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JEE Main 2 April 2025 Shift 1 Question Paper with Answers

2 April 2025 · April session · 75 questions

The complete JEE Main 2 April 2025 Shift 1 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 2 April 2025 Shift 1

Q1·Physics·Wave OpticsSingle correct
A light wave is propagating with plane wave fronts of the type x+y+z=x+y+z=x+y+z= constant. The angle made by the direction of wave propagation with the x-axis is:
  1. (A)cos⁡−1(13)\cos^{-1}\left(\dfrac{1}{\sqrt{3}}\right)cos−1(3​1​)
  2. (B)cos⁡−1(23)\cos^{-1}\left(\dfrac{2}{3}\right)cos−1(32​)
  3. (C)cos⁡−1(13)\cos^{-1}\left(\dfrac{1}{3}\right)cos−1(31​)
  4. (D)cos⁡−1(23)\cos^{-1}\left(\sqrt{\dfrac{2}{3}}\right)cos−1(32​​)

Correct answer: (A)

Step-by-step solution →
Q2·Physics·Units and MeasurementsSingle correct
The equation for real gas is given by (P+aV2)(V−b)=RT\left(P+\dfrac{a}{V^2}\right)(V-b)=RT(P+V2a​)(V−b)=RT, where P,V,TP,V,TP,V,T and RRR are the pressure, volume, temperature and gas constant, respectively. The dimension of ab−2ab^{-2}ab−2 is equivalent to that of:
  1. (A)Planck’s constant
  2. (B)Compressibility
  3. (C)Strain
  4. (D)Energy density

Correct answer: (D)

Step-by-step solution →
Q3·Physics·Rotational MotionSingle correct
A cord of negligible mass is wound around the rim of a wheel supported by spokes with negligible mass. The mass of wheel is 10 kg and radius is 10 cm and it can freely rotate without any friction. Initially the wheel is at rest. If a steady pull of 20 N is applied on the cord, the angular velocity of the wheel, after the cord is unwound by 1 m, would be:
  1. (A)20 rad/s
  2. (B)30 rad/s
  3. (C)10 rad/s
  4. (D)0 rad/s

Correct answer: (A)

Step-by-step solution →
Q4·Physics·Geometrical OpticsSingle correct
A slanted object AB is placed on one side of a convex lens as shown in the diagram. The image is formed on the opposite side. Angle made by the image with the principal axis is:
  1. (A)−α2-\dfrac{\alpha}{2}−2α​
  2. (B)−45∘-45^\circ−45∘
  3. (C)+45∘+45^\circ+45∘
  4. (D)−α-\alpha−α

Correct answer: (B)

Step-by-step solution →
Q5·Physics·Electric Field and Coulomb's LawSingle correct
Consider two infinitely large plane parallel conducting plates as shown below. The plates are uniformly charged with a surface charge density +σ+\sigma+σ and −2σ-2\sigma−2σ. The force experienced by a point charge +q+q+q placed at the mid point between two plates will be:
  1. (A)σq4ε0\dfrac{\sigma q}{4\varepsilon_0}4ε0​σq​
  2. (B)3σq2ε0\dfrac{3\sigma q}{2\varepsilon_0}2ε0​3σq​
  3. (C)3σq4ε0\dfrac{3\sigma q}{4\varepsilon_0}4ε0​3σq​
  4. (D)σq2ε0\dfrac{\sigma q}{2\varepsilon_0}2ε0​σq​

Correct answer: (B)

Step-by-step solution →
Q6·Physics·KinematicsSingle correct
A river is flowing from west to east direction with speed of 9 km h−1^{-1}−1. If a boat capable of moving at a maximum speed of 27 km h−1^{-1}−1 in still water, crosses the river in half a minute, while moving with maximum speed at an angle of 150∘150^\circ150∘ to direction of river flow, then the width of the river is:
  1. (A)300 m
  2. (B)112.5 m
  3. (C)75 m
  4. (D)112.5×3112.5\times\sqrt{3}112.5×3​ m

Correct answer: (B)

Step-by-step solution →
Q7·Physics·Electric Field and Coulomb's LawSingle correct
A point charge +q+q+q is placed at the origin. A second point charge +9q+9q+9q is placed at (d,0,0)(d,0,0)(d,0,0) in Cartesian coordinate system. The point in between them where the electric field vanishes is:
  1. (A)(4d3,0,0)\left(\dfrac{4d}{3},0,0\right)(34d​,0,0)
  2. (B)(d4,0,0)\left(\dfrac{d}{4},0,0\right)(4d​,0,0)
  3. (C)(3d4,0,0)\left(\dfrac{3d}{4},0,0\right)(43d​,0,0)
  4. (D)(d3,0,0)\left(\dfrac{d}{3},0,0\right)(3d​,0,0)

Correct answer: (B)

Step-by-step solution →
Q8·Physics·Current ElectricitySingle correct
The battery of a mobile phone is rated as 4.2 V, 5800 mAh. How much energy is stored in it when fully charged?
  1. (A)43.8 kJ
  2. (B)48.7 kJ
  3. (C)87.7 kJ
  4. (D)24.4 kJ

Correct answer: (C)

Step-by-step solution →
Q9·Physics·OscillationsSingle correct
A particle is subjected to two simple harmonic motions as: x1=7sin⁡5tx_1=\sqrt{7}\sin 5tx1​=7​sin5t cm and x2=27sin⁡(5t+π3)x_2=2\sqrt{7}\sin\left(5t+\dfrac{\pi}{3}\right)x2​=27​sin(5t+3π​) cm, where xxx is displacement and ttt is time in seconds. The maximum acceleration of the particle is x×10−2x\times10^{-2}x×10−2 ms−2^{-2}−2. The value of xxx is:
  1. (A)175
  2. (B)25725\sqrt{7}257​
  3. (C)575\sqrt{7}57​
  4. (D)125

Correct answer: (A)

Step-by-step solution →
Q10·Physics·Magnetism and MatterSingle correct
The relationship between the magnetic susceptibility (χ\chiχ) and the magnetic permeability (μ\muμ) is given by: (μ0\mu_0μ0​ is the permeability of free space and μr\mu_rμr​ is relative permeability)
  1. (A)χ=μμ0−1\chi=\dfrac{\mu}{\mu_0}-1χ=μ0​μ​−1
  2. (B)χ=μrμ0+1\chi=\dfrac{\mu_r}{\mu_0}+1χ=μ0​μr​​+1
  3. (C)χ=μr+1\chi=\mu_r+1χ=μr​+1
  4. (D)χ=1−μμ0\chi=1-\dfrac{\mu}{\mu_0}χ=1−μ0​μ​

Correct answer: (A)

