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JEE Main 5 April 2026 Shift 1 Question Paper with Answers

5 April 2026 · April session · 73 questions

73 of the 75 questions from the JEE Main 5 April 2026 Shift 1 paper, each with its correct answer and tagged to the chapter it tests. Free to read, no account needed.

2 questions are 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
24

Physics — JEE Main 5 April 2026 Shift 1

Q1·PhysicsSingle correct
In a Vernier calipers, when both jaws touch each other, zero of the Vernier scale is shifted to the right of zero of the main scale and 7th^{\text{th}}th Vernier division coincides with a main scale reading. If the value of 1 main scale division is 1 mm and there are 10 Vernier scale divisions, then the Vernier caliper has
  1. (A)0.07 cm negative zero error
  2. (B)0.7 cm negative zero error
  3. (C)0.07 cm positive zero error
  4. (D)0.7 cm positive zero error

Correct answer: (C)

Step-by-step solution →
Q2·PhysicsSingle correct
LLL, CCC and RRR represents physical quantities inductance, capacitance and resistance respectively. The dimensional formula ML2T−4A−2ML^2T^{-4}A^{-2}ML2T−4A−2 corresponds to __________.
  1. (A)RLC\frac{R}{\sqrt{LC}}LC​R​
  2. (B)RLC\frac{R}{LC}LCR​
  3. (C)CLR\frac{C}{\sqrt{LR}}LR​C​
  4. (D)1RLC\frac{1}{R}\sqrt{\frac{L}{C}}R1​CL​​

Correct answer: (A)

Step-by-step solution →
Q3·PhysicsSingle correct
When one moves from a point 16 km below the earth's surface to a point 16 km above the earth's surface. The change in ggg is approximately α\alphaα %. The value of α\alphaα is ______. (Take radius of the earth === 6400 km.)
  1. (A)0.12
  2. (B)0.25
  3. (C)0.50
  4. (D)0.75

Correct answer: (B)

Step-by-step solution →
Q4·PhysicsSingle correct
Three masses m1=4m_1 = 4m1​=4 kg, m2=4m_2 = 4m2​=4 kg and m3=6m_3 = 6m3​=6 kg are suspended from a fixed smooth frictionless pulley as shown in the figure below. The value of T1/T2T_1/T_2T1​/T2​ is ______. (take g=10g = 10g=10 m/s2^22)
  1. (A)5/35/35/3
  2. (B)2/32/32/3
  3. (C)3/53/53/5
  4. (D)2/52/52/5

Correct answer: (A)

Step-by-step solution →
Q5·PhysicsSingle correct
A wedge YYY with mass of 10 kg and all frictionless surfaces and the inclined surface making 37° with horizontal. A block XXX with mass 2 kg is placed at the highest point of the wedge as shown in figure is at rest. At t=0t = 0t=0 wedge (Y)(Y)(Y) is pulled toward right with constant force (f)(f)(f) of 24 N. Taking the block XXX at rest at t=0t = 0t=0, the time taken by it to slide down 8.8 m on the slope, while YYY is on the move, is ______ s. (take tan⁡(37°)=3/4\tan(37°) = 3/4tan(37°)=3/4 and g=10g = 10g=10 m/s2^22)
  1. (A)2
  2. (B)4
  3. (C)2\sqrt{2}2​
  4. (D)222\sqrt{2}22​

Correct answer: (A)

Step-by-step solution →
Q6·PhysicsSingle correct
The Young's modulus of steel wire of radius rrr and length LLL is YYY. If the radius rrr and length LLL of the wire are doubled then the value of YYY
  1. (A)increases by two times
  2. (B)reduces by half
  3. (C)remains unchanged
  4. (D)becomes one fourth

Correct answer: (C)

Step-by-step solution →
Q7·PhysicsSingle correct
Consider the following statements: A. Zeroth law of thermodynamics gives concept of temperature B. First law of thermodynamics gives concept of internal energy C. In isothermal expansion of ideal gas, ΔQ≠ΔW\Delta Q \neq \Delta WΔQ=ΔW D. Product of intensive and extensive variables is extensive E. The ratio of any extensive variable to mass will be an extensive variable Choose the correct combination of statements from the options given below:
  1. (A)C, D and E Only
  2. (B)A, B and C Only
  3. (C)A, B and D Only
  4. (D)B, C and D Only

Correct answer: (C)

Step-by-step solution →
Q8·PhysicsSingle correct
Refer to the figure given below. The values of I1I_1I1​, I2I_2I2​ and I3I_3I3​ are __________.
  1. (A)I1=2.5I_1 = 2.5I1​=2.5 A, I2=1.875I_2 = 1.875I2​=1.875 A, I3=1.875I_3 = 1.875I3​=1.875 A
  2. (B)I1=1.875I_1 = 1.875I1​=1.875 A, I2=2.5I_2 = 2.5I2​=2.5 A, I3=1.875I_3 = 1.875I3​=1.875 A
  3. (C)I1=1.875I_1 = 1.875I1​=1.875 A, I2=1.875I_2 = 1.875I2​=1.875 A, I3=2.5I_3 = 2.5I3​=2.5 A
  4. (D)I1=2.5I_1 = 2.5I1​=2.5 A, I2=2.5I_2 = 2.5I2​=2.5 A, I3=1.875I_3 = 1.875I3​=1.875 A

Correct answer: (A)

