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JEE Main 24 January 2025 Shift 1 Question Paper with Answers

24 January 2025 · January session · 75 questions

The complete JEE Main 24 January 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 24 January 2025 Shift 1

Q1·Physics·Capacitors and DielectricsSingle correct
Consider a parallel plate capacitor of area A (of each plate) and separation ‘d’ between the plates. If E is the electric field and ε0\varepsilon_0ε0​ is the permittivity of free space between the plates, then potential energy stored in the capacitor is :-
  1. (A)12ε0E2Ad\dfrac{1}{2}\varepsilon_0 E^2 Ad21​ε0​E2Ad
  2. (B)34ε0E2Ad\dfrac{3}{4}\varepsilon_0 E^2 Ad43​ε0​E2Ad
  3. (C)14ε0E2Ad\dfrac{1}{4}\varepsilon_0 E^2 Ad41​ε0​E2Ad
  4. (D)ε0E2Ad\varepsilon_0 E^2 Adε0​E2Ad

Correct answer: (A)

Step-by-step solution →
Q2·Physics·Geometrical OpticsSingle correct
What is the relative decrease in focal length of a lens for an increase in optical power by 0.1 D from 2.5 D ? [‘D’ stands for dioptre]
  1. (A)0.04
  2. (B)0.40
  3. (C)0.1
  4. (D)0.01

Correct answer: (A)

Step-by-step solution →
Q3·Physics·Properties of Solids and LiquidsSingle correct
An air bubble of radius 0.1 cm lies at a depth of 20 cm below the free surface of a liquid of density 1000 kg/m3^33. If the pressure inside the bubble is 2100 N/m2^22 greater than the atmospheric pressure, then the surface tension of the liquid in SI unit is (use g=10g=10g=10 m/s2^22)
  1. (A)0.02
  2. (B)0.1
  3. (C)0.25
  4. (D)0.05

Correct answer: (D)

Step-by-step solution →
Q4·Physics·Units and MeasurementsSingle correct
For an experimental expression y=32.3×112527.4y=\dfrac{32.3\times1125}{27.4}y=27.432.3×1125​, where all the digits are significant. Then to report the value of y we should write :-
  1. (A)y=1326.2y=1326.2y=1326.2
  2. (B)y=1326.19y=1326.19y=1326.19
  3. (C)y=1326.186y=1326.186y=1326.186
  4. (D)y=1330y=1330y=1330

Correct answer: (D)

Step-by-step solution →
Q5·Physics·Atoms and NucleiSingle correct
During the transition of electron from state A to state C of a Bohr atom, the wavelength of emitted radiation is 2000 Å and it becomes 6000 Å when the electron jumps from state B to state C. Then the wavelength of the radiation emitted during the transition of electrons from state A to state B is :-
  1. (A)3000 Å
  2. (B)6000 Å
  3. (C)4000 Å
  4. (D)2000 Å

Correct answer: (A)

Step-by-step solution →
Q6·Physics·Electronic DevicesSingle correct
Consider the following statements : A. The junction area of solar cell is made very narrow compared to a photo diode. B. Solar cells are not connected with any external bias. C. LED is made of lightly doped p-n junction. D. Increase of forward current results in continuous increase of LED light intensity. E. LEDs have to be connected in forward bias for emission of light. Choose the correct answer from the options given below :
  1. (A)B, D, E Only
  2. (B)A, C Only
  3. (C)A, C, E Only
  4. (D)B, E Only

Correct answer: (D)

Step-by-step solution →
Q7·Physics·Properties of Solids and LiquidsSingle correct
The amount of work done to break a big water drop of radius ‘R’ into 27 small drops of equal radius is 10 J. The work done required to break the same big drop into 64 small drops of equal radius will be :-
  1. (A)15 J
  2. (B)10 J
  3. (C)20 J
  4. (D)5 J

Correct answer: (A)

Step-by-step solution →
Q8·Physics·Rotational MotionSingle correct
An object of mass ‘m’ is projected from origin in a vertical x-y plane at an angle 45° with the x-axis with an initial velocity v0v_0v0​. The magnitude and direction of the angular momentum of the object with respect to origin, when it reaches at the maximum height, is [g is acceleration due to gravity]
  1. (A)mv0322 g\dfrac{mv_0^3}{2\sqrt2\,g}22​gmv03​​ along negative z-axis
  2. (B)mv0322 g\dfrac{mv_0^3}{2\sqrt2\,g}22​gmv03​​ along positive z-axis
  3. (C)mv0342 g\dfrac{mv_0^3}{4\sqrt2\,g}42​gmv03​​ along positive z-axis
  4. (D)mv0342 g\dfrac{mv_0^3}{4\sqrt2\,g}42​gmv03​​ along negative z-axis

Correct answer: (D)

Step-by-step solution →
Q9·Physics·Wave OpticsSingle correct
The Young’s double slit interference experiment is performed using light consisting of 480 nm and 600 nm wavelengths to form interference fringes. The least number of the bright fringes of 480 nm light that are required for the first coincidence with the bright fringes formed by 600 nm light is :-
  1. (A)4
  2. (B)8
  3. (C)6
  4. (D)5

Correct answer: (D)

Step-by-step solution →
Q10·Physics·Laws of MotionSingle correct
A car of mass ‘m’ moves on a banked road having radius ‘r’ and banking angle θ. To avoid slipping from banked road, the maximum permissible speed of the car is v0v_0v0​. The coefficient of friction μ between the wheels of the car and the banked road is :-
  1. (A)v02+rgtan⁡θrg−v02tan⁡θ\dfrac{v_0^2+rg\tan\theta}{rg-v_0^2\tan\theta}rg−v02​tanθv02​+rgtanθ​
  2. (B)v02+rgtan⁡θrg+v02tan⁡θ\dfrac{v_0^2+rg\tan\theta}{rg+v_0^2\tan\theta}rg+v02​tanθv02​+rgtanθ​
  3. (C)v02−rgtan⁡θrg+v02tan⁡θ\dfrac{v_0^2-rg\tan\theta}{rg+v_0^2\tan\theta}rg+v02​tanθv02​−rgtanθ​
  4. (D)v02−rgtan⁡θrg−v02tan⁡θ\dfrac{v_0^2-rg\tan\theta}{rg-v_0^2\tan\theta}rg−v02​tanθv02​−rgtanθ​

