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JEE Main 9 April 2024 Shift 2 Question Paper with Answers

9 April 2024 · April session · 90 questions

The complete JEE Main 9 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 9 April 2024 Shift 2

Q1·PhysicsSingle correct
A nucleus at rest disintegrates into two smaller nuclei with their masses in the ratio of 2:1. After disintegration they will move:
  1. (A)In opposite directions with speed in the ratio of 1:2 respectively
  2. (B)In opposite directions with speed in the ratio of 2:1 respectively
  3. (C)In the same direction with same speed
  4. (D)In opposite directions with speed in the ratio of 3:2 respectively

Correct answer: (A)

Step-by-step solution →
Q2·PhysicsSingle correct
The figure represents two biconvex lenses L1L_{1}L1​ and L2L_{2}L2​ having focal lengths 10 cm and 15 cm respectively. The distance between L1L_{1}L1​ and L2L_{2}L2​ is:
  1. (A)10 cm
  2. (B)15 cm
  3. (C)5 cm
  4. (D)35 cm

Correct answer: (C)

Step-by-step solution →
Q3·PhysicsSingle correct
The temperature of a gas is −78 ∘-78\,^{\circ}−78∘C and the average translational kinetic energy of its molecules is K. The temperature at which the average translational kinetic energy of the molecules of the same gas becomes 2K is:
  1. (A)−39 ∘-39\,^{\circ}−39∘C
  2. (B)117 ∘117\,^{\circ}117∘C
  3. (C)127 ∘127\,^{\circ}127∘C
  4. (D)−78 ∘-78\,^{\circ}−78∘C

Correct answer: (B)

Step-by-step solution →
Q4·PhysicsSingle correct
A hydrogen atom in ground state is given an energy of 10.2 eV. How many spectral lines will be emitted due to transition of electrons?
  1. (A)6
  2. (B)3
  3. (C)10
  4. (D)1

Correct answer: (D)

Step-by-step solution →
Q5·PhysicsSingle correct
The magnetic field in a plane electromagnetic wave is By=(3.5×10−7)sin⁡(1.5×103x+0.5×1011t)B_{y}=(3.5\times10^{-7})\sin(1.5\times10^{3}x+0.5\times10^{11}t)By​=(3.5×10−7)sin(1.5×103x+0.5×1011t) T. The corresponding electric field will be:
  1. (A)Ez=1.17sin⁡(1.5×103x+0.5×1011t)E_{z}=1.17\sin(1.5\times10^{3}x+0.5\times10^{11}t)Ez​=1.17sin(1.5×103x+0.5×1011t) V m−1^{-1}−1
  2. (B)Ez=105sin⁡(1.5×103x+0.5×1011t)E_{z}=105\sin(1.5\times10^{3}x+0.5\times10^{11}t)Ez​=105sin(1.5×103x+0.5×1011t) V m−1^{-1}−1
  3. (C)Ez=1.17sin⁡(1.5×103x+0.5×1011t)E_{z}=1.17\sin(1.5\times10^{3}x+0.5\times10^{11}t)Ez​=1.17sin(1.5×103x+0.5×1011t) nV m−1^{-1}−1
  4. (D)Ez=10.5sin⁡(1.5×103x+0.5×1011t)E_{z}=10.5\sin(1.5\times10^{3}x+0.5\times10^{11}t)Ez​=10.5sin(1.5×103x+0.5×1011t) nV m−1^{-1}−1

Correct answer: (B)

Step-by-step solution →
Q6·PhysicsSingle correct
A square loop of side 15 cm is being moved towards right at a constant speed of 2 cm/s as shown in the figure. The front edge enters the 50 cm wide magnetic field at t=0t=0t=0. The value of induced emf in the loop at t=10t=10t=10 s will be:
  1. (A)0.3 mV
  2. (B)4.5 mV
  3. (C)0 mV
  4. (D)3 mV

Correct answer: (C)

Step-by-step solution →
Q7·PhysicsSingle correct
Two cars are travelling towards each other at speed of 20 m s−1^{-1}−1 each. When the cars are 300 m apart, both the drivers apply brakes and the cars retard at the rate of 2 m s−2^{-2}−2. The distance between them when they come to rest is:
  1. (A)200 m
  2. (B)50 m
  3. (C)100 m
  4. (D)25 m

Correct answer: (C)

Step-by-step solution →
Q8·PhysicsSingle correct
The III-VVV characteristics of an electronic device is shown in the figure. The device is:
  1. (A)a solar cell
  2. (B)a transistor which can be used as an amplifier
  3. (C)a zener diode which can be used as a voltage regulator
  4. (D)a diode which can be used as a rectifier

Correct answer: (C)

Step-by-step solution →
Q9·PhysicsSingle correct
The excess pressure inside a soap bubble is thrice the excess pressure inside a second soap bubble. The ratio between the volume of the first and the second bubble is:
  1. (A)1:9
  2. (B)1:3
  3. (C)1:81
  4. (D)1:27

Correct answer: (D)

Step-by-step solution →
Q10·PhysicsSingle correct
The de-Broglie wavelength associated with a particle of mass mmm and energy EEE is h/2mEh/\sqrt{2mE}h/2mE​. The dimensional formula for Planck's constant is:
  1. (A)[ML−1T−1][ML^{-1}T^{-1}][ML−1T−1]
  2. (B)[ML2T−1][ML^{2}T^{-1}][ML2T−1]
  3. (C)[MLT−1][MLT^{-1}][MLT−1]
  4. (D)[M4L2T−3][M^{4}L^{2}T^{-3}][M4L2T−3]

Correct answer: (B)

Step-by-step solution →
Q11·PhysicsSingle correct
A satellite of 10310^{3}103 kg mass is revolving in a circular orbit of radius 2R2R2R. If 104R6\dfrac{10^{4}R}{6}6104R​ J energy is supplied to the satellite, it would revolve in a new circular orbit of radius: (use g=10g=10g=10 m/s2^{2}2, R=R=R= radius of earth)
  1. (A)2.5 R
  2. (B)3 R
  3. (C)4 R
  4. (D)6 R

Correct answer: (D)

Step-by-step solution →
Q12·PhysicsSingle correct
The effective resistance between AAA and BBB, if resistance of each resistor is RRR, will be:
  1. (A)23R\dfrac{2}{3}R32​R
  2. (B)8R3\dfrac{8R}{3}38R​
  3. (C)5R3\dfrac{5R}{3}35R​
  4. (D)4R3\dfrac{4R}{3}34R​

Correct answer: (B)

