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JEE Main 28 July 2022 Shift 1 Question Paper with Answers

28 July 2022 · July session · 89 questions

89 of the 90 questions from the JEE Main 28 July 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 28 July 2022 Shift 1

Q1·PhysicsSingle correct
The dimensions of (B2μ0)\left(\frac{B^{2}}{\mu_{0}}\right)(μ0​B2​) will be : (if μ0_{0}0​ : permeability of free space and B : magnetic field)
  1. (A)[M L2^{2}2 T−2^{-2}−2]
  2. (B)[M L T−2^{-2}−2]
  3. (C)[M L−1^{-1}−1 T−2^{-2}−2]
  4. (D)[M L2^{2}2 T−2^{-2}−2 A−1^{-1}−1]

Correct answer: (C)

Step-by-step solution →
Q2·PhysicsSingle correct
A NCC parade is going at a uniform speed of 9 km/h under a mango tree on which a monkey is sitting at a height of 19.6 m. At any particular instant, the monkey drops a mango. A cadet will receive the mango whose distance from the tree at time of drop is : (Given g = 9.8 m/s2^{2}2)
  1. (A)5 m
  2. (B)10 m
  3. (C)19.8 m
  4. (D)24.5 m

Correct answer: (A)

Step-by-step solution →
Q3·PhysicsSingle correct
In two different experiments, an object of mass 5 kg moving with a speed of 25 ms−1^{-1}−1 hits two different walls and comes to rest within (i) 3 second, (ii) 5 seconds, respectively. Choose the correct option out of the following :
  1. (A)Impulse and average force acting on the object will be same for both the cases.
  2. (B)Impulse will be same for both the cases but the average force will be different.
  3. (C)Average force will be same for both the cases but the impulse will be different.
  4. (D)Average force and impulse will be different for both the cases.

Correct answer: (B)

Step-by-step solution →
Q4·PhysicsSingle correct
A balloon has mass of 10 g in air. The air escapes from the balloon at a uniform rate with velocity 4.5 cm/s. If the balloon shrinks in 5 s completely. Then, the average force acting on that balloon will be (in dyne).
  1. (A)3
  2. (B)9
  3. (C)12
  4. (D)18

Correct answer: (B)

Step-by-step solution →
Q5·PhysicsSingle correct
If the radius of earth shrinks by 2% while its mass remains same. The acceleration due to gravity on the earth's surface will approximately :
  1. (A)decrease by 2%
  2. (B)decrease by 4%
  3. (C)increase by 2%
  4. (D)increase by 4%

Correct answer: (D)

Step-by-step solution →
Q6·PhysicsSingle correct
The force required to stretch a wire of cross-section 1 cm2^{2}2 to double its length will be : (Given Yong's modulus of the wire = 2 × 1011^{11}11 N/m2^{2}2)
  1. (A)1 × 107^{7}7 N
  2. (B)1.5 × 107^{7}7 N
  3. (C)2 × 107^{7}7 N
  4. (D)2.5 × 107^{7}7 N

Correct answer: (C)

Step-by-step solution →
Q7·PhysicsSingle correct
A Carnot engine has efficiency of 50%. If the temperature of sink is reduced by 40°C, its efficiency increases by 30%. The temperature of the source will be :
  1. (A)166.7 K
  2. (B)255.1 K
  3. (C)266.7 K
  4. (D)367.7 K

Correct answer: (C)

Step-by-step solution →
Q8·PhysicsSingle correct
Given below are two statements : Statement I :The average momentum of a molecule in a sample of an ideal gas depends on temperature. Statement II : The rms speed of oxygen molecules in a gas is v. If the temperature is doubled and the oxygen molecules dissociate into oxygen atoms, the rms speed will become 2v. In the light of the above statements, choose the correct answer from the options given below :
  1. (A)Both Statement I and Statement II are true
  2. (B)Both Statement I and Statement II are false
  3. (C)Statement I is true but Statement II is false
  4. (D)Statement I is false but Statement II is true

Correct answer: (D)

Step-by-step solution →
Q9·PhysicsSingle correct
In the wave equation y=0.5sin⁡2πλ(400t−x)my = 0.5\sin\frac{2\pi}{\lambda}\left(400t - x\right)my=0.5sinλ2π​(400t−x)m the velocity of the wave will be :
  1. (A)200 m/s
  2. (B)2002\sqrt{2}2​ m / s
  3. (C)400 m/s
  4. (D)4002\sqrt{2}2​ m / s

Correct answer: (C)

Step-by-step solution →
Q10·PhysicsSingle correct
Two capacitors, each having capacitance 40 μF are connected in series. The space between one of the capacitors is filled with dielectric material of dielectric constant K such that the equivalence capacitance of the system became 24 μF. The value of K will be :
  1. (A)1.5
  2. (B)2.5
  3. (C)1.2
  4. (D)3

Correct answer: (A)

Step-by-step solution →
Q11·PhysicsSingle correct
A wire of resistance R1_{1}1​ is drawn out so that its length is increased by twice of its original length. The ratio of new resistance to original resistance is:
  1. (A)9 : 1
  2. (B)1 : 9
  3. (C)4 : 1
  4. (D)3 : 1

Correct answer: (A)

Step-by-step solution →
Q12·Physics·Magnetic Field of CurrentSingle correct
The current sensitivity of a galvanometer can be increased by : (A) decreasing the number of turns (B) increasing the magnetic field (C) decreasing the area of the coil (D) decreasing the torsional constant of the spring Choose the most appropriate answer from the options given below :
  1. (A)(B) and (C) only
  2. (B)(C) and (D) only
  3. (C)(A) and (C) only
  4. (D)(B) and (D) only

Correct answer: (D)

Step-by-step solution →
Q13·Physics·Magnetic Field of CurrentSingle correct
As shown in the figure, a metallic rod of linear density 0.45 kg m−1^{-1}−1 is lying horizontally on a smooth incline plane which makes an angle of 45° with the horizontal. The minimum current flowing in the rod required to keep it stationary, when 0.15 T magnetic field is acting on it in the vertical upward direction, will be : {Use g = 10 m/s2^{2}2}
  1. (A)30 A
  2. (B)15 A
  3. (C)10 A
  4. (D)3 A

Correct answer: (A)