Correct answer: (C)

Step-by-step solution →
Q11·Physics·Rotational MotionSingle correct
A uniform solid cylinder of mass ‘m’ and radius ‘r’ rolls along an inclined rough plane of inclination 45°. If it starts to roll from rest from the top of the plane then the linear acceleration of the cylinder axis will be :-
  1. (A)12g\dfrac{1}{\sqrt2}g2​1​g
  2. (B)132g\dfrac{1}{3\sqrt2}g32​1​g
  3. (C)2g3\dfrac{\sqrt2 g}{3}32​g​
  4. (D)2g\sqrt2 g2​g

Correct answer: (C)

Step-by-step solution →
Q12·Physics·Geometrical OpticsSingle correct
A thin plano convex lens made of glass of refractive index 1.5 is immersed in a liquid of refractive index 1.2. When the plane side of the lens is silver coated for complete reflection, the lens immersed in the liquid behaves like a concave mirror of focal length 0.2 m. The radius of curvature of the curved surface of the lens is :-
  1. (A)0.15 m
  2. (B)0.10 m
  3. (C)0.20 m
  4. (D)0.25 m

Correct answer: (B)

Step-by-step solution →
Q13·Physics·OscillationsSingle correct
A particle is executing simple harmonic motion with time period 2s and amplitude 1 cm. If D and d are the total distance and displacement covered by the particle in 12.5 s, then Dd\dfrac{D}{d}dD​ is :-
  1. (A)154\dfrac{15}{4}415​
  2. (B)25
  3. (C)10
  4. (D)165\dfrac{16}{5}516​

Correct answer: (B)

Step-by-step solution →
Q14·Physics·GravitationSingle correct
A satellite is launched into a circular orbit of radius ‘R’ around the earth. A second satellite is launched into an orbit of radius 1.03 R. The time period of revolution of the second satellite is larger than the first one approximately by :-
  1. (A)3%
  2. (B)4.5%
  3. (C)9%
  4. (D)2.5%

Correct answer: (B)

Step-by-step solution →
Q15·Physics·Geometrical OpticsSingle correct
A plano-convex lens having radius of curvature of first surface 2 cm exhibits focal length of f1f_1f1​ in air. Another plano-convex lens with first surface radius of curvature 3 cm has focal length of f2f_2f2​ when it is immersed in a liquid of refractive index 1.2. If both the lenses are made of same glass of refractive index 1.5, the ratio of f1f_1f1​ and f2f_2f2​ is :-
  1. (A)3 : 5
  2. (B)1 : 3
  3. (C)1 : 2
  4. (D)2 : 3

Correct answer: (B)

Step-by-step solution →
Q16·Physics·Alternating CurrentsSingle correct
An alternating current is given by I=IAsin⁡ωt+IBcos⁡ωtI=I_A\sin\omega t+I_B\cos\omega tI=IA​sinωt+IB​cosωt. The r.m.s. current will be :-
  1. (A)IA2+IB2\sqrt{I_A^2+I_B^2}IA2​+IB2​​
  2. (B)IA2+IB22\dfrac{\sqrt{I_A^2+I_B^2}}{2}2IA2​+IB2​​​
  3. (C)IA2+IB22\sqrt{\dfrac{I_A^2+I_B^2}{2}}2IA2​+IB2​​​
  4. (D)∣IA+IB∣2\dfrac{|I_A+I_B|}{\sqrt2}2​∣IA​+IB​∣​

Correct answer: (C)

Step-by-step solution →
Q17·Physics·Dual Nature of Matter and RadiationSingle correct
An electron of mass ‘m’ with an initial velocity v⃗=v0i^ (v0>0)\vec v=v_0\hat i\ (v_0>0)v=v0​i^ (v0​>0) enters an electric field E⃗=−E0k^\vec E=-E_0\hat kE=−E0​k^. If the initial de Broglie wavelength is λ0\lambda_0λ0​, the value after time t would be :-
  1. (A)λ01+e2E02t2m2v02\dfrac{\lambda_0}{\sqrt{1+\dfrac{e^2E_0^2t^2}{m^2v_0^2}}}1+m2v02​e2E02​t2​​λ0​​
  2. (B)λ01−e2E02t2m2v02\dfrac{\lambda_0}{\sqrt{1-\dfrac{e^2E_0^2t^2}{m^2v_0^2}}}1−m2v02​e2E02​t2​​λ0​​
  3. (C)λ0\lambda_0λ0​
  4. (D)λ01+e2E02t2m2v02\lambda_0\sqrt{1+\dfrac{e^2E_0^2t^2}{m^2v_0^2}}λ0​1+m2v02​e2E02​t2​​

Correct answer: (A)

Step-by-step solution →
Q18·Physics·Capacitors and DielectricsSingle correct
A parallel plate capacitor was made with two rectangular plates, each with a length of l=3l=3l=3 cm and breadth of b=1b=1b=1 cm. The distance between the plates is 3 μm. Out of the following, which are the ways to increase the capacitance by a factor of 10 ? A. l=30l=30l=30 cm, b=1b=1b=1 cm, d=1d=1d=1 μm. B. l=3l=3l=3 cm, b=1b=1b=1 cm, d=30d=30d=30 μm. C. l=6l=6l=6 cm, b=5b=5b=5 cm, d=3d=3d=3 μm. D. l=1l=1l=1 cm, b=1b=1b=1 cm, d=10d=10d=10 μm. E. l=5l=5l=5 cm, b=2b=2b=2 cm, d=1d=1d=1 μm. Choose the correct answer from the options given below :
  1. (A)C and E only
  2. (B)B and D only
  3. (C)A only
  4. (D)C only

Correct answer: (A)

Step-by-step solution →
Q19·Physics·Work, Energy and PowerSingle correct
A force F=α+βx2F=\alpha+\beta x^2F=α+βx2 acts on an object in the x-direction. The work done by the force is 5J when the object is displaced by 1 m. If the constant α=1\alpha=1α=1N then β will be
  1. (A)15 N/m2^22
  2. (B)10 N/m2^22
  3. (C)12 N/m2^22
  4. (D)8 N/m2^22

