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

JEE Main 27 June 2022 Shift 2 Question Paper with Answers

27 June 2022 · June session · 87 questions

87 of the 90 questions from the JEE Main 27 June 2022 Shift 2 paper, each with its correct answer and tagged to the chapter it tests. Free to read, no account needed.

3 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
29
Chemistry
30
Mathematics
28

Physics — JEE Main 27 June 2022 Shift 2

Q1·PhysicsSingle correct
The SI unit of a physical quantity is pascal-second. The dimensional formula of this quantity will be
  1. (A)[ML−1T−1][ML^{-1}T^{-1}][ML−1T−1]
  2. (B)[ML−1T−2][ML^{-1}T^{-2}][ML−1T−2]
  3. (C)[ML2T−1][ML^{2}T^{-1}][ML2T−1]
  4. (D)[M−1L3T0][M^{-1}L^{3}T^{0}][M−1L3T0]

Correct answer: (A)

Step-by-step solution →
Q2·PhysicsSingle correct
The distance of the Sun from earth is 1.5×10111.5\times10^{11}1.5×1011 m and its angular diameter is (2000) s when observed from the earth. The diameter of the Sun will be :
  1. (A)2.45×10102.45\times10^{10}2.45×1010 m
  2. (B)1.45×10101.45\times10^{10}1.45×1010 m
  3. (C)1.45×1091.45\times10^{9}1.45×109 m
  4. (D)0.14×1090.14\times10^{9}0.14×109 m

Correct answer: (C)

Step-by-step solution →
Q3·PhysicsSingle correct
When a ball is dropped into a lake from a height 4.9 m above the water level, it hits the water with a velocity v and then sinks to the bottom with the constant velocity v. It reaches the bottom of the lake 4.0 s after it is dropped. The approximate depth of the lake is :
  1. (A)19.6 m
  2. (B)29.4 m
  3. (C)39.2 m
  4. (D)73.5 m

Correct answer: (B)

Step-by-step solution →
Q4·PhysicsSingle correct
A stone tide to a string of length L is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is at its lowest position and has a speed u. The magnitude of change in its velocity, as it reaches a position where the string is horizontal, is x(u2−gL)\sqrt{x(u^{2}-gL)}x(u2−gL)​ . The value of x is
  1. (A)3
  2. (B)2
  3. (C)1
  4. (D)5

Correct answer: (B)

Step-by-step solution →
Q5·PhysicsSingle correct
Four spheres each of mass m form a square of side d (as shown in figure). A fifth sphere of mass M is situated at the centre of square. The total gravitational potential energy of the system is :
  1. (A)−Gmd[(4+2)m+42M]-\dfrac{Gm}{d}\left[(4+\sqrt{2})m+4\sqrt{2}M\right]−dGm​[(4+2​)m+42​M]
  2. (B)−Gmd[(4+2)M+42m]-\dfrac{Gm}{d}\left[(4+\sqrt{2})M+4\sqrt{2}m\right]−dGm​[(4+2​)M+42​m]
  3. (C)−Gmd[3m2+42M]-\dfrac{Gm}{d}\left[3m^{2}+4\sqrt{2}M\right]−dGm​[3m2+42​M]
  4. (D)−Gmd[6m2+42M]-\dfrac{Gm}{d}\left[6m^{2}+4\sqrt{2}M\right]−dGm​[6m2+42​M]

Correct answer: (A)

Step-by-step solution →
Q6·PhysicsSingle correct
For a perfect gas, two pressures P1P_{1}P1​ and P2P_{2}P2​ are shown in figure. The graph shows:
  1. (A)P1>P2P_{1}>P_{2}P1​>P2​
  2. (B)P1<P2P_{1}<P_{2}P1​<P2​
  3. (C)P1=P2P_{1}=P_{2}P1​=P2​
  4. (D)Insufficient data to draw any conclusion

Correct answer: (A)

Step-by-step solution →
Q7·PhysicsSingle correct
According to kinetic theory of gases, A. The motion of the gas molecules freezes at 0∘0^{\circ}0∘C B. The mean free path of gas molecules decreases if the density of molecules is increased. C. The mean free path of gas molecules increases if temperature is increased keeping pressure constant. D. Average kinetic energy per molecule per degree of freedom is 32kBT\dfrac{3}{2}k_{B}T23​kB​T (for monoatomic gases) Choose the most appropriate answer from the options given below:
  1. (A)A and C only
  2. (B)B and C only
  3. (C)A and B only
  4. (D)C and D only

Correct answer: (B)

Step-by-step solution →
Q8·PhysicsSingle correct
A lead bullet penetrates into a solid object and melts. Assuming that 40% of its kinetic energy is used to heat it, the initial speed of bullet is: (Given, initial temperature of the bullet = 127∘127^{\circ}127∘C, Melting point of the bullet = 327∘327^{\circ}327∘C, Latent heat of fusion of lead = 2.5×1042.5\times10^{4}2.5×104J Kg−1^{-1}−1, Specific heat capacity of lead = 125J/kg K)
  1. (A)125125125 ms−1^{-1}−1
  2. (B)500500500 ms−1^{-1}−1
  3. (C)250250250 ms−1^{-1}−1
  4. (D)600600600 ms−1^{-1}−1

Correct answer: (B)

Step-by-step solution →
Q9·PhysicsSingle correct
The equation of a particle executing simple harmonic motion is given by x=sin⁡π(t+13)x=\sin\pi\left(t+\dfrac{1}{3}\right)x=sinπ(t+31​) m . At t = 1s, the speed of particle will be (Given : π=3.14\pi=3.14π=3.14)
  1. (A)000 cm s−1^{-1}−1
  2. (B)157157157 cm s−1^{-1}−1
  3. (C)272272272 cm s−1^{-1}−1
  4. (D)314314314 cm s−1^{-1}−1

Correct answer: (B)

