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JEE Main 25 June 2022 Shift 1 Question Paper with Answers

25 June 2022 · June session · 89 questions

89 of the 90 questions from the JEE Main 25 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 25 June 2022 Shift 1

Q1·PhysicsSingle correct
If Z=A2B3C4Z=\frac{A^2B^3}{C^4}Z=C4A2B3​ , then the relative error in Z will be :
  1. (A)ΔAA+ΔBB+ΔCC\frac{\Delta A}{A}+\frac{\Delta B}{B}+\frac{\Delta C}{C}AΔA​+BΔB​+CΔC​
  2. (B)2ΔAA+3ΔBB−4ΔCC\frac{2\Delta A}{A}+\frac{3\Delta B}{B}-\frac{4\Delta C}{C}A2ΔA​+B3ΔB​−C4ΔC​
  3. (C)2ΔAA+3ΔBB+4ΔCC\frac{2\Delta A}{A}+\frac{3\Delta B}{B}+\frac{4\Delta C}{C}A2ΔA​+B3ΔB​+C4ΔC​
  4. (D)ΔAA+ΔBB−ΔCC\frac{\Delta A}{A}+\frac{\Delta B}{B}-\frac{\Delta C}{C}AΔA​+BΔB​−CΔC​

Correct answer: (C)

Step-by-step solution →
Q2·PhysicsSingle correct
A⃗\vec{A}A is a vector quantity such that ∣A⃗∣|\vec{A}|∣A∣ = non-zero constant. Which of the following expressions is true for A⃗\vec{A}A ?
  1. (A)A⃗.A⃗=0\vec{A}.\vec{A}=0A.A=0
  2. (B)A⃗×A⃗<0\vec{A}\times\vec{A}<0A×A<0
  3. (C)A⃗×A⃗=0\vec{A}\times\vec{A}=0A×A=0
  4. (D)A⃗×A⃗>0\vec{A}\times\vec{A}>0A×A>0

Correct answer: (C)

Step-by-step solution →
Q3·PhysicsSingle correct
Which of the following relations is true for two unit vectors A^\hat{A}A^ and B^\hat{B}B^ making an angle θ to each other?
  1. (A)∣A^+B^∣=∣A^−B^∣tan⁡θ2|\hat{A}+\hat{B}|=|\hat{A}-\hat{B}|\tan\frac{\theta}{2}∣A^+B^∣=∣A^−B^∣tan2θ​
  2. (B)∣A^−B^∣=∣A^+B^∣tan⁡θ2|\hat{A}-\hat{B}|=|\hat{A}+\hat{B}|\tan\frac{\theta}{2}∣A^−B^∣=∣A^+B^∣tan2θ​
  3. (C)∣A^+B^∣=∣A^−B^∣cos⁡θ2|\hat{A}+\hat{B}|=|\hat{A}-\hat{B}|\cos\frac{\theta}{2}∣A^+B^∣=∣A^−B^∣cos2θ​
  4. (D)∣A^−B^∣=∣A^+B^∣cos⁡θ2|\hat{A}-\hat{B}|=|\hat{A}+\hat{B}|\cos\frac{\theta}{2}∣A^−B^∣=∣A^+B^∣cos2θ​

Correct answer: (B)

Step-by-step solution →
Q4·PhysicsSingle correct
If force F⃗=3i^+4j^−2k^\vec{F}=3\hat{i}+4\hat{j}-2\hat{k}F=3i^+4j^​−2k^ acts on a particle having position vector 2i^+j^+2k^2\hat{i}+\hat{j}+2\hat{k}2i^+j^​+2k^ then, the torque about the origin will be :-
  1. (A)3i^+4j^−2k^3\hat{i}+4\hat{j}-2\hat{k}3i^+4j^​−2k^
  2. (B)−10i^+10j^+5k^-10\hat{i}+10\hat{j}+5\hat{k}−10i^+10j^​+5k^
  3. (C)10i^+5j^−10k^10\hat{i}+5\hat{j}-10\hat{k}10i^+5j^​−10k^
  4. (D)10i^+j^−5k^10\hat{i}+\hat{j}-5\hat{k}10i^+j^​−5k^

Correct answer: (B)

Step-by-step solution →
Q5·PhysicsSingle correct
The height of any point P above the surface of earth is equal to diameter of earth. The value of acceleration due to gravity at point P will be : (Given g = acceleration due to gravity at the surface of earth)
  1. (A)g/2
  2. (B)g/4
  3. (C)g/3
  4. (D)g/9

Correct answer: (D)

Step-by-step solution →
Q6·PhysicsSingle correct
The terminal velocity (vt)(v_t)(vt​) of the spherical rain drop depends on the radius (r) of the spherical rain drop as:-
  1. (A)r1/2r^{1/2}r1/2
  2. (B)r
  3. (C)r2r^2r2
  4. (D)r3r^3r3

Correct answer: (C)

Step-by-step solution →
Q7·PhysicsSingle correct
The relation between root mean square speed (vrms)(v_{rms})(vrms​) and most probable speed (vp)(v_p)(vp​) for the molar mass M of oxygen gas molecule at the temperature of 300 K will be :-
  1. (A)vrms=23vpv_{rms}=\sqrt{\frac{2}{3}}v_pvrms​=32​​vp​
  2. (B)vrms=32vpv_{rms}=\sqrt{\frac{3}{2}}v_pvrms​=23​​vp​
  3. (C)vrms=vpv_{rms}=v_pvrms​=vp​
  4. (D)vrms=13vpv_{rms}=\sqrt{\frac{1}{3}}v_pvrms​=31​​vp​

Correct answer: (B)

Step-by-step solution →
Q8·PhysicsSingle correct
In the figure, a very large plane sheet of positive charge is shown. P1P_1P1​ and P2P_2P2​ are two points at distance lll and 2l2l2l from the charge distribution. If σ is the surface charge density, then the magnitude of electric fields E1E_1E1​ and E2E_2E2​ at P1P_1P1​ and P2P_2P2​ respectively are :
  1. (A)E1=σ/ε0E_1=\sigma/\varepsilon_0E1​=σ/ε0​, E2=σ/2ε0E_2=\sigma/2\varepsilon_0E2​=σ/2ε0​
  2. (B)E1=2σ/ε0E_1=2\sigma/\varepsilon_0E1​=2σ/ε0​, E2=σ/ε0E_2=\sigma/\varepsilon_0E2​=σ/ε0​
  3. (C)E1=E2=σ/2ε0E_1=E_2=\sigma/2\varepsilon_0E1​=E2​=σ/2ε0​
  4. (D)E1=E2=σ/ε0E_1=E_2=\sigma/\varepsilon_0E1​=E2​=σ/ε0​

