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JEE Main 28 July 2022 Shift 2 Question Paper with Answers

28 July 2022 · July session · 88 questions

88 of the 90 questions from the JEE Main 28 July 2022 Shift 2 paper, each with its correct answer and tagged to the chapter it tests. Free to read, no account needed.

2 questions are held back from this paper while we re-check the transcription or the answer key. We would rather show you nothing than show you an answer we are not confident is right.

Physics
28
Chemistry
30
Mathematics
30

Physics — JEE Main 28 July 2022 Shift 2

Q1·PhysicsSingle correct
Consider the efficiency of Carnot's engine is given by η=αβsin⁡θlog⁡eβxkT\eta = \dfrac{\alpha\beta}{\sin\theta}\log_e \dfrac{\beta x}{kT}η=sinθαβ​loge​kTβx​, where α\alphaα and β\betaβ are constants. If T is temperature, k is Boltzman constant, θ\thetaθ is angular displacement and x has the dimensions of length. Then, choose the incorrect option.
  1. (A)Dimensions of β\betaβ is same as that of force.
  2. (B)Dimensions of α−1\alpha^{-1}α−1 x is same as that of energy.
  3. (C)Dimensions of η−1sin⁡θ\eta^{-1}\sin\thetaη−1sinθ is same as that of αβ\alpha\betaαβ
  4. (D)Dimensions of α\alphaα is same as that of β\betaβ

Correct answer: (D)

Step-by-step solution →
Q2·PhysicsSingle correct
At time t = 0 a particle starts travelling from a height 7z^7\hat{z}7z^ cm in a plane keeping z coordinate constant. At any instant of time it's position along the x and y directions are defined as 3t and 5t35t^35t3 respectively. At t = 1s acceleration of the particle will be
  1. (A)−30y-30y−30y
  2. (B)30y30y30y
  3. (C)3x+15y3x + 15y3x+15y
  4. (D)3x+15y+7z^3x + 15y + 7\hat{z}3x+15y+7z^

Correct answer: (B)

Step-by-step solution →
Q3·PhysicsSingle correct
A pressure-pump has a horizontal tube of cross-sectional area 10 cm2cm^2cm2 for the outflow of water at a speed of 20 m/s. The force exerted on the vertical wall just in front of the tube which stops water horizontally flowing out of the tube, is: [given : density of water = 1000 kg/m3kg/m^3kg/m3]
  1. (A)300 N
  2. (B)500 N
  3. (C)250 N
  4. (D)400 N

Correct answer: (D)

Step-by-step solution →
Q4·PhysicsSingle correct
A uniform metal chain of mass m and length 'L' passes over a massless and frictionless pulley. It is released from rest with a part of its length 'lll' is hanging on one side and rest of its length 'L −-− lll' is hanging on the other side of the pulley. At a certain point of time, when l=Lxl = \dfrac{L}{x}l=xL​, the acceleration of the chain is g2\dfrac{g}{2}2g​. The value of x is ………
  1. (A)6
  2. (B)2
  3. (C)1.5
  4. (D)4

Correct answer: (D)

Step-by-step solution →
Q5·PhysicsSingle correct
A bullet of mass 200 g having initial kinetic energy 90 J is shot inside a long swimming pool as shown in the figure. If it's kinetic energy reduces to 40 J within 1s, the minimum length of the pool, the bullet has a to travel so that it completely comes to rest is
  1. (A)45 m
  2. (B)90 m
  3. (C)125 m
  4. (D)25 m

Correct answer: (A)

Step-by-step solution →
Q6·PhysicsSingle correct
Assume there are two identical simple pendulum Clocks-1 is placed on the earth and Clock-2 is placed on a space station located at a height h above the earth surface. Clock-1 and Clock-2 operate at time periods 4s and 6s respectively. Then the value of h is −-− (consider radius of earth RER_ERE​ = 6400 km and g on earth 10 m/s2m/s^2m/s2)
  1. (A)1200 km
  2. (B)1600 km
  3. (C)3200 km
  4. (D)4800 km

Correct answer: (C)

Step-by-step solution →
Q7·PhysicsSingle correct
A vessel contains 14 g of nitrogen gas at a temperature of 27∘^\circ∘C. The amount of heat to be transferred to the gap to double the r.m.s. speed of its molecules will be : (Take R = 8.32 J mol−1k−1mol^{-1}k^{-1}mol−1k−1)
  1. (A)2229 J
  2. (B)5616 J
  3. (C)9360 J
  4. (D)13,104 J

Correct answer: (C)

Step-by-step solution →
Q8·PhysicsSingle correct
A slab of dielectric constant K has the same cross-sectional area as the plates of a parallel plate capacitor and thickness 34d\dfrac{3}{4}d43​d, where d is the separation of the plates. The capacitance of the capacitor when the slab is inserted between the plates will be : (Given CoC_oCo​ = capacitance of capacitor with air as medium between plates.)
  1. (A)4KC03+K\dfrac{4KC_0}{3+K}3+K4KC0​​
  2. (B)3KC03+K\dfrac{3KC_0}{3+K}3+K3KC0​​
  3. (C)3+K4KC0\dfrac{3+K}{4KC_0}4KC0​3+K​
  4. (D)K4+K\dfrac{K}{4+K}4+KK​

Correct answer: (A)

Step-by-step solution →
Q9·PhysicsSingle correct
A uniform electric field E = (8m/e) V/m is created between two parallel plates of length 1m as shown in figure, (where m = mass of electron and e = charge of electron). An electron enters the field symmetrically between the plates with a speed of 2m/s. The angle of the deviation (θ\thetaθ) of the path of the electron as it comes out of the field will be ……….
  1. (A)tan⁡−1(4)\tan^{-1}(4)tan−1(4)
  2. (B)tan⁡−1(2)\tan^{-1}(2)tan−1(2)
  3. (C)tan⁡−1(13)\tan^{-1}\left(\dfrac{1}{3}\right)tan−1(31​)
  4. (D)tan⁡−1(3)\tan^{-1}(3)tan−1(3)

Correct answer: (B)

Step-by-step solution →
Q10·PhysicsSingle correct
Given below are two statements : Statement I : A uniform wire of resistance 80Ω\OmegaΩ is cut into four equal parts. These parts are now connected in parallel. The equivalent resistance of the combination will be 5 Ω\OmegaΩ. Statement II : Two resistance 2R and 3R are connected in parallel in a electric circuit. The value of thermal energy developed in 3R and 2R will be in the ratio 3 : 2. 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: (C)