Step-by-step solution →
Q14·PhysicsSingle correct
The equation of current in a purely inductive circuit is 5sin⁡(49πt−30°)5\sin\left(49\pi t - 30°\right)5sin(49πt−30°). If the inductance is 30 mH then the equation for the voltage across the inductor, will be : {Let π=227}\left\{\text{Let } \pi = \frac{22}{7}\right\}{Let π=722​}
  1. (A)1.47sin⁡(49πt−30°)1.47\sin(49\pi t - 30°)1.47sin(49πt−30°)
  2. (B)1.47sin⁡(49πt+60°)1.47\sin(49\pi t + 60°)1.47sin(49πt+60°)
  3. (C)23.1sin⁡(49πt−30°)23.1\sin(49\pi t - 30°)23.1sin(49πt−30°)
  4. (D)23.1sin⁡(49πt+60°)23.1\sin(49\pi t + 60°)23.1sin(49πt+60°)

Correct answer: (D)

Step-by-step solution →
Q15·PhysicsSingle correct
As shown in the figure, after passing through the medium 1. The speed of light v2_{2}2​ in medium 2 will be : (Given c = 3 × 108^{8}8 ms−1^{-1}−1)
  1. (A)1.0 × 108^{8}8 ms−1^{-1}−1
  2. (B)0.5 × 108^{8}8 ms−1^{-1}−1
  3. (C)1.5 × 108^{8}8 ms−1^{-1}−1
  4. (D)3.0 × 108^{8}8 ms−1^{-1}−1

Correct answer: (A)

Step-by-step solution →
Q16·PhysicsSingle correct
In normal adjustment, for a refracting telescope, the distance between objective and eye piece is 30 cm. The focal length of the objective, when the angular magnification of the telescope is 2, will be:
  1. (A)20 cm
  2. (B)30 cm
  3. (C)10 cm
  4. (D)15 cm

Correct answer: (A)

Step-by-step solution →
Q17·PhysicsSingle correct
The equation λ=1.227x\lambda = \frac{1.227}{x}λ=x1.227​ nm can be used to find the de-Brogli wavelength of an electron. In this equation x stands for : Where, m = mass of electron P = momentum of electron K = Kinetic energy of electron V = Accelerating potential in volts for electron
  1. (A)mK\sqrt{mK}mK​
  2. (B)P\sqrt{P}P​
  3. (C)K\sqrt{K}K​
  4. (D)V\sqrt{V}V​

Correct answer: (D)

Step-by-step solution →
Q18·PhysicsSingle correct
The half life period of a radioactive substance is 60 days. The time taken for 78\frac{7}{8}87​th of its original mass to disintegrate will be :
  1. (A)120 days
  2. (B)130 days
  3. (C)180 days
  4. (D)20 days

Correct answer: (C)

Step-by-step solution →
Q19·PhysicsSingle correct
Identify the solar cell characteristics from the following options :
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (B)

Step-by-step solution →
Q20·PhysicsSingle correct
In the case of amplitude modulation to avoid distortion the modulation index (μ) should be :
  1. (A)μ ≤ 1
  2. (B)μ ≥ 1
  3. (C)μ = 2
  4. (D)μ = 0

Correct answer: (A)

Step-by-step solution →
Q21·PhysicsNumerical
If the projection of 2i^+4j^−2k^2\hat{i}+4\hat{j}-2\hat{k}2i^+4j^​−2k^ on i^+2j^+αk^\hat{i}+2\hat{j}+\alpha\hat{k}i^+2j^​+αk^ is zero. Then, the value of α will be

Correct answer: 5

Step-by-step solution →
Q22·PhysicsNumerical
A freshly prepared radioactive source of half life 2 hours 30 minutes emits radiation which is 64 times the permissible safe level. The minimum time, after which it would be possible to work safely with source, will be ________ hours.

Correct answer: 15

Step-by-step solution →
Q23·PhysicsNumerical
In a Young's double slit experiment, a laser light of 560 nm produces an interference pattern with consecutive bright fringes' separation of 7.2 mm. Now another light is used to produce an interference pattern with consecutive bright fringes' separation of 8.1 mm. The wavelength of second light is ________ nm.

Correct answer: 630

Step-by-step solution →
Q24·PhysicsNumerical
The frequencies at which the current amplitude in an LCR series circuit becomes 12\frac{1}{\sqrt{2}}2​1​ times its maximum value, are 212 rad s−1^{-1}−1 and 232 rad s−1^{-1}−1. The value of resistance in the circuit is R = 5Ω. The self inductance in the circuit is ________ mH.

Correct answer: 250

Step-by-step solution →
Q25·PhysicsNumerical
As shown in the figure, a potentiometer wire of resistance 20Ω and length 300 cm is connected with resistance box (R.B.) and a standard cell of emf 4 V. For a resistance 'R' of resistance box introduced into the circuit, the null point for a cell of 20 mV is found to be 60 cm. The value of 'R' is ________ Ω .

Correct answer: 780

Step-by-step solution →
Q26·Physics·Electric Field and Coulomb's LawNumerical
Two electric dipoles of dipole moments 1.2 × 10−30^{-30}−30 cm and 2.4 × 10−30^{-30}−30 cm are placed in two difference uniform electric fields of strengths 5 × 104^{4}4 NC−1^{-1}−1 and 15 × 104^{4}4 NC−1^{-1}−1 respectively. The ratio of maximum torque experienced by the electric dipoles will be 1x\frac{1}{x}x1​. The value of x is ________.

Correct answer: 6

Step-by-step solution →
Q27·PhysicsNumerical
The frequency of echo will be ________ Hz if the train blowing a whistle of frequency 320 Hz is moving with a velocity of 36 km/h towards a hill from which an echo is heard by the train driver. Velocity of sound in air is 330 m/s.

Correct answer: 340

Step-by-step solution →
Q28·PhysicsNumerical
The diameter of an air bubble which was initially 2 mm, rises steadily through a solution of density 1750 kg m−3^{-3}−3 at the rate of 0.35 cms−1^{-1}−1. The coefficient of viscosity of the solution is ________ poise (in nearest integer). (the density of air is negligible).

Correct answer: 11

Step-by-step solution →
Q29·PhysicsNumerical
A block of mass 'm' (as shown in figure) moving with kinetic energy E compresses a spring through a distance 25 cm when, its speed is halved. The value of spring constant of used spring will be nE Nm−1^{-1}−1 for n = ________.

