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JEE Main 5 April 2026 Shift 2 Question Paper with Answers

5 April 2026 · April session · 74 questions

74 of the 75 questions from the JEE Main 5 April 2026 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
24
Chemistry
25
Mathematics
25

Physics — JEE Main 5 April 2026 Shift 2

Q1·PhysicsSingle correct
Match List-I with List-II. where h (Planck's constant), G (gravitational constant) and c (speed of light in vacuum) as fundamental units. Choose the correct answer from the options given below :
List-IList-II
A.Meter (L)I.hcG\sqrt{\dfrac{hc}{G}}Ghc​​
B.Second (S)II.Ghc5\sqrt{\dfrac{Gh}{c^{5}}}c5Gh​​
C.Kilogram (M)III.K2L2c3Gh\sqrt{\dfrac{K^{2}L^{2}c^{3}}{Gh}}GhK2L2c3​​
D.Kelvin (K)IV.Ghc3\sqrt{\dfrac{Gh}{c^{3}}}c3Gh​​
  1. (A)A-II, B-IV, C-I, D-III
  2. (B)A-IV, B-II, C-I, D-III
  3. (C)A-IV, B-I, C-II, D-III
  4. (D)A-III, B-I, C-II, D-IV

Correct answer: (B)

Step-by-step solution →
Q2·PhysicsSingle correct
In an experiment to determine the resistance of a given wire using Ohm's law, the voltmeter and ammeter readings are noted as 10 V and 5 A, respectively. The least counts of voltmeter and ammeter are 500 mV and 200 mA, respectively. The estimated error in the resistance measurement is ________ Ω.
  1. (A)0.25
  2. (B)2
  3. (C)2.5
  4. (D)0.18

Correct answer: (D)

Step-by-step solution →
Q3·PhysicsSingle correct
A mass of 1 kg is kept on an inclined plane with 30° inclination with respect to horizontal plane and it is at rest initially. Then the whole assembly is moved up with constant velocity of 4 m/s. The work done by the frictional force in time 2 s is ________ J. (Take g = 10 m/s²)
  1. (A)20
  2. (B)25
  3. (C)30
  4. (D)10

Correct answer: (A)

Step-by-step solution →
Q4·PhysicsSingle correct
The velocity (v) versus time (t) plot of a particle is shown in the figure, for a time interval of 40 s. The total distance travelled by the particle and the average velocity during this period are, respectively ________.
  1. (A)25 m and zero
  2. (B)50 m and zero
  3. (C)100 m and zero
  4. (D)100 m and 2.5 m/s

Correct answer: (C)

Step-by-step solution →
Q5·PhysicsSingle correct
A wheel initially at rest is subjected to a uniform angular acceleration about its axis. In the first 2 s it rotates through an angle θ₁ and in the next 2 s it rotates through an angle θ₂. The ratio θ2θ1\dfrac{\theta_{2}}{\theta_{1}}θ1​θ2​​ is ________.
  1. (A)6
  2. (B)3
  3. (C)4
  4. (D)13\dfrac{1}{3}31​

Correct answer: (B)

Step-by-step solution →
Q6·PhysicsSingle correct
An object of uniform density rolls up the curved path with the initial velocity v₀ as shown in the figure. If the maximum height attained by an object is 7v0210g\dfrac{7v_{0}^{2}}{10g}10g7v02​​ (g = acceleration due to gravity), the object is a ________.
  1. (A)solid cylinder
  2. (B)ring
  3. (C)disc
  4. (D)solid sphere

Correct answer: (D)

Step-by-step solution →
Q7·PhysicsSingle correct
A body of mass m is taken from the surface of earth to a height equal to twice the radius of earth (Re_ee​). The increase in potential energy will be ________. (g is acceleration due to gravity at the surface of earth)
  1. (A)12mgRe\dfrac{1}{2}mgR_{e}21​mgRe​
  2. (B)34mgRe\dfrac{3}{4}mgR_{e}43​mgRe​
  3. (C)14mgRe\dfrac{1}{4}mgR_{e}41​mgRe​
  4. (D)23mgRe\dfrac{2}{3}mgR_{e}32​mgRe​

Correct answer: (D)

Step-by-step solution →
Q8·PhysicsSingle correct
Eight mercury drops, each of radius r, coalesce to form a bigger drop. The surface energy released in this process is ________. (S is the surface tension of mercury).
  1. (A)8πr²S
  2. (B)16πr²S
  3. (C)64πr²S
  4. (D)4πr²S

Correct answer: (B)

Step-by-step solution →
Q9·PhysicsSingle correct
An ideal gas at pressure P and temperature T is expanding such that PT³ = constant. The coefficient of volume expansion of the gas is ________.
  1. (A)2T\dfrac{2}{T}T2​
  2. (B)1T\dfrac{1}{T}T1​
  3. (C)4T\dfrac{4}{T}T4​
  4. (D)3T\dfrac{3}{T}T3​

Correct answer: (C)

Step-by-step solution →
Q10·PhysicsSingle correct
Match List-I with List-II. Choose the correct answer from the options given below :
List-IList-II
A.sin² ωtI.Periodic with time period T=πωT = \dfrac{\pi}{\omega}T=ωπ​ but not simple harmonic motion (SHM)
B.sin³(2ωt)II.Periodic with time period T=2πωT = \dfrac{2\pi}{\omega}T=ω2π​ but Not SHM
C.sin(ωt) + cos(πωt)III.Periodic with time period T=πωT = \dfrac{\pi}{\omega}T=ωπ​ and SHM
D.cos ωt + cos 2ωtIV.Non-periodic
  1. (A)A-III, B-I, C-IV, D-II
  2. (B)A-II, B-I, C-III, D-IV
  3. (C)A-III, B-II, C-IV, D-I
  4. (D)A-II, B-I, C-IV, D-III

