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JEE Main 27 January 2024 Shift 2 Question Paper with Answers

27 January 2024 · January session · 89 questions

89 of the 90 questions from the JEE Main 27 January 2024 Shift 2 paper, each with its correct answer and tagged to the chapter it tests. Free to read, no account needed.

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

Physics
29
Chemistry
30
Mathematics
30

Physics — JEE Main 27 January 2024 Shift 2

Q1·Physics·Kinetic Theory of GasesSingle correct
The equation of state of a real gas is given by (P+aV2)(V−b)=RT\left(P+\dfrac{a}{V^2}\right)(V-b)=RT(P+V2a​)(V−b)=RT, where P, V and T are pressure, volume and temperature respectively and R is the universal gas constant. The dimensions of ab2\dfrac{a}{b^2}b2a​ is similar to that of :
  1. (A)PV
  2. (B)P
  3. (C)RT
  4. (D)R

Correct answer: (B)

Step-by-step solution →
Q2·Physics·Current ElectricitySingle correct
Wheatstone bridge principle is used to measure the specific resistance (Sl)(S_l)(Sl​) of given wire, having length L, radius r. If X is the resistance of wire, then specific resistance is Sl=X(πr2L)S_l=X\left(\dfrac{\pi r^2}{L}\right)Sl​=X(Lπr2​). If the length of the wire gets doubled then the value of specific resistance will be :
  1. (A)Sl4\dfrac{S_l}{4}4Sl​​
  2. (B)2Sl2S_l2Sl​
  3. (C)Sl2\dfrac{S_l}{2}2Sl​​
  4. (D)SlS_lSl​

Correct answer: (D)

Step-by-step solution →
Q3·Physics·GravitationSingle correct
Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : The angular speed of the moon in its orbit about the earth is more than the angular speed of the earth in its orbit about the sun. Reason (R) : The moon takes less time to move around the earth than the time taken by the earth to move around the sun. In the light of the above statements, choose the most appropriate answer from the options given below : (1) (A) is correct but (R) is not correct (2) Both (A) and (R) are correct and (R) is the correct explanation of (A) (3) (A) is not correct but (R) is correct (4) (A) is correct but (R) is not the correct explanation of (A)
  1. (A)(A) is correct but (R) is not correct
  2. (B)Both (A) and (R) are correct and (R) is the correct explanation of (A)
  3. (C)(A) is not correct but (R) is correct
  4. (D)(A) is correct but (R) is not the correct explanation of (A)

Correct answer: (B)

Step-by-step solution →
Q4·Physics·Laws of MotionSingle correct
Given below are two statements : Statement (I) : The limiting force of static friction depends on the area of contact and independent of materials. Statement (II) : The limiting force of kinetic friction is independent of the area of contact and depends on materials. 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)Statement I is incorrect but Statement II is correct
  3. (C)Both Statement I and Statement II are incorrect
  4. (D)Both Statement I and Statement II are correct

Correct answer: (B)

Step-by-step solution →
Q5·Physics·Electronic DevicesSingle correct
The truth table of the given circuit diagram is :
  1. (A)ABY000010100111\begin{array}{ccc} A&B&Y\\ 0&0&0\\ 0&1&0\\ 1&0&0\\ 1&1&1\end{array}A0011​B0101​Y0001​
  2. (B)ABY000011101110\begin{array}{ccc} A&B&Y\\ 0&0&0\\ 0&1&1\\ 1&0&1\\ 1&1&0\end{array}A0011​B0101​Y0110​
  3. (C)ABY000011101111\begin{array}{ccc} A&B&Y\\ 0&0&0\\ 0&1&1\\ 1&0&1\\ 1&1&1\end{array}A0011​B0101​Y0111​
  4. (D)ABY001011101110\begin{array}{ccc} A&B&Y\\ 0&0&1\\ 0&1&1\\ 1&0&1\\ 1&1&0\end{array}A0011​B0101​Y1110​

Correct answer: (B)

Step-by-step solution →
Q6·Physics·Magnetic Field of CurrentSingle correct
A current of 200 μ\muμA deflects the coil of a moving coil galvanometer through 60∘60^\circ60∘. The current to cause deflection through π10\dfrac{\pi}{10}10π​ radian is :
  1. (A)60 μ60\,\mu60μA
  2. (B)120 μ120\,\mu120μA
  3. (C)30 μ30\,\mu30μA
  4. (D)180 μ180\,\mu180μA

Correct answer: (A)

Step-by-step solution →
Q7·Physics·Atoms and NucleiSingle correct
The atomic mass of 6C12_6\text{C}^{12}6​C12 is 12.000000 u and that of 6C13_6\text{C}^{13}6​C13 is 13.003354 u. The required energy to remove a neutron from 6C13_6\text{C}^{13}6​C13, if mass of neutron is 1.008665 u, will be :
  1. (A)62.5 MeV
  2. (B)6.25 MeV
  3. (C)4.95 MeV
  4. (D)49.5 MeV

Correct answer: (C)

Step-by-step solution →
Q8·Physics·OscillationsSingle correct
A ball suspended by a thread swings in a vertical plane so that its magnitude of acceleration in the extreme position and lowest position are equal. The angle θ\thetaθ of thread deflection in the extreme position will be :
  1. (A)tan⁡−1(2)\tan^{-1}(\sqrt2)tan−1(2​)
  2. (B)2tan⁡−1(12)2\tan^{-1}\left(\dfrac12\right)2tan−1(21​)
  3. (C)tan⁡−1(12)\tan^{-1}\left(\dfrac12\right)tan−1(21​)
  4. (D)2tan⁡−1(15)2\tan^{-1}\left(\dfrac{1}{\sqrt5}\right)2tan−1(5​1​)

Correct answer: (B)

Step-by-step solution →
Q9·Physics·Current ElectricitySingle correct
Three voltmeters, all having different internal resistances are joined as shown in figure. When some potential difference is applied across A and B, their readings are V1V_1V1​, V2V_2V2​ and V3V_3V3​. Choose the correct option.
  1. (A)V1=V2V_1=V_2V1​=V2​
  2. (B)V1=V3−V2V_1=V_3-V_2V1​=V3​−V2​
  3. (C)V1+V2>V3V_1+V_2>V_3V1​+V2​>V3​
  4. (D)V1+V2=V3V_1+V_2=V_3V1​+V2​=V3​

Correct answer: (D)