Step-by-step solution →
Q10·Physics·Electric Field and Coulomb's LawSingle correct
If a charge q is placed at the centre of a closed hemispherical non-conducting surface, the total flux passing through the flat surface would be :
  1. (A)qε0\dfrac{q}{\varepsilon_{0}}ε0​q​
  2. (B)q2ε0\dfrac{q}{2\varepsilon_{0}}2ε0​q​
  3. (C)q4ε0\dfrac{q}{4\varepsilon_{0}}4ε0​q​
  4. (D)q2πε0\dfrac{q}{2\pi\varepsilon_{0}}2πε0​q​

Correct answer: (B)

Step-by-step solution →
Q11·PhysicsSingle correct
Three identical charged balls each of charge 2C are suspended from a common point P by silk threads of 2m each (as shown in figure). They form an equilateral triangle of side 1m. The ratio of net force on a charged ball to the force between any two charged balls will be :
  1. (A)1 : 1
  2. (B)1 : 4
  3. (C)3\sqrt{3}3​ : 2
  4. (D)3\sqrt{3}3​ : 1

Correct answer: (D)

Step-by-step solution →
Q12·PhysicsSingle correct
Two long parallel conductors S1S_{1}S1​ and S2S_{2}S2​ are separated by a distance 10 cm and carrying currents of 4A and 2A respectively. The conductors are placed along x-axis in X-Y plane. There is a point P located between the conductors (as shown in figure). A charge particle of 3π3\pi3π coulomb is passing through the point P with velocity v⃗=(2i^+3j^) m/s\vec{v} = (2\hat{i} + 3\hat{j})\,\text{m}/\text{s}v=(2i^+3j^​)m/s; where i^ & j^\hat{i}\ \& \ \hat{j}i^ & j^​ represents unit vector along x & y axis respectively. The force acting on the charge particle is 4π×10−5(−xi^+2j^) N4\pi \times 10^{-5}(-x\hat{i} + 2\hat{j})\,\text{N}4π×10−5(−xi^+2j^​)N. The value of x is :
  1. (A)2
  2. (B)1
  3. (C)3
  4. (D)−3-3−3

Correct answer: (C)

Step-by-step solution →
Q13·PhysicsSingle correct
If L, C and R are the self inductance, capacitance and resistance respectively, which of the following does not have the dimension of time ?
  1. (A)RC
  2. (B)LR\dfrac{L}{R}RL​
  3. (C)LC\sqrt{LC}LC​
  4. (D)LC\dfrac{L}{C}CL​

Correct answer: (D)

Step-by-step solution →
Q14·PhysicsSingle correct
Given below are two statements: Statement I : A time varying electric field is a source of changing magnetic field and vice-versa. Thus a disturbance in electric or magnetic field creates EM waves. Statement II : In a material medium. The EM wave travels with speed v=1μ0ε0v = \dfrac{1}{\sqrt{\mu_{0}\varepsilon_{0}}}v=μ0​ε0​​1​ . 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 correct but statement II is false.
  4. (D)Statement I is incorrect but statement II is true.

Correct answer: (C)

Step-by-step solution →
Q15·PhysicsSingle correct
A convex lens has power P. It is cut into two halves along its principal axis. Further one piece (out of the two halves) is cut into two halves perpendicular to the principal axis (as shown in figure). Choose the incorrect option for the reported pieces.
  1. (A)Power of L1=P2L_{1} = \dfrac{P}{2}L1​=2P​
  2. (B)Power of L2=P2L_{2} = \dfrac{P}{2}L2​=2P​
  3. (C)Power of L3=P2L_{3} = \dfrac{P}{2}L3​=2P​
  4. (D)Power of L1=PL_{1} = PL1​=P

Correct answer: (A)

Step-by-step solution →
Q16·PhysicsSingle correct
If a wave gets refracted into a denser medium, then which of the following is true?
  1. (A)wavelength speed and frequency decreases.
  2. (B)wavelength increases, speed decreases and frequency remains constant.
  3. (C)wavelength and speed decreases but frequency remains constant.
  4. (D)wavelength, speed and frequency increases.

Correct answer: (C)

Step-by-step solution →
Q17·PhysicsSingle correct
Given below are two statements: Statement I : In hydrogen atom, the frequency of radiation emitted when an electron jumps from lower energy orbit (E1)(E_{1})(E1​) to higher energy orbit (E2)(E_{2})(E2​), is given as hf=E1−E2hf = E_{1} - E_{2}hf=E1​−E2​. Statement-II : The jumping of electron from higher energy orbit (E2)(E_{2})(E2​) to lower energy orbit (E1)(E_{1})(E1​) is associated with frequency of radiation given as f=(E2−E1)/hf = (E_{2} - E_{1})/hf=(E2​−E1​)/h This condition is Bohr's frequency condition. 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 correct but statement II is false
  4. (D)Statement I is incorrect but statement II is true.

Correct answer: (D)

Step-by-step solution →
Q18·PhysicsSingle correct
For a transistor to act as a switch, it must be operated in
  1. (A)Active region
  2. (B)Saturation state only
  3. (C)Cut-off state only
  4. (D)Saturation and cut-off state

Correct answer: (D)

Step-by-step solution →
Q19·PhysicsSingle correct
We do not transmit low frequency signal to long distances because (a) The size of the antenna should be comparable to signal wavelength which is unreal solution for a signal of longer wavelength. (b) Effective power radiated by a long wavelength baseband signal would be high. (c) We want to avoid mixing up signals transmitted by different transmitter simultaneously. (d) Low frequency signal can be sent to long distances by superimposing with a high frequency wave as well. Therefore, the most suitable options will be :
  1. (A)All statements are true
  2. (B)(a), (b) and (c) are true only
  3. (C)(a), (c) and (d) are true only
  4. (D)(b), (c) and (d) are true only

Correct answer: (C)

Step-by-step solution →
Q20·PhysicsNumerical
A mass of 10 kg is suspended vertically by a rope of length 5m from the roof. A force of 30 N is applied at the middle point of rope in horizontal direction. The angle made by upper half of the rope with vertical is θ=tan⁡−1(x×10−1)\theta = \tan^{-1}(x \times 10^{-1})θ=tan−1(x×10−1). The value of x is _______ . (Given g = 10 m/s2^{2}2)

Correct answer: 3

Step-by-step solution →
Q21·PhysicsNumerical
A rolling wheel of 12 kg is on an inclined plane at position P and connected to a mass of 3 kg through a string of fixed length and pulley as shown in figure. Consider PR as friction free surface. The velocity of centre of mass of the wheel when it reaches at the bottom Q of the inclined plane PQ will be 12xgh\dfrac{1}{2}\sqrt{xgh}21​xgh​ m/s. The value of x is _______ .

