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JEE Main 26 June 2022 Shift 1 Question Paper with Answers

26 June 2022 · June session · 86 questions

86 of the 90 questions from the JEE Main 26 June 2022 Shift 1 paper, each with its correct answer and tagged to the chapter it tests. Free to read, no account needed.

4 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
27
Chemistry
30
Mathematics
29

Physics — JEE Main 26 June 2022 Shift 1

Q1·PhysicsSingle correct
An expression for a dimensionless quantity P is given by P=αβlog⁡e(ktβx)P = \frac{\alpha}{\beta}\log_{e}\left(\frac{kt}{\beta x}\right)P=βα​loge​(βxkt​); where α and β are constants, x is distance ; k is Boltzmann constant and t is the temperature. Then the dimensions of α will be :
  1. (A)[M0L−1T0][M^{0}L^{-1}T^{0}][M0L−1T0]
  2. (B)[ML0T−2][ML^{0}T^{-2}][ML0T−2]
  3. (C)[MLT−2][MLT^{-2}][MLT−2]
  4. (D)[ML2T−2][ML^{2}T^{-2}][ML2T−2]

Correct answer: (C)

Step-by-step solution →
Q2·PhysicsSingle correct
A person is standing in an elevator. In which situation, he experiences weight loss ?
  1. (A)When the elevator moves upward with constant acceleration
  2. (B)When the elevator moves downward with constant acceleration
  3. (C)When the elevator moves upward with uniform velocity
  4. (D)When the elevator moves downward with uniform velocity

Correct answer: (B)

Step-by-step solution →
Q3·PhysicsSingle correct
An object is thrown vertically upwards. At its maximum height, which of the following quantity becomes zero ?
  1. (A)Momentum
  2. (B)Potential energy
  3. (C)Acceleration
  4. (D)Force

Correct answer: (A)

Step-by-step solution →
Q4·PhysicsSingle correct
A ball is released from rest from point P of a smooth semi-spherical vessel as shown in figure. The ratio of the centripetal force and normal reaction on the ball at point Q is A while angular position of point Q is α with respect to point P. Which of the following graphs represent the correct relation between A and α when ball goes from Q to R ?
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (C)

Step-by-step solution →
Q5·PhysicsSingle correct
A thin circular ring of mass M and radius R is rotating with a constant angular velocity 2 rads−1^{-1}−1 in a horizontal plane about an axis vertical to its plane and passing through the center of the ring. If two objects each of mass m be attached gently to the opposite ends of a diameter of ring, the ring will then rotate with an angular velocity (in rads−1^{-1}−1).
  1. (A)M(M+m)\frac{M}{\left(M+m\right)}(M+m)M​
  2. (B)(M+2m)2M\frac{\left(M+2m\right)}{2M}2M(M+2m)​
  3. (C)2M(M+2m)\frac{2M}{\left(M+2m\right)}(M+2m)2M​
  4. (D)2(M+2m)M\frac{2\left(M+2m\right)}{M}M2(M+2m)​

Correct answer: (C)

Step-by-step solution →
Q6·PhysicsSingle correct
The variation of acceleration due to gravity (g) with distance (r) from the center of the earth is correctly represented by : (Given R = radius of earth)
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (A)

Step-by-step solution →
Q7·PhysicsSingle correct
The efficiency of a Carnot's engine, working between steam point and ice point, will be :
  1. (A)26.81%
  2. (B)37.81%
  3. (C)47.81%
  4. (D)57.81%

Correct answer: (A)

Step-by-step solution →
Q8·PhysicsSingle correct
Time period of a simple pendulum in a stationary lift is 'T'. If the lift accelerates with g6\frac{g}{6}6g​ vertically upwards then the time period will be : (where g = acceleration due to gravity)
  1. (A)65T\sqrt{\frac{6}{5}}T56​​T
  2. (B)56T\sqrt{\frac{5}{6}}T65​​T
  3. (C)67T\sqrt{\frac{6}{7}}T76​​T
  4. (D)76T\sqrt{\frac{7}{6}}T67​​T

Correct answer: (C)

Step-by-step solution →
Q9·PhysicsSingle correct
A thermally insulated vessel contains an ideal gas of molecular mass M and ratio of specific heats 1.4. Vessel is moving with speed v and is suddenly brought to rest. Assuming no heat is lost to the surrounding and vessel temperature of the gas increases by : (R = universal gas constant)
  1. (A)Mv27R\frac{Mv^{2}}{7R}7RMv2​
  2. (B)Mv25R\frac{Mv^{2}}{5R}5RMv2​
  3. (C)2Mv27R2\frac{Mv^{2}}{7R}27RMv2​
  4. (D)7Mv25R7\frac{Mv^{2}}{5R}75RMv2​

Correct answer: (B)

Step-by-step solution →
Q10·PhysicsSingle correct
Two capacitors having capacitance C1C_{1}C1​ and C2C_{2}C2​ respectively are connected as shown in figure. Initially, capacitor C1C_{1}C1​ is charged to a potential difference V volt by a battery. The battery is then removed and the charged capacitor C1C_{1}C1​ is now connected to uncharged capacitor C2C_{2}C2​ by closing the switch S. The amount of charge on the capacitor C2C_{2}C2​, after equilibrium is :
  1. (A)C1C2(C1+C2)V\frac{C_{1}C_{2}}{\left(C_{1}+C_{2}\right)}V(C1​+C2​)C1​C2​​V
  2. (B)(C1+C2)C1C2V\frac{\left(C_{1}+C_{2}\right)}{C_{1}C_{2}}VC1​C2​(C1​+C2​)​V
  3. (C)(C1+C2)V(C_{1}+C_{2})V(C1​+C2​)V
  4. (D)(C1−C2)V(C_{1}-C_{2})V(C1​−C2​)V

Correct answer: (A)

Step-by-step solution →
Q11·Physics·Capacitors and DielectricsSingle correct
Assertion (A) : Non-polar amterials do not have my permanent dipole moment. Reason (R) : When an non-polar material is placed in a electric field. the centre of the positive charge distribution of it's individual atom or molecule coinsides with the centre of the negative charge distribution. In the light of 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 and (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: (C)