Correct answer: (C)

Step-by-step solution →
Q20·Physics·ThermodynamicsSingle correct
An ideal gas goes from an initial state to final state. During the process, the pressure of gas increases linearly with temperature. A. The work done by gas during the process is zero. B. The heat added to gas is different from change in its internal energy. C. The volume of the gas is increased. D. The internal energy of the gas is increased. E. The process is isochoric (constant volume process). Choose the correct answer from the options given below :-
  1. (A)A, B, C, D Only
  2. (B)A, D, E Only
  3. (C)E Only
  4. (D)A, C Only

Correct answer: (B)

Step-by-step solution →
Q21·Physics·Electric Field and Coulomb's LawInteger
A square loop of sides a = 1 m is held normally in front of a point charge q = 1C. The flux of the electric field through the shaded region is 5p×1ε0\dfrac{5}{p}\times\dfrac{1}{\varepsilon_0}p5​×ε0​1​ Nm2^22/C, where the value of p is _______.

Correct answer: 48

Step-by-step solution →
Q22·Physics·Experimental SkillsInteger
The least count of a screw gauge is 0.01 mm. If the pitch is increased by 75% and number of divisions on the circular scale is reduced by 50%, the new least count will be _______ ×10−3\times10^{-3}×10−3 mm.

Correct answer: 35

Step-by-step solution →
Q23·Physics·Current ElectricityInteger
A wire of resistance 9 Ω is bent to form an equilateral triangle. Then the equivalent resistance across any two vertices will be _______ ohm.

Correct answer: 2

Step-by-step solution →
Q24·Physics·Magnetic Field of CurrentInteger
A current of 5A exists in a square loop of side 12\dfrac{1}{\sqrt2}2​1​ m. Then the magnitude of the magnetic field B at the centre of the square loop will be p×10−6p\times10^{-6}p×10−6 T, where value of p is _______. [Take μ0=4π×10−7\mu_0=4\pi\times10^{-7}μ0​=4π×10−7 T m A−1^{-1}−1].

Correct answer: 8

Step-by-step solution →
Q25·Physics·Kinetic Theory of GasesInteger
The temperature of 1 mole of an ideal monoatomic gas is increased by 50°C at constant pressure. The total heat added and change in internal energy are E1E_1E1​ and E2E_2E2​, respectively. If E1E2=x9\dfrac{E_1}{E_2}=\dfrac{x}{9}E2​E1​​=9x​ then the value of x is _______.

Correct answer: 15

Step-by-step solution →

Chemistry — JEE Main 24 January 2025 Shift 1

Q26·Chemistry·Redox Reactions and ElectrochemistrySingle correct
For the given cell Fe2+(aq)+Ag+(aq)→Fe3+(aq)+Ag(s)Fe^{2+}(aq)+Ag^+(aq)\to Fe^{3+}(aq)+Ag(s)Fe2+(aq)+Ag+(aq)→Fe3+(aq)+Ag(s). The standard cell potential of the above reaction is. Given : Ag++e−→AgAg^++e^-\to AgAg++e−→Ag, E0=xE^0=xE0=x V; Fe2++2e−→FeFe^{2+}+2e^-\to FeFe2++2e−→Fe, E0=yE^0=yE0=y V; Fe3++3e−→FeFe^{3+}+3e^-\to FeFe3++3e−→Fe, E0=zE^0=zE0=z V.
  1. (A)x+y−zx+y-zx+y−z
  2. (B)x+2y−3zx+2y-3zx+2y−3z
  3. (C)y−2xy-2xy−2x
  4. (D)x+2yx+2yx+2y

Correct answer: (B)

Step-by-step solution →
Q27·Chemistry·HydrocarbonsSingle correct
Following are the four molecules "P", "Q", "R" and "S" (shown in the figure). Which one among the four molecules will react with H-Br(aq) at the fastest rate ?
  1. (A)S
  2. (B)Q
  3. (C)R
  4. (D)P

Correct answer: (B)

Step-by-step solution →
Q28·Chemistry·Coordination CompoundsSingle correct
One mole of the octahedral complex compound Co(NH₃)₅Cl₃ gives 3 moles of ions on dissolution in water. One mole of the same complex reacts with excess of AgNO₃ solution to yield two moles of AgCl(s). The structure of the complex is :
  1. (A)[Co(NH₃)₅Cl]Cl₂
  2. (B)[Co(NH₃)₄Cl]Cl₂·NH₃
  3. (C)[Co(NH₃)₄Cl₂]Cl·NH₃
  4. (D)[Co(NH₃)₅Cl₃]·2NH₃

Correct answer: (A)

Step-by-step solution →
Q29·Chemistry·Electronic Effects and StabilitySingle correct
Which one of the carbocations from the following is most stable ?
  1. (A)C+H2−CH=CH−CH2−O−CH3\overset{+}{C}H_2-CH=CH-CH_2-O-CH_3C+H2​−CH=CH−CH2​−O−CH3​
  2. (B)C+H2−CH=CH−O−CH3\overset{+}{C}H_2-CH=CH-O-CH_3C+H2​−CH=CH−O−CH3​
  3. (C)C+H2−CH=CH−O−CO−CH3\overset{+}{C}H_2-CH=CH-O-CO-CH_3C+H2​−CH=CH−O−CO−CH3​
  4. (D)C+H2−CH=CH−F\overset{+}{C}H_2-CH=CH-FC+H2​−CH=CH−F

Correct answer: (B)

Step-by-step solution →
Q30·Chemistry·Chemical Bonding and Molecular StructureSingle correct
Which of the following linear combination of atomic orbitals will lead to formation of molecular orbitals in homonuclear diatomic molecules [internuclear axis in z-direction] ? A. 2pz2p_z2pz​ and 2px2p_x2px​ B. 2s and 2px2p_x2px​ C. 3dxy3d_{xy}3dxy​ and 3dx2−y23d_{x^2-y^2}3dx2−y2​ D. 2s and 2pz2p_z2pz​ E. 2pz2p_z2pz​ and 3dz23d_{z^2}3dz2​. Choose the correct answer from the options given below :
  1. (A)E Only
  2. (B)A and B Only
  3. (C)D Only
  4. (D)C and D Only