Step-by-step solution →
Q9·PhysicsSingle correct
An electron of mass mmm is moving in an electric field E⃗=−2E0i^\vec{E} = -2E_0\hat{i}E=−2E0​i^ (E0=E_0 =E0​= constant >0> 0>0), with an initial velocity V⃗=v0i^\vec{V} = v_0\hat{i}V=v0​i^ (v0=v_0 =v0​= constant >0> 0>0). If λ0=h4mv0\lambda_0 = \frac{h}{4mv_0}λ0​=4mv0​h​, its de Broglie wavelength at time ttt is __________. (e=e =e= charge of electron)
  1. (A)4λ0[1−E0e2mtv0]\frac{4\lambda_0}{\left[1 - \frac{E_0e}{2m}\frac{t}{v_0}\right]}[1−2mE0​e​v0​t​]4λ0​​
  2. (B)4λ0[1+E0e2mtv0]\frac{4\lambda_0}{\left[1 + \frac{E_0e}{2m}\frac{t}{v_0}\right]}[1+2mE0​e​v0​t​]4λ0​​
  3. (C)4λ0[1+2E0emtv0]\frac{4\lambda_0}{\left[1 + \frac{2E_0e}{m}\frac{t}{v_0}\right]}[1+m2E0​e​v0​t​]4λ0​​
  4. (D)4λ0[1−2E0emtv0]\frac{4\lambda_0}{\left[1 - \frac{2E_0e}{m}\frac{t}{v_0}\right]}[1−m2E0​e​v0​t​]4λ0​​

Correct answer: (C)

Step-by-step solution →
Q10·PhysicsSingle correct
In the hydrogen atom, the electron makes a transition from the higher orbit (i)(i)(i) to a lower orbit (f)(f)(f). The ratio of the radius of the orbits in given by ri:rf=16:4r_i : r_f = 16 : 4ri​:rf​=16:4. The wavelength of photon emitted due to this transition is ______ nm. (Given Rydberg constant =1.0973×107= 1.0973 \times 10^7=1.0973×107 /m)
  1. (A)121
  2. (B)242
  3. (C)486
  4. (D)974

Correct answer: (C)

Step-by-step solution →
Q11·PhysicsSingle correct
A displacement current of 4.0 A can be set up in the space between two parallel plates of 6 μF capacitor. The rate of change of potential difference across the plates of the capacitor is nearly α×106\alpha \times 10^6α×106 V/s. The value of α\alphaα is __________.
  1. (A)0.58
  2. (B)0.67
  3. (C)0.82
  4. (D)0.75

Correct answer: (B)

Step-by-step solution →
Q12·PhysicsSingle correct
Refer to the figure given below, current between terminals AAA and BBB is ______ A.
  1. (A)12.5
  2. (B)1.25
  3. (C)7.5
  4. (D)5

Correct answer: (B)

Step-by-step solution →
Q13·PhysicsSingle correct
In Young's double slit experiment, the fringe width of the interference pattern produced on the screen is 2.4 μm. If the experiment is carried out in another medium having refractive index 1.2, the fringe width will be _____ μm.
  1. (A)1.2
  2. (B)2
  3. (C)2.4
  4. (D)2.88

Correct answer: (B)

Step-by-step solution →
Q14·PhysicsSingle correct
A ray of light passing through an equilateral prism is having velocity 2.12×1082.12 \times 10^{8}2.12×108 m/s in the prism material, then the minimum angle of deviation is _____ degrees.
  1. (A)45
  2. (B)30
  3. (C)28
  4. (D)58

Correct answer: (B)

Step-by-step solution →
Q15·PhysicsSingle correct
Light source having wavelength 331 nm is used to generate photo-electrons whose stopping potential is 0.2 V. The work function of the used metal in the experiment is α×10−19\alpha \times 10^{-19}α×10−19 J. The value of α\alphaα is _____. (h=6.62×10−34h = 6.62 \times 10^{-34}h=6.62×10−34 J s, e=1.6×10−19e = 1.6 \times 10^{-19}e=1.6×10−19 C and c=3×108c = 3 \times 10^{8}c=3×108 m/s)
  1. (A)3.68
  2. (B)4.68
  3. (C)5.68
  4. (D)2.68

Correct answer: (C)

Step-by-step solution →
Q16·PhysicsSingle correct
A compound microscope is designed with two symmetric biconvex lenses. The objective lens is cut vertically, creating two identical plano-convex lenses. One of them is used in place of original objective lens. To retain same magnification keeping the object distance unchanged, the tube length has to be
  1. (A)increased two times
  2. (B)increased 32\frac{3}{2}23​ times
  3. (C)decreased two times
  4. (D)decreased 32\frac{3}{2}23​ times

Correct answer: (A)

Step-by-step solution →
Q17·PhysicsSingle correct
Two wires as shown in the figure below, made of steel and have breaking stress of 12×10812 \times 10^{8}12×108 N/m2^{2}2. Area of cross-section of upper wire is 0.008 cm2^{2}2 and of lower wire is 0.004 cm2^{2}2. The maximum mass that can be added to pan without breaking any wire is _____ kg. (take g=10g = 10g=10 m/s2^{2}2)
  1. (A)56
  2. (B)38
  3. (C)96
  4. (D)5.6

Correct answer: (B)

