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JEE Main 29 July 2022 Shift 1 Question Paper with Answers

29 July 2022 · July session · 88 questions

88 of the 90 questions from the JEE Main 29 July 2022 Shift 1 paper, each with its correct answer and tagged to the chapter it tests. Free to read, no account needed.

2 questions are 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
28
Mathematics
30

Physics — JEE Main 29 July 2022 Shift 1

Q1·PhysicsSingle correct
Given below are two statements : One is labelled as Assertion (A) and other is labelled as Reason (R). Assertion (A) : Time period of oscillation of a liquid drop depends on surface tension (S), if density of the liquid is p and radius of the drop is r, then T=kpr3S3/2T = k\sqrt{\frac{pr^{3}}{S^{3/2}}}T=kS3/2pr3​​ is dimensionally correct, where K is dimensionless. Reason (R) : Using dimensional analysis we get R.H.S. having different dimension than that of time period. In the light of above statements, choose the correct answer from the options given below.
  1. (A)Both (A) and (R) are true and (R) is the correct explanation of (A)
  2. (B)Both (A) and (R) are true but (R) is not 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: (D)

Step-by-step solution →
Q2·PhysicsSingle correct
A ball is thrown up vertically with a certain velocity so that, it reaches a maximum height h. Find the ratio of the times in which it is at height h3\frac{h}{3}3h​ while going up and coming down respectively.
  1. (A)2−12+1\frac{\sqrt{2}-1}{\sqrt{2}+1}2​+12​−1​
  2. (B)3−23+2\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}3​+2​3​−2​​
  3. (C)3−13+1\frac{\sqrt{3}-1}{\sqrt{3}+1}3​+13​−1​
  4. (D)13\frac{1}{3}31​

Correct answer: (B)

Step-by-step solution →
Q3·PhysicsSingle correct
If t=x+4t = \sqrt{x} + 4t=x​+4, then (dxdt)t=4\left(\frac{dx}{dt}\right)_{t=4}(dtdx​)t=4​ is:
  1. (A)4
  2. (B)Zero
  3. (C)8
  4. (D)16

Correct answer: (B)

Step-by-step solution →
Q4·PhysicsSingle correct
A smooth circular groove has a smooth vertical wall as shown in figure. A block of mass m moves against the wall with a speed v. Which of the following curve represents the correct relation between the normal reaction on the block by the wall (N) and speed of the block (v) ?
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (A)

Step-by-step solution →
Q5·PhysicsSingle correct
A ball is projected with kinetic energy E, at an angle of 60060^{0}600 to the horizontal. The kinetic energy of this ball at the highest point of its flight will become :
  1. (A)Zero
  2. (B)E2\frac{E}{2}2E​
  3. (C)E4\frac{E}{4}4E​
  4. (D)E

Correct answer: (C)

Step-by-step solution →
Q6·PhysicsSingle correct
Two bodies of mass 1 kg and 3 kg have position vectors i^+2j^+k^\hat{i} + 2\hat{j} + \hat{k}i^+2j^​+k^ and −3i^−2j^+k^-3\hat{i} - 2\hat{j} + \hat{k}−3i^−2j^​+k^ respectively. The magnitude of position vector of centre of mass of this system will be similar to the magnitude of vector :
  1. (A)i^−2j^+k^\hat{i} - 2\hat{j} + \hat{k}i^−2j^​+k^
  2. (B)−3i^−2j^+k^-3\hat{i} - 2\hat{j} + \hat{k}−3i^−2j^​+k^
  3. (C)−2i^+2k^-2\hat{i} + 2\hat{k}−2i^+2k^
  4. (D)−2i^−j^+2k^-2\hat{i} - \hat{j} + 2\hat{k}−2i^−j^​+2k^

Correct answer: (A)

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) : Clothes containing oil or grease stains cannot be cleaned by water wash. Reason (R) : Because the angle of contact between the oil/ grease and water is obtuse. In the light of the above statements, choose the correct answer from the option given below.
  1. (A)Both (A) and (R) are true and (R) is the correct explanation of (A)
  2. (B)Both (A) and (R) are true but (R) is not the correct explanation of (A)
  3. (C)(A) is true but (R) is false
  4. (D)(A) is true but (R) is true

Correct answer: (A)

Step-by-step solution →
Q8·PhysicsSingle correct
If the length of a wire is made double and radius is halved of its respective values. Then, the Young’s modules of the material of the wire will :
  1. (A)Remains same
  2. (B)Become 8 times its initial value
  3. (C)Become 1th4\frac{1^{th}}{4}41th​ of its initial value
  4. (D)Become 4 times its initial value

Correct answer: (A)

Step-by-step solution →
Q9·PhysicsSingle correct
The time period of oscillation of a simple pendulum of length L suspended from the roof of a vehicle, which moves without friction down an inclined plane of inclination α\alphaα, is given by :
  1. (A)2πL/(gcos⁡α)2\pi\sqrt{L/(g\cos\alpha)}2πL/(gcosα)​
  2. (B)2πL/(gsin⁡α)2\pi\sqrt{L/(g\sin\alpha)}2πL/(gsinα)​
  3. (C)2πL/g2\pi\sqrt{L/g}2πL/g​
  4. (D)2πL/(gtan⁡α)2\pi\sqrt{L/(g\tan\alpha)}2πL/(gtanα)​

Correct answer: (A)

Step-by-step solution →
Q10·PhysicsSingle correct
A spherically symmetric charge distribution is considered with charge density varying as ρ(r)={ρ0(34−rR)for r≤RZerofor r>R\rho(r) = \begin{cases} \rho_{0}\left(\frac{3}{4} - \frac{r}{R}\right) & \text{for } r \leq R \\ \text{Zero} & \text{for } r > R \end{cases}ρ(r)={ρ0​(43​−Rr​)Zero​for r≤Rfor r>R​ Where, r(r < R) is the distance from the centre O (as shown in figure). The electric field at point P will be :
  1. (A)ρ0r4ε0(34−rR)\frac{\rho_{0}r}{4\varepsilon_{0}}\left(\frac{3}{4} - \frac{r}{R}\right)4ε0​ρ0​r​(43​−Rr​)
  2. (B)ρ0r3ε0(34−rR)\frac{\rho_{0}r}{3\varepsilon_{0}}\left(\frac{3}{4} - \frac{r}{R}\right)3ε0​ρ0​r​(43​−Rr​)
  3. (C)ρ0r4ε0(1−rR)\frac{\rho_{0}r}{4\varepsilon_{0}}\left(1 - \frac{r}{R}\right)4ε0​ρ0​r​(1−Rr​)
  4. (D)ρ0r5ε0(1−rR)\frac{\rho_{0}r}{5\varepsilon_{0}}\left(1 - \frac{r}{R}\right)5ε0​ρ0​r​(1−Rr​)