Correct answer: (A)

Step-by-step solution →
Q11·PhysicsSingle correct
A metal rod of length L rotates about one end at origin with a uniform angular velocity ω. The magnetic field radially falls off as B(r) = B₀e−λr^{-\lambda r}−λr; λ being a positive constant. The emf induced (neglecting the centripetal force on electrons in the rod) is :
  1. (A)B0ω[1λ2−e−λL(1λ2+Lλ)]B_{0}\omega\left[\dfrac{1}{\lambda^{2}} - e^{-\lambda L}\left(\dfrac{1}{\lambda^{2}} + \dfrac{L}{\lambda}\right)\right]B0​ω[λ21​−e−λL(λ21​+λL​)]
  2. (B)B0ω[1λ2+e−λL(1λ2+Lλ)]B_{0}\omega\left[\dfrac{1}{\lambda^{2}} + e^{-\lambda L}\left(\dfrac{1}{\lambda^{2}} + \dfrac{L}{\lambda}\right)\right]B0​ω[λ21​+e−λL(λ21​+λL​)]
  3. (C)B0ω[4λ2−e−2λL(1λ2+2Lλ)]B_{0}\omega\left[\dfrac{4}{\lambda^{2}} - e^{-2\lambda L}\left(\dfrac{1}{\lambda^{2}} + \dfrac{2L}{\lambda}\right)\right]B0​ω[λ24​−e−2λL(λ21​+λ2L​)]
  4. (D)B0ω[3λ2−e−3λL(3λ2+Lλ)]B_{0}\omega\left[\dfrac{3}{\lambda^{2}} - e^{-3\lambda L}\left(\dfrac{3}{\lambda^{2}} + \dfrac{L}{\lambda}\right)\right]B0​ω[λ23​−e−3λL(λ23​+λL​)]

Correct answer: (A)

Step-by-step solution →
Q12·PhysicsSingle correct
Under steady state condition the potential difference across the capacitor in the circuit is ________ V.
  1. (A)0.5
  2. (B)1.5
  3. (C)0
  4. (D)2

Correct answer: (A)

Step-by-step solution →
Q13·Physics·Magnetic Field of CurrentSingle correct
A particle of charge q and mass m is projected from origin with an initial velocity v⃗=(v02x^+v02y^)\vec{v} = \left(\dfrac{v_{0}}{\sqrt{2}}\hat{x} + \dfrac{v_{0}}{\sqrt{2}}\hat{y}\right)v=(2​v0​​x^+2​v0​​y^​). There exists a uniform magnetic field B⃗=B0z^\vec{B} = B_{0}\hat{z}B=B0​z^ and a space varying electric field E⃗=E0e−λxx^\vec{E} = E_{0}e^{-\lambda x}\hat{x}E=E0​e−λxx^ within the region 0 ≤ x ≤ L. After travelling a distance such that x-coordinate has changed from x = 0 to x = L, the change in the kinetic energy is ________.
  1. (A)qE0λ[1−e−λL]\dfrac{qE_{0}}{\lambda}[1 - e^{-\lambda L}]λqE0​​[1−e−λL]
  2. (B)(v0qB02λ)[2−e−2λL]\left(\dfrac{v_{0}qB_{0}}{2\lambda}\right)[2 - e^{-2\lambda L}](2λv0​qB0​​)[2−e−2λL]
  3. (C)qE0λ[1+e−λL]\dfrac{qE_{0}}{\lambda}[1 + e^{-\lambda L}]λqE0​​[1+e−λL]
  4. (D)q(E0+v0B0λ)[1−e−λL/2]q\left(\dfrac{E_{0} + v_{0}B_{0}}{\lambda}\right)[1 - e^{-\lambda L/2}]q(λE0​+v0​B0​​)[1−e−λL/2]

Correct answer: (A)

Step-by-step solution →
Q14·PhysicsSingle correct
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): The electromagnetic wave exerts pressure on the surface on which they are allowed to fall. Reason (R): There is no mass associated with the electromagnetic waves. In the light of the above statements, choose the correct answer from the options given below :
  1. (A)Both (A) and (R) are true and (R) is the correct explanation of (A)
  2. (B)Both (A) and (R) are true 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 →
Q15·PhysicsSingle correct
A thin convex lens and a thin concave lens are kept in contact and are co-axial. Which of the following statements is correct for this combination of two lenses ?
  1. (A)behaves as concave lens if ∣fconvex∣>∣fconcave∣|f_\mathrm{convex}| > |f_\mathrm{concave}|∣fconvex​∣>∣fconcave​∣
  2. (B)behaves as concave lens if ∣fconvex∣<∣fconcave∣|f_\mathrm{convex}| < |f_\mathrm{concave}|∣fconvex​∣<∣fconcave​∣
  3. (C)behaves as convex lens if ∣fconvex∣>∣fconcave∣|f_\mathrm{convex}| > |f_\mathrm{concave}|∣fconvex​∣>∣fconcave​∣
  4. (D)Focal length of the lens system will change if the positions of two lenses are interchanged

Correct answer: (A)

Step-by-step solution →
Q16·PhysicsSingle correct
An object ABABAB is placed 15 cm on the left of a convex lens PPP of focal length 10 cm. Another convex lens QQQ is now placed 15 cm right of lens PPP. If the focal length of lens QQQ is 15 cm, the final image is ________.
  1. (A)virtual, formed at 7.5 cm right of lens QQQ, with a size bigger than that of ABABAB
  2. (B)real, formed at 7.5 cm right of lens QQQ, with a size same as that of ABABAB
  3. (C)formed at infinity.
  4. (D)real, formed at 7 cm right of lens QQQ, with a size smaller than that of ABABAB

Correct answer: (B)