Step-by-step solution →
Q18·PhysicsSingle correct
An a.c. source of angular frequency ω\omegaω is connected across a resistor RRR and a capacitor CCC in series. The current is observed as III. Now the frequency of the source is changed to ω/4\omega/4ω/4, (keeping the voltage unchanged) the current is found to be I/3I/3I/3. The ratio of resistance to reactance at frequency ω\omegaω is
  1. (A)67\sqrt{\frac{6}{7}}76​​
  2. (B)35\sqrt{\frac{3}{5}}53​​
  3. (C)78\sqrt{\frac{7}{8}}87​​
  4. (D)34\sqrt{\frac{3}{4}}43​​

Correct answer: (C)

Step-by-step solution →
Q19·PhysicsSingle correct
For the given logic circuit, which of the following inputs combination will make both LED−1 and LED−2 to glow?
  1. (A)A=0,B=1,C=1A = 0, B = 1, C = 1A=0,B=1,C=1
  2. (B)A=1,B=0,C=0A = 1, B = 0, C = 0A=1,B=0,C=0
  3. (C)A=1,B=0,C=1A = 1, B = 0, C = 1A=1,B=0,C=1
  4. (D)A=1,B=1,C=0A = 1, B = 1, C = 0A=1,B=1,C=0

Correct answer: (C)

Step-by-step solution →
Q20·PhysicsNumerical
A cube has side length 5 cm and modulus of rigidity 10510^{5}105 N/m2^{2}2. The displacement produced by a force of 10 N in the upper face of cube is _____ mm.

Correct answer: 2

Step-by-step solution →
Q21·PhysicsNumerical
From 18 m height above the ground a ball is dropped from rest. The height above the ground at which the magnitude of velocity equal to the magnitude of acceleration (in the same set of units) due to gravity is _____ m. (Take g=10g = 10g=10 m/s2^{2}2 and neglect the air resistance)

Correct answer: 13

Step-by-step solution →
Q22·PhysicsNumerical
A transverse wave on a string is described by y=3sin⁡(36t+0.018x+π/4)y = 3\sin(36t + 0.018x + \pi/4)y=3sin(36t+0.018x+π/4), where x,yx, yx,y are in cm and ttt in seconds. The least distance between the two successive crests in the wave is _____ cm. (Nearest integer) (π=3.14\pi = 3.14π=3.14)

Correct answer: 349

Step-by-step solution →
Q23·Physics·Magnetic Field of CurrentNumerical
The charged particle moving in a uniform magnetic field of (3i^+2j^)(3\hat{i} + 2\hat{j})(3i^+2j^​) T has an acceleration (4i^−x2j^)\left(4\hat{i} - \frac{x}{2}\hat{j}\right)(4i^−2x​j^​) m/s2^{2}2. The value of xxx is _____.

Correct answer: 12

Step-by-step solution →
Q24·PhysicsNumerical
In the given circuit below inductance values of L1L_1L1​, L2L_2L2​ and L3L_3L3​ are same. The magnetic energy stored in the entire circuit is (Ut)(U_t)(Ut​) and that stored in the L2L_2L2​ inductor is (Ul)(U_l)(Ul​). Ut/UlU_t/U_lUt​/Ul​ is _____. (Ignore the mutual inductance if any)

Correct answer: 6

Step-by-step solution →

Chemistry — JEE Main 5 April 2026 Shift 1

Q25·ChemistrySingle correct
How many grams of residue is obtained by heating 2.76 g of silver carbonate? (Given: Molar mass of C, O and Ag are 12, 16 and 108 g mol−1^{-1}−1 respectively)
  1. (A)1.08 g
  2. (B)2.16 g
  3. (C)3.24 g
  4. (D)4.32 g

Correct answer: (B)

Step-by-step solution →
Q26·ChemistrySingle correct
Arrange the following atomic orbitals of multi electron atoms in order of increasing energy. A. n=3,l=2,m=+1n = 3, l = 2, m = +1n=3,l=2,m=+1 B. n=4,l=0,m=0n = 4, l = 0, m = 0n=4,l=0,m=0 C. n=6,l=1,m=0n = 6, l = 1, m = 0n=6,l=1,m=0 D. n=5,l=1,m=+1n = 5, l = 1, m = +1n=5,l=1,m=+1 E. n=2,l=1,m=+1n = 2, l = 1, m = +1n=2,l=1,m=+1 Choose the correct answer from the options given below:
  1. (A)C<D<B<A<EC < D < B < A < EC<D<B<A<E
  2. (B)B<A<E<C<DB < A < E < C < DB<A<E<C<D
  3. (C)E<C<D<B<AE < C < D < B < AE<C<D<B<A
  4. (D)E<B<A<D<CE < B < A < D < CE<B<A<D<C

Correct answer: (D)

Step-by-step solution →
Q27·ChemistrySingle correct
Identify the correct statements from the following: A. Heisenberg uncertainty principle is applicable to electrons. B. The size of 2px2p_x2px​ orbital is less than the size of 3px3p_x3px​ orbital. C. The energy of 2s2s2s orbital of H atom is equal to the energy of 2s2s2s orbital of Li. D. The electronic configuration of Cr is [Ar]3d54s1[\mathrm{Ar}]3d^5 4s^1[Ar]3d54s1 Choose the correct answer from the options given below:
  1. (A)A, B and C Only
  2. (B)A, B and D Only
  3. (C)B, C and D Only
  4. (D)A, C and D Only

Correct answer: (B)

Step-by-step solution →
Q28·ChemistrySingle correct
What is the mole fraction of water in 10% by weight (w/w) of aqueous urea solution? [Given: Molar mass of H, O, C and N are 1, 16, 12 and 14 g mol−1^{-1}−1 respectively.]
  1. (A)0.825
  2. (B)0.032
  3. (C)0.867
  4. (D)0.967