Step-by-step solution →
Q13·PhysicsSingle correct
Five charges +q+q+q, +5q+5q+5q, −2q-2q−2q, +3q+3q+3q and −4q-4q−4q are situated as shown in the figure. The electric flux due to this configuration through the surface S is:
  1. (A)5qε0\dfrac{5q}{\varepsilon_{0}}ε0​5q​
  2. (B)4qε0\dfrac{4q}{\varepsilon_{0}}ε0​4q​
  3. (C)3qε0\dfrac{3q}{\varepsilon_{0}}ε0​3q​
  4. (D)qε0\dfrac{q}{\varepsilon_{0}}ε0​q​

Correct answer: (B)

Step-by-step solution →
Q14·PhysicsSingle correct
A proton and a deuteron (q=+eq=+eq=+e, m=2.0m=2.0m=2.0 u) having same kinetic energies enter a region of uniform magnetic field B⃗\vec{B}B, moving perpendicular to B⃗\vec{B}B. The ratio of the radius rdr_{d}rd​ of the deuteron path to the radius rpr_{p}rp​ of the proton path is:
  1. (A)1:1
  2. (B)1:21:\sqrt{2}1:2​
  3. (C)2:1\sqrt{2}:12​:1
  4. (D)1:2

Correct answer: (C)

Step-by-step solution →
Q15·PhysicsSingle correct
UV light of 4.13 eV is incident on a photosensitive metal surface having work function 3.13 eV. The maximum kinetic energy of ejected photoelectrons will be:
  1. (A)1.13 eV
  2. (B)1 eV
  3. (C)3.13 eV
  4. (D)7.26 eV

Correct answer: (B)

Step-by-step solution →
Q16·PhysicsSingle correct
The energy released in the fusion of 2 kg of hydrogen deep in the sun is EHE_{H}EH​ and the energy released in the fission of 2 kg of 235^{235}235U is EUE_{U}EU​. The ratio EHEU\dfrac{E_{H}}{E_{U}}EU​EH​​ is approximately: (Consider the fusion reaction as 4 11H+2e−→ 24He+2ν+6γ+26.74\,^{1}_{1}H+2e^{-}\rightarrow\,^{4}_{2}He+2\nu+6\gamma+26.7411​H+2e−→24​He+2ν+6γ+26.7 MeV; energy released in the fission reaction of 235^{235}235U is 200 MeV per fission and NA=6.023×1023N_{A}=6.023\times10^{23}NA​=6.023×1023)
  1. (A)9.13
  2. (B)5.04
  3. (C)7.62
  4. (D)25.6

Correct answer: (C)

Step-by-step solution →
Q17·PhysicsSingle correct
A real gas within a closed chamber at 27 ∘27\,^{\circ}27∘C undergoes the cyclic process as shown in the figure. The gas obeys PV3=RTPV^{3}=RTPV3=RT equation for the path A to B. The net work done in the complete cycle is (assuming R=8R=8R=8 J/mol K):
  1. (A)225 J
  2. (B)205 J
  3. (C)20 J
  4. (D)−20-20−20 J

Correct answer: (B)

Step-by-step solution →
Q18·PhysicsSingle correct
A 1 kg mass is suspended from the ceiling by a rope of length 4 m. A horizontal force FFF is applied at the mid point of the rope so that the rope makes an angle of 45∘45^{\circ}45∘ with respect to the vertical axis as shown in the figure. The magnitude of FFF is: (use g=10g=10g=10 m/s2^{2}2)
  1. (A)102\dfrac{10}{\sqrt{2}}2​10​ N
  2. (B)1 N
  3. (C)1102\dfrac{1}{10\sqrt{2}}102​1​ N
  4. (D)10 N

Correct answer: (D)

Step-by-step solution →
Q19·PhysicsSingle correct
A spherical ball of radius 1×10−41\times10^{-4}1×10−4 m and density 10510^{5}105 kg/m3^{3}3 falls freely under gravity through a distance hhh before entering a tank of water. If after entering in water the velocity of the ball does not change, then the value of hhh is approximately: (The coefficient of viscosity of water is 9.8×10−69.8\times10^{-6}9.8×10−6 N s/m2^{2}2, g=9.8g=9.8g=9.8 m/s2^{2}2)
  1. (A)2296 m
  2. (B)2249 m
  3. (C)2518 m
  4. (D)2396 m

Correct answer: (C)

Step-by-step solution →
Q20·PhysicsSingle correct
In the truth table of the logic circuit shown in the figure, the values of X and Y are:
  1. (A)1, 1
  2. (B)1, 0
  3. (C)0, 1
  4. (D)0, 0

Correct answer: (B)

Step-by-step solution →
Q21·PhysicsNumerical
A straight magnetic strip has a magnetic moment of 44 A m2^{2}2. If the strip is bent in a semicircular shape, its magnetic moment will be _______ A m2^{2}2. (Given π=227\pi=\dfrac{22}{7}π=722​)

Correct answer: 28

Step-by-step solution →
Q22·PhysicsNumerical
A particle of mass 0.50 kg executes simple harmonic motion under force F=−50 (N/m) xF=-50\,(\text{N/m})\,xF=−50(N/m)x. The time period of oscillation is x35\dfrac{x}{35}35x​ s. The value of x is _______. (Given π=227\pi=\dfrac{22}{7}π=722​)

Correct answer: 22

Step-by-step solution →
Q23·PhysicsNumerical
A capacitor of reactance 43 Ω4\sqrt{3}\,\Omega43​Ω and a resistor of resistance 4 Ω4\,\Omega4Ω are connected in series with an ac source of peak value 828\sqrt{2}82​ V. The power dissipation in the circuit is _______ W.

Correct answer: 4

Step-by-step solution →
Q24·PhysicsNumerical
An electric field E⃗=(2xi^)\vec{E}=(2x\hat{i})E=(2xi^) N/C exists in space. A cube of side 2 m is placed in the space with its edges parallel to the coordinate axes and one corner at the origin. The electric flux through the cube is _______ N m2^{2}2/C.

Correct answer: 16

Step-by-step solution →
Q25·PhysicsNumerical
A circular disc reaches from top to bottom of an inclined plane of length lll. When it slips down the plane, it takes time t1t_{1}t1​. When it rolls down the plane then it takes (α2)1/2t1\left(\dfrac{\alpha}{2}\right)^{1/2}t_{1}(2α​)1/2t1​ s, where α\alphaα is _______.

Correct answer: 3

Step-by-step solution →
Q26·PhysicsNumerical
To determine the resistance (RRR) of a wire, a circuit is designed as shown in the figure. The V-I characteristic curve for this circuit is plotted for the voltmeter and the ammeter readings as shown. The value of RRR is _______ Ω\OmegaΩ.

