Jarvis OS
PYQ papersPricingSign inGet started
  1. Home
  2. /JEE Main PYQs
  3. /2022
  4. /27 Jun Shift 1

JEE Main 27 June 2022 Shift 1 Question Paper with Answers

27 June 2022 · June session · 89 questions

89 of the 90 questions from the JEE Main 27 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 27 June 2022 Shift 1

Q1·PhysicsSingle correct
A projectile is launched at an angle 'α' with the horizontal with a velocity 20 ms−1^{-1}−1 . After 10 s, its inclination with horizontal is 'β'. The value of tanβ will be : (g = 10 ms−2^{-2}−2)
  1. (A)tan⁡α+5sec⁡α\tan\alpha + 5\sec\alphatanα+5secα
  2. (B)tan⁡α−5sec⁡α\tan\alpha - 5\sec\alphatanα−5secα
  3. (C)2tan⁡α−5sec⁡α2\tan\alpha - 5\sec\alpha2tanα−5secα
  4. (D)2tan⁡α+5sec⁡α2\tan\alpha + 5\sec\alpha2tanα+5secα

Correct answer: (B)

Step-by-step solution →
Q2·PhysicsSingle correct
A girl standing on road holds her umbrella at 45º with the vertical to keep the rain away. If she starts running without umbrella with a speed of 152 kmh−115\sqrt{2}\,kmh^{-1}152​kmh−1, the rain drops hit her head vertically. The speed of rain drops with respect to the moving girl is :
  1. (A)30 kmh−130\,kmh^{-1}30kmh−1
  2. (B)252kmh−1\frac{25}{\sqrt{2}}kmh^{-1}2​25​kmh−1
  3. (C)302kmh−1\frac{30}{\sqrt{2}}kmh^{-1}2​30​kmh−1
  4. (D)25 kmh−125\,kmh^{-1}25kmh−1

Correct answer: (C)

Step-by-step solution →
Q3·PhysicsSingle correct
A sliver wire has mass (0.6±0.006)(0.6 \pm 0.006)(0.6±0.006) g, radius (0.5±0.005)(0.5 \pm 0.005)(0.5±0.005) mm and length (4±0.04)(4 \pm 0.04)(4±0.04) cm. The maximum percentage error in the measurement of its density will be :
  1. (A)4%
  2. (B)3%
  3. (C)6%
  4. (D)7%

Correct answer: (A)

Step-by-step solution →
Q4·PhysicsSingle correct
A system of two blocks of masses m = 2 kg and M = 8 kg is placed on a smooth table as shown in figure. The coefficient of static friction between two blocks is 0.5. The maximum horizontal force F that can be applied to the block of mass M so that the blocks move together will be :
  1. (A)9.8 N
  2. (B)39.2 N
  3. (C)49 N
  4. (D)78.4 N

Correct answer: (C)

Step-by-step solution →
Q5·PhysicsSingle correct
Two blocks of masses 10 kg and 30 kg are placed on the same straight line with coordinates (0, 0) cm and (x, 0) cm respectively. The block of 10 kg is moved on the same line through a distance of 6 cm towards the other block. The distance through which the block of 30 kg must be moved to keep the position of centre of mass of the system unchanged is :
  1. (A)4 cm towards the 10 kg block
  2. (B)2 cm away from the 10 kg block
  3. (C)2 cm towards the 10 kg block
  4. (D)4 cm away from the 10 kg block

Correct answer: (C)

Step-by-step solution →
Q6·PhysicsSingle correct
A 72 Ω galvanometer is shunted by a resistance of 8 Ω. The percentage of the total current which passes through the galvanometer is :
  1. (A)0.1%
  2. (B)10 %
  3. (C)25%
  4. (D)0.25%

Correct answer: (B)

Step-by-step solution →
Q7·PhysicsSingle correct
Given below are two statements : Statement I : The law of gravitation holds good for any pair of bodies in the universe. Statement II : The weight of any person becomes zero when the person is at the centre of the earth. 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 are false
  4. (D)Statement I is false but Statement II is true

Correct answer: (A)

Step-by-step solution →
Q8·PhysicsSingle correct
What percentage of kinetic energy of a moving particle is transferred to a stationary particle when it strikes the stationary particle of 5 times its mass? (Assume the collision to be head-on elastic collision)
  1. (A)50.0%
  2. (B)66.6%
  3. (C)55.5%
  4. (D)33.3%

Correct answer: (C)

Step-by-step solution →
Q9·PhysicsSingle correct
The velocity of a small ball of mass 'm' and density d1d_{1}d1​, when dropped in a container filled with glycerine, becomes constant after some time. If the density of glycerine is d2d_{2}d2​, then the viscous force acting on the ball, will be :
  1. (A)mg(1−d1d2)mg\left(1-\frac{d_{1}}{d_{2}}\right)mg(1−d2​d1​​)
  2. (B)mg(1−d2d1)mg\left(1-\frac{d_{2}}{d_{1}}\right)mg(1−d1​d2​​)
  3. (C)mg(d1d2−1)mg\left(\frac{d_{1}}{d_{2}}-1\right)mg(d2​d1​​−1)
  4. (D)mg(d2d1−1)mg\left(\frac{d_{2}}{d_{1}}-1\right)mg(d1​d2​​−1)

Correct answer: (B)

Step-by-step solution →
Q10·PhysicsSingle correct
The susceptibility of a paramagnetic material is 99. The permeability of the material in Wb/A-m is : [Permeability of free space μ0=4π×10−7 Wb/A−m\mu_{0} = 4\pi \times 10^{-7}\,Wb/A-mμ0​=4π×10−7Wb/A−m ]
  1. (A)4π×10−74\pi \times 10^{-7}4π×10−7
  2. (B)4π×10−44\pi \times 10^{-4}4π×10−4
  3. (C)4π×10−54\pi \times 10^{-5}4π×10−5
  4. (D)4π×10−64\pi \times 10^{-6}4π×10−6

Correct answer: (C)

