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JEE Main 29 January 2023 Shift 2 Question Paper with Answers

29 January 2023 · January session · 90 questions

The complete JEE Main 29 January 2023 Shift 2 paper — all 90 questions with the correct answer for each, tagged to the chapter it tests. Free to read, no account needed.

Physics
30
Chemistry
30
Mathematics
30

Physics — JEE Main 29 January 2023 Shift 2

Q1·PhysicsSingle correct
Substance AAA has atomic mass number 161616 and half-life of 111 day. Another substance BBB has atomic mass number 323232 and half life of 12\dfrac1221​ day. If both AAA and BBB simultaneously start undergo radio activity at the same time with initial mass 320 g320\,g320g each, how many total atoms of AAA and BBB combined would be left after 222 days.
  1. (A)3.38×10243.38\times10^{24}3.38×1024
  2. (B)1.69×10241.69\times10^{24}1.69×1024
  3. (C)6.76×10246.76\times10^{24}6.76×1024
  4. (D)6.76×10236.76\times10^{23}6.76×1023

Correct answer: (A)

Step-by-step solution →
Q2·PhysicsSingle correct
For the given logic gates combination, the correct truth table will be:
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (C)

Step-by-step solution →
Q3·PhysicsSingle correct
The time taken by an object to slide down 45∘45^\circ45∘ rough inclined plane is nnn times as it takes to slide down a perfectly smooth 45∘45^\circ45∘ incline plane. The coefficient of kinetic friction between the object and the incline plane is:
  1. (A)1−1n2\sqrt{1-\dfrac{1}{n^2}}1−n21​​
  2. (B)1+1n21+\dfrac{1}{n^2}1+n21​
  3. (C)1−1n21-\dfrac{1}{n^2}1−n21​
  4. (D)11−n2\sqrt{\dfrac{1}{1-n^2}}1−n21​​

Correct answer: (C)

Step-by-step solution →
Q4·PhysicsSingle correct
Heat energy of 184 kJ184\,kJ184kJ is given to ice of mass 600 g600\,g600g at −12∘C-12^\circ C−12∘C. Specific heat of ice is 2222.3 J kg−1 ∘C−12222.3\,J\,kg^{-1}\,^\circ C^{-1}2222.3Jkg−1∘C−1 and latent heat of ice is 336 kJ kg−1336\,kJ\,kg^{-1}336kJkg−1. A. Final temperature of system will be 0∘C0^\circ C0∘C. B. Final temperature of the system will be greater than 0∘C0^\circ C0∘C. C. The final system will have a mixture of ice and water in the ratio of 5:15:15:1. D. The final system will have a mixture of ice and water in the ratio of 1:51:51:5. E. The final system will have water only. Choose the correct answer from the options given below:
  1. (A)A and D Only
  2. (B)A and E Only
  3. (C)B and C Only
  4. (D)B and D only

Correct answer: (A)

Step-by-step solution →
Q5·PhysicsSingle correct
Identify the correct statements from the following: A. Work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket is negative. B. Work done by gravitational force in lifting a bucket out of a well by a rope tied to the bucket is negative. C. Work done by friction on a body sliding down an inclined plane is positive. D. Work done by an applied force on a body moving on a rough horizontal plane with uniform velocity is zero. E. Work done by air resistance on an oscillating pendulum is negative. Choose the correct answer from the options given below:
  1. (A)B, D and E only
  2. (B)A and C Only
  3. (C)B and D only
  4. (D)B and E only

Correct answer: (D)

Step-by-step solution →
Q6·PhysicsSingle correct
A scientist is observing a bacteria through a compound microscope. For better analysis and to improve its resolving power he should. (Select the best option)
  1. (A)Increase the refractive index of the medium between the object and objective lens
  2. (B)Decrease the diameter of the objective lens
  3. (C)Decrease the wave length of the light
  4. (D)Decrease the focal length of the eye piece

Correct answer: (A)

Step-by-step solution →
Q7·PhysicsSingle correct
With the help of potentiometer, we can determine the value of emf of a given cell. The sensitivity of the potentiometer is (A) directly proportional to the length of the potentiometer wire (B) directly proportional to the potential gradient of the wire (C) inversely proportional to the potential gradient of the wire (D) inversely proportional to the length of the potentiometer wire Choose the correct option for the above statements:
  1. (A)A only
  2. (B)C only
  3. (C)A and C only
  4. (D)B and D only

Correct answer: (C)

Step-by-step solution →
Q8·PhysicsSingle correct
A force acts for 20 s20\,s20s on a body of mass 20 kg20\,kg20kg, starting from rest, after which the force ceases and then body describes 50 m50\,m50m in the next 10 s10\,s10s. The value of force will be:
  1. (A)40 N40\,N40N
  2. (B)5 N5\,N5N
  3. (C)20 N20\,N20N
  4. (D)10 N10\,N10N

Correct answer: (B)

Step-by-step solution →
Q9·PhysicsSingle correct
The modulation index for an A.M. wave having maximum and minimum peak-to-peak voltages of 14 mV14\,mV14mV and 6 mV6\,mV6mV respectively is:
  1. (A)0.40.40.4
  2. (B)0.60.60.6
  3. (C)0.20.20.2
  4. (D)1.41.41.4

Correct answer: (A)

Step-by-step solution →
Q10·PhysicsSingle correct
Given below are two statements: Statement I: Electromagnetic waves are not deflected by electric and magnetic field. Statement II: The amplitude of electric field and the magnetic field in electromagnetic waves are related to each other as E0=μ0ε0B0E_0=\sqrt{\dfrac{\mu_0}{\varepsilon_0}}B_0E0​=ε0​μ0​​​B0​. In the light of the above statements, choose the correct answer from the options given below:
  1. (A)Statement I is true but statement II is false
  2. (B)Both Statement I and Statement II are false
  3. (C)Statement I is false but Statement II is true
  4. (D)Both Statement I and Statement II are true

Correct answer: (A)

Step-by-step solution →
Q11·PhysicsSingle correct
A square loop of area 25 cm225\,cm^225cm2 has a resistance of 10 Ω10\,\Omega10Ω. The loop is placed in uniform magnetic field of magnitude 40.0 T40.0\,T40.0T. The plane of loop is perpendicular to the magnetic field. The work done in pulling the loop out of the magnetic field slowly and uniformly in 1.0 sec1.0\,sec1.0sec, will be:
  1. (A)1.0×10−3 J1.0\times10^{-3}\,J1.0×10−3J
  2. (B)2.5×10−3 J2.5\times10^{-3}\,J2.5×10−3J
  3. (C)5×10−3 J5\times10^{-3}\,J5×10−3J
  4. (D)1.0×10−4 J1.0\times10^{-4}\,J1.0×10−4J

Correct answer: (A)

Step-by-step solution →
Q12·PhysicsSingle correct
For the given figures, choose the correct options:
  1. (A)At resonance, current in (b) is less than that in (a)
  2. (B)The rms current in circuit (b) can never be larger than that in (a)
  3. (C)The rms current in circuit (b) is always equal to that in (a)
  4. (D)The rms current in circuit (b) can be larger than that in (a)

Correct answer: (B)