Step-by-step solution →
Q11·Physics·Magnetic Field of CurrentSingle correct
A triangular shaped wire carrying 10A current is placed in a uniform magnetic field of 0.5T, as shown in figure. The magnetic force on segment CD is (Given BC = CD = BD = 5 cm).
  1. (A)0.126 N
  2. (B)0.312 N
  3. (C)0.216 N
  4. (D)0.245 N

Correct answer: (C)

Step-by-step solution →
Q12·PhysicsSingle correct
The magnetic field at the center of current carrying circular loop is B1B_1B1​. The magnetic field at a distance of 3\sqrt{3}3​ times radius of the given circular loop from the center on its axis is B2B_2B2​. The value of B1/B2B_1/B_2B1​/B2​ will be
  1. (A)9:49 : 49:4
  2. (B)12:512 : \sqrt{5}12:5​
  3. (C)8:18 : 18:1
  4. (D)5:35 : \sqrt{3}5:3​

Correct answer: (C)

Step-by-step solution →
Q13·PhysicsSingle correct
A transformer operating at primary voltage 8 kV and secondary voltage 160 V serves a load of 80 kW. Assuming the transformer to be ideal with purely resistive load and working on unity power factor, the loads in the primary and secondary circuit would be
  1. (A)800 Ω\OmegaΩ and 1.06 Ω\OmegaΩ
  2. (B)10 Ω\OmegaΩ and 500 Ω\OmegaΩ
  3. (C)800 Ω\OmegaΩ and 0.32 Ω\OmegaΩ
  4. (D)1.06 Ω\OmegaΩ and 500 Ω\OmegaΩ

Correct answer: (C)

Step-by-step solution →
Q14·PhysicsSingle correct
Sun light falls normally on a surface of area 36 cm2cm^2cm2 and exerts an average force of 7.2×10−97.2\times10^{-9}7.2×10−9 N within a time period of 20 minutes. Considering a case of complete absorption, the energy flux of incident light is
  1. (A)25.92×10225.92\times10^{2}25.92×102 W/cm2W/cm^2W/cm2
  2. (B)8.64×10−68.64\times10^{-6}8.64×10−6 W/cm2W/cm^2W/cm2
  3. (C)6.0 W/cm2W/cm^2W/cm2
  4. (D)0.06 W/cm2W/cm^2W/cm2

Correct answer: (D)

Step-by-step solution →
Q15·PhysicsSingle correct
The power of a lens (biconvex) is 1.25 m−1m^{-1}m−1 in particular medium. Refractive index of the lens is 1.5 and radii of curvature are 20 cm and 40 cm respectively. The refractive index of surrounding medium :
  1. (A)1.0
  2. (B)97\dfrac{9}{7}79​
  3. (C)32\dfrac{3}{2}23​
  4. (D)43\dfrac{4}{3}34​

Correct answer: (B)

Step-by-step solution →
Q16·PhysicsSingle correct
Two streams of photons, possessing energies to five and ten times the work function of metal are incident on the metal surface successively. The ratio of the maximum velocities of the photoelectron emitted, in the two cases respectively, will be
  1. (A)1:21 : 21:2
  2. (B)1:31 : 31:3
  3. (C)2:32 : 32:3
  4. (D)3:23 : 23:2

Correct answer: (C)

Step-by-step solution →
Q17·PhysicsSingle correct
An n.p.n transistor with current gain β\betaβ = 100 in common emitter configuration is shown in figure. The output voltage of the amplifier will be
  1. (A)0.1 V
  2. (B)1.0 V
  3. (C)10 V
  4. (D)100 V

Correct answer: (B)

Step-by-step solution →
Q18·PhysicsSingle correct
A FM Broad cast transmitter, using modulating signal of frequency 20 kHz has a deviation ratio of 10. The Bandwidth required for transmission is :
  1. (A)220 kHz
  2. (B)180 kHz
  3. (C)360 kHz
  4. (D)440 kHz

Correct answer: (D)

Step-by-step solution →
Q19·PhysicsNumerical
A ball is thrown vertically upwards with a velocity of 19.6 ms−1\mathrm{ms}^{-1}ms−1 from the top of a tower. The ball strikes the ground after 6 s. The height from the ground up to which the ball can rise will be (k5)\left(\dfrac{k}{5}\right)(5k​) m. The value of k is ..... (use g = 9.8 m/s2^22)

Correct answer: 392

Step-by-step solution →
Q20·PhysicsNumerical
The distance of centre of mass from end A of a one dimensional rod (AB) having mass density ρ=ρ0(1−x2L2)\rho = \rho_0\left(1 - \dfrac{x^2}{L^2}\right)ρ=ρ0​(1−L2x2​) kg/m and length L (in meter) is 3Lα\dfrac{3L}{\alpha}α3L​ m. The value of α\alphaα is ......... (where x is the distance form end A)

Correct answer: 8

Step-by-step solution →
Q21·PhysicsNumerical
A string of area of cross-section 4 mm2^22 and length 0.5 is connected with a rigid body of mass 2 kg. The body is rotated in a vertical circular path of radius 0.5 m. The body acquires a speed of 5 m/s at the bottom of the circular path. Strain produced in the string when the body is at the bottom of the circle is ...... ×10−5\times 10^{-5}×10−5. (Use Young's modulus 101110^{11}1011 N/m2^22 and g = 10 m/s2^22)

Correct answer: 30

Step-by-step solution →
Q22·PhysicsNumerical
At a certain temperature, the degrees of freedom per molecule for gas is 8. The gas performs 150 J of work when it expands under constant pressure. The amount of heat absorbed by the gas will be ........ J.

Correct answer: 750

Step-by-step solution →
Q23·PhysicsNumerical
The potential energy of a particle of mass 4 kg in motion along the x-axis is given by U=4(1−cos⁡4x)U = 4(1 - \cos 4x)U=4(1−cos4x) J. The time period of the particle for small oscillation (sin⁡θ≃θ)(\sin\theta \simeq \theta)(sinθ≃θ) is (πK)\left(\dfrac{\pi}{K}\right)(Kπ​) s. The value of K is ........

Correct answer: 2

Step-by-step solution →
Q24·PhysicsNumerical
An electrical bulb rated 220 V, 100 W, is connected in series with another bulb rated 220 V, 60 W. If the voltage across combination is 220 V, the power consumed by the 100 W bulb will be about ......... W.

Correct answer: 14

Step-by-step solution →
Q25·PhysicsNumerical
For the given circuit the current through battery of 6 V just after closing the switch 'S' will be ......... A.