Step-by-step solution →
Q11·PhysicsSingle correct
The current flowing through an ac circuit is given by I=5sin⁡(120πt) AI = 5\sin(120\pi t)\,AI=5sin(120πt)A How long will the current take to reach the peak value starting from zero?
  1. (A)160s\frac{1}{60}s601​s
  2. (B)60s
  3. (C)1120s\frac{1}{120}s1201​s
  4. (D)1240s\frac{1}{240}s2401​s

Correct answer: (D)

Step-by-step solution →
Q12·PhysicsSingle correct
Mach List-I with List – II : Choose the correct answer from the options given below :
List-IList-Ii
a.Ultraviolet raysi.Study crystal structure
b.Microwavesii.Greenhouse effect
c.Infrared wavesiii.Sterilizing surgical instrument
d.X-raysiv.Radar system
  1. (A)(a) – (iii), (b) – (iv), (c) – (ii), (d) – (i)
  2. (B)(a) – (iii), (b) – (i), (c) – (ii), (d) – (iv)
  3. (C)(a) – (iv), (b) – (iii), (c) – (ii), (d) – (i)
  4. (D)(a) – (iii), (b) – (iv), (c) – (i), (d) – (ii)

Correct answer: (A)

Step-by-step solution →
Q13·PhysicsSingle correct
An α particle and a carbon 12 atom has same kinetic energy K. The ratio of their de-Broglie wavelength (λa:λC12)(\lambda_{a} : \lambda_{C12})(λa​:λC12​) is :
  1. (A)1:31:\sqrt{3}1:3​
  2. (B)3:1\sqrt{3}:13​:1
  3. (C)3:13:13:1
  4. (D)2:32:\sqrt{3}2:3​

Correct answer: (B)

Step-by-step solution →
Q14·PhysicsSingle correct
A force of 10N acts on a charged particle placed between two plates of a charged capacitor. If one plate of capacitor is removed, then the force acting on that particle will be :
  1. (A)5 N
  2. (B)10 N
  3. (C)20 N
  4. (D)Zero

Correct answer: (A)

Step-by-step solution →
Q15·PhysicsSingle correct
The displacement of simple harmonic oscillator after 3 seconds starting from its mean position is equal to half of its amplitude. The time period of harmonic motion is :
  1. (A)6 s
  2. (B)8 s
  3. (C)12s
  4. (D)36 s

Correct answer: (D)

Step-by-step solution →
Q16·PhysicsSingle correct
An observer moves towards a stationary source of sound with a velocity equal to one-fifth of the velocity of sound. The percentage change in the frequency will be :
  1. (A)20%
  2. (B)10%
  3. (C)5%
  4. (D)0%

Correct answer: (A)

Step-by-step solution →
Q17·PhysicsSingle correct
Consider a light ray travelling in air is incident into a medium of refractive index 2n\sqrt{2n}2n​ . The incident angle is twice that of refracting angle. Then, the angle of incidence will be :
  1. (A)sin⁡−1(n)\sin^{-1}\left(\sqrt{n}\right)sin−1(n​)
  2. (B)cos⁡−1(n2)\cos^{-1}\left(\sqrt{\frac{n}{2}}\right)cos−1(2n​​)
  3. (C)sin⁡−1(2n)\sin^{-1}\left(\sqrt{2n}\right)sin−1(2n​)
  4. (D)2cos⁡−1(n2)2\cos^{-1}\left(\sqrt{\frac{n}{2}}\right)2cos−1(2n​​)

Correct answer: (D)

Step-by-step solution →
Q18·PhysicsSingle correct
A hydrogen atom in is ground state absorbs 10.2 eV of energy. The angular momentum of electron of the hydrogen atom will increase by the value of : (Given, Plank's constant =6.6×10−34= 6.6 \times 10^{-34}=6.6×10−34 Js)
  1. (A)2.10×10−34 Js2.10 \times 10^{-34}\ Js2.10×10−34 Js
  2. (B)1.05×10−34 Js1.05 \times 10^{-34}\ Js1.05×10−34 Js
  3. (C)3.15×10−34 Js3.15 \times 10^{-34}\ Js3.15×10−34 Js
  4. (D)4.2×10−34 Js4.2 \times 10^{-34}\ Js4.2×10−34 Js

Correct answer: (B)

Step-by-step solution →
Q19·PhysicsSingle correct
Identify the correct Logic Gate for the following output (Y) of two inputs A and B.
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (B)

Step-by-step solution →
Q20·PhysicsSingle correct
A mixture of hydrogen and oxygen has volume 2000 cm3cm^{3}cm3, temperature 300 K, pressure 100 kPa and mass 0.76 g The ratio of number of moles of hydrogen to number of moles of oxygen in the mixture will be :
  1. (A)13\frac{1}{3}31​
  2. (B)31\frac{3}{1}13​
  3. (C)116\frac{1}{16}161​
  4. (D)161\frac{16}{1}116​

Correct answer: (B)

Step-by-step solution →
Q21·PhysicsNumerical
In a carnot engine, the temperature of reservoir is 527∘^{\circ}∘C and that of sink is 200 K. If the workdone by the engine when it transfers heat from reservoir to sink is 12000 kJ, the quantity of heat absorbed by the engine from reservoir is ________ ×106\times 10^{6}×106 J.

Correct answer: 16

Step-by-step solution →
Q22·PhysicsNumerical
A 220 V, 50 Hz AC source is connected to a 25 V, 5 W lamp and an additional resistance R in series (as shown in figure) to run the lamp at its peak brightness, then the value of R (in ohm) will be _______ .

Correct answer: 975

Step-by-step solution →
Q23·PhysicsNumerical
In Young's double slit experiment the two slits are 0.6 mm distance apart. Interference pattern is observed on a screen at a distance 80 cm from the slits. The first dark fringe is observed on the screen directly opposite to one of the slits. The wavelength of light will be ______ nm.

Correct answer: 450

Step-by-step solution →
Q24·PhysicsNumerical
A beam of monochromatic light is used to excite the electron in Li++Li^{++}Li++ from the first orbit to the third orbit. The wavelength of monochromatic light is found to be x×10−10mx \times 10^{-10} mx×10−10m . The value of x is _______ . [Given hc = 1242 eV nm]

Correct answer: 114

Step-by-step solution →
Q25·PhysicsNumerical
A cell, shunted by a 8 Ω\OmegaΩ resistance, is balanced across a potentiometer wire of length 3m. The balancing length is 2 m when the cell is shunted by 4Ω\OmegaΩ resistance. The value of internal resistance of the cell will be ______ Ω\OmegaΩ .