Correct answer: 3

Step-by-step solution →
Q22·PhysicsNumerical
A diatomic gas (γ=1.4)(\gamma = 1.4)(γ=1.4) does 400 J of work when it is expanded isobarically. The heat given to the gas in the process is ________ J.

Correct answer: 1400

Step-by-step solution →
Q23·PhysicsNumerical
A particle executes simple harmonic motion. Its amplitude is 8 cm and time period is 6s. The time it will take to travel from its position of maximum displacement to the point corresponding to half of its amplitude, is _________ s.

Correct answer: 1

Step-by-step solution →
Q24·PhysicsNumerical
A paralle plate capacitor is made up of stair like structure with a palte area A of each stair and that is connected with a wire of length b, as shown in the figure. The capacitance of the arrangement is x15ε0Ab\dfrac{x}{15}\dfrac{\varepsilon_0 A}{b}15x​bε0​A​ . The value of x is ________ .

Correct answer: 23

Step-by-step solution →
Q25·PhysicsNumerical
The current density in a cylindrical wire of radius r = 4.0 mm is 1.0×1061.0 \times 10^{6}1.0×106 A/m2^{2}2. The current through the outer portion of the wire between radial distances r/2 and r is xπ\piπ A; where x is ________ .

Correct answer: 12

Step-by-step solution →
Q26·PhysicsNumerical
In the given circuit 'a' is an arbitrary constant. The value of m for which the equivalent circuit resistance is minimum, will be x2\sqrt{\dfrac{x}{2}}2x​​ . The value of x is _____ .

Correct answer: 3

Step-by-step solution →
Q27·Physics·Magnetic Field of CurrentNumerical
A deuteron and a proton moving with equal kinetic energy enter into to a uniform magnetic field at right angle to the field. If rd_{d}d​ and rp_{p}p​ are the radii of their circular paths respectively, then the ratio rdrp\dfrac{r_d}{r_p}rp​rd​​ will be x\sqrt{x}x​ : 1 where x is ________ .

Correct answer: 2

Step-by-step solution →
Q28·PhysicsNumerical
A metallic rod of length 20 cm is palced in North-South direction and is moved at a constant speed of 20 m/s towards East. The horizontal component of the Earth's magnetic field at that place is 4×10−34 \times 10^{-3}4×10−3 T and the angle of dip is 45°. The emf induced in the rod is ________ mV.

Correct answer: 16

Step-by-step solution →
Q29·PhysicsNumerical
The cut-off voltage of the diodes (shown in figure) in forward bias is 0.6 V. The current through the resister of 40 Ω\OmegaΩ is ________ mA.

Correct answer: 4

Step-by-step solution →

Chemistry — JEE Main 27 June 2022 Shift 2

Q30·ChemistrySingle correct
Which amongst the given plots is the correct plot for pressure (p) vs density (d) for an ideal gas ?
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (B)

Step-by-step solution →
Q31·ChemistrySingle correct
Identify the incorrect statement for PCl5_55​ from the following.
  1. (A)In this molecule, orbitals of phosphorous are assumed to undergo sp3^{3}3d hybridization.
  2. (B)The geometry of PCl5_55​ is trigonal bipyramidal.
  3. (C)PCl5_55​ has two axial bonds stronger than three equatorial bonds.
  4. (D)The three equatorial bonds of PCl5_55​ lie in a plane.

Correct answer: (C)

Step-by-step solution →
Q32·ChemistrySingle correct
Statement I : Leaching of gold with cyanide ion in absence of air / O2_22​ leads to cyano complex of Au(III). Statement II : Zinc is oxidized during the displacement reaction carried out for gold extraction. 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 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 →
Q33·ChemistrySingle correct
The correct order of increasing intermolecular hydrogen bond strength is
  1. (A)HCN < H2_22​O < NH3_33​
  2. (B)HCN < CH4_44​ < NH3_33​
  3. (C)CH4_44​ < HCN < NH3_33​
  4. (D)CH4_44​ < NH3_33​ < HCN

Correct answer: (C)

Step-by-step solution →
Q34·ChemistrySingle correct
The correct order of increasing ionic radii is
  1. (A)Mg2+^{2+}2+ < Na+^{+}+ < F−^{-}− < O2−^{2-}2− < N3−^{3-}3−
  2. (B)N3−^{3-}3− < O2−^{2-}2− < F−^{-}− < Na+^{+}+ < Mg2+^{2+}2+
  3. (C)F−^{-}− < Na+^{+}+ < O2−^{2-}2− < Mg2+^{2+}2+ < N3−^{3-}3−
  4. (D)Na+^{+}+ < F−^{-}− < Mg2+^{2+}2+ < O2−^{2-}2− < N3−^{3-}3−

Correct answer: (A)

Step-by-step solution →
Q35·ChemistrySingle correct
The gas produced by treating an aqueous solution of ammonium chloride with sodium nitrite is
  1. (A)NH3_33​
  2. (B)N2_22​
  3. (C)N2_22​O
  4. (D)Cl2_22​

Correct answer: (B)

Step-by-step solution →
Q36·ChemistrySingle correct
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A : Flourine forms one oxoacid. Reason R : Flourine has smallest size amongst all halogens and is highly electronegative In the light of the above statements, choose the most appropriate answer from the options given below.
  1. (A)Both A and R are correct and R is the correct explanation of A.
  2. (B)Both A and R are correct but R is NOT the correct explanation of A.
  3. (C)A is correct but R is not correct.
  4. (D)A is not correct but R is correct

Correct answer: (A)

Step-by-step solution →
Q37·ChemistrySingle correct
In 3d series, the metal having the highest M2+^{2+}2+/M standard electrode potential is
  1. (A)Cr
  2. (B)Fe
  3. (C)Cu
  4. (D)Zn

Correct answer: (C)