Correct answer: (D)

Step-by-step solution →
Q29·ChemistrySingle correct
M3A2M_3A_2M3​A2​ is a sparingly soluble salt of molar mass yyy g mol−1^{-1}−1 and solubility xxx g L−1^{-1}−1. The ratio of the molar concentration of the anion (A3−A^{3-}A3−) to the solubility product of the salt is
  1. (A)154⋅y4x4\frac{1}{54} \cdot \frac{y^4}{x^4}541​⋅x4y4​
  2. (B)y5108x4\frac{y^5}{108x^4}108x4y5​
  3. (C)108⋅x5y5108 \cdot \frac{x^5}{y^5}108⋅y5x5​
  4. (D)1108⋅y4x4\frac{1}{108} \cdot \frac{y^4}{x^4}1081​⋅x4y4​

Correct answer: (A)

Step-by-step solution →
Q30·ChemistrySingle correct
Arrange the following resultant mixtures in increasing order of their pH values A. 10 mL 0.2 M Ca(OH)2_22​ + 25 mL 0.1 M HCl B. 10 mL 0.01 M H2_22​SO4_44​ + 10 mL 0.01 M Ca(OH)2_22​ C. 10 mL 0.1 M H2_22​SO4_44​ + 10 mL 0.1 M KOH Choose the correct answer from the options given below:
  1. (A)B<C<AB < C < AB<C<A
  2. (B)C<A<BC < A < BC<A<B
  3. (C)C<B<AC < B < AC<B<A
  4. (D)A<C<BA < C < BA<C<B

Correct answer: (C)

Step-by-step solution →
Q31·ChemistrySingle correct
First order gas phase reaction A→B+CA \rightarrow B + CA→B+C pip_ipi​ = initial pressure of gas A, ptp_tpt​ = total pressure of the reaction mixture at time ttt Expression of rate constant (kkk) is
  1. (A)1tln⁡pi2pi−pt\frac{1}{t} \ln \frac{p_i}{2p_i - p_t}t1​ln2pi​−pt​pi​​
  2. (B)1tln⁡2pipi−pt\frac{1}{t} \ln \frac{2p_i}{p_i - p_t}t1​lnpi​−pt​2pi​​
  3. (C)1tln⁡pi3pi−2pt\frac{1}{t} \ln \frac{p_i}{3p_i - 2p_t}t1​ln3pi​−2pt​pi​​
  4. (D)1tln⁡3pi4pi−pt\frac{1}{t} \ln \frac{3p_i}{4p_i - p_t}t1​ln4pi​−pt​3pi​​

Correct answer: (A)

Step-by-step solution →
Q32·ChemistrySingle correct
Given below are two statements: Statement I: The correct order of electronegativity of fluorine, oxygen and nitrogen is F>O>NF > O > NF>O>N. Statement II: The oxidation state of oxygen in OF2_22​ is +2+2+2 and in Na2_22​O is −2-2−2. 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
Correct statements from the following are: A. Nitrogen in oxidation states from +1+1+1 to +4+4+4 disproportionates in acid medium. B. Nitrogen has the ability to form dπ−pπd\pi - p\pidπ−pπ multiple bonds with itself and other elements with small size and high electronegativity. C. N-N single bond is stronger than P-P single bond. D. Nitrogen has highest density in its group due to small size. E. The maximum covalency of nitrogen is four since it has only four valence orbitals for bonding. Choose the correct answer from the options given below:
  1. (A)B, C and D Only
  2. (B)C, D and E Only
  3. (C)A, C and E Only
  4. (D)A and E Only

Correct answer: (D)

Step-by-step solution →
Q34·ChemistrySingle correct
Which of the following is NOT a physical or chemical characteristics of interstitial compounds?
  1. (A)They have high melting points, higher than those of pure metals.
  2. (B)They are very soft and ionic in nature.
  3. (C)They retain metallic conductivity.
  4. (D)They are chemically inert and usually non-stoichiometric.

Correct answer: (B)

Step-by-step solution →
Q35·ChemistrySingle correct
The correct statements about metal carbonyls are A. The metal-carbon bonds in metal carbonyls possess both σ\sigmaσ and π\piπ-character. B. Due to synergic bonding interactions between metal and CO ligand, the metal-carbon bond becomes weak. C. The metal-carbon σ\sigmaσ bond is formed by the donation of lone pair of electrons on the carbonyl carbon into a vacant orbital of metal. D. The metal-carbon π\piπ bond is formed by the donation of electrons from filled d-orbital of metal into vacant π∗\pi^*π∗ orbital of CO. Choose the correct answer from the options given below:
  1. (A)A and B Only
  2. (B)A, C and D Only
  3. (C)B and C Only
  4. (D)A and D Only

Correct answer: (B)

Step-by-step solution →
Q36·ChemistrySingle correct
Given below are two statements: Statement I: Each electron in ege_geg​ orbitals destabilizes the orbitals by +0.6Δo+0.6\Delta_o+0.6Δo​ and each electron in the t2gt_{2g}t2g​ orbitals stabilizes the orbitals by −0.4Δo-0.4\Delta_o−0.4Δo​ in an octahedral field on the basis of crystal field theory. Statement II: All the d-orbitals of the transition metals have the same energy in their free atomic state but when a complex is formed the ligands destroy the degeneracy of these orbitals on the basis of crystal field theory. 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 →
Q37·Chemistry·Electronic Effects and StabilitySingle correct
Given below are two statements: Statement I: On the basis of inductive effect, the order of stability of alkyl carbanions is CH3−>CH3-CH2−>(CH3)2CH−>(CH3)3C−\mathrm{CH_3^-} > \mathrm{CH_3\text{-}CH_2^-} > \mathrm{(CH_3)_2CH^-} > \mathrm{(CH_3)_3C^-}CH3−​>CH3​-CH2−​>(CH3​)2​CH−>(CH3​)3​C−. Statement II: Allyl and benzyl carbanions are more stabilised by inductive effect and not by resonance effect. 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: (C)