Step-by-step solution →
Q13·PhysicsSingle correct
A fully loaded boeing aircraft has a mass of 5.4×105 kg5.4\times10^5\,kg5.4×105kg. Its total wing area is 500 m2500\,m^2500m2. It is in level flight with a speed of 1080 km/h1080\,km/h1080km/h. If the density of air ρ\rhoρ is 1.2 kg m−31.2\,kg\,m^{-3}1.2kgm−3, the fractional increase in the speed of the air on the upper surface of the wing relative to the lower surface in percentage will be. (g=10 m/s2)(g=10\,m/s^2)(g=10m/s2)
  1. (A)161616
  2. (B)101010
  3. (C)888
  4. (D)666

Correct answer: (B)

Step-by-step solution →
Q14·PhysicsSingle correct
The ratio of de-Broglie wavelength of an α\alphaα particle and a proton accelerated from rest by the same potential is 1m\dfrac{1}{\sqrt{m}}m​1​, the value of mmm is:
  1. (A)161616
  2. (B)444
  3. (C)222
  4. (D)888

Correct answer: (D)

Step-by-step solution →
Q15·PhysicsSingle correct
The time period of a satellite of earth is 242424 hours. If the separation between the earth and the satellite is decreased to one fourth of the previous value, then its new time period will become.
  1. (A)444 hours
  2. (B)666 hours
  3. (C)333 hours
  4. (D)121212 hours

Correct answer: (C)

Step-by-step solution →
Q16·PhysicsSingle correct
The electric current in a circular coil of four turns produces a magnetic induction 32 T32\,T32T at its centre. The coil is unwound and is rewound into a circular coil of single turn, the magnetic induction at the centre of the coil by the same current will be:
  1. (A)16 T16\,T16T
  2. (B)2 T2\,T2T
  3. (C)8 T8\,T8T
  4. (D)4 T4\,T4T

Correct answer: (B)

Step-by-step solution →
Q17·PhysicsSingle correct
A point charge 2×10−2 C2\times10^{-2}\,C2×10−2C is moved from PPP to SSS in a uniform electric field of 30 NC−130\,NC^{-1}30NC−1 directed along positive x-axis. If coordinates of PPP and SSS are (1,2,0)(1,2,0)(1,2,0) and (0,0,0)(0,0,0)(0,0,0) respectively, the work done by electric field will be:
  1. (A)1200 mJ1200\,mJ1200mJ
  2. (B)−1200 mJ-1200\,mJ−1200mJ
  3. (C)−600 mJ-600\,mJ−600mJ
  4. (D)600 mJ600\,mJ600mJ

Correct answer: (C)

Step-by-step solution →
Q18·PhysicsSingle correct
An object moves at a constant speed along a circular path in a horizontal plane with center at the origin. When the object is at x=+2 mx=+2\,mx=+2m, its velocity is −4j^ m/s-4\hat j\,m/s−4j^​m/s. The object's velocity (v⃗)(\vec v)(v) and acceleration (a⃗c)(\vec a_c)(ac​) at x=−2 mx=-2\,mx=−2m will be:
  1. (A)v⃗=−4i^ m/s, a⃗=−8j^ m/s2\vec v=-4\hat i\,m/s,\ \vec a=-8\hat j\,m/s^2v=−4i^m/s, a=−8j^​m/s2
  2. (B)v⃗=4i^ m/s, a⃗=8j^ m/s2\vec v=4\hat i\,m/s,\ \vec a=8\hat j\,m/s^2v=4i^m/s, a=8j^​m/s2
  3. (C)v⃗=4j^ m/s, a⃗=8i^ m/s2\vec v=4\hat j\,m/s,\ \vec a=8\hat i\,m/s^2v=4j^​m/s, a=8i^m/s2
  4. (D)v⃗=−4j^ m/s, a⃗=8i^ m/s2\vec v=-4\hat j\,m/s,\ \vec a=8\hat i\,m/s^2v=−4j^​m/s, a=8i^m/s2

Correct answer: (C)

Step-by-step solution →
Q19·PhysicsSingle correct
At 300 K300\,K300K the rms speed of oxygen molecules is α+5α\sqrt{\dfrac{\alpha+5}{\alpha}}αα+5​​ times to that of its average speed in the gas. Then, the value of α\alphaα will be: (used π=227)\left(\text{used }\pi=\dfrac{22}{7}\right)(used π=722​)
  1. (A)282828
  2. (B)242424
  3. (C)323232
  4. (D)272727

Correct answer: (A)

Step-by-step solution →
Q20·PhysicsSingle correct
The equation of a circle is given by x2+y2=a2x^2+y^2=a^2x2+y2=a2, where aaa is the radius. If the equation is modified to change the origin other than (0,0)(0,0)(0,0), then find out the correct dimensions of AAA and BBB in a new equation: (x−At)2+(y−tB)2=a2(x-At)^2+\left(y-\dfrac{t}{B}\right)^2=a^2(x−At)2+(y−Bt​)2=a2. The dimensions of ttt is given as [T−1][T^{-1}][T−1].
  1. (A)A=[LT], B=[L−1T−1]A=[LT],\,B=[L^{-1}T^{-1}]A=[LT],B=[L−1T−1]
  2. (B)A=[L−1T−1], B=[LT]A=[L^{-1}T^{-1}],\,B=[LT]A=[L−1T−1],B=[LT]
  3. (C)A=[L−1T], B=[LT−1]A=[L^{-1}T],\,B=[LT^{-1}]A=[L−1T],B=[LT−1]
  4. (D)A=[L−1T−1], B=[LT−1]A=[L^{-1}T^{-1}],\,B=[LT^{-1}]A=[L−1T−1],B=[LT−1]

Correct answer: (A)

Step-by-step solution →
Q21·PhysicsNumerical
A particle of mass 100 g100\,g100g is projected at time t=0t=0t=0 with a speed 20 ms−120\,ms^{-1}20ms−1 at an angle 45∘45^\circ45∘ to the horizontal as given in the figure. The magnitude of the angular momentum of the particle about the starting point at time t=2 st=2\,st=2s is found to be K kg m2/s\sqrt K\,kg\,m^2/sK​kgm2/s. The value of KKK is _____. (Take g=10 ms−2g=10\,ms^{-2}g=10ms−2)

Correct answer: 800

Step-by-step solution →
Q22·PhysicsNumerical
Unpolarised light is incident on the boundary between two dielectric media, whose dielectric constants are 2.82.82.8 (medium −1-1−1) and 6.86.86.8 (medium −2-2−2), respectively. To satisfy the condition, so that the reflected and refracted rays are perpendicular to each other, the angle of incidence should be tan⁡−1(1+10n)1/2\tan^{-1}\left(1+\dfrac{10}{n}\right)^{1/2}tan−1(1+n10​)1/2 the value of θ\thetaθ is _____. (Given for dielectric media, μr=1\mu_r=1μr​=1)

Correct answer: 7

Step-by-step solution →
Q23·PhysicsNumerical
A particle of mass 250 g250\,g250g executes a simple harmonic motion under a periodic force F=(−25x)NF=(-25x)NF=(−25x)N. The particle attains a maximum speed of 4 m/s4\,m/s4m/s during its oscillation. The amplitude of the motion is _____ cm.

