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JEE Main 30 January 2023 Shift 1 Question Paper with Answers

30 January 2023 · January session · 90 questions

The complete JEE Main 30 January 2023 Shift 1 paper — all 90 questions with the correct answer for each, tagged to the chapter it tests. Free to read, no account needed.

Physics
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Chemistry
30
Mathematics
30

Physics — JEE Main 30 January 2023 Shift 1

Q1·Physics·Magnetic Field of CurrentSingle correct
The magnetic moments associated with two closely wound circular coils A and B of radius rA=10r_A=10rA​=10 cm and rB=20r_B=20rB​=20 cm respectively are equal if: (Where NA,IAN_A,I_ANA​,IA​ and NB,IBN_B,I_BNB​,IB​ are number of turn and current of A and B respectively)
  1. (A)4NAIA=NBIB4N_A I_A=N_B I_B4NA​IA​=NB​IB​
  2. (B)NA=2NBN_A=2N_BNA​=2NB​
  3. (C)NAIA=4NBIBN_A I_A=4N_B I_BNA​IA​=4NB​IB​
  4. (D)2NAIA=NBIB2N_A I_A=N_B I_B2NA​IA​=NB​IB​

Correct answer: (C)

Step-by-step solution →
Q2·PhysicsSingle correct
The figure represents the momentum time (p-t) curve for a particle moving along an axis under the influence of the force. Identify the regions on the graph where the magnitude of the force is maximum and minimum respectively? If (t3−t2)<t1(t_3-t_2)<t_1(t3​−t2​)<t1​
  1. (A)c and b
  2. (B)b and c
  3. (C)aaa and bbb
  4. (D)c and a

Correct answer: (A)

Step-by-step solution →
Q3·Physics·Electric PotentialSingle correct
Two isolated metallic solid spheres of radii RRR and 2R2R2R are charged such that both have same charge density σ\sigmaσ. The spheres are then connected by a thin conducting wire. If the new charge density of the bigger sphere is σ′\sigma'σ′, the ratio σ′σ\dfrac{\sigma'}{\sigma}σσ′​ is:
  1. (A)43\dfrac4334​
  2. (B)53\dfrac5335​
  3. (C)56\dfrac5665​
  4. (D)94\dfrac9449​

Correct answer: (C)

Step-by-step solution →
Q4·Physics·Geometrical OpticsSingle correct
A person has been using spectacles of power −1.0-1.0−1.0 dioptre for distant vision and a separate reading glass of power 2.02.02.0 dioptres. What is the least distance of distinct vision for this person:
  1. (A)404040 cm
  2. (B)303030 cm
  3. (C)101010 cm
  4. (D)505050 cm

Correct answer: (D)

Step-by-step solution →
Q5·PhysicsSingle correct
A small object at rest, absorbs a light pulse of power 202020 mW and duration 300300300 ns. Assuming speed of light as 3×1083\times10^83×108 m/s, the momentum of the object becomes equal to:
  1. (A)3×10−173\times10^{-17}3×10−17 kg m/s
  2. (B)2×10−172\times10^{-17}2×10−17 kg m/s
  3. (C)1×10−171\times10^{-17}1×10−17 kg m/s
  4. (D)0.5×10−170.5\times10^{-17}0.5×10−17 kg m/s

Correct answer: (B)

Step-by-step solution →
Q6·PhysicsSingle correct
Match Column-I (x-t graphs) with Column-II (v-t graphs). Choose the correct answer from the options given below:
Column-I (x-t graphs)Column-II (v-t graphs)
A.see figureI.see figure
B.see figureII.see figure
C.see figureIII.see figure
D.see figureIV.see figure
  1. (A)A-I, B-II, C-III, D-IV
  2. (B)A-II, B-III, C-IV, D-I
  3. (C)A-I, B-III, C-IV, D-II
  4. (D)A-II, B-IV, C-III, D-I

Correct answer: (D)

Step-by-step solution →
Q7·PhysicsSingle correct
The pressure (P) and temperature (T) relationship of an ideal gas obeys the equation PT2=PT^2=PT2= constant. The volume expansion coefficient of the gas will be:
  1. (A)3T2\dfrac{3}{T^2}T23​
  2. (B)3T3\dfrac{3}{T^3}T33​
  3. (C)3T23T^23T2
  4. (D)3T\dfrac{3}{T}T3​

Correct answer: (D)

Step-by-step solution →
Q8·PhysicsSingle correct
Heat is given to an ideal gas in an isothermal process. A. Internal energy of the gas will decrease. B. Internal energy of the gas will increase. C. Internal energy of the gas will not change. D. The gas will do positive work. E. The gas will do negative work. Choose the correct answer from the options given below:
  1. (A)C and D only
  2. (B)C and E only
  3. (C)A and E only
  4. (D)B and D only

Correct answer: (A)

Step-by-step solution →
Q9·PhysicsSingle correct
If the gravitational field in the space is given as (−Kr2)\left(-\dfrac{K}{r^2}\right)(−r2K​). Taking the reference point to be at r=2r=2r=2 cm with gravitational potential V=10V=10V=10 J/kg. Find the gravitational potential at r=3r=3r=3 cm in SI unit (Given that K=6K=6K=6 J cm/kg):
  1. (A)999
  2. (B)101010
  3. (C)111111
  4. (D)121212

Correct answer: (C)

Step-by-step solution →
Q10·Physics·Alternating CurrentsSingle correct
In a series LR circuit with XL=RX_L=RXL​=R, power factor is P1P_1P1​. If a capacitor of capacitance C with XC=XLX_C=X_LXC​=XL​ is added to the circuit the power factor becomes P2P_2P2​. The ratio of P1P_1P1​ to P2P_2P2​ will be:
  1. (A)1:31:31:3
  2. (B)1:21:21:2
  3. (C)1:21:\sqrt21:2​
  4. (D)1:11:11:1

Correct answer: (C)

Step-by-step solution →
Q11·PhysicsSingle correct
As per the given figure, a small ball P slides down the quadrant of a circle and hits the other ball Q of equal mass which is initially at rest. Neglecting the effect of friction and assume the collision to be elastic, the velocity of ball Q after collision will be: (g=10 m/s2)(g=10\,m/s^2)(g=10m/s2)
  1. (A)000
  2. (B)444 m/s
  3. (C)222 m/s
  4. (D)0.250.250.25 m/s

Correct answer: (C)