Step-by-step solution →
Q38·ChemistrySingle correct
"P" is a hydrocarbon of molecular formula: C8_88​H14_{14}14​. On ozonolysis, "P" forms "Q". "Q" on treatment with alkali under reflux condition produces "R", which on treatment with I2_22​/NaOH gives a yellow precipitate. Acidification of the solution gives "S". The structure of "S" is given below: The correct structure of "P" is
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (D)

Step-by-step solution →
Q39·ChemistrySingle correct
For the following Friedel Craft's alkylation reaction, which of the statements are correct? A. Major product is n-propyl benzene. B. iso-propyl carbocation intermediate is also generated. C. Multiple substitution is inevitable. D. Introducing electron-donating substituent on benzene will not produce any alkyl benzene. Choose the correct answer from the options given below:
  1. (A)A and D only
  2. (B)B and C only
  3. (C)A and C only
  4. (D)B and D only

Correct answer: (B)

Step-by-step solution →
Q40·ChemistrySingle correct
Benzyl isocyanide can be obtained from Choose the correct answer from the options given below:
  1. (A)A and B Only
  2. (B)A and C Only
  3. (C)B and D Only
  4. (D)D and E Only

Correct answer: (A)

Step-by-step solution →
Q41·ChemistrySingle correct
Consider compounds A, B and C with following structural formulae A = CH3_33​ - CH2_22​ - CH2_22​ - CH2_22​ - CH2_22​ - OH B = CH2_22​ = CH - CH2_22​ - CH2_22​ - CH3_33​ C = HO - CH2_22​ - CH2_22​ - CH(OH) - CH3_33​ For the conversion of B from A, reagent (D) required is __________ and structural formula of product (E) obtained when C undergoes same reaction using excess reagent (D) is __________.
  1. (A)D: Conc. H2_22​SO4_44​; E: CH2_22​ = CH −-− CH(OH)CH3_33​
  2. (B)D: PCC; E: HO −-− CH2_22​ −-− CH2_22​ −-− CH = CH2_22​
  3. (C)D: PCC; E: CH2_22​ = CH −-− CH = CH2_22​
  4. (D)D: Conc. H2_22​SO4_44​ or H3_33​PO4_44​; E: CH2_22​ = CH −-− CH = CH2_22​

Correct answer: (D)

Step-by-step solution →
Q42·ChemistrySingle correct
Identify the incorrect statements. Choose the correct answer from the options given below:
  1. (A)A and D Only
  2. (B)A and C Only
  3. (C)B and C Only
  4. (D)A and B Only

Correct answer: (C)

Step-by-step solution →
Q43·ChemistrySingle correct
Identify the correct statements. A. Glucose exists in two anomeric forms. B. Anomers of glucose differ in configuration at C−-−1 in cyclic hemiacetal structure. C. Melting point of α\alphaα-anomer of glucose is greater than β\betaβ-anomer. D. Specific rotation of α\alphaα-anomer is +19∘+19^\circ+19∘ while for β\betaβ-anomer is +112∘+112^\circ+112∘ E. α\alphaα and β\betaβ-anomers of glucose are prepared by crystallization of saturated glucose solution at 303 K and 371 K respectively. Choose the correct answer from the options given below:
  1. (A)A and B Only
  2. (B)B and C Only
  3. (C)A, B and D Only
  4. (D)A, B and E Only

Correct answer: (D)

Step-by-step solution →
Q44·ChemistrySingle correct
Given below are two statements: Statement I: Sodium dichromate and potassium dichromate are classified as primary standards in titrimetric analysis. Statement II: Phenolphthalein is a weak base, therefore it dissociates in acidic medium. 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: (B)

Step-by-step solution →
Q45·ChemistryNumerical
Consider the following species: BrF5_55​, XeF5−_5^-5−​, BF4−_4^-4−​, ICl4−_4^-4−​, XeF4_44​, SF4_44​, NH4+_4^+4+​, ClF3_33​, XeF2_22​, ICl2−_2^-2−​ Number of species having sp3^33d hybridized central atom is __________.

Correct answer: 4

Step-by-step solution →
Q46·ChemistryNumerical
In an estimation of sulphur by Carius method 0.2 g of the substance gave 0.6 g of BaSO4_44​. The percentage of sulphur in the substance is _____ %. (Given molar mass in g mol−1^{-1}−1 S : 32, BaSO4_44​ : 233)

Correct answer: 41

Step-by-step solution →
Q47·Chemistry·Alcohols and EthersNumerical
One mole of phenol is treated with dilute HNO3_33​ at 298 K to give a mixture of products. The mixture is separated by steam distillation. The steam volatile compound (X) is separated. The increase in percentage of oxygen in (X) with respect to phenol is _____ ×10−1\times 10^{-1}×10−1 % (Given molar mass in g mol−1^{-1}−1 H:1, C:12, N:14, O:16)