Step-by-step solution →
Q17·PhysicsSingle correct
The maximum intensity in a Young's double slit experiment is I0I_0I0​. Distance between the slits (ddd) is 5λ, where λ is the wavelength of light used. The intensity of the fringe, exactly opposite to one of the slits on the screen, placed at D=10dD = 10dD=10d is ________.
  1. (A)I04\frac{I_0}{4}4I0​​
  2. (B)I02\frac{I_0}{2}2I0​​
  3. (C)I0I_0I0​
  4. (D)3I04\frac{3I_0}{4}43I0​​

Correct answer: (B)

Step-by-step solution →
Q18·PhysicsSingle correct
An electron is travelling with a velocity vvv in free space and when it enters a medium, its velocity is reduced by 20%. The de Broglie wavelength of electron in the medium is αλ0\alpha\lambda_0αλ0​, where λ0\lambda_0λ0​ is its de Broglie wavelength in free space. The value of α\alphaα is ________.
  1. (A)1.20
  2. (B)1.0
  3. (C)1.25
  4. (D)0.75

Correct answer: (C)

Step-by-step solution →
Q19·PhysicsSingle correct
Assuming the experimental mass of 612C^{12}_{6}C612​C as 12 u, the mass defect of 612C^{12}_{6}C612​C atom is ________ MeV/c2c^2c2. (Mass of proton =1.00727= 1.00727=1.00727 u, mass of neutron =1.00866= 1.00866=1.00866 u, 1 u =931.5= 931.5=931.5 MeV/c2c^2c2 and ccc is the speed of the light in vacuum).
  1. (A)127.5
  2. (B)89.03
  3. (C)272.0
  4. (D)92.0

Correct answer: (B)

Step-by-step solution →
Q20·PhysicsSingle correct
In a semiconductor p-n diode, the doping concentrations on p-side and n-side are 1015^{15}15 atoms/cm3^{3}3 and 1018^{18}18 atoms/cm3^{3}3, respectively. Which one of the following statements is true ?
  1. (A)Widths of depletion region on either side of the interface are equal
  2. (B)The depletion region width is more on p-side compared to that in n-side
  3. (C)The depletion region width is more on n-side compared to that in p-side
  4. (D)No depletion region forms because of unequal doping concentrations on p and n-sides

Correct answer: (B)

Step-by-step solution →
Q21·PhysicsNumerical
A copper wire of length 3 m is stretched by 3 mm by applying an external force. The volume of the wire is 600×10−6600 \times 10^{-6}600×10−6 m3^{3}3. The elastic potential energy stored in the wire in stretched condition would be ________ J. (Given Young modulus of copper =1.1×1011= 1.1 \times 10^{11}=1.1×1011 N/m2^{2}2)

Correct answer: 33

Step-by-step solution →
Q22·PhysicsNumerical
A series LCR circuit with R=20R = 20R=20 Ω, L=1.6L = 1.6L=1.6 H and C=40C = 40C=40 μF is connected to a variable frequency a.c. source. The inductive reactance at resonant frequency is ________ Ω.

Correct answer: 200

Step-by-step solution →
Q23·PhysicsNumerical
When an external resistance of 5 Ω is connected across terminals of a cell, a current of 0.25 A flows through it. When the 5 Ω resistor is replaced by a 2 Ω resistor, a current of 0.5 A flows through it. The internal resistance of the cell is ________ Ω.

Correct answer: 1

Step-by-step solution →
Q24·PhysicsNumerical
A circular loop of radius 20 cm and resistance 2 Ω is placed in a time varying magnetic field B⃗=(2t2+2t+3)\vec{B} = (2t^2 + 2t + 3)B=(2t2+2t+3) T. At t=0t = 0t=0, for the plane of the loop being perpendicular to the magnetic field and, the induced current in the loop at t=3t = 3t=3 s is α50\frac{\alpha}{50}50α​ A. The value of α\alphaα is ________. (Take π=22/7\pi = 22/7π=22/7)

Correct answer: 44

Step-by-step solution →

Chemistry — JEE Main 5 April 2026 Shift 2

Q25·ChemistrySingle correct
What volume of hydrogen gas at STP would be liberated by action of 50 mL of H₂SO₄ of 50% purity (density = 1.3 g mL⁻¹) on 20 g of zinc ? Given : Molar mass of H, O, S, Zn are 1, 16, 32, 65 g mol⁻¹ respectively.
  1. (A)5.824 L
  2. (B)7.428 L
  3. (C)6.892 L
  4. (D)8.375 L

Correct answer: (C)

Step-by-step solution →
Q26·ChemistrySingle correct
Which of the following statement(s) is/are true ? A. If two orbitals have the same value of (n + l), the orbital with lower value of n will have lower energy. B. Energies of the orbitals in the same subshell increase with increase in atomic number. C. The size of 2px2p_x2px​ orbital is less than the size of 3px3p_x3px​ orbital. D. Among 5f, 6s, 4d, 5p and 5d orbitals, none of the orbitals have 2 radial nodes. Choose the correct answer from the options given below :
  1. (A)A, B and C only
  2. (B)A and C only
  3. (C)C and D only
  4. (D)A only

Correct answer: (B)

Step-by-step solution →
Q27·ChemistrySingle correct
The covalent radii of atoms A and B are rAr_ArA​ and rBr_BrB​, respectively. The covalent bond length and total length of AB molecule are respectively
  1. (A)(rA+rB), 2(rA+rB)(r_A + r_B),\ 2(r_A + r_B)(rA​+rB​), 2(rA​+rB​)
  2. (B)12(rA+rB), (rA+rB)\frac{1}{2}(r_A + r_B),\ (r_A + r_B)21​(rA​+rB​), (rA​+rB​)
  3. (C)(rA+rB), (rA+rB)(r_A + r_B),\ (r_A + r_B)(rA​+rB​), (rA​+rB​)
  4. (D)2(rA+rB), 12(rA+rB)2(r_A + r_B),\ \frac{1}{2}(r_A + r_B)2(rA​+rB​), 21​(rA​+rB​)

