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JEE Main 6 April 2024 Shift 2 Question Paper with Answers

6 April 2024 · April session · 90 questions

The complete JEE Main 6 April 2024 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 6 April 2024 Shift 2

Q1·PhysicsSingle correct
The longest wavelength associated with Paschen series is : (Given RH=1.097×107R_H = 1.097 \times 10^7RH​=1.097×107 SI unit)
  1. (A)1.094×10−61.094 \times 10^{-6}1.094×10−6 m
  2. (B)2.973×10−62.973 \times 10^{-6}2.973×10−6 m
  3. (C)3.646×10−63.646 \times 10^{-6}3.646×10−6 m
  4. (D)1.876×10−61.876 \times 10^{-6}1.876×10−6 m

Correct answer: (D)

Step-by-step solution →
Q2·PhysicsSingle correct
A total of 48 J heat is given to one mole of helium kept in a cylinder. The temperature of helium increases by 2∘2^\circ2∘C. The work done by the gas is : (Given, R=8.3R = 8.3R=8.3 J K−1^{-1}−1mol−1^{-1}−1.)
  1. (A)72.9 J
  2. (B)24.9 J
  3. (C)48 J
  4. (D)23.1 J

Correct answer: (D)

Step-by-step solution →
Q3·PhysicsSingle correct
In finding out refractive index of glass slab the following observations were made through travelling microscope 50 vernier scale division = 49 MSD; 20 divisions on main scale in each cm For mark on paper MSR = 8.45 cm, VC = 26 For mark on paper seen through slab MSR = 7.12 cm, VC = 41 For powder particle on the top surface of the glass slab MSR = 4.05 cm, VC = 1 (MSR = Main Scale Reading, VC = Vernier Coincidence) Refractive index of the glass slab is :
  1. (A)1.42
  2. (B)1.52
  3. (C)1.24
  4. (D)1.35

Correct answer: (A)

Step-by-step solution →
Q4·PhysicsSingle correct
In the given electromagnetic wave Ey=600sin⁡(ωt−kx)E_y = 600 \sin(\omega t - kx)Ey​=600sin(ωt−kx) Vm−1^{-1}−1, intensity of the associated light beam is (in W/m2^22); (Given ϵ0=9×10−12\epsilon_0 = 9 \times 10^{-12}ϵ0​=9×10−12 C2^22N−1^{-1}−1m−2^{-2}−2)
  1. (A)486
  2. (B)243
  3. (C)729
  4. (D)972

Correct answer: (A)

Step-by-step solution →
Q5·PhysicsSingle correct
Assuming the earth to be a sphere of uniform mass density, a body weighed 300 N on the surface of earth. How much it would weigh at R/4R/4R/4 depth under surface of earth ?
  1. (A)75 N
  2. (B)375 N
  3. (C)300 N
  4. (D)225 N

Correct answer: (D)

Step-by-step solution →
Q6·PhysicsSingle correct
The acceptor level of a p-type semiconductor is 6eV. The maximum wavelength of light which can create a hole would be : Given hc=1242hc = 1242hc=1242 eV nm.
  1. (A)407 nm
  2. (B)414 nm
  3. (C)207 nm
  4. (D)103.5 nm

Correct answer: (C)

Step-by-step solution →
Q7·PhysicsSingle correct
A car of 800 kg is taking turn on a banked road of radius 300 m and angle of banking 30∘30^\circ30∘. If coefficient of static friction is 0.2 then the maximum speed with which car can negotiate the turn safely : (g=10g = 10g=10 m/s2^22, 3=1.73\sqrt{3} = 1.733​=1.73)
  1. (A)70.4 m/s
  2. (B)51.4 m/s
  3. (C)264 m/s
  4. (D)102.8 m/s

Correct answer: (B)

Step-by-step solution →
Q8·PhysicsSingle correct
Two identical conducting spheres P and S with charge Q on each, repel each other with a force 16N. A third identical uncharged conducting sphere R is successively brought in contact with the two spheres. The new force of repulsion between P and S is :
  1. (A)4 N
  2. (B)6 N
  3. (C)1 N
  4. (D)12 N

Correct answer: (B)

Step-by-step solution →
Q9·PhysicsSingle correct
In a coil, the current changes form −2-2−2 A to +2+2+2A in 0.2 s and induces an emf of 0.1 V. The self-inductance of the coil is :
  1. (A)5 mH
  2. (B)1 mH
  3. (C)2.5 mH
  4. (D)4 mH

Correct answer: (A)

Step-by-step solution →
Q10·PhysicsSingle correct
For the thin convex lens, the radii of curvature are at 15 cm and 30 cm respectively. The focal length the lens is 20 cm. The refractive index of the material is :
  1. (A)1.2
  2. (B)1.4
  3. (C)1.5
  4. (D)1.8

Correct answer: (C)

Step-by-step solution →
Q11·PhysicsSingle correct
Energy of 10 non rigid diatomic molecules at temperature T is :
  1. (A)72\frac{7}{2}27​ RT
  2. (B)70KBT70 K_B T70KB​T
  3. (C)35 RT
  4. (D)35KBT35 K_B T35KB​T

Correct answer: (D)

Step-by-step solution →
Q12·PhysicsSingle correct
A body of weight 200 N is suspended form a tree branch thought a chain of mass 10 kg. The branch pulls the chain by a force equal to (if g=10g = 10g=10 m/s2^22):
  1. (A)150 N
  2. (B)300 N
  3. (C)200 N
  4. (D)100 N

Correct answer: (B)

Step-by-step solution →
Q13·PhysicsSingle correct
When UV light of wavelength 300 nm is incident on the metal surface having work function 2.13 eV, electron emission takes place. The stopping potential is : (Given hc=1240hc = 1240hc=1240 eV nm)
  1. (A)4 V
  2. (B)4.1 V
  3. (C)2 V
  4. (D)1.5 V

Correct answer: (C)

Step-by-step solution →
Q14·PhysicsSingle correct
The number of electrons flowing per second in the filament of a 110 W bulb operating at 220 V is : (Given e=1.6×10−19e = 1.6 \times 10^{-19}e=1.6×10−19 C)
  1. (A)31.25×101731.25 \times 10^{17}31.25×1017
  2. (B)6.25×10186.25 \times 10^{18}6.25×1018
  3. (C)6.25×10176.25 \times 10^{17}6.25×1017
  4. (D)1.25×10191.25 \times 10^{19}1.25×1019

Correct answer: (A)

Step-by-step solution →
Q15·PhysicsSingle correct
When kinetic energy of a body becomes 36 times of its original value, the percentage increase in the momentum of the body will be :
  1. (A)500%
  2. (B)600%
  3. (C)6%
  4. (D)60%