Correct answer: 2500

Step-by-step solution →
Q27·PhysicsNumerical
The resultant of two vectors A⃗\vec{A}A and B⃗\vec{B}B is perpendicular to A⃗\vec{A}A and its magnitude is half of that of B⃗\vec{B}B. The angle between vectors A⃗\vec{A}A and B⃗\vec{B}B is _______ degrees.

Correct answer: 150

Step-by-step solution →
Q28·PhysicsNumerical
Monochromatic light of wavelength 500 nm is used in Young's double slit experiment. An interference pattern is obtained on a screen. When one of the slits is covered with a very thin glass plate (refractive index = 1.5), the central maximum is shifted to a position previously occupied by the 4th4^{th}4th bright fringe. The thickness of the glass plate is _______ μ\muμm.

Correct answer: 4

Step-by-step solution →
Q29·PhysicsNumerical
A force (3x2+2x−5)(3x^{2}+2x-5)(3x2+2x−5) N displaces a body from x=2x=2x=2 m to x=4x=4x=4 m. Work done by this force is _______ J.

Correct answer: 58

Step-by-step solution →
Q30·PhysicsNumerical
At room temperature (27 ∘27\,^{\circ}27∘C), the resistance of a heating element is 50 Ω50\,\Omega50Ω. The temperature coefficient of the material is 2.4×10−4 ∘2.4\times10^{-4}\ ^{\circ}2.4×10−4 ∘C−1^{-1}−1. The temperature of the element, when its resistance is 62 Ω62\,\Omega62Ω, is _______ ∘^{\circ}∘C.

Correct answer: 1027

Step-by-step solution →

Chemistry — JEE Main 9 April 2024 Shift 2

Q31·ChemistrySingle correct
The candela is the luminous intensity, in a given direction, of a source that emits monochromatic radiation of frequency 'A' ×1012\times10^{12}×1012 hertz and that has a radiant intensity in that direction of 1′B′\dfrac{1}{'B'}′B′1​ watt per steradian. 'A' and 'B' are respectively:
  1. (A)540 and 1683\dfrac{1}{683}6831​
  2. (B)540 and 683
  3. (C)450 and 1683\dfrac{1}{683}6831​
  4. (D)450 and 683

Correct answer: (B)

Step-by-step solution →
Q32·Chemistry·Electronic Effects and StabilitySingle correct
The correct stability order of the following resonance structures of CH3−CH=CH−CHOCH_3-CH=CH-CHOCH3​−CH=CH−CHO (labelled I, II and III in the figure) is:
  1. (A)II > III > I
  2. (B)III > II > I
  3. (C)I > II > III
  4. (D)II > I > III

Correct answer: (B)

Step-by-step solution →
Q33·Chemistry·IsomerismSingle correct
Total number of stereoisomers possible for the structure shown in the figure is:
  1. (A)8
  2. (B)2
  3. (C)4
  4. (D)3

Correct answer: (A)

Step-by-step solution →
Q34·ChemistrySingle correct
The correct increasing order for bond angles among BF3BF_3BF3​, PF3PF_3PF3​ and ClF3ClF_3ClF3​ is:
  1. (A)PF3<ClF3<BF3PF_3 < ClF_3 < BF_3PF3​<ClF3​<BF3​
  2. (B)BF3<PF3<ClF3BF_3 < PF_3 < ClF_3BF3​<PF3​<ClF3​
  3. (C)ClF3<PF3<BF3ClF_3 < PF_3 < BF_3ClF3​<PF3​<BF3​
  4. (D)BF3=PF3=ClF3BF_3 = PF_3 = ClF_3BF3​=PF3​=ClF3​

Correct answer: (C)

Step-by-step solution →
Q35·ChemistrySingle correct
Match List-I (Test) with List-II (Observation). Choose the correct answer from the options given below:
List-I (Test)List-II (Observation)
A.Br2Br_2Br2​ water testI.Yellow orange or orange red precipitate formed
B.Ceric ammonium nitrate testII.Reddish orange colour disappears
C.Ferric chloride testIII.Red colour appears
D.2,4-DNP testIV.Blue, Green, Violet or Red colour appear
  1. (A)A-I, B-II, C-III, D-IV
  2. (B)A-II, B-III, C-IV, D-I
  3. (C)A-III, B-IV, C-I, D-II
  4. (D)A-IV, B-III, C-II, D-I

Correct answer: (B)

Step-by-step solution →
Q36·ChemistrySingle correct
Match List-I (Cell) with List-II (Use / Property / Reaction). Choose the correct answer from the options given below:
List-I (Cell)List-II (Use / Property / Reaction)
A.Leclanche cellI.Converts energy of combustion into electrical energy
B.Ni-Cd cellII.Does not involve any ion in solution and is used in hearing aids
C.Fuel cellIII.Rechargeable
D.Mercury cellIV.Reaction at anode Zn→Zn2++2e−Zn \rightarrow Zn^{2+} + 2e^{-}Zn→Zn2++2e−
  1. (A)A-I, B-II, C-III, D-IV
  2. (B)A-III, B-I, C-IV, D-II
  3. (C)A-IV, B-III, C-I, D-II
  4. (D)A-II, B-III, C-IV, D-I

Correct answer: (C)

Step-by-step solution →
Q37·ChemistrySingle correct
Match List-I (Complex) with List-II (Hybridisation). Choose the correct answer from the options given below:
List-I (Complex)List-II (Hybridisation)
A.K2[Ni(CN)4]K_2[Ni(CN)_4]K2​[Ni(CN)4​]I.sp3sp^3sp3
B.[Ni(CO)4][Ni(CO)_4][Ni(CO)4​]II.sp3d2sp^3d^2sp3d2
C.[Co(NH3)6]Cl3[Co(NH_3)_6]Cl_3[Co(NH3​)6​]Cl3​III.dsp2dsp^2dsp2
D.Na3[CoF6]Na_3[CoF_6]Na3​[CoF6​]IV.d2sp3d^2sp^3d2sp3
  1. (A)A-III, B-I, C-II, D-IV
  2. (B)A-III, B-II, C-IV, D-I
  3. (C)A-I, B-III, C-II, D-IV
  4. (D)A-III, B-I, C-IV, D-II

Correct answer: (D)

Step-by-step solution →
Q38·ChemistrySingle correct
The coordination environment of Ca2+Ca^{2+}Ca2+ ion in its complex with EDTA4−EDTA^{4-}EDTA4− is:
  1. (A)tetrahedral
  2. (B)octahedral
  3. (C)square planar
  4. (D)trigonal prismatic

