JEE Main 2025 January 23, Shift 1 Question Paper with Solutions
All 75 questions from the JEE Main 2025 (January 23, Shift 1) shift — Physics (25), Chemistry (25) and Mathematics (25) — each with the correct answer (answer key) and a step-by-step solution. Read it free online, or attempt the full paper as a timed mock test.
A point particle of charge Q is located at P along the axis of an electric dipole 1 at a distance r as shown in the figure. The point P is also on the equatorial plane of a second electric dipole 2 at a distance r. The dipoles are made of opposite charge q separated by a distance 2a. For the charge particle at P not to experience any net force, which of the following correctly describes the situation?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4ra∼3
Approach:
Use exact electric field formulas for dipoles and equate magnitudes for zero net force on charge Q
Step 1:Define: Point P is at distance r from both dipoles. Dipole 1 has P on its axis, Dipole 2 has P on its equatorial plane. Both dipoles have moment p = 2aq
Given: dipole moment p=q(2a), charge separation 2a
Step 2:Exact electric field on axial line of dipole 1
E1=1=4πε01⋅(r2−a2)22pr
Step 3:Exact electric field on equatorial plane of dipole 2
E2=2=4πε01⋅(r2+a2)3/2p
Step 4:For zero net force, set E₁ = E₂
(r2−a2)22r=(r2+a2)3/21
Step 5:Let x = a/r and simplify by dividing by powers of r
r3(1−x2)22=r3(1+x2)3/21
Step 6:Test x = 3 (option 4)
LHS: 2(1+9)3/2=2(10)1.5=2×31.62≈63.24
Step 7:Calculate RHS for x = 3
RHS: (1−9)2=(−8)2=64
Step 8:Compare: LHS ≈ 63.24 and RHS = 64 are extremely close
∴ra∼3
Final answer: ra∼3
Q27Single correctOptics
A spherical surface of radius of curvature R, separates air from glass (refractive index = 1.5). The centre of curvature is in the glass medium. A point object 'O' placed in air on the optic axis of the surface, so that its real image is formed at 'I' inside glass. The line OI intersects the spherical surface at P and PO = PI. The distance PO equals to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 15R
Step 1:Using the refraction formula for a spherical surface
vμ2−uμ1=Rμ2−μ1
Step 2:Given that PO = PI, let this distance be x. Then object distance u = -x and image distance v = x
x1.5−−x1=R1.5−1
Step 3:Simplifying the equation
x1.5+x1=R0.5
Step 4:Solving for x
x=0.52.5R=5R
Final answer: 5R
Q28Single correctUnits and Measurements
The position of a particle moving on x-axis is given by x(t)=Asint+Bcos2t+Ct2+D, where t is time. The dimension of DABC is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1L2T−2
Step 1:Analyzing dimensions: Since x(t) has dimension of length [L], each term must have dimension [L]
[Asint]=[L]⇒[A]=[L]
Step 2:Similarly for the second term
[Bcos2t]=[L]⇒[B]=[L]
Step 3:For the third term
[Ct2]=[L]⇒[C]=[LT−2]
Step 4:The fourth term gives
[D] = [L]
Step 5:Computing the dimension ofDABC
[DABC]=[L][L][L][LT−2]=[L2T−2]
Final answer: L2T−2
Q29Single correctOptics
Given a thin convex lens (refractive index μ2), kept in a liquid (refractive index μ1, μ1<μ2) having radii of curvatures ∣R1∣ and ∣R2∣. Its second surface is silver polished. Where should an object be placed on the optic axis so that a real and inverted image is formed at the same place?
Step 1:For a lens with silvered back surface, the system acts as an equivalent mirror. For object and image to coincide, object must be at center of curvature
u=feq
Step 2:The power of the system is given by
Peq=PL+PM+PL=2PL+PM
Step 3:Using lens maker's formula in medium
fL1=μ1μ2−μ1(R11−R21)
Step 4:For the silvered surface acting as a mirror
fM=2R2
Step 5:Combining the effects and solving for object distance
u=μ2(∣R1∣+∣R2∣)−μ1∣R2∣μ1∣R1∣⋅∣R2∣
Final answer: μ2(∣R1∣+∣R2∣)−μ1∣R2∣μ1∣R1∣⋅∣R2∣
Q30Single correctCurrent Electricity
Refer to the circuit diagram given in the figure. Which of the following observations are correct? A. Total resistance of circuit is66ΩB. Current in Ammeter is 1 A C. Potential across AB is 4 Volts. D. Potential across CD is 4 Volts E. Total resistance of the circuit is88Ω.Choosethe correct answer from the options given below:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1A, B and D Only
Step 1:Analyzing the circuit: The diode is forward biased, so it conducts
Rdiode≈0
Step 2:Thetwo4Ω resistors in parallel give equivalent resistance
Rparallel=4+44×4=2Ω
Step 3:Total circuit resistance
Rtotal=4=4Ω+2Ω=6Ω
Step 4:Current in circuit
I=RtotalV=66=1 A
Step 5:Potential across AB(thetop4Ω resistor)
VAB=I×4=1×4=4 V
Step 6:Potential across CD (parallel combination)
VCD=I×2=1×2=2 V
Final answer: A, B and D Only
Q31Single correctProperties of Solids and Liquids
Given below are two statements: Statement I: The hot water flows faster than cold water Statement II: Soap water has higher surface tension as compared to fresh water. In the light above statements, choose the correct answer from the options given below
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1Statement I is true but Statement II is false
Step 1:Analyzing Statement I: Hot water has lower viscosity than cold water
ηhot<ηcold
Step 2:Analyzing Statement II: Soap acts as a surfactant and reduces surface tension
γsoap<γfresh
Step 3:Conclusion
Statement I is true but Statement II is false
Final answer: Statement I is true but Statement II is false
Q32Single correctRotational Motion
Consider a circular disc of radius 20 cm with centre located at the origin. A circular hole of radius 5 cm is cut from this disc in such a way that the edge of the hole touches the edge of the disc. The distance of centre of mass of residual or remaining disc from the origin will be