Correct answer: 40

Step-by-step solution →
Q24·PhysicsNumerical
A car is moving on a circular path of radius 600 m600\,m600m such that the magnitudes of the tangential acceleration and centripetal acceleration are equal. The time taken by the car to complete first quarter of revolution, if it is moving with an initial speed of 54 km/hr54\,km/hr54km/hr is t(1−e−π/2)st\left(1-e^{-\pi/2}\right)st(1−e−π/2)s. The value of ttt is _____.

Correct answer: 40

Step-by-step solution →
Q25·PhysicsNumerical
When two resistances R1R_1R1​ and R2R_2R2​ connected in series and introduced into the left gap of a meter bridge and a resistance of 10 Ω10\,\Omega10Ω is introduced into the right gap, a null point is found at 60 cm60\,cm60cm from left side. When R1R_1R1​ is introduced in parallel and introduced into the left gap, a resistance of 3 Ω3\,\Omega3Ω is introduced into the right gap, a null point is found at 40 cm40\,cm40cm from left end. The product of R1R2R_1R_2R1​R2​ is _____ Ω2\Omega^2Ω2.

Correct answer: 30

Step-by-step solution →
Q26·PhysicsNumerical
In an experiment of measuring the refractive index of a glass slab using travelling microscope in physics lab, a student measures real thickness of the glass slab as 5.25 mm5.25\,mm5.25mm and apparent thickness of the glass slab as 5.00 mm5.00\,mm5.00mm. Travelling microscope has 202020 divisions in one cm on main scale and 505050 divisions on vernier scale is equal to 494949 divisions on main scale. The estimated uncertainty in the measurement of refractive index of the slab is x10×10−3\dfrac{x}{10}\times10^{-3}10x​×10−3, where xxx is _____.

Correct answer: 41

Step-by-step solution →
Q27·PhysicsNumerical
An inductor of inductance 2 μH2\,\mu H2μH is connected in series with a resistance, a variable capacitor and an AC source of frequency 7 kHz7\,kHz7kHz. The value of capacitance for which maximum current is drawn into the circuit is 1x F\dfrac{1}{x}\,Fx1​F, where the value of xxx is _____. (Take π=227)\left(\text{Take }\pi=\dfrac{22}{7}\right)(Take π=722​)

Correct answer: 3872

Step-by-step solution →
Q28·PhysicsNumerical
A null point is found at 200 cm200\,cm200cm in potentiometer when cell in secondary circuit is shunted by 5 Ω5\,\Omega5Ω. When a resistance of 15 Ω15\,\Omega15Ω is used for shunting, null point moves to 300 cm300\,cm300cm. The internal resistance of the cell is _____ Ω\OmegaΩ.

Correct answer: 5

Step-by-step solution →
Q29·PhysicsNumerical
For a charged spherical ball, electrostatic potential inside the ball varies with rrr as V=2ar2+bV=2ar^2+bV=2ar2+b. Here, aaa and bbb are constant and rrr is the distance from the center. The volume charge density inside the ball is −λaε-\lambda a\varepsilon−λaε. The value of λ\lambdaλ is _____. ε=\varepsilon=ε= permittivity of the medium

Correct answer: 12

Step-by-step solution →
Q30·PhysicsNumerical
A metal block of base area 0.20 m20.20\,m^20.20m2 is placed on a table, as shown in figure. A liquid film of thickness 0.25 mm0.25\,mm0.25mm is inserted between the block and the table. The block is pushed by a horizontal force of 0.1 N0.1\,N0.1N and moves with a constant speed. If the viscosity of the liquid is 5.0×10−3 Pl5.0\times10^{-3}\,Pl5.0×10−3Pl, the speed of block is _____ ×10−3 m/s\times10^{-3}\,m/s×10−3m/s.

Correct answer: 25

Step-by-step solution →

Chemistry — JEE Main 29 January 2023 Shift 2

Q31·ChemistrySingle correct
According to MO theory the bond orders for O22−O_2^{2-}O22−​, CO and NO+NO^+NO+ respectively, are:
  1. (A)1,21,21,2 and 333
  2. (B)1,31,31,3 and 222
  3. (C)2,32,32,3 and 333
  4. (D)1,31,31,3 and 333

Correct answer: (D)

Step-by-step solution →
Q32·ChemistrySingle correct
A doctor prescribed the drug Equanil to a patient. The patient was likely to have symptoms of which disease?
  1. (A)Hyperacidity
  2. (B)Anxiety and stress
  3. (C)Depression and hypertension
  4. (D)Stomach ulcers

Correct answer: (C)

Step-by-step solution →
Q33·ChemistrySingle correct
Reaction of propanamide with Br2/KOH (aq)Br_2/KOH\,(aq)Br2​/KOH(aq) produces:
  1. (A)Propylamine
  2. (B)Ethylnitrile
  3. (C)Propanenitrile
  4. (D)Ethylamine

Correct answer: (D)

Step-by-step solution →
Q34·ChemistrySingle correct
The one giving maximum number of isomeric alkenes on dehydrohalogenation reaction is (excluding rearrangement):
  1. (A)2-Bromopropane
  2. (B)2-Bromo-3,3-dimethylpentane
  3. (C)1-Bromo-2-methylbutane
  4. (D)2-Bromopentane

Correct answer: (D)

Step-by-step solution →
Q35·ChemistrySingle correct
An indicator ′X′'X'′X′ is used for studying the effect of variation in concentration of iodide on the rate of reaction of iodide ion with H2O2H_2O_2H2​O2​ at room temp. The indicator ′X′'X'′X′ forms blue colored complex with compound ′A′'A'′A′ present in the solution. The indicator ′X′'X'′X′ and compound ′A′'A'′A′ respectively are:
  1. (A)Methyl orange and H2O2H_2O_2H2​O2​
  2. (B)Starch and iodide
  3. (C)Starch and H2O2H_2O_2H2​O2​
  4. (D)Methyl orange and iodine

Correct answer: (B)

Step-by-step solution →
Q36·ChemistrySingle correct
The major component of which of the following ores is sulphide based mineral?
  1. (A)Siderite
  2. (B)Sphalerite
  3. (C)Malachite
  4. (D)Calamine

Correct answer: (B)

Step-by-step solution →
Q37·ChemistrySingle correct
A solution of CrO5CrO_5CrO5​ in amyl alcohol has a _______ colour.
  1. (A)Green
  2. (B)Orange-Red
  3. (C)Yellow
  4. (D)Blue

Correct answer: (D)

Step-by-step solution →
Q38·ChemistrySingle correct
The set of correct statements is: (i) Manganese exhibits +7+7+7 oxidation state in its oxide. (ii) Ruthenium and Osmium exhibit +8+8+8 oxidation in their oxides. (iii) Sc shows +4+4+4 oxidation state which is oxidizing in nature. (iv) Cr shows oxidising nature in +6+6+6 oxidation state.
  1. (A)(ii) and (iii)
  2. (B)(i), (ii) and (iv)
  3. (C)(ii), (iii) and (iv)
  4. (D)(i) and (iii)

Correct answer: (B)

Step-by-step solution →
Q39·ChemistrySingle correct
Following tetrapeptide can be represented as: (F, L, D, Y, I, Q, P are one letter codes for amino acids)
  1. (A)PLDY
  2. (B)FIQY
  3. (C)YQLF
  4. (D)FLDY