Correct answer: (B)

Step-by-step solution →
Q39·ChemistrySingle correct
The incorrect statement about Glucose is:
  1. (A)Glucose is soluble in water because of having aldehyde functional group
  2. (B)Glucose remains in multiple isomeric form in its aqueous solution
  3. (C)Glucose is an aldohexose
  4. (D)Glucose is one of the monomer unit in sucrose

Correct answer: (A)

Step-by-step solution →
Q40·ChemistrySingle correct
In the reaction shown in the figure, the product 'P' is:
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (A)

Step-by-step solution →
Q41·ChemistrySingle correct
Which of the following compounds (shown in the figure) can give a positive iodoform test when treated with aqueous KOH solution followed by potassium hypoiodite?
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (B)

Step-by-step solution →
Q42·ChemistrySingle correct
For a sparingly soluble salt AB2AB_2AB2​, the equilibrium concentrations of A2+A^{2+}A2+ ions and B−B^{-}B− ions are 1.2×10−41.2\times10^{-4}1.2×10−4 M and 0.24×10−30.24\times10^{-3}0.24×10−3 M, respectively. The solubility product of AB2AB_2AB2​ is:
  1. (A)0.069×10−120.069\times10^{-12}0.069×10−12
  2. (B)6.91×10−126.91\times10^{-12}6.91×10−12
  3. (C)0.276×10−120.276\times10^{-12}0.276×10−12
  4. (D)27.65×10−1227.65\times10^{-12}27.65×10−12

Correct answer: (B)

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

Correct answer: (B)

Step-by-step solution →
Q44·ChemistrySingle correct
Given below are two statements: Statement I: The higher oxidation states are more stable down the group among transition elements unlike p-block elements. Statement II: Copper can not liberate hydrogen from weak acids. In the light of the above statements, choose the correct answer from the options given below:
  1. (A)Both Statement I and Statement II are false
  2. (B)Statement I is false but Statement II is true
  3. (C)Both Statement I and Statement II are true
  4. (D)Statement I is true but Statement II is false

Correct answer: (C)

Step-by-step solution →
Q45·ChemistrySingle correct
The incorrect statement regarding ethyne is:
  1. (A)The C-C bond in ethyne is shorter than that in ethene
  2. (B)Both carbons are sp hybridised
  3. (C)Ethyne is linear
  4. (D)The carbon-carbon bond in ethyne is weaker than that in ethene

Correct answer: (D)

Step-by-step solution →
Q46·ChemistrySingle correct
Match List-I (Element) with List-II (Electronic Configuration). Choose the correct answer from the options given below:
List-I (Element)List-II (Electronic Configuration)
A.NI.[Ar]3d104s24p5[Ar]3d^{10}4s^24p^5[Ar]3d104s24p5
B.SII.[Ne]3s23p4[Ne]3s^23p^4[Ne]3s23p4
C.BrIII.[He]2s22p3[He]2s^22p^3[He]2s22p3
D.KrIV.[Ar]3d104s24p6[Ar]3d^{10}4s^24p^6[Ar]3d104s24p6
  1. (A)A-IV, B-III, C-II, D-I
  2. (B)A-III, B-II, C-I, D-IV
  3. (C)A-IV, B-IV, C-III, D-II
  4. (D)A-II, B-I, C-IV, D-III

Correct answer: (B)

Step-by-step solution →
Q47·ChemistrySingle correct
Match List-I (Property) with List-II (Order). Choose the correct answer from the options given below:
List-I (Property)List-II (Order)
A.Melting point [K]I.Tl > In > Ga > Al > B
B.Ionic Radius [M3+M^{3+}M3+/pm]II.B > Tl > Al > Ga > In
C.ΔiH1\Delta_i H_1Δi​H1​ [kJ mol−1^{-1}−1]III.Tl > In > Al > Ga > B
D.Atomic Radius [pm]IV.B > Al > Tl > In > Ga
  1. (A)A-III, B-IV, C-I, D-II
  2. (B)A-II, B-III, C-IV, D-I
  3. (C)A-IV, B-I, C-II, D-III
  4. (D)A-I, B-II, C-III, D-IV

Correct answer: (C)

Step-by-step solution →
Q48·ChemistrySingle correct
Which of the following compounds will give silver mirror with ammoniacal silver nitrate? (A) Formic acid (B) Formaldehyde (C) Benzaldehyde (D) Acetone. Choose the correct answer from the options given below:
  1. (A)C and D only
  2. (B)A, B and C only
  3. (C)A only
  4. (D)B and C only

Correct answer: (B)

Step-by-step solution →
Q49·ChemistrySingle correct
Which out of the following is a correct equation to show change in molar conductivity with respect to concentration of a weak electrolyte, if the symbols carry their usual meaning:
  1. (A)Λm2C−Ka(Λm0)2+KaΛmΛm0=0\Lambda_m^{2}C - K_a(\Lambda_m^{0})^{2} + K_a\Lambda_m\Lambda_m^{0} = 0Λm2​C−Ka​(Λm0​)2+Ka​Λm​Λm0​=0
  2. (B)Λm=Λm0+AC1/2=0\Lambda_m = \Lambda_m^{0} + AC^{1/2} = 0Λm​=Λm0​+AC1/2=0
  3. (C)Λm=Λm0−AC1/2=0\Lambda_m = \Lambda_m^{0} - AC^{1/2} = 0Λm​=Λm0​−AC1/2=0
  4. (D)Λm2C+Ka(Λm0)2−KaΛmΛm0=0\Lambda_m^{2}C + K_a(\Lambda_m^{0})^{2} - K_a\Lambda_m\Lambda_m^{0} = 0Λm2​C+Ka​(Λm0​)2−Ka​Λm​Λm0​=0

Correct answer: (A)

Step-by-step solution →
Q50·ChemistrySingle correct
The electronic configuration of Einsteinium is: (Given atomic number of Einsteinium = 99)
  1. (A)[Rn]5f126d07s2[Rn]5f^{12}6d^{0}7s^{2}[Rn]5f126d07s2
  2. (B)[Rn]5f116d07s2[Rn]5f^{11}6d^{0}7s^{2}[Rn]5f116d07s2
  3. (C)[Rn]5f136d07s2[Rn]5f^{13}6d^{0}7s^{2}[Rn]5f136d07s2
  4. (D)[Rn]5f116d17s2[Rn]5f^{11}6d^{1}7s^{2}[Rn]5f116d17s2

Correct answer: (B)

Step-by-step solution →
Q51·ChemistryNumerical
Number of oxygen atoms present in the chemical formula of fuming sulphuric acid is _______.