Correct answer: (A)

Step-by-step solution →
Q28·ChemistrySingle correct
Consider the following data for the reaction X₂(g) + Y₂(g) ⇌ 2XY(g) at 600 K. The ΔrG° (in kJ mol⁻¹) for the reaction is :
CompoundΔfH°(600K) (kJ mol⁻¹)S°(600K) (J mol⁻¹ K⁻¹)
XY(g)42200
X₂(g)8140
Y₂(g)80250
  1. (A)−21000
  2. (B)−10
  3. (C)−1000
  4. (D)−9.012

Correct answer: (B)

Step-by-step solution →
Q29·ChemistrySingle correct
The correct order of molar heat capacities measured at 298 K and 1 bar is :
  1. (A)Copper(s) > Bromine(l) > Helium(g)
  2. (B)Bromine(l) > Copper(s) > Helium(g)
  3. (C)Helium(g) > Bromine(l) > Copper(s)
  4. (D)Helium(g) > Bromine(l) = Copper(s)

Correct answer: (B)

Step-by-step solution →
Q30·ChemistrySingle correct
The reaction A(g) ⇌ B(g) + C(g) was initiated with the amount 'a' of A(g). At equilibrium it is found that the amount of A(g) remaining is (a − x) at a total pressure of p. The equilibrium constant KpK_pKp​ of the reaction can be calculated from the expression :
  1. (A)x2a2+x2×p\frac{x^{2}}{a^{2} + x^{2}} \times pa2+x2x2​×p
  2. (B)x2a2−x2×p\frac{x^{2}}{a^{2} - x^{2}} \times pa2−x2x2​×p
  3. (C)a+x2x2×p\frac{a + x^{2}}{x^{2}} \times px2a+x2​×p
  4. (D)a2−x2x2×p\frac{a^{2} - x^{2}}{x^{2}} \times px2a2−x2​×p

Correct answer: (B)

Step-by-step solution →
Q31·ChemistrySingle correct
One half cell in a voltaic cell is constructed by dipping silver rod in AgNO₃ solution of unknown concentration, other half cell is Zn rod dipped in 1 molar solution of ZnSO₄. A voltage of 1.60 V is measured at 298 K for this cell. What is the concentration of Ag⁺ ions used in terms of log x (x = [Ag⁺]) ? EZn2+/Zn∘E^{\circ}_{Zn^{2+}/Zn}EZn2+/Zn∘​ = −0.76 V, EAg+/Ag∘E^{\circ}_{Ag^{+}/Ag}EAg+/Ag∘​ = +0.80 V, 2.303RTF\frac{2.303RT}{F}F2.303RT​ = 0.059 V
  1. (A)23.9\frac{2}{3.9}3.92​
  2. (B)45.9\frac{4}{5.9}5.94​
  3. (C)2.92\frac{2.9}{2}22.9​
  4. (D)5.94\frac{5.9}{4}45.9​

Correct answer: (B)

Step-by-step solution →
Q32·ChemistrySingle correct
Given below are two statements : Statement I: The number of pairs among [Al₂O₃, Cr₂O₃], [Cl₂O₇, Mn₂O₇], [Na₂O, V₂O₃] and [CO, N₂O] that contain oxides of same nature (acidic, basic, neutral or amphoteric) is 4. Statement II: Among Na₂O, Al₂O₃, CO and Cl₂O₇, the most basic and acidic oxides are Na₂O and Cl₂O₇, respectively. 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 true
  2. (B)Both Statement I and Statement II are false
  3. (C)Statement I is true but Statement II is false
  4. (D)Statement I is false but Statement II is true

Correct answer: (A)

Step-by-step solution →
Q33·ChemistrySingle correct
Given below are two statements : Statement I: Aluminium upon reaction with NaOH forms [Al(OH)₆]³⁻ ion. Statement II: The geometry of ICl₄⁻, ClO₃⁻ and IBr₂⁻ is square planar, pyramidal and linear respectively. 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 true
  2. (B)Both Statement I and Statement II are false
  3. (C)Statement I is true but Statement II is false
  4. (D)Statement I is false but Statement II is true

Correct answer: (D)

Step-by-step solution →
Q34·ChemistrySingle correct
Given below are two statements : Statement I: Presence of large number of unpaired electrons in transition metal atoms results in higher enthalpies of their atomisation. Statement II: dxy=dxz=dyz<dx2−y2=dz2d_{xy} = d_{xz} = d_{yz} < d_{x^{2}-y^{2}} = d_{z^{2}}dxy​=dxz​=dyz​<dx2−y2​=dz2​ and dx2−y2=dz2<dxy=dxz=dyzd_{x^{2}-y^{2}} = d_{z^{2}} < d_{xy} = d_{xz} = d_{yz}dx2−y2​=dz2​<dxy​=dxz​=dyz​ are the d-orbital splittings in [Fe(H₂O)₆]³⁺ and [Ni(Cl)₄]²⁻ complex ions respectively. 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 correct
  2. (B)Both Statement I and Statement II are incorrect
  3. (C)Statement I is correct but Statement II is incorrect
  4. (D)Statement I is incorrect but Statement II is correct

Correct answer: (A)

