JEE Main 2025 January 29, Shift 2 Question Paper with Solutions
All 75 questions from the JEE Main 2025 (January 29, Shift 2) 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.
The difference of temperature in a material can convert heat energy into electrical energy. To harvest the heat energy, the material should have
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3
Approach:
Apply the principles of thermoelectric effect (Seebeck effect) to determine optimal material properties for heat harvesting.
Step 1:Identify the physical phenomenon
Thermoelectric effect (Seebeck effect) converts temperature difference into electrical voltage
Step 2:Understand the role of thermal conductivity
Low thermal conductivity (κ) helps maintain the temperature gradient across the material
Step 3:Understand the role of electrical conductivity
High electrical conductivity (σ) allows efficient transport of charge carriers
Step 4:Apply the figure of merit concept
Thermoelectric efficiency depends on ZT=κS2σT
Step 5:Select the correct option
Material should have: Low thermal conductivity + High electrical conductivity
Final answer: Option 3
Q27Single correct
Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): With the increase in the pressure of an ideal gas, the volume falls off more rapidly in an isothermal process in comparison to the adiabatic process.
Reason (R): In isothermal process, PV = constant, while in adiabatic process PV^γ = constant. Here γ is the ratio of specific heats, P is the pressure and V is the volume of the ideal gas.
In the light of the above statements, choose the correct answer from the options given below:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4
Approach:
Compare the slopes of isothermal and adiabatic curves on a P-V diagram to determine which process shows more rapid volume change with pressure.
R states: PV=const (isothermal) and PVγ=const (adiabatic)
Step 3:Calculate the slope for isothermal process
From PV=C: dVdP=−VP
Step 4:Calculate the slope for adiabatic process
From PVγ=C: dVdP=−γVP
Step 5:Compare the magnitudes of slopes
dVdPadi=γdVdPiso
Step 6:Interpret the physical meaning
Steeper slope means larger ∣dV∣ for same dP, i.e., volume changes more rapidly
Step 7:Verify Assertion (A)
A claims volume falls more rapidly in isothermal - this is INCORRECT
Step 8:Determine the final answer
A is false, R is true
Final answer: Option 4
Q28Single correct
An electric dipole is placed at a distance of 2 cm from an infinite plane sheet having positive charge density σ0. Choose the correct option from the following.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3
Approach:
Analyze the behavior of an electric dipole in the uniform electric field of an infinite charged plane sheet.
Step 1:Determine the electric field of the infinite plane sheet
E=2ϵ0σ0, directed perpendicular to the sheet (away from it for positive σ0)
Step 2:Analyze the net force on the dipole
In a uniform field: Fnet=qE+(−q)E=0
Step 3:Identify the stable equilibrium position
Dipole aligns parallel to E (along the field direction, θ=0)
Step 4:Calculate torque at equilibrium
τ=pEsin(0°)=0
Step 5:Calculate potential energy at equilibrium
U=−pEcos(0°)=−pE (minimum value)
Step 6:Select the correct option
At equilibrium: Potential energy is minimum AND torque is zero
Final answer: Option 3
Q29Single correct
In an experiment with photoelectric effect, the stopping potential:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3
Approach:
Analyze each option using Einstein's photoelectric equation and the definition of stopping potential.
Step 1:Write Einstein's photoelectric equation
hν=W+Kmax where W is work function, Kmax is maximum KE
Step 2:Relate stopping potential to maximum KE
eVs=Kmax⇒Vs=eKmax
Step 3:Analyze Option 1: Effect of wavelength
Kmax=hν−W=λhc−W; as λ increases, Kmax decreases
Step 4:Analyze Option 2: Effect of intensity
Intensity affects number of photons, not energy per photon
Step 5:Analyze Option 3: Relationship between Vs and Kmax
Vs=eKmax=e1×Kmax
Step 6:Analyze Option 4: Another claim about intensity
Again, intensity does not affect Vs or Kmax
Step 7:Select the correct answer
Only Option 3 is correct: Vs=eKmax
Final answer: Option 3
Q30Single correct
A point charge causes an electric flux of -2 × 104 Nm2C−1 to pass through a spherical Gaussian surface of 8.0 cm radius, centred on the charge. The value of the point charge is:
(Given ε0 = 8.85 × 10−12C2N−1m−2)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1
Approach:
Apply Gauss's law to find the enclosed charge from the given electric flux through the spherical Gaussian surface.
Step 1:List the given data
ϕ=−2×104 Nm2/C, r=8.0 cm, ϵ0=8.85×10−12C2N−1m−2
Step 2:Apply Gauss's law
ϕ=ϵ0qenc
Step 3:Rearrange to find enclosed charge
q=ϕ×ϵ0
Step 4:Substitute the values
q=(−2×104)×(8.85×10−12)
Step 5:Calculate the result
q=−17.7×10−8 C =−1.77×10−7 C
Step 6:Note about the radius
The radius of the Gaussian surface does not affect the flux calculation (flux depends only on enclosed charge)
Final answer: Option 1
Q31Single correct
A poly-atomic molecule (CV = 3R, CP = 4R, where R is gas constant) goes from phase space point A(PA = 105 Pa, VA = 4 × 10−6m3) to point B (PB = 5 × 104 Pa, VB = 6 × 10−6m3) to point C (PC = 104 Pa, VC = 8 × 10−6m3). A to B is an adiabatic path and B to C is an isothermal path. The net heat absorbed per unit mole by the system is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2
Approach:
Calculate heat absorbed in each process: zero for adiabatic (A→B) and nRTln(VC/VB) for isothermal (B→C).
Step 3:Find temperature at point B using ideal gas law
TB=nRPBVB=nR(5×104)(6×10−6)=nR0.3
Step 4:Calculate temperature at point B
PBVB=nRTB⇒TB=nRPBVB=1×R5×104×6×10−6=R0.3
Step 5:Confirm isothermal nature of B→C
Process B→C is isothermal, so TB=TC=450 K
Step 6:Calculate heat for process B→C (Isothermal)
QB→C=nRTln(VBVC)=1×R×450×ln(6×10−68×10−6)
Step 7:Simplify the logarithm
ln(68)=ln(34)=ln4−ln3
Step 8:Calculate total heat absorbed
Qtotal=QA→B+QB→C=0+450R(ln4−ln3)
Step 9:Select the correct option
Net heat absorbed per mole =450R(ln4−ln3)
Final answer: Option 2
Q32Single correct
Two identical symmetric double convex lenses of focal length f are cut into two equal parts L1, L2 by AB plane and L3, L4 by XY plane as shown in figure respectively. The ratio of focal lengths of lenses L1 and L3 is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4
Approach:
Apply the lens maker's formula to analyze how focal length changes when a symmetric double convex lens is cut in different planes.
Step 1:Write the lens maker's formula for original lens
For symmetric double convex lens: f1=(μ−1)(R1−−R1)=(μ−1)R2
Step 2:Analyze cut by AB plane (horizontal through optical axis)
Horizontal cut creates two identical plano-convex lenses L1 and L2
Step 3:Calculate focal length of L1 (plano-convex)
fL11=(μ−1)(R1−∞1)=Rμ−1
Step 4:Express fL1 in terms of original f
Since f=2(μ−1)R, we get fL1=μ−1R=2⋅2(μ−1)R=2f
Step 5:Analyze cut by XY plane (vertical through optical axis)
Vertical cut creates two half-lenses L3 and L4, each retaining both curved surfaces with same R
Step 6:Apply lens maker's formula to L3
fL31=(μ−1)(R1−−R1)=(μ−1)R2
Step 7:Account for reduced aperture effect
Vertical cut halves the lens aperture. For a lens with reduced width, effective power decreases by factor of 2
Step 8:Calculate the ratio fL1:fL3
fL1:fL3=2f:4f=1:2
Final answer: Option 4
Q33Single correct
A plane electromagnetic wave propagates along the +x direction in free space. The components of the electric field, E⃗ and magnetic field, B⃗ vectors associated with the wave in Cartesian frame are:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2
Approach:
Apply the properties of electromagnetic waves: direction of propagation is given by c=E×B, with E, B mutually perpendicular and both perpendicular to direction of propagation.