Correct answer: (D)

Step-by-step solution →
Q40·ChemistrySingle correct
Find out the major product for the following reaction.
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (D)

Step-by-step solution →
Q41·ChemistrySingle correct
Match List I with List II. Choose the correct answer from the options given below:
List IList II
A.van't Hoff factor, iiiI.Cryoscopic constant
B.kfk_fkf​II.Isotonic solutions
C.Solution with same osmotic pressureIII.Normal molar massAbnormal molar mass\dfrac{\text{Normal molar mass}}{\text{Abnormal molar mass}}Abnormal molar massNormal molar mass​
D.AzeotropesIV.Solutions with same composition of vapour above it
  1. (A)A-I, B-III, C-II, D-IV
  2. (B)A-III, B-I, C-IV, D-II
  3. (C)A-III, B-I, C-II, D-IV
  4. (D)A-III, B-II, C-I, D-IV

Correct answer: (C)

Step-by-step solution →
Q42·ChemistrySingle correct
Correct order of spin only magnetic moment of the following complex ions is: (Given At.no. Fe: 26, Co:27)
  1. (A)[FeF6]3−>[Co(C2O4)3]3−>[CoF6]3−[FeF_6]^{3-}>[Co(C_2O_4)_3]^{3-}>[CoF_6]^{3-}[FeF6​]3−>[Co(C2​O4​)3​]3−>[CoF6​]3−
  2. (B)[FeF6]3−>[CoF6]3−>[Co(C2O4)3]3−[FeF_6]^{3-}>[CoF_6]^{3-}>[Co(C_2O_4)_3]^{3-}[FeF6​]3−>[CoF6​]3−>[Co(C2​O4​)3​]3−
  3. (C)[Co(C2O4)3]3−>[CoF6]3−>[FeF6]3−[Co(C_2O_4)_3]^{3-}>[CoF_6]^{3-}>[FeF_6]^{3-}[Co(C2​O4​)3​]3−>[CoF6​]3−>[FeF6​]3−
  4. (D)[CoF6]3−>[FeF6]3−>[Co(C2O4)3]3−[CoF_6]^{3-}>[FeF_6]^{3-}>[Co(C_2O_4)_3]^{3-}[CoF6​]3−>[FeF6​]3−>[Co(C2​O4​)3​]3−

Correct answer: (B)

Step-by-step solution →
Q43·ChemistrySingle correct
Match List I with List II. Choose the correct answer from the options given below:
List IList II
A.Elastomeric polymerI.Urea formaldehyde resin
B.Fibre PolymerII.Polystyrene
C.Thermosetting PolymerIII.Polyester
D.Thermoplastic PolymerIV.Neoprene
  1. (A)A-II, B-III, C-I, D-IV
  2. (B)A-IV, B-III, C-I, D-II
  3. (C)A-IV, B-I, C-III, D-II
  4. (D)A-II, B-I, C-IV, D-III

Correct answer: (B)

Step-by-step solution →
Q44·ChemistrySingle correct
The concentration of dissolved Oxygen in water for growth of fish should be more than XXX ppm and Biochemical Oxygen Demand in clean water should be less than YYY ppm. XXX and YYY in ppm are, respectively.
  1. (A)X=4, Y=8X=4,\,Y=8X=4,Y=8
  2. (B)X=6, Y=5X=6,\,Y=5X=6,Y=5
  3. (C)X=4, Y=15X=4,\,Y=15X=4,Y=15
  4. (D)X=6, Y=12X=6,\,Y=12X=6,Y=12

Correct answer: (B)

Step-by-step solution →
Q45·ChemistrySingle correct
Find out the major products from the following reaction sequence.
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (D)

Step-by-step solution →
Q46·ChemistrySingle correct
When a hydrocarbon A undergoes combustion in the presence of air, it requires 9.59.59.5 equivalents of oxygen and produces 333 equivalents of water. What is the molecular formula of A?
  1. (A)C8H9C_8H_9C8​H9​
  2. (B)C8H6C_8H_6C8​H6​
  3. (C)C5H6C_5H_6C5​H6​
  4. (D)C6H6C_6H_6C6​H6​

Correct answer: (B)

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Q47·ChemistrySingle correct
Given below are two statements: Statement I: Nickel is being used as the catalyst for producing syn gas and edible fats. Statement II: Silicon forms both electron rich and electron deficient hydrides. In the light of the above statements, choose the most appropriate answer from the options given below:
  1. (A)Statement I is correct but statement II is incorrect
  2. (B)Both the statements I and II are correct
  3. (C)Statement I is incorrect but statement II is correct
  4. (D)Both the statements I and II are incorrect

Correct answer: (A)

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Q48·ChemistrySingle correct
Which of the following relations are correct? (A) ΔU=q+pΔV\Delta U=q+p\Delta VΔU=q+pΔV (B) ΔG=ΔH−TΔS\Delta G=\Delta H-T\Delta SΔG=ΔH−TΔS (C) ΔS=qrevT\Delta S=\dfrac{q_{rev}}{T}ΔS=Tqrev​​ (D) ΔH=ΔU−ΔnRT\Delta H=\Delta U-\Delta nRTΔH=ΔU−ΔnRT Choose the most appropriate answer from the options given below:
  1. (A)B and D Only
  2. (B)A and B Only
  3. (C)B and C Only
  4. (D)C and D Only

Correct answer: (C)

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Q49·ChemistrySingle correct
Given below are two statements: Statement I: The decrease in first ionization enthalpy from B to Al is much larger than that from Al to Ga. Statement II: The d orbitals in Ga are completely filled. In the light of the above statements, choose the most appropriate answer from the options given below:
  1. (A)Statement I is incorrect but statement II is correct
  2. (B)Both the statements I and II are correct
  3. (C)Both the statements I and II are incorrect
  4. (D)Statement I is correct but statement II is incorrect

Correct answer: (A)

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Q50·ChemistrySingle correct
Match List I and List II. Choose the correct answer from the options given below:
List IList II
A.OsmosisI.Solvent molecules pass through semi permeable membrane towards solvent side.
B.Reverse osmosisII.Movement of charged colloidal particles under the influence of applied electric potential towards oppositely charged electrodes.
C.Electro osmosisIII.Solvent molecules pass through semi permeable membrane towards solution side.
D.ElectrophoresisIV.Dispersion medium moves in an electric field.
  1. (A)A-I, B-III, C-IV, D-II
  2. (B)A-III, B-I, C-IV, D-II
  3. (C)A-III, B-I, C-II, D-IV
  4. (D)A-I, B-III, C-II, D-IV

Correct answer: (B)

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Q51·ChemistryNumerical
The radius of the first Bohr orbit of hydrogen atom is 0.6 A˚0.6\,\mathring{A}0.6A˚. The radius of the third Bohr orbit of He+He^+He+ is _____ picometer. (Nearest Integer)

Correct answer: 270

Step-by-step solution →
Q52·ChemistryNumerical
Total number of acidic oxides among N2O3, NO2, N2O, Cl2O7, SO2, CO, CaO, Na2ON_2O_3,\,NO_2,\,N_2O,\,Cl_2O_7,\,SO_2,\,CO,\,CaO,\,Na_2ON2​O3​,NO2​,N2​O,Cl2​O7​,SO2​,CO,CaO,Na2​O and NONONO is _____.