Step-by-step solution →
Q10·Physics·Kinetic Theory of GasesSingle correct
The total kinetic energy of 1 mole of oxygen at 27∘27^\circ27∘C is : [Use universal gas constant (R) = 8.31 J/mole K]
  1. (A)6845.5 J
  2. (B)5942.0 J
  3. (C)6232.5 J
  4. (D)5670.5 J

Correct answer: (C)

Step-by-step solution →
Q11·Physics·Experimental SkillsSingle correct
Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : In Vernier calliper if positive zero error exists, then while taking measurements, the reading taken will be more than the actual reading. Reason (R) : The zero error in Vernier Calliper might have happened due to manufacturing defect or due to rough handling. In the light of the above statements, choose the correct 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 true but (R) is false
  4. (D)(A) is false but (R) is true

Correct answer: (B)

Step-by-step solution →
Q12·Physics·Alternating CurrentsSingle correct
Primary side of a transformer is connected to 230 V, 50 Hz supply. Turns ratio of primary to secondary winding is 10:110:110:1. Load resistance connected to secondary side is 46 Ω\OmegaΩ. The power consumed in it is :
  1. (A)12.5 W
  2. (B)10.0 W
  3. (C)11.5 W
  4. (D)12.0 W

Correct answer: (C)

Step-by-step solution →
Q13·Physics·ThermodynamicsSingle correct
During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its absolute temperature. The ratio CPCV\dfrac{C_P}{C_V}CV​CP​​ for the gas is :
  1. (A)53\dfrac5335​
  2. (B)73\dfrac7337​
  3. (C)75\dfrac7557​
  4. (D)97\dfrac9779​

Correct answer: (B)

Step-by-step solution →
Q14·Physics·Atoms and NucleiSingle correct
The threshold frequency of a metal with work function 6.63 eV is :
  1. (A)16×101516\times10^{15}16×1015 Hz
  2. (B)16×101216\times10^{12}16×1012 Hz
  3. (C)1.6×10121.6\times10^{12}1.6×1012 Hz
  4. (D)1.6×10151.6\times10^{15}1.6×1015 Hz

Correct answer: (D)

Step-by-step solution →
Q15·Physics·Properties of Solids and LiquidsSingle correct
Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : The property of body, by virtue of which it tends to regain its original shape when the external force is removed, is Elasticity. Reason (R) : The restoring force depends on the bonded inter atomic and inter molecular force of solid. In the light of the above statements, choose the correct answer from the options given below :
  1. (A)(A) is true but (R) is false
  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)Both (A) and (R) are true but (R) is not the correct explanation of (A)

Correct answer: (C)

Step-by-step solution →
Q16·Physics·Wave OpticsSingle correct
When a polaroid sheet is rotated between two crossed polaroids then the transmitted light intensity will be maximum for a rotation of :
  1. (A)60∘60^\circ60∘
  2. (B)30∘30^\circ30∘
  3. (C)90∘90^\circ90∘
  4. (D)45∘45^\circ45∘

Correct answer: (D)

Step-by-step solution →
Q17·Physics·Electromagnetic WavesSingle correct
An object is placed in a medium of refractive index 3. An electromagnetic wave of intensity 6×1076\times10^76×107 W/m2^22 falls normally on the object and it is absorbed completely. The radiation pressure on the object would be (speed of light in free space =3×108=3\times10^8=3×108 m/s) :
  1. (A)363636 Nm−2^{-2}−2
  2. (B)181818 Nm−2^{-2}−2
  3. (C)666 Nm−2^{-2}−2
  4. (D)222 Nm−2^{-2}−2

Correct answer: (C)

Step-by-step solution →
Q18·Physics·Electric PotentialSingle correct
Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : Work done by electric field on moving a positive charge on an equipotential surface is always zero. Reason (R) : Electric lines of forces are always perpendicular to equipotential surfaces. 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 but (R) is not the correct explanation of (A)
  2. (B)(A) is correct but (R) is not correct
  3. (C)(A) is not correct but (R) is correct
  4. (D)Both (A) and (R) are correct and (R) is the correct explanation of (A)

Correct answer: (D)

Step-by-step solution →
Q19·Physics·Laws of MotionSingle correct
A heavy iron bar of weight 12 kg is having its one end on the ground and the other on the shoulder of a man. The rod makes an angle 60∘60^\circ60∘ with the horizontal, the weight experienced by the man is :
  1. (A)6 kg
  2. (B)12 kg
  3. (C)3 kg
  4. (D)636\sqrt363​ kg

Correct answer: (C)

Step-by-step solution →
Q20·Physics·Magnetic Field of CurrentNumerical
The magnetic field at the centre O of a wire loop formed by two semicircular wires of radii R1=2πR_1=2\piR1​=2π m and R2=4πR_2=4\piR2​=4π m carrying current I=4I=4I=4 A as that figure given below is α×10−7\alpha\times10^{-7}α×10−7 T. The value of α\alphaα is ________. (Centre O is common for all segments)

Correct answer: 3.00

Step-by-step solution →
Q21·Physics·Electric Field and Coulomb's LawNumerical
Two charges of −4 μ-4\,\mu−4μC and +4 μ+4\,\mu+4μC are placed at the points A(1,0,4) m and B(2,-1,5) m located in an electric field E⃗=0.20 i^\vec E=0.20\,\hat iE=0.20i^ V/cm. The magnitude of the torque acting on the dipole is 8α×10−58\sqrt\alpha\times10^{-5}8α​×10−5 Nm. Where α\alphaα is ________.

Correct answer: 2.00

Step-by-step solution →
Q22·Physics·WavesNumerical
A closed organ pipe 150 cm long gives 7 beats per second with an open organ pipe of length 350 cm, both vibrating in fundamental mode. The velocity of sound is ________ m/s.

Correct answer: 294.00

Step-by-step solution →
Q23·Physics·KinematicsNumerical
A body falling under gravity covers two points A and B separated by 80 m in 2 s. The distance of upper point A from the starting point is ________ m. (use g=10g=10g=10 ms−2^{-2}−2)

Correct answer: 45.00

Step-by-step solution →
Q24·Physics·Properties of Solids and LiquidsNumerical
The reading of pressure metre attached with a closed pipe is 4.5×1054.5\times10^54.5×105 N/m2^22. On opening the valve, water starts flowing and the reading of pressure metre falls to 2.0×1052.0\times10^52.0×105 N/m2^22. The velocity of water is found to be V\sqrt VV​ m/s. The value of V is ________.

Correct answer: 50.00

Step-by-step solution →
Q25·Physics·Rotational MotionNumerical
A ring and a solid sphere roll down the same inclined plane without slipping. They start from rest. The radii of both bodies are identical and the ratio of their kinetic energies is 7x\dfrac{7}{x}x7​ where x is ________.