Correct answer: 1

Step-by-step solution →
Q26·PhysicsNumerical
An object 'o' is placed at a distance of 100 cm in front of a concave mirror of radius of curvature 200 cm as shown in the figure. The object starts moving towards the mirror at a speed 2 cm/s. The position of the image from the mirror after 10s will be at ....... cm.

Correct answer: 400

Step-by-step solution →
Q27·PhysicsNumerical
In an experiment with a convex lens. The plot of the image distance (v') against the object distance (μ\muμ') measured from the focus gives a curve v' μ\muμ' = 225. If all the distances are measured in cm. The magnitude of the focal length of the lens is .......... cm.

Correct answer: 15

Step-by-step solution →
Q28·PhysicsNumerical
In an experiment to find acceleration due to gravity (g) using simple pendulum, time period of 0.5 s is measured from time of 100 oscillation with a watch of 1s resolution. If measured value of length is 10 cm known to 1mm accuracy. The accuracy in the determination of g is found to be x %. The value of x is

Correct answer: 5

Step-by-step solution →

Chemistry — JEE Main 28 July 2022 Shift 2

Q29·ChemistrySingle correct
Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R Assertion A : Zero orbital overlap is an out of phase overlap. Reason : It results due to different orientation/direction of approach of orbitals. In the light of the above statements. Choose the correct answer from the options given below
  1. (A)Both A and R are true and R is the correct explanation of A
  2. (B)Both A and R are true but R is NOT the correct explanation of A
  3. (C)A is true but R is false
  4. (D)A is false but R is true

Correct answer: (A)

Step-by-step solution →
Q30·ChemistrySingle correct
The correct decreasing order for metallic character is
  1. (A)Na>Mg>Be>Si>P\mathrm{Na > Mg > Be > Si > P}Na>Mg>Be>Si>P
  2. (B)P>Si>Be>Mg>Na\mathrm{P > Si > Be > Mg > Na}P>Si>Be>Mg>Na
  3. (C)Si>P>Be>Na>Mg\mathrm{Si > P > Be > Na > Mg}Si>P>Be>Na>Mg
  4. (D)Be>Na>Mg>Si>P\mathrm{Be > Na > Mg > Si > P}Be>Na>Mg>Si>P

Correct answer: (A)

Step-by-step solution →
Q31·ChemistrySingle correct
Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R Assertion A : The reduction of a metal oxide is easier if the metal formed is in liquid state than solid state. Reason R : The value of ΔG⊖\Delta \mathrm{G}^{\ominus}ΔG⊖ becomes more on negative side as entropy is higher in liquid state than solid state. 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 →
Q32·ChemistrySingle correct
The products obtained during treatment of hard water using Clark's method are:
  1. (A)CaCO3\mathrm{CaCO_3}CaCO3​ and MgCO3\mathrm{MgCO_3}MgCO3​
  2. (B)Ca(OH)2\mathrm{Ca(OH)_2}Ca(OH)2​ and Mg(OH)2\mathrm{Mg(OH)_2}Mg(OH)2​
  3. (C)CaCO3\mathrm{CaCO_3}CaCO3​ and Mg(OH)2\mathrm{Mg(OH)_2}Mg(OH)2​
  4. (D)Ca(OH)2\mathrm{Ca(OH)_2}Ca(OH)2​ and MgCO3\mathrm{MgCO_3}MgCO3​

Correct answer: (C)

Step-by-step solution →
Q33·ChemistrySingle correct
Statement I: An alloy of lithium and magnesium is used to make aircraft plates. Statement II : The magnesium ions are important for cell-membrane integrity. In the light 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: (B)

Step-by-step solution →
Q34·ChemistrySingle correct
White phosphorus reacts with thionyl chloride to give
  1. (A)PCl5\mathrm{PCl_5}PCl5​, SO2\mathrm{SO_2}SO2​ and S2Cl2\mathrm{S_2Cl_2}S2​Cl2​
  2. (B)PCl3\mathrm{PCl_3}PCl3​. SO2\mathrm{SO_2}SO2​ and S2Cl2\mathrm{S_2Cl_2}S2​Cl2​
  3. (C)PCl3\mathrm{PCl_3}PCl3​, SO2\mathrm{SO_2}SO2​ and Cl2\mathrm{Cl_2}Cl2​
  4. (D)PCl5\mathrm{PCl_5}PCl5​, SO2\mathrm{SO_2}SO2​ and Cl2\mathrm{Cl_2}Cl2​

Correct answer: (B)

Step-by-step solution →
Q35·ChemistrySingle correct
Concentrated HNO3\mathrm{HNO_3}HNO3​ reacts with Iodine to give
  1. (A)HI\mathrm{HI}HI, NO2\mathrm{NO_2}NO2​ and H2O\mathrm{H_2O}H2​O
  2. (B)HIO2\mathrm{HIO_2}HIO2​, N2O\mathrm{N_2O}N2​O and H2O\mathrm{H_2O}H2​O
  3. (C)HIO3\mathrm{HIO_3}HIO3​, NO2\mathrm{NO_2}NO2​ and H2O\mathrm{H_2O}H2​O
  4. (D)HIO4\mathrm{HIO_4}HIO4​, N2O\mathrm{N_2O}N2​O and H2O\mathrm{H_2O}H2​O

Correct answer: (C)

Step-by-step solution →
Q36·ChemistrySingle correct
Which of the following pair is not isoelectronic species? (At. no. Sm, 62; Er, 68: Yb, 70: Lu, 71; Eu, 63: Tb, 65; Tm, 69)
  1. (A)Sm2+\mathrm{Sm^{2+}}Sm2+ and Er3+\mathrm{Er^{3+}}Er3+
  2. (B)Yb2+\mathrm{Yb^{2+}}Yb2+ and Lu3+\mathrm{Lu^{3+}}Lu3+
  3. (C)Eu2+\mathrm{Eu^{2+}}Eu2+ and Tb4+\mathrm{Tb^{4+}}Tb4+
  4. (D)Tb2+\mathrm{Tb^{2+}}Tb2+ and Tm4+\mathrm{Tm^{4+}}Tm4+

Correct answer: (D)

Step-by-step solution →
Q37·ChemistrySingle correct
Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R Assertion A : Permanganate titrations are not performed in presence of hydrochloric acid. Reason R : Chlorine is formed as a consequence of oxidation of hydrochloric acid. In the light of the above statements, choose the correct answer from the options given below
  1. (A)Both A and R are true and R is the correct explanation of A
  2. (B)Both A and R are true but R is NOT the correct explanation of A
  3. (C)A is true but R is false
  4. (D)A is false but R is true