Correct answer: 175

Step-by-step solution →
Q48·ChemistryNumerical
The values of pressure equilibrium constant recorded at different temperatures for the following equilibrium reaction have been given below A(g)⇌B(g)+C(g)A(g) \rightleftharpoons B(g) + C(g)A(g)⇌B(g)+C(g) The magnitude of ΔH∘R\frac{\Delta H^\circ}{R}RΔH∘​ calculated from the above data is _____. (Nearest integer)
1T\frac{1}{T}T1​ (K−1^{-1}−1)log⁡10Kp\log_{10} K_plog10​Kp​
0.053.5
0.062.5
0.071.5

Correct answer: 230

Step-by-step solution →
Q49·ChemistryNumerical
If the half life of a first order reaction is 6.93 minutes then the time required for completion of 99% of the reaction will be _____ minutes. (Given : log⁡2=0.3010\log 2 = 0.3010log2=0.3010)

Correct answer: 46.06

Step-by-step solution →

Mathematics — JEE Main 5 April 2026 Shift 1

Q50·MathematicsSingle correct
Let a,b∈Ca, b \in \mathbb{C}a,b∈C. Let α,β\alpha, \betaα,β be the roots of the equation x2+ax+b=0x^2 + ax + b = 0x2+ax+b=0. If β−α=11\beta - \alpha = \sqrt{11}β−α=11​ and β2−α2=3i11\beta^2 - \alpha^2 = 3i\sqrt{11}β2−α2=3i11​, then (β3−α3)2(\beta^3 - \alpha^3)^2(β3−α3)2 is equal to:
  1. (A)160
  2. (B)176
  3. (C)194
  4. (D)187

Correct answer: (B)

Step-by-step solution →
Q51·MathematicsSingle correct
Let the sum of the first nnn terms of an A.P. be 3n2+5n3n^2 + 5n3n2+5n. Then the sum of squares of the first 10 terms of the A.P. is:
  1. (A)10220
  2. (B)12860
  3. (C)15220
  4. (D)19780

Correct answer: (C)

Step-by-step solution →
Q52·MathematicsSingle correct
Let AAA be a 3×33 \times 33×3 matrix such that AT[101]=[522]A^T \begin{bmatrix}1\\0\\1\end{bmatrix} = \begin{bmatrix}5\\2\\2\end{bmatrix}AT​101​​=​522​​, AT[001]=[311]A^T \begin{bmatrix}0\\0\\1\end{bmatrix} = \begin{bmatrix}3\\1\\1\end{bmatrix}AT​001​​=​311​​, A[101]=[344]A \begin{bmatrix}1\\0\\1\end{bmatrix} = \begin{bmatrix}3\\4\\4\end{bmatrix}A​101​​=​344​​ and A[001]=[131]A \begin{bmatrix}0\\0\\1\end{bmatrix} = \begin{bmatrix}1\\3\\1\end{bmatrix}A​001​​=​131​​. If det⁡(A)=1\det(A) = 1det(A)=1, then det⁡(adj(A2+A))\det(\mathrm{adj}(A^2 + A))det(adj(A2+A)) is equal to:
  1. (A)16
  2. (B)25
  3. (C)49
  4. (D)64

Correct answer: (D)

Step-by-step solution →
Q53·MathematicsSingle correct
Consider the system of linear equations in xxx, yyy, zzz: x+2y+tz=0x + 2y + tz = 0x+2y+tz=0, 6x+y+5tz=06x + y + 5tz = 06x+y+5tz=0, 3x+t2y+f(t)z=03x + t^2 y + f(t)z = 03x+t2y+f(t)z=0, where f:R→Rf : \mathbb{R} \to \mathbb{R}f:R→R is a differentiable function. If this system has infinitely many solutions for all t∈Rt \in \mathbb{R}t∈R, then fff
  1. (A)is a constant function
  2. (B)is strictly increasing on R\mathbb{R}R
  3. (C)is strictly decreasing on R\mathbb{R}R
  4. (D)has two critical points

Correct answer: (B)

Step-by-step solution →
Q54·MathematicsSingle correct
∑n=110(528n(n+1)(n+2))\sum_{n=1}^{10}\left(\frac{528}{n(n+1)(n+2)}\right)∑n=110​(n(n+1)(n+2)528​) is equal to:
  1. (A)65
  2. (B)130
  3. (C)220
  4. (D)440

Correct answer: (B)

Step-by-step solution →
Q55·MathematicsSingle correct
Let tan⁡A\tan AtanA, tan⁡B\tan BtanB, where A,B∈(−π2,π2)A, B \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)A,B∈(−2π​,2π​), be the roots of the quadratic equation x2−2x−5=0x^2 - 2x - 5 = 0x2−2x−5=0. Then 20sin⁡2(A+B2)20\sin^2\left(\frac{A+B}{2}\right)20sin2(2A+B​) is equal to:
  1. (A)10+1010 + \sqrt{10}10+10​
  2. (B)10−21010 - 2\sqrt{10}10−210​
  3. (C)10−31010 - 3\sqrt{10}10−310​
  4. (D)10−1010 - \sqrt{10}10−10​

Correct answer: (C)

Step-by-step solution →
Q56·MathematicsSingle correct
A letter is known to have arrived by post either from KANPUR or from ANANTPUR. On the envelope just two consecutive letters AN are visible. The probability, that the letter came from ANANTPUR, is:
  1. (A)710\frac{7}{10}107​
  2. (B)1017\frac{10}{17}1710​
  3. (C)1219\frac{12}{19}1912​
  4. (D)719\frac{7}{19}197​

