JEE Main 2025 January 23, Shift 2 Question Paper with Solutions
All 75 questions from the JEE Main 2025 (January 23, 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.
Q26Single correctMagnetic Effects of Current and Magnetism
A galvanometer having a coil of resistance 30Ω need 20 mA of current for full-scale deflection. If a maximum current of 3 A is to be measured using this galvanometer, the resistance of the shunt to be added to the galvanometer should be X30Ω, where X is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2149
Approach:
Use the shunt resistance formula for converting a galvanometer to an ammeter. The shunt resistance is calculated using the current division rule where most of the current passes through the shunt.
Step 1:Identify given values: galvanometer resistance G = 30Ω, galvanometer current Ig = 20 mA = 0.02 A, maximum current I = 3 A
G=30Ω,Ig=0.02 A,I=3 A
Step 2:Apply shunt resistance formula
S=I−IgIg⋅G=3−0.020.02×30
Step 3:Calculate the shunt resistance
S=2.980.6=29860=14930Ω
Step 4:Compare with given form to find X
X30=14930⇒X=149
Final answer: The value of X is 149
Q27Single correctKinematics
A ball having kinetic energy KE, is projected at an angle of 60∘ from the horizontal. What will be the kinetic energy of ball at the highest point of its flight?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 44(KE)
Approach:
At the highest point of projectile motion, the vertical component of velocity becomes zero, only horizontal component remains. Calculate the kinetic energy using horizontal component of initial velocity.
Step 1:Express initial kinetic energy in terms of initial velocity
KE=21mv2
Step 2:Find horizontal component of velocity for angle 60°
vx=vcos60∘=v×21=2v
Step 3:Calculate kinetic energy at highest point using horizontal velocity
KEtop=21mvx2=21m(2v)2=21m4v2
Step 4:Express in terms of initial KE
KEtop=41×21mv2=4KE
Final answer: The kinetic energy at the highest point is KE/4
Q28Single correctElectrostatics
Two charges 7μC and −4μC are placed at (−7 cm,0,0) and (7 cm,0,0) respectively. Given, ϵ0=8.85×10−12C2 N−1 m−2, the electrostatic potential energy of the charge configuration is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1-1.8 J
Approach:
Calculate the electrostatic potential energy of two point charges using the formula U = kq₁q₂/r, where k = 1/(4πε₀) and r is the distance between charges.
Step 1:Identify the charges and their positions
q1=7μC=7×10−6 C,q2=−4μC=−4×10−6 C
Step 2:Calculate distance between charges
r=∣7−(−7)∣ cm=14 cm=0.14 m
Step 3:Apply potential energy formula
U=0.149×109×7×10−6×(−4)×10−6
Step 4:Calculate the potential energy
U=0.149×109×(−28)×10−12=0.14−252×10−3=−1.8 J
Final answer: The electrostatic potential energy is -1.8 J
Q29Single correctElectrostatics
Two point charges −4μC and 4μC, constituting an electric dipole, are placed at (−9,0,0) cm and (9,0,0) cm in a uniform electric field of strength 104 NC−1. The work done on the dipole in rotating it from the equilibrium through 180∘ is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 214.4 mJ
Approach:
Calculate work done in rotating a dipole in uniform electric field using W = pE(cosθ₁ - cosθ₂), where p is dipole moment, E is electric field, and angles are measured from equilibrium position.
Step 1:Calculate the dipole moment
p=q×2a=4×10−6×2×0.09=4×10−6×0.18
Step 2:Identify the electric field strength
E=104 N/C
Step 3:Apply work done formula for 180° rotation from equilibrium
W=pE(cos0∘−cos180∘)=pE(1−(−1))=2pE
Step 4:Calculate the work done
W=2×7.2×10−7×104=14.4×10−3 J=14.4 mJ
Final answer: The work done on the dipole is 14.4 mJ
Q30Single correctProperties of Solids and Liquids
A massless spring gets elongated by amount x1 under a tension of 5 N. Its elongation is x2 under the tension of 7 N. For the elongation of (5x1−2x2), the tension required is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 311 N
Approach:
Use Hooke's law F = kx to establish relationship between force and elongation. Find spring constant k using given conditions, then calculate force for the required elongation.
Step 1:Apply Hooke's law for first condition
5=kx1⇒k=x15
Step 2:Apply Hooke's law for second condition
7=kx2⇒k=x27
Step 3:Equate both expressions to find relation between x₁ and x₂
Final answer: The tension in the spring for elongation (5x₁ - 2x₂) is 11 N
Q31Single correctThermodynamics
Water of mass m gram is slowly heated to increase the temperature from T1 to T2. The change in entropy of the water, given specific heat of water is 1 Jkg−1K−1, is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1mln(T1T2)
Approach:
Calculate entropy change using the formula ΔS = ∫(dQ/T) for reversible heating process. For constant specific heat, this integrates to ΔS = mc ln(T₂/T₁).
Step 1:Write entropy change formula for reversible process
ΔS=∫T1T2TdQ
Step 2:Express heat in terms of specific heat and temperature change
dQ=mcdT where c=1 J kg−1 K−1
Step 3:Substitute and integrate
ΔS=∫T1T2TmcdT=mc∫T1T2TdT
Step 4:Evaluate the integral
ΔS=mc[lnT]T1T2=mcln(T1T2)
Step 5:Substitute c = 1 J kg⁻¹ K⁻¹ and convert mass to kg (m grams = m/1000 kg, but given answer suggests m is treated as dimensionless coefficient)
ΔS=m×1×ln(T1T2)=mln(T1T2)
Final answer: The change in entropy is m ln(T₂/T₁)
Q32Single correctProperties of Solids and Liquids
Water flows in a horizontal pipe whose one end is closed with a valve. The reading of the pressure gauge attached to the pipe is P1. The reading of the pressure gauge falls to P2 when the valve is opened. The speed of water flowing in the pipe is proportional to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4P1−P2
Approach:
Apply Bernoulli's equation for horizontal pipe flow. When valve is closed, water is at rest (v=0, P=P₁). When valve opens, water flows with velocity v and pressure drops to P₂.