Correct answer: 7

Step-by-step solution →
Q52·ChemistryNumerical
A transition metal 'M' among Sc, Ti, V, Cr, Mn and Fe has the highest second ionisation enthalpy. The spin-only magnetic moment value of M+M^{+}M+ ion is _______ BM (nearest integer). (Given atomic number Sc: 21, Ti: 22, V: 23, Cr: 24, Mn: 25, Fe: 26)

Correct answer: 6

Step-by-step solution →
Q53·ChemistryNumerical
The vapour pressure of pure benzene and methyl benzene at 27 ∘27\,^{\circ}27∘C is given as 80 Torr and 24 Torr, respectively. The mole fraction of methyl benzene in vapour phase, in equilibrium with an equimolar mixture of these two liquids (ideal solution) at the same temperature is _______ ×10−2\times10^{-2}×10−2 (nearest integer).

Correct answer: 23

Step-by-step solution →
Q54·ChemistryNumerical
Consider the following test for a group-IV cation: M2++H2S→AM^{2+}+H_2S\rightarrow AM2++H2​S→A (Black precipitate) + byproduct; A+A+A+ aqua regia →B+NOCl+S+H2O\rightarrow B+NOCl+S+H_2O→B+NOCl+S+H2​O; B+KNO2+CH3COOH→CB+KNO_2+CH_3COOH\rightarrow CB+KNO2​+CH3​COOH→C + byproduct. The spin-only magnetic moment value of the metal complex C is _______ BM (nearest integer).

Correct answer: 0

Step-by-step solution →
Q55·ChemistryNumerical
Consider the following first order gas phase reaction at constant temperature: A(g)→2B(g)+C(g)A(g)\rightarrow 2B(g)+C(g)A(g)→2B(g)+C(g). If the total pressure of the gases is found to be 200 torr after 23 sec and 300 torr upon the complete decomposition of A after a very long time, then the rate constant of the given reaction is _______ ×10−2\times10^{-2}×10−2 s−1^{-1}−1 (nearest integer). [Given: log⁡10(2)=0.301\log_{10}(2)=0.301log10​(2)=0.301]

Correct answer: 3

Step-by-step solution →
Q56·ChemistryNumerical
In thin layer chromatography (TLC), the distance of spot A and B are 5 cm and 7 cm from the bottom of the TLC plate, respectively (as shown in the figure). RfR_fRf​ value of B is x×10−1x\times10^{-1}x×10−1 times more than A. The value of x is _______.

Correct answer: 15

Step-by-step solution →
Q57·ChemistryNumerical
Based on Heisenberg's uncertainty principle, the uncertainty in the velocity of the electron to be found within an atomic nucleus of diameter 10−1510^{-15}10−15 m is _______ ×109\times10^{9}×109 m s−1^{-1}−1 (nearest integer). [Given: mass of electron =9.1×10−31=9.1\times10^{-31}=9.1×10−31 kg, Planck's constant (hhh) =6.626×10−34=6.626\times10^{-34}=6.626×10−34 J s]

Correct answer: 58

Step-by-step solution →
Q58·ChemistryNumerical
Number of compounds from the following which cannot undergo Friedel-Crafts reactions is _______: toluene, nitrobenzene, xylene, cumene, aniline, chlorobenzene, m-nitroaniline, m-dinitrobenzene.

Correct answer: 4

Step-by-step solution →
Q59·ChemistryNumerical
Total number of electrons present in (π∗)(\pi^{*})(π∗) molecular orbitals of O2O_2O2​, O2+O_2^{+}O2+​ and O2−O_2^{-}O2−​ is _______.

Correct answer: 6

Step-by-step solution →
Q60·ChemistryNumerical
When ΔHvap=30\Delta H_{vap}=30ΔHvap​=30 kJ/mol and ΔSvap=75\Delta S_{vap}=75ΔSvap​=75 J mol−1^{-1}−1 K−1^{-1}−1, then the temperature of vapour, at one atmosphere is _______ K.

Correct answer: 400

Step-by-step solution →

Mathematics — JEE Main 9 April 2024 Shift 2

Q61·Mathematics·Limits and ContinuitySingle correct
lim⁡x→0e−(1+2x)12xx\displaystyle\lim_{x\to 0}\dfrac{e-(1+2x)^{\frac{1}{2x}}}{x}x→0lim​xe−(1+2x)2x1​​ is equal to:
  1. (A)eee
  2. (B)−2e\dfrac{-2}{e}e−2​
  3. (C)000
  4. (D)−e2-e^{2}−e2

Correct answer: (A)

Step-by-step solution →
Q62·MathematicsSingle correct
Consider the line L passing through the points (1,2,3)(1, 2, 3)(1,2,3) and (2,3,5)(2, 3, 5)(2,3,5). The distance of the point (113,113,193)\left(\dfrac{11}{3}, \dfrac{11}{3}, \dfrac{19}{3}\right)(311​,311​,319​) from the line L along the line 3x−112=3y−111=3z−192\dfrac{3x-11}{2}=\dfrac{3y-11}{1}=\dfrac{3z-19}{2}23x−11​=13y−11​=23z−19​ is equal to:
  1. (A)333
  2. (B)555
  3. (C)444
  4. (D)666

Correct answer: (A)

Step-by-step solution →
Q63·MathematicsSingle correct
Let ∫0x1−(y′(t))2 dt=∫0xy(t) dt\displaystyle\int_{0}^{x}\sqrt{1-(y'(t))^{2}}\,dt=\int_{0}^{x}y(t)\,dt∫0x​1−(y′(t))2​dt=∫0x​y(t)dt, 0≤x≤30\le x\le 30≤x≤3, y≥0y\ge 0y≥0, y(0)=0y(0)=0y(0)=0. Then at x=2x=2x=2, y′′+y+1y''+y+1y′′+y+1 is equal to:
  1. (A)111
  2. (B)222
  3. (C)2\sqrt{2}2​
  4. (D)12\dfrac{1}{2}21​

Correct answer: (A)

Step-by-step solution →
Q64·MathematicsSingle correct
Let z be a complex number such that the real part of z−2iz+2i\dfrac{z-2i}{z+2i}z+2iz−2i​ is zero. Then, the maximum value of ∣z−(6+8i)∣|z-(6+8i)|∣z−(6+8i)∣ is equal to:
  1. (A)121212
  2. (B)∞\infty∞
  3. (C)101010
  4. (D)888

Correct answer: (A)