Step-by-step solution →
Q11·Physics·Electronic DevicesSingle correct
A zener diode with 5V zener voltage is used to regulate an unregulated dc voltage input of 25 V. For a 400 Ω\OmegaΩ resistor connected in series, the zener current is found to be 4 times load current. The load current (ILI_LIL​) and load resistance (RLR_LRL​) are:
  1. (A)IL=20I_L=20IL​=20 mA; RL=250 ΩR_L=250\ \OmegaRL​=250 Ω
  2. (B)IL=10I_L=10IL​=10 mA; RL=0.5 ΩR_L=0.5\ \OmegaRL​=0.5 Ω
  3. (C)IL=0.02I_L=0.02IL​=0.02 mA; RL=250 ΩR_L=250\ \OmegaRL​=250 Ω
  4. (D)IL=10I_L=10IL​=10 mA; RL=500 ΩR_L=500\ \OmegaRL​=500 Ω

Correct answer: (D)

Step-by-step solution →
Q12·Physics·ThermodynamicsSingle correct
In an adiabatic process, which of the following statements is true?
  1. (A)The molar heat capacity is infinite
  2. (B)Work done by the gas equals the increase in internal energy
  3. (C)The molar heat capacity is zero
  4. (D)The internal energy of the gas decreases as the temperature increases

Correct answer: (C)

Step-by-step solution →
Q13·Physics·Rotational MotionSingle correct
A square Lamina OABC of length 10 cm is pivoted at ‘O’. Forces act at the lamina as shown in figure. If the lamina remains stationary, then the magnitude of FFF is:
  1. (A)20 N
  2. (B)0 (zero)
  3. (C)10 N
  4. (D)10210\sqrt{2}102​ N

Correct answer: (C)

Step-by-step solution →
Q14·Physics·Magnetic Field of CurrentSingle correct
Let B1B_1B1​ be the magnitude of magnetic field at the centre of a circular coil of radius RRR carrying current III. Let B2B_2B2​ be the magnitude of magnetic field at an axial distance xxx from the centre. For x:R=3:4x:R=3:4x:R=3:4, B2B1\dfrac{B_2}{B_1}B1​B2​​ is:
  1. (A)4 : 5
  2. (B)16 : 25
  3. (C)64 : 125
  4. (D)25 : 16

Correct answer: (C)

Step-by-step solution →
Q15·Physics·Atoms and NucleiSingle correct
Considering Bohr’s atomic model for hydrogen atom: (A) the energy of H atom in ground state is same as energy of He+^++ ion in its first excited state. (B) The energy of H atom in ground state is same as that for Li2+^{2+}2+ ion in its second excited state. (C) The energy of H atom in its ground state is same as that of He+^++ ion for its ground state. (D) the energy of He+^++ ion in its first excited state is same as that for Li2+^{2+}2+ ion in its ground state. Choose the correct answer from the options given below:
  1. (A)(B), (D) only
  2. (B)(A), (B) only
  3. (C)(A), (D) only
  4. (D)(A), (C) only

Correct answer: (B)

Step-by-step solution →
Q16·Physics·Rotational MotionSingle correct
Moment of inertia of a rod of mass ‘M’ and length ‘L’ about an axis passing through its centre and normal to its length is ‘α\alphaα’. Now the rod is cut into two equal parts and these parts are joined symmetrically to form a cross shape. Moment of inertia of this cross about an axis passing through its centre and normal to the plane containing the cross is:
  1. (A)α\alphaα
  2. (B)α4\dfrac{\alpha}{4}4α​
  3. (C)α8\dfrac{\alpha}{8}8α​
  4. (D)α2\dfrac{\alpha}{2}2α​

Correct answer: (B)

Step-by-step solution →
Q17·Physics·Geometrical OpticsSingle correct
A spherical surface separates two media of refractive indices 1 and 1.5 as shown in the figure. Distance of the image of an object ‘O’, is: (CCC is the centre of curvature of the spherical surface and RRR is the radius of curvature)
  1. (A)0.24 m right to the spherical surface
  2. (B)0.4 m left to the spherical surface
  3. (C)0.24 m left to the spherical surface
  4. (D)0.4 m right to the spherical surface

Correct answer: (B)

Step-by-step solution →
Q18·Physics·Units and MeasurementsSingle correct
Match List-I (physical quantities) with List-II (dimensional formulae). Choose the correct answer from the options given below:
List-IList-II
A.Coefficient of viscosityI.[ML0T−3][ML^0T^{-3}][ML0T−3]
B.Intensity of waveII.[ML−2T−2][ML^{-2}T^{-2}][ML−2T−2]
C.Pressure gradientIII.[M−1LT2][M^{-1}LT^2][M−1LT2]
D.CompressibilityIV.[ML−1T−1][ML^{-1}T^{-1}][ML−1T−1]
  1. (A)(A)-(I), (B)-(IV), (C)-(III), (D)-(II)
  2. (B)(A)-(IV), (B)-(I), (C)-(II), (D)-(III)
  3. (C)(A)-(IV), (B)-(II), (C)-(I), (D)-(III)
  4. (D)(A)-(II), (B)-(III), (C)-(IV), (D)-(I)

Correct answer: (B)

Step-by-step solution →
Q19·Physics·Electric Field and Coulomb's LawSingle correct
A small bob of mass 100 mg and charge +10 μC+10\ \mu C+10 μC is connected to an insulating string of length 1 m. It is brought near to an infinitely long non-conducting sheet of charge density ‘σ\sigmaσ’ as shown in figure. If the string subtends an angle of 45∘45^\circ45∘ with the sheet at equilibrium, the charge density of the sheet will be: (Given ε0=8.85×10−12\varepsilon_0=8.85\times10^{-12}ε0​=8.85×10−12 F/m and acceleration due to gravity g=10g=10g=10 m/s2^22)
  1. (A)0.885 nC/m2^22
  2. (B)17.7 nC/m2^22
  3. (C)885 nC/m2^22
  4. (D)1.77 nC/m2^22

Correct answer: (D)

Step-by-step solution →
Q20·Physics·Dual Nature of Matter and RadiationSingle correct
A monochromatic light is incident on a metallic plate having work function ϕ\phiϕ. An electron, emitted normally to the plate from a point A with maximum kinetic energy, enters a constant magnetic field, perpendicular to the initial velocity of the electron. The electron passes through a curve and hits back the plate at a point B. The distance between A and B is: (Given: the magnitude of charge of an electron is eee and mass is mmm, hhh is Planck’s constant, ccc is velocity of light, magnetic field BBB exists throughout the path of the electron)
  1. (A)2m(hcλ−ϕ)eB\dfrac{\sqrt{2m\left(\dfrac{hc}{\lambda}-\phi\right)}}{eB}eB2m(λhc​−ϕ)​​
  2. (B)m(hcλ−ϕ)eB\dfrac{\sqrt{m\left(\dfrac{hc}{\lambda}-\phi\right)}}{eB}eBm(λhc​−ϕ)​​
  3. (C)8m(hcλ−ϕ)eB\dfrac{\sqrt{8m\left(\dfrac{hc}{\lambda}-\phi\right)}}{eB}eB8m(λhc​−ϕ)​​
  4. (D)2m(hcλ−ϕ)eB\dfrac{2\sqrt{m\left(\dfrac{hc}{\lambda}-\phi\right)}}{eB}eB2m(λhc​−ϕ)​​