Correct answer: (C)

Step-by-step solution →
Q9·Physics·Electromagnetic InductionSingle correct
Match List-I with List-II List-I List-II (A) AC generator (I) Detects the presence of current in the circuit (B) Galvanometer (II) Converts mechanical energy into electrical energy (C) Transformer (III) Works on the principle of resonance in AC circuit (D) Metal detector (IV) Changes an alternating voltage for smaller or greater value Choose the correct answer from the options given below :-
  1. (A)(A)–(II), B–(I), (C)–(IV), (D)–(III)
  2. (B)(A)–(II), B–(I), (C)–(III), (D)–(IV)
  3. (C)(A)–(III), B–(IV), (C)–(II), (D)–(I)
  4. (D)(A)–(III), B–(I), (C)–(II), (D)–(IV)

Correct answer: (A)

Step-by-step solution →
Q10·PhysicsSingle correct
A long straight wire with a circular cross-section having radius R, is carrying a steady current I. The current I is uniformly distributed across this cross-section. Then the variation of magnetic field due to current I with distance r (r < R) from its centre will be :-
  1. (A)B∝r2B\propto r^2B∝r2
  2. (B)B∝rB\propto rB∝r
  3. (C)B∝1r2B\propto\frac{1}{r^2}B∝r21​
  4. (D)B∝1rB\propto\frac{1}{r}B∝r1​

Correct answer: (B)

Step-by-step solution →
Q11·PhysicsSingle correct
If wattless current flows in the AC circuit, then the circuit is
  1. (A)Purely Resistive circuit
  2. (B)Purely Inductive circuit
  3. (C)LCR series circuit
  4. (D)RC series circuit only

Correct answer: (B)

Step-by-step solution →
Q12·PhysicsSingle correct
The electric field in an electromagnetic wave is given by E=56.5sin⁡ω(t−x/c) NC−1E=56.5\sin\omega(t-x/c)\,NC^{-1}E=56.5sinω(t−x/c)NC−1. Find the intensity of the wave if it is propagating along x-axis in the free space. (Given ε0=8.85×10−12C2N−1m−2\varepsilon_0=8.85\times10^{-12}C^2N^{-1}m^{-2}ε0​=8.85×10−12C2N−1m−2)
  1. (A)5.65 Wm−2Wm^{-2}Wm−2
  2. (B)4.24 Wm−2Wm^{-2}Wm−2
  3. (C)1.9×10−71.9\times10^{-7}1.9×10−7 Wm−2Wm^{-2}Wm−2
  4. (D)56.5 Wm−2Wm^{-2}Wm−2

Correct answer: (B)

Step-by-step solution →
Q13·PhysicsSingle correct
The two light beams having intensities I and 9I interfere to produce a fringe pattern on a screen. The phase difference between the beams is π2\frac{\pi}{2}2π​ at point P and π at point Q. Then the difference between the resultant intensities at P and Q will be :
  1. (A)2 I
  2. (B)6 I
  3. (C)5 I
  4. (D)7 I

Correct answer: (B)

Step-by-step solution →
Q14·PhysicsSingle correct
A light wave travelling linearly in a medium of dielectric constant 4, incident on the horizontal interface separating medium with air. The angle of incidence for which the total intensity of incident wave will be reflected back into the same medium will be (Given : relative permeability of medium μr\mu_rμr​ = 1)
  1. (A)10°
  2. (B)20°
  3. (C)30°
  4. (D)60°

Correct answer: (D)

Step-by-step solution →
Q15·PhysicsSingle correct
Given below are two statements :- Statement I : Davisson-Germer experiment establishes the wave nature of electrons. Statement II : If electrons have wave nature, they can interfere and show diffraction. 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: (A)

Step-by-step solution →
Q16·PhysicsSingle correct
The ratio for the speed of the electron in the 3rd orbit of He+^{+}+ to the speed of the electron in the 3rd orbit of hydrogen atom will be :-
  1. (A)1 : 1
  2. (B)1 : 2
  3. (C)4 : 1
  4. (D)2 : 1

Correct answer: (D)

Step-by-step solution →
Q17·PhysicsSingle correct
The photodiode is used to detect the optiocal signals. These diodes are preferably operated in reverse biased mode because.
  1. (A)fractional change in majority carriers produce higher forward bias current
  2. (B)fractional change in majority carriers produce higher reverse bias current
  3. (C)fractional change in minority carriers produce higher forward bias current
  4. (D)fractional change in minority carriers produce higher reverse bias current

Correct answer: (D)

Step-by-step solution →
Q18·PhysicsSingle correct
A signal of 100 THz frequency can be transmitted with maximum efficiency by :
  1. (A)Coaxial cable
  2. (B)Optical fibre
  3. (C)Twisted pair of copper wires
  4. (D)Water

Correct answer: (B)

Step-by-step solution →
Q19·PhysicsSingle correct
The difference of speed of light in the two media A and B (vA−vB)(v_{A} - v_{B})(vA​−vB​) is 2.6×1072.6 \times 10^{7}2.6×107 m/s. If the refractive index of medium B is 1.47, then the ratio of refractive index of medium B to medium A is : (Given : speed of light in vacuum c=3×108 ms−1c = 3 \times 10^{8}\ \mathrm{ms}^{-1}c=3×108 ms−1)
  1. (A)1.303
  2. (B)1.318
  3. (C)1.13
  4. (D)0.12

Correct answer: (C)

Step-by-step solution →
Q20·PhysicsSingle correct
A teacher in his physics laboratory allotted an experiment to determine the resistance (G) of a galvanometer. Students took the observations for 13\frac{1}{3}31​ deflection in the galvanometer. Which of the below is true for measuring value of G?
  1. (A)13\frac{1}{3}31​ deflection method cannot be used for determining the resistance of the galvanometer.
  2. (B)13\frac{1}{3}31​ deflection method can be used and in this case the G equals to twice the value of shunt resistance(s).
  3. (C)13\frac{1}{3}31​ deflection method can be used and in this case, the G equals to three times the value of shunt resistance(s)
  4. (D)13\frac{1}{3}31​ deflection method can be used and in this case the G value equals to the shunt resistance(s).

Correct answer: (B)

Step-by-step solution →
Q21·PhysicsNumerical
A uniform chain of 6 m length is placed on a table such that a part of its length is hanging over the edge of the table. The system is at rest. The co-efficient of static friction between the chain and the surface of the table is 0.5, the maximum length of the chain hanging from the table is __________ m.

