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JEE Main 8 April 2025 Shift 2 Question Paper with Answers

8 April 2025 · April session · 75 questions

The complete JEE Main 8 April 2025 Shift 2 paper — all 75 questions with the correct answer for each, tagged to the chapter it tests. Free to read, no account needed.

Physics
25
Chemistry
25
Mathematics
25

Physics — JEE Main 8 April 2025 Shift 2

Q1·Physics·Electric PotentialSingle correct
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: Work done in moving a test charge between two points inside a uniformly charged spherical shell is zero, no matter which path is chosen. Reason R: Electrostatic potential inside a uniformly charged spherical shell is constant and is same as that on the surface of the shell. In the light of the above statements, choose the correct answer from the options given below:
  1. (A)A is true but R is false
  2. (B)Both A and R are true and R is the correct explanation of A
  3. (C)A is false but R is true
  4. (D)Both A and R are true but R is NOT the correct explanation of A

Correct answer: (B)

Step-by-step solution →
Q2·Physics·Rotational MotionSingle correct
A rod of linear mass density 'λ\lambdaλ' and length 'LLL' is bent to form a ring of radius 'RRR'. Moment of inertia of ring about any of its diameter is:
  1. (A)λL316π2\dfrac{\lambda L^3}{16\pi^2}16π2λL3​
  2. (B)λL312\dfrac{\lambda L^3}{12}12λL3​
  3. (C)λL34π2\dfrac{\lambda L^3}{4\pi^2}4π2λL3​
  4. (D)λL38π2\dfrac{\lambda L^3}{8\pi^2}8π2λL3​

Correct answer: (D)

Step-by-step solution →
Q3·Physics·Properties of Solids and LiquidsSingle correct
A 3 m long wire of radius 3 mm shows an extension of 0.1 mm when loaded vertically by a mass of 50 kg in an experiment to determine Young’s modulus. The value of Young’s modulus of the wire as per this experiment is P×1010 Nm−2P\times10^{10}\ Nm^{-2}P×1010 Nm−2, where the value of PPP is: (Take g=3π m/s2g=3\pi\ m/s^2g=3π m/s2)
  1. (A)555
  2. (B)101010
  3. (C)252525
  4. (D)2.52.52.5

Correct answer: (A)

Step-by-step solution →
Q4·Physics·Electric Field and Coulomb's LawSingle correct
Electric charge is transferred to an irregular metallic disk as shown in figure. If σ1\sigma_1σ1​, σ2\sigma_2σ2​, σ3\sigma_3σ3​ and σ4\sigma_4σ4​ are charge densities at given points then, choose the correct answer from the options given below: (A) σ1>σ3\sigma_1>\sigma_3σ1​>σ3​ ; σ2=σ4\sigma_2=\sigma_4σ2​=σ4​ (B) σ1>σ2\sigma_1>\sigma_2σ1​>σ2​ ; σ3>σ4\sigma_3>\sigma_4σ3​>σ4​ (C) σ1>σ3>σ2=σ4\sigma_1>\sigma_3>\sigma_2=\sigma_4σ1​>σ3​>σ2​=σ4​ (D) σ1<σ3<σ2=σ4\sigma_1<\sigma_3<\sigma_2=\sigma_4σ1​<σ3​<σ2​=σ4​ (E) σ1=σ2=σ3=σ4\sigma_1=\sigma_2=\sigma_3=\sigma_4σ1​=σ2​=σ3​=σ4​
  1. (A)A, B and C Only
  2. (B)A and C Only
  3. (C)D and E Only
  4. (D)B and C Only

Correct answer: (A)

Step-by-step solution →
Q5·Physics·ThermodynamicsSingle correct
Water falls from a height of 200 m into a pool. Calculate the rise in temperature of the water assuming no heat dissipation from the water in the pool. (Take g=10 m/s2g=10\ m/s^2g=10 m/s2, specific heat of water =4200 J/(kg⋅K)=4200\ J/(kg\cdot K)=4200 J/(kg⋅K))
  1. (A)0.23 K0.23\ K0.23 K
  2. (B)0.36 K0.36\ K0.36 K
  3. (C)0.14 K0.14\ K0.14 K
  4. (D)0.48 K0.48\ K0.48 K

Correct answer: (D)

Step-by-step solution →
Q6·Physics·Geometrical OpticsSingle correct
A concave-convex lens of refractive index 1.5 and the radii of curvature of its surfaces are 30 cm and 20 cm, respectively. The concave surface is upwards and is filled with a liquid of refractive index 1.3. The focal length of the liquid-glass combination will be:
  1. (A)50011\dfrac{500}{11}11500​ cm
  2. (B)80011\dfrac{800}{11}11800​ cm
  3. (C)70011\dfrac{700}{11}11700​ cm
  4. (D)60011\dfrac{600}{11}11600​ cm

Correct answer: (D)

Step-by-step solution →
Q7·Physics·Electric Field and Coulomb's LawSingle correct
An infinitely long wire has uniform linear charge density λ=2 nC/m\lambda=2\ nC/mλ=2 nC/m. The net flux through a Gaussian cube of side length 3\sqrt33​ cm, if the wire passes through any two corners of the cube, that are maximally displaced from each other, would be x Nm2C−1x\ Nm^2C^{-1}x Nm2C−1, where xxx is: [Neglect any edge effects and use 14πε0=9×109\dfrac{1}{4\pi\varepsilon_0}=9\times10^94πε0​1​=9×109 SI units]
  1. (A)0.72π0.72\pi0.72π
  2. (B)1.44π1.44\pi1.44π
  3. (C)6.48π6.48\pi6.48π
  4. (D)2.16π2.16\pi2.16π

Correct answer: (D)

Step-by-step solution →
Q8·Physics·Electronic DevicesSingle correct
The output voltage in the following circuit is (Consider ideal diode case):
  1. (A)101010 V
  2. (B)000 V
  3. (C)+5+5+5 V
  4. (D)−5-5−5 V

Correct answer: (B)

Step-by-step solution →
Q9·Physics·Electric Field and Coulomb's LawSingle correct
Two metal spheres of radius RRR and 3R3R3R have same surface charge density σ\sigmaσ. If they are brought in contact and then separated, the surface charge density on smaller and bigger sphere becomes σ1\sigma_1σ1​ and σ2\sigma_2σ2​, respectively. The ratio σ1σ2\dfrac{\sigma_1}{\sigma_2}σ2​σ1​​ is:
  1. (A)19\dfrac{1}{9}91​
  2. (B)999
  3. (C)13\dfrac{1}{3}31​
  4. (D)333

Correct answer: (D)