Correct answer: (A)

Step-by-step solution →
Q16·PhysicsSingle correct
Pressure inside a soap bubble is greater than the pressure outside by an amount : (given : RRR = Radius of bubble, SSS = Surface tension of bubble)
  1. (A)4SR\frac{4S}{R}R4S​
  2. (B)4RS\frac{4R}{S}S4R​
  3. (C)SR\frac{S}{R}RS​
  4. (D)2SR\frac{2S}{R}R2S​

Correct answer: (A)

Step-by-step solution →
Q17·PhysicsSingle correct
Match List-I (a quantity Y plotted against a quantity X in magnetism) with List-II (the shape of the corresponding graph). Choose the correct answer:
List-I (Y vs X)List-II (Shape of Graph)
A.Y = magnetic susceptibility, X = magnetising fieldI.see figure
B.Y = magnetic field, X = distance from centre of a current carrying wire for x < a (where a = radius of wire)II.see figure
C.Y = magnetic field, X = distance from centre of a current carrying wire for x > a (where a = radius of wire)III.see figure
D.Y = magnetic field inside solenoid, X = distance from centerIV.see figure
  1. (A)(A)-(III), (B)-(I), (C)-(IV), (D)-(II)
  2. (B)(A)-(I), (B)-(III), (C)-(II), (D)-(IV)
  3. (C)(A)-(IV), (B)-(I), (C)-(III), (D)-(II)
  4. (D)(A)-(III), (B)-(IV), (C)-(I), (D)-(II)

Correct answer: (D)

Step-by-step solution →
Q18·PhysicsSingle correct
In a vernier calliper, when both jaws touch each other, zero of the vernier scale shifts towards left and its 4th4^{th}4th division coincides exactly with a certain division on main scale. If 50 vernier scale divisions equal to 49 main scale divisions and zero error in the instrument is 0.040.040.04 mm then how many main scale divisions are there in 1 cm ?
  1. (A)40
  2. (B)5
  3. (C)20
  4. (D)10

Correct answer: (C)

Step-by-step solution →
Q19·PhysicsSingle correct
Given below are two statements : Statement (I) : Dimensions of specific heat is [L2T−2K−1][L^{2}T^{-2}K^{-1}][L2T−2K−1]. Statement (II) : Dimensions of gas constant is [ML2T−1K−1][M L^{2}T^{-1}K^{-1}][ML2T−1K−1].
  1. (A)Statement (I) is incorrect but statement (II) is correct
  2. (B)Both statement (I) and statement (II) are incorrect
  3. (C)Statement (I) is correct but statement (II) is incorrect
  4. (D)Both statement (I) and statement (II) are correct

Correct answer: (C)

Step-by-step solution →
Q20·PhysicsSingle correct
A body projected vertically upwards with a certain speed from the top of a tower reaches the ground in t1t_{1}t1​. If it is projected vertically downwards from the same point with the same speed, it reaches the ground in t2t_{2}t2​. Time required to reach the ground, if it is dropped from the top of the tower, is :
  1. (A)t1t2\sqrt{t_{1}t_{2}}t1​t2​​
  2. (B)t1−t2\sqrt{t_{1}-t_{2}}t1​−t2​​
  3. (C)t1t2\sqrt{\frac{t_{1}}{t_{2}}}t2​t1​​​
  4. (D)t1+t2\sqrt{t_{1}+t_{2}}t1​+t2​​

Correct answer: (A)

Step-by-step solution →
Q21·PhysicsNumerical
In Franck-Hertz experiment, the first dip in the current-voltage graph for hydrogen is observed at 10.210.210.2 V. The wavelength of light emitted by hydrogen atom when excited to the first excitation level is ______ nm. (Given hc=1245hc = 1245hc=1245 eV nm, e=1.6×10−19e = 1.6 \times 10^{-19}e=1.6×10−19 C).

Correct answer: 122

Step-by-step solution →
Q22·PhysicsNumerical
For a given series LCR circuit it is found that maximum current is drawn when value of variable capacitance is 2.52.52.5 nF. If resistance of 200 Ω200\,\Omega200Ω and 100100100 mH inductor is being used in the given circuit. The frequency of ac source is ______ ×103\times 10^{3}×103 Hz. (given π2=10\pi^{2} = 10π2=10)

Correct answer: 10

Step-by-step solution →
Q23·PhysicsNumerical
A particle moves in a straight line so that its displacement xxx at any time ttt is given by x2=1+t2x^{2}=1+t^{2}x2=1+t2. Its acceleration at any time ttt is x−nx^{-n}x−n where n=n =n= ______ .

Correct answer: 3

Step-by-step solution →
Q24·PhysicsNumerical
Three balls of masses 2kg, 4kg and 6kg respectively are arranged at centre of the edges of an equilateral triangle of side 2 m. The moment of inertia of the system about an axis through the centroid and perpendicular to the plane of triangle, will be ______ kg m2^{2}2.

Correct answer: 4

Step-by-step solution →
Q25·PhysicsNumerical
A coil having 100 turns, area of 5×10−35 \times 10^{-3}5×10−3 m2^{2}2, carrying current of 1 mA is placed in uniform magnetic field of 0.200.200.20 T such a way that plane of coil is perpendicular to the magnetic field. The work done in turning the coil through 90∘90^{\circ}90∘ is ______ μ\muμJ.

Correct answer: 100

Step-by-step solution →
Q26·PhysicsNumerical
In the given figure an ammeter A consists of a 240 Ω240\,\Omega240Ω coil connected in parallel to a 10 Ω10\,\Omega10Ω shunt. The reading of the ammeter is ______ mA.

Correct answer: 160

Step-by-step solution →
Q27·PhysicsNumerical
A wire of cross sectional area A, modulus of elasticity 2×10112 \times 10^{11}2×1011 Nm−2^{-2}−2 and length 2 m is stretched between two vertical rigid supports. When a mass of 2 kg is suspended at the middle it sags lower from its original position making angle θ=1100\theta = \frac{1}{100}θ=1001​ radian on the points of support. The value of A is ______ ×10−4\times 10^{-4}×10−4 m2^{2}2 (consider x<<L). (given : g=10g = 10g=10 m/s2^{2}2)

Correct answer: 1

Step-by-step solution →
Q28·PhysicsNumerical
Two coherent monochromatic light beams of intensities III and 4I4I4I are superimposed. The difference between maximum and minimum possible intensities in the resulting beam is x Ix\,IxI. The value of x is ______ .