Step-by-step solution →
Q65·MathematicsSingle correct
The area (in square units) of the region enclosed by the ellipse x2+3y2=18x^{2}+3y^{2}=18x2+3y2=18 in the first quadrant below the line y=xy=xy=x is:
  1. (A)3π+34\sqrt{3}\pi+\dfrac{3}{4}3​π+43​
  2. (B)3π\sqrt{3}\pi3​π
  3. (C)3π−34\sqrt{3}\pi-\dfrac{3}{4}3​π−43​
  4. (D)3π+1\sqrt{3}\pi+13​π+1

Correct answer: (B)

Step-by-step solution →
Q66·MathematicsSingle correct
Let the foci of a hyperbola H coincide with the foci of the ellipse E:(x−1)2100+(y−1)275=1E:\dfrac{(x-1)^{2}}{100}+\dfrac{(y-1)^{2}}{75}=1E:100(x−1)2​+75(y−1)2​=1 and the eccentricity of the hyperbola H be the reciprocal of the eccentricity of the ellipse E. If the length of the transverse axis of H is α\alphaα and the length of its conjugate axis is β\betaβ, then 3α2+2β23\alpha^{2}+2\beta^{2}3α2+2β2 is equal to:
  1. (A)242242242
  2. (B)225225225
  3. (C)237237237
  4. (D)205205205

Correct answer: (B)

Step-by-step solution →
Q67·MathematicsSingle correct
Two vertices of a triangle ABC are A(3,−1)A(3, -1)A(3,−1) and B(−2,3)B(-2, 3)B(−2,3), and its orthocentre is P(1,1)P(1, 1)P(1,1). If the coordinates of the point C are (α,β)(\alpha, \beta)(α,β) and the centre of the circle circumscribing the triangle PAB is (h,k)(h, k)(h,k), then the value of (α+β)+2(h+k)(\alpha+\beta)+2(h+k)(α+β)+2(h+k) equals:
  1. (A)515151
  2. (B)818181
  3. (C)555
  4. (D)151515

Correct answer: (C)

Step-by-step solution →
Q68·MathematicsSingle correct
If the variance of the frequency distribution given by xxx: ccc, 2c2c2c, 3c3c3c, 4c4c4c, 5c5c5c, 6c6c6c with corresponding frequencies fff: 222, 111, 111, 111, 111, 111 is 160, then the value of c∈Nc\in\mathbb{N}c∈N is:
  1. (A)555
  2. (B)888
  3. (C)777
  4. (D)666

Correct answer: (C)

Step-by-step solution →
Q69·MathematicsSingle correct
Let the range of the function f(x)=12+sin⁡3x+cos⁡3xf(x)=\dfrac{1}{2+\sin 3x+\cos 3x}f(x)=2+sin3x+cos3x1​, x∈Rx\in\mathbb{R}x∈R be [a,b][a, b][a,b]. If α\alphaα and β\betaβ are respectively the A.M. and the G.M. of aaa and bbb, then αβ\dfrac{\alpha}{\beta}βα​ is equal to:
  1. (A)2\sqrt{2}2​
  2. (B)222
  3. (C)π\sqrt{\pi}π​
  4. (D)π\piπ

Correct answer: (A)

Step-by-step solution →
Q70·MathematicsSingle correct
Between the following two statements: Statement-I: Let a⃗=i^+2j^−3k^\vec{a}=\hat{i}+2\hat{j}-3\hat{k}a=i^+2j^​−3k^ and b⃗=2i^+j^−k^\vec{b}=2\hat{i}+\hat{j}-\hat{k}b=2i^+j^​−k^. Then the vector r⃗\vec{r}r satisfying a⃗×r⃗=a⃗×b⃗\vec{a}\times\vec{r}=\vec{a}\times\vec{b}a×r=a×b and a⃗⋅r⃗=0\vec{a}\cdot\vec{r}=0a⋅r=0 is of magnitude 10\sqrt{10}10​. Statement-II: In a triangle ABC, cos⁡2A+cos⁡2B+cos⁡2C≥−32\cos 2A+\cos 2B+\cos 2C\ge -\dfrac{3}{2}cos2A+cos2B+cos2C≥−23​. In the light of the above statements, choose the correct answer:
  1. (A)Both Statement-I and Statement-II are incorrect
  2. (B)Statement-I is incorrect but Statement-II is correct
  3. (C)Both Statement-I and Statement-II are correct
  4. (D)Statement-I is correct but Statement-II is incorrect

Correct answer: (B)

Step-by-step solution →
Q71·Mathematics·Limits and ContinuitySingle correct
lim⁡x→π2∫x3(π/2)3(sin⁡(2t1/3)+cos⁡(t1/3))dt(x−π2)2\displaystyle\lim_{x\to\frac{\pi}{2}}\dfrac{\int_{x^{3}}^{(\pi/2)^{3}}\left(\sin\left(2t^{1/3}\right)+\cos\left(t^{1/3}\right)\right)dt}{\left(x-\dfrac{\pi}{2}\right)^{2}}x→2π​lim​(x−2π​)2∫x3(π/2)3​(sin(2t1/3)+cos(t1/3))dt​ is equal to:
  1. (A)9π28\dfrac{9\pi^{2}}{8}89π2​
  2. (B)11π210\dfrac{11\pi^{2}}{10}1011π2​
  3. (C)3π22\dfrac{3\pi^{2}}{2}23π2​
  4. (D)4π29\dfrac{4\pi^{2}}{9}94π2​

Correct answer: (A)

Step-by-step solution →
Q72·MathematicsSingle correct
The sum of the coefficient of x2/3x^{2/3}x2/3 and x−2/5x^{-2/5}x−2/5 in the binomial expansion of (x2/3+12x−2/5)9\left(x^{2/3}+\dfrac{1}{2}x^{-2/5}\right)^{9}(x2/3+21​x−2/5)9 is:
  1. (A)214\dfrac{21}{4}421​
  2. (B)6916\dfrac{69}{16}1669​
  3. (C)6316\dfrac{63}{16}1663​
  4. (D)194\dfrac{19}{4}419​

Correct answer: (A)

Step-by-step solution →
Q73·MathematicsSingle correct
Let B=[1315]B=\begin{bmatrix}1&3\\1&5\end{bmatrix}B=[11​35​] and A be a 2×22\times 22×2 matrix such that AB−1=A−1AB^{-1}=A^{-1}AB−1=A−1. If BCB−1=ABCB^{-1}=ABCB−1=A and C4+αC2+βI=OC^{4}+\alpha C^{2}+\beta I=OC4+αC2+βI=O, then 2β−α2\beta-\alpha2β−α is equal to:
  1. (A)161616
  2. (B)222
  3. (C)888
  4. (D)101010

