·ChemistryNumerical
Geraniol, a volatile organic compound, is a component of rose oil. The density of the vapour is 0.46 gL−1gL^{-1}gL−1 at 257°C and 100 mm Hg. The molar mass of geraniol is _______ (Nearest Integer) [Given R = 0.082 L atm K−1K^{-1}K−1 mol−1mol^{-1}mol−1]

Correct answer: 152

Step-by-step solution →
Q52·ChemistryNumerical
17.0 g of NH3NH_3NH3​ completely vapourises at − 33.42°C and 1 bar pressure and the enthalpy change in the process is 23.4 kJ mol−1mol^{-1}mol−1. The enthalpy change for the vapourisation of 85 g of NH3NH_3NH3​ under the same conditions is _______ kJ.

Correct answer: 117

Step-by-step solution →
Q53·ChemistryNumerical
1.2 mL of acetic acid is dissolved in water to make 2.0 L of solution. The depression in freezing point observed for this strength of acid is 0.0198°C. The percentage of dissociation of the acid is _______ . (Nearest integer) [Given : Density of acetic acid is 1.02 g mL−1mL^{-1}mL−1 Molar mass of acetic acid is 60 g mol−1mol^{-1}mol−1 KfK_fKf​ (H2OH_2OH2​O) = 1.85 K kg mol−1mol^{-1}mol−1]

Correct answer: 5

Step-by-step solution →
Q54·Chemistry·Redox Reactions and ElectrochemistryNumerical
A dilute solution of sulphuric acid is electrolysed using a current of 0.10 A for 2 hours to produce hydrogen and oxygen gas. The total volume of gases produced at STP is _______ cm3cm^3cm3. (Nearest integer) [Given : Faraday constant F = 96500 C mol−1mol^{-1}mol−1 at STP, molar volume of an ideal gas is 22.7 L mol−1mol^{-1}mol−1]

Correct answer: 127

Step-by-step solution →
Q55·ChemistryNumerical
The activation energy of one of the reactions in a biochemical process is 532611 J mol−1mol^{-1}mol−1. When the temperature falls from 310 K to 300 K, the change in rate constant observed is k300k_{300}k300​ = x × 10−310^{-3}10−3 k310k_{310}k310​. The value of x is _______ . [Given: ln10 = 2.3 R = 8.3 J K−1K^{-1}K−1 mol−1mol^{-1}mol−1]

Correct answer: 1

Step-by-step solution →
Q56·ChemistryNumerical
The number of terminal oxygen atoms present in the product B obtained from the following reaction is _______ . FeCr2O4FeCr_2O_4FeCr2​O4​ + Na2CO3Na_2CO_3Na2​CO3​ + O2O_2O2​ → A + Fe2O3Fe_2O_3Fe2​O3​ + CO2CO_2CO2​ A + H+H^+H+ → B + H2OH_2OH2​O + Na+Na^+Na+

Correct answer: 6

Step-by-step solution →
Q57·ChemistryNumerical
An acidified manganate solution undergoes disproportionation reaction. The spin-only magnetic moment value of the product having manganese in higher oxidation state is _______ B.M. (Nearest integer)

Correct answer: 0

Step-by-step solution →
Q58·ChemistryNumerical
Kjeldahl's method was used for the estimation of nitrogen in an organic compound. The ammonia evolved from 0.55 g of the compound neutralised 12.5 mL of 1 M H2SO4H_2SO_4H2​SO4​ solution. The percentage of nitrogen in the compound is _______ . (Nearest integer)

Correct answer: 64

Step-by-step solution →
Q59·Chemistry·IsomerismNumerical
Observe structures of the following compounds The total number of structures/compounds which possess asymmetric carbon atoms is _______ .

Correct answer: 3

Step-by-step solution →
Q60·ChemistryNumerical
C6H12O6C_6H_{12}O_6C6​H12​O6​ →[Zymase] A →[NaOI, Δ] B + CHI3CHI_3CHI3​ The number of carbon atoms present in the product B is _______ .