Correct answer: (A)

Step-by-step solution →
Q38·ChemistrySingle correct
Match List I with List II Choose the correct answer from the options given below:
List I (Complex)List II (Hybridization)
A.Ni(CO)4\mathrm{Ni(CO)_4}Ni(CO)4​I.sp3\mathrm{sp^3}sp3
B.[Ni (CN)4]2−\mathrm{[Ni\,(CN)_4]^{2-}}[Ni(CN)4​]2−II.sp3d2\mathrm{sp^3d^2}sp3d2
C.[Co (CN)6]3−\mathrm{[Co\,(CN)_6]^{3-}}[Co(CN)6​]3−III.d2sp3\mathrm{d^2sp^3}d2sp3
D.[CoF6]3−\mathrm{[CoF_6]^{3-}}[CoF6​]3−IV.dsp2\mathrm{dsp^2}dsp2
  1. (A)A-IV, B-I, C-III, D-II
  2. (B)A-I. B-IV, C-III, D-II
  3. (C)A-I. B-IV, C-II, D-III
  4. (D)A-IV, B-I, C-II. D-III

Correct answer: (B)

Step-by-step solution →
Q39·ChemistrySingle correct
Dinitrogen and dioxygen. the main constituents of air do not react with each other in atmosphere to form oxides of nitrogen because
  1. (A)N2\mathrm{N_2}N2​ is unreactive in the condition of atmosphere.
  2. (B)Oxides of nitrogen are unstable.
  3. (C)Reaction between them can occur in the presence of a catalyst.
  4. (D)The reaction is endothermic and require very high temperature.

Correct answer: (D)

Step-by-step solution →
Q40·ChemistrySingle correct
The major product in the given reaction is
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (C)

Step-by-step solution →
Q41·ChemistrySingle correct
Arrange the following in increasing order of reactivity towards nitration A. p-xylene B. bromobenzene C. mesitylene D, nitrobenzene E. benzene Choose the correct answer from the options given below
  1. (A)C<D<E<A<B\mathrm{C < D < E < A < B}C<D<E<A<B
  2. (B)D<B<E<A<C\mathrm{D < B < E < A < C}D<B<E<A<C
  3. (C)D<C<E<A<B\mathrm{D < C < E < A < B}D<C<E<A<B
  4. (D)C<D<E<B<A\mathrm{C < D < E < B < A}C<D<E<B<A

Correct answer: (B)

Step-by-step solution →
Q42·ChemistrySingle correct
Compound I is heated with Conc. HI to give a hydroxy compound A which is further heated with Zn dust to give compound B. Identify A and B.
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (D)

Step-by-step solution →
Q43·ChemistrySingle correct
Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R Assertion A : Aniline on nitration yields ortho, meta & para nitro derivatives of aniline. Reason R: Nitrating mixture is a strong acidic mixture. In the light of the above statements, choose the correct answer from the options given below
  1. (A)Both A and R are true and R is the correct explanation of A
  2. (B)Both A and R are true but R is NOT the correct explanation of A
  3. (C)A is true but R is false
  4. (D)A is false but R is true

Correct answer: (A)

Step-by-step solution →
Q44·ChemistrySingle correct
Match List I with List II List II (Nature) I. Thermosetting polymer II. Fibers III. Elastomer IV. Thermoplastic polymer Choose the correct answer from the options given below:
  1. (A)A-II, B-III, C-IV, D-I
  2. (B)A-III, B-II, C-IV, D-I
  3. (C)A-III, B-I, C-IV, D-II
  4. (D)A-I. B-III, C-IV, D-II

Correct answer: (B)

Step-by-step solution →
Q45·ChemistrySingle correct
Two statements in respect of drug-enzyme interaction are given below Statement I : Action of an enzyme can be blocked only when an inhibitor blocks the active site of the enzyme. Statement II : An inhibitor can form a strong covalent bond with the enzyme. 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: (D)

Step-by-step solution →
Q46·ChemistrySingle correct
Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R Assertion A : Thin layer chromatography is an adsorption chromatography. Reason : A thin layer of silica gel is spread over a glass plate of suitable size in thin layer chromatography which acts as an adsorbent. In the light of the above statements, choose the correct answer from the options given below
  1. (A)Both A and R are true and R is the correct explanation of A
  2. (B)Both A and R are true but R is NOT the correct explanation of A
  3. (C)A is true but R is false
  4. (D)A is false but R is true

Correct answer: (A)

Step-by-step solution →
Q47·ChemistrySingle correct
The formulas of A and B for the following reaction sequence are
  1. (A)A=C7H14O8\mathrm{A = C_7H_{14}O_8}A=C7​H14​O8​, B=C6H14\mathrm{B = C_6H_{14}}B=C6​H14​
  2. (B)A=C7H13O7\mathrm{A = C_7H_{13}O_7}A=C7​H13​O7​, B=C7H14O\mathrm{B = C_7H_{14}O}B=C7​H14​O
  3. (C)A=C7H12O8\mathrm{A = C_7H_{12}O_8}A=C7​H12​O8​, B=C6H14\mathrm{B = C_6H_{14}}B=C6​H14​
  4. (D)A=C7H14O8\mathrm{A = C_7H_{14}O_8}A=C7​H14​O8​, B=C6H14O6\mathrm{B = C_6H_{14}O_6}B=C6​H14​O6​

Correct answer: (A)

Step-by-step solution →
Q48·ChemistrySingle correct
Find out the major product for the above reaction.
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (C)

Step-by-step solution →
Q49·ChemistryNumerical
2L of 0.2 M H2SO4\mathrm{H_2SO_4}H2​SO4​ is reacted with 2L of 0.1 M NaOH solution, the molarity of the resulting product Na2SO4\mathrm{Na_2SO_4}Na2​SO4​ in the solution is _____ millimolar. (Nearest integer).