Step-by-step solution →
Q10·Physics·Units and MeasurementsSingle correct
A quantity QQQ is formulated as X−2Y3/2Z−2/5X^{-2}Y^{3/2}Z^{-2/5}X−2Y3/2Z−2/5. XXX, YYY and ZZZ are independent parameters which have fractional errors of 0.1, 0.2 and 0.5, respectively in measurement. The maximum fractional error of QQQ is:
  1. (A)0.10.10.1
  2. (B)0.80.80.8
  3. (C)0.70.70.7
  4. (D)0.60.60.6

Correct answer: (C)

Step-by-step solution →
Q11·Physics·ThermodynamicsSingle correct
A monoatomic gas having γ=53\gamma=\dfrac{5}{3}γ=35​ is stored in a thermally insulated container and the gas is suddenly compressed to (18)th\left(\dfrac{1}{8}\right)^{th}(81​)th of its initial volume. The ratio of final pressure and initial pressure is: (γ\gammaγ is the ratio of specific heats of the gas at constant pressure and at constant volume)
  1. (A)161616
  2. (B)404040
  3. (C)323232
  4. (D)282828

Correct answer: (C)

Step-by-step solution →
Q12·Physics·Geometrical OpticsSingle correct
A convex lens of focal length 30 cm is placed in contact with a concave lens of focal length 20 cm. An object is placed at 20 cm to the left of this lens system. The distance of the image from the lens in cm is ___
  1. (A)303030
  2. (B)454545
  3. (C)607\dfrac{60}{7}760​
  4. (D)151515

Correct answer: (D)

Step-by-step solution →
Q13·Physics·WavesSingle correct
Two strings with circular cross section and made of same material, are stretched to have same amount of tension. A transverse wave is then made to pass through both the strings. The velocity of the wave in the first string having the radius of cross section RRR is v1v_1v1​ and that in the other string having radius of cross section R/2R/2R/2 is v2v_2v2​. Then v2v1=\dfrac{v_2}{v_1}=v1​v2​​=
  1. (A)2\sqrt22​
  2. (B)222
  3. (C)888
  4. (D)444

Correct answer: (B)

Step-by-step solution →
Q14·Physics·Magnetic Field of CurrentSingle correct
Figure shows a current carrying square loop ABCD of edge length is 'aaa' lying in a plane. If the resistance of the ABC part is rrr and that of ADC part is 2r2r2r, then the magnitude of the resultant magnetic field at centre of the square loop is:
  1. (A)3πμ0I2 a\dfrac{3\pi\mu_0 I}{\sqrt2\,a}2​a3πμ0​I​
  2. (B)μ0I2πa\dfrac{\mu_0 I}{2\pi a}2πaμ0​I​
  3. (C)2 μ0I3πa\dfrac{\sqrt2\,\mu_0 I}{3\pi a}3πa2​μ0​I​
  4. (D)2μ0I3πa\dfrac{2\mu_0 I}{3\pi a}3πa2μ0​I​

Correct answer: (C)

Step-by-step solution →
Q15·Physics·Laws of MotionSingle correct
A body of mass 2 kg moving with velocity v⃗in=3i^+4j^ ms−1\vec v_{in}=3\hat i+4\hat j\ ms^{-1}vin​=3i^+4j^​ ms−1 enters into a constant force field of 6 N directed along positive z-axis. If the body remains in the field for a period of 53\dfrac{5}{3}35​ seconds, then velocity of the body when it emerges from force field is:
  1. (A)4i^+3j^+5k^4\hat i+3\hat j+5\hat k4i^+3j^​+5k^
  2. (B)3i^+4j^+5k^3\hat i+4\hat j+5\hat k3i^+4j^​+5k^
  3. (C)3i^+4j^−5k^3\hat i+4\hat j-5\hat k3i^+4j^​−5k^
  4. (D)3i^+4j^+5 k^3\hat i+4\hat j+\sqrt5\,\hat k3i^+4j^​+5​k^

Correct answer: (B)

Step-by-step solution →
Q16·Physics·KinematicsSingle correct
Two balls with same mass and initial velocity, are projected at different angles in such a way that maximum height reached by first ball is 8 times higher than that of the second ball. T1T_1T1​ and T2T_2T2​ are the total flying times of first and second ball, respectively, then the ratio of T1T_1T1​ and T2T_2T2​ is:
  1. (A)22:12\sqrt2:122​:1
  2. (B)2:12:12:1
  3. (C)2:1\sqrt2:12​:1
  4. (D)4:14:14:1

Correct answer: (A)

Step-by-step solution →
Q17·Physics·WavesSingle correct
The amplitude and phase of a wave that is formed by the superposition of two harmonic travelling waves, y1(x,t)=4sin⁡(kx−ωt)y_1(x,t)=4\sin(kx-\omega t)y1​(x,t)=4sin(kx−ωt) and y2(x,t)=2sin⁡(kx−ωt+2π3)y_2(x,t)=2\sin\left(kx-\omega t+\dfrac{2\pi}{3}\right)y2​(x,t)=2sin(kx−ωt+32π​), are: (Take the angular frequency of initial waves same as ω\omegaω)
  1. (A)[6,2π3]\left[6,\dfrac{2\pi}{3}\right][6,32π​]
  2. (B)[6,π3]\left[6,\dfrac{\pi}{3}\right][6,3π​]
  3. (C)[3,π6]\left[\sqrt3,\dfrac{\pi}{6}\right][3​,6π​]
  4. (D)[23,π6]\left[2\sqrt3,\dfrac{\pi}{6}\right][23​,6π​]

Correct answer: (D)

Step-by-step solution →
Q18·Physics·Wave OpticsSingle correct
In a Young’s double slit experiment, the source is white light. One of the slits is covered by red filter and another by a green filter. In this case (1) There shall be an interference pattern for red distinct from that for green. (2) There shall be no interference fringes. (3) There shall be alternate interference fringes of red and greens. (4) There shall be an interference pattern, where each fringe’s pattern center is green and outer edges is red.
  1. (A)There shall be an interference pattern for red distinct from that for green.
  2. (B)There shall be no interference fringes.
  3. (C)There shall be alternate interference fringes of red and greens.
  4. (D)There shall be an interference pattern, where each fringe’s pattern center is green and outer edges is red.