Correct answer: 7.00

Step-by-step solution →
Q26·Physics·Geometrical OpticsNumerical
A parallel beam of monochromatic light of wavelength 5000 A˚\mathring{A}A˚ is incident normally on a single narrow slit of width 0.001 mm. The light is focused by convex lens on screen, placed on its focal plane. The first minima will be formed for the angle of diffraction of ________ (degree).

Correct answer: 30.00

Step-by-step solution →
Q27·Physics·Electric PotentialNumerical
The electric potential at the surface of an atomic nucleus (Z=50)(Z=50)(Z=50) of radius 9×10−139\times10^{-13}9×10−13 cm is ________ ×106\times10^6×106 V.

Correct answer: 8.00

Step-by-step solution →
Q28·Physics·Atoms and NucleiNumerical
If Rydberg's constant is R, the longest wavelength of radiation in Paschen series will be α7R\dfrac{\alpha}{7R}7Rα​, where α\alphaα is ________.

Correct answer: 144.00

Step-by-step solution →
Q29·Physics·Alternating CurrentsNumerical
A series LCR circuit with L=100πL=\dfrac{100}{\pi}L=π100​ mH, C=10−3πC=\dfrac{10^{-3}}{\pi}C=π10−3​ F and R=10 ΩR=10\,\OmegaR=10Ω, is connected across an ac source of 220 V, 50 Hz supply. The power factor of the circuit would be ________.

Correct answer: 1.00

Step-by-step solution →

Chemistry — JEE Main 27 January 2024 Shift 2

Q30·Chemistry·Electronic Effects and StabilitySingle correct
The order of relative stability of the contributing structures (I, II and III) is :
  1. (A)I > II > III
  2. (B)II > I > III
  3. (C)I = II = III
  4. (D)III > II > I

Correct answer: (A)

Step-by-step solution →
Q31·Chemistry·Haloalkanes and Haloarenes (Class 12)Single correct
Which among the following halide/s will not show SN1S_N1SN​1 reaction ? Choose the most appropriate answer from the options given below :
  1. (A)(A), (B) and (D) only
  2. (B)(A) and (B) only
  3. (C)(B) and (C) only
  4. (D)(B) only

Correct answer: (D)

Step-by-step solution →
Q32·Chemistry·Electrochemistry (Class 12)Single correct
Which of the following statements is not correct about rusting of iron ?
  1. (A)Coating of iron surface by tin prevents rusting, even if the tin coating is peeling off.
  2. (B)When pH lies above 9 or 10, rusting of iron does not take place.
  3. (C)Dissolved acidic oxides SO2SO_2SO2​, NO2NO_2NO2​ in water act as catalyst in the process of rusting.
  4. (D)Rusting of iron is envisaged as setting up of electrochemical cell on the surface of iron object.

Correct answer: (A)

Step-by-step solution →
Q33·Chemistry·d- and f-Block ElementsSingle correct
Given below are two statements : Statement (I) : In the Lanthanoids, the formation of Ce+4Ce^{+4}Ce+4 is favoured by its noble gas configuration. Statement (II) : Ce+4Ce^{+4}Ce+4 is a strong oxidant reverting to the common +3 state. In the light of the above statements, choose the most appropriate answer from the options given below :
  1. (A)Statement I is false but Statement II is true
  2. (B)Both Statement I and Statement II are true
  3. (C)Statement I is true but Statement II is false
  4. (D)Both Statement I and Statement II are false

Correct answer: (B)

Step-by-step solution →
Q34·Chemistry·d- and f-Block ElementsSingle correct
Choose the correct option having all the elements with d10d^{10}d10 electronic configuration from the following :
  1. (A)27_{27}27​Co, 28_{28}28​Ni, 24_{24}24​Cr
  2. (B)29_{29}29​Cu, 30_{30}30​Zn, 48_{48}48​Cd, 47_{47}47​Ag
  3. (C)46_{46}46​Pd, 28_{28}28​Ni, 26_{26}26​Fe, 24_{24}24​Cr
  4. (D)28_{28}28​Ni, 24_{24}24​Cr, 26_{26}26​Fe, 29_{29}29​Cu

Correct answer: (B)

Step-by-step solution →
Q35·Chemistry·Alcohols, Phenols and Ethers (Class 12)Single correct
Phenolic group can be identified by a positive :
  1. (A)Phthalein dye test
  2. (B)Lucas test
  3. (C)Tollen's test
  4. (D)Carbylamine test

Correct answer: (A)

Step-by-step solution →
Q36·Chemistry·Aldehydes, Ketones and Carboxylic Acids (Class 12)Single correct
The molecular formula of second homologue in the homologous series of mono carboxylic acids is _________.
  1. (A)C2H4O2C_2H_4O_2C2​H4​O2​
  2. (B)C4H8O2C_4H_8O_2C4​H8​O2​
  3. (C)CH2O2CH_2O_2CH2​O2​
  4. (D)C3H6O2C_3H_6O_2C3​H6​O2​

Correct answer: (A)

Step-by-step solution →
Q37·Chemistry·Purification and Characterisation of Organic CompoundsSingle correct
The technique used for purification of steam volatile water immiscible substance is :
  1. (A)Fractional distillation
  2. (B)Fractional distillation under reduced pressure
  3. (C)Distillation
  4. (D)Steam distillation

Correct answer: (D)

Step-by-step solution →
Q38·Chemistry·Alcohols, Phenols and Ethers (Class 12)Single correct
The final product B, formed in the following reaction sequence is : Ph–CH–CH3→(ii) HBr(i) BH3→(iv) Mg, ether, then HCHO/H3O+B\text{Ph–CH–CH}_3 \xrightarrow[\text{(ii) HBr}]{\text{(i) BH}_3} \xrightarrow[\text{(iv) Mg, ether, then HCHO/H}_3\text{O}^+]{} BPh–CH–CH3​(i) BH3​(ii) HBr​(iv) Mg, ether, then HCHO/H3​O+​B
  1. (A)Ph–CH–CH3\text{Ph–CH–CH}_3Ph–CH–CH3​ (with −CH3-CH_3−CH3​)
  2. (B)Ph–CH–CH3\text{Ph–CH–CH}_3Ph–CH–CH3​ (with −CH2OH-CH_2OH−CH2​OH)
  3. (C)Ph–CH2–CH2OH\text{Ph–CH}_2\text{–CH}_2\text{OH}Ph–CH2​–CH2​OH
  4. (D)Ph–CH2–CH2–OH\text{Ph–CH}_2\text{–CH}_2\text{–OH}Ph–CH2​–CH2​–OH