Correct answer: (C)

Step-by-step solution →
Q21·Physics·Properties of Solids and LiquidsInteger
A vessel with square cross-section and height of 6 m is vertically partitioned. A small window of 100 cm2^22 with a hinged door is fitted at a depth of 3 m in the partition wall. One part of the vessel is filled completely with water and the other side is filled with the liquid having density 1.5×1031.5\times10^31.5×103 kg/m3^33. What force one needs to apply on the hinged door so that it does not get opened? (Acceleration due to gravity =10=10=10 m/s2^22)

Correct answer: 150

Step-by-step solution →
Q22·Physics·Properties of Solids and LiquidsInteger
A steel wire of length 2 m and Young’s modulus 2.0×10112.0\times10^{11}2.0×1011 Nm−2^{-2}−2 is stretched by a force. If Poisson ratio and transverse strain for the wire are 0.2 and 10−310^{-3}10−3 respectively, then the elastic potential energy density of the wire is ______ ×105\times10^5×105 (in SI units).

Correct answer: 25

Step-by-step solution →
Q23·Physics·Wave OpticsInteger
If the measured angular separation between the second minimum to the left of the central maximum and the third minimum to the right of the central maximum is 30∘30^\circ30∘ in a single slit diffraction pattern recorded using 628 nm light, then the width of the slit is ______ μ\muμm.

Correct answer: 6

Step-by-step solution →
Q24·Physics·Kinetic Theory of GasesInteger
γA\gamma_AγA​ is the specific heat ratio of monoatomic gas A having 3 translational degrees of freedom. γB\gamma_BγB​ is the specific heat ratio of polyatomic gas B having 3 translational, 3 rotational degrees of freedom and 1 vibrational mode. If γAγB=(1+1n)\dfrac{\gamma_A}{\gamma_B}=\left(1+\dfrac{1}{n}\right)γB​γA​​=(1+n1​), then the value of nnn is ______.

Correct answer: 3

Step-by-step solution →
Q25·Physics·KinematicsInteger
A person travelling on a straight line moves with a uniform velocity v1v_1v1​ for a distance xxx and with a uniform velocity v2v_2v2​ for the next 32x\dfrac{3}{2}x23​x distance. The average velocity in this motion is 507\dfrac{50}{7}750​ m/s. If v1v_1v1​ is 5 m/s then v2=v_2=v2​= ______ m/s.

Correct answer: 10

Step-by-step solution →

Chemistry — JEE Main 2 April 2025 Shift 1

Q26·Chemistry·AromaticitySingle correct
Designate whether each of the following compounds (shown in the figure, labelled a–h) is aromatic or not aromatic.
  1. (A)e, g aromatic and a, b, c, d, f, h not aromatic
  2. (B)b, e, f, g aromatic and a, c, d, h not aromatic
  3. (C)a, b, c, d aromatic and e, f, g, h not aromatic
  4. (D)a, c, d, e, h aromatic and b, f, g not aromatic

Correct answer: (D)

Step-by-step solution →
Q27·Chemistry·Organic Compounds Containing HalogensSingle correct
An optically active alkyl halide C4_44​H9_99​Br [A] reacts with hot KOH dissolved in ethanol and forms alkene [B] as major product which reacts with bromine to give dibromide [C]. The compound [C] is converted into a gas [D] upon reacting with alcoholic NaOH. During hydration 18 gram of water is added to 1 mole of gas [D] on warming with mercuric sulphate and dilute acid at 333 K to form compound [E]. The IUPAC name of compound [E] is:
  1. (A)But-2-yne
  2. (B)Butan-2-ol
  3. (C)Butan-2-one
  4. (D)Butan-1-al

Correct answer: (C)

Step-by-step solution →
Q28·Chemistry·p-Block ElementsSingle correct
The property/properties that show irregularity in the first four elements of group-17 is/are: (A) Covalent radius (B) Electron affinity (C) Ionic radius (D) First ionization energy. Choose the correct answer from the options given below:
  1. (A)B and D only
  2. (B)A and C only
  3. (C)B only
  4. (D)A, B, C and D

Correct answer: (C)

Step-by-step solution →
Q29·Chemistry·SolutionsSingle correct
Which of the following graph (shown in the figure) correctly represents the plots of KHK_HKH​ at 1 bar of gases in water versus temperature?
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (D)

Step-by-step solution →
Q30·Chemistry·Atomic StructureSingle correct
According to Bohr’s model of hydrogen atom, which of the following statement is incorrect?
  1. (A)Radius of 3rd orbit is nine times larger than that of 1st orbit.
  2. (B)Radius of 8th orbit is four times larger than that of 4th orbit.
  3. (C)Radius of 6th orbit is three times larger than that of 4th orbit.
  4. (D)Radius of 4th orbit is four times larger than that of 2nd orbit.

Correct answer: (C)

Step-by-step solution →
Q31·Chemistry·Chemical ThermodynamicsSingle correct
Two vessels A and B are connected via a stopcock. The vessel A is filled with a gas at a certain pressure. The vessel B is empty and is allowed to come to thermal equilibrium with water and no change in temperature is observed in the thermometer. After opening the stopcock the gas from vessel A expands into vessel B and no change in temperature is observed. Which of the following statement is true?
  1. (A)dw≠0dw\ne0dw=0
  2. (B)dq≠0dq\ne0dq=0
  3. (C)dU≠0dU\ne0dU=0
  4. (D)The pressure in the vessel B before opening the stopcock is zero.