Correct answer: 8

Step-by-step solution →
Q29·PhysicsNumerical
Two open organ pipes of length 60 cm and 90 cm resonate at 6th6^{th}6th and 5th5^{th}5th harmonics respectively. The difference of frequencies for the given modes is ______ Hz. (Velocity of sound in air = 333 m/s)

Correct answer: 740

Step-by-step solution →
Q30·PhysicsNumerical
A capacitor of 10 μ10\,\mu10μF capacitance whose plates are separated by 10 mm through air and each plate has area 4 cm2^{2}2 is now filled equally with two dielectric media of K1=2K_{1} = 2K1​=2, K2=3K_{2} = 3K2​=3 respectively as shown in figure. If new force between the plates is 8 N. The supply voltage is ______ V.

Correct answer: 80

Step-by-step solution →

Chemistry — JEE Main 6 April 2024 Shift 2

Q31·Chemistry·AromaticitySingle correct
Given are four aromatic compounds labelled (I), (II), (III) and (IV) (shown in the figure). The correct arrangement for decreasing order of reactivity towards electrophilic substitution for these compounds is:
  1. (A)(IV) > (I) > (II) > (III)
  2. (B)(III) > (I) > (II) > (IV)
  3. (C)(II) > (IV) > (III) > (I)
  4. (D)(III) > (IV) > (II) > (I)

Correct answer: (B)

Step-by-step solution →
Q32·ChemistrySingle correct
Molality (m) of 3 M aqueous solution of NaCl is: (Given : Density of solution = 1.25 g mL−1mL^{-1}mL−1, Molar mass in g mol−1mol^{-1}mol−1 : Na-23, Cl-35.5)
  1. (A)2.90 m
  2. (B)2.79 m
  3. (C)1.90 m
  4. (D)3.85 m

Correct answer: (B)

Step-by-step solution →
Q33·ChemistrySingle correct
The incorrect statements regarding enzymes are: (A) Enzymes are biocatalysts. (B) Enzymes are non-specific and can catalyse different kinds of reactions. (C) Most Enzymes are globular proteins. (D) Enzyme - oxidase catalyses the hydrolysis of maltose into glucose. Choose the correct answer from the option given below:
  1. (A)(B) and (C)
  2. (B)(B), (C) and (D)
  3. (C)(B) and (D)
  4. (D)(A), (B) and (C)

Correct answer: (C)

Step-by-step solution →
Q34·ChemistrySingle correct
Consider the reaction shown in the figure. The product 'A' is:
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (B)

Step-by-step solution →
Q35·ChemistrySingle correct
During the detection of acidic radical present in a salt, a student gets a pale yellow precipitate soluble with difficulty in NH4OHNH_4OHNH4​OH solution when sodium carbonate extract was first acidified with dil. HNO3HNO_3HNO3​ and then AgNO3AgNO_3AgNO3​ solution was added. This indicates presence of:
  1. (A)Br−Br^-Br−
  2. (B)CO32−CO_3^{2-}CO32−​
  3. (C)I−I^-I−
  4. (D)Cl−Cl^-Cl−

Correct answer: (A)

Step-by-step solution →
Q36·ChemistrySingle correct
How can an electrochemical cell be converted into an electrolytic cell?
  1. (A)Applying an external opposite potential greater than Ecell0E^0_{cell}Ecell0​
  2. (B)Reversing the flow of ions in salt bridge.
  3. (C)Applying an external opposite potential lower than Ecell0E^0_{cell}Ecell0​.
  4. (D)Exchanging the electrodes at anode and cathode.

Correct answer: (A)

Step-by-step solution →
Q37·ChemistrySingle correct
Arrange the following elements in the increasing order of number of unpaired electrons in it. (A) Sc (B) Cr (C) V (D) Ti (E) Mn. Choose the correct answer from the options given below:
  1. (A)(C) < (E) < (B) < (A) < (D)
  2. (B)(B) < (C) < (D) < (E) < (A)
  3. (C)(A) < (D) < (C) < (B) < (E)
  4. (D)(A) < (D) < (C) < (E) < (B)

Correct answer: (D)

Step-by-step solution →
Q38·ChemistrySingle correct
Match List-I (alkali metals) with List-II (their characteristic emission wavelengths). Choose the correct answer:
List-I (Alkali Metal)List-II (Emission Wavelength in nm)
A.LiI.589.2
B.NaII.455.5
C.RbIII.670.8
D.CsIV.780.0
  1. (A)(A)-(I), (B)-(IV), (C)-(III), (D)-(II)
  2. (B)(A)-(III), (B)-(I), (C)-(IV), (D)-(II)
  3. (C)(A)-(IV), (B)-(II), (C)-(I), (D)-(III)
  4. (D)(A)-(II), (B)-(IV), (C)-(III), (D)-(I)

Correct answer: (B)

Step-by-step solution →
Q39·ChemistrySingle correct
In the reaction sequence shown in the figure, the major products 'A' and 'B' respectively are:
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (B)

Step-by-step solution →
Q40·Chemistry·IsomerismSingle correct
The incorrect statement regarding the geometrical isomers of 2-butene is:
  1. (A)cis-2-butene and trans-2-butene are not interconvertible at room temperature.
  2. (B)cis-2-butene has less dipole moment than trans-2-butene.
  3. (C)trans-2-butene is more stable than cis-2-butene.
  4. (D)cis-2-butene and trans-2-butene are stereoisomers.

Correct answer: (B)

Step-by-step solution →
Q41·ChemistrySingle correct
Given below are two statements: Statement I: PF5PF_5PF5​ and BrF5BrF_5BrF5​ both exhibit sp3dsp^3dsp3d hybridisation. Statement II: Both SF6SF_6SF6​ and [Co(NH3)6]3+[Co(NH_3)_6]^{3+}[Co(NH3​)6​]3+ exhibit sp3d2sp^3d^2sp3d2 hybridisation. 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 true
  3. (C)Both Statement I and Statement II are false
  4. (D)Statement I is false but Statement II is true

Correct answer: (C)

Step-by-step solution →
Q42·ChemistrySingle correct
The number of ions from the following that are expected to behave as oxidising agent is: Sn4+Sn^{4+}Sn4+, Sn2+Sn^{2+}Sn2+, Pb2+Pb^{2+}Pb2+, Tl3+Tl^{3+}Tl3+, Pb4+Pb^{4+}Pb4+, Tl+Tl^+Tl+
  1. (A)3
  2. (B)4
  3. (C)1
  4. (D)2

Correct answer: (D)

Step-by-step solution →
Q43·ChemistrySingle correct
Identify the major product (A) in the reaction sequence shown in the figure.
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (B)

Step-by-step solution →
Q44·ChemistrySingle correct
The correct statements among the following, for a "chromatography" purification method is:
  1. (A)Organic compounds run faster than solvent in the thin layer chromatographic plate.
  2. (B)Non-polar compounds are retained at top and polar compounds come down in column chromatography.
  3. (C)RfR_fRf​ of a polar compound is smaller than that of a non-polar compound.
  4. (D)RfR_fRf​ is an integral value.