Correct answer: 8

Step-by-step solution →
Q26·PhysicsNumerical
The current density in a cylindrical wire of radius 4 mm is 4×106Am−24 \times 10^{6} Am^{-2}4×106Am−2. The current through the outer portion of the wire between radial distance R2\frac{R}{2}2R​ and R is ____ π\piπ A.

Correct answer: 48

Step-by-step solution →
Q27·PhysicsNumerical
A capacitor of capacitance 50 pF is charged by 100 V source. It is then connected to another uncharged identical capacitor. Electrostatic energy loss in the process is ____ nJ.

Correct answer: 125

Step-by-step solution →
Q28·PhysicsNumerical
The height of a transmitting antenna at the top of a tower is 25 m and that of receiving antenna is, 49 m. The maximum distance between them, for satisfactory communication in LOS (Line-Of-Sight) is K5×102mK\sqrt{5} \times 10^{2} mK5​×102m . The value of K is _______ . [Assume radius of Earth is 64×10+5m64 \times 10^{+5} m64×10+5m] (Calculate upto nearest integer value)

Correct answer: 192

Step-by-step solution →
Q29·PhysicsNumerical
The area of cross-section of a large tank is 0.5 m2m^{2}m2. It has a narrow opening near the bottom having area of cross-section 1 cm2cm^{2}cm2. A load of 25 kg is applied on the water at the top in the tank. Neglecting the speed of water in the tank, the velocity of the water, coming out of the opening at the time when the height of water level in the tank is 40 cm above the bottom, will be ________ cms−1cms^{-1}cms−1. [Take g = 10 ms−2ms^{-2}ms−2]

Correct answer: 300

Step-by-step solution →
Q30·PhysicsNumerical
A pendulum of length 2 m consists of a wooden bob of mass 50 g. A bullet of mass 75 g is fired towards the stationary bob with a speed v. The bullet emerges out of the bob with a speed v3\frac{v}{3}3v​ and the bob just completes the vertical circle. The value of v is ________ ms−1ms^{-1}ms−1. (if g = 10 m/s2m/s^{2}m/s2)

Correct answer: 10

Step-by-step solution →

Chemistry — JEE Main 27 June 2022 Shift 1

Q31·ChemistrySingle correct
Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R) Assertion (A) : At 10°C, the density of a 5M solution of KCl [atomic masses of K and Cl are 39 & 35.5 g mol−1mol^{-1}mol−1]. The solution is cooled to -21°C. The molality of the solution will remain unchanged. Reason (R) : The molality of a solution does not change with temperature as mass remains unaffected with temperature. In the light of the 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: (A)

Step-by-step solution →
Q32·ChemistrySingle correct
Based upon VSEPR theory, match the shape (geometry) of the molecules in List-I with the molecules in List-II and select the most appropriate option List-I (Shape) (A) T-shaped (B) Trigonal planar (C) Square planar (D) See-saw List-II (Molecules) (I) XeF4XeF_{4}XeF4​ (II) SF4SF_{4}SF4​ (III) ClF3ClF_{3}ClF3​ (IV) BF3BF_{3}BF3​
  1. (A)(A) – I, (B) – (II), (C) – (III), (D) – (IV)
  2. (B)(A) – (III), (B) – (IV), (C) – (I), (D) – (II)
  3. (C)(A) – (III), (B) – (IV), (C) – (II), (D) – (I)
  4. (D)(A) – (IV), (B) – (III), (C) – (I), (D) – (II)

Correct answer: (B)

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.Spontaneous processI.ΔH<0\Delta H < 0ΔH<0
B.Process with ΔP=0\Delta P = 0ΔP=0, ΔT=0\Delta T = 0ΔT=0II.ΔGT,P<0\Delta G_{T,P} < 0ΔGT,P​<0
C.ΔHreaction\Delta H_{reaction}ΔHreaction​III.Isothermal and isobaric process
D.Exothermic processIV.[Bond energies of molecules in reactants] - [Bond energies of product molecules
  1. (A)(A) – (III), (B) – (II), (C) – (IV), (D) – (I)
  2. (B)(A) – (II), (B) – (III), (C) – (IV), (D) – (I)
  3. (C)(A) – (II), (B) – (III), (C) – (I), (D) – (IV)
  4. (D)(A) – (II), (B) – (I), (C) – (III), (D) – (IV)

Correct answer: (B)

Step-by-step solution →
Q34·ChemistrySingle correct
Match List-I with List-II List-I (A) Lyophilic colloid (B) Emulsion (C) Positively charged (D) Negatively charged colloid List-II (I) Liquid-liquid colloid (II) protective colloid (III) FeCl3FeCl_{3}FeCl3​ + NaOH (IV) FeCl3FeCl_{3}FeCl3​ + hot water Choose the correct answer from the options given below:
  1. (A)(A) – (II), (B) – (I), (C) – (IV), (D) – (III)
  2. (B)(A) – (III), (B) – (I), (C) – (IV), (D) – (II)
  3. (C)(A) – (II), (B) – (I), (C) – (III), (D) – (IV)
  4. (D)(A) – (III), (B) – (II), (C) – (I), (D) – (IV)

Correct answer: (A)

Step-by-step solution →
Q35·ChemistrySingle correct
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason(R) Assertion (A): The ionic radii of O2−O^{2-}O2− and Mg2+Mg^{2+}Mg2+ are same. Reason (R) : Both O2−O^{2-}O2− and Mg2+Mg^{2+}Mg2+ are isoelectronic species In the light of the 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 →
Q36·ChemistrySingle correct
Match List-I with List-II List-I (A) Concentration of gold ore (B) Leaching of alumina (C) Froth stabiliser (D) Blister copper List-II (I) Aniline (II) NaOH (III) SO2SO_{2}SO2​ (IV) NaCN Choose the correct answer from the options given below.
  1. (A)(A) – (IV), (B) – (III), (C) – (II), (D) – (I)
  2. (B)(A) – (IV), (B) – (II), (C) – (I), (D) – (III)
  3. (C)(A) – (III), (B) – (II), (C) – (I), (D) – (IV)
  4. (D)(A) – (II), (B) – (IV), (C) – (III), (D) – (I)