Correct answer: (C)

Step-by-step solution →
Q31·Chemistry·d- and f-Block ElementsSingle correct
Which of the following ions is the strongest oxidizing agent ? (Atomic Number of Ce = 58, Eu = 63, Tb = 65, Lu = 71)
  1. (A)Lu3+Lu^{3+}Lu3+
  2. (B)Eu2+Eu^{2+}Eu2+
  3. (C)Tb4+Tb^{4+}Tb4+
  4. (D)Ce3+Ce^{3+}Ce3+

Correct answer: (C)

Step-by-step solution →
Q32·Chemistry·EquilibriumSingle correct
Ksp for Cr(OH)₃ is 1.6×10−301.6\times10^{-30}1.6×10−30. What is the molar solubility of this salt in water?
  1. (A)1.6×10−30274\sqrt[4]{\dfrac{1.6\times10^{-30}}{27}}4271.6×10−30​​
  2. (B)1.8×10−101.8\times10^{-10}1.8×10−10
  3. (C)1.8×10−303\sqrt[3]{1.8\times10^{-30}}31.8×10−30​
  4. (D)1.6×10−30\sqrt{1.6\times10^{-30}}1.6×10−30​

Correct answer: (A)

Step-by-step solution →
Q33·Chemistry·Chemical ThermodynamicsSingle correct
Let us consider an endothermic reaction which is non-spontaneous at the freezing point of water. However, the reaction is spontaneous at boiling point of water. Choose the correct option.
  1. (A)Both ΔH and ΔS are (+ve)
  2. (B)ΔH is (−ve) but ΔS is (+ve)
  3. (C)ΔH is (+ve) but ΔS is (−ve)
  4. (D)Both ΔH and ΔS are (−ve)

Correct answer: (A)

Step-by-step solution →
Q34·Chemistry·Purification and Characterisation of Organic CompoundsSingle correct
Given below are two statements I and II. Statement I : Dumas method is used for estimation of "Nitrogen" in an organic compound. Statement II : Dumas method involves the formation of ammonium sulphate by heating the organic compound with conc. H₂SO₄. 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 →
Q35·Chemistry·Classification of Elements and Periodicity in PropertiesSingle correct
Which of the following Statements are NOT true about the periodic table? A. The properties of elements are function of atomic weights. B. The properties of elements are function of atomic numbers. C. Elements having similar outer electronic configuration are arranged in same period. D. An element's location reflects the quantum numbers of the last filled orbital. E. The number of elements in a period is same as the number of atomic orbitals available in energy level that is being filled. Choose the correct answer from the options given below :
  1. (A)A, C and E Only
  2. (B)D and E Only
  3. (C)A and E Only
  4. (D)B, C and E Only

Correct answer: (A)

Step-by-step solution →
Q36·Chemistry·BiomoleculesSingle correct
The carbohydrate "Ribose" present in DNA, is A. A pentose sugar B. present in pyranose form C. in "D" configuration D. a reducing sugar, when free E. in α-anomeric form. Choose the correct answer from the options given below :
  1. (A)A, C and D Only
  2. (B)A, B and E Only
  3. (C)B, D and E Only
  4. (D)A, D and E Only

Correct answer: (A)

Step-by-step solution →
Q37·Chemistry·d- and f-Block ElementsSingle correct
Preparation of potassium permanganate from MnO₂ involves two step process in which the 1st step is a reaction with KOH and KNO₃ to produce
  1. (A)K4[Mn(OH)6]K_4[Mn(OH)_6]K4​[Mn(OH)6​]
  2. (B)K3MnO4K_3MnO_4K3​MnO4​
  3. (C)KMnO4KMnO_4KMnO4​
  4. (D)K2MnO4K_2MnO_4K2​MnO4​

Correct answer: (D)

Step-by-step solution →
Q38·Chemistry·p-Block ElementsSingle correct
The large difference between the melting and boiling points of oxygen and sulphur may be explained on the basis of
  1. (A)Atomic size
  2. (B)Atomicity
  3. (C)Electronegativity
  4. (D)Electron gain enthalpy

Correct answer: (B)

Step-by-step solution →
Q39·Chemistry·Chemical KineticsSingle correct
For a reaction, N2O5(g)→2NO2(g)+12O2(g)N_2O_5(g)\to 2NO_2(g)+\dfrac{1}{2}O_2(g)N2​O5​(g)→2NO2​(g)+21​O2​(g) in a constant volume container, no products were present initially. The final pressure of the system when 50% of reaction gets completed is
  1. (A)72\dfrac{7}{2}27​ times of initial pressure
  2. (B)5 times of initial pressure
  3. (C)52\dfrac{5}{2}25​ times of initial pressure
  4. (D)74\dfrac{7}{4}47​ times of initial pressure

Correct answer: (D)

Step-by-step solution →
Q40·Chemistry·Aldehydes and KetonesSingle correct
Which of the following arrangements with respect to their reactivity in nucleophilic addition reaction is correct?
  1. (A)benzaldehyde < acetophenone < p-nitrobenzaldehyde < p-tolualdehyde
  2. (B)acetophenone < benzaldehyde < p-tolualdehyde < p-nitrobenzaldehyde
  3. (C)acetophenone < p-tolualdehyde < benzaldehyde < p-nitrobenzaldehyde
  4. (D)p-nitrobenzaldehyde < benzaldehyde < p-tolualdehyde < acetophenone

Correct answer: (C)

Step-by-step solution →
Q41·Chemistry·HydrocarbonsSingle correct
Aman has been asked to synthesise the molecule (x) shown in the figure. He thought of preparing the molecule using an aldol condensation reaction. He found a few cyclic alkenes in his laboratory. He thought of performing ozonolysis reaction on alkene to produce a dicarbonyl compound followed by aldol reaction to prepare "x". Predict the suitable alkene that can lead to the formation of "x".
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (A)

Step-by-step solution →
Q42·Chemistry·SolutionsSingle correct
Consider the given plots of vapour pressure (VP) vs temperature (T/K). Which amongst the following options is correct graphical representation showing ΔTf\Delta T_fΔTf​, depression in the freezing point of solvent in a solution ?
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (C)