Step-by-step solution →
Q12·PhysicsSingle correct
A ball of mass 200200200 g rests on a vertical post of height 202020 m. A bullet of mass 101010 g, travelling in horizontal direction, hits the centre of the ball. After collision both travels independently. The ball hits the ground at a distance 303030 m and the bullet at a distance of 120120120 m from the foot of the post. The value of initial velocity of the bullet will be (if g=10 m/s2g=10\,m/s^2g=10m/s2):
  1. (A)360360360 m/s
  2. (B)400400400 m/s
  3. (C)606060 m/s
  4. (D)120120120 m/s

Correct answer: (A)

Step-by-step solution →
Q13·PhysicsSingle correct
The output waveform of the given logical circuit for the following inputs A and B as shown below, is
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (C)

Step-by-step solution →
Q14·PhysicsSingle correct
The charge flowing in a conductor changes with time as Q(t)=αt−βt2+γt3Q(t)=\alpha t-\beta t^2+\gamma t^3Q(t)=αt−βt2+γt3, where α,β\alpha,\betaα,β and γ\gammaγ are constants. Minimum value of current is:
  1. (A)α−3β2γ\alpha-\dfrac{3\beta^2}{\gamma}α−γ3β2​
  2. (B)α−γ23β\alpha-\dfrac{\gamma^2}{3\beta}α−3βγ2​
  3. (C)α−β23γ\alpha-\dfrac{\beta^2}{3\gamma}α−3γβ2​
  4. (D)β−α23γ\beta-\dfrac{\alpha^2}{3\gamma}β−3γα2​

Correct answer: (C)

Step-by-step solution →
Q15·PhysicsSingle correct
Choose the correct relationship between Poisson ratio (σ)(\sigma)(σ), bulk modulus (K) and modulus of rigidity (η)(\eta)(η) of a given solid object:
  1. (A)σ=3K+2η6K+2η\sigma=\dfrac{3K+2\eta}{6K+2\eta}σ=6K+2η3K+2η​
  2. (B)σ=3K−2η6K+2η\sigma=\dfrac{3K-2\eta}{6K+2\eta}σ=6K+2η3K−2η​
  3. (C)σ=6K+2η3K−2η\sigma=\dfrac{6K+2\eta}{3K-2\eta}σ=3K−2η6K+2η​
  4. (D)σ=6K−2η3K−2η\sigma=\dfrac{6K-2\eta}{3K-2\eta}σ=3K−2η6K−2η​

Correct answer: (B)

Step-by-step solution →
Q16·PhysicsSingle correct
Speed of an electron in Bohr's 7th7^{th}7th orbit for Hydrogen atom is 3.6×1063.6\times10^63.6×106 m/s. The corresponding speed of the electron in 3rd3^{rd}3rd orbit, in m/s is:
  1. (A)1.8×1061.8\times10^61.8×106
  2. (B)3.6×1063.6\times10^63.6×106
  3. (C)7.5×1067.5\times10^67.5×106
  4. (D)8.4×1068.4\times10^68.4×106

Correct answer: (D)

Step-by-step solution →
Q17·Physics·Magnetic Field of CurrentSingle correct
A massless square loop, of wire of resistance 10 Ω10\,\Omega10Ω, supporting a mass of 111 g, hangs vertically with one of its sides in a uniform magnetic field of 10310^3103 G, directed outwards in the shaded region. A dc voltage V is applied to the loop. For what value of V, the magnetic force will exactly balance the weight of the supporting mass of 111 g? (If sides of the loop =10=10=10 cm, g=10 ms−2g=10\,ms^{-2}g=10ms−2)
  1. (A)110\dfrac{1}{10}101​ V
  2. (B)100100100 V
  3. (C)101010 V
  4. (D)111 V

Correct answer: (C)

Step-by-step solution →
Q18·Physics·Electric Field and Coulomb's LawSingle correct
Electric field in a certain region is given by E⃗=(Ax2i^+By3j^)\vec E=\left(\dfrac{A}{x^2}\hat i+\dfrac{B}{y^3}\hat j\right)E=(x2A​i^+y3B​j^​). The SI unit of A and B are:
  1. (A)Nm3C−1; Nm2C−1Nm^3C^{-1};\ Nm^2C^{-1}Nm3C−1; Nm2C−1
  2. (B)Nm2C−1; Nm3C−1Nm^2C^{-1};\ Nm^3C^{-1}Nm2C−1; Nm3C−1
  3. (C)Nm3C; Nm2CNm^3C;\ Nm^2CNm3C; Nm2C
  4. (D)Nm2C; Nm3CNm^2C;\ Nm^3CNm2C; Nm3C

Correct answer: (B)

Step-by-step solution →
Q19·PhysicsSingle correct
The height of liquid column raised in a capillary tube of certain radius when dipped in liquid A vertically is 555 cm. If the tube is dipped in a similar manner in another liquid B of surface tension and density double the values of liquid A, the height of liquid column raised in liquid B would be _________ m
  1. (A)0.050.050.05
  2. (B)0.100.100.10
  3. (C)0.200.200.20
  4. (D)0.50.50.5

Correct answer: (A)

Step-by-step solution →
Q20·PhysicsSingle correct
A sinusoidal carrier voltage is amplitude modulated. The resultant amplitude modulated wave has maximum and minimum amplitude of 120120120 V and 808080 V respectively. The amplitude of each sideband is:
  1. (A)202020 V
  2. (B)151515 V
  3. (C)101010 V
  4. (D)555 V

Correct answer: (C)

Step-by-step solution →
Q21·Physics·OscillationsNumerical
The general displacement of a simple harmonic oscillator is x=Asin⁡ωtx=A\sin\omega tx=Asinωt. Let T be its time period. The slope of its potential energy (U) - time (t) curve will be maximum when t=Tβt=\dfrac{T}{\beta}t=βT​. The value of β\betaβ is

Correct answer: 8

Step-by-step solution →
Q22·PhysicsNumerical
A thin uniform rod of length 222 m, cross sectional area 'AAA' and density 'ddd' is rotated about an axis passing through the centre and perpendicular to its length with angular velocity ω\omegaω. If value of ω\omegaω in terms of its rotational kinetic energy EEE is αEAd\sqrt{\dfrac{\alpha E}{Ad}}AdαE​​ then value of α\alphaα is

Correct answer: 3

Step-by-step solution →
Q23·PhysicsNumerical
A horse rider covers half the distance with 555 m/s speed. The remaining part of the distance was travelled with speed 101010 m/s for half the time and with speed 151515 m/s for other half of the time. The mean speed of the rider averaged over the whole time of motion is x7\dfrac{x}{7}7x​ m/s. The value of xxx is

Correct answer: 50

Step-by-step solution →
Q24·Physics·Electromagnetic InductionNumerical
As per the given figure, if dIdt=−1\dfrac{dI}{dt}=-1dtdI​=−1 A/s then the value of VABV_{AB}VAB​ at this instant will be _________ V.