Step 1:Identify the key properties of EM waves
For EM waves: E, B, and c are mutually perpendicular (transverse wave)
Step 2:Write the direction of propagation formula
c=E×B (Poynting vector direction)
Step 3:Given condition
Wave propagates along +x direction, so c∥i^
Step 4:Analyze Option 1: Ey, Bx
j^×i^=−k^ (along −z, not +x)
Step 5:Analyze Option 2: Ey, Bz
j^×k^=i^ (along +x) ✓
Step 6:Analyze Option 3: Ex, By
Ex means E∥i^, but E must be ⊥ to propagation direction
Step 7:Analyze Option 4: Ez, By
k^×j^=−i^ (along −x, not +x)
Step 8:Conclude the answer
Only Ey, Bz gives E×B=i^
Final answer: Option 2
Q34Single correctOptics
Two concave refracting surfaces of equal radii of curvature and refractive index 1.5 face each other in air as shown in figure. A point object O is placed midway, between P and B. The separation between the images of O, formed by each refracting surface is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 20.114R
Approach:
Apply the refraction formula at curved surfaces for each refracting surface separately, then find the separation between the two images formed.
Step 1:Identify the given data
μglass=1.5, μair=1, both surfaces have radius R, object O at midpoint
Step 2:Set up refraction at surface B (closer to O)
Step 5:Set up refraction at surface A (farther from O)
For surface A: u=−23R (object distance from A), μ1=1, μ2=1.5, RA=−R
Step 6:Apply refraction formula for surface A
vA1.5−−3R/21=−R0.5
Step 7:Solve for image position from surface A
vA1.5=−R0.5−3R2=−6R7, so vA=−79R
Step 8:Calculate separation between images
Separation =∣vB−vA+2R∣=∣0.6R+79R−2R∣
Step 9:Compute final answer
Separation =∣0.6R+1.286R−2R∣=∣−0.114R∣=0.114R
Final answer: 0.114R
Q35Single correctSimple Harmonic Motion
Two bodies A and B of equal mass are suspended from two massless springs of spring constant k1 and k2, respectively. If the bodies oscillate vertically such that their amplitudes are equal, the ratio of the maximum velocity of A to the maximum velocity of B is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1k1/k2
Approach:
Use the SHM relations for maximum velocity and angular frequency to find the ratio of maximum velocities given equal amplitudes.
Step 1:Write the expression for maximum velocity in SHM
Vmax=Aω where A = amplitude, ω = angular frequency
Step 2:Write angular frequency for spring-mass system
ω=mk
Step 3:Find angular frequencies for both bodies
ωA=mk1 and ωB=mk2
Step 4:Write maximum velocities for A and B
VA=AAωA and VB=ABωB
Step 5:Apply given condition
Given: AA=AB=A (equal amplitudes)
Step 6:Calculate the ratio of maximum velocities
VBVA=A⋅ωBA⋅ωA=ωBωA
Step 7:Substitute angular frequency expressions
VBVA=k2/mk1/m=k2k1
Step 8:State the final answer
VBVA=k2k1
Final answer: k2k1
Q36Single correctCollision
Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): Three identical spheres of same mass undergo one dimensional motion as shown in figure with initial velocities vA = 5 m/s, vB = 2 m/s, vC = 4 m/s. If we wait sufficiently long for elastic collision to happen, then vA = 4 m/s, vB = 2 m/s, vC = 5 m/s will be the final velocities.
Reason (R): In an elastic collision between identical masses, two objects exchange their velocities.
In the light of the above statements, choose the correct answer from the options given below:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4(A) is false but (R) is true
Approach:
Apply the property of elastic collision between identical masses (velocity exchange) step by step to verify the assertion, then evaluate both A and R.
Step 1:State the given initial velocities
vA=5 m/s, vB=2 m/s, vC=4 m/s (all moving in same direction)
Step 2:Verify Reason (R)
In elastic collision between identical masses, velocities ARE exchanged
Step 3:Analyze first collision: A catches up with B
Since vA>vB, A collides with B first
Step 4:Apply velocity exchange for A-B collision
After collision: vA′=2 m/s, vB′=5 m/s
Step 5:Analyze second collision: B catches up with C
Now vB′=5 m/s >vC=4 m/s, so B collides with C
Step 6:Apply velocity exchange for B-C collision
After collision: vB′′=4 m/s, vC′=5 m/s
Step 7:Determine final velocities after all collisions
vA=2 m/s, vB=4 m/s, vC=5 m/s (no more collisions as vA<vB<vC)
Step 8:Compare with Assertion (A)
A claims: vA=4 m/s, vB=2 m/s, vC=5 m/s
Step 9:Evaluate Assertion
Assertion (A) is FALSE (wrong final velocities for A and B)
Step 10:Final conclusion
(A) is false but (R) is true
Final answer: Option 4
Q37Single correctPower and Momentum
A sand dropper drops sand of mass m(t) on a conveyor belt at a rate proportional to the square root of speed (v) of the belt, i.e. dtdm∝v. If P is the power delivered to run the belt at constant speed then which of the following relationship is true?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4P2∝v5
Approach:
Apply Newton's second law for variable mass systems to find force, then calculate power using P = Fv, and determine the relationship between P and v.
Step 1:Write the given condition
dtdm∝v, so dtdm=Cv where C is constant
Step 2:Apply force equation for variable mass system
For sand landing on belt at constant speed: F=vdtdm
Step 3:Substitute the mass rate expression
F=v⋅Cv=Cv⋅v1/2=Cv3/2
Step 4:Calculate power delivered to the belt
P=Fv=Cv3/2⋅v=Cv5/2
Step 5:Find the relationship between P2 and v
Squaring: P2=C2v5
Step 6:Select the correct option
From P2∝v5, Option 4 is correct
Final answer: Option 4: P2∝v5
Q38Single correctOptics - Lens
A convex lens made of glass (refractive index = 1.5) has focal length 24 cm in air. When it is totally immersed in water (refractive index = 1.33), its focal length changes to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 296 cm
Approach:
Apply lens maker's formula in air and water medium, then find the ratio of focal lengths to determine the new focal length.
Step 1:List the given data
μg=1.5 (glass), μw=1.33 (water), f=24 cm (in air)
Step 2:Write lens maker's formula in air
f1=(μg−1)(R11−R21)=(1.5−1)(R11−R21)
Step 3:Write lens maker's formula in water
f′1=(μwμg−1)(R11−R21)
Step 4:Calculate the relative refractive index
μwμg=1.331.5=1.128
Step 5:Substitute in the formula
f′1=(1.128−1)(R11−R21)=0.128(R11−R21)
Step 6:Divide the two equations
1/f′1/f=ff′=0.1280.5=0.170.5×1.33
Step 7:Calculate the new focal length
f′=4×f=4×24=96 cm
Step 8:State the final answer
The focal length changes from 24 cm to 96 cm when immersed in water
Final answer: 96 cm
Q39Single correctCapacitors
A capacitor, C1 = 6μF is charged to a potential difference of V0 = 5V using a 5V battery. The battery is removed and another capacitor, C2 = 12μF is inserted in place of the battery. When the switch 'S' is closed, the charge flows between the capacitors for some time until equilibrium condition is reached. What are the charges (q1 and q2) on the capacitors C1 and C2 when equilibrium condition is reached?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3q1=10μC,q2=20μC
Approach:
Apply charge conservation and the condition that at equilibrium both capacitors have the same potential when connected in parallel.
Step 1:List the given data
C1=6μF, C2=12μF, Initial voltage on C1: V0=5V
Step 2:Calculate initial charge on C1
Q1initial=C1V0=6×5=30μC
Step 3:Initial charge on C2
Q2initial=0 (uncharged)
Step 4:Apply charge conservation
Qtotal=Q1+Q2=30μC (charge is conserved)
Step 5:Apply equilibrium condition
At equilibrium: V1=V2=VC (common potential)
Step 6:Express charges in terms of common potential
q1=C1VC=6VC and q2=C2VC=12VC
Step 7:Apply charge conservation equation
q1+q2=30⇒6VC+12VC=30
Step 8:Solve for common potential
VC=1830=35 V
Step 9:Calculate final charge on C1
q1=6×35=10μC
Step 10:Calculate final charge on C2
q2=12×35=20μC
Step 11:State the final answer
q1=10μC, q2=20μC
Final answer: q1=10μC, q2=20μC
Q40Single correctAngular Momentum
Three equal masses m are kept at vertices (A, B, C) of an equilateral triangle of side a in free space. At t = 0, they are given an initial velocity V⃗A = V0AC̅, V⃗B = V0BA̅ and V⃗C = V0CB̅. Here, AC̅, CB̅ and BA̅ are unit vectors along the edges of the triangle. If the three masses interact gravitationally, then the magnitude of the net angular momentum of the system at the point of collision is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 323amV0
Approach:
Calculate the angular momentum of each mass about the centroid (collision point), using the perpendicular distance from centroid to the velocity direction.