Correct answer: (B)

Step-by-step solution →
Q19·Physics·Atoms and NucleiSingle correct
For a nucleus of mass number AAA and radius RRR, the mass density of nucleus can be represented as:
  1. (A)A3A^{3}A3
  2. (B)A1/3A^{1/3}A1/3
  3. (C)A2/3A^{2/3}A2/3
  4. (D)Independent of AAA

Correct answer: (D)

Step-by-step solution →
Q20·Physics·OscillationsSingle correct
A block of mass 2 kg is attached to one end of a massless spring whose other end is fixed at a wall. The spring-mass system moves on a frictionless horizontal table. The spring’s natural length is 2 m and spring constant is 200 N/m. The block is pushed such that the length of the spring becomes 1 m and then released. At distance xxx m (x<2x<2x<2) from the wall, the speed of the block will be:
  1. (A)10[1−(2−x)]3/2 m/s10[1-(2-x)]^{3/2}\ m/s10[1−(2−x)]3/2 m/s
  2. (B)10[1−(2−x)2]1/2 m/s10[1-(2-x)^2]^{1/2}\ m/s10[1−(2−x)2]1/2 m/s
  3. (C)10[1−(2−x)2] m/s10[1-(2-x)^2]\ m/s10[1−(2−x)2] m/s
  4. (D)10[1−(2−x)2]2 m/s10[1-(2-x)^2]^{2}\ m/s10[1−(2−x)2]2 m/s

Correct answer: (B)

Step-by-step solution →
Q21·Physics·Dual Nature of Matter and RadiationInteger
An electron is released from rest near an infinite non-conducting sheet of uniform charge density '−σ-\sigma−σ'. The rate of change of de-Broglie wave length associated with the electron varies inversely as nthn^{th}nth power of time. The numerical value of nnn is ___

Correct answer: 2

Step-by-step solution →
Q22·Physics·Properties of Solids and LiquidsInteger
A sample of a liquid is kept at 1 atm. It is compressed to 5 atm which leads to change of volume of 0.8 cm30.8\ cm^30.8 cm3. If the bulk modulus of the liquid is 2 GPa, the initial volume of the liquid was ___ litre. (Take 1 atm =105=10^5=105 Pa)

Correct answer: 4

Step-by-step solution →
Q23·Physics·Capacitors and DielectricsInteger
Space between the plates of a parallel plate capacitor of plate area 4 cm24\ cm^24 cm2 and separation of (d) 1.77 mm, is filled with uniform dielectric materials with dielectric constants (3 and 5) as shown in figure. Another capacitor of capacitance 7.5 pF is connected in parallel with it. The effective capacitance of this combination is ___ pF. (Given ε0=8.85×10−12\varepsilon_0=8.85\times10^{-12}ε0​=8.85×10−12 F/m)

Correct answer: 15

Step-by-step solution →
Q24·Physics·Rotational MotionInteger
A thin solid disk of 1 kg is rotating along its diameter axis at the speed of 1800 rpm. By applying an external torque of 25π25\pi25π Nm for 40 s, the speed increases to 2100 rpm. The diameter of the disk is ___ m.

Correct answer: 40

Step-by-step solution →
Q25·Physics·Properties of Solids and LiquidsInteger
A cube having a side of 10 cm with unknown mass and 200 gm mass were hung at two ends of an uniform rigid rod of 27 cm long. The rod along with masses was placed on a wedge keeping the distance between wedge point and 200 gm weight as 25 cm. Initially the masses were not at balance. A beaker is placed beneath the unknown mass and water is added slowly to it. At given point the masses were in balance and half volume of the unknown mass was inside the water. (Take the density of unknown mass is more than that of the water, the mass did not absorb water and water density is 1 gm/cm31\ gm/cm^31 gm/cm3.) The unknown mass is ___ kg.

Correct answer: 3

Step-by-step solution →

Chemistry — JEE Main 8 April 2025 Shift 2

Q26·Chemistry·Chemical KineticsSingle correct
In a first order decomposition reaction, the time taken for the decomposition of reactant to one fourth and one eighth of its initial concentration are t1t_1t1​ and t2t_2t2​ (s), respectively. The ratio t1t2\dfrac{t_1}{t_2}t2​t1​​ will:
  1. (A)43\dfrac{4}{3}34​
  2. (B)32\dfrac{3}{2}23​
  3. (C)34\dfrac{3}{4}43​
  4. (D)23\dfrac{2}{3}32​

Correct answer: (D)

Step-by-step solution →
Q27·Chemistry·Some Basic Principles of Organic ChemistrySingle correct
Match the LIST-I with LIST-II. Choose the correct answer from the options given below:
LIST-ILIST-II
A.CarbocationI.Species that can supply a pair of electrons.
B.C-Free radicalII.Species that can receive a pair of electrons.
C.NucleophileIII.sp2sp^2sp2 hybridized carbon with empty p-orbital.
D.ElectrophileIV.sp2/sp3sp^2/sp^3sp2/sp3 hybridized carbon with one unpaired electron.
  1. (A)A-IV, B-II, C-III, D-I
  2. (B)A-II, B-III, C-I, D-IV
  3. (C)A-III, B-IV, C-II, D-I
  4. (D)A-III, B-IV, C-I, D-II

Correct answer: (D)

Step-by-step solution →
Q28·Chemistry·Carboxylic Acids and DerivativesSingle correct
A→(i) NaOH (ii) H3O+B→(i) EtOH (ii) H2SO4, ΔCA\xrightarrow{\text{(i) NaOH (ii) }H_3O^+}B\xrightarrow{\text{(i) EtOH (ii) }H_2SO_4,\,\Delta}CA(i) NaOH (ii) H3​O+​B(i) EtOH (ii) H2​SO4​,Δ​C. 'A' shows positive Lassaigne's test for N and its molar mass is 121. 'B' gives effervescence with aq. NaHCO3NaHCO_3NaHCO3​. 'C' gives fruity smell. Identify A, B and C from the following:
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (A)

Step-by-step solution →
Q29·Chemistry·HydrocarbonsSingle correct
Choose the correct set of reagents for the following conversion: Ethyl benzene →\rightarrow→ (product shown in figure)
  1. (A)Br2/FeBr_2/FeBr2​/Fe; Cl2,ΔCl_2,\DeltaCl2​,Δ; alc. KOH
  2. (B)Cl2/FeCl_2/FeCl2​/Fe; Br2/Br_2/Br2​/anhy.AlCl3AlCl_3AlCl3​; aq. KOH
  3. (C)Br2/Br_2/Br2​/anhy.AlCl3AlCl_3AlCl3​; Cl2,ΔCl_2,\DeltaCl2​,Δ; aq. KOH
  4. (D)Cl2/Cl_2/Cl2​/anhy.AlCl3AlCl_3AlCl3​; Br2/FeBr_2/FeBr2​/Fe; alc. KOH

Correct answer: (A)

Step-by-step solution →
Q30·Chemistry·HydrocarbonsSingle correct
1,2-dibromocyclooctane →(i) KOH (alc.) (ii) NaNH2 (iii) Hg2+/H+ (iv) Zn-Hg/H+P\xrightarrow{\text{(i) KOH (alc.) (ii) }NaNH_2\text{ (iii) }Hg^{2+}/H^+\text{ (iv) Zn-Hg}/H^+}P(i) KOH (alc.) (ii) NaNH2​ (iii) Hg2+/H+ (iv) Zn-Hg/H+​P (Major product). 'P' is:
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (B)

Step-by-step solution →
Q31·Chemistry·Coordination CompoundsSingle correct
Given below are two statements: Statement I: A homoleptic octahedral complex, formed using monodentate ligands, will not show stereoisomerism. Statement II: cis- and trans- platin are heteroleptic complexes of Pd. 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)Statement I is false but Statement II is true.
  3. (C)Both statement I and Statement II are false.
  4. (D)Statement I is true but Statement II is false.