Correct answer: 30

Step-by-step solution →
Q25·PhysicsNumerical
A point source of light is placed at the centre of curvature of a hemispherical surface. The source emits a power of 242424 W. The radius of curvature of hemisphere is 101010 cm and the inner surface is completely reflecting. The force on the hemisphere due to the light falling on it is _________ ×10−8\times10^{-8}×10−8 N

Correct answer: 4

Step-by-step solution →
Q26·PhysicsNumerical
In the following circuit, the magnitude of current I1I_1I1​ is _________ A.

Correct answer: 1.5

Step-by-step solution →
Q27·PhysicsNumerical
In a screw gauge, there are 100100100 divisions on the circular scale and the main scale moves by 0.50.50.5 mm on a complete rotation of the circular scale. The zero of circular scale lies 666 divisions below the line of graduation when two studs are brought in contact with each other. When a wire is placed between the studs, 444 linear scale divisions are clearly visible while 46th46^{th}46th division the circular scale coincide with the reference line. The diameter of the wire is _________ ×10−2\times10^{-2}×10−2 mm

Correct answer: 220

Step-by-step solution →
Q28·Physics·Wave OpticsNumerical
In Young's double slit experiment, two slits S1S_1S1​ and S2S_2S2​ are 'ddd' distance apart and the separation from slits to screen is D (as shown in figure). Now if two transparent slabs of equal thickness 0.10.10.1 mm but refractive index 1.511.511.51 and 1.551.551.55 are introduced in the path of beam (λ=4000\lambda=4000λ=4000 Å) from S1S_1S1​ and S2S_2S2​ respectively. The central bright fringe spot will shift by _________ number of fringes.

Correct answer: 10

Step-by-step solution →
Q29·Physics·Capacitors and DielectricsNumerical
A capacitor of capacitance 900 μF900\,\mu F900μF is charged by a 100100100 V battery. The capacitor is disconnected from the battery and connected to another uncharged identical capacitor such that one plate of uncharged capacitor connected to positive plate and another plate of uncharged capacitor connected to negative plate of the charged capacitor. The loss of energy in this process is measured as x×10−2x\times10^{-2}x×10−2 J. The value of xxx is

Correct answer: 225

Step-by-step solution →
Q30·PhysicsNumerical
In an experiment for estimating the value of focal length of converging mirror, image of an object placed at 404040 cm from the pole of the mirror is formed at distance 120120120 cm from the pole of the mirror. These distances are measured with a modified scale in which there are 202020 small divisions in 111 cm. The value of error in measurement of focal length of the mirror is 1K\dfrac{1}{K}K1​ cm. The value of K is

Correct answer: 32

Step-by-step solution →

Chemistry — JEE Main 30 January 2023 Shift 1

Q31·ChemistrySingle correct
Lithium aluminium hydride can be prepared from the reaction of
  1. (A)LiH and Al(OH)3_33​
  2. (B)LiH and Al2_22​Cl6_66​
  3. (C)LiCl and Al2_22​H6_66​
  4. (D)LiCl, Al and H2_22​

Correct answer: (B)

Step-by-step solution →
Q32·ChemistrySingle correct
Amongst the following compounds, which one is an antacid?
  1. (A)Terfenadine
  2. (B)Meprobamate
  3. (C)Brompheniramine
  4. (D)Ranitidine

Correct answer: (D)

Step-by-step solution →
Q33·ChemistrySingle correct
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): In expensive scientific instruments, silica gel is kept in watch-glasses or in semipermeable membrane bags. Reason (R): Silica gel adsorbs moisture from air via adsorption, thus protects the instrument from water corrosion (rusting) and/or prevents malfunctioning. In the light of the above statements, choose the correct answer from the options given below:
  1. (A)Both (A) and (R) are true but (R) is not the correct explanation of (A)
  2. (B)(A) is false but (R) is true
  3. (C)Both (A) and (R) are true and (R) is the correct explanation of (A)
  4. (D)(A) is true but (R) is false

Correct answer: (C)

Step-by-step solution →
Q34·ChemistrySingle correct
Match List I (atomic number) with List II (block of periodic table). Choose the correct answer from the options given below:
List I (Atomic number)List II (Block of periodic table)
A.37I.p-block
B.78II.d-block
C.52III.f-block
D.65IV.s-block
  1. (A)A-IV, B-III, C-II, D-I
  2. (B)A-II, B-IV, C-I, D-III
  3. (C)A-IV, B-II, C-I, D-III
  4. (D)A-I, B-III, C-IV, D-II

Correct answer: (C)

Step-by-step solution →
Q35·Chemistry·Electronic Effects and StabilitySingle correct
What is the correct order of acidity of the protons marked A-D in the given compounds?
  1. (A)HC>HA>HD>HBH_C>H_A>H_D>H_BHC​>HA​>HD​>HB​
  2. (B)HD>HC>HB>HAH_D>H_C>H_B>H_AHD​>HC​>HB​>HA​
  3. (C)HC>HD>HB>HAH_C>H_D>H_B>H_AHC​>HD​>HB​>HA​
  4. (D)HC>HD>HA>HBH_C>H_D>H_A>H_BHC​>HD​>HA​>HB​

Correct answer: (D)

Step-by-step solution →
Q36·ChemistrySingle correct
Which of the following compounds would give the following set of qualitative analysis? (i) Fehling's Test: Positive (ii) Na fusion extract upon treatment with sodium nitroprusside gives a blood red colour but not prussian blue.
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (D)

Step-by-step solution →
Q37·ChemistrySingle correct
The major products 'A' and 'B', respectively formed when the alkene shown reacts with cold dilute H2SO4H_2SO_4H2​SO4​ (giving A) and with H2SO4H_2SO_4H2​SO4​ at 80 ∘C80\,^\circ C80∘C (giving B), are
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (B)