Correct answer: (C)

Step-by-step solution →
Q11·PhysicsSingle correct
Given below are two statements. Statement I : Electric potential is constant within and at the surface of each conductor. Statement II : Electric field just outside a charged conductor is perpendicular to the surface of the conductor at every point. In the light of the above statements, choose the most appropriate answer from the options give below.
  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 and statement II is correct

Correct answer: (A)

Step-by-step solution →
Q12·PhysicsSingle correct
Two metallic wires of identical dimensions are connected is series. If σ1\sigma_{1}σ1​ and σ2\sigma_{2}σ2​ are the conductivities of the these wires respectively, the effective conductivity of the combination is :
  1. (A)σ1σ2σ1+σ2\frac{\sigma_{1}\sigma_{2}}{\sigma_{1}+\sigma_{2}}σ1​+σ2​σ1​σ2​​
  2. (B)2σ1σ2σ1+σ2\frac{2\sigma_{1}\sigma_{2}}{\sigma_{1}+\sigma_{2}}σ1​+σ2​2σ1​σ2​​
  3. (C)σ1+σ22σ1σ2\frac{\sigma_{1}+\sigma_{2}}{2\sigma_{1}\sigma_{2}}2σ1​σ2​σ1​+σ2​​
  4. (D)σ1+σ2σ1σ2\frac{\sigma_{1}+\sigma_{2}}{\sigma_{1}\sigma_{2}}σ1​σ2​σ1​+σ2​​

Correct answer: (B)

Step-by-step solution →
Q13·PhysicsSingle correct
An alternating emf E = 440 sin 100πt is applited to a circuit containing an inductance of 2π\frac{\sqrt{2}}{\pi}π2​​ H. If an a.c. ammeter is connected in the circuit, its reading will be :
  1. (A)4.4 A
  2. (B)1.55 A
  3. (C)2.2 A
  4. (D)3.11 A

Correct answer: (C)

Step-by-step solution →
Q14·PhysicsSingle correct
A coil of inductance 1 H and resistance 100 Ω is connected to a battery of 6 V. Determine approximately : (a) The time elapsed before the current acquires half of its steady – state value (b) The energy stored in the magnetic field associated with the coil at an instant 15 ms after the circuit is switched on. (Given In2 = 0.693, e−3/2=0.25e^{-3/2} = 0.25e−3/2=0.25)
  1. (A)t = 10 ms; U = 2 mJ
  2. (B)t = 10 ms; U = 1 mJ
  3. (C)t = 7 ms; U = 1 mJ
  4. (D)t = 7 ms; U = 2 mJ

Correct answer: (C)

Step-by-step solution →
Q15·PhysicsSingle correct
Match List – I with List – II Choose the correct answer from the options given below :
List – IList - II
a.UV raysi.Diagnostic tool in medicine
b.X-raysii.Water purification
c.Microwaveiii.Communication, Radar
d.Infrared waveiv.Improving visibility in foggy days
  1. (A)(a)–(iii), (b)-(ii), (c)-(i), (d)-(iv)
  2. (B)(a)–(ii), (b)-(i), (c)-(iii), (d)-(iv)
  3. (C)(a)–(ii), (b)-(iv), (c)-(iii), (d)-(i)
  4. (D)(a)–(iii), (b)-(i), (c)-(ii), (d)-(iv)

Correct answer: (B)

Step-by-step solution →
Q16·PhysicsSingle correct
The kinetic energy of emitted electron is E when the light incident on the metal has wavelength λ\lambdaλ. To double the kinetic energy, the incident light must have wavelength :
  1. (A)hcEλ−hc\frac{hc}{E\lambda - hc}Eλ−hchc​
  2. (B)hcλEλ+hc\frac{hc\lambda}{E\lambda + hc}Eλ+hchcλ​
  3. (C)hλEλ+hc\frac{h\lambda}{E\lambda + hc}Eλ+hchλ​
  4. (D)hcλEλ−hc\frac{hc\lambda}{E\lambda - hc}Eλ−hchcλ​

Correct answer: (B)

Step-by-step solution →
Q17·PhysicsSingle correct
Find the ratio of energies of photons produced due to transition of an election of hydrogen atom from its(i) second permitted energy level to the first level, and (ii) the highest permitted energy level to the first permitted level.
  1. (A)3 : 4
  2. (B)4 : 3
  3. (C)1 : 4
  4. (D)4 : 1

Correct answer: (A)

Step-by-step solution →
Q18·PhysicsSingle correct
Find the modulation index of an AM wave having 8 V variation where maximum amplitude of the AM wave is 9 V.
  1. (A)0.8
  2. (B)0.5
  3. (C)0.2
  4. (D)0.1

Correct answer: (A)

Step-by-step solution →
Q19·PhysicsSingle correct
A travelling microscope has 20 divisions per cm on the main scale while its Vernier scale has total 50 divisions and 25 Vernier scale divisions are equal to 24 main scale divisions, what is the least count of the travelling microscope ?
  1. (A)0.001 cm
  2. (B)0.002 mm
  3. (C)0.002 cm
  4. (D)0.005 cm

Correct answer: (C)

Step-by-step solution →
Q20·PhysicsSingle correct
In an experiment to find out the diameter of wire using screw gauge, the following observation were noted : (a) Screw moves 0.5 mm on main scale in one complete rotation (b) Total divisions on circular scale = 50 (c) Main scale reading is 2.5 mm (d) 45th45^{th}45th division of circular scale is in the pitch line (e) Instrument has 0.03 mm negative error Then the diameter of wire is :
  1. (A)2.92 mm
  2. (B)2.54 mm
  3. (C)2.98 mm
  4. (D)3.45 mm

Correct answer: (C)

Step-by-step solution →
Q21·PhysicsNumerical
An object is projected in the air with initial velocity u at an angle θ\thetaθ. The projectile motion is such that the horizontal range R, is maximum. Another object is projected in the air with a horizontal range half of the range of first object. The initial velocity remains same in both the case. The value of the angle of projection, at which the second object is projected, will be ________degree.