Correct answer: 2

Step-by-step solution →
Q22·PhysicsNumerical
A 0.5 kg block moving at a speed of 12 ms−112\ \mathrm{ms}^{-1}12 ms−1 compresses a spring through a distance 30 cm when its speed is halved. The spring constant of the spring will be __________ Nm−1\mathrm{Nm}^{-1}Nm−1.

Correct answer: 600

Step-by-step solution →
Q23·PhysicsNumerical
The velocity of upper layer of water in a river is 36 kmh−136\ \mathrm{kmh}^{-1}36 kmh−1. Shearing stress between horizontal layers of water is 10−3 Nm−210^{-3}\ \mathrm{Nm}^{-2}10−3 Nm−2. Depth of the river is __________ m. (Co-efficiency of viscosity of water is 10−210^{-2}10−2 Pa.s)

Correct answer: 100

Step-by-step solution →
Q24·PhysicsNumerical
A steam engine intakes 50g of steam at 100°C per minute and cools it down to 20°C. If latent heat of vaporization of steam is 540 cal g−1540\ \mathrm{cal\ g}^{-1}540 cal g−1, then the heat rejected by the steam engine per minute is ______ ×103\times 10^{3}×103 cal.

Correct answer: 31

Step-by-step solution →
Q25·PhysicsNumerical
The first overtone frequency of an open organ pipe is equal to the fundamental frequency of a closed organ pipe. If the length of the closed organ pipe is 20 cm. The length of the open organ pipe is ______ cm.

Correct answer: 80

Step-by-step solution →
Q26·PhysicsNumerical
The equivalent capacitance between points A and B in below shown figure will be _____ μF.

Correct answer: 6

Step-by-step solution →
Q27·PhysicsNumerical
A resistor develops 300 J of thermal energy in 15s, when a current of 2A is passed through it. If the current increases to 3A, the energy developed in 10s is _________ J.

Correct answer: 450

Step-by-step solution →
Q28·PhysicsNumerical
The total current supplied to the circuit as shown in figure by the 5V battery is _________ A

Correct answer: 2

Step-by-step solution →
Q29·PhysicsNumerical
The current in a coil of self inductance 2.0 H is increasing according to I=2sin⁡(t2)I = 2\sin(t^{2})I=2sin(t2) A. The amount of energy spent during the period when current changes from 0 to 2A is ______ J.

Correct answer: 4

Step-by-step solution →
Q30·PhysicsNumerical
A force on an object of mass 100g is (10i^+5j^)(10\hat{i} + 5\hat{j})(10i^+5j^​) N. The position of that object at t=2t = 2t=2s is (ai^+bj^)(a\hat{i} + b\hat{j})(ai^+bj^​) m after starting from rest. The value of ab\frac{a}{b}ba​ will be _________

Correct answer: 2

Step-by-step solution →

Chemistry — JEE Main 25 June 2022 Shift 1

Q31·ChemistrySingle correct
Bonding in which of the following diatomic molecule(s) become(s) stronger, on the basis of MO Theory, by removal of an electron ? (A) NO (B) N2N_2N2​ (C) O2O_2O2​ (D) C2C_2C2​ (E) B2B_2B2​ Choose the most appropriate answer from the options given below :-
  1. (A)(A), (B), (C) only
  2. (B)(B), (C), (E) only
  3. (C)(A), (C) only
  4. (D)(D) only

Correct answer: (C)

Step-by-step solution →
Q32·ChemistrySingle correct
Incorrect statement for Tyndall effect is :-
  1. (A)The refractive indices of the dispersed phase and the dispersion medium differ greatly in magnitude.
  2. (B)The diameter of the dispersed particles is much smaller than the wavelength of the light used.
  3. (C)During projection of movies in the cinemas hall, Tyndall effect is noticed.
  4. (D)It is used to distinguish a true solution from a colloidal solution.

Correct answer: (B)

Step-by-step solution →
Q33·ChemistrySingle correct
The pair, in which ions are isoelectronic with Al3+Al^{3+}Al3+ is :-
  1. (A)Br−Br^-Br− and Be2+Be^{2+}Be2+
  2. (B)Cl−Cl^-Cl− and Li+Li^+Li+
  3. (C)S2−S^{2-}S2− and K+K^+K+
  4. (D)O2−O^{2-}O2− and Mg2+Mg^{2+}Mg2+

Correct answer: (D)

Step-by-step solution →
Q34·ChemistrySingle correct
Leaching of gold with dilute aqueous solution of NaCN in presence of oxygen gives complex [A], which on reaction with zinc forms the elemental gold and another complex [B]. [A] and [B], respectively are :-
  1. (A)[Au(CN)4]−[Au(CN)_4]^-[Au(CN)4​]− and [Zn(CN)2(OH)2]2−[Zn(CN)_2(OH)_2]^{2-}[Zn(CN)2​(OH)2​]2−
  2. (B)[Au(CN)2]−[Au(CN)_2]^-[Au(CN)2​]− and [Zn(OH)4]2−[Zn(OH)_4]^{2-}[Zn(OH)4​]2−
  3. (C)[Au(CN)2]−[Au(CN)_2]^-[Au(CN)2​]− and [Zn(CN)4]2−[Zn(CN)_4]^{2-}[Zn(CN)4​]2−
  4. (D)[Au(CN)4]2−[Au(CN)_4]^{2-}[Au(CN)4​]2− and [Zn(CN)6]4−[Zn(CN)_6]^{4-}[Zn(CN)6​]4−

Correct answer: (C)

Step-by-step solution →
Q35·ChemistrySingle correct
Number of electron deficient molecules among the following PH3PH_3PH3​, B2H6B_2H_6B2​H6​, CCl4CCl_4CCl4​, NH3NH_3NH3​, LiH and BCl3BCl_3BCl3​ is
  1. (A)0
  2. (B)1
  3. (C)2
  4. (D)3

Correct answer: (C)

Step-by-step solution →
Q36·ChemistrySingle correct
Which one of the following alkaline earth metal ions has the highest ionic mobility in its aqueous solution?
  1. (A)Be2+Be^{2+}Be2+
  2. (B)Mg2+Mg^{2+}Mg2+
  3. (C)Ca2+Ca^{2+}Ca2+
  4. (D)Sr2+Sr^{2+}Sr2+

Correct answer: (D)

Step-by-step solution →
Q37·ChemistrySingle correct
White precipitate of AgCl dissolves in aqueous ammonia solution due to formation of :
  1. (A)[Ag(NH3)4]Cl2[Ag(NH_3)_4]Cl_2[Ag(NH3​)4​]Cl2​
  2. (B)[Ag(Cl)2(NH3)2][Ag(Cl)_2(NH_3)_2][Ag(Cl)2​(NH3​)2​]
  3. (C)[Ag(NH3)2]Cl[Ag(NH_3)_2]Cl[Ag(NH3​)2​]Cl
  4. (D)[Ag(NH3)Cl]Cl[Ag(NH_3)Cl]Cl[Ag(NH3​)Cl]Cl

Correct answer: (C)

Step-by-step solution →
Q38·ChemistrySingle correct
Cerium (IV) has a noble gas configuration. Which of the following is correct statement about it?
  1. (A)It will not prefer to undergo redox reactions.
  2. (B)It will prefer to gain electron and act as an oxidizing agent
  3. (C)It will prefer to give away an electron and behave as reducing agent
  4. (D)It acts as both, oxidizing and reducing agent.