Correct answer: (C)

Step-by-step solution →
Q45·ChemistrySingle correct
Evaluate the following statements related to group 14 elements for their correctness. (A) Covalent radius decreases down the group from C to Pb in a regular manner. (B) Electronegativity decreases from C to Pb down the group gradually. (C) Maximum covalence of C is 4 whereas other elements can expand their covalence due to presence of d orbitals. (D) Heavier elements do not form pπ−pπp\pi-p\pipπ−pπ bonds. (E) Carbon can exhibit negative oxidation states. Choose the correct answer from the options given below:
  1. (A)(C), (D) and (E) Only
  2. (B)(A) and (B) Only
  3. (C)(A), (B) and (C) Only
  4. (D)(C) and (D) Only

Correct answer: (A)

Step-by-step solution →
Q46·ChemistrySingle correct
Match List-I (redox reactions) with List-II (the type of redox reaction). Choose the correct answer:
List-I (Reaction)List-II (Type of redox reaction)
A.N2(g)+O2(g)→2NO(g)N_{2(g)} + O_{2(g)} \rightarrow 2NO_{(g)}N2(g)​+O2(g)​→2NO(g)​I.Decomposition
B.2Pb(NO3)2(s)→2PbO(s)+4NO2(g)+O2(g)2Pb(NO_3)_{2(s)} \rightarrow 2PbO_{(s)} + 4NO_{2(g)} + O_{2(g)}2Pb(NO3​)2(s)​→2PbO(s)​+4NO2(g)​+O2(g)​II.Displacement
C.2Na(s)+2H2O(l)→2NaOH(aq.)+H2(g)2Na_{(s)} + 2H_2O_{(l)} \rightarrow 2NaOH_{(aq.)} + H_{2(g)}2Na(s)​+2H2​O(l)​→2NaOH(aq.)​+H2(g)​III.Disproportionation
D.2NO2(g)+2 −OH(aq.)→NO2(aq.)−+NO3(aq.)−+H2O(l)2NO_{2(g)} + 2\,^{-}OH_{(aq.)} \rightarrow NO_{2(aq.)}^{-} + NO_{3(aq.)}^{-} + H_2O_{(l)}2NO2(g)​+2−OH(aq.)​→NO2(aq.)−​+NO3(aq.)−​+H2​O(l)​IV.Combination
  1. (A)(A)-(I), (B)-(II), (C)-(III), (D)-(IV)
  2. (B)(A)-(III), (B)-(II), (C)-(I), (D)-(IV)
  3. (C)(A)-(II), (B)-(III), (C)-(IV), (D)-(I)
  4. (D)(A)-(IV), (B)-(I), (C)-(II), (D)-(III)

Correct answer: (D)

Step-by-step solution →
Q47·ChemistrySingle correct
Consider the given reaction, identify the major product P. CH3−COOHCH_3-COOHCH3​−COOH undergoes (i) LiAlH4LiAlH_4LiAlH4​ (ii) PCC (iii) HCN/−OHHCN/^{-}OHHCN/−OH (iv) H2O/−OH,ΔH_2O/^{-}OH, \DeltaH2​O/−OH,Δ to give "P".
  1. (A)CH3−CH2−CH2−OHCH_3-CH_2-CH_2-OHCH3​−CH2​−CH2​−OH
  2. (B)CH3−CH2−CO−NH2CH_3-CH_2-CO-NH_2CH3​−CH2​−CO−NH2​
  3. (C)CH3−CO−CH2CH3CH_3-CO-CH_2CH_3CH3​−CO−CH2​CH3​
  4. (D)CH3−CH(OH)−COOHCH_3-CH(OH)-COOHCH3​−CH(OH)−COOH

Correct answer: (D)

Step-by-step solution →
Q48·ChemistrySingle correct
The correct IUPAC name of [PtBr2(PMe3)2][PtBr_2(PMe_3)_2][PtBr2​(PMe3​)2​] is:
  1. (A)bis(trimethylphosphine)dibromoplatinum(II)
  2. (B)bis[bromo(trimethylphosphine)]platinum(II)
  3. (C)dibromobis(trimethylphosphine)platinum(II)
  4. (D)dibromodi(trimethylphosphine)platinum(II)

Correct answer: (C)

Step-by-step solution →
Q49·ChemistrySingle correct
Match List-I (tetrahedral complexes) with List-II (their d-electron configurations). Choose the correct answer:
List-I (Tetrahedral Complex)List-II (Electronic configuration)
A.TiCl4TiCl_4TiCl4​I.e2,t20e^2, t_2^0e2,t20​
B.[FeO4]2−[FeO_4]^{2-}[FeO4​]2−II.e4,t23e^4, t_2^3e4,t23​
C.[FeCl4]−[FeCl_4]^{-}[FeCl4​]−III.e0,t20e^0, t_2^0e0,t20​
D.[CoCl4]2−[CoCl_4]^{2-}[CoCl4​]2−IV.e2,t23e^2, t_2^3e2,t23​
  1. (A)(A)-(I), (B)-(III), (C)-(IV), (D)-(II)
  2. (B)(A)-(IV), (B)-(III), (C)-(I), (D)-(II)
  3. (C)(A)-(III), (B)-(IV), (C)-(II), (D)-(I)
  4. (D)(A)-(III), (B)-(I), (C)-(IV), (D)-(II)

Correct answer: (D)

Step-by-step solution →
Q50·ChemistrySingle correct
The ratio KPKC\dfrac{K_P}{K_C}KC​KP​​ for the reaction: CO(g)+12O2(g)⇌CO2(g)CO_{(g)} + \dfrac{1}{2}O_{2(g)} \rightleftharpoons CO_{2(g)}CO(g)​+21​O2(g)​⇌CO2(g)​ is:
  1. (A)(RT)1/2(RT)^{1/2}(RT)1/2
  2. (B)RTRTRT
  3. (C)111
  4. (D)1RT\dfrac{1}{\sqrt{RT}}RT​1​

Correct answer: (D)