Correct answer: (D)

Step-by-step solution →
Q74·Mathematics·DifferentiabilitySingle correct
If log⁡ey=3sin⁡−1x\log_{e}y=3\sin^{-1}xloge​y=3sin−1x, then (1−x2)y′′−xy′(1-x^{2})y''-xy'(1−x2)y′′−xy′ at x=12x=\dfrac{1}{2}x=21​ is equal to:
  1. (A)9e5π/69e^{5\pi/6}9e5π/6
  2. (B)3e5π/63e^{5\pi/6}3e5π/6
  3. (C)3eπ/23e^{\pi/2}3eπ/2
  4. (D)9eπ/29e^{\pi/2}9eπ/2

Correct answer: (D)

Step-by-step solution →
Q75·MathematicsSingle correct
The integral ∫1/43/4cos⁡(2cot⁡−11−x1+x)dx\displaystyle\int_{1/4}^{3/4}\cos\left(2\cot^{-1}\sqrt{\dfrac{1-x}{1+x}}\right)dx∫1/43/4​cos(2cot−11+x1−x​​)dx is equal to:
  1. (A)−12-\dfrac{1}{2}−21​
  2. (B)14\dfrac{1}{4}41​
  3. (C)12\dfrac{1}{2}21​
  4. (D)−14-\dfrac{1}{4}−41​

Correct answer: (D)

Step-by-step solution →
Q76·MathematicsSingle correct
Let aaa, ararar, ar2ar^{2}ar2, ... be an infinite G.P. If ∑n=0∞arn=57\displaystyle\sum_{n=0}^{\infty}ar^{n}=57n=0∑∞​arn=57 and ∑n=0∞a3r3n=9747\displaystyle\sum_{n=0}^{\infty}a^{3}r^{3n}=9747n=0∑∞​a3r3n=9747, then a+18ra+18ra+18r is equal to:
  1. (A)272727
  2. (B)464646
  3. (C)383838
  4. (D)313131

Correct answer: (D)

Step-by-step solution →
Q77·MathematicsSingle correct
If an unbiased dice is rolled thrice, then the probability of getting a greater number in the ithi^{th}ith roll than the number obtained in the (i−1)th(i-1)^{th}(i−1)th roll, i=2,3i=2, 3i=2,3, is equal to:
  1. (A)354\dfrac{3}{54}543​
  2. (B)254\dfrac{2}{54}542​
  3. (C)554\dfrac{5}{54}545​
  4. (D)154\dfrac{1}{54}541​

Correct answer: (C)

Step-by-step solution →
Q78·MathematicsSingle correct
The value of the integral ∫−12log⁡e(x+x2+1)dx\displaystyle\int_{-1}^{2}\log_{e}\left(x+\sqrt{x^{2}+1}\right)dx∫−12​loge​(x+x2+1​)dx is:
  1. (A)5−2+log⁡e(9+451+2)\sqrt{5}-\sqrt{2}+\log_{e}\left(\dfrac{9+4\sqrt{5}}{1+\sqrt{2}}\right)5​−2​+loge​(1+2​9+45​​)
  2. (B)2−5+log⁡e(9+451+2)\sqrt{2}-\sqrt{5}+\log_{e}\left(\dfrac{9+4\sqrt{5}}{1+\sqrt{2}}\right)2​−5​+loge​(1+2​9+45​​)
  3. (C)5−2+log⁡e(7+451+2)\sqrt{5}-\sqrt{2}+\log_{e}\left(\dfrac{7+4\sqrt{5}}{1+\sqrt{2}}\right)5​−2​+loge​(1+2​7+45​​)
  4. (D)2−5+log⁡e(7+451+2)\sqrt{2}-\sqrt{5}+\log_{e}\left(\dfrac{7+4\sqrt{5}}{1+\sqrt{2}}\right)2​−5​+loge​(1+2​7+45​​)

Correct answer: (B)

Step-by-step solution →
Q79·MathematicsSingle correct
Let α,β\alpha, \betaα,β (α>β\alpha>\betaα>β) be the roots of the equation x2−2x−3=0x^{2}-\sqrt{2}x-\sqrt{3}=0x2−2​x−3​=0. Let Pn=αn−βnP_{n}=\alpha^{n}-\beta^{n}Pn​=αn−βn, n∈Nn\in\mathbb{N}n∈N. Then (113−102)P10+(112+10)P11−11P12\left(11\sqrt{3}-10\sqrt{2}\right)P_{10}+\left(11\sqrt{2}+10\right)P_{11}-11P_{12}(113​−102​)P10​+(112​+10)P11​−11P12​ is equal to:
  1. (A)102 P910\sqrt{2}\,P_{9}102​P9​
  2. (B)103 P910\sqrt{3}\,P_{9}103​P9​
  3. (C)112 P911\sqrt{2}\,P_{9}112​P9​
  4. (D)113 P911\sqrt{3}\,P_{9}113​P9​

Correct answer: (B)

Step-by-step solution →
Q80·MathematicsSingle correct
Let a⃗=2i^+αj^+k^\vec{a}=2\hat{i}+\alpha\hat{j}+\hat{k}a=2i^+αj^​+k^, b⃗=−i^+k^\vec{b}=-\hat{i}+\hat{k}b=−i^+k^, c⃗=βj^−k^\vec{c}=\beta\hat{j}-\hat{k}c=βj^​−k^, where α\alphaα and β\betaβ are integers and αβ=−6\alpha\beta=-6αβ=−6. Let the values of the ordered pair (α,β)(\alpha, \beta)(α,β) for which the area of the parallelogram with diagonals a⃗+b⃗\vec{a}+\vec{b}a+b and b⃗+c⃗\vec{b}+\vec{c}b+c is 212\dfrac{\sqrt{21}}{2}221​​, be (α1,β1)(\alpha_{1}, \beta_{1})(α1​,β1​) and (α2,β2)(\alpha_{2}, \beta_{2})(α2​,β2​). Then α12+β12−α2β2\alpha_{1}^{2}+\beta_{1}^{2}-\alpha_{2}\beta_{2}α12​+β12​−α2​β2​ is equal to:
  1. (A)171717
  2. (B)242424
  3. (C)212121
  4. (D)191919

Correct answer: (D)

Step-by-step solution →
Q81·MathematicsNumerical
Consider the circle C:x2+y2=4C:x^{2}+y^{2}=4C:x2+y2=4 and the parabola P:y2=8xP:y^{2}=8xP:y2=8x. If the set of all values of α\alphaα, for which three chords of the circle C on three distinct lines passing through the point (α,0)(\alpha, 0)(α,0) are bisected by the parabola P is the interval (p,q)(p, q)(p,q), then (2q−p)2(2q-p)^{2}(2q−p)2 is equal to _______.