Correct answer: 1

Step-by-step solution →

Mathematics — JEE Main 29 June 2022 Shift 1

Q61·MathematicsSingle correct
The probability that a randomly chosen 2 × 2 matrix with all the entries from the set of first 10 primes, is singular, is equal to :
  1. (A)133104\frac{133}{10^{4}}104133​
  2. (B)18103\frac{18}{10^{3}}10318​
  3. (C)19103\frac{19}{10^{3}}10319​
  4. (D)271104\frac{271}{10^{4}}104271​

Correct answer: (C)

Step-by-step solution →
Q62·MathematicsSingle correct
Let the solution curve of the differential equation xdydx−y=y2+16x2x\frac{dy}{dx} - y = \sqrt{y^{2} + 16x^{2}}xdxdy​−y=y2+16x2​, y(1)=3y(1) = 3y(1)=3 be y=y(x)y = y(x)y=y(x). Then y(2)y(2)y(2) is equal to :
  1. (A)15
  2. (B)11
  3. (C)13
  4. (D)17

Correct answer: (A)

Step-by-step solution →
Q63·MathematicsSingle correct
If the mirror image of the point (2, 4, 7) in the plane 3x−y+4z=23x - y + 4z = 23x−y+4z=2 is (a, b, c), the 2a+b+2c2a + b + 2c2a+b+2c is equal to :
  1. (A)54
  2. (B)50
  3. (C)−6
  4. (D)−42

Correct answer: (C)

Step-by-step solution →
Q64·MathematicsSingle correct
Let f:R→Rf : R \to Rf:R→R be a function defined by : f(x)={max⁡t≤x{t3−3t} ;x≤2x2+2x−6 ;2<x<3[x−3]+9 ;3≤x≤52x+1 ;x>5f(x) = \begin{cases} \max\limits_{t \le x}\{t^{3} - 3t\}\ ; & x \le 2 \\ x^{2} + 2x - 6\ ; & 2 < x < 3 \\ [x - 3] + 9\ ; & 3 \le x \le 5 \\ 2x + 1\ ; & x > 5 \end{cases}f(x)=⎩⎨⎧​t≤xmax​{t3−3t} ;x2+2x−6 ;[x−3]+9 ;2x+1 ;​x≤22<x<33≤x≤5x>5​ Where [t] is the greatest integer less than or equal to t. Let m be the number of points where f is not differentiable and I=∫−22f(x)dxI = \int_{-2}^{2} f(x)dxI=∫−22​f(x)dx. Then the ordered pair (m, I) is equal to :
  1. (A)(3,274)\left(3, \frac{27}{4}\right)(3,427​)
  2. (B)(3,234)\left(3, \frac{23}{4}\right)(3,423​)
  3. (C)(4,274)\left(4, \frac{27}{4}\right)(4,427​)
  4. (D)(4,234)\left(4, \frac{23}{4}\right)(4,423​)

Correct answer: (C)

Step-by-step solution →
Q65·MathematicsSingle correct
Let a⃗=αi^+3j^−k^,b⃗=3i^−βj^+4k^\vec{a} = \alpha\hat{i} + 3\hat{j} - \hat{k}, \vec{b} = 3\hat{i} - \beta\hat{j} + 4\hat{k}a=αi^+3j^​−k^,b=3i^−βj^​+4k^ and c⃗=i^+2j^−2k^\vec{c} = \hat{i} + 2\hat{j} - 2\hat{k}c=i^+2j^​−2k^ where α, β ∈ R, be three vectors. If the projection of a⃗\vec{a}a on c⃗\vec{c}c is 103\frac{10}{3}310​ and b⃗×c⃗=−6i^+10j^+7k^\vec{b} \times \vec{c} = -6\hat{i} + 10\hat{j} + 7\hat{k}b×c=−6i^+10j^​+7k^, then the value of α + β equal to :
  1. (A)3
  2. (B)4
  3. (C)5
  4. (D)6

Correct answer: (A)