Correct answer: (D)

Step-by-step solution →
Q32·Chemistry·SolutionsSingle correct
A solution is made by mixing one mole of volatile liquid A with 3 moles of volatile liquid B. The vapour pressure of pure A is 200 mm Hg and that of the solution is 500 mm Hg. The vapour pressure of pure B and the least volatile component of the solution, respectively, are:
  1. (A)1400 mm Hg, A
  2. (B)1400 mm Hg, B
  3. (C)600 mm Hg, B
  4. (D)600 mm Hg, A

Correct answer: (D)

Step-by-step solution →
Q33·Chemistry·Some Basic Concepts in ChemistrySingle correct
CaCO3(s)+2HCl(aq)→CaCl2(aq)+CO2(g)+H2O(l)CaCO_3(s)+2HCl(aq)\to CaCl_2(aq)+CO_2(g)+H_2O(l)CaCO3​(s)+2HCl(aq)→CaCl2​(aq)+CO2​(g)+H2​O(l). Consider the above reaction, what mass of CaCl2CaCl_2CaCl2​ will be formed if 250 mL of 0.76 M HCl reacts with 1000 g of CaCO3CaCO_3CaCO3​? (Given: molar mass of Ca, C, O, H and Cl are 40, 12, 16, 1 and 35.5 g mol−1^{-1}−1, respectively)
  1. (A)3.908 g
  2. (B)2.636 g
  3. (C)10.545 g
  4. (D)5.272 g

Correct answer: (C)

Step-by-step solution →
Q34·Chemistry·EquilibriumSingle correct
If equal volumes of AB2AB_2AB2​ and XYXYXY (both are salts) aqueous solutions are mixed, which of the following combination will give a precipitate of AY2AY_2AY2​ at 300 K? (Given KspK_{sp}Ksp​ at 300 K for AY2=5.2×10−5AY_2=5.2\times10^{-5}AY2​=5.2×10−5)
  1. (A)3.6×10−33.6\times10^{-3}3.6×10−3 M AB2AB_2AB2​, 5.0×10−45.0\times10^{-4}5.0×10−4 M XYXYXY
  2. (B)2.0×10−42.0\times10^{-4}2.0×10−4 M AB2AB_2AB2​, 0.8×10−40.8\times10^{-4}0.8×10−4 M XYXYXY
  3. (C)2.0×10−22.0\times10^{-2}2.0×10−2 M AB2AB_2AB2​, 2.0×10−22.0\times10^{-2}2.0×10−2 M XYXYXY
  4. (D)1.5×10−41.5\times10^{-4}1.5×10−4 M AB2AB_2AB2​, 1.5×10−31.5\times10^{-3}1.5×10−3 M XYXYXY

Correct answer: (C)

Step-by-step solution →
Q35·Chemistry·Chemical Bonding and Molecular StructureSingle correct
Among SO2SO_2SO2​, NF3NF_3NF3​, NH3NH_3NH3​, XeF2XeF_2XeF2​, ClF3ClF_3ClF3​ and SF4SF_4SF4​, the hybridization of the molecule with non-zero dipole moment and highest number of lone-pairs of electrons on the central atom is:
  1. (A)sp3sp^3sp3
  2. (B)dsp2dsp^2dsp2
  3. (C)sp3d2sp^3d^2sp3d2
  4. (D)sp3dsp^3dsp3d

Correct answer: (D)

Step-by-step solution →
Q36·Chemistry·Aldehydes and KetonesSingle correct
Given below are two statements: Statement (I): Vanillin will react with NaOH and also with Tollen’s reagent. Statement (II): Vanillin will undergo self aldol condensation very easily. In the light of the above statements, choose the most appropriate answer from the options given below:
  1. (A)Statement I is incorrect but Statement II is correct
  2. (B)Statement I is correct but Statement II is incorrect
  3. (C)Both Statement I and Statement II are incorrect
  4. (D)Both Statement I and Statement II are correct

Correct answer: (B)

Step-by-step solution →
Q37·Chemistry·BiomoleculesSingle correct
Identify the correct statement among the following:
  1. (A)All naturally occurring amino acids except glycine contain one chiral centre.
  2. (B)All naturally occurring amino acids are optically active.
  3. (C)Glutamic acid is the only amino acid that contains a –COOH group at the side chain.
  4. (D)Amino acid cysteine easily undergoes dimerization due to the presence of free SH group.

Correct answer: (D)

Step-by-step solution →
Q38·Chemistry·AminesSingle correct
The correct order of basic nature in aqueous solution for the bases NH3NH_3NH3​, H2N−NH2H_2N-NH_2H2​N−NH2​, CH3CH2NH2CH_3CH_2NH_2CH3​CH2​NH2​, (CH3CH2)2NH(CH_3CH_2)_2NH(CH3​CH2​)2​NH and (CH3CH2)3N(CH_3CH_2)_3N(CH3​CH2​)3​N is:
  1. (A)NH3<H2N−NH2<(CH3CH2)2NH<CH3CH2NH2<(CH3CH2)3NNH_3<H_2N-NH_2<(CH_3CH_2)_2NH<CH_3CH_2NH_2<(CH_3CH_2)_3NNH3​<H2​N−NH2​<(CH3​CH2​)2​NH<CH3​CH2​NH2​<(CH3​CH2​)3​N
  2. (B)NH3<H2N−NH2<CH3CH2NH2<(CH3CH2)3N<(CH3CH2)2NHNH_3<H_2N-NH_2<CH_3CH_2NH_2<(CH_3CH_2)_3N<(CH_3CH_2)_2NHNH3​<H2​N−NH2​<CH3​CH2​NH2​<(CH3​CH2​)3​N<(CH3​CH2​)2​NH
  3. (C)H2N−NH2<NH3<(CH3CH2)3N<CH3CH2NH2<(CH3CH2)2NHH_2N-NH_2<NH_3<(CH_3CH_2)_3N<CH_3CH_2NH_2<(CH_3CH_2)_2NHH2​N−NH2​<NH3​<(CH3​CH2​)3​N<CH3​CH2​NH2​<(CH3​CH2​)2​NH
  4. (D)H2N−NH2<NH3<CH3CH2NH2<(CH3CH2)3N<(CH3CH2)2NHH_2N-NH_2<NH_3<CH_3CH_2NH_2<(CH_3CH_2)_3N<(CH_3CH_2)_2NHH2​N−NH2​<NH3​<CH3​CH2​NH2​<(CH3​CH2​)3​N<(CH3​CH2​)2​NH

Correct answer: (D)

Step-by-step solution →
Q39·Chemistry·p-Block ElementsSingle correct
Given below are two statements: Statement (I): The metallic radius of Al is less than that of Ga. Statement (II): The ionic radius of Al3+Al^{3+}Al3+ is less than that of Ga3+Ga^{3+}Ga3+. In the light of the above statements, choose the most appropriate answer from the options given below:
  1. (A)Both Statement I and Statement II are incorrect
  2. (B)Statement I is incorrect but Statement II is correct
  3. (C)Statement I is correct but Statement II is incorrect
  4. (D)Both Statement I and Statement II are correct

Correct answer: (B)