Step 1:Understand the motion
Each mass moves along an edge towards the opposite vertex (tangential to inscribed circle)
Step 2:Identify the collision point
Due to symmetry and equal masses, all three collide at the centroid
Step 3:Find perpendicular distance from centroid to velocity direction
The velocity is along an edge; perpendicular distance = inradius r
Step 4:Calculate inradius of equilateral triangle
For equilateral triangle: r=23a=6a3
Step 5:Calculate angular momentum of one mass
L1=mV0×r=mV0×23a
Step 6:Check direction of angular momentum
All three masses rotate in the same sense about the centroid (clockwise or anticlockwise)
Match each magnetic quantity with its correct unit by analyzing the physical dimensions and standard unit systems.
Step 1:Match Magnetic induction (A) with its unit
Magnetic induction B is measured in Gauss in CGS system
Step 2:Match Magnetic intensity (B) with its unit
Magnetic intensity H=μ0B, measured in Ampere/meter
Step 3:Match Magnetic flux (C) with its unit
Magnetic flux Φ=B⋅A, measured in Weber
Step 4:Match Magnetic moment (D) with its unit
Magnetic moment M=i×A, measured in Ampere-meter2
Step 5:Compile the final matching
A-III, B-IV, C-II, D-I
Final answer: Option (2): (A)-(III), (B)-(IV), (C)-(II), (D)-(I)
Q43Single correctElectronic Devices
The truth table for the circuit given below is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1A0011B0101Y0110
Approach:
Analyze the given circuit diagram to identify the logic gate and construct its truth table. The circuit represents an XOR (Exclusive OR) gate.
Step 1:Identify the circuit as XOR gate
Y=A⊕B=A⋅B+A⋅B
Step 2:For A=0, B=0
Y=0⋅0+0⋅0=0⋅1+1⋅0=0
Step 3:For A=0, B=1
Y=0⋅1+0⋅1=0⋅0+1⋅1=1
Step 4:For A=1, B=0
Y=1⋅0+1⋅0=1⋅1+0⋅0=1
Step 5:For A=1, B=1
Y=1⋅1+1⋅1=1⋅0+0⋅1=0
Step 6:Construct truth table
XOR gate gives output 1 when inputs are different, output 0 when inputs are same
Final answer: Option (1) - XOR gate truth table
Q44Single correctHeat and Thermodynamics
A cup of coffee cools from 90°C to 80°C intminutes when the room temperature is 20°C. The time taken by the similar cup of coffee to cool from 80°C to 60°C at the same room temperature is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1513t
Approach:
Apply Newton's law of cooling using average temperature method to find the time taken for the second cooling interval.
Step 1:Apply Newton's law for first cooling (90°C to 80°C)
t90−80=k(290+80−20)
Step 2:Simplify the first equation
t10=65k
Step 3:Apply Newton's law for second cooling (80°C to 60°C)
t′80−60=k(280+60−20)
Step 4:Simplify the second equation
t′20=50k
Step 5:Divide equation (i) by equation (ii)
20/t′10/t=50k65k⇒20t10t′=5065
Step 6:Calculate final answer
t′=513t
Final answer: Option (1):513t
Q45Single correctAtoms and Nuclei
The number of spectral lines emitted by atomic hydrogen that is in the 4th energy level, is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 16
Approach:
Calculate the number of possible transitions from n=4 energy level to all lower levels using the formula for spectral lines.
Step 1:Identify the energy level
n=4(electron is in 4th energy level)
Step 2:List all possible downward transitions from n=4
Fromn=4:4→3,4→2,4→1; Fromn=3:3→2,3→1; Fromn=2:2→1
Step 3:Apply the formula for number of spectral lines
N=2n(n−1)=24(4−1)=24×3
Step 4:Verify by counting transitions
Transitions:4→3,4→2,4→1,3→2,3→1,2→1= 6 lines
Final answer: Option (1): 6 spectral lines
Q46NumericalMagnetic Effects of Current and Magnetism
The magnetic field inside a 200 turns solenoid of radius 10 cm is2.9×10−4Tesla. If the solenoid carries a current of 0.29 A, then the length of the solenoid is __________πcm.
SolutionAnswer: 8
Approach:
Use the formula for magnetic field inside a long solenoid and solve for length.
Step 1:Write the formula for magnetic field inside a solenoid
A parallel plate capacitor consisting of two circular plates of radius 10 cm is being charged by a constant current of 0.15 A. If the rate of change of potential difference between the plates is7×108V/s then the integer value of the distance between the parallel plates is _________μm. (Take,ϵ0=9×10−12mF,π=722)
SolutionAnswer: 1320
Approach:
Use the capacitor charging equation relating current, capacitance, and rate of voltage change to find the distance.
Step 1:Write the relationship between voltage, charge, and capacitance
V=CQ=ϵ0A/dit=ϵ0πr2itd
Step 2:Differentiate with respect to time
dtdV=ϵ0πr2id
Step 3:Rearrange to solve for distance d
d=iϵ0πr2(dtdV)
Step 4:Substitute the given values
d=0.15(9×10−12)(722)(0.1)2(7×108)
Step 5:Simplify the calculation
d=0.159×10−12×722×0.01×7×108
Step 6:Complete the calculation
d=0.159×22×0.01×10−4=0.151.98×10−4=13.2×10−4m
Final answer: 1320
Q48NumericalUnits and Measurements
A physical quantity Q is related to four observables a, b, c, d as follows:Q=cdab4where,a=(60±3)Pa;b=(20±0.1)m;c=(40±0.2)Nsm−2andd=(50±0.1)m, then the percentage error in Q is1000x, wherex= ______.
SolutionAnswer: 77
Approach:
Apply error propagation rules for the formula Q=cdab4 to find the percentage error.
Step 1:Write the relative error formula for the given expression
Two planets, A and B are orbiting a common star in circular orbits of radiiRAandRB, respectively, withRB=2RA. The planet B is42times more massive than planet A. The ratio(LALB)of angular momentum(LB)of planet B to that of planet A(LA)is closest to integer __________.
SolutionAnswer: 8
Approach:
Calculate angular momentum for each planet using orbital mechanics and find their ratio.
Step 1:Write angular momentum formula for circular orbit
L=mv0Rwherev0=RGM
Step 2:Express angular momentum in simplified form
L=mRGM×R=mGMR
Step 3:Write ratio of angular momenta
LALB=mARAmBRB
Step 4:Substitute the given values
LALB=mARA42mA2RA=42×2
Step 5:Simplify the result
LALB=4×2=8
Final answer: 8
Q50NumericalMotion in a Plane
Two cars P and Q are moving on a road in the same direction. Acceleration of car P increases linearly with time whereas car Q moves with a constant acceleration. Both cars cross each other at timet=0, for the first time. The maximum possible number of crossing(s) (including the crossing att=0) is ________.
SolutionAnswer: 3
Approach:
Analyze relative motion between cars P and Q with different acceleration patterns to determine maximum possible crossings.
Step 1:Define accelerations
aP=kt(linear),aQ=a(constant)
Step 2:Calculate relative acceleration
aQ/P=aQ−aP=a−kt
Step 3:Case I: Initial velocities and acceleration in same direction
IfuQ/PandaQ/Pare in same direction initially
Step 4:Case II: Initial velocity and acceleration in opposite directions
IfuQ/PandaQ/Pare in opposite directions
Step 5:Analyze Case II for maximum crossings
WithaQ/P=a−kt, the relative motion can have up to 2 turning points
Step 6:Determine maximum
Maximum crossings occur when initial conditions favor Case II
Final answer: 3
Chemistry25 questions
Q51Single correctCoordination Compounds
The calculated spin-only magnetic moments ofK3[Fe(OH)6]andK4[Fe(OH)6]respectively are :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 25.92 and 4.90 B.M.
Approach:
Calculate the spin-only magnetic moment using the formula μ = n(n+2) B.M., where n is the number of unpaired electrons. Determine the oxidation state of Fe and electronic configuration in each complex.
Step 1:Determine oxidation state of Fe in K3[Fe(OH)6]
3(+1)+x+6(−1)=0⇒x=+3
Step 2:Electronic configuration of Fe3+
Fe3+:[Ar]3d5
Step 3:Calculate magnetic moment for K3[Fe(OH)6]
μ=5(5+2)=35=5.92B.M.
Step 4:Determine oxidation state of Fe in K4[Fe(OH)6]
4(+1)+x+6(−1)=0⇒x=+2
Step 5:Electronic configuration of Fe2+
Fe2+:[Ar]3d6
Step 6:Calculate magnetic moment for K4[Fe(OH)6]
μ=4(4+2)=24=4.90B.M.
Final answer: 5.92 and 4.90 B.M.