Correct answer: 4

Step-by-step solution →
Q53·ChemistryNumerical
The denticity of the ligand present in the Fehling's reagent is _____.

Correct answer: 4

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Q54·Chemistry·Redox Reactions and ElectrochemistryNumerical
The equilibrium constant for the reaction Zn(s)+Sn2+(aq)⇌Zn2+(aq)+Sn(s)Zn(s)+Sn^{2+}(aq)\rightleftharpoons Zn^{2+}(aq)+Sn(s)Zn(s)+Sn2+(aq)⇌Zn2+(aq)+Sn(s) is 1×10201\times10^{20}1×1020 at 298 K298\,K298K. The magnitude of standard electrode potential of Sn/Sn2+Sn/Sn^{2+}Sn/Sn2+ if EZn2+/Zn∘=−0.76 VE^\circ_{Zn^{2+}/Zn}=-0.76\,VEZn2+/Zn∘​=−0.76V is _____ ×10−2 V\times10^{-2}\,V×10−2V (Nearest integer). Given: 2.303RTF=0.059 V\dfrac{2.303RT}{F}=0.059\,VF2.303RT​=0.059V.

Correct answer: 17

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Q55·ChemistryNumerical
The volume of HCl, containing 73 g L−173\,g\,L^{-1}73gL−1, required to completely neutralise NaOH obtained by reacting 0.69 g0.69\,g0.69g of metallic sodium with water, is _____ mL. (Nearest Integer) (Given: molar Masses of Na, Cl, O, HNa,\,Cl,\,O,\,HNa,Cl,O,H are 23, 35.5, 1623,\,35.5,\,1623,35.5,16 and 1 g mol−11\,g\,mol^{-1}1gmol−1 respectively)

Correct answer: 15

Step-by-step solution →
Q56·ChemistryNumerical
For conversion of compound A→BA\to BA→B, the rate constant of the reaction was found to be 4.6×10−5 L mol−1 s−14.6\times10^{-5}\,L\,mol^{-1}\,s^{-1}4.6×10−5Lmol−1s−1. The order of the reaction is _____.

Correct answer: 2

Step-by-step solution →
Q57·ChemistryNumerical
On heating, LiNO3LiNO_3LiNO3​ gives how many compounds among the following? _____ Li2O, N2, O2, LiNO2, NO2Li_2O,\,N_2,\,O_2,\,LiNO_2,\,NO_2Li2​O,N2​,O2​,LiNO2​,NO2​

Correct answer: 3

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Q58·ChemistryNumerical
When 0.010.010.01 mol of an organic compound containing 60%60\%60% carbon was burnt completely, 4.4 g4.4\,g4.4g of CO2CO_2CO2​ was produced. The molar mass of compound is _____ g mol−1g\,mol^{-1}gmol−1 (Nearest integer).

Correct answer: 200

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Q59·ChemistryNumerical
At 298 K298\,K298K N2(g)+3H2(g)⇌2NH3(g), K1=4×105N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g),\,K_1=4\times10^5N2​(g)+3H2​(g)⇌2NH3​(g),K1​=4×105 N2(g)+O2(g)⇌2NO(g), K2=1.6×1012N_2(g)+O_2(g)\rightleftharpoons 2NO(g),\,K_2=1.6\times10^{12}N2​(g)+O2​(g)⇌2NO(g),K2​=1.6×1012 H2(g)+12O2(g)⇌H2O(g), K3=1.0×10−13H_2(g)+\dfrac12 O_2(g)\rightleftharpoons H_2O(g),\,K_3=1.0\times10^{-13}H2​(g)+21​O2​(g)⇌H2​O(g),K3​=1.0×10−13 Based on above equilibria, the equilibrium constant of the reaction, 2NH3(g)+52O2(g)⇌2NO(g)+3H2O(g)2NH_3(g)+\dfrac52 O_2(g)\rightleftharpoons 2NO(g)+3H_2O(g)2NH3​(g)+25​O2​(g)⇌2NO(g)+3H2​O(g) is _____ ×10−33\times10^{-33}×10−33 (Nearest integer).

Correct answer: 4

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Q60·ChemistryNumerical
A metal MMM forms hexagonal close-packed structure. The total number of voids in 0.020.020.02 mol of it is _____ ×1021\times10^{21}×1021 (Nearest integer). (Given NA=6.02×1023N_A=6.02\times10^{23}NA​=6.02×1023)

Correct answer: 36

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Mathematics — JEE Main 29 January 2023 Shift 2

Q61·MathematicsSingle correct
The statement B⇒((∼A)∨B)B\Rightarrow ((\sim A)\lor B)B⇒((∼A)∨B) is equivalent to:
  1. (A)A⇒(A⇔B)A\Rightarrow (A\Leftrightarrow B)A⇒(A⇔B)
  2. (B)A⇒((∼A)⇒B)A\Rightarrow ((\sim A)\Rightarrow B)A⇒((∼A)⇒B)
  3. (C)B⇒(A⇒B)B\Rightarrow (A\Rightarrow B)B⇒(A⇒B)
  4. (D)B⇒((∼A)⇒B)B\Rightarrow ((\sim A)\Rightarrow B)B⇒((∼A)⇒B)

Correct answer: (A)

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Q62·MathematicsSingle correct
The value of the integral ∫12(t4+1t6+1)dt\displaystyle\int_{1}^{2}\left(\dfrac{t^{4}+1}{t^{6}+1}\right)dt∫12​(t6+1t4+1​)dt is:
  1. (A)tan⁡−12−13tan⁡−18+π4\tan^{-1}2-\dfrac13\tan^{-1}8+\dfrac{\pi}{4}tan−12−31​tan−18+4π​
  2. (B)tan⁡−112+13tan⁡−18−π3\tan^{-1}\dfrac12+\dfrac13\tan^{-1}8-\dfrac{\pi}{3}tan−121​+31​tan−18−3π​
  3. (C)tan⁡−112−13tan⁡−18+π3\tan^{-1}\dfrac12-\dfrac13\tan^{-1}8+\dfrac{\pi}{3}tan−121​−31​tan−18+3π​
  4. (D)tan⁡−12+13tan⁡−18−π3\tan^{-1}2+\dfrac13\tan^{-1}8-\dfrac{\pi}{3}tan−12+31​tan−18−3π​

Correct answer: (D)

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Q63·MathematicsSingle correct
The set of all values of λ\lambdaλ for which the equation cos⁡22x−2sin⁡4x−2cos⁡2x=λ\cos^2 2x-2\sin^4 x-2\cos^2 x=\lambdacos22x−2sin4x−2cos2x=λ has a real solution xxx, is:
  1. (A)[−2,−1][-2,-1][−2,−1]
  2. (B)[−1,−12]\left[-1,-\dfrac12\right][−1,−21​]
  3. (C)[−32,−1]\left[-\dfrac32,-1\right][−23​,−1]
  4. (D)[−2,−32]\left[-2,-\dfrac32\right][−2,−23​]

Correct answer: (C)