Correct answer: 80

Step-by-step solution →
Q82·MathematicsNumerical
Let the set of all values of ppp, for which f(x)=(p2−6p+8)(sin⁡22x−cos⁡22x)+2(2−p)x+cos⁡21f(x)=(p^{2}-6p+8)\left(\sin^{2}2x-\cos^{2}2x\right)+2(2-p)x+\cos^{2}1f(x)=(p2−6p+8)(sin22x−cos22x)+2(2−p)x+cos21 does not have any critical point, be the interval (a,b)(a, b)(a,b). Then 16ab16ab16ab is equal to _______.

Correct answer: 252

Step-by-step solution →
Q83·MathematicsNumerical
For a differentiable function f:R→Rf:\mathbb{R}\to\mathbb{R}f:R→R, f′(x)=3f(x)+αf'(x)=3f(x)+\alphaf′(x)=3f(x)+α, where α∈R\alpha\in\mathbb{R}α∈R, f(0)=1f(0)=1f(0)=1 and lim⁡x→−∞f(x)=7\displaystyle\lim_{x\to-\infty}f(x)=7x→−∞lim​f(x)=7. Then 9f(−log⁡e3)9f(-\log_{e}3)9f(−loge​3) is equal to _______.

Correct answer: 61

Step-by-step solution →
Q84·MathematicsNumerical
The number of integers, between 100 and 1000 having the sum of their digits equal to 14, is _______.

Correct answer: 70

Step-by-step solution →
Q85·MathematicsNumerical
Let A={(x,y):2x+3y=23, x,y∈N}A=\{(x, y):2x+3y=23,\ x, y\in\mathbb{N}\}A={(x,y):2x+3y=23, x,y∈N} and B={x:(x,y)∈A}B=\{x:(x, y)\in A\}B={x:(x,y)∈A}. Then the number of one-one functions from A to B is equal to _______.

Correct answer: 24

Step-by-step solution →
Q86·MathematicsNumerical
Let A, B and C be three points on the parabola y2=6xy^{2}=6xy2=6x and let the line segment AB meet the line L through C parallel to the x-axis at the point D. Let M and N respectively be the feet of the perpendiculars from A and B on L. Then (AM⋅BNCD)2\left(\dfrac{AM\cdot BN}{CD}\right)^{2}(CDAM⋅BN​)2 is equal to _______.

Correct answer: 36

Step-by-step solution →
Q87·MathematicsNumerical
The square of the distance of the image of the point (6,1,5)(6, 1, 5)(6,1,5) in the line x−13=y2=z−24\dfrac{x-1}{3}=\dfrac{y}{2}=\dfrac{z-2}{4}3x−1​=2y​=4z−2​, from the origin is _______.

Correct answer: 62

Step-by-step solution →
Q88·MathematicsNumerical
If (1α+1+1α+2+⋯+1α+1012)−(12⋅1+14⋅3+16⋅5+⋯+12024⋅2023)=12024\left(\dfrac{1}{\alpha+1}+\dfrac{1}{\alpha+2}+\cdots+\dfrac{1}{\alpha+1012}\right)-\left(\dfrac{1}{2\cdot 1}+\dfrac{1}{4\cdot 3}+\dfrac{1}{6\cdot 5}+\cdots+\dfrac{1}{2024\cdot 2023}\right)=\dfrac{1}{2024}(α+11​+α+21​+⋯+α+10121​)−(2⋅11​+4⋅31​+6⋅51​+⋯+2024⋅20231​)=20241​, then α\alphaα is equal to _______.

Correct answer: 1011

Step-by-step solution →
Q89·MathematicsNumerical
Let the inverse trigonometric functions take principal values. The number of real solutions of the equation 2sin⁡−1x+3cos⁡−1x=2π52\sin^{-1}x+3\cos^{-1}x=\dfrac{2\pi}{5}2sin−1x+3cos−1x=52π​, is _______.

Correct answer: 0

Step-by-step solution →
Q90·MathematicsNumerical
Consider the matrices A=[2−53m]A=\begin{bmatrix}2&-5\\3&m\end{bmatrix}A=[23​−5m​], B=[20m]B=\begin{bmatrix}20\\m\end{bmatrix}B=[20m​] and X=[xy]X=\begin{bmatrix}x\\y\end{bmatrix}X=[xy​]. Let the set of all mmm, for which the system of equations AX=BAX=BAX=B has a negative solution (i.e., x<0x<0x<0 and y<0y<0y<0), be the interval (a,b)(a, b)(a,b). Then 8∫ab∣A∣ dm8\displaystyle\int_{a}^{b}|A|\,dm8∫ab​∣A∣dm is equal to _______.

Correct answer: 450

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
  • 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
  • Aldehydes and Ketones 135/186
  • Equilibrium 163/186
  • Solutions 158/186
  • Laws of Motion 130/186
  • Chemical Thermodynamics 165/186
  • Units and Measurements 149/186
  • Hydrocarbons 126/186
  • Complex Numbers 165/186
  • Trigonometric Functions 144/186
  • Biomolecules 162/186
  • Chemical Kinetics 169/186
  • Gravitation 152/186
  • Electric Field and Coulomb's Law 133/186
  • Atomic Structure 161/186
  • Electronic Devices 145/186
  • Dual Nature of Matter and Radiation 155/186
  • Quadratic Equations 148/186
  • Work, Energy and Power 132/186
  • Some Basic Concepts in Chemistry 129/186
  • Electromagnetic Induction 120/186
  • Wave Optics 130/186
  • Area Under Curves 139/186
  • Kinetic Theory of Gases 135/186
  • Purification and Characterisation of Organic Compounds 116/186
  • Oscillations 117/186
  • Alternating Currents 108/186
  • Nuclei 116/186
  • Organic Compounds Containing Halogens 109/186
  • Straight Lines 114/186
  • Atoms 112/186
  • Classification of Elements and Periodicity in Properties 113/186
  • Parabola 101/186
  • Statistics 118/186
  • Differentiability 91/186
  • Inverse Trigonometric Functions 93/186
  • Electronic Effects and Stability 74/186
  • Hyperbola 77/186
  • Principles of Qualitative Analysis 58/186
  • Isomerism 51/186
  • Magnetism and Matter 50/186
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