Correct answer: (B)

Step-by-step solution →
Q39·ChemistrySingle correct
Among the following, which is the strongest oxidizing agent ?
  1. (A)Mn3+Mn^{3+}Mn3+
  2. (B)Fe3+Fe^{3+}Fe3+
  3. (C)Ti3+Ti^{3+}Ti3+
  4. (D)Cr3+Cr^{3+}Cr3+

Correct answer: (A)

Step-by-step solution →
Q40·ChemistrySingle correct
The eutrophication of water body results in :
  1. (A)loss of Biodiversity
  2. (B)breakdown of organic matter
  3. (C)increase in biodiversity
  4. (D)decrease in BOD.

Correct answer: (A)

Step-by-step solution →
Q41·ChemistrySingle correct
Phenol on reaction with dilute nitric acid, gives two products. Which method will be most effective for large scale separation ?
  1. (A)Chromatographic separation
  2. (B)Fractional Crystallisation
  3. (C)Steam distillation
  4. (D)Sublimation

Correct answer: (C)

Step-by-step solution →
Q42·ChemistrySingle correct
In the following structures, which one is having staggered conformation with maximum dihedral angle?
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (C)

Step-by-step solution →
Q43·ChemistrySingle correct
The products formed in the following reaction.
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (B)

Step-by-step solution →
Q44·ChemistrySingle correct
The IUPAC name of ethylidene chloride is :-
  1. (A)1-Chloroethene
  2. (B)1-Chloroethyne
  3. (C)1,2-Dichloroethane
  4. (D)1,1-Dichloroethane

Correct answer: (D)

Step-by-step solution →
Q45·ChemistrySingle correct
The major product in the reaction
  1. (A)t-Butyl ethyl ether
  2. (B)2,2-Dimethyl butane
  3. (C)2-Methyl pent-1-ene
  4. (D)2-Methyl prop-1-ene

Correct answer: (D)

Step-by-step solution →
Q46·ChemistrySingle correct
The intermediate X, in the reaction is :
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (C)

Step-by-step solution →
Q47·ChemistrySingle correct
In the following reaction : The compounds A and B respectively are :-
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (C)

Step-by-step solution →
Q48·ChemistrySingle correct
The reaction of R−C∥O−NH2R-\underset{\substack{\parallel \\ O}}{C}-NH_2R−∥O​C​−NH2​ with bromine and KOH gives RNH2RNH_2RNH2​ as the end product. Which one of the following is the intermediate product formed in this reaction ?
  1. (A)R−C∥O−NH−BrR-\underset{\substack{\parallel \\ O}}{C}-NH-BrR−∥O​C​−NH−Br
  2. (B)R−NH−BrR-NH-BrR−NH−Br
  3. (C)R−N=C=OR-N=C=OR−N=C=O
  4. (D)R−C∥O−NBr2R-\underset{\substack{\parallel \\ O}}{C}-NBr_2R−∥O​C​−NBr2​

Correct answer: (C)

Step-by-step solution →
Q49·ChemistrySingle correct
Using very little soap while washing clothes, does not serve the purpose of cleaning of clothes because
  1. (A)soap particles remain floating in water as ions
  2. (B)the hydrophobic part of soap is not able to take away grease
  3. (C)the micelles are not formed due to concentration of soap, below its CMC value
  4. (D)colloidal structure of soap in water is completely disturbed.

Correct answer: (C)

Step-by-step solution →
Q50·ChemistrySingle correct
Which one of the following is an example of artificial sweetner ?
  1. (A)Bithional
  2. (B)Alitame
  3. (C)Salvarsan
  4. (D)Lactose

Correct answer: (B)

Step-by-step solution →
Q51·ChemistryNumerical
The number of N atoms is 681 g of C7H5N3O6C_7H_5N_3O_6C7​H5​N3​O6​ is x × 102110^{21}1021. The value of x is ______ (NAN_ANA​ = 6.02 × 102310^{23}1023 mol−1mol^{-1}mol−1) (Nearest Integer)

Correct answer: 5418

Step-by-step solution →
Q52·ChemistryNumerical
The distance between Na+Na^+Na+ and Cl−Cl^-Cl− ions in solid NaCl of density 43.1 g cm−3cm^{-3}cm−3 is ________ × 10−1010^{-10}10−10m. (Nearest Integer) (Given : NAN_ANA​ = 6.02 × 102310^{23}1023 mol−1mol^{-1}mol−1)

Correct answer: 1

Step-by-step solution →
Q53·ChemistryNumerical
The longest wavelength of light that can be used for the ionisation of lithium atom (Li) in its ground state is x × 10−810^{-8}10−8 m. The value of x is ____________. (Nearest Integer) (Given : Energy of the electron in the first shell of the hydrogen atom is −2.2 × 10−1810^{-18}10−18 J; h = 6.63 × 10−3410^{-34}10−34 Js and c = 3 × 10810^{8}108 ms−1ms^{-1}ms−1)

Correct answer: 4

Step-by-step solution →
Q54·ChemistryNumerical
The standard entropy change for the reaction 4Fe(s)+3O2(g)→2Fe2O3(s)4Fe(s) + 3O_2(g) \rightarrow 2Fe_2O_3(s)4Fe(s)+3O2​(g)→2Fe2​O3​(s) is −550 JK−1JK^{-1}JK−1 at 298 K. [Given : The standard enthalpy change for the reaction is −165 kJ mol−1mol^{-1}mol−1]. The temperature in K at which the reaction attains equilibrium is _________. (Nearest Integer)

Correct answer: 300

Step-by-step solution →
Q55·ChemistryNumerical
1 L aqueous solution of H2SO4H_2SO_4H2​SO4​ contains 0.02 m mol H2SO4H_2SO_4H2​SO4​. 50% of this solution is diluted with deionized water to give 1 L solution (A). In solution (A), 0.01 m mol of H2SO4H_2SO_4H2​SO4​ are added. Total m mols of H2SO4H_2SO_4H2​SO4​ in the final solution is ___________ × 10310^{3}103 m mols.