Step-by-step solution →
Q35·ChemistrySingle correct
Identify the correct statements from the following A. [Fe(C₂O₄)₃]³⁻ is the most stable complex among [Fe(OH)₆]³⁻, [Fe(C₂O₄)₃]³⁻ and [Fe(SCN)₆]³⁻ B. The stability of [Cu(NH₃)₄]²⁺ is greater than that of [Cu(en)₂]²⁺ C. The hybridization of Fe in K₄[Fe(CN)₆] is d²sp³ D. [Fe(NO₂)₃Cl₃]³⁻ exhibits linkage isomerism E. NO₂⁻ and SCN⁻ ligands are NOT ambidentate ligands Choose the correct answer from the options given below :
  1. (A)A, B, C, D and E
  2. (B)B, C and D only
  3. (C)A, C and D only
  4. (D)A, C and E only

Correct answer: (C)

Step-by-step solution →
Q36·ChemistrySingle correct
Match List-I (Purification technique) with List-II (Used to separate). Choose the correct answer from the options given below :
List-I (Purification technique)List-II (Used to separate)
A.Simple distillationI.Steam volatile compound
B.Fractional distillationII.Two liquids with large difference in boiling points
C.Steam distillationIII.Liquid decomposing at its boiling point
D.Distillation under reduced pressureIV.Two liquids with close boiling points
  1. (A)A-II, B-III, C-I, D-IV
  2. (B)A-II, B-IV, C-I, D-III
  3. (C)A-II, B-IV, C-III, D-I
  4. (D)A-IV, B-III, C-II, D-I

Correct answer: (B)

Step-by-step solution →
Q37·ChemistrySingle correct
IUPAC name of the some alkenes are given below. Find out the correct stability order. A. 2-Methylbut-2-ene B. cis-But-2-ene C. 2,3-Dimethylbut-2-ene D. Prop-1-ene Choose the correct answer from the options given below :
  1. (A)C > A > B > D
  2. (B)C > A > D > B
  3. (C)B > D > A > C
  4. (D)A > B > C > D

Correct answer: (A)

Step-by-step solution →
Q38·ChemistrySingle correct
Identify the correct IUPAC name of hydrocarbon (x) containing three primary carbon atoms and with molar mass 72 g mol⁻¹.
  1. (A)1,1-Dimethylcyclopropane
  2. (B)2,2-Dimethylpropane
  3. (C)2-Methylbutane
  4. (D)n-pentane

Correct answer: (C)

Step-by-step solution →
Q39·ChemistrySingle correct
Complete the following reaction sequence and give the name of major product 'P'. CH₃ − CH₂ − C ≡ N →(ii) H3O+(i) OH−/H2O/Δ\xrightarrow[\text{(ii) } H_3O^+]{\text{(i) } OH^-/H_2O/\Delta}(i) OH−/H2​O/Δ(ii) H3​O+​ →(iv) H2O(iii) Cl2/Red P\xrightarrow[\text{(iv) } H_2O]{\text{(iii) } Cl_2/\text{Red P}}(iii) Cl2​/Red P(iv) H2​O​ P (Major product)
  1. (A)2-Chloropropanoic acid
  2. (B)3-Chloropropanoic acid
  3. (C)1-Chloropropane
  4. (D)2-Chloropropane

Correct answer: (A)

Step-by-step solution →
Q40·ChemistrySingle correct
Given below are two statements : Statement I: The condensation reaction between CH₃ − CH = O and H₂N − NH − C(= O) − NH₂ under optimum pH will produce CH₃ − CH = N − N(H) − C(= O) − NH₂. Statement II: The molecule, Ph − CH(O − H)(O − CH₃) will generate Ph − CH = O in the presence of dilute acid. 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 true
  2. (B)Both Statement I and Statement II are false
  3. (C)Statement I is true but Statement II is false
  4. (D)Statement I is false but Statement II is true

Correct answer: (A)

Step-by-step solution →
Q41·ChemistrySingle correct
Given below are two statements : Statement I: Heating benzamide with bromine in an ethanolic solution of sodium hydroxide will give benzylamine. Statement II: Nitration of aniline with HNO₃/H₂SO₄ at 288 K produces m-nitroaniline in higher amount than o-nitroaniline (pH adjusted). 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 true
  2. (B)Both Statement I and Statement II are false
  3. (C)Statement I is true but Statement II is false
  4. (D)Statement I is false but Statement II is true

Correct answer: (D)

Step-by-step solution →
Q42·ChemistrySingle correct
Identify the incorrect statement about tertiary structure of proteins.
  1. (A)They can be fibrous or globular in structure.
  2. (B)The main forces that stabilize the structure are hydrogen bonding, disulphide links, van der Waals and electrostatic forces of attraction.
  3. (C)The structure remains intact when exposed to pH changes.
  4. (D)A linear polypeptide chain will convert to a secondary structure and then further folding of the secondary structure will convert to tertiary structure.

Correct answer: (C)

Step-by-step solution →
Q43·ChemistrySingle correct
Given below are two statements : Statement I: The two cyclic forms of D-(+)-glucose (α and β anomers differing at C1 hydroxyl orientation) are two anomers of D-(+)-glucose. Statement II: The open chain forms of D-glucose and D-fructose contain three similar chiral carbons at C₃, C₄ and C₅. 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 true
  2. (B)Both Statement I and Statement II are false
  3. (C)Statement I is true but Statement II is false
  4. (D)Statement I is false but Statement II is true

Correct answer: (A)

Step-by-step solution →
Q44·ChemistrySingle correct
A paper dipped in a dil. H₂SO₄ solution of 'X' upon treatment with SO₂ gas turns into green. The compound 'X' is :
  1. (A)KI-starch
  2. (B)KMnO₄
  3. (C)Pb(CH₃COO)₂
  4. (D)K₂Cr₂O₇

Correct answer: (D)

Step-by-step solution →
Q45·ChemistryNumerical
The total number of unpaired electrons present in the d³, d⁴ (low spin), d⁵ (high spin), d⁶ (high spin) and d⁷ (low spin) octahedral complex systems is _______.