Correct answer: (D)

Step-by-step solution →
Q39·Chemistry·Alcohols, Phenols and Ethers (Class 12)Single correct
Match List-I with List-II. Choose the correct answer from the options given below :
List-I (Reaction)List-II (Reagent)
A.see figureI.Na2Cr2O7Na_2Cr_2O_7Na2​Cr2​O7​; H2SO4H_2SO_4H2​SO4​
B.see figureII.(i) NaOH (ii) CH3ClCH_3ClCH3​Cl
C.see figureIII.(i) NaOH, CHCl3CHCl_3CHCl3​ (ii) NaOH (iii) HCl
D.see figureIV.(i) CO2CO_2CO2​ (ii) HCl
  1. (A)(A)-(IV), (B)-(I), (C)-(II), (D)-(III)
  2. (B)(A)-(IV), (B)-(III), (C)-(I), (D)-(II)
  3. (C)(A)-(I), (B)-(II), (C)-(III), (D)-(IV)
  4. (D)(A)-(IV), (B)-(III), (C)-(II), (D)-(I)

Correct answer: (B)

Step-by-step solution →
Q40·Chemistry·HydrocarbonsSingle correct
Major product formed in the following reaction is a mixture of :
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (D)

Step-by-step solution →
Q41·Chemistry·IUPAC NomenclatureSingle correct
The bond line formula of HOCH(CN)2HOCH(CN)_2HOCH(CN)2​ is :
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (D)

Step-by-step solution →
Q42·Chemistry·p-Block ElementsSingle correct
Given below are two statements : Statement (I) : Oxygen being the first member of group 16 exhibits only −2-2−2 oxidation state. Statement (II) : Down the group 16 stability of +4+4+4 oxidation state decreases and +6+6+6 oxidation state increases. 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 Statement I and Statement II are incorrect
  3. (C)Both Statement I and Statement II are correct
  4. (D)Statement I is incorrect but Statement II is correct

Correct answer: (B)

Step-by-step solution →
Q43·Chemistry·Coordination CompoundsSingle correct
Identify from the following species in which d2sp3d^2sp^3d2sp3 hybridization is shown by central atom :
  1. (A)[Co(NH3)6]3+[Co(NH_3)_6]^{3+}[Co(NH3​)6​]3+
  2. (B)BrF5BrF_5BrF5​
  3. (C)[Pt(Cl)4]2−[Pt(Cl)_4]^{2-}[Pt(Cl)4​]2−
  4. (D)SF6SF_6SF6​

Correct answer: (A)

Step-by-step solution →
Q44·Chemistry·Amines (Class 12)Single correct
Identify B formed in the reaction. Cl–(CH2)4–Cl+excess NH3⟶A→NaOHB+H2O+NaClCl–(CH_2)_4–Cl + \text{excess } NH_3 \longrightarrow A \xrightarrow{NaOH} B + H_2O + NaClCl–(CH2​)4​–Cl+excess NH3​⟶ANaOH​B+H2​O+NaCl
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (A)

Step-by-step solution →
Q45·Chemistry·SolutionsSingle correct
Which among the following changes with temperature is :
  1. (A)Molarity
  2. (B)Mass percentage
  3. (C)Molality
  4. (D)Mole fraction

Correct answer: (A)

Step-by-step solution →
Q46·Chemistry·BiomoleculesSingle correct
Which structure of protein remains intact after coagulation of egg white on boiling ?
  1. (A)Primary
  2. (B)Tertiary
  3. (C)Secondary
  4. (D)Quaternary

Correct answer: (A)

Step-by-step solution →
Q47·Chemistry·p-Block ElementsSingle correct
Which of the following cannot function as an oxidising agent ?
  1. (A)N3−N^{3-}N3−
  2. (B)SO42−SO_4^{2-}SO42−​
  3. (C)BrO3−BrO_3^-BrO3−​
  4. (D)MnO4−MnO_4^-MnO4−​

Correct answer: (A)

Step-by-step solution →
Q48·Chemistry·IsomerismSingle correct
The incorrect statement regarding conformations of ethane is : (1) the infinite number of conformations of ethane. (2) the dihedral angle in staggered conformation is 60∘60^\circ60∘. (3) eclipsed conformation is the most stable conformation. (4) the conformations of ethane are inter-convertible to one-another.
  1. (A)the infinite number of conformations of ethane
  2. (B)the dihedral angle in staggered conformation is 60∘60^\circ60∘
  3. (C)eclipsed conformation is the most stable conformation
  4. (D)the conformations of ethane are inter-convertible to one-another

Correct answer: (C)

Step-by-step solution →
Q49·Chemistry·d- and f-Block ElementsSingle correct
Identify the incorrect pair from the following : (1) Photography – AgBr (2) Polythene preparation – TiCl4TiCl_4TiCl4​, Al(CH3)3Al(CH_3)_3Al(CH3​)3​ (3) Rubber process – Pt (4) Wacker process – PtCl2PtCl_2PtCl2​
  1. (A)Photography – AgBr
  2. (B)Polythene preparation – TiCl4TiCl_4TiCl4​, Al(CH3)3Al(CH_3)_3Al(CH3​)3​
  3. (C)Rubber process – Pt
  4. (D)Wacker process – PtCl2PtCl_2PtCl2​

Correct answer: (D)

Step-by-step solution →
Q50·Chemistry·Atomic StructureNumerical
Total number of ions from the following that have noble gas electronic configuration is __________. Sr2+Sr^{2+}Sr2+ (Z=38), Cs+Cs^+Cs+ (Z=55), La3+La^{3+}La3+ (Z=57), Pb2+Pb^{2+}Pb2+ (Z=82), Yb2+Yb^{2+}Yb2+ (Z=70), Fe2+Fe^{2+}Fe2+ (Z=26).

Correct answer: 3

Step-by-step solution →
Q51·Chemistry·Chemical Bonding and Molecular StructureNumerical
The number of non-polar molecules from the following is __________. HF, H2OH_2OH2​O, SO2SO_2SO2​, H2H_2H2​, CO2CO_2CO2​, CH4CH_4CH4​, HCl, CHCl3CHCl_3CHCl3​, BF3BF_3BF3​.

Correct answer: 4

Step-by-step solution →
Q52·Chemistry·Chemical KineticsNumerical
Time required for completion of 99.9% of a First order reaction is __________ times of half life (t1/2t_{1/2}t1/2​) of the reaction.