Step-by-step solution →
Q38·ChemistrySingle correct
The 'f' orbitals are half and completely filled, respectively in lanthanide ions (Given: Atomic no. Eu, 63; Sm, 62; Tm, 69; Tb, 65; Yb, 70; Dy, 66]
  1. (A)Eu2+^{2+}2+ and Tm2+^{2+}2+
  2. (B)Sm2+^{2+}2+ and Tm3+^{3+}3+
  3. (C)Tb4+^{4+}4+ and Yb2+^{2+}2+
  4. (D)Dy3+^{3+}3+ and Yb3+^{3+}3+

Correct answer: (C)

Step-by-step solution →
Q39·ChemistrySingle correct
Arrange the following coordination compounds in the increasing order of magnetic moments. (Atomic numbers: Mn = 25; Fe = 26) (A) [FeF6_66​]3−^{3-}3− (B) [Fe(CN)6_66​]3−^{3-}3− (C) [MnCl6_66​]3−^{3-}3− (high spin) (D) [Mn(CN)6_66​]3−^{3-}3−
  1. (A)A < B < D < C
  2. (B)B < D < C < A
  3. (C)A < C < D < B
  4. (D)B < D < A < C

Correct answer: (B)

Step-by-step solution →
Q40·ChemistrySingle correct
On the surface of polar stratospheric clouds, hydrolysis of chlorine nitrate gives A and B while its reaction with HCl produces B and C. A, B and C are, respectively
  1. (A)HOCl, HNO3_33​, Cl2_22​
  2. (B)Cl2_22​, HNO3_33​, HOCl
  3. (C)HClO2_22​, HNO2_22​, HOCl
  4. (D)HOCl, HNO2_22​, Cl2_22​O

Correct answer: (A)

Step-by-step solution →
Q41·Chemistry·Electronic Effects and StabilitySingle correct
Which of the following is most stable?
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (A)

Step-by-step solution →
Q42·ChemistrySingle correct
What will be the major product of following sequence of reactions? Printed reaction scheme (words and formulae only): n – Bu– ≡\equiv≡ ; above the arrow: (i) n – BuLi, n – C5_55​H11_{11}11​Cl ; below the arrow: (ii) Lindlar cat, H2_22​
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (C)

Step-by-step solution →
Q43·ChemistrySingle correct
Product 'A' of following sequence of reactions is Printed reaction scheme (words and formulae only): Ethylbenzene →\rightarrow→ 'A'(Major product) ; on the arrow: (a) Br2_22​ .Fe (b) Cl2_22​ , Δ\DeltaΔ (c) alc. KOH
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (D)

Step-by-step solution →
Q44·ChemistrySingle correct
Match List I with List II List I A. (reaction scheme shown as a drawn figure) B. (reaction scheme shown as a drawn figure) C. (reaction scheme shown as a drawn figure) D. (reaction scheme shown as a drawn figure) List II I. Br2_22​ in CS2_22​ II. Na2_22​Cr2_22​O7_77​/H2_22​SO4_44​ III. Zn IV. CHCl3_33​/NaOH Choose the correct answer from the options given below:
  1. (A)A-IV, B-III, C-II, D-I
  2. (B)A-IV, B-III, C-I, D-II
  3. (C)A-II, B-III, C-I, D-IV
  4. (D)A-IV, B-II, C-III, D-I

Correct answer: (A)

Step-by-step solution →
Q45·ChemistrySingle correct
Decarboxylation of all six possible forms of diaminobenzoic acids C6_66​H3_33​(NH2_22​)2_22​COOH yields three products A, B and C. Three acids give a product 'A', two acids gives a product 'B' and one acid give a product 'C'. The melting point of product 'C' is
  1. (A)63∘^{\circ}∘C
  2. (B)90∘^{\circ}∘C
  3. (C)104∘^{\circ}∘C
  4. (D)142∘^{\circ}∘C

Correct answer: (D)

Step-by-step solution →
Q46·ChemistrySingle correct
Which is true about Buna-N?
  1. (A)It is a linear polymer of 1, 3-butadiene.
  2. (B)It is obtained by copolymerization of 1, 3-butadiene and styrene.
  3. (C)It is obtained by copolymerization of 1, 3-butadiene and acrylonitrile.
  4. (D)The suffix N in Buna-N stands for its natural occurrence

Correct answer: (C)

Step-by-step solution →
Q47·ChemistrySingle correct
Given below are two statements. Statments I: Maltose has two α\alphaα-D-glucose units linked at C1_11​ and C4_44​ and is a reducing sugar. Statement II: Maltose has two monosaccharides: α\alphaα-D-glucose and β\betaβ-D-glucose linked at C1_11​ and C6_66​ and it is a non-reducing sugar. 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: (C)

Step-by-step solution →
Q48·ChemistrySingle correct
Match List I with List Ii List I A. Antipyretic B. Analgesic C. Tranquilizer D. Antacid List II I. Reduces pain II. Reduces stress III. Reduces fever IV. Reduces acidity (Stomach) Choose the correct answer from the options given below:
  1. (A)A-III, B-I, C-II, D-IV
  2. (B)A-III, B-I, C-IV, D-II
  3. (C)A-I, B-IV, C-II, D-III
  4. (D)A-I, B-III, C-II, D-IV

Correct answer: (A)

Step-by-step solution →
Q49·ChemistrySingle correct
Match List I with List II List I (Anion) A. CO32−_3^{2-}32−​ B. S2−^{2-}2− C. SO32−_3^{2-}32−​ D. NO2−_2^-2−​ List II (Gas evolved on reaction with dil. H2_22​SO4_44​) I. Colourless gas which turns lead acetate paper black II. Colourless gas which turns acidified potassium dichromate solution green. III. Brown fumes which turns acidified KI solution containing starch blue. IV. Colourless gas evolved with brisk effervescence, which turns lime water milky. Choose the correct answer from the options given below:
  1. (A)A-III, B-I, C-II, D-IV
  2. (B)A-II, B-I, C-IV, D-III
  3. (C)A-IV, B-I, C-III, D-II
  4. (D)A-IV, B-I, C-II, D-III

Correct answer: (D)