Step-by-step solution →
Q51·ChemistryNumerical
An amine (X) is prepared by ammonolysis of benzyl chloride. On adding p-toluenesulphonyl chloride to it the solution remains clear. Molar mass of the amine (X) formed is _______ g mol−1^{-1}−1. (Given molar mass in g mol−1^{-1}−1 C : 12, H : 1, O : 16, N : 14)

Correct answer: 287

Step-by-step solution →
Q52·ChemistryNumerical
Consider the following reactions: NiS+HNO3+HCl→A+NO+S+H2ONiS + HNO_3 + HCl \rightarrow A + NO + S + H_2ONiS+HNO3​+HCl→A+NO+S+H2​O and A+NH4OH+H3C−C(=N−OH)−C(=N−OH)−CH3→B+NH4Cl+H2OA + NH_4OH + H_3C-C(=N-OH)-C(=N-OH)-CH_3 \rightarrow B + NH_4Cl + H_2OA+NH4​OH+H3​C−C(=N−OH)−C(=N−OH)−CH3​→B+NH4​Cl+H2​O. The number of protons that do not involve in hydrogen bonding in the product B is _______.

Correct answer: 12

Step-by-step solution →
Q53·ChemistryNumerical
When 'x' ×10−2\times 10^{-2}×10−2 mL methanol (molar mass = 32 g; density = 0.792 g/cm3^33) is added to 100 mL water (density = 1 g/cm3^33), the following diagram is obtained. x = _______ (nearest integer). [Given: Molal freezing point depression constant of water at 273.15 K is 1.86 K kg mol−1^{-1}−1]

Correct answer: 543

Step-by-step solution →
Q54·ChemistryNumerical
In the reaction sequence shown in the figure, the ratio of the number of oxygen atoms to bromine atoms in the product Q is _______ ×10−1\times 10^{-1}×10−1.

Correct answer: 15

Step-by-step solution →
Q55·Chemistry·Electronic Effects and StabilityNumerical
Number of carbocation from the following (shown in the figure) that are not stabilized by hyperconjugation is _______.

Correct answer: 5

Step-by-step solution →
Q56·ChemistryNumerical
For the reaction at 298 K, 2A+B→C2A + B \rightarrow C2A+B→C. ΔH=400\Delta H = 400ΔH=400 kJ mol−1^{-1}−1 and ΔS=0.2\Delta S = 0.2ΔS=0.2 kJ mol−1^{-1}−1 K−1^{-1}−1. The reaction will become spontaneous above _______ K.

Correct answer: 2000

Step-by-step solution →
Q57·ChemistryNumerical
Total number of species from the following with central atom utilising 2p22p^22p2 hybrid orbitals for bonding is _______. NH3NH_3NH3​, SO2SO_2SO2​, SiO2SiO_2SiO2​, BeCl2BeCl_2BeCl2​, C2H2C_2H_2C2​H2​, C2H4C_2H_4C2​H4​, BCl3BCl_3BCl3​, HCHOHCHOHCHO, C6H6C_6H_6C6​H6​, BF3BF_3BF3​, C2H4Cl2C_2H_4Cl_2C2​H4​Cl2​

Correct answer: 6

Step-by-step solution →
Q58·ChemistryNumerical
Consider the two different first order reactions given below: A+B→CA + B \rightarrow CA+B→C (Reaction 1) and P→QP \rightarrow QP→Q (Reaction 2). The ratio of the half life of Reaction 1 : Reaction 2 is 5 : 2. If t1t_1t1​ and t2t_2t2​ represent the time taken to complete 23rd\frac{2}{3}^{rd}32​rd of Reaction 1 and 45th\frac{4}{5}^{th}54​th of Reaction 2, respectively, then the value of the ratio t1:t2t_1 : t_2t1​:t2​ is _______ ×10−1\times 10^{-1}×10−1 (nearest integer). [Given: log⁡10(3)=0.477\log_{10}(3) = 0.477log10​(3)=0.477 and log⁡10(5)=0.699\log_{10}(5) = 0.699log10​(5)=0.699]

Correct answer: 17

Step-by-step solution →
Q59·ChemistryNumerical
For hydrogen atom, energy of an electron in first excited state is −3.4-3.4−3.4 eV, K.E. of the same electron of hydrogen atom is x eV. Value of x is _______ ×10−1\times 10^{-1}×10−1 eV. (Nearest integer)

Correct answer: 34

Step-by-step solution →
Q60·ChemistryNumerical
Among VO2+VO_2^{+}VO2+​, MnO4−MnO_4^{-}MnO4−​ and Cr2O72−Cr_2O_7^{2-}Cr2​O72−​, the spin-only magnetic moment value of the species with least oxidising ability is _______ BM (Nearest integer). (Given atomic number V = 23, Mn = 25, Cr = 24)

Correct answer: 0

Step-by-step solution →

Mathematics — JEE Main 6 April 2024 Shift 2

Q61·MathematicsSingle correct
Let ABC be an equilateral triangle. A new triangle is formed by joining the middle points of all sides of the triangle ABC and the same process is repeated infinitely many times. If P is the sum of perimeters and Q is be the sum of areas of all the triangles formed in this process, then:
  1. (A)P2=363 QP^2 = 36\sqrt{3}\,QP2=363​Q
  2. (B)P2=63 QP^2 = 6\sqrt{3}\,QP2=63​Q
  3. (C)P=363 Q2P = 36\sqrt{3}\,Q^2P=363​Q2
  4. (D)P2=723 QP^2 = 72\sqrt{3}\,QP2=723​Q

Correct answer: (A)

Step-by-step solution →
Q62·MathematicsSingle correct
Let A={1,2,3,4,5}A = \{1, 2, 3, 4, 5\}A={1,2,3,4,5}. Let R be a relation on A defined by xRyxRyxRy if and only if 4x≤5y4x \le 5y4x≤5y. Let m be the number of elements in R and n be the minimum number of elements from A×AA \times AA×A that are required to be added to R to make it a symmetric relation. Then m+nm + nm+n is equal to:
  1. (A)24
  2. (B)23
  3. (C)25
  4. (D)26

Correct answer: (C)

Step-by-step solution →
Q63·MathematicsSingle correct
If three letters can be posted to any one of the 5 different addresses, then the probability that the three letters are posted to exactly two addresses is:
  1. (A)1225\tfrac{12}{25}2512​
  2. (B)1825\tfrac{18}{25}2518​
  3. (C)425\tfrac{4}{25}254​
  4. (D)625\tfrac{6}{25}256​

Correct answer: (A)