Correct answer: (B)

Step-by-step solution →
Q57·MathematicsSingle correct
The mean deviation about the mean for the data is equal to:
xix_ixi​fif_ifi​
58
76
92
102
122
156
  1. (A)4013\frac{40}{13}1340​
  2. (B)4213\frac{42}{13}1342​
  3. (C)4413\frac{44}{13}1344​
  4. (D)4613\frac{46}{13}1346​

Correct answer: (C)

Step-by-step solution →
Q58·MathematicsSingle correct
Let a focus of the ellipse E:x2a2+y2b2=1E : \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1E:a2x2​+b2y2​=1 be S(4,0)S(4,0)S(4,0) and its eccentricity be 45\frac{4}{5}54​. If the point P(3,α)P(3, \alpha)P(3,α) lies on EEE and OOO is the origin, then the area of △POS\triangle POS△POS is equal to:
  1. (A)12/5
  2. (B)14/5
  3. (C)24/5
  4. (D)48/5

Correct answer: (C)

Step-by-step solution →
Q59·MathematicsSingle correct
Let PPP be a moving point on the circle x2+y2−6x−8y+21=0x^2 + y^2 - 6x - 8y + 21 = 0x2+y2−6x−8y+21=0. Then, the maximum distance of PPP from the vertex of the parabola x2+6x+y+13=0x^2 + 6x + y + 13 = 0x2+6x+y+13=0 is equal to:
  1. (A)8
  2. (B)10
  3. (C)12
  4. (D)9

Correct answer: (C)

Step-by-step solution →
Q60·MathematicsSingle correct
In an equilateral triangle PQRPQRPQR, let the vertex PPP be at (3,5)(3,5)(3,5) and the side QRQRQR be along the line x+y=4x + y = 4x+y=4. If the orthocentre of the triangle PQRPQRPQR is (α,β)(\alpha, \beta)(α,β), then 9(α+β)9(\alpha + \beta)9(α+β) is equal to:
  1. (A)16
  2. (B)27
  3. (C)36
  4. (D)48

Correct answer: (D)

Step-by-step solution →
Q61·MathematicsSingle correct
The sum of all the integral values of ppp such that the equation 3sin⁡2x+12cos⁡x−3=p3\sin^2 x + 12\cos x - 3 = p3sin2x+12cosx−3=p, x∈Rx \in \mathbb{R}x∈R, has at least one solution, is:
  1. (A)−54-54−54
  2. (B)−60-60−60
  3. (C)−75-75−75
  4. (D)−84-84−84

Correct answer: (C)

Step-by-step solution →
Q62·MathematicsSingle correct
The square of the distance of the point P(5,6,7)P(5,6,7)P(5,6,7) from the line x−22=y−53=z−24\frac{x-2}{2} = \frac{y-5}{3} = \frac{z-2}{4}2x−2​=3y−5​=4z−2​ is equal to:
  1. (A)3
  2. (B)5
  3. (C)6
  4. (D)8

Correct answer: (C)

Step-by-step solution →
Q63·MathematicsSingle correct
Let a⃗=7i^+j^−k^\vec{a} = \sqrt{7}\hat{i} + \hat{j} - \hat{k}a=7​i^+j^​−k^ and b⃗=j^+2k^\vec{b} = \hat{j} + 2\hat{k}b=j^​+2k^. If r⃗\vec{r}r is a vector such that r⃗×a⃗+a⃗×b⃗=0⃗\vec{r} \times \vec{a} + \vec{a} \times \vec{b} = \vec{0}r×a+a×b=0 and r⃗⋅a⃗=0\vec{r} \cdot \vec{a} = 0r⋅a=0, then ∣3r⃗∣2|3\vec{r}|^2∣3r∣2 is equal to:
  1. (A)44
  2. (B)54
  3. (C)86
  4. (D)132

Correct answer: (A)

Step-by-step solution →
Q64·MathematicsSingle correct
The square of the distance of the point of intersection of the lines r⃗=(i^+j^−k^)+λ(ai^−j^)\vec{r} = (\hat{i} + \hat{j} - \hat{k}) + \lambda(a\hat{i} - \hat{j})r=(i^+j^​−k^)+λ(ai^−j^​), a≠0a \neq 0a=0 and r⃗=(4i^−k^)+μ(2i^+ak^)\vec{r} = (4\hat{i} - \hat{k}) + \mu(2\hat{i} + a\hat{k})r=(4i^−k^)+μ(2i^+ak^) from the origin is:
  1. (A)5
  2. (B)10
  3. (C)17
  4. (D)26

Correct answer: (C)

Step-by-step solution →
Q65·MathematicsSingle correct
The product of all possible values of α\alphaα, for which lim⁡x→0(1−cos⁡(αx)cos⁡((α+1)x)cos⁡((α+2)x)sin⁡2((α+1)x))=2\lim\limits_{x \to 0} \left( \frac{1 - \cos(\alpha x)\cos((\alpha + 1)x)\cos((\alpha + 2)x)}{\sin^2((\alpha + 1)x)} \right) = 2x→0lim​(sin2((α+1)x)1−cos(αx)cos((α+1)x)cos((α+2)x)​)=2, is:
  1. (A)−2-2−2
  2. (B)1
  3. (C)−1-1−1
  4. (D)54\frac{5}{4}45​

Correct answer: (C)