Step-by-step solution →
Q12·PhysicsSingle correct
If electric field intensity of a uniform plane electro magnetic wave is given as E=−301.6 sin⁡(kz−ωt)a^x+452.4 sin⁡(kz−ωt)a^y VmE = -301.6\ \sin(kz - \omega t)\hat{a}_{x} + 452.4\ \sin(kz - \omega t)\hat{a}_{y}\ \frac{V}{m}E=−301.6 sin(kz−ωt)a^x​+452.4 sin(kz−ωt)a^y​ mV​ Then, magnetic intensity H of this wave in Am−1Am^{-1}Am−1 will be:' [Given: Speed of light in vacuum c=3×108 ms−1c = 3 \times 10^{8}\ ms^{-1}c=3×108 ms−1, permeability of vacuum μ0=4π×10−7 NA−2\mu_{0} = 4\pi \times 10^{-7}\ NA^{-2}μ0​=4π×10−7 NA−2]
  1. (A)+0.8sin⁡(kz−ωt)a^y+0.8sin⁡(kz−ωt)a^x+0.8\sin(kz - \omega t)\hat{a}_{y} + 0.8\sin(kz - \omega t)\hat{a}_{x}+0.8sin(kz−ωt)a^y​+0.8sin(kz−ωt)a^x​
  2. (B)+1.0×10−6sin⁡(kz−ωt)a^y+1.5×10−6(kz−ωt)a^x+1.0 \times 10^{-6}\sin(kz - \omega t)\hat{a}_{y} + 1.5 \times 10^{-6}(kz - \omega t)\hat{a}_{x}+1.0×10−6sin(kz−ωt)a^y​+1.5×10−6(kz−ωt)a^x​
  3. (C)−0.8sin⁡(kz−ωt)a^y−1.2sin⁡(kz−ωt)a^x-0.8\sin(kz - \omega t)\hat{a}_{y} - 1.2\sin(kz - \omega t)\hat{a}_{x}−0.8sin(kz−ωt)a^y​−1.2sin(kz−ωt)a^x​
  4. (D)−1.0×10−6sin⁡(kz−ωt)a^y−1.5×10−6sin⁡(kz−ωt)a^x-1.0 \times 10^{-6}\sin(kz - \omega t)\hat{a}_{y} - 1.5 \times 10^{-6}\sin(kz - \omega t)\hat{a}_{x}−1.0×10−6sin(kz−ωt)a^y​−1.5×10−6sin(kz−ωt)a^x​

Correct answer: (C)

Step-by-step solution →
Q13·PhysicsSingle correct
In free space, an electromagnetic wave of 3 GHz of 3 GHz frequency strikes over the edge of an object of size λ100\frac{\lambda}{100}100λ​, where λ\lambdaλ is the wavelength of the wave in free space. The phenomenon, which happens there will be:
  1. (A)Reflection
  2. (B)Refraction
  3. (C)Diffraction
  4. (D)Scattering

Correct answer: (D)

Step-by-step solution →
Q14·PhysicsSingle correct
An electron with speed v and a photon with speed c have the same de-Broglie wavelength. If the kinetic energy and momentum of electron are EeE_{e}Ee​ and pep_{e}pe​ and that of photon are EphE_{ph}Eph​ and pphp_{ph}pph​ respectively. Which of the following is correct?
  1. (A)EeEph=2cv\frac{E_{e}}{E_{ph}}=\frac{2c}{v}Eph​Ee​​=v2c​
  2. (B)EeEph=v2c\frac{E_{e}}{E_{ph}}=\frac{v}{2c}Eph​Ee​​=2cv​
  3. (C)pepph=2cv\frac{p_{e}}{p_{ph}}=\frac{2c}{v}pph​pe​​=v2c​
  4. (D)pepph=v2c\frac{p_{e}}{p_{ph}}=\frac{v}{2c}pph​pe​​=2cv​

Correct answer: (B)

Step-by-step solution →
Q15·PhysicsSingle correct
How many alpha and beta particles are emitted when Uranium 92U238_{92}U^{238}92​U238 decays to lead 82Pb206_{82}Pb^{206}82​Pb206 ?
  1. (A)3 alpha particles and 5 beta particles
  2. (B)6 alpha particles and 4 beta particles
  3. (C)4 alpha particles and 5 beta particles
  4. (D)8 alpha particles and 6 beta particles

Correct answer: (D)

Step-by-step solution →
Q16·PhysicsSingle correct
The I-V characteristics of a p-n junction diode in forward bias is shown in the figure. The ratio of dynamic resistance, corresponding to forward bias voltages of 2V and 4V respectively, is :
  1. (A)1 : 2
  2. (B)5 : 1
  3. (C)1 : 40
  4. (D)20 : 1

Correct answer: (B)

Step-by-step solution →
Q17·PhysicsSingle correct
Choose the correct statement for amplitude modulation:
  1. (A)Amplitude of modulating is varied in accordance with the information signal.
  2. (B)Amplitude of modulated is varied in accordance with the information signal.
  3. (C)Amplitude of carrier signal is varied in accordance with the information signal.
  4. (D)Amplitude of modulated is varied in accordance with the modulating signal.

Correct answer: (C)

Step-by-step solution →
Q18·PhysicsNumerical
A fighter jet is flying horizontally at a certain altitude with a speed of 200 ms−1ms^{-1}ms−1. When it passes directly overhead an anti-aircraft gun, bullet is fired from the gun, at an angle θ with the horizontal, to hit the jet. If the bullet speed is 400 m/s, the value of θ will be ............ °.

Correct answer: 60

Step-by-step solution →
Q19·PhysicsNumerical
A ball of mass 0.5 kg is dropped from the height of 10m. The height, at which the magnitude of velocity becomes equal to the magnitude of acceleration due to gravity, is ............... m. (Use g = 10 m/s2m/s^{2}m/s2).

Correct answer: 5

Step-by-step solution →
Q20·PhysicsNumerical
The elastic behaviour of material for linear streass and linear strain, is shown in the figure. The energy density for a linear strain of 5 × 10−410^{-4}10−4 is ............. kJ/m3kJ/m^{3}kJ/m3. Assume that material is elastic upto the linear strain of 5 × 10−410^{-4}10−4.