Step-by-step solution →
Q38·ChemistrySingle correct
During the qualitative analysis of SO32−SO_3^{2-}SO32−​ using dilute H2SO4H_2SO_4H2​SO4​, SO2SO_2SO2​ gas is evolved which turns K2Cr2O7K_2Cr_2O_7K2​Cr2​O7​ solution (acidified with dilute H2SO4H_2SO_4H2​SO4​):
  1. (A)green
  2. (B)blue
  3. (C)red
  4. (D)black

Correct answer: (A)

Step-by-step solution →
Q39·ChemistrySingle correct
In the wet tests for identification of various cations by precipitation, which transition element cation doesn't belong to group IV in qualitative inorganic analysis?
  1. (A)Ni2+^{2+}2+
  2. (B)Zn2+^{2+}2+
  3. (C)Co2+^{2+}2+
  4. (D)Fe3+^{3+}3+

Correct answer: (D)

Step-by-step solution →
Q40·ChemistrySingle correct
For OF2OF_2OF2​ molecule consider the following: A. Number of lone pairs on oxygen is 2. B. FOF angle is less than 104.5°104.5°104.5°. C. Oxidation state of O is −2-2−2. D. Molecule is bent 'V' shaped. E. Molecular geometry is linear. correct options are:
  1. (A)A, C, D only
  2. (B)C, D, E only
  3. (C)A, B, D only
  4. (D)B, E, A only

Correct answer: (C)

Step-by-step solution →
Q41·ChemistrySingle correct
Caprolactam when heated at high temperature in presence of water, gives
  1. (A)Nylon 6, 6
  2. (B)Nylon 6
  3. (C)Teflon
  4. (D)Dacron

Correct answer: (B)

Step-by-step solution →
Q42·Chemistry·AminesSingle correct
Benzyl isocyanide can be obtained by the reactions shown. Choose the correct answer from the options given below:
  1. (A)A and D
  2. (B)Only B
  3. (C)B and C
  4. (D)A and B

Correct answer: (D)

Step-by-step solution →
Q43·ChemistrySingle correct
Formation of photochemical smog involves the following reaction in which A, B and C are respectively. i. NO2→hνA+BNO_2\xrightarrow{h\nu}A+BNO2​hν​A+B ii. B+O2→CB+O_2\rightarrow CB+O2​→C iii. A+C→NO2+O2A+C\rightarrow NO_2+O_2A+C→NO2​+O2​. Choose the correct answer from the options given below:
  1. (A)O, N2O & NOO,\ N_2O\ \&\ NOO, N2​O & NO
  2. (B)O, NO & NO3−O,\ NO\ \&\ NO_3^-O, NO & NO3−​
  3. (C)NO, O & O3NO,\ O\ \&\ O_3NO, O & O3​
  4. (D)N, O2 & O3N,\ O_2\ \&\ O_3N, O2​ & O3​

Correct answer: (C)

Step-by-step solution →
Q44·ChemistrySingle correct
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): Ketoses give Seliwanoff's test faster than Aldoses. Reason (R): Ketoses undergo β\betaβ-elimination followed by formation of furfural. In the light of the above statements, choose the correct answer from the options given below:
  1. (A)(A) is false but (R) is true
  2. (B)(A) is true but (R) is false
  3. (C)Both (A) and (R) are true but (R) is not the correct explanation of (A)
  4. (D)Both (A) and (R) are true and (R) is the correct explanation of (A)

Correct answer: (B)

Step-by-step solution →
Q45·ChemistrySingle correct
Match List I (molecules/ions) with List II (number of lone pairs of electrons on the central atom). Choose the correct answer from the options given below:
List I (molecules/ions)List II (No. of lone pairs of e⁻ on central atom)
A.IF7_77​I.Three
B.ICl4−_4^-4−​II.One
C.XeF6_66​III.Two
D.XeF2_22​IV.Zero
  1. (A)A-II, B-III, C-II, D-I
  2. (B)A-II, B-I, C-IV, D-III
  3. (C)A-IV, B-I, C-II, D-III
  4. (D)A-IV, B-III, C-II, D-I

Correct answer: (D)

Step-by-step solution →
Q46·ChemistrySingle correct
To inhibit the growth of tumours, identify the compounds used from the following: A. EDTA B. Coordination Compounds of Pt C. D-Penicillamine D. Cis-Platin. Choose the correct answer from the option given below:
  1. (A)B and D Only
  2. (B)C and D Only
  3. (C)A and C Only
  4. (D)A and B Only

Correct answer: (A)

Step-by-step solution →
Q47·ChemistrySingle correct
The alkaline earth metal sulphate(s) which are readily soluble in water is/are: A. BeSO4_44​ B. MgSO4_44​ C. CaSO4_44​ D. SrSO4_44​ E. BaSO4_44​. Choose the correct answer from the options given below:
  1. (A)B Only
  2. (B)A and B
  3. (C)B and C
  4. (D)A only

Correct answer: (B)

Step-by-step solution →
Q48·ChemistrySingle correct
Which of the following is correct order of ligand field strength?
  1. (A)CO < en < NH3_33​ < C2_22​O42−_4^{2-}42−​ < S2−^{2-}2−
  2. (B)en < CO < S2−^{2-}2− < C2_22​O42−_4^{2-}42−​
  3. (C)S2−^{2-}2− < C2_22​O42−_4^{2-}42−​ < NH3_33​ < en < CO
  4. (D)S2−^{2-}2− < NH3_33​ < en < CO < C2_22​O42−_4^{2-}42−​

Correct answer: (C)

Step-by-step solution →
Q49·ChemistrySingle correct
Match List I (reactions) with List II (name of reaction). Choose the correct answer from the options given below:
List I (Reactions)List II (Name of reaction)
A.see figureI.Fittig reaction
B.see figureII.Wurtz Fittig reaction
C.see figureIII.Finkelstein reaction
D.see figureIV.Sandmeyer reaction
  1. (A)A-II, B-I, C-IV, D-III
  2. (B)A-IV, B-II, C-III, D-I
  3. (C)A-III, B-II, C-IV, D-I
  4. (D)A-II, B-I, C-III, D-IV

Correct answer: (A)

Step-by-step solution →
Q50·ChemistrySingle correct
In the extraction of copper, its sulphide ore is heated in a reverberatory furnace after mixing with silica to:
  1. (A)remove FeO as FeSiO3_33​
  2. (B)decrease the temperature needed for roasting of Cu2_22​S
  3. (C)separate CuO as CuSiO3_33​
  4. (D)remove calcium as CaSiO3_33​