Step-by-step solution →
Q43·Chemistry·Chemical Bonding and Molecular StructureSingle correct
Which of the following statement is true with respect to H₂O, NH₃ and CH₄? A. The central atoms of all the molecules are sp³ hybridized. B. The H-O-H, H-N-H and H-C-H angles in the above molecules are 104.5°, 107.5° and 109.5° respectively. C. The increasing order of dipole moment is CH₄ < NH₃ < H₂O. D. Both H₂O and NH₃ are Lewis acids and CH₄ a Lewis base. E. A solution of NH₃ in H₂O is basic. In this solution NH₃ and H₂O act as Lowry-Bronsted acid and base respectively. Choose the correct answer from the options given below :
  1. (A)A, B and C only
  2. (B)C, D and E only
  3. (C)A, D and E only
  4. (D)A, B, C and E only

Correct answer: (A)

Step-by-step solution →
Q44·Chemistry·Organic Compounds Containing HalogensSingle correct
Given below are two statements : Statement-I : The conversion proceeds well in the less polar medium. CH3CH2CH2CH2Cl→HO−CH3CH2CH2CH2OH+Cl−CH_3CH_2CH_2CH_2Cl\xrightarrow{HO^-}CH_3CH_2CH_2CH_2OH+Cl^-CH3​CH2​CH2​CH2​ClHO−​CH3​CH2​CH2​CH2​OH+Cl−. Statement-II : The conversion proceeds well in the more polar medium. CH3CH2CH2CH2Cl→R3NCH3CH2CH2CH2N+R3 Cl−CH_3CH_2CH_2CH_2Cl\xrightarrow{R_3N}CH_3CH_2CH_2CH_2\overset{+}{N}R_3\,Cl^-CH3​CH2​CH2​CH2​ClR3​N​CH3​CH2​CH2​CH2​N+R3​Cl−. In the light of the above statements, choose the correct answer 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 false but statement II is true
  4. (D)Statement I is true but statement II is false

Correct answer: (A)

Step-by-step solution →
Q45·Chemistry·Aldehydes and KetonesSingle correct
The product (A) formed in the following reaction sequence is : CH3−C≡CH→(i) Hg2+/H2SO4, (ii) HCN, (iii) H2/NiCH_3{-}C\equiv CH\xrightarrow{\text{(i) Hg}^{2+}/\text{H}_2\text{SO}_4,\ \text{(ii) HCN},\ \text{(iii) H}_2/\text{Ni}}CH3​−C≡CH(i) Hg2+/H2​SO4​, (ii) HCN, (iii) H2​/Ni​ (A)
  1. (A)(CH3)2C(NH2)−CH2OH(CH_3)_2C(NH_2){-}CH_2OH(CH3​)2​C(NH2​)−CH2​OH
  2. (B)(CH3)2C(OH)−CH2NH2(CH_3)_2C(OH){-}CH_2NH_2(CH3​)2​C(OH)−CH2​NH2​
  3. (C)CH3CH2CH(NH2)−CH2OHCH_3CH_2CH(NH_2){-}CH_2OHCH3​CH2​CH(NH2​)−CH2​OH
  4. (D)CH3CH2CH(OH)−CH2NH2CH_3CH_2CH(OH){-}CH_2NH_2CH3​CH2​CH(OH)−CH2​NH2​

Correct answer: (B)

Step-by-step solution →
Q46·Chemistry·EquilibriumInteger
37.8 g N₂O₅ was taken in a 1 L reaction vessel and allowed to undergo the following reaction at 500 K: 2N2O5(g)→2N2O4(g)+O2(g)2N_2O_5(g)\to 2N_2O_4(g)+O_2(g)2N2​O5​(g)→2N2​O4​(g)+O2​(g). The total pressure at equilibrium was found to be 18.65 bar. Then, Kp=K_p=Kp​= _______ ×10−2\times10^{-2}×10−2 [nearest integer]. Assume N₂O₅ to behave ideally under these conditions. Given : R = 0.082 bar L mol⁻¹ K⁻¹

Correct answer: 962

Step-by-step solution →
Q47·Chemistry·Chemical ThermodynamicsInteger
Standard entropies of X₂, Y₂ and XY₅ are 70, 50 and 110 J K⁻¹ mol⁻¹ respectively. The temperature in Kelvin at which the reaction 12X2+52Y2→XY5\dfrac{1}{2}X_2+\dfrac{5}{2}Y_2\to XY_521​X2​+25​Y2​→XY5​, ΔH−=−35\Delta H^-=-35ΔH−=−35 kJ mol⁻¹, will be at equilibrium is _______ (Nearest integer)

Correct answer: 700

Step-by-step solution →
Q48·Chemistry·Some Basic Concepts in ChemistryInteger
X g of benzoic acid on reaction with aq. NaHCO₃ release CO₂ that occupied 11.2 L volume at STP. X is _______ g.

Correct answer: 61

Step-by-step solution →
Q49·Chemistry·Principles of Qualitative AnalysisInteger
Among the following cations, the number of cations which will give characteristic precipitate in their identification tests with K4[Fe(CN)6]K_4[Fe(CN)_6]K4​[Fe(CN)6​] is : Cu2+, Fe3+, Ba2+, Ca2+, NH4+, Mg2+, Zn2+Cu^{2+},\ Fe^{3+},\ Ba^{2+},\ Ca^{2+},\ NH_4^+,\ Mg^{2+},\ Zn^{2+}Cu2+, Fe3+, Ba2+, Ca2+, NH4+​, Mg2+, Zn2+

Correct answer: 3

Step-by-step solution →
Q50·Chemistry·Some Basic Concepts in ChemistryInteger
Consider the following reaction occurring in the blast furnace. Fe3O4(s)+4CO(g)→3Fe(l)+4CO2(g)Fe_3O_4(s)+4CO(g)\to 3Fe(l)+4CO_2(g)Fe3​O4​(s)+4CO(g)→3Fe(l)+4CO2​(g). ‘x’ kg of iron is produced when 2.32×1032.32\times10^32.32×103 kg Fe₃O₄ and 2.8×1022.8\times10^22.8×102 kg CO are brought together in the furnace. The value of ‘x’ is _______ (nearest integer). [Given : Molar mass of Fe₃O₄ = 232 g mol⁻¹, Molar mass of CO = 28 g mol⁻¹, Molar mass of Fe = 56 g mol⁻¹]