Correct answer: 24

Step-by-step solution →
Q30·PhysicsNumerical
Four identical discs each of mass 'M' and diameter 'a' are arranged in a small plane as shown in figure. If the moment of inertia of the system about OO' is x4\frac{x}{4}4x​Ma2^{2}2. Then, the value of x will be ________.

Correct answer: 3

Step-by-step solution →

Chemistry — JEE Main 28 July 2022 Shift 1

Q31·ChemistrySingle correct
Identify the incorrect statement from the following.
  1. (A)A circular path around the nucleus in which an electron moves is proposed as Bohr's orbit.
  2. (B)An orbital is the one electron wave function (Ψ) in an atom.
  3. (C)The existence of Bohr's orbits is supported by hydrogen spectrum.
  4. (D)Atomic orbital is characterised by the quantum numbers n and lll only

Correct answer: (D)

Step-by-step solution →
Q32·ChemistrySingle correct
Which of the following relation is not correct ?
  1. (A)ΔH = ΔU − PΔV
  2. (B)ΔU = q + W
  3. (C)ΔSsys_{sys}sys​ + ΔSsurr_{surr}surr​ ≥ 0
  4. (D)ΔG = ΔH − TΔS

Correct answer: (A)

Step-by-step solution →
Q33·ChemistrySingle correct
Match List-I with List-II. Choose the correct answer from the options given below :
List-IList-II
A.Cd(s) + 2Ni(OH)3_33​(s) → CdO(s) + 2Ni(OH)2_22​(s) + H2_22​O(lll)I.Primary battery
B.Zn(Hg) + HgO(s) → ZnO(s) + Hg(lll)II.Discharging of secondary battery
C.2PbSO4_44​(s) + 2H2_22​O(lll) → Pb(s) + PbO2_22​(s) + 2H2_22​SO4_44​(aq)III.Fuel cell
D.2H2_22​(g) + O2_22​(g) → 2H2_22​O(lll)IV.Charging of secondary battery
  1. (A)(A) – (I), (B) – (II), (C) – (III), (D) – (IV)
  2. (B)(A) – (IV), (B) – (I), (C) – (II), (D) – (III)
  3. (C)(A) – (II), (B) – (I), (C) – (IV), (D) – (III)
  4. (D)(A) – (II), (B) – (I), (C) – (III), (D) – (IV)

Correct answer: (C)

Step-by-step solution →
Q34·ChemistrySingle correct
Match List-I with List-II. Choose the correct answer from the options given below :
List-I (Reaction)List-II (Catalyst)
A.4NH3_33​(g) + 5O2_22​(g) → 4NO(g) + 6H2_22​O(g)I.NO(g)
B.N2_22​(g) + 3H2_22​(g) → 2NH3_33​(g)II.H2_22​SO4_44​(lll)
C.C12_{12}12​H22_{22}22​O11_{11}11​(aq) + H2_22​O(lll) → C6_66​H12_{12}12​O6_66​ (Glucose) + C6_66​H12_{12}12​O6_66​ (Fructose)III.Pt(s)
D.2SO2_22​(g) + O2_22​(g) → 2SO3_33​(g)IV.Fe(s)
  1. (A)(A) – (II), (B) – (III), (C) – (I), (D) – (IV)
  2. (B)(A) – (III), (B) – (II), (C) – (I), (D) – (IV)
  3. (C)(A) – (III), (B) – (IV), (C) – (II), (D) – (I)
  4. (D)(A) – (III), (B) – (II), (C) – (IV), (D) – (I)

Correct answer: (C)

Step-by-step solution →
Q35·ChemistrySingle correct
In which of the following pairs, electron gain enthalpies of constituent elements are nearly the same or identical ? (A) Rb and Cs (B) Na and K (C) Ar and Kr (D) I and At Choose the correct answer from the options given below :
  1. (A)(A) and (B) only
  2. (B)(B) and (C) only
  3. (C)(A) and (C) only
  4. (D)(C) and (D) only

Correct answer: (C)

Step-by-step solution →
Q36·ChemistrySingle correct
Which of the reaction is suitable for concentrating ore by leaching process ?
  1. (A)2Cu2_22​S + 3O2_22​ → 2Cu2_22​O + 2SO2_22​
  2. (B)Fe3_33​O4_44​ + CO → 3FeO + CO2_22​
  3. (C)Al2_22​O3_33​ + 2NaOH + 3H2_22​O → 2Na[Al(OH)4_44​]
  4. (D)Al2_22​O3_33​ + 6Mg → 6MgO + 4Al

Correct answer: (C)

Step-by-step solution →
Q37·ChemistrySingle correct
The metal salts formed during softening of hardwater using Clark's method are :
  1. (A)Ca(OH)2_22​ and Mg(OH)2_22​
  2. (B)CaCO3_33​ and Mg(OH)2_22​
  3. (C)Ca(OH)2_22​ and MgCO3_33​
  4. (D)CaCO3_33​ and MgCO3_33​

Correct answer: (B)

Step-by-step solution →
Q38·ChemistrySingle correct
Which of the following statement is incorrect ?
  1. (A)Low solubility of LiF in water is due to its small hydration enthalpy.
  2. (B)KO2_22​ is paramagnetic.
  3. (C)Solution of sodium in liquid ammonia is conducting in nature.
  4. (D)Sodium metal has higher density than potassium metal

Correct answer: (A)

Step-by-step solution →
Q39·ChemistrySingle correct
Match List-I with List-II, match the gas evolved during each reaction. Choose the correct answer from the options given below :
List-IList-II
A.(NH4_44​)2_22​Cr2_22​O7_77​ →Δ\xrightarrow{\Delta}Δ​I.H2_22​
B.KMnO4_44​ + HCl →II.N2_22​
C.Al + NaOH + H2_22​O →III.O2_22​
D.NaNO3_33​ →Δ\xrightarrow{\Delta}Δ​IV.Cl2_22​
  1. (A)(A) – (II), (B) – (III), (C) – (I), (D) – (IV)
  2. (B)(A) – (III), (B) – (I), (C) – (IV), (D) – (II)
  3. (C)(A) – (II), (B) – (IV), (C) – (I), (D) – (III)
  4. (D)(A) – (III), (B) – (IV), (C) – (I), (D) – (II)