Correct answer: (A)

Step-by-step solution →
Q51·ChemistryNumerical
600600600 mL of 0.010.010.01 M HCl is mixed with 400400400 mL of 0.010.010.01 M H2_22​SO4_44​. The pH of the mixture is _________ ×10−2\times10^{-2}×10−2. (Nearest integer) [Given log 2 = 0.30, log 3 = 0.48, log 5 = 0.69, log 7 = 0.84, log 11 = 1.04]

Correct answer: 186

Step-by-step solution →
Q52·ChemistryNumerical
The energy of one mole of photons of radiation of frequency 2×10122\times10^{12}2×1012 Hz in J mol−1^{-1}−1 is _________. (Nearest integer) [Given: h = 6.626×10−346.626\times10^{-34}6.626×10−34 Js, NA_AA​ = 6.022×10236.022\times10^{23}6.022×1023 mol−1^{-1}−1]

Correct answer: 798

Step-by-step solution →
Q53·ChemistryNumerical
Consider the cell Pt(s)_{(s)}(s)​|H2_22​(g, 1 atm)|H+^++(aq, 1M)||Fe3+^{3+}3+(aq), Fe2+^{2+}2+(aq)|Pt(s). When the potential of the cell is 0.7120.7120.712 V at 298298298 K, the ratio [Fe2+^{2+}2+]/[Fe3+^{3+}3+] is _________ (Nearest integer). Given: Fe3+^{3+}3+ + e−^-− = Fe2+^{2+}2+, E0^00 = 0.771 V, 2.303RTF=0.06\dfrac{2.303RT}{F}=0.06F2.303RT​=0.06 V

Correct answer: 10

Step-by-step solution →
Q54·ChemistryNumerical
The number of electrons involved in the reduction of permanganate to manganese dioxide in acidic medium is

Correct answer: 3

Step-by-step solution →
Q55·ChemistryNumerical
A 300300300 mL bottle of soft drink has 0.20.20.2 M CO2_22​ dissolved in it. Assuming CO2_22​ behaves as an ideal gas, the volume of the dissolved CO2_22​ at STP is _________ mL. (Nearest integer) Given: At STP, molar volume of an ideal gas is 22.722.722.7 L mol−1^{-1}−1

Correct answer: 1362

Step-by-step solution →
Q56·Chemistry·Reaction MechanismNumerical
A trisubstituted compound 'A', C10_{10}10​H12_{12}12​O2_22​ gives neutral FeCl3_33​ test positive. Treatment of compound 'A' with NaOH and CH3_33​Br gives C11_{11}11​H14_{14}14​O2_22​, with hydroiodic acid gives methyl iodide and with hot conc. NaOH gives a compound B, C10_{10}10​H12_{12}12​O2_22​. Compound 'A' also decolourises alkaline KMnO4_44​. The number of π\piπ bond/s present in the compound 'A' is

Correct answer: 4

Step-by-step solution →
Q57·ChemistryNumerical
If compound A reacts with B following first order kinetics with rate constant 2.011×10−32.011\times10^{-3}2.011×10−3 s−1^{-1}−1. The time taken by A (in seconds) to reduce from 777 g to 222 g will be _________ (Nearest Integer) [log 5 = 0.698, log 7 = 0.845, log 2 = 0.301]

Correct answer: 623

Step-by-step solution →
Q58·ChemistryNumerical
A solution containing 222 g of a non-volatile solute in 202020 g of water boils at 373.52373.52373.52 K. The molecular mass of the solute is _________ g mol−1^{-1}−1. (Nearest integer) Given, Water boils at 373373373 K, Kb_bb​ for water = 0.520.520.52 K kg mol−1^{-1}−1

Correct answer: 100

Step-by-step solution →
Q59·ChemistryNumerical
When 222 litre of ideal gas expands isothermally into vacuum to a total volume of 666 litre, the change in internal energy is _________ J. (Nearest integer)

Correct answer: 0

Step-by-step solution →
Q60·ChemistryNumerical
Some amount of dichloromethane (CH2_22​Cl2_22​) is added to 671.141671.141671.141 mL of chloroform (CHCl3_33​) to prepare 2.6×10−32.6\times10^{-3}2.6×10−3 M solution of CH2_22​Cl2_22​ (DCM). The concentration of DCM is _________ ppm (by mass). Given: atomic mass C = 12, H = 1, Cl = 35.5, density of CHCl3_33​ = 1.49 g cm−3^{-3}−3

Correct answer: 148.322

Step-by-step solution →

Mathematics — JEE Main 30 January 2023 Shift 1

Q61·MathematicsSingle correct
A straight line cuts off the intercepts OA=aOA=aOA=a and OB=bOB=bOB=b on the positive directions of x-axis and y-axis respectively. If the perpendicular from origin O to this line makes an angle of π6\dfrac{\pi}{6}6π​ with positive direction of y-axis and the area of △OAB\triangle OAB△OAB is 9833\dfrac{98}{3}\sqrt3398​3​, then a2−b2a^2-b^2a2−b2 is equal to:
  1. (A)3923\dfrac{392}{3}3392​
  2. (B)1963\dfrac{196}{3}3196​
  3. (C)989898
  4. (D)196196196

Correct answer: (A)

Step-by-step solution →
Q62·MathematicsSingle correct
The minimum number of elements that must be added to the relation R={(a,b),(b,c)}R=\{(a,b),(b,c)\}R={(a,b),(b,c)} on the set {a,b,c}\{a,b,c\}{a,b,c} so that it becomes symmetric and transitive is:
  1. (A)333
  2. (B)444
  3. (C)555
  4. (D)777

Correct answer: (D)

Step-by-step solution →
Q63·MathematicsSingle correct
If an unbiased die, marked with −2,−1,0,1,2,3-2,-1,0,1,2,3−2,−1,0,1,2,3 on its faces, is thrown five times, then the probability that the product of the outcomes is positive, is:
  1. (A)8812592\dfrac{881}{2592}2592881​
  2. (B)27288\dfrac{27}{288}28827​
  3. (C)4402592\dfrac{440}{2592}2592440​
  4. (D)5212592\dfrac{521}{2592}2592521​

Correct answer: (D)