Step-by-step solution →
Q50·ChemistryNumerical
116 g of a substance upon dissociation reaction, yields 7.5 g of hydrogen, 60g of oxygen and 48.5 g of carbon. Given that the atomic masses of H, O and C are 1, 16 and 12 respectively. The data agrees with how many formulae of the following? (A) CH3_33​COOH (B) HCHO (C) CH3_33​OOCH3_33​ (D) CH3_33​CHO

Correct answer: 2

Step-by-step solution →
Q51·ChemistryNumerical
Consider the following set of quantum numbers n l ml_ll​ A. 3 3 -3 B. 3 2 -2 C. 2 1 +1 D. 2 2 +2 The number of correct sets of quantum numbers is ________

Correct answer: 2

Step-by-step solution →
Q52·ChemistryNumerical
BeO reacts with HF in presence of ammonia to give [A] which on thermal decomposition produces [B] and ammonium fluoride. Oxidation state of Be in [A] is________

Correct answer: 2

Step-by-step solution →
Q53·ChemistryNumerical
When 5 moles of He gas expand isothermally and reversibly at 300 K from 10 litre to 20 litre, the magnitude of the maximum work obtained is ____ J. [nearest integer] (Given: R = 8.3 J K−1^{-1}−1mol−1^{-1}−1 and log 2 = 0.3010)

Correct answer: 8630

Step-by-step solution →
Q54·ChemistryNumerical
A solution containing 2.5 ×\times× 10−3^{-3}−3 kg of a solute dissolved in 75 ×\times× 10−3^{-3}−3 kg of water boils at 373.535 K. The molar mass of the solute is _______ g mol−1^{-1}−1. [nearest integer] (Given: Kb_bb​ (H2_22​O) = 0.52 K Kg mol−1^{-1}−1 , boiling point of water = 373.15K)

Correct answer: 45

Step-by-step solution →
Q55·ChemistryNumerical
pH value of 0.001 M NaOH solution is_______.

Correct answer: 11

Step-by-step solution →
Q56·ChemistryNumerical
For the reaction taking place in the cell: Pt(s) | H2_22​(g) | H+^++(aq) || Ag+^++(aq) | Ag(s) ECello^o_{Cell}Cello​ = +0.5332 V. The value of Δf\Delta_fΔf​G0^00 is _________ kJ mol−1^{-1}−1. (in nearest integer)

Correct answer: -51

Step-by-step solution →
Q57·ChemistryNumerical
It has been found that for a chemical reaction with rise in temperature by 9K the rate constant gets doubled. Assuming a reaction to be occurring at 300 K, the value of activation energy is found to be ______ kJ mol−1^{-1}−1. [nearest integer] (Given ln 10 = 2.3, R = 8.3 JK−1^{-1}−1mol−1^{-1}−1, log2 = 0.30)

Correct answer: 59

Step-by-step solution →
Q58·ChemistryNumerical
If the initial pressure of a gas is 0.03 atm, the mass of the gas adsorbed per gram of the adsorbent is ______ ×\times× 10−2^{-2}−2g.

Correct answer: 12

Step-by-step solution →
Q59·ChemistryNumerical
0.25 g of an organic compound containing chlorine gave 0.40 g of silver chloride in Carius estimation. The percentage of chlorine present in the compound is _______. [in nearest integer] (Given: Molar mass of Ag is 108 g mol−1^{-1}−1 and that of Cl is 35.5 g mol−1^{-1}−1)

Correct answer: 40

Step-by-step solution →

Mathematics — JEE Main 27 June 2022 Shift 2

Q60·MathematicsSingle correct
The number of points of intersection of ∣z−(4+3i)∣=2\left|z-(4+3i)\right|=2∣z−(4+3i)∣=2 and ∣z∣+∣z−4∣=6\left|z\right|+\left|z-4\right|=6∣z∣+∣z−4∣=6, z∈Cz \in \mathbb{C}z∈C is :
  1. (A)000
  2. (B)111
  3. (C)222
  4. (D)333

Correct answer: (C)

Step-by-step solution →
Q61·MathematicsSingle correct
Let f(x)=∣a−10axa−1ax2axa∣f(x)=\begin{vmatrix} a & -1 & 0 \\ ax & a & -1 \\ ax^{2} & ax & a \end{vmatrix}f(x)=​aaxax2​−1aax​0−1a​​, a∈Ra \in \mathbb{R}a∈R. Then the sum of which the squares of all the values of a for 2f′(10)−f′(5)+100=02f'(10)-f'(5)+100=02f′(10)−f′(5)+100=0 is :
  1. (A)117117117
  2. (B)106106106
  3. (C)125125125
  4. (D)136136136

Correct answer: (C)

Step-by-step solution →
Q62·MathematicsSingle correct
Let for some real numbers α\alphaα and β\betaβ, a=α−iβa=\alpha-i\betaa=α−iβ. If the system of equations 4ix+(1+i)y=04ix+(1+i)y=04ix+(1+i)y=0 and 8(cos⁡2π3+isin⁡2π3)x+aˉy=08\left(\cos\dfrac{2\pi}{3}+i\sin\dfrac{2\pi}{3}\right)x+\bar{a}y=08(cos32π​+isin32π​)x+aˉy=0 has more than one solution then αβ\dfrac{\alpha}{\beta}βα​ is equal to :
  1. (A)−2+3-2+\sqrt{3}−2+3​
  2. (B)2−32-\sqrt{3}2−3​
  3. (C)2+32+\sqrt{3}2+3​
  4. (D)−2−3-2-\sqrt{3}−2−3​

Correct answer: (B)

Step-by-step solution →
Q63·MathematicsSingle correct
Let A and B be two 3×33\times33×3 matrices such that AB=IAB=IAB=I and ∣A∣=18|A|=\dfrac{1}{8}∣A∣=81​ then ∣adj(B adj(2A))∣\left|\mathrm{adj}(B\,\mathrm{adj}(2A))\right|∣adj(Badj(2A))∣ is equal to
  1. (A)161616
  2. (B)323232
  3. (C)646464
  4. (D)128128128

Correct answer: (C)