Correct answer: 420

Step-by-step solution →

Mathematics — JEE Main 24 January 2025 Shift 1

Q51·Mathematics·Vector AlgebraSingle correct
Let a⃗=i^+2j^+3k^\vec a=\hat i+2\hat j+3\hat ka=i^+2j^​+3k^, b⃗=3i^+j^−k^\vec b=3\hat i+\hat j-\hat kb=3i^+j^​−k^ and c⃗\vec cc be three vectors such that c⃗\vec cc is coplanar with a⃗\vec aa and b⃗\vec bb. If the vector c⃗\vec cc is perpendicular to b⃗\vec bb and a⃗⋅c⃗=5\vec a\cdot\vec c=5a⋅c=5, then ∣c⃗∣|\vec c|∣c∣ is equal to
  1. (A)132\dfrac{1}{3\sqrt2}32​1​
  2. (B)181818
  3. (C)161616
  4. (D)116\sqrt{\dfrac{11}{6}}611​​

Correct answer: (D)

Step-by-step solution →
Q52·Mathematics·Definite IntegrationSingle correct
In I(m,n)=∫01xm−1(1−x)n−1dxI(m,n)=\int_0^1 x^{m-1}(1-x)^{n-1}dxI(m,n)=∫01​xm−1(1−x)n−1dx, m,n>0m,n>0m,n>0, then I(9,14)+I(10,13)I(9,14)+I(10,13)I(9,14)+I(10,13) is
  1. (A)I(9,1)I(9,1)I(9,1)
  2. (B)I(19,27)I(19,27)I(19,27)
  3. (C)I(1,13)I(1,13)I(1,13)
  4. (D)I(9,13)I(9,13)I(9,13)

Correct answer: (D)

Step-by-step solution →
Q53·Mathematics·Limits and ContinuitySingle correct
Let f:R−{0}→Rf:\mathbb{R}-\{0\}\to\mathbb{R}f:R−{0}→R be a function such that f(x)−6f(1x)=353x−52f(x)-6f\left(\dfrac{1}{x}\right)=\dfrac{35}{3x}-\dfrac{5}{2}f(x)−6f(x1​)=3x35​−25​. If lim⁡x→0(1αx+f(x))=β\lim_{x\to 0}\left(\dfrac{1}{\alpha x}+f(x)\right)=\betalimx→0​(αx1​+f(x))=β; α,β∈R\alpha,\beta\in\mathbb{R}α,β∈R, then α+2β\alpha+2\betaα+2β is equal to
  1. (A)333
  2. (B)555
  3. (C)444
  4. (D)666

Correct answer: (C)

Step-by-step solution →
Q54·Mathematics·Sequence and SeriesSingle correct
Let Sn=12+16+112+120+…S_n=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\ldotsSn​=21​+61​+121​+201​+… upto n terms. If the sum of the first six terms of an A.P. with first term −p-p−p and common difference ppp is 2026 S2025\sqrt{2026\,S_{2025}}2026S2025​​, then the absolute difference between 20th20^{th}20th and 15th15^{th}15th terms of the A.P. is
  1. (A)252525
  2. (B)909090
  3. (C)202020
  4. (D)454545

Correct answer: (A)

Step-by-step solution →
Q55·Mathematics·Sets, Relations and FunctionsSingle correct
Let f(x)=2x+2+1622x+1+2x+4+32f(x)=\dfrac{2^{x+2}+16}{2^{2x+1}+2^{x+4}+32}f(x)=22x+1+2x+4+322x+2+16​. Then the value of 8(f(115)+f(215)+…+f(5915))8\left(f\left(\dfrac{1}{15}\right)+f\left(\dfrac{2}{15}\right)+\ldots+f\left(\dfrac{59}{15}\right)\right)8(f(151​)+f(152​)+…+f(1559​)) is equal to
  1. (A)118118118
  2. (B)929292
  3. (C)102102102
  4. (D)108108108

Correct answer: (A)

Step-by-step solution →
Q56·Mathematics·Complex NumbersSingle correct
If α\alphaα and β\betaβ are the roots of the equation 2z2−3z−2i=02z^2-3z-2i=02z2−3z−2i=0, where i=−1i=\sqrt{-1}i=−1​, then 16⋅Re(α19+β19+α11+β11α15+β15)⋅Im(α19+β19+α11+β11α15+β15)16\cdot\mathrm{Re}\left(\dfrac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}\right)\cdot\mathrm{Im}\left(\dfrac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}\right)16⋅Re(α15+β15α19+β19+α11+β11​)⋅Im(α15+β15α19+β19+α11+β11​) is equal to
  1. (A)398398398
  2. (B)312312312
  3. (C)409409409
  4. (D)441441441

Correct answer: (D)

Step-by-step solution →
Q57·Mathematics·Limits and ContinuitySingle correct
lim⁡x→0cosec⁡x(2cos⁡2x+3cos⁡x−cos⁡2x+sin⁡x+4)\lim_{x\to 0}\operatorname{cosec}x\left(\sqrt{2\cos^2 x+3\cos x}-\sqrt{\cos^2 x+\sin x+4}\right)limx→0​cosecx(2cos2x+3cosx​−cos2x+sinx+4​) is
  1. (A)000
  2. (B)125\dfrac{1}{2\sqrt5}25​1​
  3. (C)115\dfrac{1}{\sqrt{15}}15​1​
  4. (D)−125-\dfrac{1}{2\sqrt5}−25​1​

Correct answer: (D)

Step-by-step solution →
Q58·Mathematics·Three Dimensional GeometrySingle correct
Let in a △ABC\triangle ABC△ABC, the length of the side AC be 6, the vertex B be (1,2,3)(1,2,3)(1,2,3) and the vertices A, C lie on the line x−63=y−72=z−7−2\dfrac{x-6}{3}=\dfrac{y-7}{2}=\dfrac{z-7}{-2}3x−6​=2y−7​=−2z−7​. Then the area (in sq. units) of △ABC\triangle ABC△ABC is
  1. (A)424242
  2. (B)212121
  3. (C)565656
  4. (D)171717