Step-by-step solution →
Q40·Chemistry·Coordination CompoundsSingle correct
Given below are two statements: Statement (I): In octahedral complexes, when Δ0<P\Delta_0<PΔ0​<P high spin complexes are formed. When Δ0>P\Delta_0>PΔ0​>P low spin complexes are formed. Statement (II): In tetrahedral complexes, because of Δt<P\Delta_t<PΔt​<P, low spin complexes are rarely formed. In the light of the above statements, choose the most appropriate answer from the options given below:
  1. (A)Statement I is correct but Statement II is incorrect
  2. (B)Both Statement I and Statement II are incorrect
  3. (C)Statement I is incorrect but Statement II is correct
  4. (D)Both Statement I and Statement II are correct

Correct answer: (D)

Step-by-step solution →
Q41·Chemistry·Principles of Qualitative AnalysisSingle correct
Choose the correct tests with their respective observations. (A) CuSO4CuSO_4CuSO4​ (acidified with acetic acid) + K4[Fe(CN)6]→+\,K_4[Fe(CN)_6]\to+K4​[Fe(CN)6​]→ Chocolate brown precipitate. (B) FeCl3+K4[Fe(CN)6]→FeCl_3+K_4[Fe(CN)_6]\toFeCl3​+K4​[Fe(CN)6​]→ Prussian blue precipitate. (C) ZnCl2+K4[Fe(CN)6]ZnCl_2+K_4[Fe(CN)_6]ZnCl2​+K4​[Fe(CN)6​], neutralised with NH4OH→NH_4OH\toNH4​OH→ White or bluish white precipitate. (D) MgCl2+K4[Fe(CN)6]→MgCl_2+K_4[Fe(CN)_6]\toMgCl2​+K4​[Fe(CN)6​]→ Blue precipitate. (E) BaCl2+K4[Fe(CN)6]BaCl_2+K_4[Fe(CN)_6]BaCl2​+K4​[Fe(CN)6​], neutralised with NaOH→NaOH\toNaOH→ White precipitate. Choose the correct answer from the options given below:
  1. (A)A, D and E only
  2. (B)B, D and E only
  3. (C)A, B and C only
  4. (D)C, D and E only

Correct answer: (C)

Step-by-step solution →
Q42·Chemistry·Some Basic Concepts in ChemistrySingle correct
On complete combustion 1.0 g of an organic compound (X) gave 1.46 g of CO2CO_2CO2​ and 0.567 g of H2OH_2OH2​O. The empirical formula mass of compound (X) is ______ g. (Given molar mass in g mol−1^{-1}−1: C:12, H:1, O:16)
  1. (A)30
  2. (B)45
  3. (C)60
  4. (D)15

Correct answer: (A)

Step-by-step solution →
Q43·Chemistry·Electronic Effects and StabilitySingle correct
Consider the compound (X): H−C≡C−CH2−CH(CH3)−CH3H-C\equiv C-CH_2-CH(CH_3)-CH_3H−C≡C−CH2​−CH(CH3​)−CH3​, whose C–H bonds are labelled I (at the ≡C−H\equiv C-H≡C−H), II (at the CH2CH_2CH2​), III (at the CHCHCH) and IV (at the CH3CH_3CH3​). The most stable and least stable carbon radicals, respectively, produced by homolytic cleavage of the corresponding C–H bond are:
  1. (A)II, IV
  2. (B)III, II
  3. (C)I, IV
  4. (D)II, I

Correct answer: (D)

Step-by-step solution →
Q44·Chemistry·Carboxylic Acids and DerivativesSingle correct
Consider the following molecules: (p) CH3CH2COClCH_3CH_2COClCH3​CH2​COCl (acyl chloride), (q) CH3CH2CO−O−COCH3CH_3CH_2CO-O-COCH_3CH3​CH2​CO−O−COCH3​ (anhydride), (r) CH3CH2CO−O−CH2CH3CH_3CH_2CO-O-CH_2CH_3CH3​CH2​CO−O−CH2​CH3​ (ester), (s) CH3CH2CONH2CH_3CH_2CONH_2CH3​CH2​CONH2​ (amide). The correct order of rate of hydrolysis is:
  1. (A)r > q > p > s
  2. (B)q > p > r > s
  3. (C)p > r > q > s
  4. (D)p > q > r > s

Correct answer: (D)

Step-by-step solution →
Q45·Chemistry·p-Block ElementsSingle correct
A molecule with the formula AX4YAX_4YAX4​Y has all its elements from the p-block. Element A is rarest, monoatomic, non-radioactive and has the lowest ionization enthalpy value among A, X and Y. Elements X and Y have first and second highest electronegativity values respectively among all the known elements. The shape of the molecule is:
  1. (A)Square pyramidal
  2. (B)Octahedral
  3. (C)Pentagonal planar
  4. (D)Trigonal bipyramidal

Correct answer: (A)

Step-by-step solution →
Q46·Chemistry·Coordination CompoundsInteger
A transition metal (M) among Mn, Cr, Co and Fe has the highest standard electrode potential (M3+/M2+)(M^{3+}/M^{2+})(M3+/M2+). It forms a metal complex of the type [M(CN)6]4−[M(CN)_6]^{4-}[M(CN)6​]4−. The number of electrons present in the ege_geg​ orbital of the complex is ______.

Correct answer: 1

Step-by-step solution →
Q47·Chemistry·Redox Reactions and ElectrochemistryInteger
Consider the following electrochemical cell at standard conditions: Au(s) ∣ QH2 ∣ Q ∣∣ NH4X(0.01 M) ∣∣ Ag+(1 M) ∣ Ag(s)Au(s)\,|\,QH_2\,|\,Q\,||\,NH_4X(0.01\,M)\,||\,Ag^+(1\,M)\,|\,Ag(s)Au(s)∣QH2​∣Q∣∣NH4​X(0.01M)∣∣Ag+(1M)∣Ag(s), Ecell∘=+0.4E^\circ_{cell}=+0.4Ecell∘​=+0.4 V. The couple QH2/QQH_2/QQH2​/Q represents the quinhydrone electrode, whose half-cell reaction is Q+2e−+2H+→QH2Q+2e^-+2H^+\to QH_2Q+2e−+2H+→QH2​, EQ/QH2∘=+0.7E^\circ_{Q/QH_2}=+0.7EQ/QH2​∘​=+0.7 V. Given EAg+/Ag∘=+0.8E^\circ_{Ag^+/Ag}=+0.8EAg+/Ag∘​=+0.8 V and 2.303RTF=0.06\dfrac{2.303RT}{F}=0.06F2.303RT​=0.06 V, the pKbpK_bpKb​ value of the ammonium halide salt (NH4X)(NH_4X)(NH4​X) used here is ______ (nearest integer).