Correct answer: (C)

Step-by-step solution →
Q40·ChemistrySingle correct
Which of the following has least tendency to liberate H2_22​ from mineral acids ?
  1. (A)Cu
  2. (B)Mn
  3. (C)Ni
  4. (D)Zn

Correct answer: (A)

Step-by-step solution →
Q41·ChemistrySingle correct
Given below are two statements : Statement I : In polluted water values of both dissolved oxygen and BOD are very low. Statement II : Eutrophication results in decrease in the amount of dissolved oxygen. 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)Both Statement I and Statement II are false
  3. (C)Statement I is true but Statement II is false
  4. (D)Statement I is false but Statement II is true

Correct answer: (D)

Step-by-step solution →
Q42·ChemistrySingle correct
Match List-I with List-II. Choose the correct answer from the options given below :
  1. (A)(A) – (II), (B) – (I), (C) – (IV), (D) – (III)
  2. (B)(A) – (IV), (B) – (III), (C) – (I), (D) – (II)
  3. (C)(A) – (III), (B) – (IV), (C) – (I), (D) – (II)
  4. (D)(A) – (IV), (B) – (III), (C) – (II), (D) – (I)

Correct answer: (C)

Step-by-step solution →
Q43·ChemistrySingle correct
Choose the correct option for the following reactions.
  1. (A)'A' and 'B' are both Markovnikov addition products.
  2. (B)'A' is Markovnikov product and 'B' is anti-Markovnikov product.
  3. (C)'A' and 'B' are both anti-Markovnikov products.
  4. (D)'B' is Markovnikov and 'A' is anti-Markovnikov product.

Correct answer: (B)

Step-by-step solution →
Q44·Chemistry·Electronic Effects and StabilitySingle correct
Among the following marked proton of which compound shows lowest pKa_aa​ value ?
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (C)

Step-by-step solution →
Q45·ChemistrySingle correct
Identify the major product A and B for the below given reaction sequence.
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (B)

Step-by-step solution →
Q46·ChemistrySingle correct
Identify the correct statement for the below given transformation.
  1. (A)A - CH3_33​CH2_22​CH = CH–CH3_33​, B – CH3_33​CH2_22​CH2_22​CH = CH2_22​, Saytzeff products
  2. (B)A - CH3_33​CH2_22​CH = CH–CH3_33​, B – CH3_33​CH2_22​CH2_22​CH = CH2_22​, Hafmann products
  3. (C)A - CH3_33​CH2_22​CH2_22​CH = CH2_22​, B – CH3_33​CH2_22​CH = CHCH3_33​, Hofmann products
  4. (D)A - CH3_33​CH2_22​CH2_22​CH = CH2_22​, B – CH3_33​CH2_22​CH = CHCH3_33​, Saytzeff products

Correct answer: (C)

Step-by-step solution →
Q47·ChemistrySingle correct
Terylene polymer is obtained by condensation of :
  1. (A)Ethane-1, 2-diol and Benzene-1, 3 dicarboxylic acid
  2. (B)Propane-1, 2-diol and Benzene-1, 4 dicarboxylic acid
  3. (C)Ethane-1, 2-diol and Benzene-1, 4 dicarboxylic acid
  4. (D)Ethane-1, 2-diol and Benzene-1, 2 dicarboxylic acid

Correct answer: (C)

Step-by-step solution →
Q48·ChemistrySingle correct
For the below given cyclic hemiacetal (X), the correct pyranose structure is :
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (D)

Step-by-step solution →
Q49·ChemistrySingle correct
Statements about Enzyme Inhibitor Drugs are given below : (A) There are Competitive and Non-competitive inhibitor drugs. (B) These can bind at the active sites and allosteric sites. (C) Competitive Drugs are allosteric site blocking drugs. (D) Non-competitive Drugs are active site blocking drugs. Choose the correct answer from the options given below :
  1. (A)(A), (D) only
  2. (B)(A), (C) only
  3. (C)(A), (B) only
  4. (D)(A), (B), (C) only

Correct answer: (C)

Step-by-step solution →
Q50·ChemistrySingle correct
For kinetic study of the reaction of iodide ion with H2_22​O2_22​ at room temperature : (A) Always use freshly prepared starch solution. (B) Always keep the concentration of sodium thiosulphate solution less than that of KI solution. (C) Record the time immediately after the appearance of blue colour. (D) Record the time immediately before the appearance of blue colour. (E) Always keep the concentration of sodium thiosulphate solution more than that of KI solution. Choose the correct answer from the options given below :
  1. (A)(A), (B), (C) only
  2. (B)(A), (D), (E) only
  3. (C)(D), (E) only
  4. (D)(A), (B), (E) only

Correct answer: (A)

Step-by-step solution →
Q51·ChemistryNumerical
In the given reaction, X + Y + 3Z ⇌ XYZ3_33​ if one mole of each of X and Y with 0.05 mol of Z gives compound XYZ3_33​. (Given : Atomic masses of X, Y and Z are 10, 20 and 30 amu, respectively). The yield of XYZ3_33​ is __________ g. (Nearest integer)

Correct answer: 2

Step-by-step solution →
Q52·ChemistryNumerical
An element M crystallises in a body centred cubic unit cell with a cell edge of 300 pm. The density of the element is 6.0 g cm−3^{-3}−3. The number of atoms present in 180 g of the element is _______ × 1023^{23}23. (Nearest integer)

Correct answer: 22

Step-by-step solution →
Q53·ChemistryNumerical
The number of paramagnetic species among the following is __________. B2_22​, Li2_22​, C2_22​, C2−_2^-2−​, O22−_2^{2-}22−​, O2+_2^+2+​ and He2+_2^+2+​

Correct answer: 4

Step-by-step solution →
Q54·ChemistryNumerical
150 g of acetic acid was contaminated with 10.2 g ascorbic acid (C6_66​H8_88​O6_66​) to lower down its freezing point by (x × 10−1^{-1}−1)°C. The value of x is _________. (Nearest integer) [Given Kf_ff​ = 3.9 K kg mol−1^{-1}−1; Molar mass of ascorbic acid = 176 g mol−1^{-1}−1]