Correct answer: 25

Step-by-step solution →
Q21·PhysicsNumerical
The elongation of a wire on the surface of the earth is 10−410^{-4}10−4 m. The same wire of same dimensions is elongated by 6 × 10−510^{-5}10−5m on another planet. The acceleration due to gravity on the planet will be ............ ms−2ms^{-2}ms−2. (Take acceleration due to gravity on the surface of earth = 10 m/s−2m/s^{-2}m/s−2)

Correct answer: 6

Step-by-step solution →
Q22·PhysicsNumerical
A 10Ω, 20 mH coil carrying constant current is connected to a battery of 20 V through a switch is opened current becomes zero in 100μs. The average emf induced in the coil is ............ V.

Correct answer: 400

Step-by-step solution →
Q23·PhysicsNumerical
A light ray is incident, at an incident angle θ1\theta_{1}θ1​, on the system of two plane mirrors M1M_{1}M1​ and M2M_{2}M2​ having an inclination angle 75° between them (as shown in figure). After reflecting from mirror M1M_{1}M1​ it gets reflected back by the mirror M2M_{2}M2​ with an angle of reflection 30°. The total deviation of the ray will be .............. degree.

Correct answer: 210

Step-by-step solution →
Q24·PhysicsNumerical
In a vernier callipers, each cm on the main scale is divided into 20 equal parts. If tenth vernier scale division coincides with nineth main scale division. Then the value of vernier constant will be .............. × 10−210^{-2}10−2 mm.

Correct answer: 5

Step-by-step solution →
Q25·PhysicsNumerical
As per the given circuit, the value of current through the battery will be ........... A.

Correct answer: 1

Step-by-step solution →
Q26·PhysicsNumerical
A 110 V , 50 Hz, AC source is connected in the circuit (as shown in figure). The current through the resistance 55 Ω, at resonance in the circuit, will be ............... A.

Correct answer: 0

Step-by-step solution →
Q27·PhysicsNumerical
An ideal fluid of density 800 kgm−3kgm^{-3}kgm−3, flows smoothly through a bent pipe (as shown in figure) that tapers in cross-sectional area from a to a2\frac{a}{2}2a​. The pressure difference between the wide and narrow sections of pipe is 4100 Pa. At wider section, the velocity of fluid is x6\frac{\sqrt{x}}{6}6x​​ms−1ms^{-1}ms−1 for x = .................. (Given g = 10 m−2m^{-2}m−2)

Correct answer: 363

Step-by-step solution →

Chemistry — JEE Main 26 June 2022 Shift 1

Q28·ChemistrySingle correct
A commercially sold conc. HCl is 35% HCl by mass. If the density of this commercial acid is 1.46 g/mL, the molarity of this solution is : (Atomic mass : Cl = 35.5 amu, H = 1 amu)
  1. (A)10.2 M
  2. (B)12.5 M
  3. (C)14.0 M
  4. (D)18.2 M

Correct answer: (C)

Step-by-step solution →
Q29·ChemistrySingle correct
An evacuated glass vessel weighs 40.0 g when empty, 135.0 g when filled with a liquid of density 0.95 g mL−1mL^{-1}mL−1 and 40.5 g when filled with an ideal gas at 0.82 atm at 250 K. The molar mass of the gas in g mol−1mol^{-1}mol−1 is : (Given : R = 0.082 L atm K−1K^{-1}K−1 mol−1mol^{-1}mol−1)
  1. (A)35
  2. (B)50
  3. (C)75
  4. (D)125

Correct answer: (D)

Step-by-step solution →
Q30·ChemistrySingle correct
If the radius of the 3rd3^{rd}3rd Bohr's orbit of hydrogen atom is r3r_{3}r3​ and the radius of 4th4^{th}4th Bohr's orbit is r4r_{4}r4​. Then :
  1. (A)r4=916r3r_{4} = \frac{9}{16} r_{3}r4​=169​r3​
  2. (B)r4=169r3r_{4} = \frac{16}{9} r_{3}r4​=916​r3​
  3. (C)r4=34r3r_{4} = \frac{3}{4} r_{3}r4​=43​r3​
  4. (D)r4=43r3r_{4} = \frac{4}{3} r_{3}r4​=34​r3​

Correct answer: (B)

Step-by-step solution →
Q31·ChemistrySingle correct
Consider the ions/molecule O2+O_{2}^{+}O2+​, O2O_{2}O2​, O2−O_{2}^{-}O2−​, O22−O_{2}^{2-}O22−​ For increasing bond order the correct option is :
  1. (A)O22−<O2−<O2<O2+O_{2}^{2-} < O_{2}^{-} < O_{2} < O_{2}^{+}O22−​<O2−​<O2​<O2+​
  2. (B)O2−<O22−<O2<O2+O_{2}^{-} < O_{2}^{2-} < O_{2} < O_{2}^{+}O2−​<O22−​<O2​<O2+​
  3. (C)O2−<O22−<O2+<O2O_{2}^{-} < O_{2}^{2-} < O_{2}^{+} < O_{2}O2−​<O22−​<O2+​<O2​
  4. (D)O2−<O2+<O22−<O2O_{2}^{-} < O_{2}^{+} < O_{2}^{2-} < O_{2}O2−​<O2+​<O22−​<O2​

Correct answer: (A)

Step-by-step solution →
Q32·ChemistrySingle correct
The (∂E∂T)P\left(\frac{\partial E}{\partial T}\right)_{P}(∂T∂E​)P​ of different types of half cells are as follows : A B C D 1×10−41 \times 10^{-4}1×10−4 2×10−42 \times 10^{-4}2×10−4 0.1×10−40.1 \times 10^{-4}0.1×10−4 0.2×10−40.2 \times 10^{-4}0.2×10−4 (Where E is the electromotive force) Which of the above half cells would be preferred to be used as reference electrode ?
  1. (A)A
  2. (B)B
  3. (C)C
  4. (D)D

Correct answer: (C)

Step-by-step solution →
Q33·ChemistrySingle correct
Choose the correct stability order of group 13 elements in their +1 oxidation state.
  1. (A)Al < Ga < In < Tl
  2. (B)Tl < In < Ga < Al
  3. (C)Al < Ga < Tl < In
  4. (D)Al < Tl < Ga < In