Correct answer: (D)

Step-by-step solution →
Q32·Chemistry·Classification of Elements and Periodicity in PropertiesSingle correct
The atomic number of the element from the following with lowest 1st1^{st}1st ionisation enthalpy is:
  1. (A)323232
  2. (B)353535
  3. (C)878787
  4. (D)191919

Correct answer: (C)

Step-by-step solution →
Q33·Chemistry·SolutionsSingle correct
Which of the following binary mixture does not show the behaviour of minimum boiling azeotropes?
  1. (A)H2O+CH3COC2H5H_2O+CH_3COC_2H_5H2​O+CH3​COC2​H5​
  2. (B)C2H5OH+C2H5NH2C_2H_5OH+C_2H_5NH_2C2​H5​OH+C2​H5​NH2​
  3. (C)CS2+CH3COCH3CS_2+CH_3COCH_3CS2​+CH3​COCH3​
  4. (D)CH3OH+CHCl3CH_3OH+CHCl_3CH3​OH+CHCl3​

Correct answer: (B)

Step-by-step solution →
Q34·Chemistry·SolutionsSingle correct
HA(aq)⇌H+(aq)+A−(aq)HA(aq)\rightleftharpoons H^+(aq)+A^-(aq)HA(aq)⇌H+(aq)+A−(aq). The freezing point depression of a 0.1 m aqueous solution of a monobasic weak acid HA is 0.20∘C0.20^\circ C0.20∘C. The dissociation constant for the acid is: Given: Kf(H2O)=1.8 K kg mol−1K_f(H_2O)=1.8\ K\,kg\,mol^{-1}Kf​(H2​O)=1.8 Kkgmol−1, molality === molarity.
  1. (A)1.38×10−31.38\times10^{-3}1.38×10−3
  2. (B)1.1×10−21.1\times10^{-2}1.1×10−2
  3. (C)1.90×10−31.90\times10^{-3}1.90×10−3
  4. (D)1.89×10−11.89\times10^{-1}1.89×10−1

Correct answer: (A)

Step-by-step solution →
Q35·Chemistry·IUPAC NomenclatureSingle correct
What is the correct IUPAC name of the following compound?
  1. (A)4-Ethyl-1-hydroxycyclopent-2-ene
  2. (B)1-Ethyl-3-hydroxycyclopent-2-ene
  3. (C)1-Ethylcyclopent-2-en-3-ol
  4. (D)4-Ethylcyclopent-2-en-1-ol

Correct answer: (D)

Step-by-step solution →
Q36·Chemistry·d- and f-Block ElementsSingle correct
The correct decreasing order of spin only magnetic moment (BM) of Cu+Cu^+Cu+, Cu2+Cu^{2+}Cu2+, Cr2+Cr^{2+}Cr2+ and Cr3+Cr^{3+}Cr3+ ions is:
  1. (A)Cu+>Cu2+>Cr2+>Cr3+Cu^+>Cu^{2+}>Cr^{2+}>Cr^{3+}Cu+>Cu2+>Cr2+>Cr3+
  2. (B)Cu2+>Cu+>Cr2+>Cr3+Cu^{2+}>Cu^+>Cr^{2+}>Cr^{3+}Cu2+>Cu+>Cr2+>Cr3+
  3. (C)Cr2+>Cr3+>Cu2+>Cu+Cr^{2+}>Cr^{3+}>Cu^{2+}>Cu^+Cr2+>Cr3+>Cu2+>Cu+
  4. (D)Cr3+>Cr2+>Cu2+>Cu+Cr^{3+}>Cr^{2+}>Cu^{2+}>Cu^+Cr3+>Cr2+>Cu2+>Cu+

Correct answer: (C)

Step-by-step solution →
Q37·Chemistry·Alcohols and EthersSingle correct
Which one of the following reactions will not lead to the desired ether formation in major proportion? (iso-Bu ⇒\Rightarrow⇒ isobutyl, sec-Bu ⇒\Rightarrow⇒ sec-butyl, nPr ⇒\Rightarrow⇒ n-propyl, tBu⇒^tBu\RightarrowtBu⇒ tert-butyl, Et ⇒\Rightarrow⇒ ethyl)
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (D)

Step-by-step solution →
Q38·Chemistry·Purification and Characterisation of Organic CompoundsSingle correct
On combustion 0.210 g of an organic compound containing C, H and O gave 0.127 g H2OH_2OH2​O and 0.307 g CO2CO_2CO2​. The percentages of hydrogen and oxygen in the given organic compound respectively are:
  1. (A)53.41, 39.653.41,\ 39.653.41, 39.6
  2. (B)6.72, 53.416.72,\ 53.416.72, 53.41
  3. (C)7.55, 43.857.55,\ 43.857.55, 43.85
  4. (D)6.72, 39.876.72,\ 39.876.72, 39.87

Correct answer: (B)

Step-by-step solution →
Q39·Chemistry·BiomoleculesSingle correct
For the amino acid shown in the figure, AAA and BBB represent its structures at pH=2pH=2pH=2 and pH=10pH=10pH=10, respectively. Choose the correct option for structures of A and B respectively:
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (A)

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Q40·Chemistry·Atomic StructureSingle correct
Correct statements for an element with atomic number 9 are: A. There can be 5 electrons for which ms=+12m_s=+\dfrac{1}{2}ms​=+21​ and 4 electrons for which ms=−12m_s=-\dfrac{1}{2}ms​=−21​. B. There is only one electron in pxp_xpx​ orbital. C. The last electron goes to orbital with n=2n=2n=2 and l=1l=1l=1. D. The sum of angular nodes of all the atomic orbitals is 1. Choose the correct answer from the options given below:
  1. (A)C and D Only
  2. (B)A and C Only
  3. (C)A, C and D Only
  4. (D)A and B Only

Correct answer: (B)