Correct answer: 15

Step-by-step solution →
Q55·ChemistryNumerical
Ka_aa​ for butyric acid (C3_33​H7_77​COOH) is 2 × 10−5^{-5}−5. The pH of 0.2 M solution of butyric acid is ___ × 10−1^{-1}−1. (Nearest integer) [Given log 2 = 0.30]

Correct answer: 27

Step-by-step solution →
Q56·ChemistryNumerical
For the given first order reaction A → B the half life of the reaction is 0.3010 min. The ratio of the initial concentration of reactant to the concentration of reactant at time 2.0 min will be equal to __________. (Nearest integer)

Correct answer: 100

Step-by-step solution →
Q57·ChemistryNumerical
The number of interhalogens from the following having square pyramidal structure is : ClF3_33​, IF7_77​, BrF5_55​, BrF3_33​, I2_22​Cl6_66​, IF5_55​, ClF, ClF5_55​

Correct answer: 3

Step-by-step solution →
Q58·ChemistryNumerical
The disproportionation of MnO42−_4^{2-}42−​ in acidic medium resulted in the formation of two manganese compounds A and B. If the oxidation state of Mn in B is smaller than that of A, then the spin-only magnetic moment (μ) value of B in BM is __________. (Nearest integer)

Correct answer: 4

Step-by-step solution →
Q59·ChemistryNumerical
Total number of relatively more stable isomer(s) possible for octahedral complex [Cu(en)2_22​(SCN)2_22​] will be __________.

Correct answer: 3

Step-by-step solution →
Q60·ChemistryNumerical
On complete combustion of 0.492 g of an organic compound containing C, H and O, 0.7938 g of CO2_22​ and 0.4428 g of H2_22​O was produced. The % composition of oxygen in the compound is _____.

Correct answer: 46

Step-by-step solution →

Mathematics — JEE Main 28 July 2022 Shift 1

Q61·MathematicsSingle correct
Let the solution curve of the differential equation xdy=(x2+y2+y)dxxdy = \left(\sqrt{x^{2}+y^{2}} + y\right)dxxdy=(x2+y2​+y)dx, x>0x > 0x>0, intersect the line x=1x = 1x=1 at y=0y = 0y=0 and the line x=2x = 2x=2 at y=αy = \alphay=α. Then the value of α\alphaα is :
  1. (A)12\frac{1}{2}21​
  2. (B)32\frac{3}{2}23​
  3. (C)−32-\frac{3}{2}−23​
  4. (D)52\frac{5}{2}25​

Correct answer: (B)

Step-by-step solution →
Q62·MathematicsSingle correct
Considering only the principal values of the inverse trigonometric functions, the domain of the function f(x)=cos⁡−1(x2−4x+2x2+3)f(x) = \cos^{-1}\left(\frac{x^{2}-4x+2}{x^{2}+3}\right)f(x)=cos−1(x2+3x2−4x+2​) is :
  1. (A)(−∞,14]\left(-\infty, \frac{1}{4}\right](−∞,41​]
  2. (B)[−14,∞)\left[-\frac{1}{4}, \infty\right)[−41​,∞)
  3. (C)(−13,∞)\left(-\frac{1}{3}, \infty\right)(−31​,∞)
  4. (D)(−∞,13]\left(-\infty, \frac{1}{3}\right](−∞,31​]

Correct answer: (B)

Step-by-step solution →
Q63·MathematicsSingle correct
Let the vectors a⃗=(1+t)i^+(1−t)j^+k^\vec{a} = (1+t)\hat{i} + (1-t)\hat{j} + \hat{k}a=(1+t)i^+(1−t)j^​+k^, b⃗=(1−t)i^+(1+t)j^+2k^\vec{b} = (1-t)\hat{i} + (1+t)\hat{j} + 2\hat{k}b=(1−t)i^+(1+t)j^​+2k^ and c⃗=ti^−tj^+k^\vec{c} = t\hat{i} - t\hat{j} + \hat{k}c=ti^−tj^​+k^, t∈Rt \in Rt∈R be such that for α,β,γ∈R\alpha, \beta, \gamma \in Rα,β,γ∈R, αa⃗+βb⃗+γc⃗=0⃗⇒α=β=γ=0\alpha\vec{a} + \beta\vec{b} + \gamma\vec{c} = \vec{0} \Rightarrow \alpha = \beta = \gamma = 0αa+βb+γc=0⇒α=β=γ=0. Then, the set of all values of t is :
  1. (A)a non-empty finite set
  2. (B)equal to N
  3. (C)equal to R − {0}
  4. (D)equal to R

Correct answer: (C)

Step-by-step solution →
Q64·MathematicsSingle correct
Considering the principal values of the inverse trigonometric functions, the sum of all the solutions of the equation cos⁡−1(x)−2sin⁡−1(x)=cos⁡−1(2x)\cos^{-1}(x) - 2\sin^{-1}(x) = \cos^{-1}(2x)cos−1(x)−2sin−1(x)=cos−1(2x) is equal to :
  1. (A)0
  2. (B)1
  3. (C)12\frac{1}{2}21​
  4. (D)−12-\frac{1}{2}−21​

Correct answer: (A)

Step-by-step solution →
Q65·MathematicsSingle correct
Let the operations ∗,⊙∈{∧,∨}*, \odot \in \{\wedge, \vee\}∗,⊙∈{∧,∨}. If (p∗q)⊙(p⊙∼q)(p * q) \odot (p \odot \sim q)(p∗q)⊙(p⊙∼q) is a tautology, then the ordered pair (∗,⊙)(*, \odot)(∗,⊙) is :
  1. (A)(∨,∧)(\vee, \wedge)(∨,∧)
  2. (B)(∨,∨)(\vee, \vee)(∨,∨)
  3. (C)(∧,∧)(\wedge, \wedge)(∧,∧)
  4. (D)(∧,∨)(\wedge, \vee)(∧,∨)

Correct answer: (B)

Step-by-step solution →
Q66·MathematicsSingle correct
Let a vector a⃗\vec{a}a has a magnitude 9. Let a vector b⃗\vec{b}b be such that for every (x,y)∈R×R−{(0,0)}(x, y) \in R \times R - \{(0,0)\}(x,y)∈R×R−{(0,0)}, the vector (xa⃗+yb⃗)(x\vec{a} + y\vec{b})(xa+yb) is perpendicular to the vector (6y a⃗−18x b⃗)(6y\,\vec{a} - 18x\,\vec{b})(6ya−18xb). Then the value of ∣a⃗×b⃗∣\left|\vec{a} \times \vec{b}\right|​a×b​ is equal to:
  1. (A)939\sqrt{3}93​
  2. (B)27327\sqrt{3}273​
  3. (C)9
  4. (D)81