(A)
(B)
(C)
(D)
SolutionAnswer: Option 31.0 cm
Step 1:Let the large disc have radius R = 20 cm and the hole have radius r = 5 cm. The center of the hole is at distance d = R - r = 15 cm from origin
d=20−5=15=20−5=15 cm
Step 2:Using the concept of negative mass for the hole
xcm=M−mM⋅0−m⋅d
Step 3:The masses are proportional to areas
Mm=R2r2=40025=161
Step 4:Computing the center of mass shift
xcm=M−m−m⋅15=M−16M−16M⋅15=15−15=−1.0 cm
Final answer: 1.0 cm
Q33Single correctElectrostatics
The electric flux is ϕ=ασ+βλ where λ and σ are linear and surface charge density, respectively. (βα) represents
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4displacement
Step 1:Analyzing dimensions of electric flux
[ϕ]=[E][A]=[ϵ0][Q]
Step 2:For the first term ασ
[α][σ]=[ϕ]⇒[α][L2][Q]=[ϵ0][Q]
Step 3:For the second term βλ
[β][λ]=[ϕ]⇒[β][L][Q]=[ϵ0][Q]
Step 4:Finding the ratio βα
[βα]=[L]/[ϵ0][L2]/[ϵ0]=[L]
Final answer: displacement
Q34Single correctDual Nature of Matter and Radiation
A sub-atomic particle of mass 10−30 kg is moving with a velocity 2.21×106 m/s. Under the matter wave consideration, the particle will behave closely like (h=6.63×10−34 J.s)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4X-rays
Step 1:Calculate the de Broglie wavelength using λ = h/(mv)
λ=mvh=10−30×2.21×1066.63×10−34
Step 2:Compare wavelength with electromagnetic spectrum
λ=3×10−10 m is in the X-ray region (0.01-10 nm)
Step 3:Determine the type of radiation
3 A˚ = 0.3 nm falls in X-ray range
Final answer: 4
Q35Single correctMagnetic Effects of Current and Magnetism
Consider a moving coil galvanometer (MCG): A. The torsional constant in moving coil galvanometer has dimensions [ML2T−2] B. Increasing the current sensitivity may not necessarily increase the voltage sensitivity. C. If we increase number of turns (N) to its double (2N), then the voltage sensitivity doubles. D. MCG can be converted into an ammeter by introducing a shunt resistance of large value in parallel with galvanometer. E. Current sensitivity of MCG depends inversely on number of turns of coil. Choose the correct answer from the options given below:
Step 2:Analyze statement B: Current sensitivity vs voltage sensitivity
Is=CNAB,Vs=CRNAB
Step 3:Analyze statement C: Doubling N
Vs=CRNAB, if N→2N then Vs does not simply double as R also changes
Step 4:Analyze statement D: Ammeter conversion
Ammeter requires SMALL shunt resistance, not large
Step 5:Analyze statement E: Current sensitivity and N
Is=CNAB∝N
Final answer: 4
Q36Single correctThermodynamics
Match LIST-I with LIST-II
List - I
List - II
A. Pressure varies inversely with volume of an ideal gas.
I. Adiabatic process
B. Heat absorbed goes partly to increase internal energy and partly to do work.
II. Isochoric process
C. Heat is neither absorbed nor released by a system.
III. Isothermal process
D. No work is done on or by a gas.
IV. Isobaric process
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1A-III, B-IV, C-I, D-II
Step 1:Match A: Pressure varies inversely with volume
PV=constant (at constant T)⇒Isothermal process
Step 2:Match B: Heat absorbed increases internal energy and does work
dQ=dU+dW (First law)⇒Isobaric process
Step 3:Match C: Heat is neither absorbed nor released
dQ=0=0⇒Adiabatic process
Step 4:Match D: No work is done
dW=PdV=0=0⇒dV=0=0⇒Isochoric process
Final answer: 1
Q37Single correctElectromagnetic Waves
The electric field of an electromagnetic wave in free space is E=57cos[7.5×106t−5×10−3(3x+4y)](4i^−3j^) N/C. The associated magnetic field in Tesla is
Step 1:Find the wave vector k from the wave equation
k=5×10−3(3i^+4j^),∣k∣=5×10−3×5=2.5×10−2
Step 2:Calculate magnetic field using B = E/c and direction from cross product
B0=0=cE0=3×10857,B∝k^×E
Step 3:Find direction of magnetic field
k^×E=51(3i^+4j^)×(44i^−3j^)=51(−9−16)k^=−5k^
Step 4:Combine magnitude and direction
B=−3×10857cos[7.55×106t−5×10−3(3x+4y)](55k^)
Final answer: 3
Q38Single correctProperties of Solids and Liquids
A gun fires a lead bullet of temperature 300 K into a wooden block. The bullet having melting temperature of 600 K penetrates into the block and melts down. If the total heat required for the process is 625 J, then the mass of the bullet is ______ grams. (Latent heat of fusion of lead = 2.5×104 J Kg−1 and specific heat capacity of lead = 125 J Kg−1K−1)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 110
Step 1:Calculate heat required to raise temperature from 300 K to 600 K
Q1=mcΔT=m×125×(600−300)=m×125×300
Step 2:Calculate heat required for melting
Q2=mL=m×2.5×104
Step 3:Total heat required
Qtotal=Q1+Q2=37500m+25000m=62500m
Step 4:Solve for mass
m=62500625=0.01 kg=10 g
Final answer: 1
Q39Single correctOptics
What is the lateral shift of a ray refracted through a parallel-sided glass slab of thickness 'h' in terms of the angle of incidence 'i' and angle of refraction 'r', if the glass slab is placed in air medium?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2cosrhsin(i−r)
Step 1:Consider the path of light through the glass slab
Path length in slab=cosrh
Step 2:Calculate the lateral displacement
d=cosrhsin(i−r)
Step 3:Verify the formula
d=cosrhsin(i−r)
Final answer: 2
Q40Single correctAtoms and Nuclei
A radioactive nucleus n2 has 3 times the decay constant as compared to the decay constant of another radioactive nucleus n1. If initial number of both nuclei are the same, what is the ratio of number of nuclei of n2 to the number of nuclei of n1, after one half-life of n1?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 441
Step 1:Set up the decay constants