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Q41·Chemistry·Coordination CompoundsSingle correct
The number of species from the following that are involved in sp3d2sp^3d^2sp3d2 hybridization: [Co(NH3)6]3+[Co(NH_3)_6]^{3+}[Co(NH3​)6​]3+, SF6SF_6SF6​, [CrF6]3−[CrF_6]^{3-}[CrF6​]3−, [CoF6]3−[CoF_6]^{3-}[CoF6​]3−, [Mn(CN)6]3−[Mn(CN)_6]^{3-}[Mn(CN)6​]3− and [MnCl6]3−[MnCl_6]^{3-}[MnCl6​]3− are:
  1. (A)555
  2. (B)666
  3. (C)444
  4. (D)333

Correct answer: (D)

Step-by-step solution →
Q42·Chemistry·Purification and Characterisation of Organic CompoundsSingle correct
Match the LIST-I (Reagent) with LIST-II (Functional Group detected). Choose the correct answer from the options given below:
LIST-I (Reagent)LIST-II (Functional Group detected)
A.Sodium bicarbonate solutionI.double bond/unsaturation
B.Neutral ferric chlorideII.carboxylic acid
C.ceric ammonium nitrateIII.phenolic - OH
D.alkaline KMnO4KMnO_4KMnO4​IV.alcoholic - OH
  1. (A)A-II, B-III, C-IV, D-I
  2. (B)A-II, B-III, C-I, D-IV
  3. (C)A-III, B-II, C-IV, D-I
  4. (D)A-II, B-IV, C-III, D-I

Correct answer: (A)

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Q43·Chemistry·Aldehydes and KetonesSingle correct
When the compound shown in the figure undergoes intramolecular aldol condensation, the major product formed is:
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (A)

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Q44·Chemistry·Coordination CompoundsSingle correct
Match the LIST-I (Complex/Species) with LIST-II (Shape & magnetic moment). Choose the correct answer from the options given below:
LIST-I (Complex/Species)LIST-II (Shape & magnetic moment)
A.[Ni(CO)4][Ni(CO)_4][Ni(CO)4​]I.Tetrahedral, 2.8 BM
B.[Ni(CN)4]2−[Ni(CN)_4]^{2-}[Ni(CN)4​]2−II.Square planar, 0 BM
C.[NiCl4]2−[NiCl_4]^{2-}[NiCl4​]2−III.Tetrahedral, 0 BM
D.[MnBr4]2−[MnBr_4]^{2-}[MnBr4​]2−IV.Tetrahedral, 5.9 BM
  1. (A)A-III, B-IV, C-II, D-I
  2. (B)A-I, B-II, C-III, D-IV
  3. (C)A-III, B-II, C-I, D-IV
  4. (D)A-IV, B-I, C-III, D-II

Correct answer: (C)

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Q45·Chemistry·p-Block ElementsSingle correct
Given below are two statements: Statement I: H2SeH_2SeH2​Se is more acidic than H2TeH_2TeH2​Te. Statement II: H2SeH_2SeH2​Se has higher bond enthalpy for dissociation than H2TeH_2TeH2​Te. 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)Both statement I and Statement II are true.
  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 →
Q46·Chemistry·Chemical Bonding and Molecular StructureInteger
Resonance in X2YX_2YX2​Y can be represented as shown in the figure. The enthalpy of formation of X2YX_2YX2​Y (X≡X(g)+12Y=Y(g)→X2Y(g))\left(X\equiv X(g)+\dfrac{1}{2}Y=Y(g)\to X_2Y(g)\right)(X≡X(g)+21​Y=Y(g)→X2​Y(g)) is 80 kJ mol−1^{-1}−1. The magnitude of resonance energy of X2YX_2YX2​Y is ___ kJ mol−1^{-1}−1 (nearest integer value). Given: Bond energies of X≡XX\equiv XX≡X, X=XX=XX=X, Y=YY=YY=Y and X=YX=YX=Y are 940, 410, 500 and 602 kJ mol−1^{-1}−1 respectively. valence X : 3, Y : 2

Correct answer: 98

Step-by-step solution →
Q47·Chemistry·Atomic StructureInteger
The energy of an electron in first Bohr orbit of H-atom is −13.6-13.6−13.6 eV. The magnitude of energy value of electron in the first excited state of Be3+Be^{3+}Be3+ is ___ eV. (nearest integer value)

Correct answer: 54

Step-by-step solution →
Q48·Chemistry·Some Basic Concepts in ChemistryInteger
20 mL of sodium iodide solution gave 4.74 g silver iodide when treated with excess of silver nitrate solution. The molarity of the sodium iodide solution is ___ M. (Nearest Integer value). (Given: Na = 23, I = 127, Ag = 108, N = 14, O = 16 g mol−1^{-1}−1)

Correct answer: 1

Step-by-step solution →
Q49·Chemistry·EquilibriumInteger
The equilibrium constant for decomposition of H2O(g)H_2O(g)H2​O(g) (H2O(g)⇌H2(g)+12O2(g), ΔG∘=92.34 kJ mol−1)\left(H_2O(g)\rightleftharpoons H_2(g)+\dfrac{1}{2}O_2(g),\ \Delta G^\circ=92.34\ kJ\,mol^{-1}\right)(H2​O(g)⇌H2​(g)+21​O2​(g), ΔG∘=92.34 kJmol−1) is 8.0×10−38.0\times10^{-3}8.0×10−3 at 2300 K and total pressure at equilibrium is 1 bar. Under this condition, the degree of dissociation (α\alphaα) of water is ___ ×10−2\times10^{-2}×10−2 (nearest integer value). [Assume α\alphaα is negligible with respect to 1]

Correct answer: 5

Step-by-step solution →
Q50·Chemistry·Redox Reactions and ElectrochemistryInteger
Consider the following half cell reaction: Cr2O72−(aq)+6e−+14H+(aq)→2Cr3+(aq)+7H2O(l)Cr_2O_7^{2-}(aq)+6e^-+14H^+(aq)\to2Cr^{3+}(aq)+7H_2O(l)Cr2​O72−​(aq)+6e−+14H+(aq)→2Cr3+(aq)+7H2​O(l). The reaction was conducted with the ratio of [Cr3+]2[Cr2O72−]=10−4\dfrac{[Cr^{3+}]^2}{[Cr_2O_7^{2-}]}=10^{-4}[Cr2​O72−​][Cr3+]2​=10−4. The pH value at which the EMF of the half cell will become zero is ___ (nearest integer value). [Given: standard half cell reduction potential ECr2O72−/Cr3+∘=1.33E^\circ_{Cr_2O_7^{2-}/Cr^{3+}}=1.33ECr2​O72−​/Cr3+∘​=1.33 V, 2.303RTF=0.059\dfrac{2.303RT}{F}=0.059F2.303RT​=0.059 V]