Correct answer: (A)

Step-by-step solution →
Q34·ChemistrySingle correct
Given below are two statements : Statement I : According to the Ellingham diagram, any metal oxide with higher ΔG° is more stable than the one with lower ΔG°. Statement II : The metal involved in the formation of oxide placed lower in the Ellingham diagram can reduce the oxide of a metal placed higher in the diagram. In the light of the above statements, choose the most appropriate answer from the options given below :
  1. (A)Both Statement I and Statement II are 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 →
Q35·ChemistrySingle correct
Consider the following reaction : 2HSO4−(aq)→(1) Electrolysis (2) Hydrolysis2HSO4−+2H++A2HSO_{4}^{-}(aq) \xrightarrow{(1)\ Electrolysis\ (2)\ Hydrolysis} 2HSO_{4}^{-} + 2H^{+} + A2HSO4−​(aq)(1) Electrolysis (2) Hydrolysis​2HSO4−​+2H++A The dihedral angle in product A in its solid phase at 110 K is :
  1. (A)104°
  2. (B)111.5°
  3. (C)90.2°
  4. (D)111.0°

Correct answer: (C)

Step-by-step solution →
Q36·ChemistrySingle correct
The correct order of melting point is :
  1. (A)Be > Mg > Ca > Sr
  2. (B)Sr > Ca > Mg > Be
  3. (C)Be > Ca > Mg > Sr
  4. (D)Be > Ca > Sr > Mg

Correct answer: (D)

Step-by-step solution →
Q37·ChemistrySingle correct
The correct order of melting points of hydrides of group 16 elements is :
  1. (A)H2S<H2Se<H2Te<H2OH_{2}S < H_{2}Se < H_{2}Te < H_{2}OH2​S<H2​Se<H2​Te<H2​O
  2. (B)H2O<H2S<H2Se<H2TeH_{2}O < H_{2}S < H_{2}Se < H_{2}TeH2​O<H2​S<H2​Se<H2​Te
  3. (C)H2S<H2Te<H2Se<H2OH_{2}S < H_{2}Te < H_{2}Se < H_{2}OH2​S<H2​Te<H2​Se<H2​O
  4. (D)H2Se<H2S<H2Te<H2OH_{2}Se < H_{2}S < H_{2}Te < H_{2}OH2​Se<H2​S<H2​Te<H2​O

Correct answer: (A)

Step-by-step solution →
Q38·ChemistrySingle correct
Consider the following reaction : A + alkali → B (Major Product) If B is an oxoacid of phosphorus with no P–H bond, then A is :
  1. (A)White P4P_{4}P4​
  2. (B)Red P4P_{4}P4​
  3. (C)P2O3P_{2}O_{3}P2​O3​
  4. (D)H3PO3H_{3}PO_{3}H3​PO3​

Correct answer: (B)

Step-by-step solution →
Q39·ChemistrySingle correct
Polar stratospheric clouds facilitate the formation of :
  1. (A)ClONO2ClONO_{2}ClONO2​
  2. (B)HOCl
  3. (C)ClO
  4. (D)CH4CH_{4}CH4​

Correct answer: (B)

Step-by-step solution →
Q40·ChemistrySingle correct
Given below are two statements : Statement I : In 'Lassaigne's Test, when both nitrogen and sulphur are present in an organic compound, sodium thiocyanate is formed. Statement II : If both nitrogen and sulphur are present in an organic compound, then the excess of sodium used in sodium fusion will decompose the sodium thiocyanate formed to give NaCN and Na2SNa_{2}SNa2​S. In the light of the above statements, choose the most appropriate answer from the options given below :
  1. (A)Both Statement I and Statement II are 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 →
Q41·Chemistry·Reaction MechanismSingle correct
(C7H5O2)2→hν[X]+2C˙6H5+2CO2(C_{7}H_{5}O_{2})_{2} \xrightarrow{h\nu} [X] + 2\dot{C}_{6}H_{5} + 2CO_{2}(C7​H5​O2​)2​hν​[X]+2C˙6​H5​+2CO2​ Consider the above reaction and identify the intermediate 'X'
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (D)

Step-by-step solution →
Q42·ChemistrySingle correct
Consider the above reaction sequence and identify the product B.
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (A)

Step-by-step solution →
Q43·ChemistrySingle correct
Which will have the highest enol content ?
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (C)

Step-by-step solution →
Q44·ChemistrySingle correct
Among the following structures, which will show the most stable enamine formation ? (Where Me is −CH3-CH_{3}−CH3​)
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (C)

Step-by-step solution →
Q45·ChemistrySingle correct
Which of the following sets are correct regarding polymer ? (A) Copolymer : Buna–S (B) Condensation polymer : Nylon–6,6 (C) Fibre : Nylon–6,6 (D) Thermosetting polymer : Terylene (E) Homopolymer : Buna–N Choose the correct answer from given options below:
  1. (A)(A), (B) and (C) are correct
  2. (B)(B), (C) and (D) are correct
  3. (C)(A), (C) and (E) are correct
  4. (D)(A), (B) and (D) are correct

Correct answer: (A)

Step-by-step solution →
Q46·ChemistrySingle correct
A chemical which stimulates the secretion of pepsin is :
  1. (A)Anti histamine
  2. (B)Cimetidine
  3. (C)Histamine
  4. (D)Zantac

Correct answer: (C)

Step-by-step solution →
Q47·ChemistrySingle correct
Which statement is not true with respect to nitrate ion test ?
  1. (A)A dark brown ring is formed at the junction of two solutions.
  2. (B)Ring is formed due to nitroferrous sulphate complex.
  3. (C)The brown complex is [Fe(H2O)5(NO)]SO4[Fe(H_{2}O)_{5} (NO)]SO_{4}[Fe(H2​O)5​(NO)]SO4​.
  4. (D)Heating the nitrate salt with conc. H2SO4H_{2}SO_{4}H2​SO4​, light brown fumes are evolved.