Step-by-step solution →
Q64·MathematicsSingle correct
If a⃗,b⃗,c⃗\vec a,\vec b,\vec ca,b,c are three non-zero vectors and n^\hat nn^ is a unit vector perpendicular to c⃗\vec cc such that a⃗=αb⃗−n^\vec a=\alpha\vec b-\hat na=αb−n^, (α≠0)(\alpha\neq0)(α=0) and b⃗⋅c⃗=12\vec b\cdot\vec c=12b⋅c=12, then ∣c⃗×(a⃗×b⃗)∣|\vec c\times(\vec a\times\vec b)|∣c×(a×b)∣ is equal to:
  1. (A)999
  2. (B)151515
  3. (C)666
  4. (D)121212

Correct answer: (D)

Step-by-step solution →
Q65·MathematicsSingle correct
Among the statements: (S1) ((p∨q)⇒r)⇔(p⇒r)((p\vee q)\Rightarrow r)\Leftrightarrow(p\Rightarrow r)((p∨q)⇒r)⇔(p⇒r) and (S2) ((p∨q)⇒r)⇔((p⇒r)∨(q⇒r))((p\vee q)\Rightarrow r)\Leftrightarrow((p\Rightarrow r)\vee(q\Rightarrow r))((p∨q)⇒r)⇔((p⇒r)∨(q⇒r))
  1. (A)only (S2) is a tautology
  2. (B)only (S1) is a tautology
  3. (C)neither (S1) nor (S2) is a tautology
  4. (D)both (S1) and (S2) are tautologies

Correct answer: (C)

Step-by-step solution →
Q66·MathematicsSingle correct
If P(h,k)P(h,k)P(h,k) be a point on the parabola x=4y2x=4y^2x=4y2, which is nearest to the point Q(0,33)Q(0,33)Q(0,33), then the distance of PPP from the directrix of the parabola y2=4(x+y)y^2=4(x+y)y2=4(x+y) is equal to:
  1. (A)222
  2. (B)666
  3. (C)888
  4. (D)444

Correct answer: (B)

Step-by-step solution →
Q67·MathematicsSingle correct
Let y=x+2y=x+2y=x+2, 4y=3x+64y=3x+64y=3x+6 and 3y=4x+13y=4x+13y=4x+1 be three tangent lines to the circle (x−h)2+(y−k)2=r2(x-h)^2+(y-k)^2=r^2(x−h)2+(y−k)2=r2. Then h+kh+kh+k is equal to:
  1. (A)5(1+2)5(1+\sqrt2)5(1+2​)
  2. (B)525\sqrt252​
  3. (C)666
  4. (D)555

Correct answer: (D)

Step-by-step solution →
Q68·MathematicsSingle correct
The number of points on the curve y=54x5−135x4−70x3+180x2+210xy=54x^5-135x^4-70x^3+180x^2+210xy=54x5−135x4−70x3+180x2+210x at which the normal lines are parallel to x+90y+2=0x+90y+2=0x+90y+2=0 is:
  1. (A)444
  2. (B)222
  3. (C)000
  4. (D)333

Correct answer: (A)

Step-by-step solution →
Q69·MathematicsSingle correct
If an=−24n2−16n+15a_n=\dfrac{-2}{4n^2-16n+15}an​=4n2−16n+15−2​, then a1+a2+⋯+a25a_1+a_2+\cdots+a_{25}a1​+a2​+⋯+a25​ is equal to:
  1. (A)52147\dfrac{52}{147}14752​
  2. (B)49138\dfrac{49}{138}13849​
  3. (C)50141\dfrac{50}{141}14150​
  4. (D)51144\dfrac{51}{144}14451​

Correct answer: (C)

Step-by-step solution →
Q70·MathematicsSingle correct
If tan⁡15°+1tan⁡75°+1tan⁡105°+tan⁡195°=2a\tan15°+\dfrac{1}{\tan75°}+\dfrac{1}{\tan105°}+\tan195°=2atan15°+tan75°1​+tan105°1​+tan195°=2a, then the value of (a+1a)\left(a+\dfrac1a\right)(a+a1​) is:
  1. (A)222
  2. (B)4−234-2\sqrt34−23​
  3. (C)5−3235-\dfrac32\sqrt35−23​3​
  4. (D)444

Correct answer: (D)

Step-by-step solution →
Q71·MathematicsSingle correct
If the solution of the equation log⁡cos⁡xcot⁡x+4log⁡sin⁡xtan⁡x=1\log_{\cos x}\cot x+4\log_{\sin x}\tan x=1logcosx​cotx+4logsinx​tanx=1, x∈(0,π2)x\in\left(0,\dfrac{\pi}{2}\right)x∈(0,2π​), is sin⁡−1(α+β2)\sin^{-1}\left(\dfrac{\alpha+\sqrt\beta}{2}\right)sin−1(2α+β​​), where α\alphaα and β\betaβ are integers, then α+β\alpha+\betaα+β is equal to:
  1. (A)555
  2. (B)666
  3. (C)444
  4. (D)333

Correct answer: (C)

Step-by-step solution →
Q72·MathematicsSingle correct
Let the system of linear equations x+y+kz=2x+y+kz=2x+y+kz=2, 2x+3y−z=12x+3y-z=12x+3y−z=1, 3x+4y+2z=k3x+4y+2z=k3x+4y+2z=k have infinitely many solutions. Then the system (k+1)x+(2k−1)y=7(k+1)x+(2k-1)y=7(k+1)x+(2k−1)y=7, (2k+1)x+(k+5)y=10(2k+1)x+(k+5)y=10(2k+1)x+(k+5)y=10 has:
  1. (A)infinitely many solutions
  2. (B)unique solution satisfying x−y=1x-y=1x−y=1
  3. (C)unique solution satisfying x+y=1x+y=1x+y=1
  4. (D)no solution

Correct answer: (C)

Step-by-step solution →
Q73·MathematicsSingle correct
The line l1l_1l1​ passes through the point (2,6,2)(2,6,2)(2,6,2) and is perpendicular to the plane 2x+y−2z=102x+y-2z=102x+y−2z=10. Then the shortest distance between the line l1l_1l1​ and the line x+12=y+4−3=z2\dfrac{x+1}{2}=\dfrac{y+4}{-3}=\dfrac{z}{2}2x+1​=−3y+4​=2z​ is:
  1. (A)113\dfrac{11}{3}311​
  2. (B)193\dfrac{19}{3}319​
  3. (C)777
  4. (D)999

Correct answer: (D)