Correct answer: 15

Step-by-step solution →
Q22·PhysicsNumerical
If the acceleration due to gravity experienced by a point mass at a height h above the surface of earth is same as that of the acceleration due to gravity at a depth α\alphaαh (h << ReR_eRe​) from the earth surface. The value of α\alphaα will be __________. (use ReR_eRe​ = 6400 km)

Correct answer: 2

Step-by-step solution →
Q23·PhysicsNumerical
The pressure P1P_1P1​ and density d1d_1d1​ of diatomic gas (γ=75)\left(\gamma = \frac{7}{5}\right)(γ=57​) changes suddenly to P2(>P1)P_2(>P_1)P2​(>P1​) and d2d_2d2​ respectively during an adiabatic process. The temperature of the gas increases and becomes __________ times of its initial temperature. (given d2d1\frac{d_2}{d_1}d1​d2​​ = 32)

Correct answer: 4

Step-by-step solution →
Q24·PhysicsNumerical
One mole of a monoatomic gas is mixed with three moles of a diatomic gas. The molecular specific heat of mixture at constant volume is α24\frac{\alpha^2}{4}4α2​R J/mol K; then the value of α\alphaα will be ________. (Assume that the given diatomic gas has no vibrational mode.)

Correct answer: 3

Step-by-step solution →
Q25·PhysicsNumerical
The current I flowing through the given circuit will be __________ A.

Correct answer: 2

Step-by-step solution →
Q26·PhysicsNumerical
A closely wounded circular coil of radius 5 cm produces a magnetic field of 37.68 x 10−410^{-4}10−4 T at its center. The current through the coil is __________ A. [Given, number of turns in the coil is 100 and π\piπ = 3.14]

Correct answer: 3

Step-by-step solution →
Q27·PhysicsNumerical
Two light beams of intensities 4I and 9I interfere on a screen. The phase difference between these beams on the screen at point A is zero and at point B is π\piπ. The difference of resultant intensities, at the point A and B, will be __________ I.

Correct answer: 24

Step-by-step solution →
Q28·PhysicsNumerical
A wire of length 314 cm carrying current of 14 A is bent to form a circle. The magnetic moment of the coil is ________ A-m2m^2m2. [Given π\piπ = 3.14]

Correct answer: 11

Step-by-step solution →
Q29·PhysicsNumerical
The X-Y plane be taken as the boundary between two transparent media M1M_1M1​ and M2M_2M2​. M1M_1M1​ in Z≥0Z \geq 0Z≥0 has a refractive index of 2\sqrt{2}2​ and M2M_2M2​ with Z<0Z < 0Z<0 has a refractive index of 3\sqrt{3}3​. A ray of light travelling in M1M_1M1​ along the direction given by the vector A⃗=43i^−33j^−5k^\vec{A} = 4\sqrt{3}\hat{i} - 3\sqrt{3}\hat{j} - 5\hat{k}A=43​i^−33​j^​−5k^, is incident on the plane of separation. The value of difference between the angle of incident in M1M_1M1​ and the angle of refraction in M2M_2M2​ will be ______________ degree.

Correct answer: 15

Step-by-step solution →
Q30·PhysicsNumerical
If the potential barrier across a p-n junction is 0.6 V. Then the electric field intensity, in the depletion region having the width of 6×10−66 \times 10^{-6}6×10−6m, will be __________ ×105\times 10^{5}×105 N/C.

Correct answer: 1

Step-by-step solution →

Chemistry — JEE Main 29 July 2022 Shift 1

Q31·ChemistrySingle correct
Which of the following pair of molecules contain odd electron molecule and an expanded octet molecule?
  1. (A)BCl3BCl_{3}BCl3​ and SF6SF_{6}SF6​
  2. (B)NO and H2SO4H_{2}SO_{4}H2​SO4​
  3. (C)SF6SF_{6}SF6​ and H2SO4H_{2}SO_{4}H2​SO4​
  4. (D)BCl3BCl_{3}BCl3​ and NO

Correct answer: (B)

Step-by-step solution →
Q32·ChemistrySingle correct
N2(g)N_{2(g)}N2(g)​ + 3H2(g)3H_{2(g)}3H2(g)​ ⇌ 2NH3(g)2NH_{3(g)}2NH3(g)​ 20 g 5 g Consider the above reaction, the limiting reagent of the reaction and number of moles of NH3NH_{3}NH3​ formed respectively are:
  1. (A)H2H_{2}H2​, 1.42 moles
  2. (B)H2H_{2}H2​, 0.71 moles
  3. (C)N2N_{2}N2​, 1.42 moles
  4. (D)N2N_{2}N2​, 0.71 moles

Correct answer: (C)

Step-by-step solution →
Q33·ChemistrySingle correct
100 mL of 5% (w/v) solution of NaCl in water was prepared in 250 mL beaker. Albumin from the egg was poured into NaCl solution and stirred well. This resulted in a/ an :
  1. (A)Lyophilic sol
  2. (B)Lyophobic sol
  3. (C)Emulsion
  4. (D)Precipitate

Correct answer: (A)

Step-by-step solution →
Q34·ChemistrySingle correct
The first ionization enthalpy of Na, Mg and Si, respectively, are: 496, 737 and 786 kJ mo1−1^{-1}−1. The first ionization enthalpy (kJ mol−1^{-1}−1) of Al is:
  1. (A)487
  2. (B)768
  3. (C)577
  4. (D)856

Correct answer: (C)

Step-by-step solution →
Q35·ChemistrySingle correct
In metallurgy the term "gangue" is used for:
  1. (A)Contamination of undesired earthy materials.
  2. (B)Contamination of metals, other than desired metal
  3. (C)Minerals which are naturally occuring in pure form
  4. (D)Magnetic impurities in an ore.