Correct answer: (B)

Step-by-step solution →
Q48·ChemistryNumerical
For complete combustion of methanol CH3OH(l)+32O2(g)→CO2(g)+2H2O(l)CH_{3}OH(l) + \frac{3}{2}O_{2}(g) \rightarrow CO_{2}(g) + 2H_{2}O(l)CH3​OH(l)+23​O2​(g)→CO2​(g)+2H2​O(l) the amount of heat produced as measured by bomb calorimeter is 726 kJ mol−1mol^{-1}mol−1 at 27°C. The enthalpy of combustion for the reaction is –x kJ mol−1mol^{-1}mol−1, where x is ______. (Nearest integer) (Given : R = 8.3 JK−1JK^{-1}JK−1 mol−1mol^{-1}mol−1)

Correct answer: 727

Step-by-step solution →
Q49·ChemistryNumerical
A 0.5 percent solution of potassium chloride was found to freeze at –0.24°C. The percentage dissociation of potassium chloride is ______. (Nearest integer) (Molal depression constant for water is 1.80 K kg mol−1mol^{-1}mol−1 and molar mass of KCl is 74.6 g mol−1mol^{-1}mol−1)

Correct answer: 98

Step-by-step solution →
Q50·ChemistryNumerical
50 mL of 0.1 M CH3COOHCH_{3}COOHCH3​COOH is being titrated against 0.1 M NaOH. When 25 mL of NaOH has been added, the pH of the solution will be ______ ×10−2\times 10^{-2}×10−2. (Nearest integer) (Given : pKapK_{a}pKa​ (CH3COOHCH_{3}COOHCH3​COOH) = 4.76) log 2 = 0.30 log 3 = 0.48 log 5 = 0.69 log 7 = 0.84 log 11 = 1.04

Correct answer: 476

Step-by-step solution →
Q51·ChemistryNumerical
A flask is filled with equal moles of A and B. The half lives of A and B are 100 s and 50 s respectively and are independent of the initial concentration. The time required for the concentration of A to be four times that of B is ______s. (Given : ln 2 = 0.693)

Correct answer: 200

Step-by-step solution →
Q52·ChemistryNumerical
2.0 g of H2H_{2}H2​ gas is adsorbed on 2.5 g of platinum powder at 300 K and 1 bar pressure. The volume of the gas adsorbed per gram of the adsorbent is ______ mL. (Given : R = 0.083 L bar K−1K^{-1}K−1 mol−1mol^{-1}mol−1)

Correct answer: 9960

Step-by-step solution →
Q53·ChemistryNumerical
The spin–only magnetic moment value of the most basic oxide of vanadium among V2O3V_{2}O_{3}V2​O3​, V2O4V_{2}O_{4}V2​O4​ and V2O5V_{2}O_{5}V2​O5​ is ______ B.M. (Nearest Integer)

Correct answer: 3

Step-by-step solution →
Q54·ChemistryNumerical
The spin–only magnetic moment value of an octahedral complex among CoCl3.4NH3CoCl_{3}.4NH_{3}CoCl3​.4NH3​, NiCl2.6H2ONiCl_{2}.6H_{2}ONiCl2​.6H2​O and PtCl4.2HClPtCl_{4}.2HClPtCl4​.2HCl, which upon reaction with excess of AgNO3AgNO_{3}AgNO3​ gives 2 moles of AgCl is ______ B.M. (Nearest Integer)

Correct answer: 3

Step-by-step solution →
Q55·ChemistryNumerical
On complete combustion 0.30 g of an organic compound gave 0.20 g of carbon dioxide and 0.10 g of water. The percentage of carbon in the given organic compound is ______ (Nearest Integer)

Correct answer: 18

Step-by-step solution →
Q56·ChemistryNumerical
Compound 'P' on nitration with dil. HNO3HNO_{3}HNO3​ yields two isomers (A) and (B). These isomers can be separated by steam distillation. Isomers (A) and (B) show the intramolecular and intermolecular hydrogen bonding respectively. Compound (P) on reaction with conc. HNO3HNO_{3}HNO3​ yields a yellow compound 'C', a strong acid. The number of oxygen atoms is present in compound 'C' ______.

Correct answer: 7

Step-by-step solution →
Q57·ChemistryNumerical
The number of oxygens present in a nucleotide formed from a base, that is present only in RNA is ______.

Correct answer: 9

Step-by-step solution →

Mathematics — JEE Main 26 June 2022 Shift 1

Q58·MathematicsSingle correct
Let f(x)=x−1x+1f(x) = \frac{x-1}{x+1}f(x)=x+1x−1​, x∈R−{0,−1,1}x \in R - \{0, -1, 1\}x∈R−{0,−1,1}. If fn+1(x)=f(fn(x))f^{n+1}(x) = f(f^{n}(x))fn+1(x)=f(fn(x)) for all n∈Nn \in Nn∈N, then f6(6)+f7(7)f^{6}(6) + f^{7}(7)f6(6)+f7(7) is equal to :
  1. (A)76\frac{7}{6}67​
  2. (B)−32-\frac{3}{2}−23​
  3. (C)712\frac{7}{12}127​
  4. (D)−1112-\frac{11}{12}−1211​

Correct answer: (B)

Step-by-step solution →
Q59·MathematicsSingle correct
Let A={z∈C:∣z+1z−1∣<1}A = \left\{ z \in C : \left| \frac{z+1}{z-1} \right| < 1 \right\}A={z∈C:​z−1z+1​​<1} and B={z∈C:arg⁡(z−1z+1)=2π3}B = \left\{ z \in C : \arg\left( \frac{z-1}{z+1} \right) = \frac{2\pi}{3} \right\}B={z∈C:arg(z+1z−1​)=32π​}. Then A∩BA \cap BA∩B is :
  1. (A)a portion of a circle centred at (0,−13)\left(0, -\frac{1}{\sqrt{3}}\right)(0,−3​1​) that lies in the second and third quadrants only
  2. (B)a portion of a circle centred at (0,−13)\left(0, -\frac{1}{\sqrt{3}}\right)(0,−3​1​) that lies in the second quadrant only
  3. (C)an empty set
  4. (D)a portion of a circle of radius 23\frac{2}{\sqrt{3}}3​2​ that lies in the third quadrant only

Correct answer: (B)