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Q64·MathematicsSingle correct
Let RRR be a relation defined on N\mathbb{N}N as a R ba\,R\,baRb if 2a+3b2a+3b2a+3b is a multiple of 5, a,b∈N5,\,a,b\in\mathbb{N}5,a,b∈N. Then RRR is:
  1. (A)an equivalence relation
  2. (B)transitive but not symmetric
  3. (C)not reflexive
  4. (D)symmetric but not transitive

Correct answer: (A)

Step-by-step solution →
Q65·MathematicsSingle correct
Consider a function f:N→Rf:\mathbb{N}\to\mathbb{R}f:N→R, satisfying f(1)+2f(2)+3f(3)+⋯+xf(x)=x(x+1)f(x); x≥2f(1)+2f(2)+3f(3)+\dots+xf(x)=x(x+1)f(x);\,x\ge 2f(1)+2f(2)+3f(3)+⋯+xf(x)=x(x+1)f(x);x≥2 with f(1)=1f(1)=1f(1)=1. Then 1f(2022)+1f(2028)\dfrac{1}{f(2022)}+\dfrac{1}{f(2028)}f(2022)1​+f(2028)1​ is equal to:
  1. (A)810081008100
  2. (B)840084008400
  3. (C)800080008000
  4. (D)820082008200

Correct answer: (A)

Step-by-step solution →
Q66·MathematicsSingle correct
If a⃗=i^+2k^, b⃗=i^+j^+k^, c⃗=7i^−3j^+4k^, r⃗×b⃗+b⃗×c⃗=0⃗\vec a=\hat i+2\hat k,\,\vec b=\hat i+\hat j+\hat k,\,\vec c=7\hat i-3\hat j+4\hat k,\,\vec r\times\vec b+\vec b\times\vec c=\vec 0a=i^+2k^,b=i^+j^​+k^,c=7i^−3j^​+4k^,r×b+b×c=0 and r⃗⋅a⃗=0\vec r\cdot\vec a=0r⋅a=0. Then r⃗⋅c⃗\vec r\cdot\vec cr⋅c is equal to:
  1. (A)323232
  2. (B)303030
  3. (C)363636
  4. (D)343434

Correct answer: (D)

Step-by-step solution →
Q67·MathematicsSingle correct
The shortest distance between the lines x−12=y−21=z−6−3\dfrac{x-1}{2}=\dfrac{y-2}{1}=\dfrac{z-6}{-3}2x−1​=1y−2​=−3z−6​ and x−12=y+8−7=z−45\dfrac{x-1}{2}=\dfrac{y+8}{-7}=\dfrac{z-4}{5}2x−1​=−7y+8​=5z−4​ is:
  1. (A)535\sqrt353​
  2. (B)232\sqrt323​
  3. (C)333\sqrt333​
  4. (D)434\sqrt343​

Correct answer: (D)

Step-by-step solution →
Q68·MathematicsSingle correct
The plane 2x−y+z=42x-y+z=42x−y+z=4 intersects the line segment joining the points A(a,−2,4)A(a,-2,4)A(a,−2,4) and B(2,b,−3)B(2,b,-3)B(2,b,−3) at the point CCC in the ratio 2:12:12:1 and the distance of the point CCC from the origin is 5\sqrt55​. If ab<0ab<0ab<0 and PPP is the point (a−b, b, 2b−a)(a-b,\,b,\,2b-a)(a−b,b,2b−a) then CP2CP^2CP2 is equal to:
  1. (A)973\dfrac{97}{3}397​
  2. (B)173\dfrac{17}{3}317​
  3. (C)163\dfrac{16}{3}316​
  4. (D)733\dfrac{73}{3}373​

Correct answer: (B)

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Q69·MathematicsSingle correct
The value of the integral ∫1/22tan⁡−1xxdx\displaystyle\int_{1/2}^{2}\dfrac{\tan^{-1}x}{x}dx∫1/22​xtan−1x​dx is equal to:
  1. (A)π2log⁡e2\dfrac{\pi}{2}\log_e 22π​loge​2
  2. (B)πlog⁡e2\pi\log_e 2πloge​2
  3. (C)12log⁡e2\dfrac12\log_e 221​loge​2
  4. (D)π4log⁡e2\dfrac{\pi}{4}\log_e 24π​loge​2

Correct answer: (A)

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Q70·MathematicsSingle correct
The letters of the word OUGHT are written in all possible ways and these words are arranged as in a dictionary, in a series. Then the serial number of the word TOUGH is:
  1. (A)848484
  2. (B)797979
  3. (C)898989
  4. (D)868686

Correct answer: (C)

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Q71·Mathematics·Matrices and DeterminantsSingle correct
The set of all values of t∈Rt\in\mathbb{R}t∈R, for which the matrix [ete−t(sin⁡t−2cos⁡t)e−t(−2sin⁡t−cos⁡t)ete−t(2sin⁡t+cos⁡t)e−t(sin⁡t−2cos⁡t)ete−tcos⁡te−tsin⁡t]\begin{bmatrix} e^t & e^{-t}(\sin t-2\cos t) & e^{-t}(-2\sin t-\cos t) \\ e^t & e^{-t}(2\sin t+\cos t) & e^{-t}(\sin t-2\cos t) \\ e^t & e^{-t}\cos t & e^{-t}\sin t \end{bmatrix}​etetet​e−t(sint−2cost)e−t(2sint+cost)e−tcost​e−t(−2sint−cost)e−t(sint−2cost)e−tsint​​ is invertible, is:
  1. (A)R\mathbb{R}R
  2. (B){kπ+π4, k∈Z}\{k\pi+\dfrac{\pi}{4},\,k\in\mathbb{Z}\}{kπ+4π​,k∈Z}
  3. (C){kπ, k∈Z}\{k\pi,\,k\in\mathbb{Z}\}{kπ,k∈Z}
  4. (D){(2k+1)π2, k∈Z}\{(2k+1)\dfrac{\pi}{2},\,k\in\mathbb{Z}\}{(2k+1)2π​,k∈Z}

Correct answer: (A)

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Q72·MathematicsSingle correct
The area of the region A={(x,y):∣cos⁡x−sin⁡x∣≤y≤sin⁡x, 0≤x≤π2}A=\{(x,y):|\cos x-\sin x|\le y\le\sin x,\,0\le x\le\dfrac{\pi}{2}\}A={(x,y):∣cosx−sinx∣≤y≤sinx,0≤x≤2π​} is:
  1. (A)5+22−4.5\sqrt5+2\sqrt2-4.55​+22​−4.5
  2. (B)1−32+451-\dfrac{3}{\sqrt2}+\dfrac{4}{\sqrt5}1−2​3​+5​4​
  3. (C)35−32+1\dfrac{3}{\sqrt5}-\dfrac{3}{\sqrt2}+15​3​−2​3​+1
  4. (D)5−22+1\sqrt5-2\sqrt2+15​−22​+1

Correct answer: (D)

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Q73·MathematicsSingle correct
The number of 3 digit numbers, that are divisible by either 333 or 444 but not divisible by 484848, is:
  1. (A)507507507
  2. (B)432432432
  3. (C)472472472
  4. (D)400400400

Correct answer: (B)