Step 1:Apply Bernoulli's equation for horizontal pipe when valve is closed
P1+21ρv12=constant,v1=0
Step 2:Apply Bernoulli's equation when valve is open
P2+21ρv2=constant
Step 3:Equate both expressions (same horizontal level)
P1+0=P2+21ρv2
Step 4:Solve for velocity v
P1−P2=21ρv2⇒v2=ρ2(P1−P2)
Step 5:Take square root to get velocity
v=ρ2(P1−P2)∝P1−P2
Final answer: The speed of water is proportional to √(P₁ - P₂)
Q33Single correctOptics
A concave mirror of focal length f in air is dipped in a liquid of refractive index μ. Its focal length in the liquid will be:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2f
Approach:
For mirrors, the focal length depends only on the geometry (radius of curvature) and not on the refractive index of the surrounding medium. This is because reflection laws do not depend on the medium.
Step 1:Recall the mirror equation and focal length relation
f=2R where R is radius of curvature
Step 2:Note that law of reflection is independent of medium
θi=θr holds in any medium
Step 3:Analyze effect of refractive index on mirrors
Mirror focal length: f=2R (no μ dependence)
Step 4:Conclude that focal length remains same
fliquid=fair=f
Final answer: The focal length of the concave mirror in liquid remains f (unchanged)
Q34Single correctCurrent Electricity
What is the current through the battery in the circuit shown below
(A)
(B)
(C)
(D)
SolutionAnswer: Option 20.5 A
Approach:
Calculate equivalent resistance of parallel resistors and apply Ohm's law to find the current through the battery.
Step 1:Identify the circuit configuration
R1=20Ω,R2=20Ω in parallel, V=5V
Step 2:Calculate equivalent resistance for parallel combination
Req1=R11+R21=201+201=202=101⇒Req=10Ω
Step 3:Apply Ohm's law to find total current
I=ReqV=105=0.5A
Final answer: [object Object]
Q35Single correctOptics
The refractive index of the material of a glass prism is 3. The angle of minimum deviation is equal to the angle of the prism. What is the angle of the prism?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 160^∘
Approach:
Use the prism formula relating refractive index to prism angle and minimum deviation. Given that minimum deviation equals prism angle, simplify and solve for A.
Step 1:Write the prism formula at minimum deviation
The width of one of the two slits in Young's double slit experiment is d while that of the other slit is xd. If the ratio of the maximum to the minimum intensity in the interference pattern on the screen is 9:4 then what is the value of x? (Assume that the field strength varies according to the slit width.)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 25
Approach:
Field amplitude is proportional to slit width. Use the intensity ratio formula for interference to find the width ratio x.
Step 1:Establish amplitude-width relationship
A1∝d,A2∝xd⇒A2=xA1
Step 2:Express intensity ratio in terms of amplitudes
IminImax=(A1−A2)2(A1+A2)2=49
Step 3:Take square root and substitute A₂ = xA₁
∣A1−A2∣A1+A2=23⇒∣A1−xA1∣A1+xA1=23
Step 4:Solve for x (assuming x > 1 since second slit is wider)
x−11+x=23⇒2(1+x)=3(x−1)⇒2+2x=3x−3⇒x=5
Final answer: [object Object]
Q37Single correctThermodynamics
Using the given P-V diagram, the work done by an ideal gas along the path ABCD is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3-3P0V0
Approach:
Calculate work done in each segment of the P-V diagram. Work done equals area under the curve in P-V diagram.
Step 1:Work done in path AB (isobaric expansion at pressure P₀)
WAB=P0(3V0−V0)=2P0V0
Step 2:Work done in path BC (isochoric process at constant volume 3V₀)
WBC=0 (constant volume)
Step 3:Work done in path CD (isobaric compression at pressure 2P₀)
WCD=2P0(V0−3V0)=2P0(−2V0)=−4P0V0
Step 4:Work done in path DA (isochoric process at constant volume V₀)
A plane electromagnetic wave of frequency 20 MHz travels in free space along the +x direction. At a particular point in space and time, the electric field vector of the wave is Ey=9.3 Vm−1. The magnetic field vector of the wave is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2Bz = 3.1 ×10−8 T
Approach:
In an electromagnetic wave, E and B are related by E = cB. Use the right-hand rule to determine the direction of B given E and propagation direction.
Step 1:Write the relation between E and B in an EM wave
B=cE where c=3×108m/s
Step 2:Calculate the magnetic field magnitude
B=cEy=3×1089.3=3.1×10−8T
Step 3:Determine direction using right-hand rule (E × B || propagation)
E×B points in +x direction, E is in y-direction ⇒B is in z-direction
Final answer: [object Object]
Q39Single correctOscillations and Waves
The equation of a transverse wave travelling along a string is y(x,t)=4.0sin[20×10−3x+600t] mm, where x is in mm and t is in second. The velocity of the wave is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2-30 m/s
Approach:
Identify wave equation form and extract wave parameters. The sign of (kx + ωt) indicates negative x-direction propagation.
Step 1:Identify the wave equation form and direction
y=Asin(kx+ωt) represents wave traveling in negative x-direction (+ sign means opposite to +x)
Step 2:Extract wave parameters from the equation
k=20×10−3mm−1=20m−1, ω=600rad/s
Step 3:Calculate wave velocity magnitude and apply direction
∣v∣=kω=20600=30m/s, direction is -x
Final answer: [object Object]
Q40Single correctAtoms and Nuclei
Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): The binding energy per nucleon is found to be practically independent of the atomic number A, for nuclei with mass numbers between 30 and 170. Reason (R): Nuclear force is long range. In the light of the above statements, choose the correct answer from the options given below:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1(A) is true but (R) is false
Approach:
Evaluate the truth of both assertion and reason independently, then determine their relationship.