Correct answer: 6

Step-by-step solution →
Q48·Chemistry·BiomoleculesInteger
0.1 mol of the antiviral compound (P) shown in the figure will weigh ______ ×10−1\times10^{-1}×10−1 g. (Given molar mass in g mol−1^{-1}−1: H:1, C:12, N:14, O:16, F:19, I:127)

Correct answer: 372

Step-by-step solution →
Q49·Chemistry·EquilibriumInteger
Consider the following equilibrium, CO(g)+2H2(g)⇌CH3OH(g)CO(g)+2H_2(g)\rightleftharpoons CH_3OH(g)CO(g)+2H2​(g)⇌CH3​OH(g). 0.1 mol of CO along with a catalyst is present in a 2 dm3^33 flask maintained at 500 K. Hydrogen is introduced into the flask until the pressure is 5 bar and 0.04 mol of CH3OHCH_3OHCH3​OH is formed. The Kp∘K_p^\circKp∘​ is ______ ×10−3\times10^{-3}×10−3 (nearest integer). (Given R=0.08R=0.08R=0.08 dm3^33 bar K−1^{-1}−1 mol−1^{-1}−1; assume only methanol is formed and the system follows ideal gas behaviour.)

Correct answer: 74

Step-by-step solution →
Q50·Chemistry·Chemical KineticsInteger
For the reaction A → products, a plot of half-life t1/2t_{1/2}t1/2​/min versus initial concentration [A]0[A]_0[A]0​/mol L−1^{-1}−1 (shown in the figure) is a straight line with slope 76.92 (appropriate units). The concentration of A at 10 minutes is ______ ×10−3\times10^{-3}×10−3 mol L−1^{-1}−1 (nearest integer). The reaction was started with 2.5 mol L−1^{-1}−1 of A.

Correct answer: 2435

Step-by-step solution →

Mathematics — JEE Main 2 April 2025 Shift 1

Q51·Mathematics·Permutations and CombinationsSingle correct
The largest n∈Nn\in Nn∈N such that 3n3^n3n divides 50!50!50! is:
  1. (A)21
  2. (B)22
  3. (C)20
  4. (D)23

Correct answer: (B)

Step-by-step solution →
Q52·Mathematics·ParabolaSingle correct
Let one focus of the hyperbola H:x2a2−y2b2=1H:\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1H:a2x2​−b2y2​=1 be at (10,0)(\sqrt{10},0)(10​,0) and the corresponding directrix be x=910x=\dfrac{9}{\sqrt{10}}x=10​9​. If eee and lll respectively are the eccentricity and the length of the latus rectum of HHH, then 9(e2+l)9(e^2+l)9(e2+l) is equal to:
  1. (A)14
  2. (B)15
  3. (C)16
  4. (D)12

Correct answer: (C)

Step-by-step solution →
Q53·Mathematics·Permutations and CombinationsSingle correct
The number of sequences of ten terms, whose terms are either 0 or 1 or 2, that contain exactly five 1s and exactly three 2s, is equal to
  1. (A)360
  2. (B)45
  3. (C)2520
  4. (D)1820

Correct answer: (C)

Step-by-step solution →
Q54·Mathematics·Sets, Relations and FunctionsSingle correct
Let f:R→Rf:R\to Rf:R→R be a twice differentiable function such that (sin⁡xcos⁡y)(f(2x+2y)−f(2x−2y))=(cos⁡xsin⁡y)(f(2x+2y)+f(2x−2y))(\sin x\cos y)(f(2x+2y)-f(2x-2y))=(\cos x\sin y)(f(2x+2y)+f(2x-2y))(sinxcosy)(f(2x+2y)−f(2x−2y))=(cosxsiny)(f(2x+2y)+f(2x−2y)), for all x,y∈Rx,y\in Rx,y∈R. If f′(0)=12f'(0)=\dfrac{1}{2}f′(0)=21​, then the value of 24 f′′(5π3)24\,f''\left(\dfrac{5\pi}{3}\right)24f′′(35π​) is:
  1. (A)2
  2. (B)−3-3−3
  3. (C)3
  4. (D)−2-2−2

Correct answer: (B)

Step-by-step solution →
Q55·Mathematics·Matrices and DeterminantsSingle correct
Let A=[α−16β]A=\begin{bmatrix}\alpha & -1\\ 6 & \beta\end{bmatrix}A=[α6​−1β​], α>0\alpha>0α>0, such that det⁡(A)=0\det(A)=0det(A)=0 and α+β=1\alpha+\beta=1α+β=1. If III denotes 2×22\times22×2 identity matrix, then the matrix (1+A)8(1+A)^8(1+A)8 is:
  1. (A)[4−16−1]\begin{bmatrix}4 & -1\\ 6 & -1\end{bmatrix}[46​−1−1​]
  2. (B)[257−64514−127]\begin{bmatrix}257 & -64\\ 514 & -127\end{bmatrix}[257514​−64−127​]
  3. (C)[1025−5112024−1024]\begin{bmatrix}1025 & -511\\ 2024 & -1024\end{bmatrix}[10252024​−511−1024​]
  4. (D)[766−2551530−509]\begin{bmatrix}766 & -255\\ 1530 & -509\end{bmatrix}[7661530​−255−509​]

Correct answer: (D)

Step-by-step solution →
Q56·Mathematics·Binomial Theorem and Its Simple ApplicationsSingle correct
The term independent of xxx in the expansion of ((x+1)x2/3+1−x1/3−(x+1)x−x1/2)10\left(\dfrac{(x+1)}{x^{2/3}+1-x^{1/3}}-\dfrac{(x+1)}{x-x^{1/2}}\right)^{10}(x2/3+1−x1/3(x+1)​−x−x1/2(x+1)​)10, x>1x>1x>1 is:
  1. (A)210
  2. (B)150
  3. (C)240
  4. (D)120

Correct answer: (A)

Step-by-step solution →
Q57·Mathematics·Trigonometric FunctionsSingle correct
If θ∈[−2π,2π]\theta\in[-2\pi,2\pi]θ∈[−2π,2π], then the number of solutions of 22cos⁡2θ+(2−6)cos⁡θ−3=02\sqrt{2}\cos^2\theta+(2-\sqrt{6})\cos\theta-\sqrt{3}=022​cos2θ+(2−6​)cosθ−3​=0, is equal to:
  1. (A)12
  2. (B)6
  3. (C)8
  4. (D)10

Correct answer: (C)

Step-by-step solution →
Q58·Mathematics·Sequence and SeriesSingle correct
Let a1,a2,a3,…a_1, a_2, a_3,\ldotsa1​,a2​,a3​,… be in an A.P. such that ∑k=112a2k−1=−725a1\displaystyle\sum_{k=1}^{12}a_{2k-1}=-\dfrac{72}{5}a_1k=1∑12​a2k−1​=−572​a1​, a1≠0a_1\ne0a1​=0. If ∑k=1nak=0\displaystyle\sum_{k=1}^{n}a_k=0k=1∑n​ak​=0, then nnn is:
  1. (A)11
  2. (B)10
  3. (C)18
  4. (D)17