Step-by-step solution →
Q66·MathematicsSingle correct
The area enclosed by y2=8xy^{2} = 8xy2=8x and y=2 xy = \sqrt{2}\,xy=2​x that lies outside the triangle formed by y=2xy = \sqrt{2}xy=2​x, x=1x = 1x=1, y=22y = 2\sqrt{2}y=22​, is equal to :
  1. (A)1626\frac{16\sqrt{2}}{6}6162​​
  2. (B)1126\frac{11\sqrt{2}}{6}6112​​
  3. (C)1326\frac{13\sqrt{2}}{6}6132​​
  4. (D)526\frac{5\sqrt{2}}{6}652​​

Correct answer: (C)

Step-by-step solution →
Q67·Mathematics·Matrices and DeterminantsSingle correct
If the system of linear equations 2x+y−z=72x + y - z = 72x+y−z=7 x−3y+2z=1x - 3y + 2z = 1x−3y+2z=1 x+4y+δz=kx + 4y + \delta z = kx+4y+δz=k, where δ, k ∈ R has infinitely many solutions, then δ + k is equal to:
  1. (A)−3
  2. (B)3
  3. (C)6
  4. (D)9

Correct answer: (B)

Step-by-step solution →
Q68·MathematicsSingle correct
Let α and β be the roots of the equation x2+(2i−1)=0x^{2} + (2i - 1) = 0x2+(2i−1)=0. Then, the value of ∣α8+β8∣|\alpha^{8} + \beta^{8}|∣α8+β8∣ is equal to :
  1. (A)50
  2. (B)250
  3. (C)1250
  4. (D)1500

Correct answer: (A)

Step-by-step solution →
Q69·MathematicsSingle correct
Let Δ∈{∧,∨,⇒,⇔}\Delta \in \{\wedge, \vee, \Rightarrow, \Leftrightarrow\}Δ∈{∧,∨,⇒,⇔} be such that (p∧q)Δ((p∨q)⇒q)(p \wedge q)\Delta((p \vee q) \Rightarrow q)(p∧q)Δ((p∨q)⇒q) is a tautology. Then Δ is equal to :
  1. (A)∧\wedge∧
  2. (B)∨\vee∨
  3. (C)⇒\Rightarrow⇒
  4. (D)⇔\Leftrightarrow⇔

Correct answer: (C)

Step-by-step solution →
Q70·Mathematics·Matrices and DeterminantsSingle correct
Let A=[aij]A = [a_{ij}]A=[aij​] be a square matrix of order 3 such that aij=2 j−ia_{ij} = 2^{\,j-i}aij​=2j−i, for all i, j = 1, 2, 3. Then, the matrix A2+A3+…+A10A^{2} + A^{3} + \ldots + A^{10}A2+A3+…+A10 is equal to :
  1. (A)(310−32)A\left(\frac{3^{10} - 3}{2}\right)A(2310−3​)A
  2. (B)(310−12)A\left(\frac{3^{10} - 1}{2}\right)A(2310−1​)A
  3. (C)(310+12)A\left(\frac{3^{10} + 1}{2}\right)A(2310+1​)A
  4. (D)(310+32)A\left(\frac{3^{10} + 3}{2}\right)A(2310+3​)A

Correct answer: (A)

Step-by-step solution →
Q71·MathematicsSingle correct
Let a set A=A1∪A2∪…∪AkA = A_{1} \cup A_{2} \cup \ldots \cup A_{k}A=A1​∪A2​∪…∪Ak​, where Ai∩Aj=ϕA_{i} \cap A_{j} = \phiAi​∩Aj​=ϕ for i≠ji \ne ji=j 1≤i,j≤k1 \le i, j \le k1≤i,j≤k. Define the relation R from A to A by R={(x,y):y∈AiR = \{(x, y) : y \in A_{i}R={(x,y):y∈Ai​ if and only if x∈Ai,1≤i≤k}x \in A_{i}, 1 \le i \le k\}x∈Ai​,1≤i≤k}. Then, R is :
  1. (A)reflexive, symmetric but not transitive
  2. (B)reflexive, transitive but not symmetric
  3. (C)reflexive but not symmetric and transitive
  4. (D)an equivalence relation

Correct answer: (D)