Q52Single correctAtomic Structure
For hydrogen like species, which of the following graphs provides the most appropriate representation of E vs Z plot for a constant n? [E : Energy of the stationary state, Z : atomic number, n = principal quantum number]
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2Parabola opening downward
Approach:
Use the energy formula for hydrogen-like species to determine the relationship between E and Z at constant n.
Step 1:Write the energy formula for hydrogen-like species
En=−13.6×n2Z2eV
Step 2:At constant n, analyze the relationship
E∝−Z2(negative quadratic relationship)
Step 3:Determine the graph shape
SinceE=−n213.6Z2, this is a parabola opening downward
Step 4:Verify key features
AtZ=0,E=0; asZincreases,Ebecomes more negative
Final answer: Parabola opening downward
Q53Single correctSurface Chemistry
Given below are two statements: Statement (I): In partition chromatography, stationary phase is thin film of liquid present in the inert support. Statement (II): In paper chromatography, the material of paper acts as a stationary phase. In the light of the above statements, choose the correct answer from the options given below:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2Statement I is true but Statement II is false
Approach:
Evaluate each statement based on the principles of partition chromatography and paper chromatography.
Step 1:Analyze Statement I about partition chromatography
In partition chromatography, the stationary phase is a thin liquid film present on an inert support
Step 2:Analyze Statement II about paper chromatography
In paper chromatography, the stationary phase is water absorbed in the cellulose fibers, NOT the paper material itself
Step 3:Determine the correct option
Statement I is true, Statement II is false
Final answer: Statement I is true but Statement II is false
Q54Single correctBiomolecules
Identify the essential amino acids from below: (A) Valine (B) Proline (C) Lysine (D) Threonine (E) Tyrosine. Choose the correct answer from the options given below:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1(A), (C) and (D) only
Approach:
Identify which amino acids are essential (cannot be synthesized by the human body and must be obtained from diet).
Step 1:Check if Valine (A) is essential
Valine is an essential amino acid
Step 2:Check if Proline (B) is essential
Proline is a non-essential amino acid (can be synthesized)
Step 3:Check if Lysine (C) is essential
Lysine is an essential amino acid
Step 4:Check if Threonine (D) is essential
Threonine is an essential amino acid
Step 5:Check if Tyrosine (E) is essential
Tyrosine is a non-essential amino acid (can be synthesized from phenylalanine)
Ph3C+has 3 phenyl groups providing extensive resonance stabilization
Step 5:Apply carbocation stability order
Stability:Ph3C+>Ph-CH2+>cyclohexyl+>phenyl+
Final answer: Triphenylmethyl bromide
Q56Single correctChemical Equilibrium
Consider the equilibriumCO(g)+3H2(g)⇌CH4(g)+H2O(g). If the pressure applied over the system increases by two fold at constant temperature then (A) Concentration of reactants and products increases. (B) Equilibrium will shift in forward direction. (C) Equilibrium constant increases since concentration of products increases. (D) Equilibrium constant remains unchanged as concentration of reactants and products remain same. Choose the correct answer from the options given below:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1(A) and (B) only
Approach:
Apply Le Chatelier's principle and analyze the effect of pressure increase on the equilibrium system.
Step 1:Count moles on each side
Reactants:1+3=4moles, Products:1+1=2moles
Step 2:Analyze statement A - concentration increase
When pressure increases, volume decreases, soC=Vnincreases for all species
Step 3:Analyze statement B - equilibrium shift
Increased pressure favors side with fewer moles (forward direction, products side)
Step 4:Analyze statement C - equilibrium constant
Kcdepends only on temperature, NOT on pressure or concentration changes
Step 5:Analyze statement D - equilibrium constant
Kcremains constant at constant temperature, but individual concentrations DO change
Final answer: (A) and (B) only
Q57Single correctSolutions
Given below are two statements: Statement (I): NaCl is added to the ice at 0°C, present in the ice cream box to prevent the melting of ice cream. Statement (II): On addition of NaCl to ice at 0°C, there is a depression in freezing point. In the light of the above statements, choose the correct answer from the options given below:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2Both Statement I and Statement II are true
Approach:
Evaluate both statements using the principle of freezing point depression.
Step 1:Analyze Statement II on freezing point depression
Adding NaCl to ice causesΔTf(depression in freezing point below 0°C)
Step 2:Understand the mechanism
The ice-salt mixture has freezing point < 0°C, creating a colder environment
Step 3:Analyze Statement I on ice cream preservation
The colder ice-salt mixture (below 0°C) prevents ice cream from melting by maintaining lower temperature
Step 4:Determine the correct option
Both statements are scientifically accurate and related
Final answer: Both Statement I and Statement II are true
Q58Single correctHydrocarbons
Given below are two statements: Statement (I): On nitration of m-xylene withHNO3,H2SO4followed by oxidation, 4-nitrobenzene-1,3-dicarboxylic acid is obtained as the major product. Statement (II):CH3group is o/p-directing whileNO2group is m-directing group. In the light of the above statements, choose the correct answer from the options given below:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3Both Statement I and Statement II are true
Approach:
Analyze the nitration of m-xylene followed by oxidation, considering the directing effects of substituents.
Step 1:Analyze Statement II on directing effects
−CH3is ortho/para directing (activating),−NO2is meta directing (deactivating)
Step 2:Consider m-xylene structure
m-xylene hasCH3groups at positions 1 and 3
Step 3:Determine nitration position
BothCH3groups directNO2to position 4 (between them)
Step 4:Analyze oxidation step
Oxidation converts bothCH3groups toCOOHgroups
Step 5:Evaluate Statement I
The product matches the description in Statement I
Final answer: Both Statement I and Statement II are true
Q59Single correctChemical Kinetics and Equilibrium
0.1 M solution of KI reacts with excess of H2SO4and KIO3solution. According to equation5I−+IO3−+6H+→3I2+3H2OIdentify the correct statements: (A) 200 mL of KI solution reacts with 0.004 mol of KIO3(B) 200 mL of KI solution reacts with 0.006 mol of H2SO4(C) 0.5 L of KI solution produced 0.005 mol of I2(D) Equivalent weight of KIO3is equal to5Molecular weight
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1(A) and (D) only
Approach:
Analyze each statement based on stoichiometry of the given reaction between KI, H2SO4, and KIO3
Step 1:From the balanced equation, determine molar ratios
5mol I−reacts with1mol IO3−to produce3mol I2
Step 2:Statement (A): 200 mL of 0.1 M KI contains how many moles of KI?
Moles of KI =0.1×0.2=0.02mol
Step 3:From stoichiometry, moles of KIO3required
Moles of KIO3=50.02=0.004mol
Step 4:Statement (D): Equivalent weight of KIO3
In the reaction IO3−gains 5 electrons. Equivalent weight =5Molecular weight
Final answer: Correct statements are (A) and (D) only, which is option (1)
Q60Single correctElectrochemistry
Match List-I with List-II:
List-I (Applications)
List-II (Batteries/Cell)
A. Transistors
I. Anode - Zn/Hg; Cathode - HgO + C
B. Hearing aids
II. Hydrogen fuel cell
C. Invertors
III. Anode - Zn; Cathode - Carbon
D. Apollo space ship
IV. Anode - Pb; Cathode - Pb | PbO2
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1A-III, B-I, C-IV, D-II
Approach:
Match each application with the appropriate battery type based on their characteristics and usage
Final answer: A-III, B-I, C-IV, D-II, which is option (1)
Q61Single correctElectrochemistry
O2gas will be evolved as a product of electrolysis of: (A) an aqueous solution of AgNO3using silver electrodes. (B) an aqueous solution of AgNO3using platinum electrodes. (C) a dilute solution of H2SO4using platinum electrodes. (D) a high concentration solution of H2SO4using platinum electrodes. Choose the correct answer from the options given below:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1(B) and (C) only
Approach:
Analyze anode reactions in each case to determine when O2gas is evolved
Step 1:Case (A): AgNO3with silver electrodes
Silver electrode is reactive. Anode: Ag→Ag++e−. No O2 evolution
Step 2:Case (B): AgNO3with platinum electrodes
Platinum is inert. At anode, OH−or H2O oxidizes:4OH−→O2+2H2O+4e−
Final answer: (B) and (C) only, which is option (1)
Q62Single correctCoordination Compounds
Identify the homoleptic complexes with odd number of d electrons in the central metal. (A) [FeO4]2− (B) [Fe(CN)6]3− (C) [Fe(CN)5NO]2− (D) [CoCl4]2− (E) [Co(H2O)3F3] Choose the correct answer from the options given below:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1(B) and (D) only
Approach:
Identify homoleptic complexes (same type of ligands) and count d electrons in central metal
Step 1:Define homoleptic complex
Homoleptic complex has ALL ligands of the SAME type
Step 2:Analyze (B): [Fe(CN)6]3−
Homoleptic (all CN ligands). Fe oxidation state:x+6(−1)=−3⇒x=+3. Fe3+: 3d5(odd)
Step 3:Analyze (D): [CoCl4]2−
Homoleptic (all Cl ligands). Co oxidation state:x+4(−1)=−2⇒x=+2. Co2+: 3d7(odd)
Final answer: (B) and (D) only, which is option (1)
Q63Single correctOrganic Chemistry
Total number of sigma (σ) _____ and pi (π) _____ bonds respectively present in hex-1-en-4-yne are:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 113 and 3
Approach:
Draw the structure of hex-1-en-4-yne and count all sigma and π bonds systematically using bond analysis rules for alkenes and alkynes.