Correct answer: (A)

Step-by-step solution →
Q36·ChemistrySingle correct
Lithium nitrate and sodium nitrate, when heated separately, respectively, give :
  1. (A)LiNO2LiNO_{2}LiNO2​ and NaNO2NaNO_{2}NaNO2​
  2. (B)Li2OLi_{2}OLi2​O and Na2ONa_{2}ONa2​O
  3. (C)Li2OLi_{2}OLi2​O and NaNO2NaNO_{2}NaNO2​
  4. (D)LiNO2LiNO_{2}LiNO2​ and Na2ONa_{2}ONa2​O

Correct answer: (C)

Step-by-step solution →
Q37·ChemistrySingle correct
In following pairs, the one in which both transition metal ions are colourless is :
  1. (A)Sc3+Sc^{3+}Sc3+, Zn2+Zn^{2+}Zn2+
  2. (B)Ti4+Ti^{4+}Ti4+, Cu2+Cu^{2+}Cu2+
  3. (C)V2+V^{2+}V2+, Ti3+Ti^{3+}Ti3+
  4. (D)Zn2+Zn^{2+}Zn2+, Mn2+Mn^{2+}Mn2+

Correct answer: (A)

Step-by-step solution →
Q38·ChemistrySingle correct
In neutral or faintly alkaline medium, KMnO4KMnO_{4}KMnO4​ being a powerful oxidant can oxidize, thiosulphate almost quantitatively, to sulphate. In this reaction overall change in oxidation state of manganese will be :
  1. (A)5
  2. (B)1
  3. (C)0
  4. (D)3

Correct answer: (D)

Step-by-step solution →
Q39·ChemistrySingle correct
Which among the following pairs has only herbicides ?
  1. (A)Aldrin and Dieldrin
  2. (B)Sodium chlorate and Aldrin
  3. (C)Sodium arsinate and Dieldrin
  4. (D)Sodium chlorate and sodium arsinite.

Correct answer: (D)

Step-by-step solution →
Q40·ChemistrySingle correct
Which among the following is the strongest Bronsted base ?
  1. (A)Aldrin and Dieldrin
  2. (B)Sodium chlorate and Aldrin
  3. (C)Sodium arsinate and Dieldrin
  4. (D)Sodium chlorate and sodium arsinite.

Correct answer: (D)

Step-by-step solution →
Q41·ChemistrySingle correct
Which among the following pairs of the structures will give different products on ozonolysis? (Consider the double bonds in the structures are rigid and not delocalized.)
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (C)

Step-by-step solution →
Q42·ChemistrySingle correct
Considering the above reactions, the compound 'A' and compound 'B' respectively are :
  1. (A)N C ,N C ,N C ,≡≡≡N CN CN C≡≡≡
  2. (B)C N ,≡C N≡
  3. (C)N C ,≡C N≡
  4. (D)C N ,≡N C≡

Correct answer: (C)

Step-by-step solution →
Q43·ChemistrySingle correct
Consider the above reaction sequence, the Product 'C' is :
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (D)

Step-by-step solution →
Q44·ChemistrySingle correct
Consider the above reaction, the compound 'A' is :
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (C)

Step-by-step solution →
Q45·ChemistrySingle correct
Which among the following represent reagent 'A'?
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (A)

Step-by-step solution →
Q46·ChemistrySingle correct
Consider the following reaction sequence : The product 'B' is :
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (B)

Step-by-step solution →
Q47·ChemistrySingle correct
Which of the following compounds is an example of hypnotic drug ?
  1. (A)Seldane
  2. (B)Amytal
  3. (C)Aspartame
  4. (D)Prontosil

Correct answer: (B)

Step-by-step solution →
Q48·ChemistrySingle correct
A compound 'X' is acidic and it is soluble in NaOH solution, but insoluble in NaHCO3NaHCO_{3}NaHCO3​ solution. Compound 'X' also gives violet colour with neutral FeCl3FeCl_{3}FeCl3​ solution. The compound 'X' is :
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (B)

Step-by-step solution →
Q49·ChemistryNumerical
Resistance of a conductivity cell (cell constant 129 m−1m^{-1}m−1) filled with 74.5 ppm solution of KCl is 100 Ω (labelled as solution 1). When the same cell is filled with KCl solution of 149 ppm, the resistance is 50 Ω (labelled as solution 2). The ratio of molar conductivity of solution 1 and solution 2 is i.e. Λ1Λ2=x×10−3\frac{\Lambda_{1}}{\Lambda_{2}} = x \times 10^{-3}Λ2​Λ1​​=x×10−3. The value of x is ______. (Nearest integer) Given, molar mass of KCl is 74.5 g mol−1mol^{-1}mol−1

Correct answer: 1000

Step-by-step solution →
Q50·ChemistryNumerical
Ionic radii of cation A+A^{+}A+ and anion B−B^{-}B− are 102 and 181 pm respectively. These ions are allowed to crystallize into an ionic solid. This crystal has cubic close packing for B−B^{-}B−. A+A^{+}A+ is present in all octahedral voids. The edge length of the unit cell of the crystal AB is ______ pm. (Nearest Integer)

Correct answer: 566

Step-by-step solution →
Q51·ChemistryNumerical
The minimum uncertainty in the speed of an electron in an one dimensional region of length 2a02a_{0}2a0​ (Where a0a_{0}a0​ = Bohr radius 52.9 pm) is ______ km s−1s^{-1}s−1. (Given : Mass of electron = 9.1 × 10−3110^{-31}10−31 kg, Planck's constant h = 6.63 × 10−3410^{-34}10−34 Js)

Correct answer: 548

Step-by-step solution →
Q52·ChemistryNumerical
When 600 mL of 0.2 M HNO3HNO_{3}HNO3​ is mixed with 400 mL of 0.1M NaOH solution in a flask, the rise in temperature of the flask is ______ × 10−210^{-2}10−2 °C. (Enthalpy of neutralisation = 57 kJ mol−1mol^{-1}mol−1 and Specific heat of water = 4.2 JK−1JK^{-1}JK−1 g−1g^{-1}g−1) (Neglect heat capacity of flask)