Correct answer: (A)

Step-by-step solution →
Q59·Mathematics·Application of DerivativesSingle correct
If the function f(x)=2x3−9ax2+12a2x+1f(x)=2x^3-9ax^2+12a^2x+1f(x)=2x3−9ax2+12a2x+1, where a>0a>0a>0, attains its local maximum and minimum values at ppp and qqq, respectively, such that p2=qp^2=qp2=q, then f(3)f(3)f(3) is equal to:
  1. (A)55
  2. (B)10
  3. (C)23
  4. (D)37

Correct answer: (D)

Step-by-step solution →
Q60·Mathematics·Complex NumbersSingle correct
Let zzz be a complex number such that ∣z∣=1|z|=1∣z∣=1. If 2+k2zk+zˉ=kz\dfrac{2+k^2z}{k+\bar{z}}=kzk+zˉ2+k2z​=kz, k∈Rk\in Rk∈R, then the maximum distance of k+ik2k+ik^2k+ik2 from the circle ∣z−(1+2i)∣=1|z-(1+2i)|=1∣z−(1+2i)∣=1 is:
  1. (A)5+1\sqrt{5}+15​+1
  2. (B)2
  3. (C)3
  4. (D)3+1\sqrt{3}+13​+1

Correct answer: (A)

Step-by-step solution →
Q61·Mathematics·Vector AlgebraSingle correct
If a⃗\vec{a}a is nonzero vector such that its projections on the vectors 2i^−j^+2k^2\hat{i}-\hat{j}+2\hat{k}2i^−j^​+2k^, i^+2j^−2k^\hat{i}+2\hat{j}-2\hat{k}i^+2j^​−2k^ and k^\hat{k}k^ are equal, then a unit vector along a⃗\vec{a}a is:
  1. (A)1155(−7i^+9j^+5k^)\dfrac{1}{\sqrt{155}}(-7\hat{i}+9\hat{j}+5\hat{k})155​1​(−7i^+9j^​+5k^)
  2. (B)1155(−7i^+9j^−5k^)\dfrac{1}{\sqrt{155}}(-7\hat{i}+9\hat{j}-5\hat{k})155​1​(−7i^+9j^​−5k^)
  3. (C)1155(7i^+9j^+5k^)\dfrac{1}{\sqrt{155}}(7\hat{i}+9\hat{j}+5\hat{k})155​1​(7i^+9j^​+5k^)
  4. (D)1155(7i^+9j^−5k^)\dfrac{1}{\sqrt{155}}(7\hat{i}+9\hat{j}-5\hat{k})155​1​(7i^+9j^​−5k^)

Correct answer: (C)

Step-by-step solution →
Q62·Mathematics·Sets, Relations and FunctionsSingle correct
Let AAA be the set of all functions f:Z→Zf:Z\to Zf:Z→Z and RRR be a relation on AAA such that R={(f,g):f(0)=g(1) and f(1)=g(0)}R=\{(f,g):f(0)=g(1)\text{ and }f(1)=g(0)\}R={(f,g):f(0)=g(1) and f(1)=g(0)}. Then RRR is:
  1. (A)Symmetric and transitive but not reflexive
  2. (B)Symmetric but neither reflexive nor transitive
  3. (C)Reflexive but neither symmetric nor transitive
  4. (D)Transitive but neither reflexive nor symmetric

Correct answer: (B)

Step-by-step solution →
Q63·Mathematics·Limits and ContinuitySingle correct
For α,β,γ∈R\alpha,\beta,\gamma\in Rα,β,γ∈R, if lim⁡x→0x2sin⁡αx+(γ−1)ex2sin⁡2x−βx=3\displaystyle\lim_{x\to0}\dfrac{x^2\sin\alpha x+(\gamma-1)e^{x^2}}{\sin 2x-\beta x}=3x→0lim​sin2x−βxx2sinαx+(γ−1)ex2​=3, then β+γ−α\beta+\gamma-\alphaβ+γ−α is equal to:
  1. (A)7
  2. (B)4
  3. (C)6
  4. (D)−1-1−1

Correct answer: (A)

Step-by-step solution →
Q64·Mathematics·Matrices and DeterminantsSingle correct
If the system of linear equations 3x+y+βz=33x+y+\beta z=33x+y+βz=3; 2x+αy−z=−32x+\alpha y-z=-32x+αy−z=−3; x+2y+z=4x+2y+z=4x+2y+z=4 has infinitely many solutions, then the value of 22β−9α22\beta-9\alpha22β−9α is:
  1. (A)49
  2. (B)31
  3. (C)43
  4. (D)37

Correct answer: (B)

Step-by-step solution →
Q65·Mathematics·Quadratic EquationsSingle correct
Let Pn=αn+βnP_n=\alpha^n+\beta^nPn​=αn+βn, n∈Nn\in Nn∈N. If P10=123P_{10}=123P10​=123, P9=76P_9=76P9​=76, P8=47P_8=47P8​=47 and P1=1P_1=1P1​=1, then the quadratic equation having roots 1α\dfrac{1}{\alpha}α1​ and 1β\dfrac{1}{\beta}β1​ is:
  1. (A)x2−x+1=0x^2-x+1=0x2−x+1=0
  2. (B)x2+x−1=0x^2+x-1=0x2+x−1=0
  3. (C)x2−x−1=0x^2-x-1=0x2−x−1=0
  4. (D)x2+x+1=0x^2+x+1=0x2+x+1=0

Correct answer: (B)

Step-by-step solution →
Q66·Mathematics·EllipseSingle correct
If SSS and S′S'S′ are the foci of the ellipse x218+y29=1\dfrac{x^2}{18}+\dfrac{y^2}{9}=118x2​+9y2​=1 and PPP be a point on the ellipse, then min⁡(SP⋅S′P)+max⁡(SP⋅S′P)\min(SP\cdot S'P)+\max(SP\cdot S'P)min(SP⋅S′P)+max(SP⋅S′P) is equal to:
  1. (A)3(1+2)3(1+\sqrt{2})3(1+2​)
  2. (B)3(6+2)3(6+\sqrt{2})3(6+2​)
  3. (C)9
  4. (D)27

Correct answer: (D)