Final answer: 13 σ bonds and 3 π bonds, which is option (1)
Q64Single correctThermodynamics
If C(diamond) → C(graphite) + X kJ mol−1, C(diamond) + O2(g) → CO2(g) + Y kJ mol−1, C(graphite) + O2(g) → CO2(g) + Z kJ mol−1 at constant temperature. Then
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4X = Y - Z
Approach:
Apply Hess's Law to relate the three given thermochemical equations
Step 1:Equation (1): Diamond to graphite transformation
Given below are two statements: Statement (I): It is impossible to specify simultaneously with arbitrary precision, both the linear momentum and the position of a particle. Statement (II): If the uncertainty in the measurement of position and uncertainty in measurement of momentum are equal for an electron, then the uncertainty in the measurement of velocity is≥πh×2m1. In the light of the above statements, choose the correct answer from the options given below:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2Both Statement I and Statement II are true.
Approach:
Verify both statements using Heisenberg Uncertainty Principle
Step 1:Verify Statement I
Δx⋅Δp≥4πh
Step 2:Given condition for Statement II
Given:Δx=Δp, From HUP:(Δx)2≥4πh
Step 3:Calculate uncertainty in velocity
Δp=mΔv=Δx, soΔv=mΔx≥2m1πh
Final answer: Both Statement I and Statement II are true, which is option (2)
Q66Single correctOrganic Chemistry - Some Basic Principles and Techniques
Which one of the following reaction sequences will give an azo dye?
Azo dyes are formed by coupling reaction of diazonium salts with phenols or aromatic amines. The sequence must include reduction of nitro group to amine, diazotization, and coupling with phenol/amine.
Step 1:Reduction of nitrobenzene to aniline using Sn/HCl
C6H5NO2+3[H]Sn/HClC6H5NH2+2H2O
Step 2:Diazotization of aniline with NaNO2/HCl at 0-5°C
Step 3:Coupling of diazonium salt with β-naphthol in alkaline medium
C6H5N2+Cl−+β-naphtholNaOHPh−N=N−β-naphthol
Step 4:Verification that this produces an azo dye
Azo dye contains -N=N- chromophore group
Final answer: Option 1 is correct: Nitrobenzene → Sn/HCl → NaNO2/HCl → β-naphthol/NaOH gives an azo dye
Q67Single correctPhysical Chemistry - Chemical Kinetics
Drug X becomes ineffective after 50% decomposition. The original concentration of drug in a bottle was 16 mg/mL which becomes 4 mg/mL in 12 months. The expiry time of the drug in months is _____. Assume that the decomposition of the drug follows first order kinetics.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 46
Approach:
For first order kinetics, use the integrated rate equation to find the rate constant, then calculate the time for 50% decomposition (expiry time).
Step 1:Identify initial and final concentrations
[A]0=16 mg/mL,[A]=4 mg/mL,t=12 months
Step 2:Calculate the rate constant using first order equation
Step 3:Calculate time for 50% decomposition (expiry time)
t50%=k2.303log50100=k2.303log2=k2.303×0.301
Step 4:Alternatively, observe that 16 → 4 represents two half-lives
16t1/28t1/24, so2×t1/2=12months
Step 5:Calculate expiry time when 50% decomposition occurs
Drug becomes ineffective at 50% decomposition=t1/2=6 months
Final answer: 6 months
Q68Single correctPeriodic Table and Properties
The type of oxide formed by the element among Li, Na, Be, Mg, B and Al that has the least atomic radius is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1A2O3
Approach:
Compare atomic radii of given elements to identify the smallest, then determine its oxide type based on oxidation state.
Step 1:List given elements with their groups and periods
Li (Group 1, Period 2), Na (Group 1, Period 3), Be (Group 2, Period 2), Mg (Group 2, Period 3), B (Group 13, Period 2), Al (Group 13, Period 3)
Step 2:Apply periodic trends for atomic radius
Across period: B < Be < Li (Period 2); Al < Mg < Na (Period 3)
Step 3:Compare B and Al (both Period 2 < Period 3)
Atomic radius: B < Al
Step 4:Determine oxidation state and oxide type of Boron
B is in Group 13, has oxidation state +3, forms B2O3
Step 5:Verify oxide formula
B3++O2−→B2O3
Final answer: A2O3 (Boron oxide - B2O3)
Q69Single correctPeriodic Table and Properties
First ionisation enthalpy values of first four group 15 elements are given below. Choose the correct value for the element that is a main component of apatite family:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 11012 kJ mol−1
Approach:
Identify the main component of apatite family minerals, then match with the correct first ionization enthalpy from Group 15 elements.
Step 1:Identify main component of apatite family
Apatite: Ca5(PO4)3(F, Cl, OH)
Step 2:List first four Group 15 elements
Group 15: N, P, As, Sb
Step 3:Apply ionization energy trend
IE: N > P > As > Sb
Step 4:Match typical first ionization enthalpy values
N≈1402,P≈1012,As≈947,Sb≈834 kJ mol−1
Step 5:Confirm phosphorus is in apatite
Ca5(P5+O4)3(F, Cl, OH)
Final answer: 1012 kJ mol−1 (Phosphorus)
Q70Single correctOrganic Chemistry - Alcohols, Phenols and Ethers
Which one of the following, with HBr will give a phenol?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2Anisole structure: Benzene ring with−O−CH3
Approach:
Identify which compound undergoes SN2 cleavage with HBr to yield phenol. Anisole (phenyl methyl ether) reacts with HBr to give phenol and methyl bromide.
Step 1:Identify anisole structure
Anisole: C6H5−O−CH3 (phenyl methyl ether)
Step 2:Protonation of oxygen atom by HBr
C6H5−O−CH3+H+→C6H5−O+H−CH3
Step 3:SN2 attack by bromide ion on methyl carbon
Br−+C6H5−O+H−CH3SN2C6H5−OH+CH3Br
Step 4:Explain selectivity of cleavage
C-O bond adjacent to less hindered CH3 is cleaved, not C-O bond adjacent to benzene ring
Step 5:Verify product is phenol
Product: C6H5−OH (phenol)
Final answer: Option 2: Anisole (C6H5-O-CH3) gives phenol with HBr
Q71NumericalCoordination Chemistry
Consider the following low-spin complexes K3[Co(NO2)6], K4[Fe(CN)6], K3[Fe(CN)6], Cu2[Fe(CN)6] and Zn2[Fe(CN)6]. The sum of the spin-only magnetic moment values of complexes having yellow colour is _______ B.M. (answer is nearest integer)
SolutionAnswer: 0
Approach:
Identify yellow colored complexes from the list, determine their electronic configurations in low-spin state, and calculate spin-only magnetic moments.
Step 1:Identify yellow colored low-spin complexes
K3[Co(NO2)6] - yellow,K4[Fe(CN)6] - yellow
Step 2:Determine oxidation state and d-electron count for Co complex
K3[Co(NO2)6]:Co3+ has 3d6 configuration
Step 3:Apply low-spin configuration for Co3+ with NO2− (strong field ligand)
Co3+(3d6):t2g6eg0 (all electrons paired in low-spin)
Step 4:Calculate magnetic moment for Co complex
μ=0(0+2)=0=0 B.M.
Step 5:Determine oxidation state and d-electron count for Fe complex
K4[Fe(CN)6]:Fe2+ has 3d6 configuration
Step 6:Apply low-spin configuration for Fe2+ with CN− (strong field ligand)
Fe2+(3d6):t2g6eg0 (all electrons paired in low-spin)
Step 7:Calculate magnetic moment for Fe complex
μ=0(0+2)=0=0 B.M.
Step 8:Sum the magnetic moments of yellow colored complexes
Total μ=0+0=0 B.M.
Final answer: 0
Q72NumericalOrganic Chemistry - Hydrocarbons
Isomeric hydrocarbons→negative Baeyer's test (Molecular formula C9H12). The total number of isomers from above with four different non-aliphatic substitution sites is -
SolutionAnswer: 2
Approach:
Identify C9H12 isomers that give negative Baeyer's test (no C=C bonds, i.e., aromatic) and have four different aromatic substitution sites.