Correct answer: (B)

Step-by-step solution →
Q67·MathematicsSingle correct
For t∈(0,2π)t \in (0, 2\pi)t∈(0,2π), if ABC is an equilateral triangle with vertices A(sint, −cost), B(cost, sint) and C(a, b) such that its orthocentre lies on a circle with centre (1,13)\left(1, \frac{1}{3}\right)(1,31​), then (a2−b2)(a^{2} - b^{2})(a2−b2) is equal to :
  1. (A)83\frac{8}{3}38​
  2. (B)8
  3. (C)779\frac{77}{9}977​
  4. (D)809\frac{80}{9}980​

Correct answer: (B)

Step-by-step solution →
Q68·MathematicsSingle correct
For α∈N\alpha \in Nα∈N, consider a relation R on N given by R={(x,y):3x+αy is a multiple of 7}R = \{(x, y) : 3x + \alpha y \text{ is a multiple of } 7\}R={(x,y):3x+αy is a multiple of 7}. The relation R is an equivalence relation if and only if :
  1. (A)α=14\alpha = 14α=14
  2. (B)α\alphaα is a multiple of 4
  3. (C)4 is the remainder when α\alphaα is divided by 10
  4. (D)4 is the remainder when α\alphaα is divided by 7

Correct answer: (D)

Step-by-step solution →
Q69·MathematicsSingle correct
If y=y(x)y = y(x)y=y(x), x∈(0,π2)x \in \left(0, \frac{\pi}{2}\right)x∈(0,2π​) be the solution curve of the differential equation (sin⁡22x)dydx+(8sin⁡22x+2sin⁡4x)y=2e−4x(2sin⁡2x+cos⁡2x)\left(\sin^{2}2x\right)\frac{dy}{dx} + \left(8\sin^{2}2x + 2\sin 4x\right)y = 2e^{-4x}\left(2\sin 2x + \cos 2x\right)(sin22x)dxdy​+(8sin22x+2sin4x)y=2e−4x(2sin2x+cos2x), with y(π4)=e−πy\left(\frac{\pi}{4}\right) = e^{-\pi}y(4π​)=e−π, then y(π6)y\left(\frac{\pi}{6}\right)y(6π​) is equal to :
  1. (A)23e−2π/3\frac{2}{\sqrt{3}}e^{-2\pi/3}3​2​e−2π/3
  2. (B)23e2π/3\frac{2}{\sqrt{3}}e^{2\pi/3}3​2​e2π/3
  3. (C)13e−2π/3\frac{1}{\sqrt{3}}e^{-2\pi/3}3​1​e−2π/3
  4. (D)13e2π/3\frac{1}{\sqrt{3}}e^{2\pi/3}3​1​e2π/3

Correct answer: (A)

Step-by-step solution →
Q70·MathematicsSingle correct
If the tangents drawn at the points P and Q on the parabola y2=2x−3y^{2} = 2x - 3y2=2x−3 intersect at the point R(0, 1), then the orthocentre of the triangle PQR is :
  1. (A)(0, 1)
  2. (B)(2, −1)
  3. (C)(6, 3)
  4. (D)(2, 1)

Correct answer: (B)

Step-by-step solution →
Q71·MathematicsSingle correct
Let C be the centre of the circle x2+y2−x+2y=114x^{2} + y^{2} - x + 2y = \frac{11}{4}x2+y2−x+2y=411​ and P be a point on the circle. A line passes through the point C, makes an angle of π4\frac{\pi}{4}4π​ with the line CP and intersects the circle at the points Q and R. Then the area of the triangle PQR (in unit2^{2}2) is :
  1. (A)2
  2. (B)222\sqrt{2}22​
  3. (C)8sin⁡(π8)8\sin\left(\frac{\pi}{8}\right)8sin(8π​)
  4. (D)8cos⁡(π8)8\cos\left(\frac{\pi}{8}\right)8cos(8π​)

Correct answer: (B)

Step-by-step solution →
Q72·MathematicsSingle correct
The remainder when 72022+320227^{2022} + 3^{2022}72022+32022 is divided by 5 is:
  1. (A)0
  2. (B)2
  3. (C)3
  4. (D)4

Correct answer: (C)

Step-by-step solution →
Q73·MathematicsSingle correct
Let the matrix A=[010001100]A = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{bmatrix}A=​001​100​010​​ and the matrix B0=A49+2A98B_{0} = A^{49} + 2A^{98}B0​=A49+2A98. If Bn=Adj(Bn−1)B_{n} = Adj(B_{n-1})Bn​=Adj(Bn−1​) for all n≥1n \geq 1n≥1, then det⁡(B4)\det(B_{4})det(B4​) is equal to :
  1. (A)3283^{28}328
  2. (B)3303^{30}330
  3. (C)3323^{32}332
  4. (D)3363^{36}336

Correct answer: (C)

Step-by-step solution →
Q74·MathematicsSingle correct
Let S1={z1∈C:∣z1−3∣=12}S_{1} = \left\{ z_{1} \in C : \left| z_{1} - 3 \right| = \frac{1}{2} \right\}S1​={z1​∈C:∣z1​−3∣=21​} and S2={z2∈C:∣z2−∣z2+1∣∣=∣z2+∣z2−1∣∣}S_{2} = \left\{ z_{2} \in C : \left| z_{2} - \left| z_{2} + 1 \right| \right| = \left| z_{2} + \left| z_{2} - 1 \right| \right| \right\}S2​={z2​∈C:∣z2​−∣z2​+1∣∣=∣z2​+∣z2​−1∣∣}. Then, for z1∈S1z_{1} \in S_{1}z1​∈S1​ and z2∈S2z_{2} \in S_{2}z2​∈S2​, the least value of ∣z2−z1∣\left| z_{2} - z_{1} \right|∣z2​−z1​∣ is :
  1. (A)0
  2. (B)12\frac{1}{2}21​
  3. (C)32\frac{3}{2}23​
  4. (D)52\frac{5}{2}25​

Correct answer: (C)