Correct answer: 15

Step-by-step solution →
Q46·ChemistryNumerical
RMgI when treated with ice cold water liberated a gas which occupied 1.4 dm³/g at STP. The gas produced is further reacted with iodine in presence of HIO₃ to give compound (X). Compound (X) in presence of Na and dry ether produced compound (Y). Molar mass of compound (Y) is _______ g mol⁻¹. (Nearest integer)

Correct answer: 30

Step-by-step solution →
Q47·ChemistryNumerical
20 g hemoglobin in a 1 L aqueous solution (A) at 300 K is separated from pure water by semi permeable membrane. At equilibrium the height of solution in a tube dipped in a solution (A) is found to be 80.0 mm higher than the tube dipped in water. The molar mass of hemoglobin is _______ kg mol⁻¹. (Nearest integer) (Given : g = 10 m s⁻², R = 8.3 kPa dm³ K⁻¹mol⁻¹, density of solution = 1000 kg m⁻³)

Correct answer: 62

Step-by-step solution →
Q48·ChemistryNumerical
At 298 K, the molar conductivity of x% (w/w) MX solution (aqueous) is 123.5 S cm² mol⁻¹. The conductance of same solution is 1.9 × 10⁻³ S. The value of x is _______ ×10⁻². (Given : cell constant = 1.3 cm⁻¹; molar mass of MX is 75 g mol⁻¹, density of aqueous solution of MX at 298 K is 1.0 g mL⁻¹)

Correct answer: 15

Step-by-step solution →
Q49·ChemistryNumerical
For a reaction A → P at T K, the half life (t₁/₂) is plotted as a function of initial concentration [A]₀ of A as given below. The value of x in the given figure is _______ s (Nearest integer)

Correct answer: 90

Step-by-step solution →

Mathematics — JEE Main 5 April 2026 Shift 2

Q50·MathematicsSingle correct
Let α, β be the roots of the equation x2−x+p=0x^2 - x + p = 0x2−x+p=0 and γ, δ be the roots the equation x2−4x+q=0x^2 - 4x + q = 0x2−4x+q=0; p, q ∈ Z. If α, β, γ, δ are in G.P., then ∣p+q∣|p + q|∣p+q∣ equals :
  1. (A)16
  2. (B)32
  3. (C)34
  4. (D)38

Correct answer: (C)

Step-by-step solution →
Q51·MathematicsSingle correct
Let z1,z2z_1, z_2z1​,z2​ ∈ ℂ be the distinct solutions of the equation z2+4z−(1+12i)=0z^2 + 4z - (1 + 12i) = 0z2+4z−(1+12i)=0. Then ∣z1∣2+∣z2∣2|z_1|^2 + |z_2|^2∣z1​∣2+∣z2​∣2 is equal to :
  1. (A)18
  2. (B)22
  3. (C)29
  4. (D)34

Correct answer: (D)

Step-by-step solution →
Q52·MathematicsSingle correct
If f:N→Zf : \mathbf{N} \to \mathbf{Z}f:N→Z is defined by f(n)=∣n−1−5−2n23(2k+1)2k+1−3n33k(2k+1)3k(k+2)+1∣f(n) = \begin{vmatrix} n & -1 & -5 \\ -2n^2 & 3(2k+1) & 2k+1 \\ -3n^3 & 3k(2k+1) & 3k(k+2)+1 \end{vmatrix}f(n)=​n−2n2−3n3​−13(2k+1)3k(2k+1)​−52k+13k(k+2)+1​​, k ∈ N, and ∑n=1kf(n)=98\sum_{n=1}^{k} f(n) = 98∑n=1k​f(n)=98, then k is equal to :
  1. (A)3
  2. (B)4
  3. (C)5
  4. (D)6

Correct answer: (A)

Step-by-step solution →
Q53·MathematicsSingle correct
Let M be a 3×33 \times 33×3 matrix such that M(100)=(123)M \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}M​100​​=​123​​, M(010)=(012)M \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix}M​010​​=​012​​ and M(001)=(−111)M \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix}M​001​​=​−111​​. If M(xyz)=(1711)M \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ 7 \\ 11 \end{pmatrix}M​xyz​​=​1711​​, then x+y+zx + y + zx+y+z equals :
  1. (A)4
  2. (B)5
  3. (C)7
  4. (D)11

Correct answer: (B)

Step-by-step solution →
Q54·MathematicsSingle correct
If the sum of the first 10 terms of the series 11+14×4+21+24×4+31+34×4+41+44×4+…\frac{1}{1 + 1^4 \times 4} + \frac{2}{1 + 2^4 \times 4} + \frac{3}{1 + 3^4 \times 4} + \frac{4}{1 + 4^4 \times 4} + \ldots1+14×41​+1+24×42​+1+34×43​+1+44×44​+… is mn\frac{m}{n}nm​, gcd(m, n) = 1, then m+nm + nm+n is equal to :
  1. (A)256
  2. (B)264
  3. (C)276
  4. (D)284

Correct answer: (C)

Step-by-step solution →
Q55·MathematicsSingle correct
Let A1,A2,A3,…,A39A_1, A_2, A_3, \ldots, A_{39}A1​,A2​,A3​,…,A39​ be 39 arithmetic means between the numbers 59 and 159. Then the mean of A25,A28,A31A_{25}, A_{28}, A_{31}A25​,A28​,A31​ and A36A_{36}A36​ is equal to :
  1. (A)129
  2. (B)136
  3. (C)131.50
  4. (D)134

Correct answer: (D)

Step-by-step solution →
Q56·MathematicsSingle correct
The coefficient of x2x^2x2 in the expansion of (2x2+1x)10\left( 2x^2 + \frac{1}{x} \right)^{10}(2x2+x1​)10, x≠0x \neq 0x=0, is :
  1. (A)3240
  2. (B)3360
  3. (C)3480
  4. (D)3600