Correct answer: (B)

Step-by-step solution →
Q37·ChemistrySingle correct
Addition of H2SO4H_{2}SO_{4}H2​SO4​ to BaO2BaO_{2}BaO2​ produces:
  1. (A)BaO, SO2SO_{2}SO2​ and H2OH_{2}OH2​O
  2. (B)BaHSO4BaHSO_{4}BaHSO4​ and O2O_{2}O2​
  3. (C)BaSO4BaSO_{4}BaSO4​, H2H_{2}H2​ and O2O_{2}O2​
  4. (D)BaSO4BaSO_{4}BaSO4​ and H2O2H_{2}O_{2}H2​O2​

Correct answer: (D)

Step-by-step solution →
Q38·ChemistrySingle correct
BeCl2BeCl_{2}BeCl2​ reacts with LiAlH4LiAlH_{4}LiAlH4​ to give
  1. (A)Be + Li[AlCl4]Li[AlCl_{4}]Li[AlCl4​] + H2H_{2}H2​
  2. (B)Be + AlH3AlH_{3}AlH3​ + LiCl + HCl
  3. (C)BeH2BeH_{2}BeH2​ + LiCl + AlCl3AlCl_{3}AlCl3​
  4. (D)BeH2BeH_{2}BeH2​ + Li[AlCl4]Li[AlCl_{4}]Li[AlCl4​]

Correct answer: (C)

Step-by-step solution →
Q39·ChemistrySingle correct
Match List-I with List-II List-I (Si-Compounds) (A) (CH3)4Si(CH_{3})_{4}Si(CH3​)4​Si (B) (CH3)Si(OH)3(CH_{3})Si(OH)_{3}(CH3​)Si(OH)3​ (C) (CH3)2Si(OH)2(CH_{3})_{2}Si(OH)_{2}(CH3​)2​Si(OH)2​ (D) (CH3)3Si(OH)(CH_{3})_{3}Si(OH)(CH3​)3​Si(OH) List-II (Si-Polymeric/other products) (I) Chain silicone (II) Dimeric silicone (III) Silane (IV) 2D – Silicone Choose the correct answer from the options given below:
  1. (A)(A) – (III), (B) – (II), (C) – (I), (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) – (III), (B) – (IV), (C) – (I), (D) – (II)

Correct answer: (D)

Step-by-step solution →
Q40·ChemistrySingle correct
Heating white phosphorus with conc. NaOH solution gives mainly
  1. (A)Na3PNa_{3}PNa3​P and H2OH_{2}OH2​O
  2. (B)H3POH_{3}POH3​PO and NaH
  3. (C)P(OH)3P(OH)_{3}P(OH)3​ and NaH2PO4NaH_{2}PO_{4}NaH2​PO4​
  4. (D)PH3PH_{3}PH3​ and NaH2PO2NaH_{2}PO_{2}NaH2​PO2​

Correct answer: (D)

Step-by-step solution →
Q41·ChemistrySingle correct
Which of the following will have maximum stabilization due to crystal field?
  1. (A)[Ti(H2O)6]3+[Ti(H_{2}O)_{6}]^{3+}[Ti(H2​O)6​]3+
  2. (B)[Co(H2O)6]2+[Co(H_{2}O)_{6}]^{2+}[Co(H2​O)6​]2+
  3. (C)[Co(CN)6]3−[Co(CN)_{6}]^{3-}[Co(CN)6​]3−
  4. (D)[Cu(NH3)4]2+[Cu(NH_{3})_{4}]^{2+}[Cu(NH3​)4​]2+

Correct answer: (C)

Step-by-step solution →
Q42·ChemistrySingle correct
Given below are two statements: Statement I: Classical smog occurs in cool humid climate. It is a reducing mixture of smoke, fog and sulphur dioxide Statement II: Photochemical smog has components, ozone, nitric oxide, acrolein, formaldehyde, PAN etc. In the light of 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 Statement II is correct

Correct answer: (A)

Step-by-step solution →
Q43·ChemistrySingle correct
Which of the following is structure of a separating funnel?
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (A)

Step-by-step solution →
Q44·ChemistrySingle correct
'A' and 'B' respectively are: A →(2)Zn−H2O(1)O3\xrightarrow[(2) Zn-H_{2}O]{(1) O_{3}}(1)O3​(2)Zn−H2​O​ Ethane-1,2-dicarbaldehyde + Glyoxal/Oxaldehyde B →(2)Zn−H2O(1)O3\xrightarrow[(2) Zn-H_{2}O]{(1) O_{3}}(1)O3​(2)Zn−H2​O​ 5-oxohexanal
  1. (A)1-methylcyclohex-1, 3-diene & cyclopentene
  2. (B)Cyclohex-1, 3-diene & cyclopentene
  3. (C)1-methylcyclohex-1,4-diene & 1-methylcyclopent-1-ene
  4. (D)Cyclohex-1,3-diene & 1-methylcyclopent-1-ene

Correct answer: (D)

Step-by-step solution →
Q45·ChemistrySingle correct
The major product of the following reaction is:
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (A)

Step-by-step solution →
Q46·ChemistrySingle correct
Which of the following reactions will yield benzaldehyde as a product?
  1. (A)(B) and (C)
  2. (B)(C) and (D)
  3. (C)(A) and (D)
  4. (D)(A) and (C)

Correct answer: (C)

Step-by-step solution →
Q47·ChemistrySingle correct
Given below are two statements: Statements-I : In Hofmann degradation reaction, the migration of only an alkyl group takes place from carbonyl carbon of the amide to the nitrogen atom. Statement-II : The group is migrated in Hofmann degradation reaction to electron deficient atom. In the light of the above statement, choose the most appropriate answer from the options given 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 Statement-II is correct

Correct answer: (D)