Correct answer: 10

Step-by-step solution →

Mathematics — JEE Main 8 April 2025 Shift 2

Q51·Mathematics·Three Dimensional GeometrySingle correct
Let the values of λ\lambdaλ for which the shortest distance between the lines x−12=y−23=z−34\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}2x−1​=3y−2​=4z−3​ and x−λ3=y−44=z−55\dfrac{x-\lambda}{3}=\dfrac{y-4}{4}=\dfrac{z-5}{5}3x−λ​=4y−4​=5z−5​ is 16\dfrac{1}{\sqrt6}6​1​ be λ1\lambda_1λ1​ and λ2\lambda_2λ2​. Then the radius of the circle passing through the points (0,0)(0,0)(0,0), (λ1,λ2)(\lambda_1,\lambda_2)(λ1​,λ2​) and (λ2,λ1)(\lambda_2,\lambda_1)(λ2​,λ1​) is:
  1. (A)523\dfrac{5\sqrt2}{3}352​​
  2. (B)444
  3. (C)23\dfrac{\sqrt2}{3}32​​
  4. (D)333

Correct answer: (A)

Step-by-step solution →
Q52·Mathematics·Matrices and DeterminantsSingle correct
Let α\alphaα be a solution of x2+x+1=0x^2+x+1=0x2+x+1=0, and for some aaa and bbb in R\mathbb{R}R, [4ab][11613−1−12−2−14−8]=[000]\begin{bmatrix}4 & a & b\end{bmatrix}\begin{bmatrix}1 & 16 & 13\\ -1 & -1 & 2\\ -2 & -14 & -8\end{bmatrix}=\begin{bmatrix}0 & 0 & 0\end{bmatrix}[4​a​b​]​1−1−2​16−1−14​132−8​​=[0​0​0​]. If 4α4+mαa+nαb=3\dfrac{4}{\alpha^4}+\dfrac{m}{\alpha^a}+\dfrac{n}{\alpha^b}=3α44​+αam​+αbn​=3, then m+nm+nm+n is equal to ___
  1. (A)333
  2. (B)111111
  3. (C)777
  4. (D)888

Correct answer: (B)

Step-by-step solution →
Q53·Mathematics·Application of DerivativesSingle correct
Let the function f(x)=x3+3x+3, x≠0f(x)=\dfrac{x}{3}+\dfrac{3}{x}+3,\ x\ne0f(x)=3x​+x3​+3, x=0 be strictly increasing in (−∞,α1)∪(α2,∞)(-\infty,\alpha_1)\cup(\alpha_2,\infty)(−∞,α1​)∪(α2​,∞) and strictly decreasing in (α3,α4)∪(α4,α5)(\alpha_3,\alpha_4)\cup(\alpha_4,\alpha_5)(α3​,α4​)∪(α4​,α5​). Then ∑i=15αi2\sum_{i=1}^{5}\alpha_i^2∑i=15​αi2​ is equal to:
  1. (A)484848
  2. (B)282828
  3. (C)404040
  4. (D)363636

Correct answer: (D)

Step-by-step solution →
Q54·Mathematics·Statistics and ProbabilitySingle correct
If AAA and BBB are two events such that P(A)=0.7P(A)=0.7P(A)=0.7, P(B)=0.4P(B)=0.4P(B)=0.4 and P(A∩Bˉ)=0.5P(A\cap\bar B)=0.5P(A∩Bˉ)=0.5, where Bˉ\bar BBˉ denotes the complement of BBB, then P(B ∣ (A∪Bˉ))P\big(B\,|\,(A\cup\bar B)\big)P(B∣(A∪Bˉ)) is equal to:
  1. (A)14\dfrac{1}{4}41​
  2. (B)12\dfrac{1}{2}21​
  3. (C)16\dfrac{1}{6}61​
  4. (D)13\dfrac{1}{3}31​

Correct answer: (A)

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Q55·Mathematics·Sequence and SeriesSingle correct
If 114+124+134+…∞=π490\dfrac{1}{1^4}+\dfrac{1}{2^4}+\dfrac{1}{3^4}+\ldots\infty=\dfrac{\pi^4}{90}141​+241​+341​+…∞=90π4​, 114+134+154+…∞=α\dfrac{1}{1^4}+\dfrac{1}{3^4}+\dfrac{1}{5^4}+\ldots\infty=\alpha141​+341​+541​+…∞=α, 124+144+164+…∞=β\dfrac{1}{2^4}+\dfrac{1}{4^4}+\dfrac{1}{6^4}+\ldots\infty=\beta241​+441​+641​+…∞=β, then αβ\dfrac{\alpha}{\beta}βα​ is equal to:
  1. (A)232323
  2. (B)181818
  3. (C)151515
  4. (D)141414

Correct answer: (C)

Step-by-step solution →
Q56·Mathematics·Quadratic EquationsSingle correct
The sum of the squares of the roots of ∣x+2∣2+∣x−2∣−2=0|x+2|^2+|x-2|-2=0∣x+2∣2+∣x−2∣−2=0 and the squares of the roots of x2−2∣x−3∣−5=0x^2-2|x-3|-5=0x2−2∣x−3∣−5=0, is:
  1. (A)262626
  2. (B)363636
  3. (C)303030
  4. (D)242424

Correct answer: (B)

Step-by-step solution →
Q57·Mathematics·Straight LinesSingle correct
Let aaa be the length of a side of a square OABCOABCOABC with OOO being the origin. Its side OAOAOA makes an acute angle α\alphaα with the positive x-axis and the equations of its diagonals are (3+1)x+(3−1)y=0(\sqrt3+1)x+(\sqrt3-1)y=0(3​+1)x+(3​−1)y=0 and (3−1)x−(3+1)y+83=0(\sqrt3-1)x-(\sqrt3+1)y+8\sqrt3=0(3​−1)x−(3​+1)y+83​=0. Then a2a^2a2 is equal to:
  1. (A)484848
  2. (B)323232
  3. (C)161616
  4. (D)242424

Correct answer: (A)

Step-by-step solution →
Q58·Mathematics·Definite IntegrationSingle correct
Let f(x)f(x)f(x) be a positive function and I1=∫−1/212x f(2x(1−2x)) dxI_1=\displaystyle\int_{-1/2}^{1}2x\,f\big(2x(1-2x)\big)\,dxI1​=∫−1/21​2xf(2x(1−2x))dx and I2=∫−12f(x(1−x)) dxI_2=\displaystyle\int_{-1}^{2}f\big(x(1-x)\big)\,dxI2​=∫−12​f(x(1−x))dx. Then the value of I2I1\dfrac{I_2}{I_1}I1​I2​​ is equal to ___
  1. (A)999
  2. (B)666
  3. (C)121212
  4. (D)444