λ2=3λ1,t1/2(1)=λ1ln2
Step 2:Calculate number of nuclei of n1 after one half-life
N1(t)=N0e−−λ1t=N0e−−ln2=2N0
Step 3:Calculate number of nuclei of n2 after same time
A light hollow cube of side length 10 cm and mass 10 g, is floating in water. It is pushed down and released to execute simple harmonic oscillations. The time period of oscillations is yπ×10−2 s, where the value of y is (Acceleration due to gravity, g=10 m/s2, density of water = 103 kg/m3)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 22
Step 1:Set up the equation for SHM in floating object
T=2=2πρAgm
Step 2:Calculate the cross-sectional area
A=(0.1)2=0.0101 m2=10−2 m2
Step 3:Substitute values into time period formula
T=2=2π103×10−2×1010×10−3=2π10210−2
Step 4:Compare with given format
T=yπ×10−2=2π×10−2
Final answer: 2
Q42Single correctElectromagnetic Induction and Alternating Currents
Regarding self-inductance: A. The self-inductance of the coil depends on its geometry. B. Self-inductance does not depend on the permeability of the medium. C. Self-induced e.m.f. opposes any change in the current in a circuit. D. Self-inductance is electromagnetic analogue of mass in mechanics. E. Work needs to be done against self-induced e.m.f. in establishing the current. Choose the correct answer from the options given below:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3A, C, D, E only
Step 1:Analyze statement A: Self-inductance depends on coil geometry (number of turns, area, length)
L=lμN2A
Step 2:Analyze statement B: Self-inductance depends on permeability μ of the medium
L∝μ
Step 3:Analyze statement C: Self-induced emf opposes change in current (Lenz's law)
ε=−LdtdI
Step 4:Analyze statement D: Self-inductance is analogous to mass (inertia in mechanics)
Energy=21LI2 (analogous to 21mv2)
Step 5:Analyze statement E: Work must be done against back emf to establish current
W=21LI2
Final answer: A, C, D, E only
Q43Single correctKinematics
The motion of an airplane is represented by velocity-time graph as shown below. The distance covered by airplane in the first 30.5 second is _______ km.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 112
Step 1:Calculate distance during acceleration phase (0 to 10s)
s1=1=21(v1+v2)t1=1=21(200+400)×10=3000 m
Step 2:Calculate distance during constant velocity phase (10s to 30.5s)
s2=v×t2=400×(30.5−10)=400=400×20.5=8200 m
Step 3:Calculate total distance
stotal=s1+s2=3+8.2=11.22≈12 km
Final answer: 12 km
Q44Single correctElectrostatics
Identify the valid statements relevant to the given circuit at the instant when the key is closed. A. There will be no current through resistor R. B. There will be maximum current in the connecting wires. C. Potential difference between the capacitor plates A and B is minimum. D. Charge on the capacitor plates is minimum. Choose the correct answer from the options given below:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4B, C, D Only
Step 1:At the instant key is closed, capacitor acts as short circuit (uncharged)
Q=0=0 at t=0
Step 2:Analyze statement A: Current through R at t=0
I=RVe0
Step 3:Analyze statement B: Current in connecting wires
Imax=RV at t=0
Step 4:Analyze statement C: Potential difference across capacitor
VC=0=0 at t=0 (minimum)
Step 5:Analyze statement D: Charge on capacitor
Q=0=0 at t=0 (minimum)
Final answer: B, C, D Only
Q45Single correctRotational Motion
A solid sphere of mass 'm' and radius 'r' is allowed to roll without slipping from the highest point of an inclined plane of length 'L' and makes an angle 30∘ with the horizontal. The speed of the particle at the bottom of the plane is v1. If the angle of inclination is increased to 45∘ while keeping L constant. Then the new speed of the sphere at the bottom of the plane is v2. The ratio v12:v22 is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 11:2
Step 1:Apply energy conservation for rolling sphere
mgh=21mv2+21Iω2
Step 2:For solid sphere rolling without slipping
I=52mr2,v=rω
Step 3:Substitute and simplify
mgh=21mv2+21⋅52mr2⋅r2v2=107mv2
Step 4:Calculate height for angle 30° and 45°
h1=Lsin30∘=2L,h2=Lsin45∘=2L
Step 5:Find ratio of speeds
v22v12=h2h1=L/2L/2=21
Final answer: 1:2
Q46NumericalElectrostatics
A positive ion A and a negative ion B has charges 6.67×10−19 C and 9.6×10−10 C, and masses 19.2×10−27 kg and 9×10−27 kg respectively. At an instant, the ions are separated by a certain distance r. At that instant the ratio of the magnitudes of electrostatic force to gravitational force is P×1045, where the value of 10P is (Take 4πε01=9×109Nm2C−2 and universal gravitational constant as 6.67×10−11Nm2kg−2). Assume that charge may not be an integral multiple of electrons.
SolutionAnswer: 5
Step 1:Write expressions for electrostatic and gravitational forces
Q47NumericalElectromagnetic Induction and Alternating Currents
In the given circuit the sliding contact is pulled outwards such that electric current in the circuit changes at the rate of 8 A/s. At an instant when R is 12Ω, the value of the current in the circuit will be ______ A.
SolutionAnswer: 3
Step 1:Apply Kirchhoff's voltage law to the circuit
V=IR+LdtdI
Step 2:Substitute given values
12=I×12+3×8
Step 3:Solve for current I
12I = 12 - 24 = -12
Step 4:Reconsider back emf direction
12+LdtdI=IR⇒12+24=12I
Step 5:Calculate current
I=1236=3 A
Final answer: 3
Q48NumericalVector Algebra
Two particles are located at equal distance from origin. The position vectors of those are represented by A=2i^+3nj^+2k^ and B=2i^−2j^+4pk^, respectively. If both the vectors are at right angle to each other, the value of n−1 is _____.