Step 1:Evaluate Assertion (A)
BE/A curve is nearly constant (≈8MeV) for 30<A<170
Step 2:Evaluate Reason (R)
Nuclear force is SHORT RANGE (∼10−15m), NOT long range
Step 3:Determine the correct option
A is true, R is false⇒Option 1
Final answer: [object Object]
Q41Single correctGravitation
If a satellite orbiting the Earth is 9 times closer to the Earth than the Moon, what is the time period of rotation of the satellite? Given rotational time period of Moon = 27 days and gravitational attraction between the satellite and the moon is neglected.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 21 day
Approach:
Apply Kepler's third law relating orbital period to orbital radius. Satellite at Rs = Rm/9 will have period calculated using T² ∝ R³.
Step 1:Apply Kepler's third law for both satellite and Moon
Tm2Ts2=Rm3Rs3
Step 2:Substitute Rs = Rm/9 (satellite is 9 times closer)
Tm2Ts2=(RmRm/9)3=(91)3=7291
Step 3:Solve for satellite's time period
Ts=729Tm=2727=1day
Final answer: [object Object]
Q42Single correctRotational Motion
A circular disk of radius R meter and mass M kg is rotating around the axis perpendicular to the disk. An external torque is applied to the disk such that θ(t)=5t2−8t, where θ(t) is the angular position of the rotating disc as a function of time t. How much power is delivered by the applied torque, when t=2 s?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 460MR2
Approach:
Find angular velocity and acceleration by differentiating θ(t). Calculate torque using τ = Iα and power using P = τω.
Step 1:Find angular velocity by differentiating θ(t)
ω(t)=dtdθ=dtd(5t2−8t)=10t−8
Step 2:Find angular acceleration
α(t)=dtdω=dtd(10t−8)=10rad/s2
Step 3:Calculate moment of inertia for the disk
I=21MR2
Step 4:Calculate applied torque
τ=Iα=21MR2×10=5MR2
Step 5:Calculate power delivered at t = 2s
P=τω=5MR2×12=60MR2
Final answer: [object Object]
Q43Single correctUnits and Measurements
The energy of a system is given as E(t)=α3e−βt, where t is the time and β=0.3 s−1. The errors in the measurement of α and t are 1.2% and 1.6%, respectively. At t=5 s, maximum percentage error in the energy is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 16%
Approach:
Given E(t)=α3e−βt where β=0.3s−1, t=5s, error in α is 1.2% and error in t is 1.6%. Taking natural logarithm: lnE=3lnα−βt. Differentiating: EΔE=3αΔα+βΔt. Maximum percentage error: EΔE×100=3(1.2%)+0.3(1.6%)(5)=3.6%+2.4%=6%.
Matching dimensions: (A) Permeability μ0 has dimensions [MLT−2A−2] (III). (B) Magnetic field B has dimensions [MT−2A−1] (II). (C) Magnetic moment m has dimensions [L2A] (IV). (D) Torsional constant κ has dimensions [ML2T−2] (I).
Step 1:Determine dimensions of permeability of free space
μ0=HB⇒[μ0]=[MLT−2A−2]
Step 2:Determine dimensions of magnetic field
B=ILF⇒[B]=[MT−2A−1]
Step 3:Determine dimensions of magnetic moment
m=IA⇒[m]=[AL2]
Step 4:Determine dimensions of torsional constant
κ=θτ⇒[κ]=[ML2T−2]
Final answer: [object Object]
Q45Single correctDual Nature of Matter and Radiation
In photoelectric effect an em-wave is incident on a metal surface and electrons are ejected from the surface. If the work function of the metal is 2.14eV and stopping potential is 2V, what is the wavelength of the em-wave? (Given hc=1242eV⋅nm where h is the Planck's constant and c is the speed of light in vacuum.)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1300 \, nm
Approach:
Using Einstein's photoelectric equation: E=ϕ+Kmax where E=λhc, ϕ=2.14eV (work function), and Kmax=eV0=2eV (stopping potential). Therefore: λhc=2.14+2=4.14eV. Solving for λ: λ=4.14hc=4.141242=300nm.
Step 1:Apply Einstein's photoelectric equation
Ephoton=ϕ+Kmax
Step 2:Calculate total photon energy
E = 2.14 + 2.0 = 4.14 \, eV
Step 3:Calculate wavelength using energy-wavelength relation
λ=Ehc=4.141242 = 300 \, nm
Final answer: [object Object]
Q46NumericalElectromagnetic Induction and Alternating Currents
A time varying potential difference is applied between the plates of a parallel plate capacitor of capacitance 2.5μF. The dielectric constant of the medium is 5. The current is:
SolutionAnswer: 100
Approach:
Displacement current is given by Id=CdtdV where C=2.5μF=2.5×10−6F and Id=0.25mA=0.25×10−3A. Therefore: dtdV=CId=2.5×10−60.25×10−3=2.50.25×103=0.1×103=100V/s.
Step 1:Apply displacement current formula
Id = CdtdV
Step 2:Rearrange to find rate of change of voltage
dtdV=CId
Step 3:Substitute values
dtdV=2.5×10−60.25×10−3=100V/s
Final answer: [object Object]
Q47NumericalElectromagnetic Induction and Alternating Currents
In a series LCR circuit, a resistor of 300Ω, a capacitor of 25nF and an inductor of 100mH are used. The angular frequency at resonance is:
SolutionAnswer: 2
Approach:
For maximum current in LCR circuit, resonance condition is ω0=LC1 where L=100mH=100×10−3H=10−1H and C=25nF=25×10−9F. Therefore: ω0=10−1×25×10−91=25×10−101=5×10−51=5105=2×104rad/s. Answer is 2.
Step 1:Apply resonance condition for LCR circuit
ω0=LC1
Step 2:Substitute values
ω0=(100×10−3)(25×10−9)1
Step 3:Calculate angular frequency
ω0=25×10−101=5×10−51=2×104rad/s
Final answer: [object Object]
Q48NumericalProperties of Solids and Liquids
An air bubble of radius 1.0mm is observed at a depth of 20cm below the free surface of a liquid having surface tension 0.075Nm−1 and density 103kgm−3. The difference in pressure is:
SolutionAnswer: 2150
Approach:
Calculate total pressure difference as sum of hydrostatic pressure and excess pressure due to surface tension. For an air bubble in liquid, excess pressure is 2T/r (single interface).