Correct answer: (B)

Step-by-step solution →
Q59·Mathematics·Differential EquationsSingle correct
Let y=y(x)y=y(x)y=y(x) be the solution of the differential equation (xy−5x21+x2)dx+(1+x2)dy=0(xy-5x^2\sqrt{1+x^2})dx+(1+x^2)dy=0(xy−5x21+x2​)dx+(1+x2)dy=0, y(0)=0y(0)=0y(0)=0. Then y(3)y(\sqrt3)y(3​) is equal to
  1. (A)532\dfrac{5\sqrt3}{2}253​​
  2. (B)143\sqrt{\dfrac{14}{3}}314​​
  3. (C)222\sqrt222​
  4. (D)152\sqrt{\dfrac{15}{2}}215​​

Correct answer: (A)

Step-by-step solution →
Q60·Mathematics·EllipseSingle correct
Let the product of the focal distances of the point (3,12)\left(\sqrt3,\dfrac{1}{2}\right)(3​,21​) on the ellipse x2a2+y2b2=1\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1a2x2​+b2y2​=1, (a>b)(a>b)(a>b), be 74\dfrac{7}{4}47​. Then the absolute difference of the eccentricities of two such ellipses is
  1. (A)3−2232\dfrac{3-2\sqrt2}{3\sqrt2}32​3−22​​
  2. (B)1−32\dfrac{1-\sqrt3}{\sqrt2}2​1−3​​
  3. (C)3−2223\dfrac{3-2\sqrt2}{2\sqrt3}23​3−22​​
  4. (D)1−23\dfrac{1-\sqrt2}{\sqrt3}3​1−2​​

Correct answer: (C)

Step-by-step solution →
Q61·Mathematics·Statistics and ProbabilitySingle correct
A and B alternately throw a pair of dice. A wins if he throws a sum of 5 before B throws a sum of 8, and B wins if he throws a sum of 8 before A throws a sum of 5. The probability, that A wins if A makes the first throw, is
  1. (A)917\dfrac{9}{17}179​
  2. (B)919\dfrac{9}{19}199​
  3. (C)817\dfrac{8}{17}178​
  4. (D)819\dfrac{8}{19}198​

Correct answer: (B)

Step-by-step solution →
Q62·Mathematics·Limits and ContinuitySingle correct
Consider the region R={(x,y):x≤y≤9−113x2, x≥0}R=\left\{(x,y):x\le y\le 9-\dfrac{11}{3}x^2,\ x\ge 0\right\}R={(x,y):x≤y≤9−311​x2, x≥0}. The area, of the largest rectangle of sides parallel to the coordinate axes and inscribed in R, is :
  1. (A)625111\dfrac{625}{111}111625​
  2. (B)730119\dfrac{730}{119}119730​
  3. (C)567121\dfrac{567}{121}121567​
  4. (D)821123\dfrac{821}{123}123821​

Correct answer: (C)

Step-by-step solution →
Q63·Mathematics·Area Under CurvesSingle correct
The area of the region {(x,y):x2+4x+2≤y≤∣x+2∣}\{(x,y):x^2+4x+2\le y\le|x+2|\}{(x,y):x2+4x+2≤y≤∣x+2∣} is equal to
  1. (A)777
  2. (B)245\dfrac{24}{5}524​
  3. (C)203\dfrac{20}{3}320​
  4. (D)555

Correct answer: (C)

Step-by-step solution →
Q64·Mathematics·Statistics and ProbabilitySingle correct
For a statistical data x1,x2,…,x10x_1,x_2,\ldots,x_{10}x1​,x2​,…,x10​ of 10 values, a student obtained the mean as 5.5 and ∑i=110xi2=371\sum_{i=1}^{10}x_i^2=371∑i=110​xi2​=371. He later found that he had noted two values in the data incorrectly as 4 and 5, instead of the correct values 6 and 8, respectively. The variance of the corrected data is
  1. (A)777
  2. (B)444
  3. (C)999
  4. (D)555

Correct answer: (A)

Step-by-step solution →
Q65·Mathematics·CirclesSingle correct
Let circle C be the image of x2+y2−2x+4y−4=0x^2+y^2-2x+4y-4=0x2+y2−2x+4y−4=0 in the line 2x−3y+5=02x-3y+5=02x−3y+5=0 and A be the point on C such that OA is parallel to x-axis and A lies on the right hand side of the centre O of C. If B(α,β)B(\alpha,\beta)B(α,β), with β<4\beta<4β<4, lies on C such that the length of the arc AB is 16\dfrac{1}{6}61​th of the perimeter of C, then β−3α\beta-\sqrt3\alphaβ−3​α is equal to
  1. (A)333
  2. (B)3+33+\sqrt33+3​
  3. (C)4−34-\sqrt34−3​
  4. (D)444

Correct answer: (D)

Step-by-step solution →
Q66·Mathematics·Binomial Theorem and Its Simple ApplicationsSingle correct
For some n≠10n\ne 10n=10, let the coefficients of the 5th5^{th}5th, 6th6^{th}6th and 7th7^{th}7th terms in the binomial expansion of (1+x)n+4(1+x)^{n+4}(1+x)n+4 be in A.P. Then the largest coefficient in the expansion of (1+x)n+4(1+x)^{n+4}(1+x)n+4 is :
  1. (A)707070
  2. (B)353535
  3. (C)202020
  4. (D)101010

Correct answer: (B)

Step-by-step solution →
Q67·Mathematics·Quadratic EquationsSingle correct
The product of all the rational roots of the equation (x2−9x+11)2−(x−4)(x−5)=3(x^2-9x+11)^2-(x-4)(x-5)=3(x2−9x+11)2−(x−4)(x−5)=3, is equal to :
  1. (A)141414
  2. (B)777
  3. (C)282828
  4. (D)212121

Correct answer: (A)