Step-by-step solution →
Q72·MathematicsSingle correct
Let {an}n=0∞\{a_{n}\}_{n=0}^{\infty}{an​}n=0∞​ be a sequence such that a0=a1=0a_{0} = a_{1} = 0a0​=a1​=0 and an+2=2an+1−an+1a_{n+2} = 2a_{n+1} - a_{n} + 1an+2​=2an+1​−an​+1 for all n≥0n \ge 0n≥0. Then, ∑n=2∞an7n\sum_{n=2}^{\infty}\frac{a_{n}}{7^{n}}∑n=2∞​7nan​​ is equal to
  1. (A)6343\frac{6}{343}3436​
  2. (B)7216\frac{7}{216}2167​
  3. (C)8343\frac{8}{343}3438​
  4. (D)49216\frac{49}{216}21649​

Correct answer: (B)

Step-by-step solution →
Q73·MathematicsSingle correct
The distance between the two points A and A′A'A′ which lie on y=2y = 2y=2 such that both the line segments AB and A′BA'BA′B (where B is the point (2, 3)) subtend angle π4\frac{\pi}{4}4π​ at the origin, is equal to :
  1. (A)10
  2. (B)485\frac{48}{5}548​
  3. (C)525\frac{52}{5}552​
  4. (D)3

Correct answer: (C)

Step-by-step solution →
Q74·MathematicsSingle correct
A wire of length 22 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into an equilateral triangle. Then, the length of the side of the equilateral triangle, so that the combined area of the square and the equilateral triangle is minimum, is :
  1. (A)229+43\frac{22}{9 + 4\sqrt{3}}9+43​22​
  2. (B)669+43\frac{66}{9 + 4\sqrt{3}}9+43​66​
  3. (C)224+93\frac{22}{4 + 9\sqrt{3}}4+93​22​
  4. (D)664+93\frac{66}{4 + 9\sqrt{3}}4+93​66​

Correct answer: (B)

Step-by-step solution →
Q75·MathematicsSingle correct
The domain of the function cos⁡−1(2sin⁡−1(14x2−1)π)\cos^{-1}\left(\frac{2\sin^{-1}\left(\frac{1}{4x^{2} - 1}\right)}{\pi}\right)cos−1(π2sin−1(4x2−11​)​) is :
  1. (A)R−{−12,12}R - \left\{-\frac{1}{2}, \frac{1}{2}\right\}R−{−21​,21​}
  2. (B)(−∞,−1]∪[1,∞)∪{0}(-\infty, -1] \cup [1, \infty) \cup \{0\}(−∞,−1]∪[1,∞)∪{0}
  3. (C)(−∞,−12)∪(12,∞)∪{0}\left(-\infty, \frac{-1}{2}\right) \cup \left(\frac{1}{2}, \infty\right) \cup \{0\}(−∞,2−1​)∪(21​,∞)∪{0}
  4. (D)(−∞,−12]∪[12,∞)∪{0}\left(-\infty, \frac{-1}{\sqrt{2}}\right] \cup \left[\frac{1}{\sqrt{2}}, \infty\right) \cup \{0\}(−∞,2​−1​]∪[2​1​,∞)∪{0}

Correct answer: (D)

Step-by-step solution →
Q76·MathematicsSingle correct
If the constant term in the expansion of (3x3−2x2+5x5)10\left(3x^3 - 2x^2 + \frac{5}{x^5}\right)^{10}(3x3−2x2+x55​)10 is 2k⋅l2^k \cdot l2k⋅l, where lll is an odd integer, then the value of k is equal to :
  1. (A)6
  2. (B)7
  3. (C)8
  4. (D)9

Correct answer: (D)

Step-by-step solution →
Q77·MathematicsSingle correct
∫05cos⁡(π(x−[x2]))dx\int_{0}^{5} \cos\left(\pi\left(x - \left[\frac{x}{2}\right]\right)\right) dx∫05​cos(π(x−[2x​]))dx, Where [t] denotes greatest integer less than or equal to t, is equal to :
  1. (A)−3
  2. (B)−2
  3. (C)2
  4. (D)0

Correct answer: (D)