Correct answer: 25

Step-by-step solution →
Q50·ChemistryNumerical
Metal M crystallizes into a FCC lattice with the edge length of 4.0×10−84.0\times10^{-8}4.0×10−8 cm. The atomic mass of the metal is _____ g/mol. (Nearest integer). (Use : NA=6.02×1023 mol−1\mathrm{N_A} = 6.02\times10^{23}\ \mathrm{mol^{-1}}NA​=6.02×1023 mol−1, density of metal, M = 9.03 g cm−3\mathrm{cm^{-3}}cm−3)

Correct answer: 87

Step-by-step solution →
Q51·ChemistryNumerical
If the wavelength for an electron emitted from H-atom is 3.3×10−103.3\times10^{-10}3.3×10−10m, then energy absorbed by the electron in its ground state compared to minimum energy required for its escape from the atom, is _____ times. (Nearest integer). [Given : h = 6.626×10−346.626\times10^{-34}6.626×10−34 Js, Mass of electron = 9.1×10−319.1\times10^{-31}9.1×10−31]

Correct answer: 2

Step-by-step solution →
Q52·ChemistryNumerical
A gaseous mixture of two substances A and B, under a total pressure of 0.8 atm is in equilibrium with an ideal liquid solution. The mole fraction of substance A is 0.5 in the vapour phase and 0.2 in the liquid phase. The vapour pressure of pure liquid A is _____ atm. (Nearest integer)

Correct answer: 2

Step-by-step solution →
Q53·ChemistryNumerical
At 600K, 2 mol of NO are mixed with 1 mol of O2\mathrm{O_2}O2​. 2NO(g)+O2(g)⇌2NO2(g)2\mathrm{NO_{(g)}} + \mathrm{O_2(g)} \rightleftharpoons 2\mathrm{NO_2(g)}2NO(g)​+O2​(g)⇌2NO2​(g) The reaction occurring as above comes to equilibrium under a total pressure of 1 atom. Analysis of the system shows that 0.6 mol of oxygen are present at equilibrium. The equilibrium constant for the reaction is _____. (Nearest integer).

Correct answer: 2

Step-by-step solution →
Q54·ChemistryNumerical
A sample of 0.125 g of an organic compound when analysed by Duma's method yields 22.78 mL of nitrogen gas collected over KOH solution at 280K and 759 mm Hg. The percentage of nitrogen in the given organic compound is _____. (Nearest integer). (a) The vapour pressure of water at 280 K is 14.2 mm Hg (b) R = 0.082 L atm K−1 mol−1\mathrm{K^{-1}\,mol^{-1}}K−1mol−1

Correct answer: 22

Step-by-step solution →
Q55·ChemistryNumerical
On reaction with stronger oxidizing agent like KIO4\mathrm{KIO_4}KIO4​, hydrogen peroxide oxidizes with the evolution of O2\mathrm{O_2}O2​. The oxidation number of I in KIO4\mathrm{KIO_4}KIO4​ changes to _____.

Correct answer: 5

Step-by-step solution →
Q56·ChemistryNumerical
For a reaction, given below is the graph of ln k vs 1T\dfrac{1}{\mathrm{T}}T1​. The activation energy for the reaction is equal to _____ cal mol−1\mathrm{mol^{-1}}mol−1. (Nearest integer). (Given : R = 2 cal K−1 mol−1\mathrm{K^{-1}\,mol^{-1}}K−1mol−1)

Correct answer: 8

Step-by-step solution →
Q57·ChemistryNumerical
Among the following the number of curves not in accordance with Freundlich adsorption isotherm is _____.

Correct answer: 3

Step-by-step solution →
Q58·ChemistryNumerical
Among the following the number of state variable is _____. Internal energy (U) Volume (V) Heat (q) Enthalpy (H)

Correct answer: 3

Step-by-step solution →

Mathematics — JEE Main 28 July 2022 Shift 2

Q59·MathematicsSingle correct
Let S={x∈[−6,3]−{−2,2}:∣x+3∣−1∣x∣−2≥0}S=\left\{x \in[-6,3]-\{-2,2\}: \frac{|x+3|-1}{|x|-2} \geq 0\right\}S={x∈[−6,3]−{−2,2}:∣x∣−2∣x+3∣−1​≥0} and T={x∈Z:x2−7∣x∣+9≤0}T=\left\{x \in Z: x^{2}-7|x|+9 \leq 0\right\}T={x∈Z:x2−7∣x∣+9≤0}. Then the number of elements in S∩TS \cap TS∩T is
  1. (A)7
  2. (B)5
  3. (C)4
  4. (D)3

Correct answer: (D)

Step-by-step solution →
Q60·MathematicsSingle correct
Let α\alphaα, β\betaβ be the roots of the equation x2−2x+6=0x^{2}-\sqrt{2}x+\sqrt{6}=0x2−2​x+6​=0 and 1α2+1,1β2+1\frac{1}{\alpha^{2}}+1, \frac{1}{\beta^{2}}+1α21​+1,β21​+1 be the roots of the equation x2+ax+b=0x^{2}+ax+b=0x2+ax+b=0. Then the roots of the equation x2−(a+b−2)x+(a+b+2)=0x^{2}-(a+b-2)x+(a+b+2)=0x2−(a+b−2)x+(a+b+2)=0 are :
  1. (A)non-real complex numbers
  2. (B)real and both negative
  3. (C)real and both positive
  4. (D)real and exactly one of them is positive

Correct answer: (B)

Step-by-step solution →
Q61·MathematicsSingle correct
Let A and B be any two 3×33 \times 33×3 symmetric and skew symmetric matrices respectively. Then which of the following is <b>NOT</b> true?
  1. (A)A4−B4A^{4}-B^{4}A4−B4 is a symmetric matrix
  2. (B)AB−BAAB-BAAB−BA is a symmetric matrix
  3. (C)B5−A5B^{5}-A^{5}B5−A5 is a skew-symmetric matrix
  4. (D)AB+BAAB+BAAB+BA is a skew-symmetric matrix

Correct answer: (C)

Step-by-step solution →
Q62·MathematicsSingle correct
Let f(x)=ax2+bx+cf(x)=ax^{2}+bx+cf(x)=ax2+bx+c be such that f(1)=3f(1)=3f(1)=3, f(−2)=λf(-2)=\lambdaf(−2)=λ and f(3)=4f(3)=4f(3)=4. If f(0)+f(1)+f(−2)+f(3)=14f(0)+f(1)+f(-2)+f(3)=14f(0)+f(1)+f(−2)+f(3)=14, then λ\lambdaλ is equal to
  1. (A)−4-4−4
  2. (B)132\frac{13}{2}213​
  3. (C)232\frac{23}{2}223​
  4. (D)4

Correct answer: (D)

Step-by-step solution →
Q63·MathematicsSingle correct
The function f:R→Rf: R \rightarrow Rf:R→R defined by f(x)=lim⁡n→∞cos⁡(2πx)−x2nsin⁡(x−1)1+x2n+1−x2nf(x)=\lim_{n \rightarrow \infty} \frac{\cos(2\pi x)-x^{2n}\sin(x-1)}{1+x^{2n+1}-x^{2n}}f(x)=limn→∞​1+x2n+1−x2ncos(2πx)−x2nsin(x−1)​ is continuous for all x in
  1. (A)R−{−1}R-\{-1\}R−{−1}
  2. (B)R−{−1,1}R-\{-1,1\}R−{−1,1}
  3. (C)R−{1}R-\{1\}R−{1}
  4. (D)R−{0}R-\{0\}R−{0}