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Q74·MathematicsSingle correct
If the lines x−11=y−22=z+31\dfrac{x-1}{1}=\dfrac{y-2}{2}=\dfrac{z+3}{1}1x−1​=2y−2​=1z+3​ and x−a2=y+23=z−31\dfrac{x-a}{2}=\dfrac{y+2}{3}=\dfrac{z-3}{1}2x−a​=3y+2​=1z−3​ intersect at the point PPP, then the distance of the point PPP from the plane z=az=az=a is:
  1. (A)282828
  2. (B)161616
  3. (C)101010
  4. (D)222222

Correct answer: (A)

Step-by-step solution →
Q75·MathematicsSingle correct
Let y=y(x)y=y(x)y=y(x) be the solution of the differential equation xlog⁡exdydx+y=x2log⁡ex, (x>1)x\log_e x\dfrac{dy}{dx}+y=x^2\log_e x,\,(x>1)xloge​xdxdy​+y=x2loge​x,(x>1). If y(2)=2y(2)=2y(2)=2, then y(e)y(e)y(e) is equal to:
  1. (A)1+e22\dfrac{1+e^2}{2}21+e2​
  2. (B)4+e24\dfrac{4+e^2}{4}44+e2​
  3. (C)2+e22\dfrac{2+e^2}{2}22+e2​
  4. (D)1+e24\dfrac{1+e^2}{4}41+e2​

Correct answer: (B)

Step-by-step solution →
Q76·MathematicsSingle correct
Let fff and ggg be twice differentiable functions on R\mathbb{R}R such that f ′ ′(x)=g ′ ′(x)+6x, f ′(1)=4g ′(1)−3=9, f(2)=3g(2)=12f\,'\,'(x)=g\,'\,'(x)+6x,\,f\,'(1)=4g\,'(1)-3=9,\,f(2)=3g(2)=12f′′(x)=g′′(x)+6x,f′(1)=4g′(1)−3=9,f(2)=3g(2)=12. Then which of the following is NOT true?
  1. (A)There exists x0∈(1,3/2)x_0\in(1,3/2)x0​∈(1,3/2) such that f(x0)=g(x0)f(x_0)=g(x_0)f(x0​)=g(x0​)
  2. (B)∣f ′(x)−g ′(x)∣<6⇒−1<x<1|f\,'(x)-g\,'(x)|<6\Rightarrow -1<x<1∣f′(x)−g′(x)∣<6⇒−1<x<1
  3. (C)If −1<x<2-1<x<2−1<x<2, then ∣f(x)−g(x)∣<8|f(x)-g(x)|<8∣f(x)−g(x)∣<8
  4. (D)g(−2)−f(−2)=20g(-2)-f(-2)=20g(−2)−f(−2)=20

Correct answer: (C)

Step-by-step solution →
Q77·MathematicsSingle correct
If the tangent at a point PPP on the parabola y2=3xy^2=3xy2=3x is parallel to the line x+2y=1x+2y=1x+2y=1 and the tangents at the points QQQ and RRR on the ellipse x24+y21=1\dfrac{x^2}{4}+\dfrac{y^2}{1}=14x2​+1y2​=1 are perpendicular to the line x−y=2x-y=2x−y=2, then the area of the triangle PQRPQRPQR is:
  1. (A)325\dfrac32\sqrt523​5​
  2. (B)353\sqrt535​
  3. (C)95\dfrac{9}{\sqrt5}5​9​
  4. (D)535\sqrt353​

Correct answer: (B)

Step-by-step solution →
Q78·MathematicsSingle correct
Let a⃗=4i^+3j^\vec a=4\hat i+3\hat ja=4i^+3j^​ and b⃗=3i^−4j^+5k^\vec b=3\hat i-4\hat j+5\hat kb=3i^−4j^​+5k^. If c⃗\vec cc is a vector such that c⃗⋅(a⃗×b⃗)+25=0, c⃗⋅(i^+j^+k^)=4\vec c\cdot(\vec a\times\vec b)+25=0,\,\vec c\cdot(\hat i+\hat j+\hat k)=4c⋅(a×b)+25=0,c⋅(i^+j^​+k^)=4, and projection of c⃗\vec cc on a⃗\vec aa is 111, then the projection of c⃗\vec cc on b⃗\vec bb equals:
  1. (A)15\dfrac1551​
  2. (B)52\dfrac{5}{\sqrt2}2​5​
  3. (C)32\dfrac{3}{\sqrt2}2​3​
  4. (D)12\dfrac{1}{\sqrt2}2​1​

Correct answer: (B)

Step-by-step solution →
Q79·MathematicsSingle correct
Let S={w1,w2,… }S=\{w_1,w_2,\dots\}S={w1​,w2​,…} be the sample space associated to a random experiment. Let P(wn)=P(wn−1)2, n≥2P(w_n)=\dfrac{P(w_{n-1})}{2},\,n\ge 2P(wn​)=2P(wn−1​)​,n≥2. Let A={2k+3l: k,l∈N}A=\{2k+3l:\,k,l\in\mathbb{N}\}A={2k+3l:k,l∈N} and B={wn: n∈A}B=\{w_n:\,n\in A\}B={wn​:n∈A}. Then P(B)P(B)P(B) is equal to:
  1. (A)364\dfrac{3}{64}643​
  2. (B)116\dfrac{1}{16}161​
  3. (C)132\dfrac{1}{32}321​
  4. (D)332\dfrac{3}{32}323​

Correct answer: (A)

Step-by-step solution →
Q80·MathematicsSingle correct
Let KKK be the sum of the coefficients of the odd powers of xxx in the expansion of (1+x)99(1+x)^{99}(1+x)99. Let aaa be the middle term in the expansion of (2+12)200\left(2+\dfrac{1}{\sqrt2}\right)^{200}(2+2​1​)200. If 200C99⋅Ka=2ℓ⋅mn\dfrac{{}^{200}C_{99}\cdot K}{a}=\dfrac{2^{\ell}\cdot m}{n}a200C99​⋅K​=n2ℓ⋅m​, where mmm and nnn are odd numbers, then the ordered pair (ℓ,n)(\ell,n)(ℓ,n) is equal to:
  1. (A)(50,51)(50,51)(50,51)
  2. (B)(50,101)(50,101)(50,101)
  3. (C)(51,99)(51,99)(51,99)
  4. (D)(51,101)(51,101)(51,101)

Correct answer: (B)

Step-by-step solution →
Q81·MathematicsNumerical
The total number of 4-digit numbers whose greatest common divisor with 545454 is 222, is _____.

Correct answer: 3000

Step-by-step solution →
Q82·MathematicsNumerical
Let a1=b1=1a_1=b_1=1a1​=b1​=1 and an=an−1+(n−1), bn=bn−1+an−1, ∀n≥2a_n=a_{n-1}+(n-1),\,b_n=b_{n-1}+a_{n-1},\,\forall n\ge 2an​=an−1​+(n−1),bn​=bn−1​+an−1​,∀n≥2. If S=∑n=110bn2nS=\sum_{n=1}^{10}\dfrac{b_n}{2^n}S=∑n=110​2nbn​​ and T=∑n=18n2n−1T=\sum_{n=1}^{8}\dfrac{n}{2^{n-1}}T=∑n=18​2n−1n​, then 27(2S−T)2^{7}(2S-T)27(2S−T) is equal to _____.