Step 1:Calculate degree of unsaturation for C9H12
DBE=22C+2−H=22(9)+2−12=28=4
Step 2:Identify that negative Baeyer's test means aromatic compound
Negative Baeyer’s test→benzene ring present, no alkene C=C bonds
Step 3:Determine possible substituent patterns on benzene
C9H12=C6H5+C3H7 (propyl groups as substituents)
Step 4:Identify isomers with four different substitution sites
Four different sites means no symmetry in substitution pattern
Step 5:Analyze 1,2,4-trimethylbenzene
1,2,4-trimethylbenzene has 4 different aromatic H positions
Step 6:Analyze 1,2,3-trimethylbenzene
1,2,3-trimethylbenzene has 4 different aromatic H positions
Q73NumericalOrganic Chemistry - Aldehydes, Ketones and Carboxylic Acids
In the Claisen-Schmidt reaction to prepare dibenzalacetone from 5.3 g benzaldehyde, a total of 3.51 g of product was obtained. The percentage yield in this reaction was ______ %.
SolutionAnswer: 60
Approach:
Calculate theoretical yield of dibenzalacetone from given benzaldehyde, then determine percentage yield using actual yield obtained.
Step 1:Calculate moles of benzaldehyde
Molar mass of benzaldehyde (C6H5CHO)=106 g/mol
Step 2:Calculate moles from given mass
Moles of benzaldehyde=1065.3=0.05 mol
Step 3:Determine stoichiometry of reaction
2 moles benzaldehyde→1 mole dibenzalacetone
Step 4:Calculate theoretical moles of product
Moles of dibenzalacetone=20.05=0.025 mol
Step 5:Calculate molar mass of dibenzalacetone
Molar mass of C17H14O=17(12)+14(1)+16=234 g/mol
Step 6:Calculate theoretical yield in grams
Theoretical yield=0.025×234=5.85 g
Step 7:Calculate percentage yield
Percentage yield=5.853.51×100=60%
Final answer: 60
Q74NumericalOrganic Chemistry - Some Basic Principles and Techniques
In the sulphur estimation, 0.20 g of a pure organic compound gave 0.40 g of barium sulphate. The percentage of sulphur in the compound is ____×10−1%. (Molar mass: O = 16, S = 32, Ba = 137 in g mol−1)
SolutionAnswer: 275
Approach:
Calculate mass of sulphur from mass of BaSO4 formed, then determine percentage of sulphur in the original compound.
Step 1:Calculate molar mass of BaSO4
MBaSO4=137+32+4(16)=137+32+64=233 g/mol
Step 2:Calculate moles of BaSO4 formed
Moles of BaSO4=2330.40 mol
Step 3:Determine moles of sulphur (1:1 ratio with BaSO4)
Moles of S=Moles of BaSO4=2330.40 mol
Step 4:Calculate mass of sulphur
Mass of S=2330.40×32=23312.8 g
Step 5:Calculate percentage of sulphur in compound
Q75NumericalChemical Bonding and Molecular Structure
Total number of non bonded electrons present in NO2−ion based on Lewis theory is _____.
SolutionAnswer: 12
Approach:
Draw Lewis structure of NO2− ion, count total valence electrons, then subtract bonding electrons to find non-bonding electrons.
Step 1:Count total valence electrons in NO2−
N: 5 electrons, O: 2×6=12 electrons, negative charge: +1 electron
Step 2:Draw Lewis structure with resonance
N is central atom bonded to two O atoms with resonance structures
Step 3:Analyze bonding in resonance structures
One N=O double bond (4e−) and one N-O single bond (2e−) in each resonance form
Step 4:Count bonding electrons
Bonding electrons=2×3=6 electrons (3 electron pairs in bonds)
Step 5:Calculate non-bonding electrons
Non-bonding electrons=18−6=12 electrons
Step 6:Verify electron distribution
Each O has 3 lone pairs (6e−), total =2×6=12 non-bonding electrons
Final answer: 12
Mathematics25 questions
Q1Single correctComplex Numbers and Quadratic Equations
If the set of alla∈R, for which the equation2x2+(a−5)x+15=3ahas no real root, is the interval(α,β), andX={x∈Z:α<x<β}, then∑x∈Xx2is equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 32139
Approach:
Use discriminant condition for no real roots, solve the quadratic inequality, find integer values in the interval, and calculate sum of squares.
Step 1:For no real roots, discriminant must be negative
(a−5)2−8(15−3a)<0
Step 2:Expand and simplify the inequality
a2+14a+25−120<0⇒a2+14a−95<0
Step 3:Factor the quadratic
(a+19)(a−5)<0
Step 4:Solve the inequality
a∈(−19,5)
Step 5:Find integers in the interval
−19<x<5⇒x∈{−18,−17,...,3,4}
Step 6:Calculate sum of squares
∑x∈Xx2=(12+22+...+42)+(12+22+...+182)
Step 7:Apply sum of squares formula
64×5×9+618×19×37=30+2109=2139
Final answer: 2139
Q2Single correctTrigonometry
Ifsinx+sin2x=1,x∈(0,2π), then(cos12x+tan12x)+3(cos10x+tan10x+cos8x+tan8x)+(cos6x+tan6x)is equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 32
Approach:
From the given condition, derive thatsinx=cos2xandtanx=cosx, then simplify the expression.
Step 1:From given condition, find relation
sinx+sin2x=1⇒sinx=1−sin2x=cos2x
Step 2:Derive another relation
tanx=cosxsinx=cosxcos2x=cosx
Step 3:Substitute in the expression
Expression =2cos12x+6[cos10x+cos8x]+2cos6x
Step 4:Factor and simplify using binomial theorem
=2[cos12x+3cos10x+3cos8x+cos6x]=2cos6x[(cosx+1)3]
Step 5:Use relationsinx=cos2x
=2sin3x(sinx+sinx)3=2[sin2x+sinx]3=2×13=2
Final answer: 2
Q3Single correctIntegral Calculus
Let the area enclosed between the curves∣y∣=1−x2andx2+y2=1beα. If9α=βπ+γ;β,γare integers, then the value of∣β−γ∣equals
(A)
(B)
(C)
(D)
SolutionAnswer: Option 433
Approach:
Calculate area using integration in first quadrant and multiply by 4 due to symmetry.
Step 1:Identify curves: parabola and circle
C1:∣y∣=1−x2,C2:x2+y2=1
Step 2:Calculate area in first quadrant
α=4[Area of circle−∫01(1−x2)dx]
Step 3:Evaluate the integral
4[4π−[x−3x3]01]=4[4π−(1−31)]
Step 4:Simplify to getα
α=π−38
Step 5:Multiply by 9
9α=9π−24=9π+(−24)
Step 6:Calculate the answer
∣β−γ∣=∣9−(−24)∣=∣9+24∣=33
Final answer: 33
Q4Single correctLimit, Continuity and Differentiability
If the domain of the functionlog5(18x−x2−77)is(α,β)and the domain of the functionlog(x−1)(x2−3x−42x2+3x−2)is(γ,δ), thenα2+β2+γ2is equal to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3186
Approach:
Find domain conditions for both logarithmic functions and calculate sum of squares.
Step 1:Domain of first function
18x−x2−77>0⇒x2−18x+77<0
Step 2:Solve the inequality
x∈(7,11)⇒α=7,β=11
Step 3:Domain conditions for second function
x−1>0,x−1=1,x2−3x−42x2+3x−2>0
Step 4:Factor and analyze the rational function
(x−4)(x+1)(2x−1)(x+2)>0
Step 5:Combine all conditions
x>1,x=2,x∈(4,∞)⇒x∈(4,∞)
Step 6:Calculate the answer
α2+β2+γ2=49+121+16=186
Final answer: 186
Q5Single correctLimit, Continuity and Differentiability
Let the functionf(x)=(x2−1)∣x2−ax+2∣+cos∣x∣be not differentiable at the two pointsx=α=2andx=β. Then the distance of the point(α,β)from the line12x+5y+10=0is equal to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 13
Approach:
Analyze non-differentiability points of f(x)=(x2−1)∣x2−ax+2∣+cos∣x∣ and find distance from the given line.