Step-by-step solution →
Q48·ChemistrySingle correct
Match List-I with List-II List-I (Polymer) (A) Bakelite (B) Glyptal (C) PVC (D) Polystyrene List-II (Used in) (I) Radio and television Cabinets (II) Electrical switches (III) Paints and Lacquers (IV) Water pipes Choose the correct answer from the options given below:
  1. (A)(A) – (II), (B) – (III), (C) – (IV), (D) – (I)
  2. (B)(A) – (I), (B) – (II), (C) – (III), (D) – (IV)
  3. (C)(A) – (IV), (B) – (III), (C) – (II), (D) – (I)
  4. (D)(A) – (II), (B) – (III), (C) – (I), (D) – (IV)

Correct answer: (A)

Step-by-step solution →
Q49·ChemistrySingle correct
L-isomer of a compound 'A' (C4H8O4C_{4}H_{8}O_{4}C4​H8​O4​) gives a positive test with [Ag(NH3)2]+[Ag(NH_{3})_{2}]^{+}[Ag(NH3​)2​]+. Treatment of 'A' with acetic anhydride yield triacetate derivative. Compound 'A' produces an optically active compound (B) and an optically inactive compound (C) on treatment with bromine water and HNO3HNO_{3}HNO3​ respectively, compound (A) is:
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (A)

Step-by-step solution →
Q50·ChemistrySingle correct
Match List-I with List-II List-I (A) (B) (C) C17H35COO−Na+C_{17}H_{35}COO^{-}Na^{+}C17​H35​COO−Na+ + Na2CO3Na_{2}CO_{3}Na2​CO3​ + Rosinate (D) CH3(CH2)16COO(CH2CH2O)nCH2CH2OHCH_{3}(CH_{2})_{16}COO(CH_{2}CH_{2}O)_{n}CH_{2}CH_{2}OHCH3​(CH2​)16​COO(CH2​CH2​O)n​CH2​CH2​OH List-II (I) Dishwashing powder (II) Toothpaste (III) Laundry soap (IV) Hair conditioner
  1. (A)(A) – (III), (B) – (II), (C) – (IV), (D) – (I)
  2. (B)(A) – (IV), (B) – (II), (C) – (III), (D) – (I)
  3. (C)(A) – (IV), (B) – (III), (C) – (II), (D) – (I)
  4. (D)(A) – (III), (B) – (IV), (C) – (I), (D) – (II)

Correct answer: (B)

Step-by-step solution →
Q51·ChemistryNumerical
Metal deficiency defect is shown by Fe0.93OFe_{0.93}OFe0.93​O. In the crystal, some Fe2+Fe^{2+}Fe2+ cations are missing and loss of positive charge is compensated by the presence of Fe3+Fe^{3+}Fe3+ ions. The percentage of Fe2+Fe^{2+}Fe2+ ions in the Fe0.93OFe_{0.93}OFe0.93​O crystals is __________ . (Nearest integer)

Correct answer: 85

Step-by-step solution →
Q52·ChemistryNumerical
If the uncertainty in velocity and position of a minute particle in space are, 2.4 × 10−2610^{-26}10−26 (ms−1^{-1}−1) and 10−710^{-7}10−7(m) respectively. The mass of the particle in g is __________ (Nearest integer) (Given : h = 6.626 × 10−3410^{-34}10−34 Js)

Correct answer: 22

Step-by-step solution →
Q53·ChemistryNumerical
2g of a non-volatile non-electrolyte solute is dissolved in 200 g of two different solvents A and B whose ebullioscopic constants are in the ratio of 1 : 8. The elevation in boiling points of A and B are in the ratio xy\frac{x}{y}yx​ (x : y). The value of y is__________ (Nearest integer)

Correct answer: 8

Step-by-step solution →
Q54·ChemistryNumerical
2NOCl(g) ⇌ 2NO(g) + Cl2Cl_{2}Cl2​(g) In an experiment, 2.0 moles of NOCl was placed in a one-litre flask and the concentration of NO after equilibrium established, was found to be 0.4 mol/L. The equilibrium constant at 30°C is _______ × 10−410^{-4}10−4.

Correct answer: 125

Step-by-step solution →
Q55·ChemistryNumerical
The limiting molar conductivities of NaI, NaNO3NaNO_{3}NaNO3​ and AgNO3AgNO_{3}AgNO3​ are 12.7, 12.0 and 13.3 mS m2^{2}2 mol−1^{-1}−1, respectively (all at 25°C). The limiting molar conductivity of AgI at this temperature is _____ mS m2^{2}2 mol−1^{-1}−1

Correct answer: 14

Step-by-step solution →
Q56·ChemistryNumerical
The rate constant for a first order reaction is given by the following equation: ln k = 33.24 - 2.0×104KT\frac{2.0 \times 10^{4} K}{T}T2.0×104K​ The Activation energy for the reaction is given by _______ kJ mol−1^{-1}−1. (In Nearest integer) (Given: R = 8.3 J K−1^{-1}−1 mol−1^{-1}−1)

Correct answer: 166

Step-by-step solution →
Q57·ChemistryNumerical
The number of statement(s) correct from the following for copper (at no. 29) is/are _______ (A) Cu(II) complexes are always paramagnetic (B) Cu(I) complexes are generally colourless (C) Cu(I) is easily oxidized (D) In Fehling solution, the active reagent has Cu(I)

Correct answer: 3

Step-by-step solution →
Q58·ChemistryNumerical
Acidified potassium permanganate solution oxidises oxalic acid. The spin-only magnetic moment of the manganese product formed from the above reaction is ______ B.M. (Nearest Integer)

Correct answer: 6

Step-by-step solution →
Q59·ChemistryNumerical
Two elements A and B which form 0.15 moles of A2BA_{2}BA2​B and AB3AB_{3}AB3​ type compounds. If both A2BA_{2}BA2​B and AB3AB_{3}AB3​ weigh equally, then the atomic weight of A is _____ times of atomic weight of B.