Correct answer: 0

Step-by-step solution →
Q56·ChemistryNumerical
The standard free energy change (ΔG°) for 50% dissociation of N2O4N_2O_4N2​O4​ into NO2NO_2NO2​ at 27°C and 1 atm pressure is −x J mol−1mol^{-1}mol−1. The value of x is _________. (Nearest Integer) [Given : R = 8.31 J K−1K^{-1}K−1 mol−1mol^{-1}mol−1, log 1.33 = 0.1239 ln 10 = 2.3]

Correct answer: 710

Step-by-step solution →
Q57·ChemistryNumerical
In a cell, the following reactions take place Fe2+→Fe3+e−Fe^{2+} \rightarrow Fe^{3+}e^{-}Fe2+→Fe3+e− EFe3+/Fe2+o=0.77E^{o}_{Fe^{3+}/Fe^{2+}} = 0.77EFe3+/Fe2+o​=0.77 V 2I−→I2+2e−2I^{-} \rightarrow I_2 + 2e^{-}2I−→I2​+2e− EI2/I−o=0.54E^{o}_{I_2/I^{-}} = 0.54EI2​/I−o​=0.54 V The standard electrode potential for the spontaneous reaction in the cell is x × 10−210^{-2}10−2V 298 K. The value of x is __________ (Nearest Integer)

Correct answer: 23

Step-by-step solution →
Q58·ChemistryNumerical
For a given chemical reaction γ1A+γ2B→γ3C+γ4D\gamma_1 A + \gamma_2 B \rightarrow \gamma_3 C + \gamma_4 Dγ1​A+γ2​B→γ3​C+γ4​D Concentration of C changes from 10 mmol dm−3dm^{-3}dm−3 to 20 mmol dm−3dm^{-3}dm−3 in 10 seconds. Rate of appearance of D is 1.5 times the rate of disappearance of B which is twice the rate of disappearance A. The rate of appearance of D has been experimentally determined to be 9 mmol dm−3dm^{-3}dm−3 s−1s^{-1}s−1. Therefore the rate of reaction is ________mmol dm−3dm^{-3}dm−3 s−1s^{-1}s−1. (Nearest Integer)

Correct answer: 1

Step-by-step solution →
Q59·ChemistryNumerical
If [Cu(H2O)4]2+[Cu(H_2O)_4]^{2+}[Cu(H2​O)4​]2+ absorbs a light of wavelength 600 nm for d–d transition, then the value of octahedral crystal field splitting energy for [Cu(H2O)6]2+[Cu(H_2O)_6]^{2+}[Cu(H2​O)6​]2+ will be__________× 10−2110^{-21}10−21 J. (Nearest Integer) (Given : h = 6.63 × 10−3410^{-34}10−34Js and c = 3.08 × 10810^{8}108 ms−1ms^{-1}ms−1)

Correct answer: 766

Step-by-step solution →
Q60·ChemistryNumerical
Number of grams of bromine that will completely react with 5.0g of pent-1-ene is _______ × 10−210^{-2}10−2g. (Atomic mass of Br = 80 g/mol) [Nearest Integer)

Correct answer: 1143

Step-by-step solution →

Mathematics — JEE Main 25 June 2022 Shift 1

Q61·MathematicsSingle correct
Let a circle C touch the lines L1:4x−3y+K1=0L_1 : 4x - 3y + K_1 = 0L1​:4x−3y+K1​=0 and L2:4x−3y+K2=0L_2 : 4x - 3y + K_2 = 0L2​:4x−3y+K2​=0, K1K_1K1​, K2∈RK_2 \in RK2​∈R. If a line passing through the centre of the circle C intersects L1L_1L1​ at (−1,2)(-1, 2)(−1,2) and L2L_2L2​ at (3,−6)(3, -6)(3,−6), then the equation of the circle C is
  1. (A)(x−1)2+(y−2)2=4(x - 1)^2 + (y - 2)^2 = 4(x−1)2+(y−2)2=4
  2. (B)(x+1)2+(y−2)2=4(x + 1)^2 + (y - 2)^2 = 4(x+1)2+(y−2)2=4
  3. (C)(x−1)2+(y+2)2=16(x - 1)^2 + (y + 2)^2 = 16(x−1)2+(y+2)2=16
  4. (D)(x−1)2+(y−2)2=16(x - 1)^2 + (y - 2)^2 = 16(x−1)2+(y−2)2=16

Correct answer: (C)

Step-by-step solution →
Q62·MathematicsSingle correct
The value of ∫0πecos⁡xsin⁡x(1+cos⁡2x)(ecos⁡x+e−cos⁡x)dx\int_0^{\pi} \frac{e^{\cos x}\sin x}{(1+\cos^2 x)(e^{\cos x} + e^{-\cos x})}dx∫0π​(1+cos2x)(ecosx+e−cosx)ecosxsinx​dx is equal to
  1. (A)π24\frac{\pi^2}{4}4π2​
  2. (B)π22\frac{\pi^2}{2}2π2​
  3. (C)π4\frac{\pi}{4}4π​
  4. (D)π2\frac{\pi}{2}2π​

Correct answer: (C)

Step-by-step solution →
Q63·MathematicsSingle correct
Let a, b and c be the length of sides of a triangle ABC such that a+b7=b+c8=c+a9\frac{a+b}{7} = \frac{b+c}{8} = \frac{c+a}{9}7a+b​=8b+c​=9c+a​. If r and R are the radius of incircle and radius of circumcircle of the triangle ABC, respectively, then the value of Rr\frac{R}{r}rR​ is equal to
  1. (A)52\frac{5}{2}25​
  2. (B)222
  3. (C)32\frac{3}{2}23​
  4. (D)111

Correct answer: (A)

Step-by-step solution →
Q64·MathematicsSingle correct
Let f:N→Rf : N \to Rf:N→R be a function such that f(x+y)=2f(x)f(y)f(x+y)=2f(x)f(y)f(x+y)=2f(x)f(y) for natural numbers x and y. If f(1)=2f(1) = 2f(1)=2, then the value of α for which ∑k=110f(α+k)=5123(220−1)\sum_{k=1}^{10} f(\alpha + k) = \frac{512}{3}(2^{20} - 1)∑k=110​f(α+k)=3512​(220−1) holds, is
  1. (A)222
  2. (B)333
  3. (C)444
  4. (D)666

Correct answer: (C)