Correct answer: (B)

Step-by-step solution →
Q57·MathematicsSingle correct
The probabilities that players A and B of a team are selected for the captaincy for a tournament are 0.6 and 0.4, respectively. If A is selected the captain, the probability that the team wins the tournament is 0.8 and if B is selected the captain, the probability that the team wins the tournament is 0.7. Then the probability, that the team wins the tournament, is :
  1. (A)0.74
  2. (B)0.76
  3. (C)0.72
  4. (D)0.78

Correct answer: (B)

Step-by-step solution →
Q58·MathematicsSingle correct
A box contains 5 blue, 6 yellow and 4 red balls. The number of ways, of drawing 8 balls containing at least two balls of each colour, is :
  1. (A)4100
  2. (B)4140
  3. (C)4230
  4. (D)4290

Correct answer: (A)

Step-by-step solution →
Q59·MathematicsSingle correct
A variable X takes values 0,0,2,6,12,20,…,n(n−1)0, 0, 2, 6, 12, 20, \ldots, n(n - 1)0,0,2,6,12,20,…,n(n−1) with frequencies nC0,nC1,nC2,nC3,nC4,nC5,…,nCn^{n}C_0, ^{n}C_1, ^{n}C_2, ^{n}C_3, ^{n}C_4, ^{n}C_5, \ldots, ^{n}C_nnC0​,nC1​,nC2​,nC3​,nC4​,nC5​,…,nCn​, respectively. If the mean of this data is 60, then its median is :
  1. (A)56
  2. (B)42
  3. (C)72
  4. (D)90

Correct answer: (A)

Step-by-step solution →
Q60·MathematicsSingle correct
Let the point P be the vertex of the parabola y=x2−6x+12y = x^2 - 6x + 12y=x2−6x+12. If a line passing through the point P intersects the circle x2+y2−2x−4y+3=0x^2 + y^2 - 2x - 4y + 3 = 0x2+y2−2x−4y+3=0 at the points R and S, then the maximum value of (PR+PS)2(PR + PS)^2(PR+PS)2 is :
  1. (A)10
  2. (B)20
  3. (C)25
  4. (D)5

Correct answer: (B)

Step-by-step solution →
Q61·MathematicsSingle correct
Let the directrix of the parabola P:y2=8xP : y^2 = 8xP:y2=8x, cut x-axis at the point A. Let B(α,β)B(α, β)B(α,β), α>1α > 1α>1, be a point on P such that the slope of AB is 3/5. If BC is a focal chord of P, then six times the area of △ABC\triangle ABC△ABC is :
  1. (A)80
  2. (B)160
  3. (C)174
  4. (D)192

Correct answer: (B)

Step-by-step solution →
Q62·MathematicsSingle correct
Let the eccentricity e of a hyperbola satisfy the equation 6e2−11e+3=06e^2 - 11e + 3 = 06e2−11e+3=0. If the foci of the hyperbola are (3, 5) and (3, −4), then the length of its latus rectum is :
  1. (A)113\frac{11}{3}311​
  2. (B)173\frac{17}{3}317​
  3. (C)152\frac{15}{2}215​
  4. (D)172\frac{17}{2}217​

Correct answer: (C)

Step-by-step solution →
Q63·MathematicsSingle correct
Let a triangle PQR be such that P and Q lie on the line x+38=y−42=z+12\frac{x+3}{8} = \frac{y-4}{2} = \frac{z+1}{2}8x+3​=2y−4​=2z+1​ and are at a distance of 6 units from R(1, 2, 3). If (α, β, γ) is the centroid of △PQR, then α + β + γ is equal to :
  1. (A)4
  2. (B)5
  3. (C)6
  4. (D)8

Correct answer: (C)

Step-by-step solution →
Q64·MathematicsSingle correct
If the distance of the point (a, 2, 5) from the image of the point (1, 2, 7) in the line x1=y−11=z−22\frac{x}{1} = \frac{y-1}{1} = \frac{z-2}{2}1x​=1y−1​=2z−2​ is 4, then the sum of all possible values of a is equal to :
  1. (A)11
  2. (B)9
  3. (C)6
  4. (D)4

Correct answer: (C)

Step-by-step solution →
Q65·MathematicsSingle correct
Let O be the origin, OP⃗=a⃗\vec{OP} = \vec{a}OP=a and OQ⃗=b⃗\vec{OQ} = \vec{b}OQ​=b. If R is the point on OP⃗\vec{OP}OP such that OP⃗=5OR⃗\vec{OP} = 5\vec{OR}OP=5OR, and M is the point such that OQ⃗=5RM⃗\vec{OQ} = 5\vec{RM}OQ​=5RM, then PM⃗\vec{PM}PM is equal to :
  1. (A)15(a⃗−4b⃗)\frac{1}{5}(\vec{a} - 4\vec{b})51​(a−4b)
  2. (B)15(b⃗−4a⃗)\frac{1}{5}(\vec{b} - 4\vec{a})51​(b−4a)
  3. (C)15(−a⃗+4b⃗)\frac{1}{5}(-\vec{a} + 4\vec{b})51​(−a+4b)
  4. (D)15(−b⃗+4a⃗)\frac{1}{5}(-\vec{b} + 4\vec{a})51​(−b+4a)

Correct answer: (B)

Step-by-step solution →
Q66·Mathematics·Limits and ContinuitySingle correct
Let f(x)=lim⁡y→0(1−cos⁡(xy))tan⁡(xy)y3f(x) = \lim_{y \to 0} \frac{(1 - \cos(xy))\tan(xy)}{y^{3}}f(x)=limy→0​y3(1−cos(xy))tan(xy)​. Then the number of solutions of the equation f(x) = sin x, x ∈ ℝ is :
  1. (A)0
  2. (B)2
  3. (C)3
  4. (D)1