Correct answer: 2

Step-by-step solution →
Q60·Chemistry·IsomerismNumerical
Total number of possible stereoisomers of dimethyl cyclopentane is ___________

Correct answer: 6

Step-by-step solution →

Mathematics — JEE Main 27 June 2022 Shift 1

Q61·MathematicsSingle correct
The area of the polygon, whose vertices are the non-real roots of the equation z‾=iz2\overline{z} = iz^{2}z=iz2 is :
  1. (A)334\frac{3\sqrt{3}}{4}433​​
  2. (B)332\frac{3\sqrt{3}}{2}233​​
  3. (C)32\frac{3}{2}23​
  4. (D)34\frac{3}{4}43​

Correct answer: (A)

Step-by-step solution →
Q62·MathematicsSingle correct
Let the system of linear equations x+2y+z=2x + 2y + z = 2x+2y+z=2, αx+3y−z=α\alpha x + 3y - z = \alphaαx+3y−z=α, −αx+y+2z=−α-\alpha x + y + 2z = -\alpha−αx+y+2z=−α be inconsistent. Then α\alphaα is equal to :
  1. (A)52\frac{5}{2}25​
  2. (B)−52-\frac{5}{2}−25​
  3. (C)72\frac{7}{2}27​
  4. (D)−72-\frac{7}{2}−27​

Correct answer: (D)

Step-by-step solution →
Q63·MathematicsSingle correct
If x=∑n=0∞anx = \sum_{n=0}^{\infty} a^{n}x=∑n=0∞​an, y=∑n=0∞bny = \sum_{n=0}^{\infty} b^{n}y=∑n=0∞​bn, z=∑n=0∞cnz = \sum_{n=0}^{\infty} c^{n}z=∑n=0∞​cn, where a, b, c are in A.P. and ∣a∣<1|a| < 1∣a∣<1, ∣b∣<1|b| < 1∣b∣<1, ∣c∣<1|c| < 1∣c∣<1, abc≠0abc \neq 0abc=0, then
  1. (A)x, y, z are in A.P.
  2. (B)x, y, z are in G.P.
  3. (C)1x,1y,1z\frac{1}{x}, \frac{1}{y}, \frac{1}{z}x1​,y1​,z1​ are in A.P.
  4. (D)1x+1y+1z=1−(a+b+c)\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1 - \left(a + b + c\right)x1​+y1​+z1​=1−(a+b+c)

Correct answer: (C)

Step-by-step solution →
Q64·MathematicsSingle correct
Let dydx=ax−by+abx+cy+a\frac{dy}{dx} = \frac{ax - by + a}{bx + cy + a}dxdy​=bx+cy+aax−by+a​, where a, b, c are constants, represent a circle passing through the point (2, 5). Then the shortest distance of the point (11, 6) from this circle is :
  1. (A)10
  2. (B)8
  3. (C)7
  4. (D)5

Correct answer: (B)

Step-by-step solution →
Q65·Mathematics·Limits and ContinuitySingle correct
Let a be an integer such that lim⁡x→718−[1−x][x−3a]\lim_{x \to 7} \frac{18 - \left[1 - x\right]}{\left[x - 3a\right]}limx→7​[x−3a]18−[1−x]​ exists, where [t] is greatest integer ≤\leq≤ t. Then a is equal to :
  1. (A)-6
  2. (B)-2
  3. (C)2
  4. (D)6

Correct answer: (A)

Step-by-step solution →
Q66·MathematicsSingle correct
The number of distinct real roots of x4−4x+1=0x^{4} - 4x + 1 = 0x4−4x+1=0 is :
  1. (A)4
  2. (B)2
  3. (C)1
  4. (D)0

Correct answer: (B)

Step-by-step solution →
Q67·MathematicsSingle correct
The lengths of the sides of a triangle are 10+x210 + x^{2}10+x2, 10+x210 + x^{2}10+x2 and 20−2x220 - 2x^{2}20−2x2. If for x = k, the area of the triangle is maximum, then 3k23k^{2}3k2 is equal to :
  1. (A)5
  2. (B)8
  3. (C)10
  4. (D)12

Correct answer: (C)

Step-by-step solution →
Q68·Mathematics·DifferentiabilitySingle correct
If cos⁡−1(y2)=log⁡e(x5)5,∣y∣<2\cos^{-1}\left(\frac{y}{2}\right) = \log_{e}\left(\frac{x}{5}\right)^{5}, |y| < 2cos−1(2y​)=loge​(5x​)5,∣y∣<2, then :
  1. (A)x2y′′+xy′−25y=0x^{2}y'' + xy' - 25y = 0x2y′′+xy′−25y=0
  2. (B)x2y′′−xy′−25y=0x^{2}y'' - xy' - 25y = 0x2y′′−xy′−25y=0
  3. (C)x2y′′−xy′+25y=0x^{2}y'' - xy' + 25y = 0x2y′′−xy′+25y=0
  4. (D)x2y′′+xy′+25y=0x^{2}y'' + xy' + 25y = 0x2y′′+xy′+25y=0

Correct answer: (D)

Step-by-step solution →
Q69·MathematicsSingle correct
∫(x2+1)ex(x+1)2dx=f(x)ex+C\int \frac{\left(x^{2}+1\right)e^{x}}{\left(x+1\right)^{2}}dx = f\left(x\right)e^{x} + C∫(x+1)2(x2+1)ex​dx=f(x)ex+C, Where C is a constant, then d3fdx3\frac{d^{3}f}{dx^{3}}dx3d3f​ at x = 1 is equal to :
  1. (A)−34-\frac{3}{4}−43​
  2. (B)34\frac{3}{4}43​
  3. (C)−32-\frac{3}{2}−23​
  4. (D)32\frac{3}{2}23​

Correct answer: (B)

Step-by-step solution →
Q70·MathematicsSingle correct
The value of the integral ∫−22∣x3+x∣(ex∣x∣+1)dx\int_{-2}^{2} \frac{\left|x^{3} + x\right|}{\left(e^{x|x|} + 1\right)}dx∫−22​(ex∣x∣+1)∣x3+x∣​dx is equal to :
  1. (A)5e25e^{2}5e2
  2. (B)3e−23e^{-2}3e−2
  3. (C)4
  4. (D)6

Correct answer: (D)