Correct answer: 10

Step-by-step solution →
Q53·Chemistry·Coordination CompoundsNumerical
The spin only magnetic moment value of square planar complex [Pt(NH3)2(NH2CH3)2]Cl2[Pt(NH_3)_2(NH_2CH_3)_2]Cl_2[Pt(NH3​)2​(NH2​CH3​)2​]Cl2​ is __________ B.M. (Nearest integer) (Given atomic number for Pt = 78).

Correct answer: 0

Step-by-step solution →
Q54·Chemistry·Chemical ThermodynamicsNumerical
For a certain thermochemical reaction M→NM\to NM→N at T=400T=400T=400 K, ΔH∘=77.2\Delta H^\circ=77.2ΔH∘=77.2 kJ mol−1^{-1}−1, ΔS∘=122\Delta S^\circ=122ΔS∘=122 JK−1^{-1}−1, log⁡K\log KlogK is __________ ×10−1\times10^{-1}×10−1.

Correct answer: -37

Step-by-step solution →
Q55·Chemistry·SolutionsNumerical
Volume of 3 M NaOH (formula weight 40 g mol−1^{-1}−1) which can be prepared from 84 g of NaOH is __________ ×10−1\times10^{-1}×10−1 dm3^33.

Correct answer: 7

Step-by-step solution →
Q56·Chemistry·Redox Reactions and ElectrochemistryNumerical
1 mole of PbS is oxidised by "X" moles of O3O_3O3​ to get "Y" moles of O2O_2O2​. X+YX+YX+Y is __________.

Correct answer: 8

Step-by-step solution →
Q57·Chemistry·Electrochemistry (Class 12)Numerical
The hydrogen electrode is dipped in a solution of pH = 3 at 25∘25^\circ25∘C. The potential of the electrode will be __________ ×10−2\times10^{-2}×10−2 V. (2.303RTF=0.059 V)\left(\dfrac{2.303RT}{F}=0.059\text{ V}\right)(F2.303RT​=0.059 V)

Correct answer: -18

Step-by-step solution →
Q58·Chemistry·Some Basic Concepts in ChemistryNumerical
9.3 g of aniline is subjected to reaction with excess of acetic anhydride to prepare acetanilide. The mass of acetanilide produced if the reaction is 100% completed is __________ ×10−1\times10^{-1}×10−1 g. (Given molar mass in g mol−1^{-1}−1 N : 14, O : 16, C : 12, H : 1).

Correct answer: 135

Step-by-step solution →
Q59·Chemistry·IsomerismNumerical
Total number of compounds from the following with Chiral carbon atoms is __________.

Correct answer: 5

Step-by-step solution →

Mathematics — JEE Main 27 January 2024 Shift 2

Q60·Mathematics·Inverse Trigonometric FunctionsSingle correct
Considering only the principal values of inverse trigonometric functions, the number of positive real values of xxx satisfying tan⁡−1(x)+tan⁡−1(2x)=π4\tan^{-1}(x)+\tan^{-1}(2x)=\dfrac{\pi}{4}tan−1(x)+tan−1(2x)=4π​ is :
  1. (A)More than 2
  2. (B)1
  3. (C)2
  4. (D)0

Correct answer: (B)

Step-by-step solution →
Q61·Mathematics·Limits and ContinuitySingle correct
Consider the function f:(0,2)→Rf:(0,2)\to\mathbb{R}f:(0,2)→R defined by f(x)=x2+2xf(x)=\dfrac{x}{2}+\dfrac{2}{x}f(x)=2x​+x2​ and the function g(x)g(x)g(x) defined by g(x)={min⁡{f(t)}, 0<t≤x and 0<x≤132+x, 1<x<2g(x)=\begin{cases}\min\{f(t)\},\ 0<t\le x\text{ and }0<x\le 1 \\ \dfrac{3}{2}+x,\ 1<x<2\end{cases}g(x)=⎩⎨⎧​min{f(t)}, 0<t≤x and 0<x≤123​+x, 1<x<2​. Then
  1. (A)ggg is continuous but not differentiable at x=1x=1x=1
  2. (B)ggg is not continuous for all x∈(0,2)x\in(0,2)x∈(0,2)
  3. (C)ggg is neither continuous nor differentiable at x=1x=1x=1
  4. (D)ggg is continuous and differentiable for all x∈(0,2)x\in(0,2)x∈(0,2)

Correct answer: (A)

Step-by-step solution →
Q62·Mathematics·Three Dimensional GeometrySingle correct
Let the image of the point (1,0,7)(1,0,7)(1,0,7) in the line x1=y−12=z−23\dfrac{x}{1}=\dfrac{y-1}{2}=\dfrac{z-2}{3}1x​=2y−1​=3z−2​ be the point (α,β,γ)(\alpha,\beta,\gamma)(α,β,γ). Then which one of the following points lies on the line passing through (α,β,γ)(\alpha,\beta,\gamma)(α,β,γ) and making angles 2π3\dfrac{2\pi}{3}32π​ and 3π4\dfrac{3\pi}{4}43π​ with y-axis and z-axis respectively and an acute angle with x-axis ?
  1. (A)(1,−2,1+2)\left(1,-2,1+\sqrt2\right)(1,−2,1+2​)
  2. (B)(1,2,1−2)\left(1,2,1-\sqrt2\right)(1,2,1−2​)
  3. (C)(3,4,3−22)\left(3,4,3-2\sqrt2\right)(3,4,3−22​)
  4. (D)(3,−4,3+22)\left(3,-4,3+2\sqrt2\right)(3,−4,3+22​)

Correct answer: (C)

Step-by-step solution →
Q63·Mathematics·Straight LinesSingle correct
Let R be the interior region between the lines 3x−y+1=03x-y+1=03x−y+1=0 and x+2y−5=0x+2y-5=0x+2y−5=0 containing the origin. The set of all values of aaa, for which the points (a2,a+1)(a^2,a+1)(a2,a+1) lie in R, is :
  1. (A)(−3,−1)∪(−13,1)(-3,-1)\cup\left(-\dfrac13,1\right)(−3,−1)∪(−31​,1)
  2. (B)(−3,0)∪(13,1)(-3,0)\cup\left(\dfrac13,1\right)(−3,0)∪(31​,1)
  3. (C)(−3,0)∪(23,1)(-3,0)\cup\left(\dfrac23,1\right)(−3,0)∪(32​,1)
  4. (D)(−3,−1)∪(13,1)(-3,-1)\cup\left(\dfrac13,1\right)(−3,−1)∪(31​,1)