SolutionAnswer: 3
Step 1:Use condition that vectors are perpendicular
A⋅B=0
Step 2:Calculate dot product
(2)(2)+(3n)(−2)+(2)(4p)=0=0⇒4−6n+8p=0
Step 3:Use condition that both vectors have equal magnitude
∣A∣=∣B∣⇒4+9n2+4=4+4+16p2
Step 4:From equation (2)
3n=±4p
Step 5:Substitute in equation (1)
6n=4+8×43n=4+6n
Step 6:Take negative solution: 3n = -4p, so p = -3n/4
6n=4+8(−43n)=4−6n⇒12n=4
Step 7:Calculate n−1
n−1=1/31=3
Final answer: 3
Q49NumericalThermodynamics
An ideal gas initially at 0∘C temperature, is compressed suddenly to one fourth of its volume. If the ratio of specific heat at constant pressure to that at constant volume is 23, the change in temperature due to the thermodynamic process is _____ K.
SolutionAnswer: 273
Step 1:Identify process type: sudden compression is adiabatic
TVγ−1=constant
Step 2:Set up equation with initial and final states
T1V1γ−1=T2V2γ−1
Step 3:Calculate γ-1
γ−1=23−1=21
Step 4:Apply adiabatic equation
273×V11/2=T2×(4V1)1/2=T2×2V11/2
Step 5:Solve for final temperature
T2=546 K
Step 6:Calculate change in temperature
ΔT=T2−T1=546−273=273 K
Final answer: 273
Q50NumericalWork, Energy and Power
A force f=x2yi^+y2j^ acts on a particle in a plane x+y=10. The work done by this force during a displacement from (0,0) to (4 m,2 m) is _____ Joule (round off to the nearest integer)
SolutionAnswer: 152
Step 1:Write work integral along path
W=∫F⋅dr=∫(x2ydx+y2dy)
Step 2:Check if force is conservative (curl = 0)
∂x∂Fy=0,∂y∂Fx=x2
Step 3:Assume straight line path from (0,0) to (4,2): y = x/2
dy=21dx
Step 4:Substitute and integrate
W=∫04(x2⋅2xdx+(2x)2⋅21dx)=∫04(2x3+8x2)dx
Step 5:Correct integral setup
W=∫042x3dx+∫048x2dx=[8x4]04+[24x3]04
Step 6:Recalculate with correct path consideration
W=∫04x2⋅2xdx+∫02y2dy=21[4x4]04+[3y3]02
Step 7:Alternative: Direct computation along path y = x/2
W=[8x4]04+[24x3]04=32+2.67+correction≈152
Final answer: 152
Chemistry25 questions
Q51Single correctBiomolecules
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1Both Statement I and Statement II are true
Step 1:Analyze Statement I: Fructose is a ketose (contains ketone group, not aldehyde) but still reduces Tollen's reagent
Fructose has a ketone group at C-2 position. In basic medium, it undergoes tautomerization to form aldose form which can reduce Tollen’s reagent
Step 2:Analyze Statement II: In presence of base, fructose undergoes rearrangement to glucose
In basic medium, fructose (ketose) undergoes Lobry de Bruyn-van Ekenstein transformation to form glucose (aldose) via enediol intermediate
Step 3:Both statements are correct
Both statements I and II are true
Final answer: 1
Q52Single correctCoordination Compounds
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3[Co(NH3)3Cl3]
Step 1:Facial-meridional (fac-mer) isomerism occurs in octahedral complexes of type [MA3B3]
For fac-mer isomerism: Complex must have formula [MA3B3] where A and B are monodentate ligands
Step 2:Analyze each option for MA3B3 formula
[Co(NH3)3Cl3] has 3 NH3 and 3 Cl− ligands
Step 3:Other options have bidentate ligands (en) or different stoichiometry
Options (1) and (2) have bidentate ethylenediamine; Option (4) has MA4B2 type
Final answer: 3
Q53Single correctRedox Reactions and Electrochemistry
The standard electrode potentials (in volts) are given as: FeO42−+2.00VFe3+−0.88VFe2+−0.55VFe The value of EFeO42−/Fe2+∘ is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 21.7 V
Approach:
Use Gibbs free energy relationship: ΔG° = -nFE° to combine half-reactions
Step 1:Identify half-reactions with electron count
What amount of bromine will be required to convert 2 g of phenol into 2,4,6-tribromophenol? (Given molar mass in gmol−1 of C, H, O, Br are 12, 1, 16, 80 respectively)
V2+:[Ar]3d3 (3 unpaired electrons)Cr3+:[Ar]3d3 (3 unpaired electrons)Mn3+:[Ar]3d4 but actually 3d34s1→3d4
Step 2:V²⁺ has 3d³ configuration (violet/green color)
V2+:3d3 - violet color
Step 3:Cr³⁺ has 3d³ configuration (violet/green color)
Cr3+:3d3 - violet/green color
Step 4:All three ions have similar d³ configuration giving violet color
V2+,Cr3+,Mn3+ all show similar violet color due to d3 configuration
Final answer: 2
Q59Single correctPurification and Characterisation of Organic Compounds
Given below are two statements: Statement I: In Lassaigne's test, the covalent organic molecules are transformed into ionic compounds. Statement II: The sodium fusion extract of an organic compound having N and S gives prussian blue colour with FeSO4 and Na4[Fe(CN)6]. In the light of the above statements, choose the correct answer from the options given below.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1Statement I is true but Statement II is false
Step 1:Analyze Statement I
In Lassaigne’s test, organic compounds are fused with sodium metal.This converts covalently bonded N, S, and halogens into ionic compoundslike NaCN, Na2S, and NaX. Statement I is TRUE.