Step-by-step solution →
Q64·MathematicsSingle correct
Suppose the solution of the differential equation dydx=(2+α)x−βy+2βx−2αy−(βγ−4α)\frac{dy}{dx} = \frac{(2+\alpha)x - \beta y + 2}{\beta x - 2\alpha y - (\beta\gamma - 4\alpha)}dxdy​=βx−2αy−(βγ−4α)(2+α)x−βy+2​ represents a circle passing through origin. Then the radius of this circle is:
  1. (A)17\sqrt{17}17​
  2. (B)12\tfrac{1}{2}21​
  3. (C)172\tfrac{\sqrt{17}}{2}217​​
  4. (D)2

Correct answer: (C)

Step-by-step solution →
Q65·MathematicsSingle correct
If the locus of the point, whose distances from the point (2,1)(2, 1)(2,1) and (1,3)(1, 3)(1,3) are in the ratio 5:45 : 45:4, is ax2+by2+cxy+dx+ey+170=0ax^2 + by^2 + cxy + dx + ey + 170 = 0ax2+by2+cxy+dx+ey+170=0, then the value of a2+2b+3c+4d+ea^2 + 2b + 3c + 4d + ea2+2b+3c+4d+e is equal to:
  1. (A)5
  2. (B)−27-27−27
  3. (C)37
  4. (D)437

Correct answer: (C)

Step-by-step solution →
Q66·Mathematics·Limits and ContinuitySingle correct
lim⁡n→∞(12−1)(n−1)+(22−2)(n−2)+⋯+((n−1)2−(n−1))⋅1(13+23+⋯+n3)−(12+22+⋯+n2)\lim_{n\to\infty} \frac{(1^2-1)(n-1) + (2^2-2)(n-2) + \cdots + ((n-1)^2-(n-1))\cdot 1}{(1^3 + 2^3 + \cdots + n^3) - (1^2 + 2^2 + \cdots + n^2)}limn→∞​(13+23+⋯+n3)−(12+22+⋯+n2)(12−1)(n−1)+(22−2)(n−2)+⋯+((n−1)2−(n−1))⋅1​ is equal to:
  1. (A)23\tfrac{2}{3}32​
  2. (B)13\tfrac{1}{3}31​
  3. (C)34\tfrac{3}{4}43​
  4. (D)12\tfrac{1}{2}21​

Correct answer: (B)

Step-by-step solution →
Q67·MathematicsSingle correct
Let 0≤r≤n0 \le r \le n0≤r≤n. If n+1Cr+1:nCr:n−1Cr−1=55:35:21^{n+1}C_{r+1} : {}^{n}C_{r} : {}^{n-1}C_{r-1} = 55 : 35 : 21n+1Cr+1​:nCr​:n−1Cr−1​=55:35:21, then 2n+5r2n + 5r2n+5r is equal to:
  1. (A)60
  2. (B)62
  3. (C)50
  4. (D)55

Correct answer: (C)

Step-by-step solution →
Q68·MathematicsSingle correct
A software company sets up m number of computer systems to finish an assignment in 17 days. If 4 computer systems crashed on the start of the second day, 4 more computer systems crashed on the start of the third day and so on, then it took 8 more days to finish the assignment. The value of m is equal to:
  1. (A)125
  2. (B)150
  3. (C)180
  4. (D)160

Correct answer: (B)

Step-by-step solution →
Q69·MathematicsSingle correct
If z1,z2z_1, z_2z1​,z2​ are two distinct complex number such that ∣z1−2z212−z1z2ˉ∣=2\left|\frac{z_1 - 2z_2}{\frac{1}{2} - z_1\bar{z_2}}\right| = 2​21​−z1​z2​ˉ​z1​−2z2​​​=2, then
  1. (A)either z1z_1z1​ lies on a circle of radius 1 or z2z_2z2​ lies on a circle of radius 12\tfrac{1}{2}21​
  2. (B)either z1z_1z1​ lies on a circle of radius 12\tfrac{1}{2}21​ or z2z_2z2​ lies on a circle of radius 1.
  3. (C)z1z_1z1​ lies on a circle of radius 12\tfrac{1}{2}21​ and z2z_2z2​ lies on a circle of radius 1.
  4. (D)both z1z_1z1​ and z2z_2z2​ lie on the same circle.

Correct answer: (A)

Step-by-step solution →
Q70·MathematicsSingle correct
If the function f(x)=(1x)2xf(x) = \left(\frac{1}{x}\right)^{2x}f(x)=(x1​)2x ; x>0x > 0x>0 attains the maximum value at x=1ex = \frac{1}{e}x=e1​ then:
  1. (A)eπ<πee^{\pi} < \pi^{e}eπ<πe
  2. (B)e2π<(2π)ee^{2\pi} < (2\pi)^{e}e2π<(2π)e
  3. (C)eπ>πee^{\pi} > \pi^{e}eπ>πe
  4. (D)(2e)π>π(2e)(2e)^{\pi} > \pi^{(2e)}(2e)π>π(2e)

Correct answer: (C)

Step-by-step solution →
Q71·MathematicsSingle correct
Let a⃗=6i^+j^−k^\vec{a} = 6\hat{i} + \hat{j} - \hat{k}a=6i^+j^​−k^ and b⃗=i^+j^\vec{b} = \hat{i} + \hat{j}b=i^+j^​. If c⃗\vec{c}c is a vector such that ∣c⃗∣≥6|\vec{c}| \ge 6∣c∣≥6, a⃗⋅c⃗=6∣c⃗∣\vec{a}\cdot\vec{c} = 6|\vec{c}|a⋅c=6∣c∣, ∣c⃗−a⃗∣=22|\vec{c} - \vec{a}| = 2\sqrt{2}∣c−a∣=22​ and the angle between a⃗×b⃗\vec{a}\times\vec{b}a×b and c⃗\vec{c}c is 60∘60^\circ60∘, then ∣(a⃗×b⃗)×c⃗∣|(\vec{a}\times\vec{b})\times\vec{c}|∣(a×b)×c∣ is equal to:
  1. (A)92(6−6)\tfrac{9}{2}(6 - \sqrt{6})29​(6−6​)
  2. (B)323\tfrac{3}{2}\sqrt{3}23​3​
  3. (C)326\tfrac{3}{2}\sqrt{6}23​6​
  4. (D)92(6+6)\tfrac{9}{2}(6 + \sqrt{6})29​(6+6​)

Correct answer: (D)

Step-by-step solution →
Q72·MathematicsSingle correct
If all the words with or without meaning made using all the letters of the word "NAGPUR" are arranged as in a dictionary, then the word at 315th315^{th}315th position in this arrangement is:
  1. (A)NRAGUP
  2. (B)NRAGPU
  3. (C)NRAPGU
  4. (D)NRAPUG