Correct answer: 54

Step-by-step solution →
Q53·ChemistryNumerical
If O2O_{2}O2​ gas is bubbled through water at 303 K, the number of millimoles of O2O_{2}O2​ gas that dissolve in 1 litre of water is________. (Nearest Integer) (Given : Henry's Law constant for O2O_{2}O2​ at 303 K is 46.82 k bar and partial pressure of O2O_{2}O2​ = 0.920 bar) (Assume solubility of O2O_{2}O2​ in water is too small, nearly negligible)

Correct answer: 1

Step-by-step solution →
Q54·ChemistryNumerical
If the solubility product of PbS is 8 × 10−2810^{-28}10−28, then the solubility of PbS in pure water at 298 K is x × 10−1610^{-16}10−16 mol L−1L^{-1}L−1. The value of x is ________. (Nearest Integer) [Given 2\sqrt{2}2​ = 1.41]

Correct answer: 282

Step-by-step solution →
Q55·ChemistryNumerical
The reaction between X and Y is first order with respect to X and zero order with respect to Y. Examine the data of table and calculate ratio of numerical values of M and L. (Nearest Inetger)
Experiment[X] mol L−1L^{-1}L−1[Y] mol L−1L^{-1}L−1Initial rate mol L−1L^{-1}L−1 min−1min^{-1}min−1
I.0.10.12×10−32 \times 10^{-3}2×10−3
II.L0.24×10−34 \times 10^{-3}4×10−3
III.0.40.4M×10−3M \times 10^{-3}M×10−3
IV.0.10.22×10−32 \times 10^{-3}2×10−3

Correct answer: 40

Step-by-step solution →
Q56·ChemistryNumerical
In a linear tetrapeptide (Constituted with different amino acids), (number of amino acids) - (number of peptide bonds) is_______.

Correct answer: 1

Step-by-step solution →
Q57·ChemistryNumerical
In bromination of Propyne, with Bromine 1, 1, 2, 2-tetrabromopropane is obtained in 27% yield. The amount of 1, 1, 2, 2 tetrabromopropane obtained from 1 g of Bromine in this reaction is ______ × 10−110^{-1}10−1 g. (Nearest integer) (Molar Mass : Bromine = 80 g/mol)

Correct answer: 3

Step-by-step solution →
Q58·ChemistryNumerical
[Fe(CN)6]3−[Fe(CN)_{6}]^{3-}[Fe(CN)6​]3− should be an inner orbital complex. Ignoring the pairing energy, the value of crystal field stabilization energy for this complex is (–) _________ Δo\Delta_{o}Δo​. (Nearest integer)

Correct answer: 2

Step-by-step solution →

Mathematics — JEE Main 29 July 2022 Shift 1

Q59·MathematicsSingle correct
Let R be a relation from the set {1,2,3………,60}\{1, 2, 3\ldots\ldots\ldots,60\}{1,2,3………,60} to itself such that R={(a,b):b=pq,R = \{(a, b) : b = pq,R={(a,b):b=pq, where p,q≥3p, q \geq 3p,q≥3 are prime numbers}\}}. Then, the number of elements in R is :
  1. (A)600
  2. (B)660
  3. (C)540
  4. (D)720

Correct answer: (B)

Step-by-step solution →
Q60·MathematicsSingle correct
If z=2+3iz = 2 + 3iz=2+3i, then z5+(zˉ)5z^{5} + (\bar{z})^{5}z5+(zˉ)5 is equal to :
  1. (A)244
  2. (B)224
  3. (C)245
  4. (D)265

Correct answer: (A)

Step-by-step solution →
Q61·MathematicsSingle correct
Let A and B be two 3×33 \times 33×3 non-zero real matrices such that AB is a zero matrix. Then
  1. (A)The system of linear equations AX=0AX = 0AX=0 has a unique solution
  2. (B)The system of linear equations AX=0AX = 0AX=0 has infinitely many solutions
  3. (C)B is an invertible matrix
  4. (D)adj (A) is an invertible matrix

Correct answer: (B)

Step-by-step solution →
Q62·MathematicsSingle correct
If 1(20−a) (40−a)+1(40−a) (60−a)+……+1(180−a) (200−a)=1256\frac{1}{(20 - a)\,(40 - a)} + \frac{1}{(40 - a)\,(60 - a)} + \ldots\ldots+ \frac{1}{(180 - a)\,(200 - a)} = \frac{1}{256}(20−a)(40−a)1​+(40−a)(60−a)1​+……+(180−a)(200−a)1​=2561​, then the maximum value of a is :
  1. (A)198
  2. (B)202
  3. (C)212
  4. (D)218

Correct answer: (C)

Step-by-step solution →
Q63·Mathematics·Limits and ContinuitySingle correct
If lim⁡x→0αex+βe−x+γsin⁡xxsin⁡2x=23\lim_{x \to 0} \frac{\alpha e^{x} + \beta e^{-x} + \gamma \sin x}{x \sin^{2} x} = \frac{2}{3}limx→0​xsin2xαex+βe−x+γsinx​=32​, where α,β,γ∈R\alpha, \beta, \gamma \in Rα,β,γ∈R, then which of the following is NOT correct ?
  1. (A)α2+β2+γ2=6\alpha^{2} + \beta^{2} + \gamma^{2} = 6α2+β2+γ2=6
  2. (B)αβ+βγ+γα+1=0\alpha\beta + \beta\gamma + \gamma\alpha + 1 = 0αβ+βγ+γα+1=0
  3. (C)αβ2+βγ2+γα2+3=0\alpha\beta^{2} + \beta\gamma^{2} + \gamma\alpha^{2} + 3 = 0αβ2+βγ2+γα2+3=0
  4. (D)α2−β2+γ2=4\alpha^{2} - \beta^{2} + \gamma^{2} = 4α2−β2+γ2=4

Correct answer: (C)