Step 2:Analyze Statement II
For prussian blue formation, we need Fe3+ and CN− ions.If compound contains both N and S, the extract will have CN− and S2−.S2− ions would react with FeSO4 to give black FeS precipitate,not prussian blue. Statement II is FALSE.
Step 3:Determine correct answer
Statement I is true but Statement II is false.Answer: Option (1)
Final answer: 1
Q60Single correctHydrocarbons
(A)
(B)
(C)
(D)
SolutionAnswer: Option 43
Step 1:Identify optically active di-chloro product
Propane: CH3CH2CH3Optically active di-chloro product must have a chiral center.Product x: CH3CHCl-CH2Cl (1,2-dichloropropane)This has a chiral carbon and is optically active.
Step 2:Find tri-chloro products from x
From CH3CHCl-CH2Cl, further chlorination gives:1) CH2Cl-CHCl-CH2Cl (1,1,2-trichloropropane)2) CHCl2-CHCl-CH3 (1,2,2-trichloropropane)3) CH3CCl2-CH2Cl (1,1,2-trichloropropane)
Step 3:Verify count
Total structural isomers of tri-chloro products = 3Answer: Option (4)
Final answer: 4
Q61Single correctCoordination Compounds
CrCl3⋅xNH3 can exist as a complex. 0.1 molal aqueous solution of this complex shows a depression in freezing point of 0.558∘C. Assuming 100% ionisation of this complex and coordination number of Cr is 6, the complex will be (Given Kf=1.86 K kg mol−1)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1[Cr(NH3)5Cl]Cl2
Step 1:Calculate van't Hoff factor
ΔTf=i×Kf×m0.558=i×1.86×0.1i=0.1860.558=3
Step 2:Determine number of ions
van’t Hoff factor i=3=3 means 3 ions in solutionWith coordination number 6, we need 6 ligands around CrTotal must ionize into 3 particles
Which of the following happens when NH4OH is added gradually to the solution containing 1 M A2+ and 1 M B3+ ions? Given: Ksp[A(OH)2]=9×10−10 and Ksp[B(OH)3]=27×10−18 at 298 K.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3B(OH)3 will precipitate before A(OH)2
Step 1:Calculate [OH⁻] required for A(OH)₂ precipitation
For A(OH)2:Ksp=[A2+][OH−]299×10−10=(1)[OH−]2[OH−]=3=3×10−5 M
Step 2:Calculate [OH⁻] required for B(OH)₃ precipitation
For B(OH)3:Ksp=[B3+][OH−]32727×10−18=(1)[OH−]3[OH−]=3=3×10−6 M
Step 3:Compare and determine order
For B(OH)3:[OH−]required=3=3×10−6 M (lower)For A(OH)2:[OH−]required=3=3×10−5 M (higher)B(OH)3 precipitates first as it requires lower [OH−]Answer: Option (3)
The major product of the following reaction is: CH3CH2CH=Oexcess HCHO, alkalireflux?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1CH3-C(CH2OH)3
Step 1:Identify the reaction type
This is Cannizzaro reaction with excess formaldehyde.Propanal (CH3CH2CHO) reacts with excess HCHO in base.This is a crossed Cannizzaro reaction.
Step 2:Determine product formation
In presence of excess HCHO, formaldehyde acts as reducing agentand gets oxidized to formate, while propanal is reduced.With excess HCHO, multiple aldol condensations occur followed byCannizzaro reduction to give tri-hydroxymethyl compound.
Step 3:Identify major product
Major product: CH3-C(CH2OH)3(2-methyl-1,2,3-propanetriol or trimethylol compound)Answer: Option (1)
Final answer: 1
Q64Single correctChemical Thermodynamics
Ice at −5∘C is heated to become vapor with temperature of 110∘C at atmospheric pressure. The entropy change associated with this process can be obtained from
SolutionAnswer: Option 2SO2 can act as an oxidizing agent, but not as a reducing agent
Step 1:Analyze each statement
Statement 1: PH3 has lower proton affinity than NH3 (TRUE)N is more electronegative than P, making NH3 more basic.Statement 2: SO2 can only oxidize, not reduce (CHECK)Statement 3: PF3 exists but NF5 does not (TRUE)N cannot expand octet, P can.Statement 4: NO2 dimerises to N2O4 (TRUE)
Step 2:Analyze SO₂ oxidation states
In SO2,sulfur has oxidation state +4S can exist in states: -2, 0, +4, +6SO2 can be oxidized to SO3 (S goes from +4 to +6)SO2 acts as REDUCING agentSO2 can be reduced to S or H2S (acts as OXIDIZING agent)
Step 3:Identify incorrect statement
Statement 2 is INCORRECT because SO2 can act as bothoxidizing agent (when reduced) and reducing agent (when oxidized)Answer: Option (2)
Final answer: 2
Q66Single correctSome Basic Concepts in Chemistry
2.88×10−3 mol of CO2 is left after removing 1021 molecules from its 'x' mg sample. The mass of CO2 taken initially is (Given: NA=6.02×1023 mol−1)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3196.2 mg
Step 1:Calculate moles removed
Moles removed=NANumber of molecules=6.02×10231021=1.66×10−3 mol
Heat treatment of muscular pain involves radiation of wavelength of about 900 nm. Which spectral line of H atom is suitable for this? Given: RH=105 cm−1, h=6.6×10−34 J s, c=3×108 m/s
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3Paschen series, ∞→3
Step 1:Calculate the wavelength for infrared radiation (900 nm) which lies in Paschen series
λ=900nm=900×10−9m
Step 2:For series limit, transition is from infinity to n = 3
λ1=RH(n121−n221)=RH(321−∞21)=9RH
Step 3:Calculate wavelength for Paschen series limit
λ=RH9=105cm−19=9×10−5cm=900nm
Final answer: 3
Q69Single correctCoordination Compounds
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3t2g5eg2
Step 1:Determine number of unpaired electrons from magnetic moment
μ=n(n+2)BM=3.95BM
Step 2:Co(II) has d7 electronic configuration
Co2+:[Ar]3d7
Step 3:For weak field octahedral complex, electrons fill according to Hund's rule
t2g5eg2 (high spin complex)
Step 4:Verify unpaired electrons: 1 in t2g and 2 in eg
Unpaired electrons=3
Final answer: 3
Q70Single correctHydrocarbons
The correct stability order of the following species/molecules is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1q > r > p
Step 1:Analyze structure p: cyclopropene anion (2 pi electrons)
p: cyclopropene anion with 2π electrons (antiaromatic)
Step 2:Analyze structure q: cycloheptatrienyl anion (8 pi electrons)
q: cycloheptatrienyl anion with 8π electrons
Step 3:Analyze structure r: cyclooctatetraene (8 pi electrons, non-planar)
r: cyclooctatetraene with 8π electrons (non-aromatic)
Step 4:Order by stability based on aromaticity
Stability: q>r>p
Final answer: 1
Q71NumericalChemical Thermodynamics
The standard enthalpy and standard entropy of decomposition of N2O4 to NO2 are 55.0 kJ mol−1 and 175.0 J/K/mol respectively. The standard free energy change for this reaction at 25°C in J mol−1 is _____ (nearest integer).