Correct answer: (C)

Step-by-step solution →
Q67·MathematicsSingle correct
Let (21−a+21+a)(2^{1-a} + 2^{1+a})(21−a+21+a), f(a), (3a+3−a)(3^{a} + 3^{-a})(3a+3−a) be in A.P. and α be the minimum value of f(a). Then the value of the integral ∫log⁡e(α−1)log⁡e(α)dx(e2x−e−2x)\int_{\log_e(\alpha-1)}^{\log_e(\alpha)} \frac{dx}{(e^{2x} - e^{-2x})}∫loge​(α−1)loge​(α)​(e2x−e−2x)dx​ is :
  1. (A)12log⁡e(43)\frac{1}{2}\log_e\left(\frac{4}{3}\right)21​loge​(34​)
  2. (B)14log⁡e(43)\frac{1}{4}\log_e\left(\frac{4}{3}\right)41​loge​(34​)
  3. (C)12log⁡e(85)\frac{1}{2}\log_e\left(\frac{8}{5}\right)21​loge​(58​)
  4. (D)14log⁡e(85)\frac{1}{4}\log_e\left(\frac{8}{5}\right)41​loge​(58​)

Correct answer: (B)

Step-by-step solution →
Q68·MathematicsSingle correct
Let f : [1, ∞) → ℝ be a differentiable function defined as f(x)=∫1xf(t) dt+(1−x)(log⁡ex−1)+ef(x) = \int_{1}^{x} f(t)\, dt + (1 - x)(\log_e x - 1) + ef(x)=∫1x​f(t)dt+(1−x)(loge​x−1)+e. Then the value of f(f(1)) is :
  1. (A)(1+ee)(1 + e^{e})(1+ee)
  2. (B)(1 + e)
  3. (C)(1+e+ee)(1 + e + e^{e})(1+e+ee)
  4. (D)1 + 2e

Correct answer: (A)

Step-by-step solution →
Q69·MathematicsSingle correct
Let f(x) and g(x) be twice differentiable functions satisfying f''(x) = g''(x) for all x ∈ ℝ, f'(1) = 2g'(1) = 4 and g(2) = 3f(2) = 9. Then f(25) − g(25) is equal to :
  1. (A)20
  2. (B)40
  3. (C)−20
  4. (D)−40

Correct answer: (B)

Step-by-step solution →
Q70·MathematicsNumerical
Let A = {1, 4, 7} and B = {2, 3, 8}. Then the number of elements, in the relation R={((a1,b1),(a2,b2))∈((A×B)×(A×B)):a1+b2 divides a2+b1}R = \{((a_1, b_1), (a_2, b_2)) \in ((A \times B) \times (A \times B)) : a_1 + b_2 \text{ divides } a_2 + b_1\}R={((a1​,b1​),(a2​,b2​))∈((A×B)×(A×B)):a1​+b2​ divides a2​+b1​} is _______.

Correct answer: 18

Step-by-step solution →
Q71·MathematicsNumerical
From the point (−1, −1), two rays are sent making angles of 45° with the line x + y = 0. These rays get reflected from the mirror x + 2y = 1. If the equations of the reflected rays are ax + by = 9 and cx + dy = 7, a, b, c, d ∈ ℤ, then the value of ad + bc is _______.

Correct answer: 7

Step-by-step solution →
Q72·MathematicsNumerical
If S={θ∈[−π,π]:cos⁡θcos⁡5θ2=cos⁡7θcos⁡7θ2}S = \left\{\theta \in [-\pi, \pi] : \cos\theta \cos\frac{5\theta}{2} = \cos 7\theta \cos\frac{7\theta}{2}\right\}S={θ∈[−π,π]:cosθcos25θ​=cos7θcos27θ​}, then n(S) is equal to _______.

Correct answer: 19

Step-by-step solution →
Q73·MathematicsNumerical
Let f : ℝ → ℝ be a function such that f(x)+3f(π2−x)=sin⁡xf(x) + 3f\left(\frac{\pi}{2} - x\right) = \sin xf(x)+3f(2π​−x)=sinx, x ∈ ℝ. Let the maximum value of f on ℝ be α. If the area of the region bounded by the curves g(x)=x2g(x) = x^{2}g(x)=x2 and h(x)=βx3h(x) = \beta x^{3}h(x)=βx3, β > 0, is α2\alpha^{2}α2, then 30β330\beta^{3}30β3 is equal to _______.

Correct answer: 16

Step-by-step solution →
Q74·MathematicsNumerical
Let y = y(x) be the solution of the differential equation (tan⁡x)1/2 dy=(sec⁡3x−(tan⁡x)3/2y) dx(\tan x)^{1/2}\, dy = (\sec^{3} x - (\tan x)^{3/2} y)\, dx(tanx)1/2dy=(sec3x−(tanx)3/2y)dx, 0<x<π20 < x < \frac{\pi}{2}0<x<2π​, y(π4)=625y\left(\frac{\pi}{4}\right) = \frac{6\sqrt{2}}{5}y(4π​)=562​​. If y(π3)=45αy\left(\frac{\pi}{3}\right) = \frac{4}{5}\alphay(3π​)=54​α, then α4\alpha^{4}α4 equals _______.

Correct answer: 48

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
  • Limits and Continuity 149/186
  • Aldehydes and Ketones 135/186
  • Equilibrium 163/186
  • Solutions 158/186
  • 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
  • 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
  • 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
  • Classification of Elements and Periodicity in Properties 113/186
  • Parabola 101/186
  • Statistics 118/186
  • Hyperbola 77/186
  • Indefinite Integration 66/186
  • Carboxylic Acids and Derivatives 54/186
  • Principles of Qualitative Analysis 58/186
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