Step-by-step solution →
Q64·MathematicsSingle correct
The integral ∫0π213+2sin⁡x+cos⁡x dx\int_{0}^{\frac{\pi}{2}} \frac{1}{3 + 2\sin x + \cos x}\,dx∫02π​​3+2sinx+cosx1​dx is equal to:
  1. (A)tan⁡−1(2)\tan^{-1}(2)tan−1(2)
  2. (B)tan⁡−1(2)−π4\tan^{-1}(2) - \frac{\pi}{4}tan−1(2)−4π​
  3. (C)12tan⁡−1(2)−π8\frac{1}{2}\tan^{-1}(2) - \frac{\pi}{8}21​tan−1(2)−8π​
  4. (D)12\frac{1}{2}21​

Correct answer: (B)

Step-by-step solution →
Q65·MathematicsSingle correct
Let the solution curve y=y(x)y = y(x)y=y(x) of the differential equation (1+e2x)(dydx+y)=1(1 + e^{2x})\left(\frac{dy}{dx} + y\right) = 1(1+e2x)(dxdy​+y)=1 pass through the point (0,π2)\left(0, \frac{\pi}{2}\right)(0,2π​). Then, lim⁡x→∞exy(x)\lim_{x \to \infty} e^{x} y(x)limx→∞​exy(x) is equal to :
  1. (A)π4\frac{\pi}{4}4π​
  2. (B)3π4\frac{3\pi}{4}43π​
  3. (C)π2\frac{\pi}{2}2π​
  4. (D)3π2\frac{3\pi}{2}23π​

Correct answer: (B)

Step-by-step solution →
Q66·MathematicsSingle correct
Let a line L pass through the point of intersection of the lines bx+10y−8=0bx + 10y - 8 = 0bx+10y−8=0 and 2x−3y=02x - 3y = 02x−3y=0, b∈R−{43}b \in R - \left\{\frac{4}{3}\right\}b∈R−{34​}. If the line L also passes through the point (1,1)(1, 1)(1,1) and touches the circle 17 (x2+y2)=1617\,(x^{2} + y^{2}) = 1617(x2+y2)=16, then the eccentricity of the ellipse x25+y2b2=1\frac{x^{2}}{5} + \frac{y^{2}}{b^{2}} = 15x2​+b2y2​=1 is :
  1. (A)25\frac{2}{\sqrt{5}}5​2​
  2. (B)35\sqrt{\frac{3}{5}}53​​
  3. (C)15\frac{1}{\sqrt{5}}5​1​
  4. (D)25\sqrt{\frac{2}{5}}52​​

Correct answer: (B)

Step-by-step solution →
Q67·MathematicsSingle correct
If the foot of the perpendicular from the point A(−1,4,3)A(-1, 4, 3)A(−1,4,3) on the plane P:2x+my+nz=4P : 2x + my + nz = 4P:2x+my+nz=4, is (−2,72,32)\left(-2, \frac{7}{2}, \frac{3}{2}\right)(−2,27​,23​), then the distance of the point A from the plane P, measured parallel to a line with direction ratios 3,−1,−43, -1, -43,−1,−4, is equal to :
  1. (A)1
  2. (B)26\sqrt{26}26​
  3. (C)222\sqrt{2}22​
  4. (D)14\sqrt{14}14​

Correct answer: (B)

Step-by-step solution →
Q68·MathematicsSingle correct
Let a⃗=3i^+j^\vec{a} = 3\hat{i} + \hat{j}a=3i^+j^​ and b⃗=i^+2j^+k^\vec{b} = \hat{i} + 2\hat{j} + \hat{k}b=i^+2j^​+k^. Let c⃗\vec{c}c be a vector satisfying a⃗×(b⃗×c⃗)=b⃗+λc⃗\vec{a} \times (\vec{b} \times \vec{c}) = \vec{b} + \lambda\vec{c}a×(b×c)=b+λc. If b⃗\vec{b}b and c⃗\vec{c}c are non-parallel, then the value of λ\lambdaλ is :
  1. (A)−5-5−5
  2. (B)5
  3. (C)1
  4. (D)−1-1−1

Correct answer: (A)

Step-by-step solution →
Q69·MathematicsSingle correct
The angle of elevation of the top of a tower from a point A due north of it is α\alphaα and from a point B at a distance of 9 units due west of A is cos⁡−1(313)\cos^{-1}\left(\frac{3}{\sqrt{13}}\right)cos−1(13​3​). If the distance of the point B from the tower is 15 units, then cot⁡α\cot \alphacotα is equal to :
  1. (A)65\frac{6}{5}56​
  2. (B)95\frac{9}{5}59​
  3. (C)43\frac{4}{3}34​
  4. (D)73\frac{7}{3}37​

Correct answer: (A)

Step-by-step solution →
Q70·MathematicsSingle correct
The statement (p∧q)⇒(p∧r)(p \wedge q) \Rightarrow (p \wedge r)(p∧q)⇒(p∧r) is equivalent to :
  1. (A)q⇒(p∧r)q \Rightarrow (p \wedge r)q⇒(p∧r)
  2. (B)p⇒(p∧r)p \Rightarrow (p \wedge r)p⇒(p∧r)
  3. (C)(p∧r)⇒(p∧q)(p \wedge r) \Rightarrow (p \wedge q)(p∧r)⇒(p∧q)
  4. (D)(p∧q)⇒r(p \wedge q) \Rightarrow r(p∧q)⇒r

Correct answer: (D)

Step-by-step solution →
Q71·MathematicsSingle correct
Let the circumcentre of a triangle with vertices A(a,3)A(a, 3)A(a,3), B(b,5)B(b, 5)B(b,5) and C(a,b)C(a, b)C(a,b), ab>0ab > 0ab>0 be P(1,1)P(1, 1)P(1,1). If the line AP intersects the line BC at the point Q(k1,k2)Q(k_{1}, k_{2})Q(k1​,k2​), then k1+k2k_{1} + k_{2}k1​+k2​ is equal to :
  1. (A)2
  2. (B)47\frac{4}{7}74​
  3. (C)27\frac{2}{7}72​
  4. (D)4

Correct answer: (B)