Correct answer: (C)

Step-by-step solution →
Q73·Mathematics·DifferentiabilitySingle correct
Suppose for a differentiable function h, h(0)=0h(0) = 0h(0)=0, h(1)=1h(1) = 1h(1)=1 and h′(0)=h′(1)=2h'(0) = h'(1) = 2h′(0)=h′(1)=2. If g(x)=h(ex) eh(x)g(x) = h(e^x)\,e^{h(x)}g(x)=h(ex)eh(x), then g′(0)g'(0)g′(0) is equal to:
  1. (A)5
  2. (B)3
  3. (C)8
  4. (D)4

Correct answer: (D)

Step-by-step solution →
Q74·MathematicsSingle correct
Let P(α,β,γ)P(\alpha, \beta, \gamma)P(α,β,γ) be the image of the point Q(3,−3,1)Q(3, -3, 1)Q(3,−3,1) in the line x−01=y−31=z−1−1\frac{x-0}{1} = \frac{y-3}{1} = \frac{z-1}{-1}1x−0​=1y−3​=−1z−1​ and R be the point (2,5,−1)(2, 5, -1)(2,5,−1). If the area of the triangle PQR is λ\lambdaλ and λ2=14K\lambda^2 = 14Kλ2=14K, then K is equal to:
  1. (A)36
  2. (B)72
  3. (C)18
  4. (D)81

Correct answer: (D)

Step-by-step solution →
Q75·MathematicsSingle correct
If P(6,1)P(6, 1)P(6,1) be the orthocentre of the triangle whose vertices are A(5,−2)A(5, -2)A(5,−2), B(8,3)B(8, 3)B(8,3) and C(h,k)C(h, k)C(h,k), then the point C lies on the circle.
  1. (A)x2+y2−65=0x^2 + y^2 - 65 = 0x2+y2−65=0
  2. (B)x2+y2−74=0x^2 + y^2 - 74 = 0x2+y2−74=0
  3. (C)x2+y2−61=0x^2 + y^2 - 61 = 0x2+y2−61=0
  4. (D)x2+y2−52=0x^2 + y^2 - 52 = 0x2+y2−52=0

Correct answer: (A)

Step-by-step solution →
Q76·MathematicsSingle correct
Let f(x)=17−sin⁡5xf(x)=\frac{1}{7-\sin 5x}f(x)=7−sin5x1​ be a function defined on RRR. Then the range of the function f(x)f(x)f(x) is equal to:
  1. (A)[18,15]\left[\frac{1}{8},\frac{1}{5}\right][81​,51​]
  2. (B)[17,16]\left[\frac{1}{7},\frac{1}{6}\right][71​,61​]
  3. (C)[17,15]\left[\frac{1}{7},\frac{1}{5}\right][71​,51​]
  4. (D)[18,16]\left[\frac{1}{8},\frac{1}{6}\right][81​,61​]

Correct answer: (D)

Step-by-step solution →
Q77·MathematicsSingle correct
Let a⃗=2i^+j^−k^\vec{a}=2\hat{i}+\hat{j}-\hat{k}a=2i^+j^​−k^, b⃗=((a⃗×(i^+j^))×i^)×i^\vec{b}=\left(\left(\vec{a}\times(\hat{i}+\hat{j})\right)\times\hat{i}\right)\times\hat{i}b=((a×(i^+j^​))×i^)×i^. Then the square of the projection of a⃗\vec{a}a on b⃗\vec{b}b is:
  1. (A)15\frac{1}{5}51​
  2. (B)222
  3. (C)13\frac{1}{3}31​
  4. (D)23\frac{2}{3}32​

Correct answer: (B)

Step-by-step solution →
Q78·MathematicsSingle correct
If the area of the region {(x,y):ax2≤y≤1x, 1≤x≤2, 0<a<1}\left\{(x,y):\frac{a}{x^{2}}\le y\le\frac{1}{x},\,1\le x\le 2,\,0<a<1\right\}{(x,y):x2a​≤y≤x1​,1≤x≤2,0<a<1} is (log⁡e2)−17(\log_{e}2)-\frac{1}{7}(loge​2)−71​ then the value of 7a−37a-37a−3 is equal to:
  1. (A)222
  2. (B)000
  3. (C)−1-1−1
  4. (D)111

Correct answer: (C)

Step-by-step solution →
Q79·MathematicsSingle correct
If ∫1a2sin⁡2x+b2cos⁡2x dx=112tan⁡−1(3tan⁡x)+\int\frac{1}{a^{2}\sin^{2}x+b^{2}\cos^{2}x}\,dx=\frac{1}{12}\tan^{-1}(3\tan x)+∫a2sin2x+b2cos2x1​dx=121​tan−1(3tanx)+ constant, then the maximum value of asin⁡x+bcos⁡xa\sin x+b\cos xasinx+bcosx, is:
  1. (A)40\sqrt{40}40​
  2. (B)39\sqrt{39}39​
  3. (C)42\sqrt{42}42​
  4. (D)41\sqrt{41}41​

Correct answer: (A)

Step-by-step solution →
Q80·MathematicsSingle correct
If AAA is a square matrix of order 333 such that det⁡(A)=3\det(A)=3det(A)=3 and det⁡(adj⁡(−4 adj⁡(−3 adj⁡(3 adj⁡((2A)−1)))))=2m3n\det\left(\operatorname{adj}\left(-4\,\operatorname{adj}\left(-3\,\operatorname{adj}\left(3\,\operatorname{adj}\left((2A)^{-1}\right)\right)\right)\right)\right)=2^{m}3^{n}det(adj(−4adj(−3adj(3adj((2A)−1)))))=2m3n, then m+2nm+2nm+2n is equal to:
  1. (A)333
  2. (B)222
  3. (C)444
  4. (D)666

Correct answer: (C)

Step-by-step solution →
Q81·MathematicsNumerical
Let [t][t][t] denote the greatest integer less than or equal to ttt. Let f:[0,∞)→Rf:[0,\infty)\to Rf:[0,∞)→R be a function defined by f(x)=[x2+3]−[x]f(x)=\left[\frac{x}{2}+3\right]-\left[\sqrt{x}\right]f(x)=[2x​+3]−[x​]. Let SSS be the set of all points in the interval [0,8][0,8][0,8] at which fff is not continuous. Then ∑a∈Sa\sum_{a\in S}a∑a∈S​a is equal to ______.