Correct answer: (B)

Step-by-step solution →
Q64·Mathematics·Sequence and SeriesSingle correct
The 20th20^{\text{th}}20th term from the end of the progression 20,1914,1812,1734,…,−1291420,19\dfrac14,18\dfrac12,17\dfrac34,\ldots,-129\dfrac1420,1941​,1821​,1743​,…,−12941​ is :
  1. (A)−118-118−118
  2. (B)−110-110−110
  3. (C)−115-115−115
  4. (D)−100-100−100

Correct answer: (C)

Step-by-step solution →
Q65·Mathematics·Sets, Relations and FunctionsSingle correct
Let f:R−{−12}→Rf:\mathbb{R}-\left\{-\dfrac12\right\}\to\mathbb{R}f:R−{−21​}→R and g:R−{−52}→Rg:\mathbb{R}-\left\{-\dfrac52\right\}\to\mathbb{R}g:R−{−25​}→R be defined as f(x)=2x+32x+1f(x)=\dfrac{2x+3}{2x+1}f(x)=2x+12x+3​ and g(x)=∣x∣+12x+5g(x)=\dfrac{|x|+1}{2x+5}g(x)=2x+5∣x∣+1​. Then the domain of the function f∘gf\circ gf∘g is :
  1. (A)R−{−52}\mathbb{R}-\left\{-\dfrac52\right\}R−{−25​}
  2. (B)R\mathbb{R}R
  3. (C)R−{−74}\mathbb{R}-\left\{-\dfrac74\right\}R−{−47​}
  4. (D)R−{−52,−74}\mathbb{R}-\left\{-\dfrac52,-\dfrac74\right\}R−{−25​,−47​}

Correct answer: (A)

Step-by-step solution →
Q66·Mathematics·Definite IntegrationSingle correct
For 0<a<10<a<10<a<1, the value of the integral ∫0πdx1−2acos⁡x+a2\displaystyle\int_{0}^{\pi}\dfrac{dx}{1-2a\cos x+a^2}∫0π​1−2acosx+a2dx​ is :
  1. (A)π2π+a2\dfrac{\pi^2}{\pi+a^2}π+a2π2​
  2. (B)π2π−a2\dfrac{\pi^2}{\pi-a^2}π−a2π2​
  3. (C)π1−a2\dfrac{\pi}{1-a^2}1−a2π​
  4. (D)π1+a2\dfrac{\pi}{1+a^2}1+a2π​

Correct answer: (C)

Step-by-step solution →
Q67·Mathematics·Application of DerivativesSingle correct
Let g(x)=3f(x3)+f(3−x)g(x)=3f\left(\dfrac{x}{3}\right)+f(3-x)g(x)=3f(3x​)+f(3−x) and f′′(x)>0f''(x)>0f′′(x)>0 for all x∈(0,3)x\in(0,3)x∈(0,3). If ggg is decreasing in (0,α)(0,\alpha)(0,α) and increasing in (α,3)(\alpha,3)(α,3), then 8α8\alpha8α is
  1. (A)24
  2. (B)0
  3. (C)18
  4. (D)20

Correct answer: (C)

Step-by-step solution →
Q68·Mathematics·Limits and ContinuitySingle correct
If lim⁡x→03+αsin⁡x+βcos⁡x+log⁡e(1−x)3tan⁡2x=13\displaystyle\lim_{x\to 0}\dfrac{3+\alpha\sin x+\beta\cos x+\log_e(1-x)}{3\tan^2 x}=\dfrac13x→0lim​3tan2x3+αsinx+βcosx+loge​(1−x)​=31​, then 2α−β2\alpha-\beta2α−β is equal to :
  1. (A)2
  2. (B)7
  3. (C)5
  4. (D)1

Correct answer: (C)

Step-by-step solution →
Q69·Mathematics·Sequence and SeriesSingle correct
If α,β\alpha,\betaα,β are the roots of the equation x2−x−1=0x^2-x-1=0x2−x−1=0 and Sn=2023αn+2024βnS_n=2023\alpha^n+2024\beta^nSn​=2023αn+2024βn, then
  1. (A)2S12=S11+S102S_{12}=S_{11}+S_{10}2S12​=S11​+S10​
  2. (B)S12=S11+S10S_{12}=S_{11}+S_{10}S12​=S11​+S10​
  3. (C)2S11=S12+S102S_{11}=S_{12}+S_{10}2S11​=S12​+S10​
  4. (D)S11=S10+S12S_{11}=S_{10}+S_{12}S11​=S10​+S12​

Correct answer: (B)

Step-by-step solution →
Q70·Mathematics·Sets, Relations and FunctionsSingle correct
Let A and B be two finite sets with mmm and nnn elements respectively. The total number of subsets of the set A is 56 more than the total number of subsets of B. Then the distance of the point P(m,n)P(m,n)P(m,n) from the point Q(−2,−3)Q(-2,-3)Q(−2,−3) is
  1. (A)10
  2. (B)6
  3. (C)4
  4. (D)8

Correct answer: (A)

Step-by-step solution →
Q71·Mathematics·Matrices and DeterminantsSingle correct
The values of α\alphaα, for which ∣132α+32113α+132α+33α+10∣=0\begin{vmatrix} 1 & \dfrac32 & \alpha+\dfrac32 \\ 1 & \dfrac13 & \alpha+\dfrac13 \\ 2\alpha+3 & 3\alpha+1 & 0\end{vmatrix}=0​112α+3​23​31​3α+1​α+23​α+31​0​​=0, lie in the interval
  1. (A)(−2,1)(-2,1)(−2,1)
  2. (B)(−3,0)(-3,0)(−3,0)
  3. (C)(−32,32)\left(-\dfrac32,\dfrac32\right)(−23​,23​)
  4. (D)(0,3)(0,3)(0,3)

Correct answer: (B)

Step-by-step solution →
Q72·Mathematics·Statistics and ProbabilitySingle correct
An urn contains 6 white and 9 black balls. Two successive draws of 4 balls are made without replacement. The probability, that the first draw gives all white balls and the second draw gives all black balls, is :
  1. (A)5256\dfrac{5}{256}2565​
  2. (B)5715\dfrac{5}{715}7155​
  3. (C)3715\dfrac{3}{715}7153​
  4. (D)3256\dfrac{3}{256}2563​

Correct answer: (C)