Step-by-step solution →
Q72·MathematicsSingle correct
Let a^\hat{a}a^ and b^\hat{b}b^ be two unit vectors such that the angle between them is π4\frac{\pi}{4}4π​. If θ\thetaθ is the angle between the vectors (a^+b^)(\hat{a} + \hat{b})(a^+b^) and (a^+2b^+2(a^×b^))(\hat{a} + 2\hat{b} + 2(\hat{a} \times \hat{b}))(a^+2b^+2(a^×b^)), then the value of 164cos⁡2θ164 \cos^{2}\theta164cos2θ is equal to :
  1. (A)90+27290 + 27\sqrt{2}90+272​
  2. (B)45+18245 + 18\sqrt{2}45+182​
  3. (C)90+3290 + 3\sqrt{2}90+32​
  4. (D)54+90254 + 90\sqrt{2}54+902​

Correct answer: (A)

Step-by-step solution →
Q73·MathematicsSingle correct
If f(α)=∫1αlog⁡10t1+t dt,α>0f(\alpha) = \int_{1}^{\alpha} \frac{\log_{10} t}{1 + t}\,dt, \alpha > 0f(α)=∫1α​1+tlog10​t​dt,α>0, then f(e3)+f(e−3)f(e^{3}) + f(e^{-3})f(e3)+f(e−3) is equal to :
  1. (A)9
  2. (B)92\frac{9}{2}29​
  3. (C)9log⁡e(10)\frac{9}{\log_{e}(10)}loge​(10)9​
  4. (D)92log⁡e(10)\frac{9}{2\log_{e}(10)}2loge​(10)9​

Correct answer: (D)

Step-by-step solution →
Q74·MathematicsSingle correct
The area of the region {(x, y):∣ x−1 ∣≤y≤5−x2}\left\{(x,\,y) : |\,x-1\,| \le y \le \sqrt{5-x^{2}}\right\}{(x,y):∣x−1∣≤y≤5−x2​} is equal to :
  1. (A)52sin⁡−1(35)−12\frac{5}{2}\sin^{-1}\left(\frac{3}{5}\right)-\frac{1}{2}25​sin−1(53​)−21​
  2. (B)5π4−32\frac{5\pi}{4}-\frac{3}{2}45π​−23​
  3. (C)3π4+32\frac{3\pi}{4}+\frac{3}{2}43π​+23​
  4. (D)5π4−12\frac{5\pi}{4}-\frac{1}{2}45π​−21​

Correct answer: (D)

Step-by-step solution →
Q75·MathematicsSingle correct
Let the focal chord of the parabola P : y2=4xy^{2}=4xy2=4x along the line L : y=mx+cy = mx + cy=mx+c, m>0m > 0m>0 meet the parabola at the points M and N. Let the line L be a tangent to the hyperbola H : x2−y2=4x^{2}-y^{2}=4x2−y2=4. If O is the vertex of P and F is the focus of H on the positive x-axis, then the area of the quadrilateral OMFN is :
  1. (A)262\sqrt{6}26​
  2. (B)2142\sqrt{14}214​
  3. (C)464\sqrt{6}46​
  4. (D)4144\sqrt{14}414​

Correct answer: (B)

Step-by-step solution →
Q76·Mathematics·DifferentiabilitySingle correct
The number of points, where the function f:R→Rf : \mathbf{R} \rightarrow \mathbf{R}f:R→R, f(x)=∣x−1∣cos⁡∣x−2∣sin⁡∣x−1∣+(x−3)∣x2−5x+4∣f(x) = |x-1|\cos|x-2|\sin|x-1| + (x-3)\left|x^{2}-5x+4\right|f(x)=∣x−1∣cos∣x−2∣sin∣x−1∣+(x−3)​x2−5x+4​, is NOT differentiable, is :
  1. (A)1
  2. (B)2
  3. (C)3
  4. (D)4

Correct answer: (B)

Step-by-step solution →
Q77·MathematicsSingle correct
Let S={1, 2, 3, …, 2022}S = \{1,\,2,\,3,\,\ldots,\,2022\}S={1,2,3,…,2022}. Then the probability, that a randomly chosen number n from the set S such that HCF (n, 2022)=1(n,\,2022) = 1(n,2022)=1, is :
  1. (A)1281011\frac{128}{1011}1011128​
  2. (B)1661011\frac{166}{1011}1011166​
  3. (C)127337\frac{127}{337}337127​
  4. (D)112337\frac{112}{337}337112​

Correct answer: (D)

Step-by-step solution →
Q78·MathematicsSingle correct
Let f(x)=3(x2−2)3+4f(x) = 3^{\left(x^{2}-2\right)^{3}+4}f(x)=3(x2−2)3+4, x∈Rx \in \mathbf{R}x∈R. Then which of the following statements are true ? P : x=0x = 0x=0 is a point of local minima of f Q : x=2x = \sqrt{2}x=2​ is a point of inflection of f R : f' is increasing for x>2x > \sqrt{2}x>2​
  1. (A)Only P and Q
  2. (B)Only P and R
  3. (C)Only Q and R
  4. (D)All, P, Q and R

Correct answer: (D)

Step-by-step solution →
Q79·MathematicsNumerical
Let S={θ∈(0, 2π):7cos⁡2θ−3sin⁡2θ−2cos⁡22θ=2}S = \left\{\theta \in (0,\,2\pi) : 7\cos^{2}\theta - 3\sin^{2}\theta - 2\cos^{2}2\theta = 2\right\}S={θ∈(0,2π):7cos2θ−3sin2θ−2cos22θ=2}. Then, the sum of roots of all the equations x2−2(tan⁡2θ+cot⁡2θ)x+6sin⁡2θ=0x^{2} - 2\left(\tan^{2}\theta + \cot^{2}\theta\right)x + 6\sin^{2}\theta = 0x2−2(tan2θ+cot2θ)x+6sin2θ=0, θ∈S\theta \in Sθ∈S, is __________.

Correct answer: 16

Step-by-step solution →
Q80·MathematicsNumerical
Let the mean and the variance of 20 observations x1, x2,…x20x_{1},\,x_{2},\ldots x_{20}x1​,x2​,…x20​ be 15 and 9, respectively. For α∈R\alpha \in Rα∈R, if the mean of (x1+α)2, (x2+α)2,…, (x20+α)2\left(x_{1}+\alpha\right)^{2},\,\left(x_{2}+\alpha\right)^{2},\ldots,\,\left(x_{20}+\alpha\right)^{2}(x1​+α)2,(x2​+α)2,…,(x20​+α)2 is 178, then the square of the maximum value of α\alphaα is equal to __________.