Step-by-step solution →
Q75·MathematicsSingle correct
The foot of the perpendicular from a point on the circle x2+y2=1x^{2} + y^{2} = 1x2+y2=1, z=0z = 0z=0 to the plane 2x+3y+z=62x + 3y + z = 62x+3y+z=6 lies on which one of the following curves ?
  1. (A)(6x+5y−12)2+4(3x+7y−8)2=1(6x + 5y - 12)^{2} + 4(3x + 7y - 8)^{2} = 1(6x+5y−12)2+4(3x+7y−8)2=1, z=6−2x−3yz = 6 - 2x - 3yz=6−2x−3y
  2. (B)(5x+6y−12)2+4(3x+5y−9)2=1(5x + 6y - 12)^{2} + 4(3x + 5y - 9)^{2} = 1(5x+6y−12)2+4(3x+5y−9)2=1, z=6−2x−3yz = 6 - 2x - 3yz=6−2x−3y
  3. (C)(6x+5y−14)2+9(3x+5y−7)2=1(6x + 5y - 14)^{2} + 9(3x + 5y - 7)^{2} = 1(6x+5y−14)2+9(3x+5y−7)2=1, z=6−2x−3yz = 6 - 2x - 3yz=6−2x−3y
  4. (D)(5x+6y−14)2+9(3x+7y−8)2=1(5x + 6y - 14)^{2} + 9(3x + 7y - 8)^{2} = 1(5x+6y−14)2+9(3x+7y−8)2=1, z=6−2x−3yz = 6 - 2x - 3yz=6−2x−3y

Correct answer: (B)

Step-by-step solution →
Q76·MathematicsSingle correct
If the minimum value of f(x)=5x22+αx5f(x) = \frac{5x^{2}}{2} + \frac{\alpha}{x^{5}}f(x)=25x2​+x5α​, x>0x > 0x>0, is 14, then the value of α\alphaα is equal to :
  1. (A)32
  2. (B)64
  3. (C)128
  4. (D)256

Correct answer: (C)

Step-by-step solution →
Q77·MathematicsSingle correct
Let α\alphaα, β\betaβ and γ\gammaγ be three positive real numbers. Let f(x)=αx5+βx3+γxf(x) = \alpha x^{5} + \beta x^{3} + \gamma xf(x)=αx5+βx3+γx, x∈Rx \in Rx∈R and g:R→Rg : R \to Rg:R→R be such that g(f(x))=xg(f(x)) = xg(f(x))=x for all x∈Rx \in Rx∈R. If a1a_{1}a1​, a2a_{2}a2​, a3a_{3}a3​,..., ana_{n}an​ be in arithmetic progression with mean zero, then the value of f(g(1n∑i=1nf(ai)))f\left( g\left( \frac{1}{n} \sum_{i=1}^{n} f\left(a_{i}\right) \right) \right)f(g(n1​∑i=1n​f(ai​))) is equal to :
  1. (A)0
  2. (B)3
  3. (C)9
  4. (D)27

Correct answer: (A)

Step-by-step solution →
Q78·MathematicsSingle correct
Consider the sequence a1a_{1}a1​, a2a_{2}a2​, a3a_{3}a3​, …… such that a1=1a_{1} = 1a1​=1, a2=2a_{2} = 2a2​=2 and an+2=2an+1+ana_{n+2} = \frac{2}{a_{n+1}} + a_{n}an+2​=an+1​2​+an​ for n = 1, 2, 3, … . If (a1+1a2a3)⋅(a2+1a3a4)⋅(a3+1a4a5)⋯(a30+1a31a32)=2α(61C31)\left( \frac{a_{1} + \frac{1}{a_{2}}}{a_{3}} \right) \cdot \left( \frac{a_{2} + \frac{1}{a_{3}}}{a_{4}} \right) \cdot \left( \frac{a_{3} + \frac{1}{a_{4}}}{a_{5}} \right) \cdots \left( \frac{a_{30} + \frac{1}{a_{31}}}{a_{32}} \right) = 2^{\alpha} \left( {}^{61}C_{31} \right)(a3​a1​+a2​1​​)⋅(a4​a2​+a3​1​​)⋅(a5​a3​+a4​1​​)⋯(a32​a30​+a31​1​​)=2α(61C31​), then α\alphaα is equal to :
  1. (A)−30
  2. (B)−31
  3. (C)−60
  4. (D)−61

Correct answer: (C)

Step-by-step solution →
Q79·MathematicsSingle correct
The minimum value of the twice differentiable function f(x)=∫0xex−tf′(t) dt−(x2−x+1)exf(x) = \int_{0}^{x} e^{x-t} f'(t)\,dt - (x^{2} - x + 1)e^{x}f(x)=∫0x​ex−tf′(t)dt−(x2−x+1)ex, x∈Rx \in Rx∈R, is :
  1. (A)−2e-\frac{2}{\sqrt{e}}−e​2​
  2. (B)−2e-2\sqrt{e}−2e​
  3. (C)−e-\sqrt{e}−e​
  4. (D)2e\frac{2}{\sqrt{e}}e​2​

Correct answer: (A)

Step-by-step solution →
Q80·MathematicsNumerical
Let S be the set of all passwords which are six to eight characters long, where each character is either an alphabet from {A, B, C, D, E} or a number from {1, 2, 3, 4, 5} with the repetition of characters allowed. If the number of passwords in S whose at least one character is a number from {1, 2, 3, 4, 5} is α×56\alpha \times 5^{6}α×56, then α\alphaα is equal to ______.

Correct answer: 7073

Step-by-step solution →
Q81·MathematicsNumerical
Let P(−2, −1, 1) and Q(5617,4317,11117)Q\left( \frac{56}{17}, \frac{43}{17}, \frac{111}{17} \right)Q(1756​,1743​,17111​) be the vertices of the rhombus PRQS. If the direction ratios of the diagonal RS are α\alphaα, −1-1−1, β\betaβ, where both α\alphaα and β\betaβ are integers of minimum absolute values, then α2+β2\alpha^{2} + \beta^{2}α2+β2 is equal to __________.