Correct answer: (B)

Step-by-step solution →
Q64·MathematicsSingle correct
The function f(x)=xex(1−x)f(x)=xe^{x(1-x)}f(x)=xex(1−x), x∈Rx \in Rx∈R, is
  1. (A)increasing in (−12,1)\left(-\frac{1}{2},1\right)(−21​,1)
  2. (B)decreasing in (12,2)\left(\frac{1}{2},2\right)(21​,2)
  3. (C)increasing in (−1,−12)\left(-1,-\frac{1}{2}\right)(−1,−21​)
  4. (D)decreasing in (−12,12)\left(-\frac{1}{2},\frac{1}{2}\right)(−21​,21​)

Correct answer: (A)

Step-by-step solution →
Q65·MathematicsSingle correct
The sum of the absolute maximum and absolute minimum values of the function f(x)=tan⁡−1(sin⁡x−cos⁡x)f(x)=\tan^{-1}(\sin x-\cos x)f(x)=tan−1(sinx−cosx) in the interval [0,π][0,\pi][0,π] is
  1. (A)0
  2. (B)tan⁡−1(12)−π4\tan^{-1}\left(\frac{1}{\sqrt{2}}\right)-\frac{\pi}{4}tan−1(2​1​)−4π​
  3. (C)cos⁡−1(13)−π4\cos^{-1}\left(\frac{1}{\sqrt{3}}\right)-\frac{\pi}{4}cos−1(3​1​)−4π​
  4. (D)−π12\frac{-\pi}{12}12−π​

Correct answer: (C)

Step-by-step solution →
Q66·MathematicsSingle correct
Let x(t)=22cos⁡tsin⁡2tx(t)=2\sqrt{2}\cos t\sqrt{\sin 2t}x(t)=22​costsin2t​ and y(t)=22sin⁡tsin⁡2ty(t)=2\sqrt{2}\sin t\sqrt{\sin 2t}y(t)=22​sintsin2t​, t∈(0,π2)t \in \left(0,\frac{\pi}{2}\right)t∈(0,2π​). Then 1+(dydx)2d2ydx2\frac{1+\left(\frac{dy}{dx}\right)^{2}}{\frac{d^{2}y}{dx^{2}}}dx2d2y​1+(dxdy​)2​ at t=π4t=\frac{\pi}{4}t=4π​ is equal to
  1. (A)−223\frac{-2\sqrt{2}}{3}3−22​​
  2. (B)23\frac{2}{3}32​
  3. (C)13\frac{1}{3}31​
  4. (D)−23\frac{-2}{3}3−2​

Correct answer: (D)

Step-by-step solution →
Q67·MathematicsSingle correct
Let In(x)=∫0x1(t2+5)ndtI_{n}(x)=\int_{0}^{x}\frac{1}{\left(t^{2}+5\right)^{n}}dtIn​(x)=∫0x​(t2+5)n1​dt, n=1,2,3,….n=1,2,3,\ldots.n=1,2,3,…. Then
  1. (A)50I6−9I5=xI5′50I_{6}-9I_{5}=xI_{5}'50I6​−9I5​=xI5′​
  2. (B)50I6−11I5=xI5′50I_{6}-11I_{5}=xI_{5}'50I6​−11I5​=xI5′​
  3. (C)50I6−9I5=I5′50I_{6}-9I_{5}=I_{5}'50I6​−9I5​=I5′​
  4. (D)50I6−11I5=I5′50I_{6}-11I_{5}=I_{5}'50I6​−11I5​=I5′​

Correct answer: (A)

Step-by-step solution →
Q68·MathematicsSingle correct
The area enclosed by the curves y=log⁡e(x+e2)y=\log_{e}\left(x+e^{2}\right)y=loge​(x+e2), x=log⁡e(2y)x=\log_{e}\left(\frac{2}{y}\right)x=loge​(y2​) and x=log⁡e2x=\log_{e}2x=loge​2, above the line y=1y=1y=1 is
  1. (A)2+e−log⁡e22+e-\log_{e}22+e−loge​2
  2. (B)1+e−log⁡e21+e-\log_{e}21+e−loge​2
  3. (C)e−log⁡e2e-\log_{e}2e−loge​2
  4. (D)1+log⁡e21+\log_{e}21+loge​2

Correct answer: (B)

Step-by-step solution →
Q69·MathematicsSingle correct
Let y=y(x)y=y(x)y=y(x) be the solution curve of the differential equation dydx+1x2−1y=(x−1x+1)12\frac{dy}{dx}+\frac{1}{x^{2}-1}y=\left(\frac{x-1}{x+1}\right)^{\frac{1}{2}}dxdy​+x2−11​y=(x+1x−1​)21​, x>1x>1x>1 passing through the point (2,13)\left(2,\sqrt{\frac{1}{3}}\right)(2,31​​). Then 7y(8)\sqrt{7}y(8)7​y(8) is equal to
  1. (A)11+6log⁡e311+6\log_{e}311+6loge​3
  2. (B)19
  3. (C)12−2log⁡e312-2\log_{e}312−2loge​3
  4. (D)19−6log⁡e319-6\log_{e}319−6loge​3

Correct answer: (D)

Step-by-step solution →
Q70·MathematicsSingle correct
The differential equation of the family of circles passing through the points (0,2)(0,2)(0,2) and (0,−2)(0,-2)(0,−2) is
  1. (A)2xydydx+(x2−y2+4)=02xy\frac{dy}{dx}+\left(x^{2}-y^{2}+4\right)=02xydxdy​+(x2−y2+4)=0
  2. (B)2xydydx+(x2+y2−4)=02xy\frac{dy}{dx}+\left(x^{2}+y^{2}-4\right)=02xydxdy​+(x2+y2−4)=0
  3. (C)2xydydx+(y2−x2+4)=02xy\frac{dy}{dx}+\left(y^{2}-x^{2}+4\right)=02xydxdy​+(y2−x2+4)=0
  4. (D)2xydydx−(x2−y2+4)=02xy\frac{dy}{dx}-\left(x^{2}-y^{2}+4\right)=02xydxdy​−(x2−y2+4)=0

Correct answer: (A)