Correct answer: 4

Step-by-step solution →
Q81·MathematicsNumerical
Let a line with direction ratios a, –4a, –7 be perpendicular to the lines with direction ratios 3, –1, 2b and b, a, –2. If the point of intersection of the line x+1a2+b2=y−2a2−b2=z1\frac{x+1}{a^{2}+b^{2}} = \frac{y-2}{a^{2}-b^{2}} = \frac{z}{1}a2+b2x+1​=a2−b2y−2​=1z​ and the plane x−y+z=0x - y + z = 0x−y+z=0 is (α, β, γ)(\alpha,\,\beta,\,\gamma)(α,β,γ), then α+β+γ\alpha + \beta + \gammaα+β+γ is equal to __________.

Correct answer: 10

Step-by-step solution →
Q82·MathematicsNumerical
Let a1, a2, a3,…a_{1},\,a_{2},\,a_{3},\ldotsa1​,a2​,a3​,… be an A.P. If ∑r=1∞ar2r=4\sum_{r=1}^{\infty}\frac{a_{r}}{2^{r}} = 4∑r=1∞​2rar​​=4, then 4a24a_{2}4a2​ is equal to __________.

Correct answer: 16

Step-by-step solution →
Q83·MathematicsNumerical
Let the ratio of the fifth term from the beginning to the fifth term from the end in the binomial expansion of (24+134)n\left(\sqrt[4]{2}+\frac{1}{\sqrt[4]{3}}\right)^{n}(42​+43​1​)n, in the increasing powers of 134\frac{1}{\sqrt[4]{3}}43​1​ be 64:1\sqrt[4]{6} : 146​:1. If the sixth term from the beginning is α34\frac{\alpha}{\sqrt[4]{3}}43​α​, then α\alphaα is equal to __________.

Correct answer: 84

Step-by-step solution →
Q84·MathematicsNumerical
The number of matrices of order 3×33 \times 33×3, whose entries are either 0 or 1 and the sum of all the entries is a prime number, is__________.

Correct answer: 282

Step-by-step solution →
Q85·MathematicsNumerical
Let p and p + 2 be prime numbers and let Δ=∣p!(p+1)!(p+2)!(p+1)!(p+2)!(p+3)!(p+2)!(p+3)!(p+4)!∣\Delta = \begin{vmatrix} p! & (p+1)! & (p+2)! \\ (p+1)! & (p+2)! & (p+3)! \\ (p+2)! & (p+3)! & (p+4)! \end{vmatrix}Δ=​p!(p+1)!(p+2)!​(p+1)!(p+2)!(p+3)!​(p+2)!(p+3)!(p+4)!​​ Then the sum of the maximum values of α\alphaα and β\betaβ, such that pαp^{\alpha}pα and (p+2)β(p+2)^{\beta}(p+2)β divide Δ\DeltaΔ, is __________.

Correct answer: 4

Step-by-step solution →
Q86·MathematicsNumerical
If 12×3×4+13×4×5+14×5×6+…+1100×101×102=k101\frac{1}{2 \times 3 \times 4} + \frac{1}{3 \times 4 \times 5} + \frac{1}{4 \times 5 \times 6} + \ldots + \frac{1}{100 \times 101 \times 102} = \frac{k}{101}2×3×41​+3×4×51​+4×5×61​+…+100×101×1021​=101k​, then 34 k is equal to _____.

Correct answer: 286

Step-by-step solution →
Q87·MathematicsNumerical
Let S={4, 6, 9}S = \{4,\,6,\,9\}S={4,6,9} and T={9, 10, 11, …, 1000}T = \{9,\,10,\,11,\,\ldots,\,1000\}T={9,10,11,…,1000}. If A={a1+a2+…+ak:k∈N,  a1, a2, a3, …,ak∈S}A = \left\{a_{1}+a_{2}+\ldots+a_{k} : k \in N,\; a_{1},\,a_{2},\,a_{3},\,\ldots,a_{k} \in S\right\}A={a1​+a2​+…+ak​:k∈N,a1​,a2​,a3​,…,ak​∈S}, then the sum of all the elements in the set T−AT - AT−A is equal to _______.

Correct answer: 11

Step-by-step solution →
Q88·MathematicsNumerical
Let the mirror image of a circle c1:x2+y2−2x−6y+α=0c_{1} : x^{2}+y^{2}-2x-6y+\alpha = 0c1​:x2+y2−2x−6y+α=0 in line y=x+1y = x + 1y=x+1 be c2:5x2+5y2+10gx+10fy+38=0c_{2} : 5x^{2}+5y^{2}+10gx+10fy+38 = 0c2​:5x2+5y2+10gx+10fy+38=0. If r is the radius of circle c2c_{2}c2​, then α+6r2\alpha + 6r^{2}α+6r2 is equal to __________

Correct answer: 12

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
  • 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
  • Trigonometric Functions 144/186
  • 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
  • Some Basic Concepts in Chemistry 129/186
  • Electromagnetic Induction 120/186
  • Wave Optics 130/186
  • Area Under Curves 139/186
  • Kinetic Theory of Gases 135/186
  • Alcohols and Ethers 106/186
  • Oscillations 117/186
  • Alternating Currents 108/186
  • Organic Compounds Containing Halogens 109/186
  • Straight Lines 114/186
  • Atoms 112/186
  • Classification of Elements and Periodicity in Properties 113/186
  • Parabola 101/186
  • Statistics 118/186
  • Ellipse 103/186
  • Isolation of Metals 106/186
  • Differentiability 91/186
  • s-Block Elements 88/186
  • Surface Chemistry 98/186
  • Environmental Chemistry 83/186
  • Experimental Skills 68/186
  • Electric Potential 63/186
  • Solid State 63/186
  • Chemistry in Everyday Life 60/186
  • Diazonium Salts and Reactions 53/186
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