Correct answer: (D)

Step-by-step solution →
Q59·Mathematics·Vector AlgebraSingle correct
Let a⃗=i^+2j^+k^\vec a=\hat i+2\hat j+\hat ka=i^+2j^​+k^ and b⃗=2i^+j^−k^\vec b=2\hat i+\hat j-\hat kb=2i^+j^​−k^. Let c^\hat cc^ be a unit vector in the plane of the vectors a⃗\vec aa and b⃗\vec bb and be perpendicular to a⃗\vec aa. Then such a vector c^\hat cc^ is:
  1. (A)12(j^−2k^)\dfrac{1}{\sqrt2}(\hat j-2\hat k)2​1​(j^​−2k^)
  2. (B)13(−i^−j^−k^)\dfrac{1}{\sqrt3}(-\hat i-\hat j-\hat k)3​1​(−i^−j^​−k^)
  3. (C)13(i^−j^+k^)\dfrac{1}{\sqrt3}(\hat i-\hat j+\hat k)3​1​(i^−j^​+k^)
  4. (D)12(−i^+k^)\dfrac{1}{\sqrt2}(-\hat i+\hat k)2​1​(−i^+k^)

Correct answer: (D)

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Q60·Mathematics·CirclesSingle correct
Let the ellipse 3x2+py2=43x^2+py^2=43x2+py2=4 pass through the centre CCC of the circle x2+y2−2x−4y−11=0x^2+y^2-2x-4y-11=0x2+y2−2x−4y−11=0 of radius rrr. Let f1,f2f_1,f_2f1​,f2​ be the focal distances of the point CCC on the ellipse. Then 6f1f2−r6f_1f_2-r6f1​f2​−r is equal to:
  1. (A)747474
  2. (B)686868
  3. (C)707070
  4. (D)787878

Correct answer: (C)

Step-by-step solution →
Q61·Mathematics·Definite IntegrationSingle correct
The integral ∫−13/2( ∣π2xsin⁡(πx)∣ ) dx\displaystyle\int_{-1}^{3/2}\big(\,|\pi^2 x\sin(\pi x)|\,\big)\,dx∫−13/2​(∣π2xsin(πx)∣)dx is equal to:
  1. (A)3+2π3+2\pi3+2π
  2. (B)4+π4+\pi4+π
  3. (C)1+3π1+3\pi1+3π
  4. (D)2+3π2+3\pi2+3π

Correct answer: (C)

Step-by-step solution →
Q62·Mathematics·Straight LinesSingle correct
A line passing through the point P(a,0)P(a,0)P(a,0) makes an acute angle α\alphaα with the positive x-axis. Let this line be rotated about the point PPP through an angle α2\dfrac{\alpha}{2}2α​ in the clock-wise direction. If in the new position, the slope of the line is 2−32-\sqrt32−3​ and its distance from the origin is 12\dfrac{1}{\sqrt2}2​1​, then the value of 3a2tan⁡2α−233a^2\tan^2\alpha-2\sqrt33a2tan2α−23​ is:
  1. (A)444
  2. (B)666
  3. (C)555
  4. (D)888

Correct answer: (A)

Step-by-step solution →
Q63·Mathematics·Permutations and CombinationsSingle correct
There are 12 points in a plane, no three of which are in the same straight line, except 5 points which are collinear. Then the total number of triangles that can be formed with the vertices at any three of these 12 points is:
  1. (A)230230230
  2. (B)220220220
  3. (C)200200200
  4. (D)210210210

Correct answer: (D)

Step-by-step solution →
Q64·Mathematics·Complex NumbersSingle correct
Let A={θ∈[0,2π]:1+10 Re(2cos⁡θ+isin⁡θcos⁡θ−3isin⁡θ)=0}A=\left\{\theta\in[0,2\pi]:1+10\,\mathrm{Re}\left(\dfrac{2\cos\theta+i\sin\theta}{\cos\theta-3i\sin\theta}\right)=0\right\}A={θ∈[0,2π]:1+10Re(cosθ−3isinθ2cosθ+isinθ​)=0}. Then ∑θ∈Aθ2\sum_{\theta\in A}\theta^2∑θ∈A​θ2 is equal to:
  1. (A)214π2\dfrac{21}{4}\pi^2421​π2
  2. (B)8π28\pi^28π2
  3. (C)274π2\dfrac{27}{4}\pi^2427​π2
  4. (D)6π26\pi^26π2

Correct answer: (A)

Step-by-step solution →
Q65·Mathematics·Sets, Relations and FunctionsSingle correct
Let A={0,1,2,3,4,5}A=\{0,1,2,3,4,5\}A={0,1,2,3,4,5}. Let RRR be a relation on AAA defined by (x,y)∈R(x,y)\in R(x,y)∈R if and only if max⁡{x,y}∈{3,4}\max\{x,y\}\in\{3,4\}max{x,y}∈{3,4}. Then among the statements (S1)(S_1)(S1​): The number of elements in RRR is 18, and (S2)(S_2)(S2​): The relation RRR is symmetric but neither reflexive nor transitive:
  1. (A)both are true
  2. (B)both are false
  3. (C)only (S2)(S_2)(S2​) is true
  4. (D)only (S1)(S_1)(S1​) is true

Correct answer: (C)

Step-by-step solution →
Q66·Mathematics·Binomial Theorem and Its Simple ApplicationsSingle correct
The number of integral terms in the expansion of (51/2+71/8)1016\left(5^{1/2}+7^{1/8}\right)^{1016}(51/2+71/8)1016 is:
  1. (A)127127127
  2. (B)130130130
  3. (C)129129129
  4. (D)128128128

Correct answer: (D)

Step-by-step solution →
Q67·Mathematics·Differential EquationsSingle correct
Let f(x)=x−1f(x)=x-1f(x)=x−1 and g(x)=exg(x)=e^xg(x)=ex for x∈Rx\in\mathbb{R}x∈R. If dydx=(e−2x g(f(f(x)))−yx)\dfrac{dy}{dx}=\left(e^{-2\sqrt x}\,g\big(f(f(x))\big)-\dfrac{y}{\sqrt x}\right)dxdy​=(e−2x​g(f(f(x)))−x​y​), y(0)=0y(0)=0y(0)=0, then y(1)y(1)y(1) is:
  1. (A)1−e2e4\dfrac{1-e^2}{e^4}e41−e2​
  2. (B)2e−1e3\dfrac{2e-1}{e^3}e32e−1​
  3. (C)e−1e4\dfrac{e-1}{e^4}e4e−1​
  4. (D)1−e3e4\dfrac{1-e^3}{e^4}e41−e3​