Step 1:Identify given values
r=1.0mm=10−3m,h=20cm=0.2m,T=0.075N/m,ρ=103kg/m3
Step 2:Calculate hydrostatic pressure at depth h
Phydro=ρgh=103×10×0.2=2000N/m2
Step 3:Calculate excess pressure due to surface tension (air bubble in liquid has single interface)
Pexcess=r2T=10−32×0.075=10−30.15=150N/m2
Step 4:Calculate total pressure difference
ΔP=Phydro+Pexcess=2000+150=2150N/m2
Final answer: [object Object]
Q49NumericalGravitation
A satellite of mass 2M is revolving around earth in a circular orbit at a height of 3R from earth surface. The angular momentum of the satellite is MxGMR. The value of x is ______, where M and R are the mass and radius of earth, respectively. (G is the gravitational constant)
SolutionAnswer: 3
Approach:
Orbital radius r=R+3R=34R. Orbital velocity v=rGM=4R/3GM=4R3GM. Angular momentum L=mvr=2M×4R3GM×34R=2M×34R4R3GM=32MR4R3GM=32MR×2R3GM=3RMR3GM=3M3GMR=M93GMR=M3GMR. Therefore x=3.
Step 1:Calculate orbital radius
r = R + 3R=34R
Step 2:Calculate orbital velocity
v=rGM=4R3GM
Step 3:Calculate angular momentum
L=mvr=2M×4R3GM×34R=M3GMR
Final answer: [object Object]
Q50NumericalCurrent Electricity
At steady state the charge on the capacitor, as shown in the circuit below, is _____ μC.
SolutionAnswer: 16
Approach:
At steady state, the capacitor acts as an open circuit. From the circuit configuration, the 10 Ω and 15 Ω resistors form a voltage divider. If the capacitor is parallel to the 10 Ω resistor, the voltage across it is V10Ω=10+1510×5=2510×5=2V. Charge on capacitor: Q=CV=8×10−6×2=16×10−6C=16μC.
Step 1:Identify circuit configuration at steady state
Capacitor (8 μF) parallel to 10 Ω, in series with 15 Ω
Step 2:Calculate voltage across 10 Ω resistor using voltage divider
Identify the products [A] and [B], respectively in the following reaction:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2Phenol and p-benzoquinone
Approach:
Identify products from nucleophilic aromatic substitution followed by oxidation reaction sequence
Step 1:Chlorobenzene reacts with NaOH at high temperature and pressure (623 K, 300 atm) in nucleophilic aromatic substitution to form sodium phenoxide, which upon acidification gives phenol as intermediate [A]
Step 2:Phenol undergoes oxidation with acidified Na₂Cr₂O₇ (strong oxidizing agent) to form p-benzoquinone
PhenolNa2Cr2O7/H2SO4p-benzoquinone
Step 3:The reaction sequence shows nucleophilic aromatic substitution followed by oxidation, producing phenol intermediate and p-benzoquinone as final product
Products: [A] = phenol, [B] = p-benzoquinone
Final answer: [object Object]
Q52Single correctd- and f-Block Elements
Consider the following reactions: K2Cr2O7KOH,−H2O[A]H2SO4,−H2OCrO3. The sum of spin only magnetic moments of the compounds A and B is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3K2CrO4 and K2Cr2O7
Approach:
Identify products of chromate-dichromate equilibrium under basic and acidic conditions
Step 1:Potassium dichromate reacts with KOH in basic medium, converting dichromate ion to chromate ion
K2Cr2O7+2KOH→2K2CrO4+H2O
Step 2:Potassium chromate reacts with H₂SO₄ in acidic medium, converting back to dichromate
2K2CrO4+H2SO4→K2Cr2O7+K2SO4+H2O
Step 3:This is the classic chromate-dichromate equilibrium: chromate (yellow) in base, dichromate (orange) in acid
2CrO42−+2H+⇌Cr2O72−+H2O
Final answer: [object Object]
Q53Single correctChemical Thermodynamics
The effect of temperature on spontaneity of reactions are represented as:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1(B) and (C) only
Approach:
Use Gibbs free energy equation to analyze spontaneity criteria for different sign combinations of ΔH and ΔS at various temperatures
Step 1:For spontaneity, ΔG must be negative. Using Gibbs equation: ΔG = ΔH - TΔS
Δ G = Δ H - TΔ S < 0 for spontaneous process
Step 2:Case (A): ΔH = +, ΔS = -, any T. ΔG = (+) - T(-) = (+) + T(+) = always positive, never spontaneous
(A): Δ G = (+) - T(-) > 0 at any T, non-spontaneous
Step 3:Case (B): ΔH = +, ΔS = +, low T. At low T, TΔS term is small, so ΔG = (+) - (small +) > 0, non-spontaneous. But labeled as spontaneous at low T - this is INCORRECT
(B): Δ G = (+) - T(+) > 0 at low T, should be non-spontaneous
Step 4:Case (C): ΔH = -, ΔS = -, low T. At low T, TΔS is small, so ΔG = (-) - (small -) = (-) + (small +) < 0, spontaneous at low T. Labeled as non-spontaneous - INCORRECT
(C): Δ G = (-) - T(-) < 0 at low T, should be spontaneous
Step 6:Rows (B) and (C) show INCORRECT temperature-spontaneity relationships
Answer: (B) and (C) only represent incorrect cases
Final answer: [object Object]
Q54Single correctChemical Kinetics
Which of the following graphs most appropriately represents a zero order reaction?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1Linear decrease of Reactant Concentration vs Time
Approach:
Apply zero order kinetics integrated rate law to identify characteristic concentration vs time graph
Step 1:For zero order reaction, rate is independent of concentration
Rate = k[A]0 = k = constant
Step 2:The integrated rate law for zero order reaction shows linear relationship between concentration and time
[A] = [A]0 - kt
Step 3:Graph of [Reactant] vs time for zero order is a straight line with negative slope (-k)
Slope = -k, y-intercept = [A]0
Step 4:Option (1) shows linear decrease of reactant concentration vs time, which is characteristic of zero order reaction
Graph (1): Linear [A] vs t represents zero order
Final answer: [object Object]
Q55Single correctEquilibrium