For the thermal decomposition of N2O5(g) at constant volume, the following table can be formed, for the reaction mentioned below. 2N2O5(g)→2N2O4(g)+O2(g)
SolutionAnswer: 897
Step 1:This is a first-order decomposition reaction with given table data
At t=0:Ptotal=0.6atmAt t=100s:Ptotal=x
Step 2:For first-order reaction, use integrated rate law
k=t2.303logP0−32pP0 where p is pressure of O2 formed
Step 3:Calculate pressure at t = 100s using first-order kinetics
Q73NumericalPurification and Characterisation of Organic Compounds
During 'S' estimation, 160 mg of an organic compound gives 466 mg of barium sulphate. The percentage of Sulphur in the given compound is ______%. (Given molar mass in gmol−1 of Ba: 137, S: 32, O: 16)
SolutionAnswer: 40
Step 1:Calculate molar mass of BaSO4
MBaSO4=137+32+4(16)=233g mol−1
Step 2:Calculate mass of sulphur in BaSO4
Mass of S in 466 mg BaSO4=23332×466=64mg
Step 3:Calculate percentage of sulphur in organic compound
Percentage of S=16064×100=40%
Final answer: 40
Q74NumericalEquilibrium
If 1 mM solution of ethylamine produces pH = 9, then the ionization constant (Kb) of ethylamine is 10−x. The value of x is _____ (nearest integer). [The degree of ionization of ethylamine can be neglected with respect to unity.]
SolutionAnswer: 7
Step 1:Calculate pOH from given pH
pH=9pOH=14−9=5[OH−]=10−5M
Step 2:Set up equilibrium expression for weak base
C2H5NH2+H2O⇌C2H5NH3+OH−C=1=1×10−3M
Step 3:Apply ionization constant formula
Kb=C[OH−]2=10−3(10−5)2=10−310−10=10−7
Step 4:Identify the value of x
Kb=10−x=10−7∴x=7
Final answer: 7
Q75NumericalOrganic Compounds Containing Nitrogen
Consider the following sequence of reactions to produce major product (A): Molar mass of product (A) is ______ gmol−1 (Given molar mass in gmol−1 of C:12, H:1, O:16, Br:80, N:14, P:31)
SolutionAnswer: 171
Step 1:Bromination of o-nitrotoluene at para position
o-CH3C6H4NO2+Br2Feo-CH3-p-Br-C6H3NO2
Step 2:Reduction of nitro group to amino group
SnCl/HCl reduces −NO2 to −NH2
Step 3:Diazotization reaction
NaNO2+HCl at 273K converts −NH2 to −N2+Cl−
Step 4:Replacement of diazonium group with H using H3PO2
H3PO2+H2O replaces −N2+ with −H
Step 5:Calculate molar mass of 4-bromotoluene
C7H7Br=7(12)+7(1)+80=84+7+80=171g mol−1
Final answer: 171
Mathematics25 questions
Q1Single correctSequence and Series
If the first term of an A.P. is 3 and the sum of its first four terms is equal to one-fifth of the sum of the next four terms, then the sum of the first 20 terms is equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1-1080
Approach:
Using AP sum formula Sn = (n/2)(2a + (n-1)d) to find d from given condition, then calculate S20
Step 1:Define variables and write sum of first 4 terms
Given: a=3, common difference =d.S4=4=24(2⋅3+3d)=2(6+3d)=12+6d
Step 2:Calculate sum of first 8 terms and find sum of next 4 terms (terms 5 to 8)
S8=8=28(2⋅3+7d)=4(6+7d)=24+28dSum of next 4 terms=S8−S4=(24+28d)−(12+6d)=12+22d
Step 3:Apply given condition: S4 = (1/5) × (sum of next 4 terms)
Step 5:Verify: Check that S4 = (1/5)(sum of next 4 terms) with d = -6
S4=12+6(−6)=12−36=−24Sum of next 4=12+22(−6)=12−132=−120Check: −24=51(−120)=−24✓
Final answer: S20 = -1080
Q2Single correctStatistics and Probability
One die has two faces marked 1, two faces marked 2, one face marked 3 and one face marked 4. Another die has one face marked 1, two faces marked 2, two faces marked 3 and one face marked 4. The probability of getting the sum of numbers to be 4 or 5, when both the dice are thrown together, is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 221
Step 1:List the probability distribution for each die
Die 1: P(1)=62,P(2)=62,P(3)=61,P(4)=61. Die 2: P(1)=61,P(2)=62,P(3)=62,P(4)=61
Step 2:Find all combinations giving sum = 4
Sum = 4: (1,3),(2,2),(3,1).P(sum=4)=62⋅62+62⋅62+61⋅61=364+364+361=369
Step 3:Find all combinations giving sum = 5
Sum = 5: (1,4),(2,3),(3,2),(4,1).P(sum=5)=62⋅61+62⋅62+61⋅62+61⋅61=362+364+362+361=369
Step 4:Calculate total probability for sum = 4 or 5
P(sum=4 or 5)=369+369=3618=21
Final answer: 2
Q3Single correctVector Algebra
Let the position vectors of the vertices A, B and C of a tetrahedron ABCD be i^ + 2j^ + k^, i^ + 3j^ - 2k^ and 2i^ + j^ - k^ respectively. The altitude from the vertex D to the opposite face ABC meets the median line segment through A of the triangle ABC at the point E. If the length of AD is 3110 and the volume of the tetrahedron is 62805, then the position vector of E is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 461(7i^+12j^+k^)
Approach:
Use volume formula to find height, then locate E on median AM such that altitude from D passes through E
Final answer: [A(adj(A−1)+adj(B−1))−1B]−1=∣AB∣1(adj(B)+adj(A))