Step-by-step solution →
Q64·MathematicsSingle correct
Let S=2+67+1272+2073+3074+.....S=2+\dfrac{6}{7}+\dfrac{12}{7^{2}}+\dfrac{20}{7^{3}}+\dfrac{30}{7^{4}}+.....S=2+76​+7212​+7320​+7430​+..... then 4S4S4S is equal to
  1. (A)(73)2\left(\dfrac{7}{3}\right)^{2}(37​)2
  2. (B)7332\dfrac{7^{3}}{3^{2}}3273​
  3. (C)(73)3\left(\dfrac{7}{3}\right)^{3}(37​)3
  4. (D)7233\dfrac{7^{2}}{3^{3}}3372​

Correct answer: (C)

Step-by-step solution →
Q65·MathematicsSingle correct
If a1,a2,a3....a_{1},a_{2},a_{3}....a1​,a2​,a3​.... and b1,b2,b3....b_{1},b_{2},b_{3}....b1​,b2​,b3​.... are A.P. and a1=2a_{1}=2a1​=2, a10=3a_{10}=3a10​=3, a1b1=1=a10b10a_{1}b_{1}=1=a_{10}b_{10}a1​b1​=1=a10​b10​ then a4b4a_{4}b_{4}a4​b4​ is equal to
  1. (A)3527\dfrac{35}{27}2735​
  2. (B)111
  3. (C)2728\dfrac{27}{28}2827​
  4. (D)2827\dfrac{28}{27}2728​

Correct answer: (D)

Step-by-step solution →
Q66·MathematicsSingle correct
If m and n respectively are the number of local maximum and local minimum points of the function f(x)=∫0x2t2−5t+42+etdtf(x)=\int_{0}^{x^{2}}\dfrac{t^{2}-5t+4}{2+e^{t}}dtf(x)=∫0x2​2+ett2−5t+4​dt, then the ordered pair (m, n) is equal to
  1. (A)(3, 2)(3,\,2)(3,2)
  2. (B)(2, 3)(2,\,3)(2,3)
  3. (C)(2, 2)(2,\,2)(2,2)
  4. (D)(3, 4)(3,\,4)(3,4)

Correct answer: (B)

Step-by-step solution →
Q67·MathematicsSingle correct
Let f be a differentiable function in (0,π2)\left(0,\dfrac{\pi}{2}\right)(0,2π​). If ∫cos⁡x1t2f(t)dt=sin⁡3x+cos⁡x\int_{\cos x}^{1}t^{2}f(t)dt=\sin^{3}x+\cos x∫cosx1​t2f(t)dt=sin3x+cosx then 13f′(13)\dfrac{1}{\sqrt{3}}f'\left(\dfrac{1}{\sqrt{3}}\right)3​1​f′(3​1​) is equal to :
  1. (A)6−926-9\sqrt{2}6−92​
  2. (B)6−926-\dfrac{9}{\sqrt{2}}6−2​9​
  3. (C)92−62\dfrac{9}{2}-6\sqrt{2}29​−62​
  4. (D)92−6\dfrac{9}{\sqrt{2}}-62​9​−6

Correct answer: (B)

Step-by-step solution →
Q68·MathematicsSingle correct
The integral ∫0117[1x]dx\int_{0}^{1}\dfrac{1}{7^{\left[\frac{1}{x}\right]}}dx∫01​7[x1​]1​dx, where [.] denotes the greatest integer function is equal to
  1. (A)1+6log⁡e(67)1+6\log_{e}\left(\dfrac{6}{7}\right)1+6loge​(76​)
  2. (B)1−6log⁡e(67)1-6\log_{e}\left(\dfrac{6}{7}\right)1−6loge​(76​)
  3. (C)log⁡e(76)\log_{e}\left(\dfrac{7}{6}\right)loge​(67​)
  4. (D)1−7log⁡e(67)1-7\log_{e}\left(\dfrac{6}{7}\right)1−7loge​(76​)

Correct answer: (A)

Step-by-step solution →
Q69·MathematicsSingle correct
If the solution curve of the differential equation ((tan⁡−1y)−x)dy=(1+y2)dx\left((\tan^{-1}y)-x\right)dy=(1+y^{2})dx((tan−1y)−x)dy=(1+y2)dx passes through the point (1, 0) then the abscissa of the point on the curve whose ordinate is tan⁡(1)\tan(1)tan(1) is :
  1. (A)2e2e2e
  2. (B)2e\dfrac{2}{e}e2​
  3. (C)222
  4. (D)1e\dfrac{1}{e}e1​

Correct answer: (B)

Step-by-step solution →
Q70·MathematicsSingle correct
If the equation of the parabola, whose vertex is at (5, 4) and the directrix is 3x+y−29=03x+y-29=03x+y−29=0, is x2+ay2+bxy+cx+dy+k=0x^{2}+ay^{2}+bxy+cx+dy+k=0x2+ay2+bxy+cx+dy+k=0 then a+b+c+d+ka+b+c+d+ka+b+c+d+k is equal to
  1. (A)575575575
  2. (B)−575-575−575
  3. (C)576576576
  4. (D)−576-576−576

Correct answer: (D)

Step-by-step solution →
Q71·MathematicsSingle correct
The set of values of k for which the circle C:  4x2+4y2−12x+8y+k=0C:\;4x^{2}+4y^{2}-12x+8y+k=0C:4x2+4y2−12x+8y+k=0 lies inside the fourth quadrant and the point (1,−13)\left(1,-\dfrac{1}{3}\right)(1,−31​) lies on or inside the circle C is :
  1. (A)An empty set
  2. (B)(6,959]\left(6,\dfrac{95}{9}\right](6,995​]
  3. (C)[809,10)\left[\dfrac{80}{9},10\right)[980​,10)
  4. (D)(9,929]\left(9,\dfrac{92}{9}\right](9,992​]

Correct answer: (D)

Step-by-step solution →
Q72·MathematicsSingle correct
Let the foot of the perpendicular from the point (1, 2, 4) on the line x+24=y−12=z+13\dfrac{x+2}{4}=\dfrac{y-1}{2}=\dfrac{z+1}{3}4x+2​=2y−1​=3z+1​ be P. Then the distance of P from the plane 3x+4y+12z+23=03x+4y+12z+23=03x+4y+12z+23=0
  1. (A)555
  2. (B)5013\dfrac{50}{13}1350​
  3. (C)444
  4. (D)6313\dfrac{63}{13}1363​