Step-by-step solution →
Q60·MathematicsSingle correct
Let A be a 3×33 \times 33×3 invertible matrix. If ∣adj(24A)∣=adj(3adj(2A))∣|adj (24A)| = adj(3adj(2A))|∣adj(24A)∣=adj(3adj(2A))∣, then ∣A∣2|A|^{2}∣A∣2 is equal to :
  1. (A)666^{6}66
  2. (B)2122^{12}212
  3. (C)262^{6}26
  4. (D)1

Correct answer: (C)

Step-by-step solution →
Q61·MathematicsSingle correct
The ordered pair (a, b), for which the system of linear equations 3x−2y+z=b3x - 2y + z = b3x−2y+z=b 5x−8y+9z=35x - 8y + 9z = 35x−8y+9z=3 2x+y+az=−12x + y + az = -12x+y+az=−1 has no solution, is :
  1. (A)(3,13)\left(3, \frac{1}{3}\right)(3,31​)
  2. (B)(−3,13)\left(-3, \frac{1}{3}\right)(−3,31​)
  3. (C)(−3,−13)\left(-3, -\frac{1}{3}\right)(−3,−31​)
  4. (D)(3,−13)\left(3, -\frac{1}{3}\right)(3,−31​)

Correct answer: (C)

Step-by-step solution →
Q62·MathematicsSingle correct
The remainder when (2021)2023(2021)^{2023}(2021)2023 is divided by 7 is :
  1. (A)1
  2. (B)2
  3. (C)5
  4. (D)6

Correct answer: (C)

Step-by-step solution →
Q63·Mathematics·Limits and ContinuitySingle correct
lim⁡x→12sin⁡(cos⁡−1x)−x1−tan⁡(cos⁡−1x)\lim_{x \to \frac{1}{\sqrt{2}}} \frac{\sin(\cos^{-1} x) - x}{1 - \tan(\cos^{-1} x)}limx→2​1​​1−tan(cos−1x)sin(cos−1x)−x​ is equal to :
  1. (A)2\sqrt{2}2​
  2. (B)−2-\sqrt{2}−2​
  3. (C)12\frac{1}{\sqrt{2}}2​1​
  4. (D)−12-\frac{1}{\sqrt{2}}−2​1​

Correct answer: (D)

Step-by-step solution →
Q64·Mathematics·Limits and ContinuitySingle correct
Let f, g : R →\to→ R be two real valued functions defined as f(x)={−∣x+3∣,x<0ex,x≥0f(x) = \begin{cases} -|x+3| & , & x < 0 \\ e^{x} & , & x \geq 0 \end{cases}f(x)={−∣x+3∣ex​,,​x<0x≥0​ and g(x)={x2+k1x,x<04x+k2,x≥0g(x) = \begin{cases} x^{2} + k_{1}x & , & x < 0 \\ 4x + k_{2} & , & x \geq 0 \end{cases}g(x)={x2+k1​x4x+k2​​,,​x<0x≥0​, where k1k_{1}k1​ and k2k_{2}k2​ are real constants. If (gof) is differentiable at x = 0, then (gof) (-4) + (gof) (4) is equal to :
  1. (A)4(e4+1)4(e^{4} + 1)4(e4+1)
  2. (B)2(2e4+1)2(2e^{4} + 1)2(2e4+1)
  3. (C)4e44e^{4}4e4
  4. (D)2(2e4−1)2(2e^{4} - 1)2(2e4−1)

Correct answer: (D)

Step-by-step solution →
Q65·MathematicsSingle correct
The sum of the absolute minimum and the absolute maximum values of the function f(x)=∣3x−x2+2∣−xf(x) = |3x - x^{2} + 2| - xf(x)=∣3x−x2+2∣−x in the interval [−1,2][-1, 2][−1,2] is :
  1. (A)17+32\frac{\sqrt{17} + 3}{2}217​+3​
  2. (B)17+52\frac{\sqrt{17} + 5}{2}217​+5​
  3. (C)5
  4. (D)9−172\frac{9 - \sqrt{17}}{2}29−17​​

Correct answer: (A)

Step-by-step solution →
Q66·MathematicsSingle correct
Let S be the set of all the natural numbers, for which the line xa+yb=2\frac{x}{a} + \frac{y}{b} = 2ax​+by​=2 is a tangent to the curve (xa)n+(yb)n=2\left(\frac{x}{a}\right)^{n} + \left(\frac{y}{b}\right)^{n} = 2(ax​)n+(by​)n=2 at the point (a, b), ab≠0ab \neq 0ab=0. Then:
  1. (A)S=ϕS = \phiS=ϕ
  2. (B)n(S)=1n(S) = 1n(S)=1
  3. (C)S={2k:k∈N}S = \{2k : k \in N\}S={2k:k∈N}
  4. (D)S=NS = NS=N

Correct answer: (D)

Step-by-step solution →
Q67·MathematicsSingle correct
Let R be the point (3, 7) and let P and Q be two points on the line x + y = 5 such that PQR is an equilateral triangle. Then the area of Δ\DeltaΔPQR is :
  1. (A)2543\frac{25}{4\sqrt{3}}43​25​
  2. (B)2532\frac{25\sqrt{3}}{2}2253​​
  3. (C)253\frac{25}{\sqrt{3}}3​25​
  4. (D)2523\frac{25}{2\sqrt{3}}23​25​

Correct answer: (D)

Step-by-step solution →
Q68·MathematicsSingle correct
Let C be a circle passing through the points A(2, -1) and B(3, 4). The line segment AB is not a diameter of C. If r is the radius of C and its centre lies on the circle (x−5)2+(y−1)2=132(x - 5)^{2} + (y - 1)^{2} = \frac{13}{2}(x−5)2+(y−1)2=213​, then r2r^{2}r2 is equal to :
  1. (A)32
  2. (B)652\frac{65}{2}265​
  3. (C)612\frac{61}{2}261​
  4. (D)30

Correct answer: (B)

Step-by-step solution →
Q69·MathematicsSingle correct
Let the normal at the point P on the parabola y2=6xy^{2} = 6xy2=6x pass through the point (5, -8). If the tangent at P to the parabola intersects its directrix at the point Q, then the ordinate of the point Q is :
  1. (A)-3
  2. (B)−94-\frac{9}{4}−49​
  3. (C)−52-\frac{5}{2}−25​
  4. (D)-2