Correct answer: 461

Step-by-step solution →
Q83·MathematicsNumerical
A triangle is formed by the tangents at the point (2,2)(2,2)(2,2) on the curves y2=2xy^2=2xy2=2x and x2+y2=4xx^2+y^2=4xx2+y2=4x, and the line x+y+2=0x+y+2=0x+y+2=0. If rrr is the radius of its circumcircle, then r2r^2r2 is equal to _____.

Correct answer: 10

Step-by-step solution →
Q84·MathematicsNumerical
Let α1,α2,…,α7\alpha_1,\alpha_2,\dots,\alpha_7α1​,α2​,…,α7​ be the roots of the equation x7+3x5−13x3−15x=0x^7+3x^5-13x^3-15x=0x7+3x5−13x3−15x=0 and ∣α1∣≥∣α2∣≥⋯≥∣α7∣|\alpha_1|\ge|\alpha_2|\ge\dots\ge|\alpha_7|∣α1​∣≥∣α2​∣≥⋯≥∣α7​∣. Then α1α2−α3α4+α5α6\alpha_1\alpha_2-\alpha_3\alpha_4+\alpha_5\alpha_6α1​α2​−α3​α4​+α5​α6​ is equal to _____.

Correct answer: 3

Step-by-step solution →
Q85·MathematicsNumerical
Let X={11,12,13,…,40,41}X=\{11,12,13,\dots,40,41\}X={11,12,13,…,40,41} and Y={61,62,63,…,90,91}Y=\{61,62,63,\dots,90,91\}Y={61,62,63,…,90,91} be the two sets of observations. If xˉ\bar xxˉ and yˉ\bar yyˉ​ are their respective means and σ2\sigma^2σ2 is the variance of all the observations in X∪YX\cup YX∪Y, then ∣xˉ+yˉ−σ2∣|\bar x+\bar y-\sigma^2|∣xˉ+yˉ​−σ2∣ is equal to _____.

Correct answer: 603

Step-by-step solution →
Q86·MathematicsNumerical
If the equation of the normal to the curve y=x−a(x+b)(x−2)y=\dfrac{x-a}{(x+b)(x-2)}y=(x+b)(x−2)x−a​ at the point (1,−3)(1,-3)(1,−3) is x−4y=13x-4y=13x−4y=13, then the value of a+ba+ba+b is equal to _____.

Correct answer: -6

Step-by-step solution →
Q87·Mathematics·Matrices and DeterminantsNumerical
Let AAA be a symmetric matrix such that ∣A∣=2|A|=2∣A∣=2 and [21332]A=[12αβ]\begin{bmatrix} 2 & 1 \\ 3 & \tfrac32 \end{bmatrix}A=\begin{bmatrix} 1 & 2 \\ \alpha & \beta \end{bmatrix}[23​123​​]A=[1α​2β​]. If the sum of the diagonal elements of AAA is sss, then βsα2\dfrac{\beta s}{\alpha^2}α2βs​ is equal to _____.

Correct answer: 5

Step-by-step solution →
Q88·MathematicsNumerical
Let α=8−14i, A={z∈C:αz−αˉzˉz2−(zˉ)2−112i=1}\alpha=8-14i,\,A=\left\{z\in\mathbb{C}:\dfrac{\alpha z-\bar\alpha\bar z}{z^{2}-(\bar z)^{2}-112i}=1\right\}α=8−14i,A={z∈C:z2−(zˉ)2−112iαz−αˉzˉ​=1} and B={z∈C: ∣z+3i∣=4}B=\{z\in\mathbb{C}:\,|z+3i|=4\}B={z∈C:∣z+3i∣=4}. Then ∑z∈A∩B(Re⁡z−Im⁡z)\displaystyle\sum_{z\in A\cap B}(\operatorname{Re}z-\operatorname{Im}z)z∈A∩B∑​(Rez−Imz) is equal to _____.

Correct answer: 14

Step-by-step solution →
Q89·MathematicsNumerical
A circle with centre (2,3)(2,3)(2,3) and radius 444 intersects the line x+y=3x+y=3x+y=3 at the points PPP and QQQ. If the tangents at PPP and QQQ intersect at the point S(α,β)S(\alpha,\beta)S(α,β), then 4α−7β4\alpha-7\beta4α−7β is equal to _____.

Correct answer: 11

Step-by-step solution →
Q90·MathematicsNumerical
Let {ak}\{a_k\}{ak​} and {bk}, k∈N\{b_k\},\,k\in\mathbb{N}{bk​},k∈N, be two G.P.s with common ratios r1r_1r1​ and r2r_2r2​ respectively such that a1=b1=4a_1=b_1=4a1​=b1​=4 and r1<r2r_1<r_2r1​<r2​. Let ck=ak+bk, k∈Nc_k=a_k+b_k,\,k\in\mathbb{N}ck​=ak​+bk​,k∈N. If c2=5c_2=5c2​=5 and c3=134c_3=\dfrac{13}{4}c3​=413​ then ∑k=1∞ck−(12a6+8b4)\sum_{k=1}^{\infty}c_k-(12a_6+8b_4)∑k=1∞​ck​−(12a6​+8b4​) is equal to _____.

Correct answer: 9

Step-by-step solution →

Chapters tested in this paper

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

  • Properties of Solids and Liquids 172/186
  • Three Dimensional Geometry 176/186
  • Matrices and Determinants 180/186
  • Coordination Compounds 176/186
  • Sets, Relations and Functions 165/186
  • Current Electricity 160/186
  • Sequence and Series 164/186
  • p-Block Elements 164/186
  • Definite Integration 168/186
  • Rotational Motion 172/186
  • Redox Reactions and Electrochemistry 177/186
  • Geometrical Optics 172/186
  • Kinematics 156/186
  • Vector Algebra 173/186
  • Differential Equations 167/186
  • Probability 176/186
  • Permutations and Combinations 162/186
  • Magnetic Field of Current 147/186
  • Binomial Theorem and Its Simple Applications 158/186
  • Electromagnetic Waves 144/186
  • Chemical Bonding and Molecular Structure 151/186
  • Application of Derivatives 139/186
  • d- and f-Block Elements 126/186
  • Aldehydes and Ketones 135/186
  • Equilibrium 163/186
  • Solutions 158/186
  • Laws of Motion 130/186
  • Chemical Thermodynamics 165/186
  • Units and Measurements 149/186
  • Hydrocarbons 126/186
  • Complex Numbers 165/186
  • Trigonometric Functions 144/186
  • Biomolecules 162/186
  • Chemical Kinetics 169/186
  • Gravitation 152/186
  • 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
  • Area Under Curves 139/186
  • Kinetic Theory of Gases 135/186
  • Purification and Characterisation of Organic Compounds 116/186
  • Oscillations 117/186
  • Alternating Currents 108/186
  • Nuclei 116/186
  • Organic Compounds Containing Halogens 109/186
  • Parabola 101/186
  • Statistics 118/186
  • Isolation of Metals 106/186
  • s-Block Elements 88/186
  • Surface Chemistry 98/186
  • Environmental Chemistry 83/186
  • Hydrogen 81/186
  • Experimental Skills 68/186
  • Electric Potential 63/186
  • Polymers 64/186
  • Solid State 63/186
  • Chemistry in Everyday Life 60/186
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