Step-by-step solution →
Q73·Mathematics·Indefinite IntegrationSingle correct
The integral ∫(x8−x2) dx(x12+3x6+1)tan⁡−1(x3+1x3)\displaystyle\int\dfrac{(x^8-x^2)\,dx}{(x^{12}+3x^6+1)\tan^{-1}\left(x^3+\dfrac{1}{x^3}\right)}∫(x12+3x6+1)tan−1(x3+x31​)(x8−x2)dx​ is equal to :
  1. (A)log⁡e(∣tan⁡−1(x3+1x3)∣1/3)+C\log_e\left(\left|\tan^{-1}\left(x^3+\dfrac{1}{x^3}\right)\right|^{1/3}\right)+Cloge​(​tan−1(x3+x31​)​1/3)+C
  2. (B)log⁡e(∣tan⁡−1(x3+1x3)∣1/2)+C\log_e\left(\left|\tan^{-1}\left(x^3+\dfrac{1}{x^3}\right)\right|^{1/2}\right)+Cloge​(​tan−1(x3+x31​)​1/2)+C
  3. (C)log⁡e(∣tan⁡−1(x3+1x3)∣)+C\log_e\left(\left|\tan^{-1}\left(x^3+\dfrac{1}{x^3}\right)\right|\right)+Cloge​(​tan−1(x3+x31​)​)+C
  4. (D)log⁡e(∣tan⁡−1(x3+1x3)∣3)+C\log_e\left(\left|\tan^{-1}\left(x^3+\dfrac{1}{x^3}\right)\right|^{3}\right)+Cloge​(​tan−1(x3+x31​)​3)+C

Correct answer: (A)

Step-by-step solution →
Q74·Mathematics·Trigonometric FunctionsSingle correct
If 2tan⁡2θ−5sec⁡θ=12\tan^2\theta-5\sec\theta=12tan2θ−5secθ=1 has exactly 7 solutions in the interval [0,nπ2]\left[0,\dfrac{n\pi}{2}\right][0,2nπ​], for the least value of n∈Nn\in\mathbb{N}n∈N, then ∑k=1nk2k\displaystyle\sum_{k=1}^{n}\dfrac{k}{2^k}k=1∑n​2kk​ is equal to :
  1. (A)1215(214−14)\dfrac{1}{2^{15}}(2^{14}-14)2151​(214−14)
  2. (B)1214(215−15)\dfrac{1}{2^{14}}(2^{15}-15)2141​(215−15)
  3. (C)1−152131-\dfrac{15}{2^{13}}1−21315​
  4. (D)1213(214−15)\dfrac{1}{2^{13}}(2^{14}-15)2131​(214−15)

Correct answer: (D)

Step-by-step solution →
Q75·Mathematics·Vector AlgebraSingle correct
The position vectors of the vertices A, B and C of a triangle are 2i^−3j^+3k^2\hat i-3\hat j+3\hat k2i^−3j^​+3k^, 2i^+2j^+3k^2\hat i+2\hat j+3\hat k2i^+2j^​+3k^ and −i^+j^+3k^-\hat i+\hat j+3\hat k−i^+j^​+3k^ respectively. Let lll denotes the length of the angle bisector AD of ∠BAC\angle BAC∠BAC where D is on the line segment BC, then 2l22l^22l2 equals :
  1. (A)49
  2. (B)42
  3. (C)50
  4. (D)45

Correct answer: (D)

Step-by-step solution →
Q76·Mathematics·Differential EquationsSingle correct
If y=y(x)y=y(x)y=y(x) is the solution curve of the differential equation (x2−4) dy−(y2−3y) dx=0(x^2-4)\,dy-(y^2-3y)\,dx=0(x2−4)dy−(y2−3y)dx=0, x>2x>2x>2, y(4)=32y(4)=\dfrac32y(4)=23​ and the slope of the curve is never zero, then the value of y(10)y(10)y(10) equals :
  1. (A)31+(8)1/4\dfrac{3}{1+(8)^{1/4}}1+(8)1/43​
  2. (B)31+22\dfrac{3}{1+2\sqrt2}1+22​3​
  3. (C)31−22\dfrac{3}{1-2\sqrt2}1−22​3​
  4. (D)31−(8)1/4\dfrac{3}{1-(8)^{1/4}}1−(8)1/43​

Correct answer: (A)

Step-by-step solution →
Q77·Mathematics·EllipseSingle correct
Let e1e_1e1​ be the eccentricity of the hyperbola x216−y29=1\dfrac{x^2}{16}-\dfrac{y^2}{9}=116x2​−9y2​=1 and e2e_2e2​ be the eccentricity of the ellipse x2a2+y2b2=1\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1a2x2​+b2y2​=1, a>ba>ba>b, which passes through the foci of the hyperbola. If e1e2=1e_1 e_2=1e1​e2​=1, then the length of the chord of the ellipse parallel to the x-axis and passing through (0,2)(0,2)(0,2) is :
  1. (A)454\sqrt545​
  2. (B)853\dfrac{8\sqrt5}{3}385​​
  3. (C)1053\dfrac{10\sqrt5}{3}3105​​
  4. (D)353\sqrt535​

Correct answer: (C)

Step-by-step solution →
Q78·Mathematics·Permutations and CombinationsSingle correct
Let α=(4!)!(4!)3!\alpha=\dfrac{(4!)!}{(4!)^{3!}}α=(4!)3!(4!)!​ and β=(5!)!(5!)4!\beta=\dfrac{(5!)!}{(5!)^{4!}}β=(5!)4!(5!)!​. Then :
  1. (A)α∈N\alpha\in\mathbb{N}α∈N and β∉N\beta\notin\mathbb{N}β∈/N
  2. (B)α∉N\alpha\notin\mathbb{N}α∈/N and β∈N\beta\in\mathbb{N}β∈N
  3. (C)α∈N\alpha\in\mathbb{N}α∈N and β∈N\beta\in\mathbb{N}β∈N
  4. (D)α∉N\alpha\notin\mathbb{N}α∈/N and β∉N\beta\notin\mathbb{N}β∈/N

Correct answer: (C)

Step-by-step solution →
Q79·Mathematics·Vector AlgebraSingle correct
Let the position vectors of the vertices A, B and C of a triangle be 2i^+2j^+k^2\hat i+2\hat j+\hat k2i^+2j^​+k^, i^+2j^+2k^\hat i+2\hat j+2\hat ki^+2j^​+2k^ and 2i^+j^+2k^2\hat i+\hat j+2\hat k2i^+j^​+2k^ respectively. Let l1l_1l1​, l2l_2l2​ and l3l_3l3​ be the lengths of perpendiculars drawn from the ortho center of the triangle on the sides AB, BC and CA respectively, then l12+l22+l32l_1^2+l_2^2+l_3^2l12​+l22​+l32​ equals :
  1. (A)15\dfrac1551​
  2. (B)12\dfrac1221​
  3. (C)14\dfrac1441​
  4. (D)13\dfrac1331​

Correct answer: (B)

Step-by-step solution →
Q80·Mathematics·Statistics and ProbabilityNumerical
The mean and standard deviation of 15 observations were found to be 12 and 3 respectively. On rechecking it was found that an observation was read as 10 in place of 12. If μ\muμ and σ2\sigma^2σ2 denote the mean and variance of the correct observations respectively, then 15(μ+μ2+σ2)15(\mu+\mu^2+\sigma^2)15(μ+μ2+σ2) is equal to __________.