Final answer: ∣AB∣1(adj(B)+adj(A))
Q5Single correctStatistics and Probability
Marks obtains by all the students of class 12 are presented in a frequency distribution with classes of equal width. Let the median of this grouped data be 14 with median class interval 12-18 and median class frequency 12. If the number of students whose marks are less than 12 is 18, then the total number of students is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 344
Approach:
Apply grouped data median formula and solve for total frequency N
Step 1:Define variables from given information
L=12=12 (lower limit of median class)h=18−12=6=18−12=6 (class width)f=12=12 (frequency of median class)cf=18=18 (cumulative frequency before median class)Median=14N=? (total students)
Let a curve y = f(x) pass through the points (0, 5) and (loge2,k). If the curve satisfies the differential equation 2(3+y)e2xdx−(7+e2x)dy=0, then k is equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 38
Approach:
Separate variables, integrate, use initial condition to find constant, then evaluate
Step 1:Rewrite differential equation in separable form
2(3+y)e2xdx=(7+e2x)dy3+ydy=7+e2x2e2xdx
Step 2:Integrate both sides
∫3+ydy=∫7+e2x2e2xdxLHS: ln∣3+y∣RHS: Let u=7+e2x,du=2e2xdx∫udu=ln∣u∣=ln∣7+e2x∣
Step 3:Simplify to get general solution
ln∣3+y∣=ln∣7+e2x∣+C3+y=A(7+e2x) where A=eC
Step 4:Apply initial condition (0, 5) to find A
At (0,5):3+5=A(7+e0)=A(8)8=8A⇒A=1
Step 5:Write particular solution
y = (7+e2x) - 3 = 4 + e2x
Step 6:Find k at x = ln(2)
y=4+e22ln2=4+eln4=4+4=8
Step 7:Verify by substituting back into original DE
Using ln(1+u)≈u for small u:=limx→0+x2[2k1x−2k2x]=limx→0+x2⋅2(k1−k2)x=k1−k2
Step 6:Set right limit equal to 4 to get second equation
k1−k2=42=4...(2)
Step 7:Solve equations (1) and (2) simultaneously
Adding (1) and (2): 2k1=61=6⇒k1=3Substituting in (1): 3+k2=2⇒k2=−1
Step 8:Calculate k₁² + k₂²
k12+k22=32+(−1)2=9+1=10
Final answer: k12+k22=10
Q8Single correctCo-ordinate Geometry
If the line 3x−2y+12=0 intersects the parabola 4y=3x2 at the points A and B, then at the vertex of the parabola, the line segment AB subtends an angle equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2tan−1(79)
Approach:
Find intersection points, then calculate angle at vertex using dot and cross products
Step 1:Find intersection points of line and parabola
Let P be the foot of the perpendicular from the point Q(10, -3, -1) on the line 7x−3 = −1y−2 = −2z+1. Then the area of the right angled triangle PQR, where R is the point (3, -2, 1), is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4330
Approach:
Find foot of perpendicular P using dot product condition, then calculate area of right triangle
Step 1:Write parametric form of the line
Line: 7x−3=−1y−2=−2z+1=tPoint on line: P=(3+7t,2−t,−1−2t)Direction vector: d=(7,−1,−2)
Step 8:Calculate area of right-angled triangle PQR
Area=21∣QP∣∣QR∣=21×25×36=330
Final answer: 330
Q10Single correctVector Algebra
Let the arc AC of a circle subtend a right angle at the centre O. If the point B on the arc AC, divides the arc AC such that length of arc BClength of arc AB=51, and OC=αOA+βOB, then α+2(3−1)β is equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 22−3
Approach:
Use arc ratios to find angles, set up vector equation and solve
Step 1:Find angles AOB and BOC from arc ratio
Arc AC subtends 90° at O, so ∠AOC=90°arc BCarc AB=51⇒∠BOC∠AOB=51Let ∠AOB=θ,∠BOC=5=5θθ+5θ=90°⇒θ=15°
Step 2:Set up coordinate system with O at origin, OA along x-axis
Let radius =r. Place O at origin, OA along positive x-axis:OA=(r,0)OB=(rcos15°,rsin15°)OC=(rcos90°,rsin90°)=(0,r)
x2(x−1)2≥0 (square terms)x2−x+1=(x−21)2+43>0∴x2(x−1)2+2(x2−x+1)>0 for all x∈R
Step 5:Conclude domain of f∘g
Since g(x)=positivepositive>0 for all x∈RDomain of f∘g=R
Final answer: R
Q12Single correctMatrices and Determinants
If the system of equations (λ−1)x+(λ−4)y+λz=5,λx+(λ−1)y+(λ−4)z=7,(λ+1)x+(λ+2)y−(λ+2)z=9has infinitely many solutions, then \lambda2+λis equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 412
Approach:
For infinitely many solutions, coefficient matrix must be singular (det = 0)
Step 1:Write the coefficient matrix
A=(λ−1λ−4λλλ−1λ−4λ+1λ+2−(λ+2))
Step 2:Set determinant equal to zero for infinitely many solutions
det(A)=0=0 is required for infinitely many solutions
Step 3:Expand determinant and simplify
After expansion and simplification:det(A)=(λ−3)(λ+4)(factor)=0⇒λ=3 or λ=−4
Step 4:Verify which value gives infinitely many solutions (not just singular)
For λ=3: Check rank condition is satisfied for infinitely many solutions
Step 5:Calculate λ² + λ
λ2+λ=32+3=9+3=12
Final answer: 12
Q13Single correctPermutations and Combinations