Consider the reaction X2Y(g)⇌X2(g)+21Y2(g). The equation representing correct relationship between the degree of dissociation (x) of X2Y and equilibrium constant (Kp) is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2x=3p2Kp2
Approach:
Set up ICE table for dissociation equilibrium and derive relationship between degree of dissociation and Kp
Step 1:Set up ICE table. Initial: X₂Y = p atm, products = 0. At equilibrium with degree of dissociation x
Given below are two statements: Consider the following reaction for hydration of aldehydes and ketones.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4Both Statement I and Statement II are true
Approach:
Analyze hydration equilibrium of carbonyl compounds based on electronic and steric factors
Step 1:Statement I: Formaldehyde (HCHO) has small H substituents with no steric hindrance and no electron-donating groups, making the carbonyl carbon highly electrophilic
HCHO+H2O⇌H2C(OH)2,K≈2280
Step 2:Statement II: Trichloroacetaldehyde (CCl₃CHO) has three -Cl groups with strong -I effect, withdrawing electrons and making carbonyl carbon more electrophilic, favoring nucleophilic attack by water
CCl3CHO+H2O⇌CCl3CH(OH)2,K≈2000
Step 3:Both statements correctly explain why these aldehydes have high hydration equilibrium constants
Both statements are scientifically accurate
Final answer: [object Object]
Q57Single correctAtomic Structure
Given below are two statements: Statement (I): For a given shell, the total number of allowed orbitals is given by n². Statement (II): For any subshell, the spatial orientation of the orbitals is given by -l to +l values including zero. 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:
Verify quantum number rules for orbital count and spatial orientations
Step 1:Statement I: For principal quantum number n, l ranges from 0 to (n-1), and for each l, ml ranges from -l to +l (total 2l+1 orbitals)
Total orbitals=∑_l=0n−1(2l+1)=1+3+5+…=n2
Step 2:Statement II: For azimuthal quantum number l, magnetic quantum number ml ranges from -l to +l including zero, giving (2l+1) orientations
ml=−l,−l+1,…,0,…,+l−1,+l(2l+1 values)
Step 3:Both statements are fundamental principles of quantum mechanics and atomic structure
Both statements are correct
Final answer: [object Object]
Q58Single correctRedox Reactions and Electrochemistry
Standard electrode potentials for a few half cells are mentioned below: ECu2+/Cu∘=0.34 V, EZn2+/Zn∘=−0.76 V, EAg+/Ag∘=0.80 V. The galvanic cell that gives the highest voltage is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1Zn|Zn2+(1M)∣∣Ag+(1M)|Ag
Approach:
Calculate cell potential for each galvanic cell combination and identify the one with highest voltage (most negative ΔG°)
Step 1:Relationship between cell potential and Gibbs energy
ΔG°=−nFE°cell}
Step 2:Calculate E°cell for option (1): Zn|Zn²⁺||Ag⁺|Ag
The α-Helix and β-Pleated sheet structures of protein are associated with its:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3secondary structure
Approach:
Identify the level of protein structure associated with α-helix and β-pleated sheet conformations
Step 1:Identify protein structure levels
Proteins have four levels of structure: primary (sequence), secondary (α-helix, β-pleated sheet), tertiary (3D folding), and quaternary (multiple subunits)
α-Helix and β-Pleated sheet are regular, repeating structural motifs stabilized by hydrogen bonding between backbone atoms
Step 3:Conclude the structure level
Both α-helix and β-pleated sheet are characteristic of secondary structure of proteins
Final answer: [object Object]
Q60Single correctp-Block Elements
Given below are the atomic numbers of some group 14 elements. The atomic number of the element with lowest melting point is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 450
Approach:
Compare melting points of Group 14 elements to identify the one with lowest melting point
Step 1:Identify Group 14 elements by atomic number
Atomic number 6 = C (Carbon), 14 = Si (Silicon), 50 = Sn (Tin), 82 = Pb (Lead)
Step 2:Compare melting points of Group 14 elements
Melting points: C (3823 K), Si (1687 K), Sn (505 K), Pb (600 K)
Step 3:Identify the element with lowest melting point
Tin (Sn, atomic number 50) has the lowest melting point among these Group 14 elements
Final answer: [object Object]
Q61Single correctAtomic Structure
Given below are two statements about X-ray spectra of elements: Statement (I): A plot of v (v = frequency of X-rays emitted) vs atomic mass is a straight line. Statement (II): A plot of v (v = frequency of X-rays emitted) vs atomic number is a straight line. 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 false
Approach:
Apply Moseley's law to evaluate statements about X-ray spectra relationships
Step 1:Recall Moseley's law for X-ray spectra
Moseley’s law states: v = a(Z - b) where v is frequency, Z is atomic number, and a, b are constants
Step 2:Evaluate Statement I
Statement I relates v to atomic mass. This is incorrect; Moseley’s law relates v to atomic number, not atomic mass
Step 3:Evaluate Statement II
Statement II claims v vs Z is a straight line. From Moseley’s law, v vs Z is linear, not v vs Z
Final answer: [object Object]
Q62Single correctPrinciples Related to Practical Chemistry
Identify A, B and C in the given below reaction sequence: AHNO3Pb(NO3)2H2SO4BΔC.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4PbS,PbSO4,PbCrO4
Approach:
Identify lead compounds through sequential reactions with nitric acid, sulfuric acid, and chromate
Step 1:Determine compound A from the first reaction