Correct answer: (C)

Step-by-step solution →
Q68·Mathematics·Inverse Trigonometric FunctionsSingle correct
The value of cot⁡−1(1+tan⁡2(2)−1tan⁡(2))−cot⁡−1(1+tan⁡2(12)+1tan⁡(12))\cot^{-1}\left(\dfrac{\sqrt{1+\tan^2(2)}-1}{\tan(2)}\right)-\cot^{-1}\left(\dfrac{\sqrt{1+\tan^2\left(\tfrac12\right)}+1}{\tan\left(\tfrac12\right)}\right)cot−1(tan(2)1+tan2(2)​−1​)−cot−1​tan(21​)1+tan2(21​)​+1​​ is equal to:
  1. (A)π−54\pi-\dfrac{5}{4}π−45​
  2. (B)π−32\pi-\dfrac{3}{2}π−23​
  3. (C)π+32\pi+\dfrac{3}{2}π+23​
  4. (D)π+52\pi+\dfrac{5}{2}π+25​

Correct answer: (A)

Step-by-step solution →
Q69·Mathematics·Matrices and DeterminantsSingle correct
Let A=[22+p2+p+q46+2p8+3p+2q612+3p20+6p+3q]A=\begin{bmatrix}2 & 2+p & 2+p+q\\ 4 & 6+2p & 8+3p+2q\\ 6 & 12+3p & 20+6p+3q\end{bmatrix}A=​246​2+p6+2p12+3p​2+p+q8+3p+2q20+6p+3q​​. If det⁡(adj(adj(3A)))=2m⋅3n\det\big(\mathrm{adj}(\mathrm{adj}(3A))\big)=2^m\cdot3^ndet(adj(adj(3A)))=2m⋅3n, m,n∈Nm,n\in\mathbb{N}m,n∈N, then m+nm+nm+n is equal to:
  1. (A)222222
  2. (B)242424
  3. (C)262626
  4. (D)202020

Correct answer: (B)

Step-by-step solution →
Q70·Mathematics·Limits and ContinuitySingle correct
Given below are two statements: Statement I: lim⁡x→0tan⁡−1x+log⁡e1+x1−x−2xx5=25\lim\limits_{x\to0}\dfrac{\tan^{-1}x+\log_e\sqrt{\dfrac{1+x}{1-x}}-2x}{x^5}=\dfrac{2}{5}x→0lim​x5tan−1x+loge​1−x1+x​​−2x​=52​. Statement II: lim⁡x→1x21−x=1e2\lim\limits_{x\to1}x^{\frac{2}{1-x}}=\dfrac{1}{e^2}x→1lim​x1−x2​=e21​. In the light of the above statements, choose the correct answer from the options given below:
  1. (A)Statement I is false but Statement II is true
  2. (B)Statement I is true but Statement II is false
  3. (C)Both Statement I and Statement II are false
  4. (D)Both Statement I and Statement II are true

Correct answer: (D)

Step-by-step solution →
Q71·Mathematics·Area Under CurvesInteger
Let the area of the bounded region {(x,y):0≤9x≤y2, y≥3x−6}\{(x,y):0\le9x\le y^2,\ y\ge3x-6\}{(x,y):0≤9x≤y2, y≥3x−6} be AAA. Then 6A6A6A is equal to ___

Correct answer: 15

Step-by-step solution →
Q72·Mathematics·Sets, Relations and FunctionsInteger
Let the domain of the function f(x)=cos⁡−1(4x+53x−7)f(x)=\cos^{-1}\left(\dfrac{4x+5}{3x-7}\right)f(x)=cos−1(3x−74x+5​) be [α,β][\alpha,\beta][α,β] and the domain of g(x)=log⁡2(2−6log⁡27(2x+5))g(x)=\log_2\big(2-6\log_{27}(2x+5)\big)g(x)=log2​(2−6log27​(2x+5)) be (γ,δ)(\gamma,\delta)(γ,δ). Then ∣7(α+β)+4(γ+δ)∣\big|7(\alpha+\beta)+4(\gamma+\delta)\big|​7(α+β)+4(γ+δ)​ is equal to ___

Correct answer: 96

Step-by-step solution →
Q73·Mathematics·Three Dimensional GeometryInteger
Let the area of the triangle formed by the lines x+2=y−1=zx+2=y-1=zx+2=y−1=z, x−35=y−1=z−11\dfrac{x-3}{5}=\dfrac{y}{-1}=\dfrac{z-1}{1}5x−3​=−1y​=1z−1​ and x−3=y−33=z−21\dfrac{x}{-3}=\dfrac{y-3}{3}=\dfrac{z-2}{1}−3x​=3y−3​=1z−2​ be AAA. Then A2A^2A2 is equal to ___

Correct answer: 56

Step-by-step solution →
Q74·Mathematics·Binomial Theorem and Its Simple ApplicationsInteger
The product of the last two digits of (1919)1919(1919)^{1919}(1919)1919 is ___

Correct answer: 63

Step-by-step solution →
Q75·Mathematics·CirclesInteger
Let rrr be the radius of the circle, which touches x-axis at point (a,0)(a,0)(a,0), a<0a<0a<0 and the parabola y2=9xy^2=9xy2=9x at the point (4,6)(4,6)(4,6). Then rrr is equal to ___

Correct answer: 30

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
  • 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
  • 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
  • Units and Measurements 149/186
  • Hydrocarbons 126/186
  • Complex Numbers 165/186
  • Biomolecules 162/186
  • Chemical Kinetics 169/186
  • Electric Field and Coulomb's Law 133/186
  • Atomic Structure 161/186
  • Electronic Devices 145/186
  • Circles 142/186
  • Dual Nature of Matter and Radiation 155/186
  • Quadratic Equations 148/186
  • Some Basic Concepts in Chemistry 129/186
  • Wave Optics 130/186
  • Area Under Curves 139/186
  • Alcohols and Ethers 106/186
  • Purification and Characterisation of Organic Compounds 116/186
  • Oscillations 117/186
  • Nuclei 116/186
  • Straight Lines 114/186
  • Waves 109/186
  • Capacitors and Dielectrics 115/186
  • Classification of Elements and Periodicity in Properties 113/186
  • Inverse Trigonometric Functions 93/186
  • Electric Potential 63/186
  • Carboxylic Acids and Derivatives 54/186
  • IUPAC Nomenclature 37/186
  • Reaction Mechanism 29/186
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