Correct answer: 17

Step-by-step solution →
Q82·MathematicsNumerical
The length of the latus rectum and directrices of a hyperbola with eccentricity eee are 999 and x=±43x=\pm\frac{4}{\sqrt{3}}x=±3​4​, respectively. Let the line y−3x+3=0y-\sqrt{3}x+\sqrt{3}=0y−3​x+3​=0 touch this hyperbola at (x0,y0)(x_{0},y_{0})(x0​,y0​). If mmm is the product of the focal distances of the point (x0,y0)(x_{0},y_{0})(x0​,y0​), then 4e2+m4e^{2}+m4e2+m is equal to ______.

Correct answer: 61

Step-by-step solution →
Q83·MathematicsNumerical
If S(x)=(1+x)+2(1+x)2+3(1+x)3+⋯+60(1+x)60S(x)=(1+x)+2(1+x)^{2}+3(1+x)^{3}+\dots+60(1+x)^{60}S(x)=(1+x)+2(1+x)2+3(1+x)3+⋯+60(1+x)60, x≠0x\neq 0x=0, and (60)2S(60)=a(b)b+b(60)^{2}S(60)=a(b)^{b}+b(60)2S(60)=a(b)b+b, where a,b∈Na,b\in Na,b∈N, then (a+b)(a+b)(a+b) equal to ______.

Correct answer: 3660

Step-by-step solution →
Q84·MathematicsNumerical
Let [t][t][t] denote the largest integer less than or equal to ttt. If ∫03([x2]+[x22])dx=a+b2−3−5+c6−7\int_{0}^{3}\left([x^{2}]+\left[\frac{x^{2}}{2}\right]\right)dx=a+b\sqrt{2}-\sqrt{3}-\sqrt{5}+c\sqrt{6}-\sqrt{7}∫03​([x2]+[2x2​])dx=a+b2​−3​−5​+c6​−7​, where a,b,c∈Za,b,c\in Za,b,c∈Z, then a+b+ca+b+ca+b+c is equal to ______.

Correct answer: 23

Step-by-step solution →
Q85·MathematicsNumerical
From a lot of 121212 items containing 333 defectives, a sample of 555 items is drawn at random. Let the random variable XXX denote the number of defective items in the sample. Let items in the sample be drawn one by one without replacement. If variance of XXX is mn\frac{m}{n}nm​, where gcd⁡(m,n)=1\gcd(m,n)=1gcd(m,n)=1, then n−mn-mn−m is equal to ______.

Correct answer: 71

Step-by-step solution →
Q86·MathematicsNumerical
In a triangle ABCABCABC, BC=7BC=7BC=7, AC=8AC=8AC=8, AB=α∈NAB=\alpha\in NAB=α∈N and cos⁡A=23\cos A=\frac{2}{3}cosA=32​. If 49cos⁡(3C)+42=mn49\cos(3C)+42=\frac{m}{n}49cos(3C)+42=nm​, where gcd⁡(m,n)=1\gcd(m,n)=1gcd(m,n)=1, then m+nm+nm+n is equal to ______.

Correct answer: 39

Step-by-step solution →
Q87·MathematicsNumerical
If the shortest distance between the lines x−λ3=y−2−1=z−11\frac{x-\lambda}{3}=\frac{y-2}{-1}=\frac{z-1}{1}3x−λ​=−1y−2​=1z−1​ and x+2−3=y+52=z−44\frac{x+2}{-3}=\frac{y+5}{2}=\frac{z-4}{4}−3x+2​=2y+5​=4z−4​ is 4430\frac{44}{\sqrt{30}}30​44​, then the largest possible value of ∣λ∣|\lambda|∣λ∣ is equal to ______.

Correct answer: 43

Step-by-step solution →
Q88·MathematicsNumerical
Let α,β\alpha,\betaα,β be roots of x2+2x−8=0x^{2}+\sqrt{2}x-8=0x2+2​x−8=0. If Un=αn+βnU_{n}=\alpha^{n}+\beta^{n}Un​=αn+βn, then U10+12 U92U8\frac{U_{10}+\sqrt{12}\,U_{9}}{2U_{8}}2U8​U10​+12​U9​​ is equal to ______.

Correct answer: 4

Step-by-step solution →
Q89·MathematicsNumerical
If the system of equations 2x+7y+λz=32x+7y+\lambda z=32x+7y+λz=3, 3x+2y+5z=43x+2y+5z=43x+2y+5z=4, x+μy+32z=−1x+\mu y+32z=-1x+μy+32z=−1 has infinitely many solutions, then (λ−μ)(\lambda-\mu)(λ−μ) is equal to ______.

Correct answer: 38

Step-by-step solution →
Q90·MathematicsNumerical
If the solution y(x)y(x)y(x) of the given differential equation (ey+1)cos⁡x dx+eysin⁡x dy=0(e^{y}+1)\cos x\,dx+e^{y}\sin x\,dy=0(ey+1)cosxdx+eysinxdy=0 passes through the point (π2,0)\left(\frac{\pi}{2},0\right)(2π​,0), then the value of ey(π6)e^{y\left(\frac{\pi}{6}\right)}ey(6π​) is equal to ______.

Correct answer: 3

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
  • Limits and Continuity 149/186
  • d- and f-Block Elements 126/186
  • Thermodynamics 154/186
  • Equilibrium 163/186
  • Solutions 158/186
  • Laws of Motion 130/186
  • Chemical Thermodynamics 165/186
  • Units and Measurements 149/186
  • Complex Numbers 165/186
  • Trigonometric Functions 144/186
  • Biomolecules 162/186
  • Chemical Kinetics 169/186
  • Gravitation 152/186
  • Electric Field and Coulomb's Law 133/186
  • Atomic Structure 161/186
  • Electronic Devices 145/186
  • Circles 142/186
  • Dual Nature of Matter and Radiation 155/186
  • Amines 133/186
  • Quadratic Equations 148/186
  • Work, Energy and Power 132/186
  • Electromagnetic Induction 120/186
  • 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
  • Alternating Currents 108/186
  • Organic Compounds Containing Halogens 109/186
  • Straight Lines 114/186
  • Waves 109/186
  • Capacitors and Dielectrics 115/186
  • Atoms 112/186
  • Differentiability 91/186
  • s-Block Elements 88/186
  • Electronic Effects and Stability 74/186
  • Hyperbola 77/186
  • Experimental Skills 68/186
  • Indefinite Integration 66/186
  • Carboxylic Acids and Derivatives 54/186
  • Principles of Qualitative Analysis 58/186
  • Diazonium Salts and Reactions 53/186
  • Isomerism 51/186
  • Magnetism and Matter 50/186
  • Aromaticity 22/186
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