Step-by-step solution →
Q71·MathematicsSingle correct
In an isosceles triangle ABC, the vertex A is (6, 1) and the equation of the base BC is 2x+y=42x + y = 42x+y=4. Let the point B lie on the line x+3y=7x + 3y = 7x+3y=7. If (α,β)(\alpha, \beta)(α,β) is the centroid ΔABC\Delta ABCΔABC, then 15(α+β)15\left(\alpha + \beta\right)15(α+β) is equal to :
  1. (A)39
  2. (B)41
  3. (C)51
  4. (D)63

Correct answer: (C)

Step-by-step solution →
Q72·MathematicsSingle correct
Let the eccentricity of an ellipse x2a2+y2b2=1,a>b,\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, a > b,a2x2​+b2y2​=1,a>b, be 14\frac{1}{4}41​. If this ellipse passes through the point (−425,3)\left(-4\sqrt{\frac{2}{5}}, 3\right)(−452​​,3), then a2+b2a^{2} + b^{2}a2+b2 is equal to :
  1. (A)29
  2. (B)31
  3. (C)32
  4. (D)34

Correct answer: (B)

Step-by-step solution →
Q73·MathematicsSingle correct
If two straight lines whose direction cosines are given by the relations l+m−n=0l + m - n = 0l+m−n=0, 3l2+m2+cnl=03l^{2} + m^{2} + cnl = 03l2+m2+cnl=0 are parallel, then the positive value of c is :
  1. (A)6
  2. (B)4
  3. (C)3
  4. (D)2

Correct answer: (A)

Step-by-step solution →
Q74·MathematicsSingle correct
Let a⃗=i^+j^−k^\vec{a} = \hat{i} + \hat{j} - \hat{k}a=i^+j^​−k^ and c⃗=2i^−3j^+2k^\vec{c} = 2\hat{i} - 3\hat{j} + 2\hat{k}c=2i^−3j^​+2k^. Then the number of vectors b⃗\vec{b}b such that b⃗×c⃗=a⃗\vec{b} \times \vec{c} = \vec{a}b×c=a and ∣b⃗∣∈{1,2,.....,10}\left|\vec{b}\right| \in \{1, 2, ....., 10\}​b​∈{1,2,.....,10} is :
  1. (A)0
  2. (B)1
  3. (C)2
  4. (D)3

Correct answer: (A)

Step-by-step solution →
Q75·MathematicsSingle correct
Five numbers x1x_1x1​, x2x_2x2​, x3x_3x3​, x4x_4x4​, x5x_5x5​ are randomly selected from the numbers 1, 2, 3,......, 18 and are arranged in the increasing order (x1<x2<x3<x4<x5x_1 < x_2 < x_3 < x_4 < x_5x1​<x2​<x3​<x4​<x5​). The probability that x2=7x_2 = 7x2​=7 and x4=11x_4 = 11x4​=11 is :
  1. (A)1136\frac{1}{136}1361​
  2. (B)172\frac{1}{72}721​
  3. (C)168\frac{1}{68}681​
  4. (D)134\frac{1}{34}341​

Correct answer: (C)

Step-by-step solution →
Q76·MathematicsSingle correct
Let X be a random variable having binomial distribution B(7, p). If P(X = 3) = 5P(X = 4), then the sum of the mean and the variance of X is :
  1. (A)10516\frac{105}{16}16105​
  2. (B)716\frac{7}{16}167​
  3. (C)7736\frac{77}{36}3677​
  4. (D)4916\frac{49}{16}1649​

Correct answer: (C)

Step-by-step solution →
Q77·MathematicsSingle correct
The value of cos⁡(2π7)+cos⁡(4π7)+cos⁡(6π7)\cos\left(\frac{2\pi}{7}\right)+\cos\left(\frac{4\pi}{7}\right)+\cos\left(\frac{6\pi}{7}\right)cos(72π​)+cos(74π​)+cos(76π​) is equal to :
  1. (A)-1
  2. (B)−12-\frac{1}{2}−21​
  3. (C)−13-\frac{1}{3}−31​
  4. (D)−14-\frac{1}{4}−41​

Correct answer: (B)

Step-by-step solution →
Q78·MathematicsSingle correct
sin⁡−1(sin⁡2π3)+cos⁡−1(cos⁡7π6)+tan⁡−1(tan⁡3π4)\sin^{-1}\left(\sin\frac{2\pi}{3}\right)+\cos^{-1}\left(\cos\frac{7\pi}{6}\right)+\tan^{-1}\left(\tan\frac{3\pi}{4}\right)sin−1(sin32π​)+cos−1(cos67π​)+tan−1(tan43π​) is equal to :
  1. (A)11π12\frac{11\pi}{12}1211π​
  2. (B)17π12\frac{17\pi}{12}1217π​
  3. (C)31π12\frac{31\pi}{12}1231π​
  4. (D)−3π4-\frac{3\pi}{4}−43π​

Correct answer: (A)

Step-by-step solution →
Q79·MathematicsSingle correct
The Boolean expression (∼(p∧q))∨q\left(\sim\left(p \wedge q\right)\right)\vee q(∼(p∧q))∨q is equivalent to :
  1. (A)q→(p∧q)q \rightarrow \left(p \wedge q\right)q→(p∧q)
  2. (B)p→qp \rightarrow qp→q
  3. (C)p→(p→q)p \rightarrow \left(p \rightarrow q\right)p→(p→q)
  4. (D)p→(p∨q)p \rightarrow \left(p \vee q\right)p→(p∨q)

Correct answer: (D)

Step-by-step solution →
Q80·MathematicsNumerical
Let f:R→Rf : R \rightarrow Rf:R→R be a function defined f(x)=2e2xe2x+ef(x)=\frac{2e^{2x}}{e^{2x}+e}f(x)=e2x+e2e2x​. Then f(1100)+f(2100)+f(3100)+.....+f(99100)f\left(\frac{1}{100}\right)+f\left(\frac{2}{100}\right)+f\left(\frac{3}{100}\right)+.....+f\left(\frac{99}{100}\right)f(1001​)+f(1002​)+f(1003​)+.....+f(10099​) is equal to________.

Correct answer: 99

Step-by-step solution →
Q81·MathematicsNumerical
If the sum of all the roots of the equation e2x−11ex−45e−x+812=0e^{2x}-11e^{x}-45e^{-x}+\frac{81}{2}=0e2x−11ex−45e−x+281​=0 is log⁡e\log_eloge​ P, then p is equal to ________.