Step-by-step solution →
Q78·MathematicsSingle correct
Let PQ be a focal chord of the parabola y2=4xy^2 = 4xy2=4x such that it subtends an angle of π2\frac{\pi}{2}2π​ at the point (3, 0). Let the line segment PQ be also a focal chord of the ellipse E: x2a2+y2b2=1,a2>b2\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, a^2 > b^2a2x2​+b2y2​=1,a2>b2. If e is the eccentricity of the ellipse E, then the value of 1e2\frac{1}{e^2}e21​ is equal to :
  1. (A)1+21 + \sqrt{2}1+2​
  2. (B)3+223 + 2\sqrt{2}3+22​
  3. (C)1+231 + 2\sqrt{3}1+23​
  4. (D)4+534 + 5\sqrt{3}4+53​

Correct answer: (B)

Step-by-step solution →
Q79·MathematicsSingle correct
Let the tangent to the circle C1:x2+y2=2C_1 : x^2 + y^2 = 2C1​:x2+y2=2 at the point M(−1,1)M(-1, 1)M(−1,1) intersect the circle C2:(x−3)2+(y−2)2=5C_2 : (x - 3)^2 + (y - 2)^2 = 5C2​:(x−3)2+(y−2)2=5, at two distinct points A and B. If the tangents to C2C_2C2​ at the points A and B intersect at N, then the area of the triangle ANB is equal to :
  1. (A)12\frac{1}{2}21​
  2. (B)23\frac{2}{3}32​
  3. (C)16\frac{1}{6}61​
  4. (D)53\frac{5}{3}35​

Correct answer: (C)

Step-by-step solution →
Q80·MathematicsSingle correct
Let the mean and the variance of 5 observations x1,x2,x3,x4,x5x_1, x_2, x_3, x_4, x_5x1​,x2​,x3​,x4​,x5​ be 245\frac{24}{5}524​ and 19425\frac{194}{25}25194​ respectively. If the mean and variance of the first 4 observation are 72\frac{7}{2}27​ and aaa respectively, then (4a+x5)(4a + x_5)(4a+x5​) is equal to:
  1. (A)13
  2. (B)15
  3. (C)17
  4. (D)18

Correct answer: (B)

Step-by-step solution →
Q81·MathematicsNumerical
Let S={z∈C:∣z−2∣≤1,z(1+i)+zˉ(1−i)≤2}S = \{z \in C : |z - 2| \leq 1, z(1 + i) + \bar{z}(1 - i) \leq 2\}S={z∈C:∣z−2∣≤1,z(1+i)+zˉ(1−i)≤2}. Let ∣z−4i∣|z - 4i|∣z−4i∣ attains minimum and maximum values, respectively, at z1∈Sz_1 \in Sz1​∈S and z2∈Sz_2 \in Sz2​∈S. If 5(∣z1∣2+∣z2∣2)=α+β55(|z_1|^2 + |z_2|^2) = \alpha + \beta\sqrt{5}5(∣z1​∣2+∣z2​∣2)=α+β5​, where α and β are integers, then the value of α + β is equal to _____.

Correct answer: 26

Step-by-step solution →
Q82·MathematicsNumerical
Let y = y(x) be the solution of the differential equation dydx+2 y2cos⁡4x−cos⁡2x=xetan⁡−1(2cot⁡2x),0<x<π/2\frac{dy}{dx} + \frac{\sqrt{2}\,y}{2\cos^4 x - \cos 2x} = x e^{\tan^{-1}(\sqrt{2}\cot 2x)}, 0 < x < \pi/2dxdy​+2cos4x−cos2x2​y​=xetan−1(2​cot2x),0<x<π/2 with y(π4)=π232y\left(\frac{\pi}{4}\right) = \frac{\pi^2}{32}y(4π​)=32π2​. If y(π3)=π218e−tan⁡−1(α)y\left(\frac{\pi}{3}\right) = \frac{\pi^2}{18} e^{-\tan^{-1}(\alpha)}y(3π​)=18π2​e−tan−1(α), then the value of 3α23\alpha^23α2 is equal to _____.