Step-by-step solution →
Q71·MathematicsSingle correct
Let the tangents at two points A and B on the circle x2+y2−4x+3=0x^{2}+y^{2}-4x+3=0x2+y2−4x+3=0 meet at origin O(0,0)O(0,0)O(0,0). Then the area of the triangle of OAB is
  1. (A)332\frac{3\sqrt{3}}{2}233​​
  2. (B)334\frac{3\sqrt{3}}{4}433​​
  3. (C)323\frac{3}{2\sqrt{3}}23​3​
  4. (D)343\frac{3}{4\sqrt{3}}43​3​

Correct answer: (B)

Step-by-step solution →
Q72·MathematicsSingle correct
Let the hyperbola H:x2a2−y2b2=1H: \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1H:a2x2​−b2y2​=1 pass through the point (22,−22)\left(2\sqrt{2},-2\sqrt{2}\right)(22​,−22​). A parabola is drawn whose focus is same as the focus of H with positive abscissa and the directrix of the parabola passes through the other focus of H. If the length of the latus rectum of the parabola is e times the length of the latus rectum of H, where e is the eccentricity of H, then which of the following points lies on the parabola?
  1. (A)(23,32)\left(2\sqrt{3},3\sqrt{2}\right)(23​,32​)
  2. (B)(33,−62)\left(3\sqrt{3},-6\sqrt{2}\right)(33​,−62​)
  3. (C)(3,−6)\left(\sqrt{3},-\sqrt{6}\right)(3​,−6​)
  4. (D)(36,62)\left(3\sqrt{6},6\sqrt{2}\right)(36​,62​)

Correct answer: (B)

Step-by-step solution →
Q73·MathematicsSingle correct
Let the lines x−1λ=y−21=z−32\frac{x-1}{\lambda}=\frac{y-2}{1}=\frac{z-3}{2}λx−1​=1y−2​=2z−3​ and x+26−2=y+183=z+28λ\frac{x+26}{-2}=\frac{y+18}{3}=\frac{z+28}{\lambda}−2x+26​=3y+18​=λz+28​ be coplanar and P be the plane containing these two lines. Then which of the following points does <b>NOT</b> lies on P?
  1. (A)(0,−2,−2)(0,-2,-2)(0,−2,−2)
  2. (B)(−5,0,−1)(-5,0,-1)(−5,0,−1)
  3. (C)(3,−1,0)(3,-1,0)(3,−1,0)
  4. (D)(0,4,5)(0,4,5)(0,4,5)

Correct answer: (D)

Step-by-step solution →
Q74·MathematicsSingle correct
A plane P is parallel to two lines whose direction ratios are −2,1,−3-2, 1, -3−2,1,−3, and −1,2,−2-1, 2, -2−1,2,−2 and it contains the point (2,2,−2)(2,2,-2)(2,2,−2). Let P intersect the co-ordinate axes at the points A, B, C making the intercepts α,β,γ\alpha, \beta, \gammaα,β,γ. If V is the volume of the tetrahedron OABC, where O is the origin and p=α+β+γp=\alpha+\beta+\gammap=α+β+γ, then the ordered pair (V,p)(V,p)(V,p) is equal to
  1. (A)(48,−13)(48,-13)(48,−13)
  2. (B)(24,−13)(24,-13)(24,−13)
  3. (C)(48,11)(48,11)(48,11)
  4. (D)(24,−5)(24,-5)(24,−5)

Correct answer: (B)

Step-by-step solution →
Q75·MathematicsSingle correct
Let S be the set of all a∈Ra \in Ra∈R for which the angle between the vectors u⃗=a(log⁡eb)i^−6j^+3k^\vec{u}=a\left(\log_{e}b\right)\hat{i}-6\hat{j}+3\hat{k}u=a(loge​b)i^−6j^​+3k^ and v⃗=(log⁡eb)i^+2j^+2a(log⁡eb)k^,(b>1)\vec{v}=\left(\log_{e}b\right)\hat{i}+2\hat{j}+2a\left(\log_{e}b\right)\hat{k}, (b>1)v=(loge​b)i^+2j^​+2a(loge​b)k^,(b>1) is acute. Then S is equal to
  1. (A)(−∞,−43)\left(-\infty,-\frac{4}{3}\right)(−∞,−34​)
  2. (B)Φ\PhiΦ
  3. (C)(−43,0)\left(-\frac{4}{3},0\right)(−34​,0)
  4. (D)(127,∞)\left(\frac{12}{7},\infty\right)(712​,∞)

Correct answer: (C)

Step-by-step solution →
Q76·MathematicsSingle correct
A horizontal park is in the shape of a triangle OAB with AB=16AB=16AB=16. A vertical lamp post OP is erected at the point O such that ∠PAO=∠PBO=15∘\angle PAO = \angle PBO = 15^{\circ}∠PAO=∠PBO=15∘ and ∠PCO=45∘\angle PCO = 45^{\circ}∠PCO=45∘, where C is the midpoint of AB. Then (OP)2(OP)^{2}(OP)2 is equal to
  1. (A)323(3−1)\frac{32}{\sqrt{3}}\left(\sqrt{3}-1\right)3​32​(3​−1)
  2. (B)323(2−3)\frac{32}{\sqrt{3}}\left(2-\sqrt{3}\right)3​32​(2−3​)
  3. (C)163(3−1)\frac{16}{\sqrt{3}}\left(\sqrt{3}-1\right)3​16​(3​−1)
  4. (D)163(2−3)\frac{16}{\sqrt{3}}\left(2-\sqrt{3}\right)3​16​(2−3​)

Correct answer: (B)

Step-by-step solution →
Q77·MathematicsSingle correct
Let A and B be two events such that P(B∣A)=25P(B|A)=\frac{2}{5}P(B∣A)=52​, P(A∣B)=17P(A|B)=\frac{1}{7}P(A∣B)=71​ and P(A∩B)=19P(A \cap B)=\frac{1}{9}P(A∩B)=91​. Consider (S1) P(A′∪B)=56P(A' \cup B)=\frac{5}{6}P(A′∪B)=65​, (S2) P(A′∩B′)=118P(A' \cap B')=\frac{1}{18}P(A′∩B′)=181​. Then
  1. (A)Both (S1) and (S2) are true
  2. (B)Both (S1) and (S2) are false
  3. (C)Only (S1) is true
  4. (D)Only (S2) is true

Correct answer: (A)