Step-by-step solution →
Q65·MathematicsSingle correct
Let A be a 3×33 \times 33×3 real matrix such that A(110)=(110)A\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}A​110​​=​110​​; A(101)=(−101)A\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix}A​101​​=​−101​​ and A(001)=(112)A\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}A​001​​=​112​​. If X=(x1,x2,x3)TX = (x_1, x_2, x_3)^TX=(x1​,x2​,x3​)T and I is an identity matrix of order 3, then the system (A−2I)X=(411)(A - 2I)X = \begin{pmatrix} 4 \\ 1 \\ 1 \end{pmatrix}(A−2I)X=​411​​ has
  1. (A)no solution
  2. (B)infinitely many solutions
  3. (C)unique solution
  4. (D)exactly two solutions

Correct answer: (B)

Step-by-step solution →
Q66·Mathematics·DifferentiabilitySingle correct
Let f:R→Rf : R \to Rf:R→R be defined as f(x)=x3+x−5f(x) = x^3 + x - 5f(x)=x3+x−5. If g(x)g(x)g(x) is a function such that f(g(x))=xf(g(x)) = xf(g(x))=x, ∀ x∈Rx \in Rx∈R, then g′(63)g'(63)g′(63) is equal to ______.
  1. (A)149\frac{1}{49}491​
  2. (B)349\frac{3}{49}493​
  3. (C)4349\frac{43}{49}4943​
  4. (D)9149\frac{91}{49}4991​

Correct answer: (A)

Step-by-step solution →
Q67·MathematicsSingle correct
Consider the following two propositions: P1 : ~(p→ ~q) P2 : (p∧ ~ q)∧((~p)∨ q) If the proposition p → ((~p)∨ q) is evaluated as FALSE, then:
  1. (A)P1 is TRUE and P2 is FALSE
  2. (B)P1 is FALSE and P2 is TRUE
  3. (C)Both P1 and P2 are FALSE
  4. (D)Both P1 and P2 are TRUE

Correct answer: (C)

Step-by-step solution →
Q68·MathematicsSingle correct
If 12⋅310+122⋅39+⋯1210⋅3=K210⋅310\frac{1}{2 \cdot 3^{10}} + \frac{1}{2^2 \cdot 3^9} + \cdots \frac{1}{2^{10} \cdot 3} = \frac{K}{2^{10} \cdot 3^{10}}2⋅3101​+22⋅391​+⋯210⋅31​=210⋅310K​, then the remainder when K is divided by 6 is
  1. (A)111
  2. (B)222
  3. (C)333
  4. (D)555

Correct answer: (D)

Step-by-step solution →
Q69·Mathematics·Limits and ContinuitySingle correct
Let f(x)f(x)f(x) be a polynomial function such that f(x)+f′(x)+f′′(x)=x5+64f(x) + f'(x) + f''(x) = x^5 + 64f(x)+f′(x)+f′′(x)=x5+64. Then, the value of lim⁡x→1f(x)x−1\lim_{x \to 1} \frac{f(x)}{x - 1}limx→1​x−1f(x)​
  1. (A)−15-15−15
  2. (B)−60-60−60
  3. (C)606060
  4. (D)151515

Correct answer: (A)

Step-by-step solution →
Q70·MathematicsSingle correct
Let E1E_1E1​ and E2E_2E2​ be two events such that the conditional probabilities P(E1∣E2)=12P(E_1 | E_2) = \frac{1}{2}P(E1​∣E2​)=21​, P(E2∣E1)=34P(E_2 | E_1) = \frac{3}{4}P(E2​∣E1​)=43​ and P(E1∩E2)=18P(E_1 \cap E_2) = \frac{1}{8}P(E1​∩E2​)=81​. Then:
  1. (A)P(E1∩E2)=P(E1)⋅P(E2)P(E_1 \cap E_2) = P(E_1)\cdot P(E_2)P(E1​∩E2​)=P(E1​)⋅P(E2​)
  2. (B)P(E1′∩E2′)=P(E1′)⋅P(E2)P(E'_1 \cap E'_2) = P(E'_1)\cdot P(E_2)P(E1′​∩E2′​)=P(E1′​)⋅P(E2​)
  3. (C)P(E1∩E2′)=P(E1)⋅P(E2)P(E_1 \cap E'_2) = P(E_1)\cdot P(E_2)P(E1​∩E2′​)=P(E1​)⋅P(E2​)
  4. (D)P(E1′∩E2)=P(E1)⋅P(E2)P(E'_1 \cap E_2) = P(E_1)\cdot P(E_2)P(E1′​∩E2​)=P(E1​)⋅P(E2​)

Correct answer: (C)

Step-by-step solution →
Q71·MathematicsSingle correct
Let A=[0−220]A = \begin{bmatrix} 0 & -2 \\ 2 & 0 \end{bmatrix}A=[02​−20​]. If M and N are two matrices given by M=∑k=110A2kM = \sum_{k=1}^{10} A^{2k}M=∑k=110​A2k and N=∑k=110A2k−1N = \sum_{k=1}^{10} A^{2k-1}N=∑k=110​A2k−1 then MN2MN^2MN2 is
  1. (A)a non-identity symmetric matrix
  2. (B)a skew-symmetric matrix
  3. (C)neither symmetric nor skew-symmetric matrix
  4. (D)an identify matrix

Correct answer: (A)

Step-by-step solution →
Q72·MathematicsSingle correct
Let g:(0,∞)→Rg : (0, \infty) \to Rg:(0,∞)→R be a differentiable function such that ∫(x(cos⁡x−sin⁡x)ex+1+g(x)(ex+1−xex)(ex+1)2)dx=xg(x)ex+1+c\int\left(\frac{x(\cos x - \sin x)}{e^x + 1} + \frac{g(x)(e^x + 1 - xe^x)}{(e^x + 1)^2}\right)dx = \frac{xg(x)}{e^x + 1} + c∫(ex+1x(cosx−sinx)​+(ex+1)2g(x)(ex+1−xex)​)dx=ex+1xg(x)​+c, for all x>0x > 0x>0, where c is an arbitrary constant. Then.
  1. (A)g is decreasing in (0,π4)\left(0, \frac{\pi}{4}\right)(0,4π​)
  2. (B)g' is increasing in (0,π4)\left(0, \frac{\pi}{4}\right)(0,4π​)
  3. (C)g + g' is increasing in (0,π2)\left(0, \frac{\pi}{2}\right)(0,2π​)
  4. (D)g − g' is increasing in (0,π2)\left(0, \frac{\pi}{2}\right)(0,2π​)

Correct answer: (D)