Correct answer: 450

Step-by-step solution →
Q82·MathematicsNumerical
Let f:[0,1]→Rf : [0, 1] \to Rf:[0,1]→R be a twice differentiable function in (0, 1) such that f(0) = 3 and f(1) = 5. If the line y = 2x + 3 intersects the graph of f at only two distinct points in (0, 1), then the least number of points x∈(0,1)x \in (0, 1)x∈(0,1), at which f′′(x)=0f''(x) = 0f′′(x)=0, is _______.

Correct answer: 2

Step-by-step solution →
Q83·MathematicsNumerical
If ∫0315x31+x2+(1+x2)3 dx=α2+β3\int_{0}^{\sqrt{3}} \frac{15x^{3}}{\sqrt{1 + x^{2} + \sqrt{\left(1 + x^{2}\right)^{3}}}}\,dx = \alpha\sqrt{2} + \beta\sqrt{3}∫03​​1+x2+(1+x2)3​​15x3​dx=α2​+β3​, where α\alphaα, β\betaβ are integers, then α+β\alpha + \betaα+β is equal to ______.

Correct answer: 10

Step-by-step solution →
Q84·MathematicsNumerical
Let A=[1−12α]A = \begin{bmatrix} 1 & -1 \\ 2 & \alpha \end{bmatrix}A=[12​−1α​] and B=[β110]B = \begin{bmatrix} \beta & 1 \\ 1 & 0 \end{bmatrix}B=[β1​10​], α,β∈R\alpha, \beta \in Rα,β∈R. Let α1\alpha_{1}α1​ be the value of α\alphaα which satisfies (A+B)2=A2+[2222](A + B)^{2} = A^{2} + \begin{bmatrix} 2 & 2 \\ 2 & 2 \end{bmatrix}(A+B)2=A2+[22​22​] and α2\alpha_{2}α2​ be the value of α\alphaα which satisfies (A+B)2=B2(A + B)^{2} = B^{2}(A+B)2=B2. Then ∣α1−α2∣\left| \alpha_{1} - \alpha_{2} \right|∣α1​−α2​∣ is equal to _______.

Correct answer: 2

Step-by-step solution →
Q85·MathematicsNumerical
For p,q∈Rp, q \in Rp,q∈R, consider the real valued function f(x)=(x−p)2−qf(x) = (x - p)^{2} - qf(x)=(x−p)2−q, x∈Rx \in Rx∈R and q > 0. Let a1a_{1}a1​, a2a_{2}a2​, a3a_{3}a3​ and a4a_{4}a4​ be in an arithmetic progression with mean p and positive common difference. If ∣f(ai)∣=500\left| f\left(a_{i}\right) \right| = 500∣f(ai​)∣=500 for all i = 1, 2, 3, 4, then the absolute difference between the roots of f(x) = 0 is ______.

Correct answer: 50

Step-by-step solution →
Q86·MathematicsNumerical
For the hyperbola H : x2−y2=1x^{2} - y^{2} = 1x2−y2=1 and the ellipse E : x2a2+y2b2=1\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1a2x2​+b2y2​=1, a > b > 0, let the (1) eccentricity of E be reciprocal of the eccentricity of H, and (2) the line y=52x+Ky = \sqrt{\frac{5}{2}}x + Ky=25​​x+K be a common tangent of E and H. Then 4(a2+b2)4(a^{2} + b^{2})4(a2+b2) is equal to ________.

Correct answer: 3

Step-by-step solution →
Q87·MathematicsNumerical
Let x1x_{1}x1​, x2x_{2}x2​, x3x_{3}x3​, ….., x20x_{20}x20​ be in geometric progression with x1=3x_{1} = 3x1​=3 and the common ration 12\frac{1}{2}21​. A new data is constructed replacing each xix_{i}xi​ by (xi−i)2\left(x_{i} - i\right)^{2}(xi​−i)2. If xˉ\bar{x}xˉ is the mean of new data, then the greatest integer less than or equal to xˉ\bar{x}xˉ is __________.

Correct answer: 142

Step-by-step solution →
Q88·MathematicsNumerical
lim⁡x→0((x+2cos⁡x)3+2(x+2cos⁡x)2+3sin⁡(x+2cos⁡x)(x+2)3+2(x+2)2+3sin⁡(x+2))100x\lim_{x \to 0} \left( \frac{\left(x + 2\cos x\right)^{3} + 2\left(x + 2\cos x\right)^{2} + 3\sin\left(x + 2\cos x\right)}{\left(x + 2\right)^{3} + 2\left(x + 2\right)^{2} + 3\sin\left(x + 2\right)} \right)^{\frac{100}{x}}limx→0​((x+2)3+2(x+2)2+3sin(x+2)(x+2cosx)3+2(x+2cosx)2+3sin(x+2cosx)​)x100​ is equal to ___________.

Correct answer: 1

Step-by-step solution →
Q89·MathematicsNumerical
The sum of all real values of x for which 3x2−9x+17x2+3x+10=5x2−7x+193x2+5x+12\frac{3x^{2} - 9x + 17}{x^{2} + 3x + 10} = \frac{5x^{2} - 7x + 19}{3x^{2} + 5x + 12}x2+3x+103x2−9x+17​=3x2+5x+125x2−7x+19​ is equal to ____.

Correct answer: 6

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
  • Permutations and Combinations 162/186
  • Magnetic Field of Current 147/186
  • Binomial Theorem and Its Simple Applications 158/186
  • 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
  • Equilibrium 163/186
  • Solutions 158/186
  • Laws of Motion 130/186
  • Chemical Thermodynamics 165/186
  • Units and Measurements 149/186
  • 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
  • Work, Energy and Power 132/186
  • Some Basic Concepts in Chemistry 129/186
  • Wave Optics 130/186
  • Kinetic Theory of Gases 135/186
  • Alcohols and Ethers 106/186
  • Purification and Characterisation of Organic Compounds 116/186
  • Alternating Currents 108/186
  • Nuclei 116/186
  • Straight Lines 114/186
  • Waves 109/186
  • Capacitors and Dielectrics 115/186
  • Classification of Elements and Periodicity in Properties 113/186
  • Parabola 101/186
  • Statistics 118/186
  • Ellipse 103/186
  • 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
  • Electronic Effects and Stability 74/186
  • Polymers 64/186
  • Solid State 63/186
  • Chemistry in Everyday Life 60/186
  • IUPAC Nomenclature 37/186
← 27 Jul Shift 2 2022All papers28 Jul Shift 2 2022 →

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