Step 1:Identify the non-differentiability source
cos∣x∣ is differentiable everywhere. Non-differentiability comes from ∣x2−ax+2∣
Step 2:Determine condition for non-differentiability at x=α=2
For non-differentiability at x=2: x2−ax+2=0 gives 4−2a+2=0
Step 3:Find the roots of x2−3x+2=0
x2−3x+2=(x−1)(x−2)=0⇒x=1 or x=2
Step 4:Check differentiability at x=1
At x=1: (x2−1)=0, so f(x)=0⋅∣0∣+cos(1)=cos(1)
Step 5:Identify the second non-differentiability point
Given α=2, and second point β must satisfy the problem constraints
Step 6:Calculate distance from (α,β)=(2,1) to line 12x+5y+10=0
Let a straight line L pass through the point P(2, -1, 3) and be perpendicular to the lines2x−1=−1y+1=2z−3and1x−3=3y−2=4z+2. If the line L intersects the yz-plane at the point Q, then the distance between the points P and Q is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 33
Approach:
Find direction ratios of L using cross product, write equation of L, find intersection with yz-plane, calculate distance.
Step 1:Find direction ratios of L perpendicular to both lines
LetS=N∪{0}. Define a relation R from S toRby:R={(x,y):logey=xloge(52),x∈S,y∈R}. Then, the sum of all the elements in the range of R is equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 235
Approach:
Express y in terms of x, identify as geometric series, and calculate sum to ∞.
Step 1:Express y in terms of x
logey=xloge(52)⇒y=(52)x
Step 2:List range elements for x = 0, 1, 2, ...
S={0,1,2,3,...}⇒Range={1,52,(52)2,...}
Step 3:Calculate sum of geometric series
Sum=1+52+(52)2+...=1−521
Step 4:Simplify the sum
Sum=531=35
Final answer: 35
Q8Single correctTrigonometry
Let the linex+y=1meet the axes of x and y at A and B, respectively. A right-angled triangle AMN is inscribed in the triangle OAB, where O is the origin and the points M and N lie on the lines OB and AB, respectively. If the area of the triangle AMN is94of the area of the triangle OAB and AN:NB =λ:1, then the sum of all possible value(s) ofλis:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 42
Approach:
Use geometry and trigonometry to relate areas and ratios, solve for angle parameter, find lambda values.
Step 1:Setup: OAB is right triangle at O with OA = OB = 1
Area of AOB=21,Area of AMN=94×21=92
Step 2:Let angle at A in AMN be(45°−θ)
AM=sec(45°−θ),AN=sec(45°−θ)cosθ,MN=sec(45°−θ)sinθ
Step 3:Calculate area of AMN
Ar(AMN)=21sec2(45°−θ)sinθcosθ=92
Step 4:Solve for theta
tanθ=21
Step 5:Calculate ratio
NBAN=λ=cotθ=2
Step 6:Sum of all values
Sum=2(only one valid value)
Final answer: 2
Q9Single correctCo-ordinate Geometry
Ifαx+βy=109is the equation of the chord of the ellipse9x2+4y2=1, whose mid point is(25,21), thenα+βis equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 358
Approach:
Use the T = S1 formula for finding the equation of a chord with given midpoint on an ellipse
Step 1:Identify ellipse parameters
9x2+4y2=1, soa2=9,b2=4
Step 2:Apply T = S1 formula with midpoint (5/2, 1/2)
925⋅x+421⋅y=9(25)2+4(21)2
Step 3:Simplify left side
185x+8y=3625+161
Step 4:Calculate right side
3625+161=144100+9=144109
Step 5:Multiply throughout by 144 to clear denominators
40x+18y=109
Step 6:Calculate α + β
α+β=40+18=58
Final answer: 58
Q10Single correctPermutations and Combinations
If all the words with or without meaning made using all the letters of the word "KANPUR" are arranged as in a dictionary, then the word at 440th position in this arrangement, is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3PRKAUN
Approach:
Count permutations systematically in dictionary order starting from each letter
Step 1:Arrange letters in alphabetical order
A, K, N, P, R, U
Step 2:Count words starting with A
Words starting with A =5!=120
Step 3:Count words starting with K
Words starting with K =5!=120
Step 4:Count words starting with N
Words starting with N =5!=120
Step 5:Count words starting with PA, PK, PN
Words starting with PA =4!=24, PK =24, PN =24
Step 6:Count words starting with PRA
Words starting with PRA =3!=6
Step 7:List next words
Position 439: PRKANU, Position 440: PRKAUN
Final answer: PRKAUN
Q11Single correctMatrices and Determinants
Letα(α=β) be the values of m, for which the equationsx+y+z=1;x+2y+4z=mandx+4y+10z=m2have infinitely many solutions. Then the value of∑n=110(nα+nβ)is equal to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1440
Approach:
For infinitely many solutions, the determinant of coefficient matrix must be zero, and consistency conditions must be satisfied
Step 1:Write coefficient matrix determinant
Δ=1111241410=1(20−16)−1(10−4)+1(4−2)=0
Step 2:For infinite solutions, set up consistency condition
Let P be the foot of the perpendicular from the point (1, 2, 2) on the line L:1x−1=−1y+1=2z−2. Let the liner=(i^+j^−2k^)+λ(i^−j^+k^),λ∈R, intersect the line L at Q. Then2(PQ)2is equal to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 127
Approach:
Find foot of perpendicular P from given point to line L, find intersection point Q of second line with L, calculate distance PQ
Step 1:Write parametric form of line L
Point on L:(μ+1,−μ−1,2μ+2)for parameterμ
Step 2:Find foot of perpendicular P using perpendicularity condition
AP⋅d=(μ,−μ−3,2μ)⋅(1,−1,2)=0⇒μ+μ+3+4μ=0⇒μ=−21
Step 3:Find coordinates of P
P=(21,−21,1)
Step 4:Find intersection point Q by equating second line with L
(1+λ,1−λ,−2+λ)=(μ+1,−μ−1,2μ+2). Solving:λ=−2,μ=−2
Step 5:Find coordinates of Q
Q=(−1,1,−2)(usingμ=−2in line L)
Step 6:Calculate PQ2
PQ2=(−1−21)2+(1+21)2+(−2−1)2=49+49+9=227
Step 7:Calculate 2(PQ)2
2(PQ)2=2×227=27
Final answer: 27
Q14Single correctCo-ordinate Geometry
Let a circle C pass through the points (4, 2) and (0, 2), and its centre lie on3x+2y+2=0. Then the length of the chord, of the circle C, whose midpoint is (1, 2), is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 223
Approach:
Find center of circle using perpendicular bisector and given line constraint, calculate radius, then use chord-midpoint relation
Step 1:Find midpoint of (4,2) and (0,2)
MAB=(2,2). Since slope of AB is 0, perpendicular bisector is vertical:x=2
Step 2:Center lies on both x=2 and 3x+2y+2=0
3(2)+2y+2=0⇒y=−4
Step 3:Calculate radius using distance from center to (4,2)
r=(2−4)2+(−4−2)2=4+36=40=210
Step 4:Calculate distance from center O(2,-4) to chord midpoint N(1,2)
ON=(2−1)2+(−4−2)2=1+36=37
Step 5:Apply chord-midpoint formula
Chord length=2r2−(ON)2=240−37=23
Final answer: 23
Q15Single correctStatistics and Probability
Let A =[aij]be a2×2matrix such thataij∈{0,1}for all i and j. Let the random variable X denote the possible values of the determinant of the matrix A. Then, the variance of X is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 283
Approach:
Enumerate all possible 2×2 matrices with entries {0,1}, calculate determinants, find probability distribution, compute variance
Step 1:Count total possible matrices
Total matrices =24=16(4 entries, each 0 or 1)
Step 2:Determinant can be -1, 0, or 1
det(A)=a11a22−a21a12takes values in{−1,0,1}
Step 3:Count matrices with det = -1 (when a11a22 = 0 and a21a12 = 1)
3 matrices have det = -1
Step 4:Count matrices with det = 0 (when a11a22 = a21a12)
10 matrices have det = 0
Step 5:Count matrices with det = 1 (when a11a22 = 1 and a21a12 = 0)
3 matrices have det = 1
Step 6:Calculate E[X]
E[X]=(−1)163+(0)1610+(1)163=0
Step 7:Calculate E[X2]
E[X2]=1⋅163+0⋅1610+1⋅163=166=83
Step 8:Calculate Var(X)
Var(X)=E[X2]−(E[X])2=83−0=83
Final answer: 83
Q16Single correctProbability
Bag 1 contains 4 white balls and 5 black balls, and Bag 2 contains n white balls and 3 black balls. One ball is drawn randomly from Bag 1 and transferred to Bag 2. A ball is then drawn randomly from Bag 2. If the probability that the ball drawn is white is4529, then n is equal to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 46
Approach:
Use total probability theorem with conditional probabilities for ball transfer scenarios
Step 1:Identify the given information
Bag 1: 4 white, 5 black balls (total 9), Bag 2: n white, 3 black balls (total n+3)
Step 2:Case 1: White ball transferred from Bag 1 to Bag 2
P(W from Bag 1)=94, After transfer: Bag 2 has (n+1) white, 3 black
Step 3:Case 2: Black ball transferred from Bag 1 to Bag 2
P(B from Bag 1)=95, After transfer: Bag 2 has n white, 4 black
Step 4:Apply total probability theorem
P(W)=94×n+4n+1+95×n+4n=4529
Step 5:Simplify the equation
9(n+4)4(n+1)+5n=4529 → 9(n+4)9n+4=4529
Step 6:Cross multiply and solve for n
45(9n+4)=29×9(n+4) → 405n+180=261n+1044
Step 7:Calculate n
n=144864=6
Final answer: 6
Q17Single correctNumber Theory
The remainder, when 7103 is divided by 23, is equal to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 114
Approach:
Use Fermat's Little Theorem and modular arithmetic to find 7103 mod 23
Step 1:Apply Fermat's Little Theorem
Since 23 is prime,722≡1(mod23)
Step 2:Express 103 in terms of 22
103=22×4+15, so7103=(722)4×715
Step 3:Calculate 72 mod 23
72=49≡3(mod23)
Step 4:Calculate 74 mod 23
74=(72)2≡32=9(mod23)
Step 5:Calculate 78 mod 23
78=(74)2≡92=81≡12(mod23)
Step 6:Calculate 715 = 78 × 74 × 72 × 7
715≡12×9×3×7(mod23)
Step 7:Final answer
7103≡14(mod23)
Final answer: 14
Q18Single correctCalculus - Integration and Maxima-Minima
Letf(x)=∫0xt(t2−9t+20)dt,1≤x≤5. If the range offis[α,β], then4(α+β)equals:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1157
Approach:
Find f'(x) to locate critical points, evaluate f(x) at boundary and critical points to find range.