Step-by-step solution →
Q74·MathematicsSingle correct
Let A=(mnpq)A=\begin{pmatrix}m&n\\p&q\end{pmatrix}A=(mp​nq​), d=∣A∣≠0d=|A|\neq0d=∣A∣=0 and ∣A−d(Adj A)∣=0|A-d(\mathrm{Adj}\,A)|=0∣A−d(AdjA)∣=0. Then
  1. (A)1+d2=m2+q21+d^2=m^2+q^21+d2=m2+q2
  2. (B)1+d2=(m+q)21+d^2=(m+q)^21+d2=(m+q)2
  3. (C)(1+d)2=m2+q2(1+d)^2=m^2+q^2(1+d)2=m2+q2
  4. (D)(1+d)2=(m+q)2(1+d)^2=(m+q)^2(1+d)2=(m+q)2

Correct answer: (D)

Step-by-step solution →
Q75·MathematicsSingle correct
If [t][t][t] denotes the greatest integer ≤t\leq t≤t, then the value of 3(e−1)e∫12x2e[x3]+[x4] dx\dfrac{3(e-1)}{e}\int_1^2 x^2 e^{[x^3]+[x^4]}\,dxe3(e−1)​∫12​x2e[x3]+[x4]dx is:
  1. (A)e8−1e^8-1e8−1
  2. (B)e7−1e^7-1e7−1
  3. (C)e8−ee^8-ee8−e
  4. (D)e9−ee^9-ee9−e

Correct answer: (C)

Step-by-step solution →
Q76·MathematicsSingle correct
Let a unit vector OP→\overrightarrow{OP}OP make angles α,β,γ\alpha,\beta,\gammaα,β,γ with the positive directions of the co-ordinate axes OX, OY, OZ respectively, where β∈(0,π2)\beta\in\left(0,\dfrac{\pi}{2}\right)β∈(0,2π​). If OP→\overrightarrow{OP}OP is perpendicular to the plane through points (1,2,3),(2,3,4)(1,2,3),(2,3,4)(1,2,3),(2,3,4) and (1,5,7)(1,5,7)(1,5,7), then which one of the following is true?
  1. (A)α∈(0,π2)\alpha\in\left(0,\dfrac{\pi}{2}\right)α∈(0,2π​) and γ∈(0,π2)\gamma\in\left(0,\dfrac{\pi}{2}\right)γ∈(0,2π​)
  2. (B)α∈(0,π2)\alpha\in\left(0,\dfrac{\pi}{2}\right)α∈(0,2π​) and γ∈(π2,π)\gamma\in\left(\dfrac{\pi}{2},\pi\right)γ∈(2π​,π)
  3. (C)α∈(π2,π)\alpha\in\left(\dfrac{\pi}{2},\pi\right)α∈(2π​,π) and γ∈(π2,π)\gamma\in\left(\dfrac{\pi}{2},\pi\right)γ∈(2π​,π)
  4. (D)α∈(π2,π)\alpha\in\left(\dfrac{\pi}{2},\pi\right)α∈(2π​,π) and γ∈(0,π2)\gamma\in\left(0,\dfrac{\pi}{2}\right)γ∈(0,2π​)

Correct answer: (C)

Step-by-step solution →
Q77·MathematicsSingle correct
The coefficient of x301x^{301}x301 in (1+x)500+x(1+x)499+x2(1+x)498+⋯+x500(1+x)^{500}+x(1+x)^{499}+x^2(1+x)^{498}+\cdots+x^{500}(1+x)500+x(1+x)499+x2(1+x)498+⋯+x500 is:
  1. (A)500C300^{500}C_{300}500C300​
  2. (B)501C200^{501}C_{200}501C200​
  3. (C)501C302^{501}C_{302}501C302​
  4. (D)500C301^{500}C_{301}500C301​

Correct answer: (B)

Step-by-step solution →
Q78·MathematicsSingle correct
Let the solution curve y=y(x)y=y(x)y=y(x) of the differential equation dydx−3x5tan⁡−1(x3)(1+x6)3/2y=2xexp⁡{x3−tan⁡−1x31+x6}\dfrac{dy}{dx}-\dfrac{3x^5\tan^{-1}(x^3)}{(1+x^6)^{3/2}}y=2x\exp\left\{\dfrac{x^3-\tan^{-1}x^3}{\sqrt{1+x^6}}\right\}dxdy​−(1+x6)3/23x5tan−1(x3)​y=2xexp{1+x6​x3−tan−1x3​} pass through the origin. Then y(1)y(1)y(1) is equal to:
  1. (A)exp⁡(4+π42)\exp\left(\dfrac{4+\pi}{4\sqrt2}\right)exp(42​4+π​)
  2. (B)exp⁡(1−π42)\exp\left(\dfrac{1-\pi}{4\sqrt2}\right)exp(42​1−π​)
  3. (C)exp⁡(π−442)\exp\left(\dfrac{\pi-4}{4\sqrt2}\right)exp(42​π−4​)
  4. (D)exp⁡(4−π42)\exp\left(\dfrac{4-\pi}{4\sqrt2}\right)exp(42​4−π​)

Correct answer: (D)

Step-by-step solution →
Q79·MathematicsSingle correct
If the coefficient of x15x^{15}x15 in the expansion of (ax3+1bx1/3)15\left(ax^3+\dfrac{1}{bx^{1/3}}\right)^{15}(ax3+bx1/31​)15 is equal to the coefficient of x−15x^{-15}x−15 in the expansion of (ax1/3−1bx3)15\left(ax^{1/3}-\dfrac{1}{bx^3}\right)^{15}(ax1/3−bx31​)15, where aaa and bbb are positive real numbers, then for each such ordered pair (a,b)(a,b)(a,b):
  1. (A)a=3a=3a=3
  2. (B)ab=1ab=1ab=1
  3. (C)a=ba=ba=b
  4. (D)a=3ba=3ba=3b

Correct answer: (B)

Step-by-step solution →
Q80·MathematicsSingle correct
Suppose f:R→(0,∞)f:\mathbb{R}\to(0,\infty)f:R→(0,∞) be a differentiable function such that 5f(x+y)=f(x)⋅f(y)5f(x+y)=f(x)\cdot f(y)5f(x+y)=f(x)⋅f(y), ∀x,y∈R\forall x,y\in\mathbb{R}∀x,y∈R. If f(3)=320f(3)=320f(3)=320, then ∑n=05f(n)\sum_{n=0}^{5}f(n)∑n=05​f(n) is equal to:
  1. (A)687568756875
  2. (B)652565256525
  3. (C)682568256825
  4. (D)657565756575