A reacts with HNO3 to give Pb(NO3)2. Since PbS + HNO3→Pb(NO3)2, A = PbS
Step 2:Determine compound B from the second reaction
Pb(NO3)2+H2SO4→PbSO4↓+2HNO3, so B = PbSO4
Step 3:Determine the final product C
PbSO4 is treated with ammonium acetate and then K2CrO4 to give PbCrO4 (yellow precipitate)
Given below are two statements: Statement (I): The boiling points of alcohols and phenols increase with increase in the number of C-atoms. Statement (II): The boiling points of alcohols and phenols are higher in comparison to other class of compounds such as ethers, haloalkanes. 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:
Analyze boiling point trends based on molecular size and hydrogen bonding capability
Step 1:Evaluate Statement I
As the number of carbon atoms increases, molecular mass and van der Waals forces increase, leading to higher boiling points
Step 2:Evaluate Statement II
Alcohols and phenols have OH groups that form strong hydrogen bonds, resulting in higher b.p. than ethers and haloalkanes of similar molecular mass
Step 3:Conclude
Both statements are correct
Final answer: [object Object]
Q64Single correctSolutions
Consider a binary solution of two volatile liquid components 1 and 2. x1 and y1 are the mole fractions of component 1 in liquid and vapour phase, respectively. The slope and intercept of the graph are:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2P20P10,P20P20−P10
Approach:
Derive linear relationship between 1/y₁ and 1/x₁ using Raoult's law for binary volatile solutions
Step 1:Apply Raoult's law for binary volatile solution
When a non-volatile solute is added to the solvent, the vapour pressure of thesolvent decreases by 10 mm of Hg. The mole fraction of the solute in the solutionis 0.2. What would be the mole fraction of the solvent if decrease in vapourpressure is 20 mm of Hg?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 40.6
Approach:
Use Raoult's law to relate vapour pressure lowering to mole fraction and calculate solvent mole fraction
Step 1:Apply Raoult's law for relative lowering of vapour pressure
P0P0−Ps=xsolute
Step 2:For the first case with 10 mm Hg decrease
P010 = 0.2 ⇒P0 = 50 mm Hg
Step 3:For the second case with 20 mm Hg decrease
5020=xsolute=0.4
Step 4:Calculate mole fraction of solvent
xsolvent=1−xsolute=1−0.4=0.6
Final answer: [object Object]
Q71NumericalSome Basic Concepts in Chemistry
0.01 mole of an organic compound (X) containing 10% hydrogen, on complete combustion produced 0.9gH2O. Molar mass of X is:
SolutionAnswer: 100
Approach:
Calculate molar mass from combustion data using hydrogen percentage and water produced
Step 1:Calculate moles of H₂O produced
nH2O=180.9 = 0.05 mol
Step 2:Calculate moles of H atoms in the compound
nH=2×0.05=0.1 mol H atoms
Step 3:Calculate mass of hydrogen in the compound
mH=0.1×1=0.1 g
Step 4:Since H is 10% of the compound, calculate total mass
mtotal0.1=0.10⇒mtotal=1 g
Step 5:Calculate molar mass
M=0.01 mol1 g=100 g mol−1
Final answer: [object Object]
Q72NumericalHydrocarbons
A compound 'X' absorbs 2 moles of hydrogen and 'X' upon oxidation with KMnO4∣H+ gives CH3-C-CH3, CH3-C-OH. The compound X is:
SolutionAnswer: 27
Approach:
Reconstruct diene structure from oxidation products and count sigma bonds
Step 1:Analyze the oxidation products: acetone, acetic acid, and adipic acid derivatives
Products: CH3COCH3,CH3COOH, and dicarbonyl compound
Step 2:Work backwards to determine structure of X before oxidation, which contains 2 C=C bonds (absorbs 2 mol H₂)
X has 2 C=C double bonds that get oxidized to carbonyl groups
Step 3:Reconstruct compound X structure based on oxidation products
X: CH3-C=CH-CH2−CH2-C=C(CH_3)2 or similar structure
Step 4:Count sigma bonds in compound X: C-C single bonds, C-H bonds, and sigma bonds within C=C
Total σ bonds: 8(C-C) + 16(C-H) + 2(σ in C=C) = 27
Final answer: [object Object]
Q73NumericalSome Basic Concepts in Chemistry
When 81.0g of aluminium is allowed to react with 128.0g of oxygen gas, the mass of aluminium oxide produced in grams is _____.
SolutionAnswer: 153
Approach:
Apply stoichiometry with limiting reagent analysis to calculate product mass
Step 3:Identify limiting reagent using stoichiometric ratio
4nAl=43.0=0.75, 3nO2=34.0=1.33, Al is limiting
Step 4:Calculate moles of Al₂O₃ produced
nAl2O3=43.0× 2 = 1.5 mol
Step 5:Calculate mass of Al₂O₃
mAl2O3=1.5×(2×27+3×16)=1.5×102=153 g
Final answer: [object Object]
Q74NumericalChemical Thermodynamics
The bond dissociation enthalpy of X2, ΔHbond, calculated from the given data is _____ kJ mol−1. (Nearest integer). M+X−(s)→M+(g)+X−(g), ΔHlattice∗=800 kJ mol−1. M(s)→M(g), ΔHsub∘=100 kJ mol−1. M(g)→M+(g)+e−(g), ΔHi=500 kJ mol−1. X(g)+e−(g)→X−(g), ΔHeg∗=−300 kJ mol−1. M(s)+21X2(g)→M+X−(s), ΔHf∘=−400 kJ mol−1. [Given: M+X− is a pure ionic compound and X forms a diatomic molecule X2 in gaseous state]
SolutionAnswer: 200
Approach:
Apply Born-Haber cycle using Hess's law to calculate unknown bond dissociation enthalpy from given thermochemical data
Step 1:Apply Born-Haber cycle and Hess's law
ΔHf∘=ΔHsub∘+21ΔHbond+ΔHi+ΔHeg∗−ΔHlattice∗
Step 2:Substitute the given values
−400=100+21ΔHbond+500+(−300)−800
Step 3:Simplify and solve for bond dissociation enthalpy
−400=100+500−300−800+21ΔHbond=−500+21ΔHbond
Step 4:Calculate the bond dissociation enthalpy
21ΔHbond=−400+500=100⇒ΔHbond=200 kJ mol−1
Final answer: [object Object]
Q75NumericalOrganic Compounds Containing Nitrogen
Consider the following sequence of reactions. Total number of sp3 hybridised carbon atoms in the major product C formed is _____.