Step-by-step solution →
Q78·MathematicsSingle correct
Let <b>p :</b> Ramesh listens to music. <b>q :</b> Ramesh is out of his village <b>r :</b> It is Sunday <b>s :</b> It is Saturday Then the statement “Ramesh listens to music only if he is in his village and it is Sunday or Saturday” can be expressed as
  1. (A)((∼q)∧(r∨s))⇒p\left((\sim q)\wedge(r \vee s)\right) \Rightarrow p((∼q)∧(r∨s))⇒p
  2. (B)(q∧(r∨s))⇒p\left(q \wedge (r \vee s)\right) \Rightarrow p(q∧(r∨s))⇒p
  3. (C)p⇒(q∧(r∨s))p \Rightarrow \left(q \wedge (r \vee s)\right)p⇒(q∧(r∨s))
  4. (D)p⇒((∼q)∧(r∨s))p \Rightarrow \left((\sim q)\wedge(r \vee s)\right)p⇒((∼q)∧(r∨s))

Correct answer: (D)

Step-by-step solution →
Q79·MathematicsNumerical
Let the coefficients of the middle terms in the expansion of (16+βx)4\left(\frac{1}{\sqrt{6}}+\beta x\right)^{4}(6​1​+βx)4, (1−3βx)2\left(1-3\beta x\right)^{2}(1−3βx)2 and (1−β2x)6\left(1-\frac{\beta}{2}x\right)^{6}(1−2β​x)6, β>0\beta>0β>0, respectively form the first three terms of an A.P. If d is the common difference of this A.P., then 50−2dβ250-\frac{2d}{\beta^{2}}50−β22d​ is equal to _____

Correct answer: 57

Step-by-step solution →
Q80·MathematicsNumerical
A class contains b boys and g girls. If the number of ways of selecting 3 boys and 2 girls from the class is 168, then b+3gb + 3gb+3g is equal to

Correct answer: 17

Step-by-step solution →
Q81·MathematicsNumerical
Let the tangents at the points P and Q on the ellipse x22+y24=1\frac{x^{2}}{2}+\frac{y^{2}}{4}=12x2​+4y2​=1 meet at the point R(2, 22−2)R\left(\sqrt{2},\,2\sqrt{2}-2\right)R(2​,22​−2). If S is the focus of the ellipse on its negative major axis, then SP2+SQ2SP^{2} + SQ^{2}SP2+SQ2 is equal to

Correct answer: 13

Step-by-step solution →
Q82·MathematicsNumerical
If 1+(2+49C1+49C2+…+49C49)(50C2+50C4+…+50C50)1+ \left(2 + {}^{49}C_{1} + {}^{49}C_{2} + \ldots + {}^{49}C_{49}\right)\left({}^{50}C_{2} + {}^{50}C_{4} + \ldots + {}^{50}C_{50}\right)1+(2+49C1​+49C2​+…+49C49​)(50C2​+50C4​+…+50C50​) is equal to 2n.m2^{n}.m2n.m, where m is odd, then n+mn + mn+m is equal to _____

Correct answer: 99

Step-by-step solution →
Q83·MathematicsNumerical
Two tangent lines l1l_{1}l1​ and l2l_{2}l2​ are drawn from the point (2, 0) to the parabola 2y2=−x2y^{2} = -x2y2=−x. If the lines l1l_{1}l1​ and l2l_{2}l2​ are also tangent to the circle (x−5)2+y2=r(x - 5)^{2} + y^{2} = r(x−5)2+y2=r, then 17r17r17r is equal to

Correct answer: 9

Step-by-step solution →
Q84·MathematicsNumerical
If 6312+10311+20310+4039+…+102403=2n⋅m\frac{6}{3^{12}}+\frac{10}{3^{11}}+\frac{20}{3^{10}}+\frac{40}{3^{9}}+\ldots+\frac{10240}{3}=2^{n}\cdot m3126​+31110​+31020​+3940​+…+310240​=2n⋅m, where m is odd, then m.nm.nm.n is equal to _____

Correct answer: 12

Step-by-step solution →
Q85·MathematicsNumerical
Let S=[−π,π2)−{−π2,−π4,−3π4,π4}S=\left[-\pi,\frac{\pi}{2}\right)-\left\{-\frac{\pi}{2},-\frac{\pi}{4},-\frac{3\pi}{4},\frac{\pi}{4}\right\}S=[−π,2π​)−{−2π​,−4π​,−43π​,4π​}. Then the number of elements in the set A={θ∈S:tan⁡θ(1+5tan⁡(2θ))=5−tan⁡(2θ)}A=\left\{\theta\in S:\tan\theta\left(1+\sqrt{5}\tan(2\theta)\right)=\sqrt{5}-\tan(2\theta)\right\}A={θ∈S:tanθ(1+5​tan(2θ))=5​−tan(2θ)} is _____

Correct answer: 5

Step-by-step solution →
Q86·MathematicsNumerical
Let z=a+ibz = a + ibz=a+ib, b≠0b \neq 0b=0 be complex numbers satisfying z2=zˉ⋅21−∣z∣z^{2}=\bar{z}\cdot 2^{1-|z|}z2=zˉ⋅21−∣z∣. Then the least value of n∈Nn \in Nn∈N, such that zn=(z+1)nz^{n} = (z+1)^{n}zn=(z+1)n, is equal to _____

Correct answer: 6

Step-by-step solution →
Q87·MathematicsNumerical
A bag contains 4 white and 6 black balls. Three balls are drawn at random from the bag. Let X be the number of white balls, among the drawn balls. If σ2\sigma^{2}σ2 is the variance of X, then 100 σ2100\,\sigma^{2}100σ2 is equal to

Correct answer: 56

Step-by-step solution →
Q88·MathematicsNumerical
The value of the integral ∫0π/260sin⁡(6x)sin⁡xdx\int_{0}^{\pi/2}60\frac{\sin(6x)}{\sin x}dx∫0π/2​60sinxsin(6x)​dx is equal to

Correct answer: 104

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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
  • Dual Nature of Matter and Radiation 155/186
  • Amines 133/186
  • Quadratic Equations 148/186
  • Work, Energy and Power 132/186
  • Some Basic Concepts in Chemistry 129/186
  • Electromagnetic Induction 120/186
  • 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
  • Capacitors and Dielectrics 115/186
  • Classification of Elements and Periodicity in Properties 113/186
  • Parabola 101/186
  • Ellipse 103/186
  • Isolation of Metals 106/186
  • Differentiability 91/186
  • s-Block Elements 88/186
  • Surface Chemistry 98/186
  • Hydrogen 81/186
  • Hyperbola 77/186
  • Polymers 64/186
  • Carboxylic Acids and Derivatives 54/186
  • Solid State 63/186
  • Chemistry in Everyday Life 60/186
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