Step-by-step solution →
Q68·Mathematics·Three Dimensional GeometrySingle correct
Let the line passing through the points (−1,2,1)(-1,2,1)(−1,2,1) and parallel to the line x−12=y+13=z4\dfrac{x-1}{2}=\dfrac{y+1}{3}=\dfrac{z}{4}2x−1​=3y+1​=4z​ intersect the line x+23=y−32=z−41\dfrac{x+2}{3}=\dfrac{y-3}{2}=\dfrac{z-4}{1}3x+2​=2y−3​=1z−4​ at the point P. Then the distance of P from the point Q(4,−5,1)Q(4,-5,1)Q(4,−5,1) is :
  1. (A)555
  2. (B)101010
  3. (C)565\sqrt656​
  4. (D)555\sqrt555​

Correct answer: (D)

Step-by-step solution →
Q69·Mathematics·Straight LinesSingle correct
Let the lines 3x−4y−α=03x-4y-\alpha=03x−4y−α=0, 8x−11y−33=08x-11y-33=08x−11y−33=0, and 2x−3y+λ=02x-3y+\lambda=02x−3y+λ=0 be concurrent. If the image of the point (1,2)(1,2)(1,2) in the line 2x−3y+λ=02x-3y+\lambda=02x−3y+λ=0 is (5713,−4013)\left(\dfrac{57}{13},\dfrac{-40}{13}\right)(1357​,13−40​), then ∣αλ∣|\alpha\lambda|∣αλ∣ is equal to :
  1. (A)848484
  2. (B)919191
  3. (C)113113113
  4. (D)101101101

Correct answer: (B)

Step-by-step solution →
Q70·Mathematics·Matrices and DeterminantsSingle correct
If the system of equations 2x−y+z=42x-y+z=42x−y+z=4, 5x+λy+3z=125x+\lambda y+3z=125x+λy+3z=12, 100x−47y+μz=212100x-47y+\mu z=212100x−47y+μz=212, has infinitely many solutions, then μ−2λ\mu-2\lambdaμ−2λ is equal to
  1. (A)565656
  2. (B)595959
  3. (C)555555
  4. (D)575757

Correct answer: (D)

Step-by-step solution →
Q71·Mathematics·Differential EquationsInteger
Let f be a differentiable function such that 2(x+2)2f(x)−3(x+2)2=10∫0x(t+2)f(t) dt2(x+2)^2 f(x)-3(x+2)^2=10\int_0^x (t+2)f(t)\,dt2(x+2)2f(x)−3(x+2)2=10∫0x​(t+2)f(t)dt, x≥0x\ge 0x≥0. Then f(2)f(2)f(2) is equal to _______

Correct answer: 19

Step-by-step solution →
Q72·Mathematics·Inverse Trigonometric FunctionsInteger
If for some α,β\alpha,\betaα,β; α≤β\alpha\le\betaα≤β, α+β=8\alpha+\beta=8α+β=8 and sec⁡2(tan⁡−1α)+cosec⁡2(cot⁡−1β)=36\sec^2(\tan^{-1}\alpha)+\operatorname{cosec}^2(\cot^{-1}\beta)=36sec2(tan−1α)+cosec2(cot−1β)=36, then α2+β\alpha^2+\betaα2+β is _______

Correct answer: 14

Step-by-step solution →
Q73·Mathematics·Permutations and CombinationsInteger
The number of 3-digit numbers, that are divisible by 2 and 3, but not divisible by 4 and 9, is _______

Correct answer: 125

Step-by-step solution →
Q74·Mathematics·Matrices and DeterminantsInteger
Let A be a 3×33\times33×3 matrix such that XTAX=OX^T AX=OXTAX=O for all nonzero 3×13\times13×1 matrices X=[xyz]X=\begin{bmatrix}x\\y\\z\end{bmatrix}X=​xyz​​. If A[111]=[14−5]A\begin{bmatrix}1\\1\\1\end{bmatrix}=\begin{bmatrix}1\\4\\-5\end{bmatrix}A​111​​=​14−5​​, A[121]=[04−8]A\begin{bmatrix}1\\2\\1\end{bmatrix}=\begin{bmatrix}0\\4\\-8\end{bmatrix}A​121​​=​04−8​​, and det⁡(adj⁡(2(A+I)))=2α3β5γ\det(\operatorname{adj}(2(A+I)))=2^{\alpha}3^{\beta}5^{\gamma}det(adj(2(A+I)))=2α3β5γ, α,β,γ∈N\alpha,\beta,\gamma\in\mathbb{N}α,β,γ∈N, then α2+β2+γ2\alpha^2+\beta^2+\gamma^2α2+β2+γ2 is _______

Correct answer: 44

Step-by-step solution →
Q75·Mathematics·Permutations and CombinationsInteger
Let S={p1,p2,…,p10}S=\{p_1,p_2,\ldots,p_{10}\}S={p1​,p2​,…,p10​} be the set of first ten prime numbers. Let A=S∪PA=S\cup PA=S∪P, where P is the set of all possible products of distinct elements of S. Then the number of all ordered pairs (x,y)(x,y)(x,y), x∈Sx\in Sx∈S, y∈Ay\in Ay∈A, such that x divides y, is _______

Correct answer: 5120

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
  • 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
  • Limits and Continuity 149/186
  • d- and f-Block Elements 126/186
  • Thermodynamics 154/186
  • Aldehydes and Ketones 135/186
  • Equilibrium 163/186
  • Solutions 158/186
  • Laws of Motion 130/186
  • Chemical Thermodynamics 165/186
  • Units and Measurements 149/186
  • Hydrocarbons 126/186
  • Complex Numbers 165/186
  • Biomolecules 162/186
  • Chemical Kinetics 169/186
  • Gravitation 152/186
  • Electric Field and Coulomb's Law 133/186
  • Electronic Devices 145/186
  • Circles 142/186
  • Dual Nature of Matter and Radiation 155/186
  • Quadratic Equations 148/186
  • Work, Energy and Power 132/186
  • Some Basic Concepts in Chemistry 129/186
  • Wave Optics 130/186
  • Area Under Curves 139/186
  • Kinetic Theory of Gases 135/186
  • Purification and Characterisation of Organic Compounds 116/186
  • Oscillations 117/186
  • Alternating Currents 108/186
  • Organic Compounds Containing Halogens 109/186
  • Straight Lines 114/186
  • Capacitors and Dielectrics 115/186
  • 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
  • Principles of Qualitative Analysis 58/186
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