Step-by-step solution →
Q66·MathematicsSingle correct
The value of the integral ∫0∞log⁡e(x)x2+4 dx\int_0^\infty \frac{\log_e(x)}{x^2 + 4}\,dx∫0∞​x2+4loge​(x)​dx is:
  1. (A)πlog⁡e(2)2\frac{\pi \log_e(2)}{2}2πloge​(2)​
  2. (B)πlog⁡e(2)4\frac{\pi \log_e(2)}{4}4πloge​(2)​
  3. (C)1+πlog⁡e(2)1 + \pi \log_e(2)1+πloge​(2)
  4. (D)2+πlog⁡e(2)2 + \pi \log_e(2)2+πloge​(2)

Correct answer: (B)

Step-by-step solution →
Q67·MathematicsSingle correct
Let f:R→Rf : \mathbb{R} \to \mathbb{R}f:R→R be a differentiable function such that f(x+y3)=f(x)+f(y)3f\left(\frac{x + y}{3}\right) = \frac{f(x) + f(y)}{3}f(3x+y​)=3f(x)+f(y)​ for all x,y∈Rx, y \in \mathbb{R}x,y∈R, and f′(0)=3f'(0) = 3f′(0)=3. Then the minimum value of the function g(x)=3+exf(x)g(x) = 3 + e^x f(x)g(x)=3+exf(x), is:
  1. (A)3(e+1e)3\left(\frac{e + 1}{e}\right)3(ee+1​)
  2. (B)3(e−1e)3\left(\frac{e - 1}{e}\right)3(ee−1​)
  3. (C)3−ee\frac{3 - e}{e}e3−e​
  4. (D)3e3e3e

Correct answer: (B)

Step-by-step solution →
Q68·MathematicsSingle correct
The value of the integral ∫π/6π/3(4−csc⁡2xcos⁡4x)dx\int_{\pi/6}^{\pi/3} \left( \frac{4 - \csc^2 x}{\cos^4 x} \right) dx∫π/6π/3​(cos4x4−csc2x​)dx is:
  1. (A)113\frac{11}{\sqrt{3}}3​11​
  2. (B)163\frac{16}{\sqrt{3}}3​16​
  3. (C)3233\frac{32}{3\sqrt{3}}33​32​
  4. (D)6433\frac{64}{3\sqrt{3}}33​64​

Correct answer: (C)

Step-by-step solution →
Q69·MathematicsNumerical
Let A={1,2,3,4,5,6}A = \{1, 2, 3, 4, 5, 6\}A={1,2,3,4,5,6}. The number of one-one functions f:A→Af : A \to Af:A→A such that f(1)≥3f(1) \geq 3f(1)≥3, f(3)≤4f(3) \leq 4f(3)≤4 and f(2)+f(3)=5f(2) + f(3) = 5f(2)+f(3)=5, is __________.

Correct answer: 72

Step-by-step solution →
Q70·MathematicsNumerical
Two players AAA and BBB play a series of games of badminton. The player, who wins 5 games first, wins the series. Assuming that no game ends in a draw, the number of ways, in which player AAA wins the series is __________.

Correct answer: 126

Step-by-step solution →
Q71·MathematicsNumerical
If the sum of the coefficients of x7x^7x7 and x14x^{14}x14 in the expansion of (1x3−x4)n\left(\frac{1}{x^3} - x^4\right)^n(x31​−x4)n, x≠0x \neq 0x=0, is zero, then the value of nnn is __________.

Correct answer: 21

Step-by-step solution →
Q72·MathematicsNumerical
If π4+∑p=111tan⁡−1(2p−11+22p−1)=α\frac{\pi}{4} + \sum\limits_{p=1}^{11} \tan^{-1}\left(\frac{2^{p-1}}{1 + 2^{2p-1}}\right) = \alpha4π​+p=1∑11​tan−1(1+22p−12p−1​)=α, then tan⁡α\tan\alphatanα is equal to __________.

Correct answer: 2048

Step-by-step solution →
Q73·MathematicsNumerical
Let y=y(x)y = y(x)y=y(x) be the solution of the differential equation xsin⁡(yx)dy=(ysin⁡(yx)−x)dxx\sin\left(\frac{y}{x}\right) dy = \left(y\sin\left(\frac{y}{x}\right) - x\right) dxxsin(xy​)dy=(ysin(xy​)−x)dx, y(1)=π2y(1) = \frac{\pi}{2}y(1)=2π​ and let α=cos⁡(y(e12)e12)\alpha = \cos\left(\frac{y(e^{12})}{e^{12}}\right)α=cos(e12y(e12)​). Then the number of integral values of ppp, for which the equation x2+y2−2px+2py+α+2=0x^2 + y^2 - 2px + 2py + \alpha + 2 = 0x2+y2−2px+2py+α+2=0 represents a circle of radius r≤6r \leq 6r≤6, is __________.

Correct answer: 6

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
  • 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
  • Units and Measurements 149/186
  • Hydrocarbons 126/186
  • 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
  • Some Basic Concepts in Chemistry 129/186
  • Electromagnetic Induction 120/186
  • Wave Optics 130/186
  • Alcohols and Ethers 106/186
  • Purification and Characterisation of Organic Compounds 116/186
  • Alternating Currents 108/186
  • Straight Lines 114/186
  • Waves 109/186
  • Atoms 112/186
  • Classification of Elements and Periodicity in Properties 113/186
  • Statistics 118/186
  • Ellipse 103/186
  • Inverse Trigonometric Functions 93/186
  • Electronic Effects and Stability 74/186
  • Experimental Skills 68/186
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