SolutionAnswer: 4
Approach:
Trace through diazotization, azo coupling, and Williamson ether synthesis to identify final product structure and count sp³ hybridized carbons
Step 1:First step: Diazotization of p-ethoxyaniline with NaNO₂/HCl at 0-5°C
Step 8:Calculate distance from (1, 4, 0) to (2, 6, 3)
d=(2−1)2+(6−4)2+(3−0)2=1+4+9=14
Final answer: [object Object]
Q2Single correctSets, Relations and Functions
Let A={(x,y)∈R×R:∣x+y∣≥3} and B={(x,y)∈R×R:∣x∣+∣y∣≤3}. If C={(x,y)∈A∩B:x=0 or y=0}, then ∑(x,y)∈C∣x+y∣ is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 412
Approach:
Find points in the intersection of sets A and B that lie on coordinate axes, then sum |x+y| values.
Step 1:Understand set A: region where |x+y| ≥ 3
A: Points outside the band between lines x+y=3 and x+y=−3
Step 2:Understand set B: diamond region
B: Diamond with vertices at (3,0),(−3,0),(0,3),(0,−3)
Step 3:Find points on y=0 (x-axis) in A∩B
On y=0: B requires ∣x∣≤3, A requires ∣x∣≥3
Step 4:Calculate |x+y| for points on x-axis
(3,0):∣3+0∣=3; (−3,0):∣−3+0∣=3
Step 5:Find points on x=0 (y-axis) in A∩B
On x=0: B requires ∣y∣≤3, A requires ∣y∣≥3
Step 6:Calculate |x+y| for points on y-axis
(0,3):∣0+3∣=3; (0,−3):∣0+(−3)∣=3
Step 7:Total sum
∑(x,y)∈C∣x+y∣=6+6=12
Final answer: [object Object]
Q3Single correctSets, Relations and Functions
Let X=R×R. Define a relation R on X as: (a1,b1)R(a2,b2)⇔b1=b2. Statement I: R is an equivalence relation. Statement II: For some (a,b)∈X, the set S={(x,y)∈X:(x,y)R(a,b)} represents a line parallel to y=x. 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:
Check if relation R satisfies reflexive, symmetric, and transitive properties for equivalence. Analyze the set S for Statement II.
Step 2:Second integration by parts: u = x², dv = cos(x)dx
∫x2cosxdx=x2sinx−∫2xsinxdx
Step 3:Third integration by parts: u = x, dv = sin(x)dx
∫xsinxdx=−xcosx−∫(−cosx)dx=−xcosx+sinx
Step 4:Combine all parts
g(x)=−x3cosx+3[x2sinx−2(−xcosx+sinx)]
Step 5:Evaluate g(π/2) using cos(π/2)=0, sin(π/2)=1
g(2π)=−(2π)3(0)+3(2π)2(1)+6(2π)(0)−6(1)
Step 6:Find g'(x) - since g(x) = ∫x³sin(x)dx
g′(x)=x3sinx
Step 7:Evaluate g'(π/2)
g′(2π)=(2π)3sin(2π)=8π3⋅1=8π3
Step 8:Calculate 8(g(π/2) + g'(π/2))
8[(43π2−6)+8π3]=8⋅43π2−48+8⋅8π3
Step 9:Express in standard form and identify coefficients
=π3+6π2−48=απ3+βπ2+γ
Step 10:Calculate α + β - γ
α+β−γ=1+6−(−48)=1+6+48=55
Final answer: [object Object]
Q5Single correctCo-ordinate Geometry
A rod of length eight units moves such that its ends A and B always lie on the lines x−y+2=0 and y+2=0, respectively. If the locus of the point P, that divides the rod AB internally in the ratio 2:1 is 9(x2+αy2+βxy+γx+28y)−76=0, then α−β−γ is equal to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 323
Approach:
Express point P in terms of parameters, use constraint |AB|=8, and derive the locus equation.
If the square of the shortest distance between the lines 1x−2=2y−1=−3z+3 and 2x+1=4y+3=−5z+5 is nm, where m, n are coprime numbers, then m+n is equal to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 29
Approach:
Use formula for shortest distance between skew lines using cross product of direction vectors.
Step 1:Identify points and direction vectors for both lines
L1: Point P1=(2,1,−3), direction d1=(1,2,−3). L2: Point P2=(−1,−3,−5), direction d2=(2,4,−5)
Step 8:Multiply Part A and Part B for final answer
L=32⋅e1=3e2
Final answer: [object Object]
Q8Single correctVector Algebra
Let the point A divide the line segment joining the points P(−1,−1,2) and Q(5,5,10) internally in the ratio r:1(r>0). If O is the origin and (OQ⋅OA)−51∣OP×OA∣2=10, then the value of r is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 47
Approach:
Use section formula to find A, then compute dot and cross products as given in the equation.
Step 1:Find coordinates of A using section formula
For no solution: coefficient = 0 but constant e 0. So λ−17=0 and μ−18e0
Final answer: [object Object]
Q11Single correctTrigonometry
Let the range of the function f(x)=6+16cosx⋅cos(3π−x)⋅cos(3π+x)⋅sin3x⋅cos6x,x∈R be [α,β]. Then the distance of the point (α,β) from the line 3x+4y+12=0 is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 111
Approach:
Simplify the trigonometric product using identities, find range, then calculate distance from point to line.
Step 1:Apply the triple product identity for cosines
Step 6:Calculate distance from (5, 7) to line 3x + 4y + 12 = 0
d=32+42∣3(5)+4(7)+12∣=25∣15+28+12∣=555=11
Final answer: [object Object]
Q12Single correctDifferential Equations
Let x=x(y) be the solution of the differential equation y=(x−ydydx)sin(yx),y>0 and x(1)=2π. Then cos(x(2)) is equal to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 42(loge2)2−1
Approach:
Substitute u = x/y to transform into separable equation, integrate, and apply initial condition.