Correct answer: 2521

Step-by-step solution →
Q81·Mathematics·Area Under CurvesNumerical
If the area of the region {(x,y):0≤y≤min⁡{2x,6x−x2}}\{(x,y):0\le y\le\min\{2x,6x-x^2\}\}{(x,y):0≤y≤min{2x,6x−x2}} is A, then 12A12A12A is equal to __________.

Correct answer: 304

Step-by-step solution →
Q82·Mathematics·Matrices and DeterminantsNumerical
Let A be a 2×22\times 22×2 real matrix and I be the identity matrix of order 2. If the roots of the equation ∣A−xI∣=0|A-xI|=0∣A−xI∣=0 be −1-1−1 and 333, then the sum of the diagonal elements of the matrix A2A^2A2 is __________.

Correct answer: 10

Step-by-step solution →
Q83·Mathematics·Straight LinesNumerical
If the sum of squares of all real values of α\alphaα, for which the lines 2x−y+3=02x-y+3=02x−y+3=0, 6x+3y+1=06x+3y+1=06x+3y+1=0 and αx+2y−2=0\alpha x+2y-2=0αx+2y−2=0 do not form a triangle is ppp, then the greatest integer less than or equal to ppp is __________.

Correct answer: 32

Step-by-step solution →
Q84·Mathematics·Binomial Theorem and Its Simple ApplicationsNumerical
The coefficient of x2012x^{2012}x2012 in the expansion of (1−x)2008(1+x+x2)2007(1-x)^{2008}(1+x+x^2)^{2007}(1−x)2008(1+x+x2)2007 is equal to __________.

Correct answer: 0

Step-by-step solution →
Q85·Mathematics·Differential EquationsNumerical
If the solution curve, of the differential equation dydx=x+y−2x−y\dfrac{dy}{dx}=\dfrac{x+y-2}{x-y}dxdy​=x−yx+y−2​ passing through the point (2,1)(2,1)(2,1) is tan⁡−1(y−1x−1)−1βlog⁡e(α+(y−1x−1)2)=log⁡e∣x−1∣\tan^{-1}\left(\dfrac{y-1}{x-1}\right)-\dfrac{1}{\beta}\log_e\left(\alpha+\left(\dfrac{y-1}{x-1}\right)^2\right)=\log_e|x-1|tan−1(x−1y−1​)−β1​loge​(α+(x−1y−1​)2)=loge​∣x−1∣, then 5β+α5\beta+\alpha5β+α is equal to __________.

Correct answer: 11

Step-by-step solution →
Q86·Mathematics·Definite IntegrationNumerical
Let f(x)=∫0xg(t)log⁡e(1−t1+t)dtf(x)=\displaystyle\int_{0}^{x} g(t)\log_e\left(\dfrac{1-t}{1+t}\right)dtf(x)=∫0x​g(t)loge​(1+t1−t​)dt, where ggg is a continuous odd function. If ∫−π/2π/2(f(x)+x2cos⁡x1+ex)dx=(πα)2−α\displaystyle\int_{-\pi/2}^{\pi/2}\left(f(x)+\dfrac{x^2\cos x}{1+e^x}\right)dx=\left(\dfrac{\pi}{\alpha}\right)^2-\alpha∫−π/2π/2​(f(x)+1+exx2cosx​)dx=(απ​)2−α, then α\alphaα is equal to __________.

Correct answer: 2

Step-by-step solution →
Q87·Mathematics·CirclesNumerical
Consider a circle (x−α)2+(y−β)2=50(x-\alpha)^2+(y-\beta)^2=50(x−α)2+(y−β)2=50, where α,β>0\alpha,\beta>0α,β>0. If the circle touches the line y+x=0y+x=0y+x=0 at the point P, whose distance from the origin is 424\sqrt242​, then (α+β)2(\alpha+\beta)^2(α+β)2 is equal to __________.

Correct answer: 100

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Q88·Mathematics·Three Dimensional GeometryNumerical
The lines x−21=y−2=z−78\dfrac{x-2}{1}=\dfrac{y}{-2}=\dfrac{z-7}{8}1x−2​=−2y​=8z−7​ and x+34=y+23=z+21\dfrac{x+3}{4}=\dfrac{y+2}{3}=\dfrac{z+2}{1}4x+3​=3y+2​=1z+2​ intersect at the point P. If the distance of P from the line x+12=y−13=z−11\dfrac{x+1}{2}=\dfrac{y-1}{3}=\dfrac{z-1}{1}2x+1​=3y−1​=1z−1​ is lll, then 14l214l^214l2 is equal to __________.

Correct answer: 108

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Q89·Mathematics·Complex NumbersNumerical
Let the complex numbers α\alphaα and 1αˉ\dfrac{1}{\bar\alpha}αˉ1​ lie on the circles ∣z−z0∣2=4|z-z_0|^2=4∣z−z0​∣2=4 and ∣z−z0∣2=16|z-z_0|^2=16∣z−z0​∣2=16 respectively, where z0=1+iz_0=1+iz0​=1+i. Then, the value of 100∣α∣2100|\alpha|^2100∣α∣2 is __________.

Correct answer: 20

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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
  • 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
  • Some Basic Concepts in Chemistry 129/186
  • Wave Optics 130/186
  • 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
  • Nuclei 116/186
  • Organic Compounds Containing Halogens 109/186
  • Straight Lines 114/186
  • Waves 109/186
  • Atoms 112/186
  • Statistics 118/186
  • Ellipse 103/186
  • Inverse Trigonometric Functions 93/186
  • Electronic Effects and Stability 74/186
  • Experimental Skills 68/186
  • Electric Potential 63/186
  • Indefinite Integration 66/186
  • Carboxylic Acids and Derivatives 54/186
  • Isomerism 51/186
  • IUPAC Nomenclature 37/186
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