Minimum =min{f(1),f(5)}=429=α. Maximum =f(4)=32=β
Step 8:Calculate final answer
4(α+β)=4(429+32)=29+128=157
Final answer: 157
Q19Single correctVector Algebra
Leta^be a unit vector perpendicular to the vectorsb=i^−2j^+3k^andc=2i^+3j^−k^, and makes an angle ofcos−1(−31)with the vectori^+j^+k^. Ifa^makes an angle of3πwith the vectori^+αj^+k^, then the value ofαis:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3−6
Approach:
Find unit vector perpendicular to b and c using cross product, then use angle conditions to find α
Leta1,a2,…,a2024be an Arithmetic Progression such thata1+(a5+a10+a15+…+a2020)+a2024=2233. Thena1+a2+a3+…+a2024is equal to _______.
SolutionAnswer: 11132
Approach:
Use the property that in an AP, sum of terms equidistant from ends is constant. Analyze the given sum structure and find a1+a2024, then compute the total sum.
Step 1:Apply the equidistant property of AP
In an AP: a1+a2024=a5+a2020=a10+a2015=…
Step 2:Identify the sequence in the middle sum
The sequence a5,a10,a15,…,a2020 is an AP with first term a5, common difference 5d
Step 3:Count the number of terms in the middle sequence
Number of terms =52020−5+1=52015+1=403+1=404
Step 4:Pair up the middle terms using equidistant property
a5+a2020=a1+a2024, a10+a2015=a1+a2024, etc.
Step 5:Count the number of pairs
Number of pairs =2404=202
Step 6:Express the middle sum
a5+a10+…+a2020=202(a1+a2024)
Step 7:Substitute in the given equation
a1+202(a1+a2024)+a2024=2233
Step 8:Solve for a1+a2024
203(a1+a2024)=2233⇒a1+a2024=2032233=11
Step 9:Calculate the sum of all 2024 terms
S2024=22024(a1+a2024)=22024×11
Step 10:Compute the final answer
S2024=1012×11=11132
Final answer: 11132
Q24NumericalComplex Numbers
Let integers a,b∈[−3,3] be such that a+b=0. Then the number of all possible ordered pairs (a, b), for which ∣z+b∣∣z−a∣=1 and z+1ωω2ωz+ω21ω21z+ω=1, z∈C, where ω and ω2 are the roots of x2+x+1=0, is equal to _______.
SolutionAnswer: 10
Approach:
Analyze two conditions: the locus condition ∣z−a∣=∣z+b∣ and the determinant condition involving cube roots of unity. Find values of z satisfying both, then count valid pairs (a,b).
Step 1:Analyze the first condition
∣z+b∣∣z−a∣=1⇒∣z−a∣=∣z+b∣
Step 2:Interpret geometrically
Perpendicular bisector passes through 2a−b and is ⊥ to real axis (if a=−b)
Step 3:Note the properties of cube roots of unity
ω,ω2 are roots of x2+x+1=0, so ω3=1 and 1+ω+ω2=0
Step 4:Evaluate the determinant
z+1ωω2ωz+ω21ω21z+ω
Step 5:Simplify using R1+R2+R3
Adding rows: (z+1+ω+ω2,ω+z+ω2+1,ω2+1+z+ω)=(z,z,z)
Step 6:Factor out z from first row and expand
Taking z common from row 1: det =z⋅1ωω21z+ω2111z+ω
Step 7:Set determinant equal to 1 and solve
z3=1⇒z=1,ω,ω2 (cube roots of unity)
Step 8:Apply condition for z = 1
∣1−a∣=∣1+b∣ for z=1 (real)
Step 9:Solve the two cases
Case 1: 1−a=1+b⇒a=−b (excluded since a+b=0); Case 2: 1−a=−(1+b)⇒a+b=2
Step 10:Count pairs with a+b=2
Valid pairs: (−1,3),(0,2),(1,1),(2,0),(3,−1)
Step 11:Apply condition for z=ω and z=ω2
For z=ω: ∣ω−a∣=∣ω+b∣ where ω=−21+23i
Step 12:Simplify using ωωˉ=1
1−a(ω+ωˉ)+a2=1+b(ω+ωˉ)+b2
Step 13:Solve the equation
a2+a−b2+b=0⇒(a−b)(a+b)+(a+b)=0⇒(a+b)(a−b+1)=0
Step 14:Count pairs with a=b−1 and a+b=0
Valid pairs: (−3,−2),(−2,−1),(0,1),(1,2),(2,3)
Step 15:Calculate total
Total pairs =5+5=10
Final answer: 10
Q25NumericalConic Sections - Parabola
Lety2=12xbe the parabola andSbe its focus. LetPQbe a focal chord of the parabola such that(SP)(SQ)=4147. LetCbe the circle described takingPQas a diameter. If the equation of circleCis64x2+64y2−αx−643y=β, thenβ−αis equal to _______.
SolutionAnswer: 1328
Approach:
Identify parabola parameters, use focal chord properties to find endpoints P and Q, then derive the circle equation with PQ as diameter.
Step 1:Identify parabola parameters
y2=12x⇒4a=12⇒a=3
Step 2:Write parametric coordinates for focal chord endpoints
Let P=(3t2,6t) and Q=(t23,−t6) since t1⋅t2=−1
Step 3:Calculate SP and SQ
SP=a+at2=3(1+t2) and SQ=a+t2a=3(1+t21)=t23(t2+1)
Step 4:Apply the given condition
SP⋅SQ=3(1+t2)⋅t23(1+t2)=t29(1+t2)2=4147
Step 5:Solve the equation
12(1+t2)2=49t2⇒12t4+24t2+12=49t2
Step 6:Solve the quadratic in t2
t2=2425±625−576=2425±7
Step 7:Choose t2=43 with t=−23 (negative for P below x-axis)
P=(3⋅43,6⋅(−23))=(49,−33)
Step 8:Find coordinates of Q using t2=−t1=32
Q=(3⋅34,6⋅32)=(4,43)
Step 9:Write circle equation with PQ as diameter
(x−x1)(x−x2)+(y−y1)(y−y2)=0
Step 10:Expand the x-terms
x2−4x−49x+9=x2−425x+9
Step 11:Expand the y-terms
y2−43y+33y−36=y2−3y−36
Step 12:Combine to get circle equation
x2+y2−425x−3y+9−36=0
Step 13:Multiply entire equation by 64
64x2+64y2−400x−643y−1728=0
Step 14:Compare with given form 64x2+64y2−αx−643y=β
How many questions are in the JEE Main 2025 January 29, Shift 2 paper?
The JEE Main 2025 January 29, Shift 2 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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JEE Main awards +4 marks for each correct answer and −1 for a wrong answer, with 0 for questions left unattempted. The paper covers Physics, Chemistry and Mathematics.
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