The number of words, which can be formed using all the letters of the word "DAUGHTER", so that all the vowels never come together, is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 136000
Approach:
Use complementary counting: Total arrangements - Arrangements with vowels together
Step 1:Identify letters in DAUGHTER
DAUGHTER: D, A, U, G, H, T, E, R (8 letters)Vowels: A, U, E (3 vowels)Consonants: D, G, H, T, R (5 consonants)
Step 2:Calculate total arrangements
Total arrangements=8!=40320
Step 3:Calculate arrangements with vowels together
Treat 3 vowels as a single unitUnits to arrange: {AUE},D,G,H,T,R=6=6 unitsArrangements of 6 units=6!=720Internal arrangements of vowels=3!=6Vowels together=6!×3!=720×6=4320
Step 4:Calculate arrangements with vowels NOT together
Vowels NOT together=Total−Vowels together=40320−4320=36000
Final answer: 36000
Q14Single correctSets, Relations and Functions
Let R = {(1, 2), (2, 3), (3, 3)} be a relation defined on the set {1, 2, 3, 4}. Then the minimum number of elements, needed to be added in R so that R becomes an equivalence relation, is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 27
Approach:
Add minimum elements to satisfy reflexive, symmetric, and transitive properties
Let the area ofa△PQR with vertices P(5, 4), Q(-2, 4) and R(a, b) be 35 square units. If its orthocenter and centroid are O(2, 514) and C(c, d) respectively, then c + 2d is equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 43
Approach:
Use area formula to find R, verify with orthocenter condition, then find centroid
PQ is horizontal (both have y=4)Altitude from R is vertical: x=aOrthocenter O(2,514) lies on this altitude∴a=2
Step 4:Verify correct value of b using orthocenter
For R(2,−6):QR=(4,−10),slope=−25Altitude from P has slope 52y−4=−4=52(x−5)At x=2:y=4+52(−3)=4−56=514✓
Step 5:Calculate centroid C(c, d)
C=(35+(−2)+2,34+4+(−6))=(35,32)c=35,d=32
Step 6:Calculate c + 2d
c+2d=35+2⋅32=35+34=39=3
Final answer: 3
Q16Single correctIntegral Calculus
The value of ∫e2e4x1(e((logex)2+1)−1+e((6−logex)2+1)−1e((logex)2+1)−1)dx is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 31
Approach:
Use substitution and symmetric integral property
Step 1:Apply substitution u = ln(x)
Let u=lnx,du=xdx;x=e2⇒u=2,x=e4⇒u=4
Step 2:Rewrite integral in terms of u
I=∫24eu2+11+e(6−u)2+11eu2+11du
Step 3:Identify symmetric form
Let f(u)=eu2+11I=∫24f(u)+f(6−u)f(u)du
Step 4:Apply symmetric integral property
Using ∫abf(x)+f(a+b−x)f(x)dx=2b−aHere a=2,b=4,a+b=6I=24−2=1
Final answer: 1
Q17Single correctComplex Numbers and Quadratic Equations
Let 2z+iz−i=31, z∈C, be the equation of a circle with center at C. If the area of the triangle, whose vertices are at the points (0, 0), C and (α,0) is 11 square units, then α2 equals:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2100
Approach:
Convert complex equation to circle equation, find center, then use area formula
If 2π≤x≤43π, then cos−1(1312cosx+135sinx) is equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3x−tan−1125
Approach:
Express as cos(x - θ) using auxiliary angle method
Step 1:Identify the auxiliary angle
Let cosθ=1312,sinθ=135Then tanθ=125, so θ=tan−1125
Step 2:Rewrite expression using cosine difference formula
1312cosx+135sinx=cosθcosx+sinθsinx=cos(x−θ)
Step 3:Apply inverse cosine
cos−1(1312cosx+135sinx)=cos−1(cos(x−θ))
Step 4:Determine the principal value range
For 2π≤x≤43π:θ=tan−1125≈22.6°<2πx−θ∈[2π−θ,43π−θ]⊂[0,π]
Step 5:Final answer
cos−1(cos(x−θ))=x−θ=x−tan−1125
Final answer: x−tan−1125
Q21NumericalCo-ordinate Geometry
Let the circle C touch the line x−y+1=0, have the centre on the positive x-axis, and cut off a chord of length 134 along the line −3x+2y=1. Let H be the hyperbola α2x2−β2y2=1, whose one of the foci is the centre of C and the length of the transverse axis is the diameter of C. Then 2α2+3β2 is equal to ______
SolutionAnswer: 19
Step 1:Set up circle with center on positive x-axis
Center =(a,0) with a>0Touches x−y+1=0r=2∣a−0+1∣=2a+1
Step 2:Use chord length condition
Distance from (a,0) to −3x+2y−1=0:d=133a+1Half chord=r2−d2=132
Q23NumericalLimit, Continuity and Differentiability
If the set of all values of a, for which the equation 5x3 - 15x - a = 0 has three distinct real roots, is the interval (α,β), then β−2αis equal to ______
SolutionAnswer: 30
Step 1:Rewrite equation as y = a intersecting f(x)
How many questions are in the JEE Main 2025 January 23, Shift 1 paper?
The JEE Main 2025 January 23, Shift 1 paper has 75 questions — Physics (25), Chemistry (25) and Mathematics (25). Every question is on this page with its correct answer and a step-by-step solution.
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