Step-by-step solution →
Q67·Mathematics·Three Dimensional GeometrySingle correct
Let the vertices QQQ and RRR of the triangle PQRPQRPQR lie on the line x+35=y−12=z+43\dfrac{x+3}{5}=\dfrac{y-1}{2}=\dfrac{z+4}{3}5x+3​=2y−1​=3z+4​, QR=5QR=5QR=5 and the coordinates of the point PPP be (0,2,3)(0,2,3)(0,2,3). If the area of the triangle PQRPQRPQR is mn\dfrac{m}{n}nm​ then:
  1. (A)m−521 n=0m-5\sqrt{21}\,n=0m−521​n=0
  2. (B)2m−521 n=02m-5\sqrt{21}\,n=02m−521​n=0
  3. (C)5m−221 n=05m-2\sqrt{21}\,n=05m−221​n=0
  4. (D)5m−212 n=05m-21\sqrt{2}\,n=05m−212​n=0

Correct answer: (B)

Step-by-step solution →
Q68·Mathematics·Vector AlgebraSingle correct
Let ABCDABCDABCD be a tetrahedron such that the edges ABABAB, ACACAC and ADADAD are mutually perpendicular. Let the areas of the triangles ABCABCABC, ACDACDACD and ADBADBADB be 5, 6 and 7 square units respectively. Then the area (in square units) of the △BCD\triangle BCD△BCD is equal to:
  1. (A)340\sqrt{340}340​
  2. (B)12
  3. (C)110\sqrt{110}110​
  4. (D)737\sqrt{3}73​

Correct answer: (C)

Step-by-step solution →
Q69·Mathematics·Matrices and DeterminantsSingle correct
Let a∈Ra\in Ra∈R and AAA be a matrix of order 3×33\times33×3 such that det⁡(A)=−4\det(A)=-4det(A)=−4 and A+I=[1a1210a12]A+I=\begin{bmatrix}1 & a & 1\\ 2 & 1 & 0\\ a & 1 & 2\end{bmatrix}A+I=​12a​a11​102​​, where III is the identity matrix of order 3×33\times33×3. If det⁡((a+1) adj((a−1)A))\det((a+1)\,\mathrm{adj}((a-1)A))det((a+1)adj((a−1)A)) is 2m 3n2^m\,3^n2m3n, m,n∈{0,1,2,…,20}m,n\in\{0,1,2,\ldots,20\}m,n∈{0,1,2,…,20}, then m+nm+nm+n is equal to:
  1. (A)14
  2. (B)17
  3. (C)15
  4. (D)16

Correct answer: (D)

Step-by-step solution →
Q70·Mathematics·ParabolaSingle correct
Let the focal chord PQPQPQ of the parabola y2=4xy^2=4xy2=4x make an angle of 60∘60^\circ60∘ with the positive x-axis, where PPP lies in the first quadrant. If the circle, whose one diameter is PSPSPS, SSS being the focus of the parabola, touches the y-axis at the point (0,α)(0,\alpha)(0,α), then 5α25\alpha^25α2 is equal to:
  1. (A)15
  2. (B)25
  3. (C)30
  4. (D)20

Correct answer: (A)

Step-by-step solution →
Q71·Mathematics·Definite IntegrationInteger
Let [⋅][\cdot][⋅] denote the greatest integer function. If ∫0e2[1ex−1]dx=α−log⁡e2\displaystyle\int_0^{e^2}\left[\dfrac{1}{e^{x-1}}\right]dx=\alpha-\log_e 2∫0e2​[ex−11​]dx=α−loge​2, then α3\alpha^3α3 is equal to ______.

Correct answer: 8

Step-by-step solution →
Q72·Mathematics·Differential EquationsInteger
Let f:R→Rf:R\to Rf:R→R be a thrice differentiable odd function satisfying f′′(x)≥0f''(x)\ge0f′′(x)≥0, f′′(x)=f(x)f''(x)=f(x)f′′(x)=f(x), f(0)=0f(0)=0f(0)=0, f′(0)=3f'(0)=3f′(0)=3. Then 9f(log⁡e3)9f(\log_e 3)9f(loge​3) is equal to ______.

Correct answer: 36

Step-by-step solution →
Q73·Mathematics·Area Under CurvesInteger
If the area of the region {(x,y):∣4−x2∣≤y≤x2, y≤4, x≥0}\{(x,y):|4-x^2|\le y\le x^2,\ y\le4,\ x\ge0\}{(x,y):∣4−x2∣≤y≤x2, y≤4, x≥0} is (802α−β)\left(\dfrac{80\sqrt{2}}{\alpha}-\beta\right)(α802​​−β), α,β∈N\alpha,\beta\in Nα,β∈N, then α+β\alpha+\betaα+β is equal to ______.

Correct answer: 22

Step-by-step solution →
Q74·Mathematics·Statistics and ProbabilityInteger
Three distinct numbers are selected randomly from the set {1,2,3,…,40}\{1,2,3,\ldots,40\}{1,2,3,…,40}. If the probability, that the selected numbers are in an increasing G.P. is mn\dfrac{m}{n}nm​, gcd⁡(m,n)=1\gcd(m,n)=1gcd(m,n)=1, then m+nm+nm+n is equal to ______.

Correct answer: 4949

Step-by-step solution →
Q75·Mathematics·CirclesInteger
The absolute difference between the squares of the radii of the two circles passing through the point (−9,4)(-9,4)(−9,4) and touching the lines x+y=3x+y=3x+y=3 and x−y=3x-y=3x−y=3, is equal to ______.

Correct answer: 768

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
  • Chemical Bonding and Molecular Structure 151/186
  • Application of Derivatives 139/186
  • Limits and Continuity 149/186
  • Thermodynamics 154/186
  • Aldehydes and Ketones 135/186
  • Equilibrium 163/186
  • Solutions 158/186
  • Chemical Thermodynamics 165/186
  • Units and Measurements 149/186
  • Complex Numbers 165/186
  • Trigonometric Functions 144/186
  • Biomolecules 162/186
  • Chemical Kinetics 169/186
  • Electric Field and Coulomb's Law 133/186
  • Atomic Structure 161/186
  • Electronic Devices 145/186
  • Circles 142/186
  • Dual Nature of Matter and Radiation 155/186
  • Amines 133/186
  • Quadratic Equations 148/186
  • Some Basic Concepts in Chemistry 129/186
  • Wave Optics 130/186
  • Area Under Curves 139/186
  • Kinetic Theory of Gases 135/186
  • Oscillations 117/186
  • Organic Compounds Containing Halogens 109/186
  • Atoms 112/186
  • Parabola 101/186
  • Ellipse 103/186
  • Electronic Effects and Stability 74/186
  • Carboxylic Acids and Derivatives 54/186
  • Principles of Qualitative Analysis 58/186
  • Magnetism and Matter 50/186
  • Aromaticity 22/186
← 29 Jan Shift 2 2025All papers2 Apr Shift 2 2025 →

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