Correct answer: 45

Step-by-step solution →
Q82·MathematicsNumerical
The positive value of the determinant of the matrix A, whose Adj(Adj(A))=(1428−14−14142828−1414)Adj\left(Adj(A)\right)=\begin{pmatrix} 14 & 28 & -14 \\ -14 & 14 & 28 \\ 28 & -14 & 14 \end{pmatrix}Adj(Adj(A))=​14−1428​2814−14​−142814​​, is __________.

Correct answer: 14

Step-by-step solution →
Q83·MathematicsNumerical
The number of ways, 16 identical cubes, of which 11 are blue and rest are red, can be placed in a row so that between any two red cubes there should be at least 2 blue cubes, is ________.

Correct answer: 56

Step-by-step solution →
Q84·MathematicsNumerical
If the coefficient of x10x^{10}x10 in the binomial expansion of (x514+5x13)60\left(\frac{\sqrt{x}}{5^{\frac{1}{4}}}+\frac{\sqrt{5}}{x^{\frac{1}{3}}}\right)^{60}(541​x​​+x31​5​​)60 is 5kl5^{k}l5kl, where lll, k ∈\in∈ N and lll is co-prime to 5, then k is equal to _____.

Correct answer: 5

Step-by-step solution →
Q85·MathematicsNumerical
Let A1={(x,y):∣x∣≤y2,∣x∣+2y≤8}A_1=\left\{(x,y):|x|\le y^2, |x|+2y\le 8\right\}A1​={(x,y):∣x∣≤y2,∣x∣+2y≤8} and A2={(x,y):∣x∣+∣y∣≤k}A_2=\left\{(x,y):|x|+|y|\le k\right\}A2​={(x,y):∣x∣+∣y∣≤k}. If 27 (Area A1A_1A1​) = 5 (Area A2A_2A2​), then k is equal to :

Correct answer: 6

Step-by-step solution →
Q86·MathematicsNumerical
If the sum of the first ten terms of the series 15+265+3325+41025+52501+....\frac{1}{5}+\frac{2}{65}+\frac{3}{325}+\frac{4}{1025}+\frac{5}{2501}+....51​+652​+3253​+10254​+25015​+.... is mn\frac{m}{n}nm​, where m and n are co-prime numbers, then m + n is equal to __________.

Correct answer: 276

Step-by-step solution →
Q87·MathematicsNumerical
A rectangle R with end points of the one of its dies as (1, 2) and (3, 6) is inscribed in a circle. If the equation of a diameter of the circle is 2x −-− y + 4 = 0, then the area of R is _________.

Correct answer: 16

Step-by-step solution →
Q88·MathematicsNumerical
A circle of radius 2 unit passes through the vertex and the focus of the parabola y2=2xy^2 = 2xy2=2x and touches the parabola y=(x−14)2+αy=\left(x-\frac{1}{4}\right)^2+\alphay=(x−41​)2+α, where α>0\alpha > 0α>0. Then (4α−8)2\left(4\alpha-8\right)^2(4α−8)2 is equal to __________.

Correct answer: 63

Step-by-step solution →
Q89·MathematicsNumerical
Let the mirror image of the point (a, b, c) with respect to the plane 3x −-− 4y + 12z + 19 = 0 be (a- 6, β\betaβ, γ\gammaγ). If a + b + c = 5, then 7 β\betaβ - 9 γ\gammaγ is equal to _____________.

Correct answer: 137

Step-by-step solution →

Chapters tested in this paper

Open any chapter to practise every past-year question on that topic across all papers, and see how often it has actually appeared.

  • Properties of Solids and Liquids 172/186
  • Three Dimensional Geometry 176/186
  • Matrices and Determinants 180/186
  • Coordination Compounds 176/186
  • Sets, Relations and Functions 165/186
  • Current Electricity 160/186
  • Sequence and Series 164/186
  • p-Block Elements 164/186
  • Definite Integration 168/186
  • Rotational Motion 172/186
  • Redox Reactions and Electrochemistry 177/186
  • Geometrical Optics 172/186
  • Kinematics 156/186
  • Vector Algebra 173/186
  • Differential Equations 167/186
  • Probability 176/186
  • Permutations and Combinations 162/186
  • 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
  • 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
  • Area Under Curves 139/186
  • Kinetic Theory of Gases 135/186
  • Purification and Characterisation of Organic Compounds 116/186
  • Oscillations 117/186
  • Alternating Currents 108/186
  • Organic Compounds Containing Halogens 109/186
  • Straight Lines 114/186
  • Waves 109/186
  • Capacitors and Dielectrics 115/186
  • Atoms 112/186
  • Classification of Elements and Periodicity in Properties 113/186
  • Parabola 101/186
  • Ellipse 103/186
  • Isolation of Metals 106/186
  • Differentiability 91/186
  • s-Block Elements 88/186
  • Surface Chemistry 98/186
  • Inverse Trigonometric Functions 93/186
  • Environmental Chemistry 83/186
  • Hydrogen 81/186
  • Indefinite Integration 66/186
  • Polymers 64/186
  • Solid State 63/186
  • Chemistry in Everyday Life 60/186
  • Isomerism 51/186
  • Magnetism and Matter 50/186
← 26 Jun Shift 2 2022All papers27 Jun Shift 2 2022 →

Attempt this paper under exam timing.

Take the 27 June 2022 Shift 1 paper as a timed mock and Jarvis marks it, then tells you which errors were conceptual gaps, which were silly mistakes, and which pattern you have now repeated. Step-by-step solutions for every question included.

Attempt this paper freeSee pricing

Free plan, no time limit · No credit card needed

Jarvis OS

AI-powered JEE preparation.

Question papers

JEE Main PYQsJEE Advanced PYQsPhysics PYQsChemistry PYQsMaths PYQs

Product

TourPricingSign inCreate account

Legal

Privacy PolicyTerms of ServiceRefund & CancellationShipping & DeliveryContact Us

Operated by

Venyou Craft Private Limited

info@jarvisos.net

© 2026 Jarvis OS