Correct answer: (B)

Step-by-step solution →
Q70·MathematicsSingle correct
If the two lines l1:x−23=y+1−2l_{1} : \frac{x-2}{3} = \frac{y+1}{-2}l1​:3x−2​=−2y+1​, z=2z = 2z=2 and l2:x−11=2y+3α=z+52l_{2} : \frac{x-1}{1} = \frac{2y+3}{\alpha} = \frac{z+5}{2}l2​:1x−1​=α2y+3​=2z+5​ perpendicular, then an angle between the lines l2l_{2}l2​ and l3:1−x3=2y−1−4=z4l_{3} : \frac{1-x}{3} = \frac{2y-1}{-4} = \frac{z}{4}l3​:31−x​=−42y−1​=4z​ is :
  1. (A)cos⁡−1(294)\cos^{-1}\left(\frac{29}{4}\right)cos−1(429​)
  2. (B)sec⁡−1(294)\sec^{-1}\left(\frac{29}{4}\right)sec−1(429​)
  3. (C)cos⁡−1(229)\cos^{-1}\left(\frac{2}{29}\right)cos−1(292​)
  4. (D)cos⁡−1(229)\cos^{-1}\left(\frac{2}{\sqrt{29}}\right)cos−1(29​2​)

Correct answer: (B)

Step-by-step solution →
Q71·MathematicsSingle correct
Let the plane 2x+3y+z+20=02x + 3y + z + 20 = 02x+3y+z+20=0 be rotated through a right angle about its line of intersection with the plane x−3y+5z=8x - 3y + 5z = 8x−3y+5z=8. If the mirror image of the point (2,−12,2)\left(2, -\frac{1}{2}, 2\right)(2,−21​,2) in the rotated plane is B(a, b, c), then :
  1. (A)a8=b5=c−4\frac{a}{8} = \frac{b}{5} = \frac{c}{-4}8a​=5b​=−4c​
  2. (B)a4=b5=c−2\frac{a}{4} = \frac{b}{5} = \frac{c}{-2}4a​=5b​=−2c​
  3. (C)a8=b−5=c4\frac{a}{8} = \frac{b}{-5} = \frac{c}{4}8a​=−5b​=4c​
  4. (D)a4=b5=c2\frac{a}{4} = \frac{b}{5} = \frac{c}{2}4a​=5b​=2c​

Correct answer: (A)

Step-by-step solution →
Q72·MathematicsSingle correct
If a⃗⋅b⃗=1\vec{a}\cdot\vec{b}=1a⋅b=1, b⃗⋅c⃗=2\vec{b}\cdot\vec{c}=2b⋅c=2 and c⃗⋅a⃗=3\vec{c}\cdot\vec{a}=3c⋅a=3, then the value of [a⃗×(b⃗×c⃗), b⃗×(c⃗×a⃗), c⃗×(b⃗×a⃗)]\left[\vec{a}\times\left(\vec{b}\times\vec{c}\right),\ \vec{b}\times\left(\vec{c}\times\vec{a}\right),\ \vec{c}\times\left(\vec{b}\times\vec{a}\right)\right][a×(b×c), b×(c×a), c×(b×a)] is :
  1. (A)000
  2. (B)−6a⃗⋅(b⃗×c⃗)-6\vec{a}\cdot\left(\vec{b}\times\vec{c}\right)−6a⋅(b×c)
  3. (C)12c⃗⋅(a⃗×b⃗)12\vec{c}\cdot\left(\vec{a}\times\vec{b}\right)12c⋅(a×b)
  4. (D)−12b⃗⋅(c⃗×a⃗)-12\vec{b}\cdot\left(\vec{c}\times\vec{a}\right)−12b⋅(c×a)

Correct answer: (A)

Step-by-step solution →
Q73·MathematicsSingle correct
Let a biased coin be tossed 5 times. If the probability of getting 4 heads is equal to the probability of getting 5 heads, then the probability of getting atmost two heads is:
  1. (A)27565\frac{275}{6^{5}}65275​
  2. (B)3654\frac{36}{5^{4}}5436​
  3. (C)18155\frac{181}{5^{5}}55181​
  4. (D)4664\frac{46}{6^{4}}6446​

Correct answer: (D)

Step-by-step solution →
Q74·MathematicsSingle correct
The mean of the numbers a, b, 8, 5, 10 is 6 and their variance is 6.8. If M is the mean deviation of the numbers about the mean, then 25 M is equal to:
  1. (A)606060
  2. (B)555555
  3. (C)505050
  4. (D)454545

Correct answer: (A)

Step-by-step solution →
Q75·MathematicsSingle correct
Let f(x)=2cos⁡−1x+4cot⁡−1x−3x2−2x+10f(x) = 2\cos^{-1}x + 4\cot^{-1}x - 3x^{2} - 2x + 10f(x)=2cos−1x+4cot−1x−3x2−2x+10, x∈[−1,1]x \in [-1, 1]x∈[−1,1]. If [a,b][a, b][a,b] is the range of the function then 4a−b4a - b4a−b is equal to:
  1. (A)111111
  2. (B)11−π11 - \pi11−π
  3. (C)11+π11 + \pi11+π
  4. (D)15−π15 - \pi15−π

Correct answer: (B)

Step-by-step solution →
Q76·MathematicsSingle correct
Let Δ,∇∈{∧,∨}\Delta, \nabla \in \{\wedge, \vee\}Δ,∇∈{∧,∨} be such that p∇q⇒((pΔq)∇r)p \nabla q \Rightarrow ((p \Delta q) \nabla r)p∇q⇒((pΔq)∇r) is a tautology. Then (p∇q)Δr(p \nabla q) \Delta r(p∇q)Δr is logically equivalent to :
  1. (A)(pΔr)∨q(p \Delta r) \vee q(pΔr)∨q
  2. (B)(pΔr)∧q(p \Delta r) \wedge q(pΔr)∧q
  3. (C)(p∧r)Δq(p \wedge r) \Delta q(p∧r)Δq
  4. (D)(p∇r)∧q(p \nabla r) \wedge q(p∇r)∧q

Correct answer: (A)

Step-by-step solution →
Q77·MathematicsNumerical
The sum of the cubes of all the roots of the equation x4−3x3−2x2+3x+1=10x^{4} - 3x^{3} - 2x^{2} + 3x + 1 = 10x4−3x3−2x2+3x+1=10 is ______.