Correct answer: 2

Step-by-step solution →
Q83·MathematicsNumerical
The number of elements in the set S={θ∈[−4π,4π]:3cos⁡22θ+6cos⁡2θ−10cos⁡2θ+5=0}S = \{\theta \in [-4\pi, 4\pi] : 3\cos^2 2\theta + 6\cos 2\theta - 10\cos^2\theta + 5 = 0\}S={θ∈[−4π,4π]:3cos22θ+6cos2θ−10cos2θ+5=0} is ________.

Correct answer: 32

Step-by-step solution →
Q84·MathematicsNumerical
The number of solutions of the equation 2θ−cos⁡2θ+2=02\theta - \cos^2\theta + \sqrt{2} = 02θ−cos2θ+2​=0 is R is equal to ______.

Correct answer: 1

Step-by-step solution →
Q85·MathematicsNumerical
50tan⁡(3tan⁡−1(12)+2cos⁡−1(15))+42tan⁡(12tan⁡−1(22))50\tan\left(3\tan^{-1}\left(\frac{1}{2}\right) + 2\cos^{-1}\left(\frac{1}{\sqrt{5}}\right)\right) + 4\sqrt{2}\tan\left(\frac{1}{2}\tan^{-1}(2\sqrt{2})\right)50tan(3tan−1(21​)+2cos−1(5​1​))+42​tan(21​tan−1(22​)) is equal to _____.

Correct answer: 29

Step-by-step solution →
Q86·MathematicsNumerical
Let c, k ∈ R. If f(x)=(c+1)x2+(1−c2)x+2kf(x) = (c + 1)x^2 + (1 - c^2)x + 2kf(x)=(c+1)x2+(1−c2)x+2k and f(x+y)=f(x)+f(y)−xyf(x + y) = f(x) + f(y) - xyf(x+y)=f(x)+f(y)−xy, for all x, y ∈ R, then the value of ∣2(f(1)+f(2)+f(3)+…+f(20))∣|2(f(1) + f(2) + f(3) + \ldots + f(20))|∣2(f(1)+f(2)+f(3)+…+f(20))∣ is equal to ________.

Correct answer: 3395

Step-by-step solution →
Q87·MathematicsNumerical
Let H:x2a2−y2b2=1H : \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1H:a2x2​−b2y2​=1, a > 0, b > 0, be a hyperbola such that the sum of lengths of the transverse and the conjugate axes is 4(22+14)4\left(2\sqrt{2} + \sqrt{14}\right)4(22​+14​). If the eccentricity H is 112\frac{\sqrt{11}}{2}211​​, then value of a2+b2a^2 + b^2a2+b2 is equal to __________.

Correct answer: 88

Step-by-step solution →
Q88·MathematicsNumerical
Let P1:r⃗⋅(2i^+j^−3k^)=4P_1 : \vec{r} \cdot (2\hat{i} + \hat{j} - 3\hat{k}) = 4P1​:r⋅(2i^+j^​−3k^)=4 be a plane. Let P2P_2P2​ be another plane which passes through the points (2, -3, 2) (2, − 2, − 3) and (1, −4, 2). If the direction ratios of the line of intersection of P1P_1P1​ and P2P_2P2​ be 16, α, β, then the value of α + β is equal to _____.

Correct answer: 28

Step-by-step solution →
Q89·MathematicsNumerical
Let b1b2b3b4b_1b_2b_3b_4b1​b2​b3​b4​ be a 4-element permutation with bi∈{1,2,3,…,100}b_i \in \{1, 2, 3, \ldots, 100\}bi​∈{1,2,3,…,100} for 1≤i≤41 \leq i \leq 41≤i≤4 and bi≠bjb_i \neq b_jbi​=bj​ for i≠ji \neq ji=j, such that either b1,b2,b3b_1, b_2, b_3b1​,b2​,b3​ are consecutive integers or b2,b3,b4b_2, b_3, b_4b2​,b3​,b4​ are consecutive integers. Then the number of such permutations b1b2b3b4b_1b_2b_3b_4b1​b2​b3​b4​ is equal to ________.

Correct answer: 18915

Step-by-step solution →

Chapters tested in this paper

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