Correct answer: (C)

Step-by-step solution →
Q81·MathematicsNumerical
Let z=1+iz=1+iz=1+i and z1=1+izˉzˉ(1−z)+1zz_1=\dfrac{1+i\bar z}{\bar z(1-z)+\dfrac1z}z1​=zˉ(1−z)+z1​1+izˉ​. Then 12πarg⁡(z1)\dfrac{12}{\pi}\arg(z_1)π12​arg(z1​) is equal to

Correct answer: 9

Step-by-step solution →
Q82·MathematicsNumerical
If λ1<λ2\lambda_1<\lambda_2λ1​<λ2​ are two values of λ\lambdaλ such that the angle between the planes P1:r⃗⋅(3i^−5j^+k^)=7P_1:\vec r\cdot(3\hat i-5\hat j+\hat k)=7P1​:r⋅(3i^−5j^​+k^)=7 and P2:r⃗⋅(λi^+j^−3k^)=9P_2:\vec r\cdot(\lambda\hat i+\hat j-3\hat k)=9P2​:r⋅(λi^+j^​−3k^)=9 is sin⁡−1(265)\sin^{-1}\left(\dfrac{2\sqrt6}{5}\right)sin−1(526​​), then the square of the length of perpendicular from the point (38λ1,10λ2,2)(38\lambda_1,10\lambda_2,2)(38λ1​,10λ2​,2) to the plane P1P_1P1​ is

Correct answer: 315

Step-by-step solution →
Q83·MathematicsNumerical
Let α\alphaα be the area of the larger region bounded by the curve y2=8xy^2=8xy2=8x and the lines y=xy=xy=x and x=2x=2x=2, which lies in the first quadrant. Then the value of 3α3\alpha3α is equal to

Correct answer: 22

Step-by-step solution →
Q84·MathematicsNumerical
Let ∑n=0∞n3((2n)!)+(2n−1)(n!)(n!)((2n)!)=ae+be+c\sum_{n=0}^{\infty}\dfrac{n^3((2n)!)+(2n-1)(n!)}{(n!)((2n)!)}=ae+\dfrac{b}{e}+c∑n=0∞​(n!)((2n)!)n3((2n)!)+(2n−1)(n!)​=ae+eb​+c, where a,b,c∈Za,b,c\in\mathbb{Z}a,b,c∈Z and e=∑n=0∞1n!e=\sum_{n=0}^{\infty}\dfrac{1}{n!}e=∑n=0∞​n!1​. Then a2−b+ca^2-b+ca2−b+c is equal to

Correct answer: 26

Step-by-step solution →
Q85·MathematicsNumerical
If the equation of the plane passing through the point (1,1,2)(1,1,2)(1,1,2) and perpendicular to the line x−3y+2z−1=0=4x−y+zx-3y+2z-1=0=4x-y+zx−3y+2z−1=0=4x−y+z is Ax+By+Cz=1Ax+By+Cz=1Ax+By+Cz=1, then 140(C−B+A)140(C-B+A)140(C−B+A) is equal to

Correct answer: 15

Step-by-step solution →
Q86·MathematicsNumerical
Number of 4-digit numbers (the repetition of digits is allowed) which are made using the digits 1,2,31,2,31,2,3 and 555, and are divisible by 151515, is equal to

Correct answer: 21

Step-by-step solution →
Q87·MathematicsNumerical
Let f1(x)=3x+22x+3f^1(x)=\dfrac{3x+2}{2x+3}f1(x)=2x+33x+2​, x∈R−{−32}x\in\mathbb{R}-\left\{\dfrac{-3}{2}\right\}x∈R−{2−3​}. For n≥2n\geq2n≥2, define fn(x)=f1 of fn−1(x)f^n(x)=f^1\text{ of }f^{n-1}(x)fn(x)=f1 of fn−1(x). If f5(x)=ax+bbx+af^5(x)=\dfrac{ax+b}{bx+a}f5(x)=bx+aax+b​, gcd⁡(a,b)=1\gcd(a,b)=1gcd(a,b)=1, then a+ba+ba+b is equal to

Correct answer: 3125

Step-by-step solution →
Q88·MathematicsNumerical
The mean and variance of 777 observations are 888 and 161616 respectively. If one observation 141414 is omitted and aaa and bbb are respectively mean and variance of remaining 666 observation, then a+3b−5a+3b-5a+3b−5 is equal to

Correct answer: 37

Step-by-step solution →
Q89·MathematicsNumerical
Let S={1,2,3,4,5,6}S=\{1,2,3,4,5,6\}S={1,2,3,4,5,6}. Then the number of one-one functions f:S→P(S)f:S\to P(S)f:S→P(S), where P(S)P(S)P(S) denote the power set of S, such that f(n)⊂f(m)f(n)\subset f(m)f(n)⊂f(m) where n<mn<mn<m is

Correct answer: 3240

Step-by-step solution →
Q90·Mathematics·Limits and ContinuityNumerical
lim⁡x→048x4∫0xt3t4+1 dt\lim_{x\to0}\dfrac{48}{x^4}\int_0^x\dfrac{t^3}{t^4+1}\,dtlimx→0​x448​∫0x​t4+1t3​dt is equal to

Correct answer: 12

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
  • Thermodynamics 154/186
  • Equilibrium 163/186
  • Solutions 158/186
  • Laws of Motion 130/186
  • Chemical Thermodynamics 165/186
  • 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
  • Organic Compounds Containing Halogens 109/186
  • Straight Lines 114/186
  • Capacitors and Dielectrics 115/186
  • Atoms 112/186
  • Classification of Elements and Periodicity in Properties 113/186
  • Parabola 101/186
  • Statistics 118/186
  • Isolation of Metals 106/186
  • s-Block Elements 88/186
  • Surface Chemistry 98/186
  • Environmental Chemistry 83/186
  • Hydrogen 81/186
  • Electronic Effects and Stability 74/186
  • Experimental Skills 68/186
  • Electric Potential 63/186
  • Polymers 64/186
  • Principles of Qualitative Analysis 58/186
  • Chemistry in Everyday Life 60/186
  • Reaction Mechanism 29/186
← 29 Jan Shift 2 2023All papers30 Jan Shift 2 2023 →

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