Correct answer: 36

Step-by-step solution →
Q78·MathematicsNumerical
There are ten boys B1B_{1}B1​, B2B_{2}B2​, ...., B10B_{10}B10​ and five girls G1G_{1}G1​, G2G_{2}G2​, ...., G5G_{5}G5​ in a class. Then the number of ways of forming a group consisting of three boys and three girls, if both B1B_{1}B1​ and B2B_{2}B2​ together should not be the members of a group, is____________.

Correct answer: 1120

Step-by-step solution →
Q79·MathematicsNumerical
Let the common tangents to the curves 4(x2+y2)=94(x^{2} + y^{2}) = 94(x2+y2)=9 and y2=4xy^{2} = 4xy2=4x intersect at the point Q. Let an ellipse, centered at the origin O, has lengths of semi-minor and semi-major axes equal to OQ and 6, respectively. If eee and lll respectively denote the eccentricity and the length of the latus rectum of this ellipse, then le2\frac{l}{e^{2}}e2l​ is equal to____________.

Correct answer: 4

Step-by-step solution →
Q80·MathematicsNumerical
Let f(x)=max⁡{∣x+1∣,∣x+2∣,...,∣x+5∣}f(x) = \max\{|x + 1|, |x + 2|, ..., |x + 5|\}f(x)=max{∣x+1∣,∣x+2∣,...,∣x+5∣}. Then ∫−60f(x) dx\int_{-6}^{0} f(x)\,dx∫−60​f(x)dx is equal to ______________.

Correct answer: 21

Step-by-step solution →
Q81·MathematicsNumerical
Let the solution curve y=y(x)y = y(x)y=y(x) of the differential equation (4+x2)dy−2x(x2+3y+4)dx=0(4 + x^{2})dy - 2x(x^{2} + 3y + 4)dx = 0(4+x2)dy−2x(x2+3y+4)dx=0 pass through the origin. Then y(2)y(2)y(2) is equal to_________.

Correct answer: 12

Step-by-step solution →
Q82·MathematicsNumerical
If sin⁡2(10∘)sin⁡(20∘)sin⁡(40∘)sin⁡(50∘)sin⁡(70∘)=α−116sin⁡(10∘)\sin^{2}(10^{\circ})\sin(20^{\circ})\sin(40^{\circ})\sin(50^{\circ})\sin(70^{\circ}) = \alpha - \frac{1}{16}\sin(10^{\circ})sin2(10∘)sin(20∘)sin(40∘)sin(50∘)sin(70∘)=α−161​sin(10∘), then 16+α−116 + \alpha^{-1}16+α−1 is equal to ___________.

Correct answer: 80

Step-by-step solution →
Q83·MathematicsNumerical
Let A={n∈N:H.C.F. (n,45)=1}A = \{n \in N : \text{H.C.F. }(n, 45) = 1\}A={n∈N:H.C.F. (n,45)=1} and Let B={2k:k∈{1,2,...,100}}B = \{2k : k \in \{1, 2, ..., 100\}\}B={2k:k∈{1,2,...,100}}. Then the sum of all the elements of A∩BA \cap BA∩B is ___________.

Correct answer: 5264

Step-by-step solution →
Q84·MathematicsNumerical
The value of the integral 48π4∫0π(3πx22−x3)sin⁡x1+cos⁡2x dx\frac{48}{\pi^{4}}\int_{0}^{\pi}\left(\frac{3\pi x^{2}}{2} - x^{3}\right)\frac{\sin x}{1 + \cos^{2} x}\,dxπ448​∫0π​(23πx2​−x3)1+cos2xsinx​dx is equal to ________.

Correct answer: 6

Step-by-step solution →
Q85·MathematicsNumerical
Let A=∑i=110∑j=110min⁡{i,j}A = \sum_{i=1}^{10}\sum_{j=1}^{10}\min\{i, j\}A=∑i=110​∑j=110​min{i,j} and B=∑i=110∑j=110max⁡{i,j}B = \sum_{i=1}^{10}\sum_{j=1}^{10}\max\{i, j\}B=∑i=110​∑j=110​max{i,j}. Then A+BA + BA+B is equal to ________.

Correct answer: 1100

Step-by-step solution →
Q86·MathematicsNumerical
Let S=(0,2π)−{π2,3π4,3π2,7π4}S = (0, 2\pi) - \left\{\frac{\pi}{2}, \frac{3\pi}{4}, \frac{3\pi}{2}, \frac{7\pi}{4}\right\}S=(0,2π)−{2π​,43π​,23π​,47π​}. Let y=y(x)y = y(x)y=y(x), x∈Sx \in Sx∈S, be the solution curve of the differential equation dydx=11+sin⁡2x\frac{dy}{dx} = \frac{1}{1 + \sin 2x}dxdy​=1+sin2x1​, y(π4)=12y\left(\frac{\pi}{4}\right) = \frac{1}{2}y(4π​)=21​. if the sum of abscissas of all the points of intersection of the curve y=y(x)y = y(x)y=y(x) with the curve y=2sin⁡xy = \sqrt{2}\sin xy=2​sinx is kπ12\frac{k\pi}{12}12kπ​, then k is equal to __________.

Correct answer: 42

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
  • 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
  • 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
  • Quadratic Equations 148/186
  • Some Basic Concepts in Chemistry 129/186
  • Electromagnetic Induction 120/186
  • 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
  • Nuclei 116/186
  • Straight Lines 114/186
  • Capacitors and Dielectrics 115/186
  • Parabola 101/186
  • Statistics 118/186
  • Ellipse 103/186
  • Isolation of Metals 106/186
  • s-Block Elements 88/186
  • Surface Chemistry 98/186
  • Inverse Trigonometric Functions 93/186
  • Environmental Chemistry 83/186
  • Hydrogen 81/186
  • Experimental Skills 68/186
  • Polymers 64/186
  • Principles of Qualitative Analysis 58/186
  • Chemistry in Everyday Life 60/186
  • Reaction Mechanism 29/186
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