Step-by-step solution →
Q73·MathematicsSingle correct
Let f:R→Rf : R \to Rf:R→R and g:R→Rg : R \to Rg:R→R be two functions defined by f(x)=log⁡e(x2+1)−e−x+1f(x) = \log_e(x^2 + 1) - e^{-x} + 1f(x)=loge​(x2+1)−e−x+1 and g(x)=1−2e2xexg(x) = \frac{1 - 2e^{2x}}{e^x}g(x)=ex1−2e2x​. Then, for which of the following range of α, the inequality f(g((α−1)23))>f(g(α−53))f\left(g\left(\frac{(\alpha - 1)^2}{3}\right)\right) > f\left(g\left(\alpha - \frac{5}{3}\right)\right)f(g(3(α−1)2​))>f(g(α−35​)) holds?
  1. (A)(2,3)(2, 3)(2,3)
  2. (B)(−2,−1)(-2, -1)(−2,−1)
  3. (C)(1,2)(1, 2)(1,2)
  4. (D)(−1,1)(-1, 1)(−1,1)

Correct answer: (A)

Step-by-step solution →
Q74·MathematicsSingle correct
Let a⃗=a1i^+a2j^+a3k^\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}a=a1​i^+a2​j^​+a3​k^ ai>0a_i > 0ai​>0, i = 1, 2, 3 be a vector which makes equal angles with the coordinates axes OX, OY and OZ. Also, let the projection of a⃗\vec{a}a on the vector 3i^+4j^3\hat{i} + 4\hat{j}3i^+4j^​ be 7. Let b⃗\vec{b}b be a vector obtained by rotating a⃗\vec{a}a with 90°. If a⃗\vec{a}a, b⃗\vec{b}b and x-axis are coplanar, then projection of a vector b⃗\vec{b}b on 3i^+4j^3\hat{i} + 4\hat{j}3i^+4j^​ is equal to
  1. (A)7\sqrt{7}7​
  2. (B)2\sqrt{2}2​
  3. (C)222
  4. (D)777

Correct answer: (B)

Step-by-step solution →
Q75·MathematicsSingle correct
Let y=y(x)y = y(x)y=y(x) be the solution of the differential equation (x+1)y′−y=e3x(x+1)2(x + 1)y' - y = e^{3x}(x + 1)^2(x+1)y′−y=e3x(x+1)2, with y(0)=13y(0) = \frac{1}{3}y(0)=31​. Then, the point x=−43x = -\frac{4}{3}x=−34​ for the curve y=y(x)y = y(x)y=y(x) is:
  1. (A)not a critical point
  2. (B)a point of local minima
  3. (C)a point of local maxima
  4. (D)a point of inflection

Correct answer: (B)

Step-by-step solution →
Q76·MathematicsSingle correct
If y=m1x+c1y = m_1x + c_1y=m1​x+c1​ and y=m2x+c2y = m_2x + c_2y=m2​x+c2​, m1≠m2m_1 \neq m_2m1​=m2​ are two common tangents of circle x2+y2=2x^2 + y^2 = 2x2+y2=2 and parabola y2=xy^2 = xy2=x, then the value of 8∣m1m2∣8|m_1m_2|8∣m1​m2​∣ is equal to
  1. (A)3+423+4\sqrt{2}3+42​
  2. (B)−5+62-5+6\sqrt{2}−5+62​
  3. (C)−4+32-4+3\sqrt{2}−4+32​
  4. (D)7+627+6\sqrt{2}7+62​

Correct answer: (C)

Step-by-step solution →
Q77·MathematicsSingle correct
Let Q be the mirror image of the point P(1, 0, 1) with respect to the plane S : x + y + z = 5. If a line L passing through (1, −1, −1), parallel to the line PQ meets the plane S at R, then QR2QR^2QR2 is equal to:
  1. (A)2
  2. (B)5
  3. (C)7
  4. (D)11

Correct answer: (B)

Step-by-step solution →
Q78·MathematicsSingle correct
If the solution curve y=y(x)y = y(x)y=y(x) of the differential equation y2dx+(x2−xy+y2)dy=0y^2dx + (x^2 - xy + y^2)dy = 0y2dx+(x2−xy+y2)dy=0, which passes through the point (1, 1) and intersects the line y=3 xy = \sqrt{3}\,xy=3​x at the point (α,3 α)(\alpha, \sqrt{3}\,\alpha)(α,3​α), then value of log⁡e(3 α)\log_e(\sqrt{3}\,\alpha)loge​(3​α) is equal to
  1. (A)π3\frac{\pi}{3}3π​
  2. (B)π2\frac{\pi}{2}2π​
  3. (C)π12\frac{\pi}{12}12π​
  4. (D)π6\frac{\pi}{6}6π​

Correct answer: (C)

Step-by-step solution →
Q79·MathematicsSingle correct
Let x=2tx = 2tx=2t, y=t23y = \frac{t^2}{3}y=3t2​ be a conic. Let S be the focus and B be the point on the axis of the conic such that SA ⊥ BA, where A is any point on the conic. If k is the ordinate of the centroid of ΔSAB, then lim⁡t→1k\lim_{t \to 1} klimt→1​k is equal to
  1. (A)1718\frac{17}{18}1817​
  2. (B)1918\frac{19}{18}1819​
  3. (C)1118\frac{11}{18}1811​
  4. (D)1318\frac{13}{18}1813​

Correct answer: (D)

Step-by-step solution →
Q80·MathematicsNumerical
Let CrC_rCr​ denote the binomial coefficient of xrx^rxr in the expansion of (1+x)10(1 + x)^{10}(1+x)10. If α, β ∈ R. C1+3⋅2C2+5⋅3C3+…C_1 + 3\cdot 2C_2 + 5\cdot 3C_3 + \ldotsC1​+3⋅2C2​+5⋅3C3​+… upto 10 terms =α×2112β−1(C0+C12+C23+… upto 10 terms)= \frac{\alpha \times 2^{11}}{2^{\beta} - 1}\left(C_0 + \frac{C_1}{2} + \frac{C_2}{3} + \ldots\ \text{upto 10 terms}\right)=2β−1α×211​(C0​+2C1​​+3C2​​+… upto 10 terms) then the value of α + β is equal to

Correct answer: 286

Step-by-step solution →
Q81·MathematicsNumerical
The number of 3-digit odd numbers, whose sum of digits is a multiple of 7, is ______.