Correct answer: (A)

Step-by-step solution →
Q73·MathematicsSingle correct
The shortest distance between the lines x−32=y−23=z−1−1\dfrac{x-3}{2}=\dfrac{y-2}{3}=\dfrac{z-1}{-1}2x−3​=3y−2​=−1z−1​ and x+32=y−61=z−53\dfrac{x+3}{2}=\dfrac{y-6}{1}=\dfrac{z-5}{3}2x+3​=1y−6​=3z−5​ is :
  1. (A)185\dfrac{18}{\sqrt{5}}5​18​
  2. (B)2235\dfrac{22}{3\sqrt{5}}35​22​
  3. (C)4635\dfrac{46}{3\sqrt{5}}35​46​
  4. (D)636\sqrt{3}63​

Correct answer: (A)

Step-by-step solution →
Q74·MathematicsSingle correct
Let a⃗\vec{a}a and b⃗\vec{b}b be the vectors along the diagonal of a parallelogram having area 222\sqrt{2}22​. Let the angle between a⃗\vec{a}a and b⃗\vec{b}b be acute. ∣a⃗∣=1|\vec{a}|=1∣a∣=1 and ∣a⃗⋅b⃗∣=∣a⃗×b⃗∣|\vec{a}\cdot\vec{b}|=|\vec{a}\times\vec{b}|∣a⋅b∣=∣a×b∣. If c⃗=22(a⃗×b⃗)−2b⃗\vec{c}=2\sqrt{2}\left(\vec{a}\times\vec{b}\right)-2\vec{b}c=22​(a×b)−2b, then an angle between b⃗\vec{b}b and c⃗\vec{c}c is :
  1. (A)π4\dfrac{\pi}{4}4π​
  2. (B)−π4-\dfrac{\pi}{4}−4π​
  3. (C)5π6\dfrac{5\pi}{6}65π​
  4. (D)3π4\dfrac{3\pi}{4}43π​

Correct answer: (D)

Step-by-step solution →
Q75·MathematicsSingle correct
The mean and variance of the data4, 5, 6, 6, 7, 8, x, y where x<yx<yx<y are 6, and 94\dfrac{9}{4}49​ respectively. Then x4+y2x^{4}+y^{2}x4+y2 is equal to
  1. (A)162
  2. (B)320
  3. (C)674
  4. (D)420

Correct answer: (B)

Step-by-step solution →
Q76·MathematicsSingle correct
If a point A(x,y)A(x, y)A(x,y) lies in the region bounded by the y-axis, straight lines 2y+x=62y+x=62y+x=6 and 5x−6y=305x-6y=305x−6y=30, then the probability that y<1y<1y<1 is :
  1. (A)16\dfrac{1}{6}61​
  2. (B)56\dfrac{5}{6}65​
  3. (C)23\dfrac{2}{3}32​
  4. (D)67\dfrac{6}{7}76​

Correct answer: (B)

Step-by-step solution →
Q77·MathematicsSingle correct
The value of cot⁡(∑n=150tan⁡−1(11+n+n2))\cot\left(\sum_{n=1}^{50}\tan^{-1}\left(\dfrac{1}{1+n+n^{2}}\right)\right)cot(∑n=150​tan−1(1+n+n21​)) is
  1. (A)2625\dfrac{26}{25}2526​
  2. (B)2526\dfrac{25}{26}2625​
  3. (C)5051\dfrac{50}{51}5150​
  4. (D)5251\dfrac{52}{51}5152​

Correct answer: (A)

Step-by-step solution →
Q78·MathematicsSingle correct
α=sin⁡36∘\alpha=\sin 36^{\circ}α=sin36∘ is a root of which of the following equation
  1. (A)10x4−10x2−5=010x^{4}-10x^{2}-5=010x4−10x2−5=0
  2. (B)16x4+20x2−5=016x^{4}+20x^{2}-5=016x4+20x2−5=0
  3. (C)16x4−20x2+5=016x^{4}-20x^{2}+5=016x4−20x2+5=0
  4. (D)16x4−10x2+5=016x^{4}-10x^{2}+5=016x4−10x2+5=0

Correct answer: (C)

Step-by-step solution →
Q79·MathematicsSingle correct
Which of the following statement is a tautology?
  1. (A)((∼q)∧p)∧q((\sim q)\wedge p)\wedge q((∼q)∧p)∧q
  2. (B)((∼q)∧p)∧(p∧(∼p))((\sim q)\wedge p)\wedge(p\wedge(\sim p))((∼q)∧p)∧(p∧(∼p))
  3. (C)((∼q)∧p)∨(p∨(∼p))((\sim q)\wedge p)\vee(p\vee(\sim p))((∼q)∧p)∨(p∨(∼p))
  4. (D)(p∧q)∧(∼(p∧q))(p\wedge q)\wedge(\sim(p\wedge q))(p∧q)∧(∼(p∧q))

Correct answer: (C)

Step-by-step solution →
Q80·MathematicsNumerical
Let S={1,2,3,4,5,6,7,8,9,10}S=\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}S={1,2,3,4,5,6,7,8,9,10}. Define f:S→Sf:S\to Sf:S→S as f(n)={2n,if n=1,2,3,4,52n−11if n=6,7,8,9,10f(n)=\begin{cases}2n, & \text{if } n=1,2,3,4,5\\ 2n-11 & \text{if } n=6,7,8,9,10\end{cases}f(n)={2n,2n−11​if n=1,2,3,4,5if n=6,7,8,9,10​. Let g:S→Sg:S\to Sg:S→S be a function such that fog(n)={n+1, if n is oddn−1, if n is even\text{fog}(n)=\begin{cases}n+1 & \text{, if } n \text{ is odd}\\ n-1 & \text{, if } n \text{ is even}\end{cases}fog(n)={n+1n−1​, if n is odd, if n is even​, then g(10) ((g(1)+g(2)+g(3)+g(4)+g(5)))g(10)\ ((g(1)+g(2)+g(3)+g(4)+g(5)))g(10) ((g(1)+g(2)+g(3)+g(4)+g(5))) is equal to:

Correct answer: 190

Step-by-step solution →
Q81·MathematicsNumerical
Let α\alphaα, β\betaβ be the roots of the equation x2−4λx+5=0x^{2}-4\lambda x+5=0x2−4λx+5=0 and α\alphaα, γ\gammaγ be the roots of the equation x2−(32+23)x+7+3λ3=0x^{2}-\left(3\sqrt{2}+2\sqrt{3}\right)x+7+3\lambda\sqrt{3}=0x2−(32​+23​)x+7+3λ3​=0. If β+γ=32\beta+\gamma=3\sqrt{2}β+γ=32​, then (α+2β+γ)2(\alpha+2\beta+\gamma)^{2}(α+2β+γ)2 is equal to :

Correct answer: 98

Step-by-step solution →
Q82·MathematicsNumerical
Let A be a matrix of order 2×22\times 22×2, whose entries are from the set {0,1,2,3,4,5}\{0, 1, 2, 3, 4, 5\}{0,1,2,3,4,5}. If the sum of all the entries of A is a prime number p, 2<p<82<p<82<p<8, then the number of such matrices A is :

Correct answer: 180

Step-by-step solution →
Q83·MathematicsNumerical
If the sum of the coefficients of all the positive powers of x, in the binomial expansion of (xn+2x5)7\left(x^{n}+\dfrac{2}{x^{5}}\right)^{7}(xn+x52​)7 is 939, then the sum of all the possible integral values of n is :

Correct answer: 57

Step-by-step solution →
Q84·Mathematics·Limits and ContinuityNumerical
Let [t] denote the greatest integer ≤\le≤ t and {t} denote the fractional part of t. Then integral value of α\alphaα for which the left hand limit of the function f(x)=[1+x]+α2[x]+{x}+[x]−12[x]+{x}f(x)=[1+x]+\dfrac{\alpha^{2[x]+\{x\}}+[x]-1}{2[x]+\{x\}}f(x)=[1+x]+2[x]+{x}α2[x]+{x}+[x]−1​ at x = 0 is equal to α−43\alpha-\dfrac{4}{3}α−34​ is ______

Correct answer: 3

Step-by-step solution →
Q85·MathematicsNumerical
If the area of the region {(x,y):x23+y23≤1 x+y≥0, y≥0}\left\{(x,y):x^{\frac{2}{3}}+y^{\frac{2}{3}}\le 1\, x+y\ge 0,\ y\ge 0\right\}{(x,y):x32​+y32​≤1x+y≥0, y≥0} is A, then 256 Aπ\dfrac{256\,A}{\pi}π256A​ is

Correct answer: 36

Step-by-step solution →
Q86·MathematicsNumerical
Let v be the solution of the differential equation (1−x2)dy=(xy+(x3+2)1−x2)dx, −1<x<1(1-x^{2})dy =\left(xy+(x^{3}+2)\sqrt{1-x^{2}}\right)dx,\ -1<x<1(1−x2)dy=(xy+(x3+2)1−x2​)dx, −1<x<1 and y (0) = 0 if ∫−12121−x2  y(x)dx=k\int_{-\frac{1}{2}}^{\frac{1}{2}}\sqrt{1-x^{2}}\;y(x)dx=k∫−21​21​​1−x2​y(x)dx=k then k−1k^{-1}k−1 is equal to :

Correct answer: 320

Step-by-step solution →
Q87·MathematicsNumerical
Let S={E,E2....E8}S=\{E, E_{2}....E_{8}\}S={E,E2​....E8​} be a sample space of random experiment such that P(En)=n36P(E_{n})=\dfrac{n}{36}P(En​)=36n​ for every n = 1, 2....8. Then the number of elements in the set {A⊂S:P(A)≥45}\left\{A\subset S:P(A)\ge\dfrac{4}{5}\right\}{A⊂S:P(A)≥54​} is ______

Correct answer: 19

Step-by-step solution →

Chapters tested in this paper

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

  • Properties of Solids and Liquids 172/186
  • Three Dimensional Geometry 176/186
  • Matrices and Determinants 180/186
  • Coordination Compounds 176/186
  • Sets, Relations and Functions 165/186
  • Current Electricity 160/186
  • Sequence and Series 164/186
  • p-Block Elements 164/186
  • Definite Integration 168/186
  • Rotational Motion 172/186
  • Redox Reactions and Electrochemistry 177/186
  • Geometrical Optics 172/186
  • Kinematics 156/186
  • Vector Algebra 173/186
  • Differential Equations 167/186
  • Probability 176/186
  • Permutations and Combinations 162/186
  • Magnetic Field of Current 147/186
  • Binomial Theorem and Its Simple Applications 158/186
  • Electromagnetic Waves 144/186
  • Chemical Bonding and Molecular Structure 151/186
  • Application of Derivatives 139/186
  • Limits and Continuity 149/186
  • d- and f-Block Elements 126/186
  • Thermodynamics 154/186
  • 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
  • Amines 133/186
  • Quadratic Equations 148/186
  • 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
  • Purification and Characterisation of Organic Compounds 116/186
  • Oscillations 117/186
  • Alternating Currents 108/186
  • Organic Compounds Containing Halogens 109/186
  • Capacitors and Dielectrics 115/186
  • Atoms 112/186
  • Classification of Elements and Periodicity in Properties 113/186
  • Parabola 101/186
  • Statistics 118/186
  • Isolation of Metals 106/186
  • s-Block Elements 88/186
  • Surface Chemistry 98/186
  • Inverse Trigonometric Functions 93/186
  • Environmental Chemistry 83/186
  • Electronic Effects and Stability 74/186
  • Polymers 64/186
  • Principles of Qualitative Analysis 58/186
  • Chemistry in Everyday Life 60/186
  • States of Matter: Gases and Liquids 52/186
← 27 Jun Shift 1 2022All papers28 Jun Shift 1 2022 →

Attempt this paper under exam timing.

Take the 27 June 2022 Shift 2 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