Correct answer: 63

Step-by-step solution →
Q82·MathematicsNumerical
Let θ be the angle between the vectors a⃗\vec{a}a and b⃗\vec{b}b, where ∣a⃗∣=4|\vec{a}| = 4∣a∣=4, ∣b⃗∣=3|\vec{b}| = 3∣b∣=3 θ∈(π4,π3)\theta \in \left(\frac{\pi}{4}, \frac{\pi}{3}\right)θ∈(4π​,3π​). Then ∣(a⃗−b⃗)×(a⃗+b⃗)∣2+4(a⃗⋅b⃗)2\left|\left(\vec{a} - \vec{b}\right) \times \left(\vec{a} + \vec{b}\right)\right|^2 + 4\left(\vec{a}\cdot\vec{b}\right)^2​(a−b)×(a+b)​2+4(a⋅b)2 is equal to ____

Correct answer: 576

Step-by-step solution →
Q83·MathematicsNumerical
Let the abscissae of the two points P and Q be the roots of 2x2−rx+p=02x^2 - rx + p = 02x2−rx+p=0 and the ordinates of P and Q be the roots of x2−sx−q=0x^2 - sx - q = 0x2−sx−q=0. If the equation of the circle described on PQ as diameter is 2(x2+y2)−11x−14y−22=02(x^2 + y^2) - 11x - 14y - 22 = 02(x2+y2)−11x−14y−22=0, then 2r+s−2q+p2r + s - 2q + p2r+s−2q+p is equal to

Correct answer: 7

Step-by-step solution →
Q84·MathematicsNumerical
The number of values of x in the interval (π4,7π4)\left(\frac{\pi}{4}, \frac{7\pi}{4}\right)(4π​,47π​) for which 14 cosec2x−2sin⁡2x=21−4cos⁡2x14\,\mathrm{cosec}^2x - 2\sin^2x = 21 - 4\cos^2x14cosec2x−2sin2x=21−4cos2x holds, is ________

Correct answer: 4

Step-by-step solution →
Q85·MathematicsNumerical
For a natural number n, let an=19n−12na_n = 19^n - 12^nan​=19n−12n. Then, the value of 31α9−α1057α8\frac{31\alpha_9 - \alpha_{10}}{57\alpha_8}57α8​31α9​−α10​​ is

Correct answer: 4

Step-by-step solution →
Q86·MathematicsNumerical
Let f : R → R be a function defined by f(x)=(2(1−x252)(2+x25))150f(x) = \left(2\left(1 - \frac{x^{25}}{2}\right)\left(2 + x^{25}\right)\right)^{\frac{1}{50}}f(x)=(2(1−2x25​)(2+x25))501​. If the function g(x)=f(f(f(x)))+f(f(x))g(x) = f\left(f\left(f(x)\right)\right) + f\left(f(x)\right)g(x)=f(f(f(x)))+f(f(x)), the the greatest integer less than or equal to g (1) is ______

Correct answer: 2

Step-by-step solution →
Q87·MathematicsNumerical
Let the lines L1:r⃗=λ(i^+2j^+3k^)L_1 : \vec{r} = \lambda(\hat{i} + 2\hat{j} + 3\hat{k})L1​:r=λ(i^+2j^​+3k^), λ∈R L2:r⃗=(i^+3j^+k^)+μ(i^+j^+5k^)L_2 : \vec{r} = (\hat{i} + 3\hat{j} + \hat{k}) + \mu\left(\hat{i} + \hat{j} + 5\hat{k}\right)L2​:r=(i^+3j^​+k^)+μ(i^+j^​+5k^); μ∈R intersect at the point S. If a plane ax + by − z + d = 0 passes through S and is parallel to both the lines L1L_1L1​ and L2L_2L2​, then the value of a + b + d is equal to ________

Correct answer: 5

Step-by-step solution →
Q88·MathematicsNumerical
Let A be a 3×33 \times 33×3 matrix having entries from. the set {−1, 0, 1}. The number of all such matrices A having sum of all the entries equal to 5, is ________

Correct answer: 414

Step-by-step solution →
Q89·MathematicsNumerical
The greatest integer less than or equal to the sum of first 100 terms of the sequence 13,59,1927,6581,……\frac{1}{3}, \frac{5}{9}, \frac{19}{27}, \frac{65}{81}, \ldots\ldots31​,95​,2719​,8165​,…… is equal to

Correct answer: 98

Step-by-step solution →

Chapters tested in this paper

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

  • Properties of Solids and Liquids 172/186
  • Three Dimensional Geometry 176/186
  • Matrices and Determinants 180/186
  • Coordination Compounds 176/186
  • Sets, Relations and Functions 165/186
  • Current Electricity 160/186
  • Sequence and Series 164/186
  • Definite Integration 168/186
  • Rotational Motion 172/186
  • Redox Reactions and Electrochemistry 177/186
  • Geometrical Optics 172/186
  • Kinematics 156/186
  • Vector Algebra 173/186
  • Differential Equations 167/186
  • Probability 176/186
  • Permutations and Combinations 162/186
  • Magnetic Field of Current 147/186
  • Binomial Theorem and Its Simple Applications 158/186
  • Electromagnetic Waves 144/186
  • Chemical Bonding and Molecular Structure 151/186
  • Application of Derivatives 139/186
  • Limits and Continuity 149/186
  • d- and f-Block Elements 126/186
  • Equilibrium 163/186
  • Laws of Motion 130/186
  • Chemical Thermodynamics 165/186
  • Units and Measurements 149/186
  • Hydrocarbons 126/186
  • Trigonometric Functions 144/186
  • Chemical Kinetics 169/186
  • Gravitation 152/186
  • Electric Field and Coulomb's Law 133/186
  • Atomic Structure 161/186
  • Electronic Devices 145/186
  • Circles 142/186
  • Dual Nature of Matter and Radiation 155/186
  • Amines 133/186
  • Work, Energy and Power 132/186
  • Some Basic Concepts in Chemistry 129/186
  • Electromagnetic Induction 120/186
  • Wave Optics 130/186
  • Kinetic Theory of Gases 135/186
  • Alcohols and Ethers 106/186
  • Purification and Characterisation of Organic Compounds 116/186
  • Alternating Currents 108/186
  • Organic Compounds Containing Halogens 109/186
  • Waves 109/186
  • Capacitors and Dielectrics 115/186
  • Atoms 112/186
  • Classification of Elements and Periodicity in Properties 113/186
  • Parabola 101/186
  • Isolation of Metals 106/186
  • Differentiability 91/186
  • s-Block Elements 88/186
  • Surface Chemistry 98/186
  • Environmental Chemistry 83/186
  • Experimental Skills 68/186
  • Indefinite Integration 66/186
  • Solid State 63/186
  • Chemistry in Everyday Life 60/186
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