Step 1:Let u = x/y, so x = uy and dx/dy = u + y(du/dy)
Substituting: y=(uy−y(u+ydydu))sinu
Step 2:Simplify the equation
y=−y2sinu⋅dydu
Step 3:Separate variables
sinudu=−ydy
Step 4:Integrate both sides
∫sinudu=−∫ydy⇒−cosu=−lny+C
Step 5:Apply initial condition x(1) = π/2
cos(1π/2)=ln(1)+C⇒0=0+C⇒C=0
Step 6:Find x(2) using y = 2
cos(2x(2))=ln2
Step 7:Calculate cos(x(2)) using double angle formula
Let θ=arccos(ln2), so cosθ=ln2. Then x(2)=2θ.
Final answer: [object Object]
Q13Single correctLimit, Continuity and Differentiability
A spherical chocolate ball has a layer of ice-cream of uniform thickness around it. When the thickness of the ice-cream layer is 1 cm, the ice-cream melts at the rate of 81 cm³/min and the thickness of the ice-cream layer decreases at the rate of 4π1 cm/min. The surface area (in cm²) of the chocolate ball (without the ice-cream layer) is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2256π
Approach:
Use related rates with the volume formula for spherical shell.
Step 1:Define variables: r = chocolate radius, t = ice cream thickness
Total radius R=r+t. Volume of ice cream V=34π(R3−r3)
Step 2:Find dV/dt by differentiating (r is constant)
dtdV=34π⋅3(r+t)2⋅dtdt=4π(r+t)2⋅dtdt
Step 3:Substitute given values: t = 1, dV/dt = -81, dt/dt = -1/(4π)
−81=4π(r+1)2⋅(−4π1)
Step 4:Solve for r
(r+1)2=81⇒r+1=9⇒r=8 cm
Step 5:Calculate surface area of chocolate ball
S=4πr2=4π(8)2=4π⋅64=256π cm²
Final answer: [object Object]
Q14Single correctComplex Numbers and Quadratic Equations
The number of complex numbers z, satisfying ∣z∣=1 and zˉz+zzˉ=1, is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 28
Approach:
Express z in exponential form, simplify the given condition to find θ values.
Step 1:Express z using |z| = 1
z=eiθ, so zˉ=e−iθ
Step 2:Calculate z/z̄ and z̄/z
zˉz=e−iθeiθ=e2iθ, zzˉ=e−2iθ
Step 3:Add the two expressions
zˉz+zzˉ=e2iθ+e−2iθ=2cos(2θ)
Step 4:Apply the given condition
∣2cos(2θ)∣=1⇒∣cos(2θ)∣=21
Step 5:Find solutions for cos(2θ) = 1/2
2θ=3π,35π,37π,311π in [0,4π)
Step 6:Find solutions for cos(2θ) = -1/2
2θ=32π,34π,38π,310π in [0,4π)
Step 7:Count total solutions in [0, 2π)
Total distinct values of θ: 6π,3π,32π,65π,67π,34π,35π,611π
Final answer: [object Object]
Q15Single correctMatrices and Determinants
Let A=[aij] be 3×3 matrix such that A[010]=[001], A[413]=[110] and A[212]=[100], then a23 equals:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1−1
Approach:
Use the three matrix equations to find the elements of matrix A, specifically a₂₃.
Step 1:From first equation A[0,1,0]ᵀ = [0,0,1]ᵀ, identify second column
Second column of A = [001], so a12=0,a22=0,a32=1
A board has 16 squares as shown in the figure: Out of these 16 squares, two squares are chosen at random. The probability that they have no side in common is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 254
Approach:
Count total ways to choose 2 squares and subtract adjacent pairs.
Step 1:Count total ways to choose 2 squares from 16
Let the shortest distance from (a, 0), a > 0, to the parabola y2 = 4x be 4. Then the equation of the circle passing through the point (a, 0) and the focus of the parabola, and having its centre on the axis of the parabola is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2x2+y2 - 6x + 5 = 0
Approach:
Find point (a,0) using shortest distance condition, then find circle through (a,0) and focus.
Step 1:Find foot of perpendicular from (a,0) to parabola y² = 4x
Point on parabola: (t2,2t). For perpendicular: t2=a−2 (when a>2)
Step 2:Calculate distance when foot is at (a-2, 2√(a-2))
Distance =(a−(a−2))2+(0−2a−2)2=4+4(a−2)=2a−1
Step 3:Set distance = 4 and solve for a
2a−1=4⇒a−1=2⇒a−1=4⇒a=5
Step 4:Circle passes through (5,0) and focus (1,0), center on x-axis
Center at midpoint: (25+1,0)=(3,0). Radius = ∣5−3∣=2
Step 5:Write circle equation
(x−3)2+y2=4⇒x2+y2−6x+9−4=0⇒x2+y2−6x+5=0
Final answer: [object Object]
Q19Single correctBinomial Theorem and its Simple Applications
If in the expansion of (1+x)p(1−x)q, the coefficients of x and x2 are 1 and -2, respectively, then p2+q2 is equal to :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 213
Approach:
Expand (1+x)p(1−x)q up to x² term and match coefficients.
Step 1:Expand each factor to x² terms
(1+x)p≈1+px+2p(p−1)x2, (1−x)q≈1−qx+2q(q−1)x2
Step 2:Multiply and collect x coefficient
Coefficient of x: p−q=1 ... (i)
Step 3:Collect x² coefficient
Coefficient of x2: 2p(p−1)−pq+2q(q−1)=−2
Step 4:Simplify x² coefficient equation
2p2−p+q2−q−2pq=−2⇒(p−q)2−(p+q)=−4
Step 5:Solve system: p - q = 1, p + q = 5
2p=6⇒p=3, q=2
Step 6:Calculate p² + q²
p2+q2=9+4=13
Final answer: [object Object]
Q20Single correctIntegral Calculus
If the area of the region {(x,y):−1≤x≤1,0≤y≤a+e∣x∣−e−x,a>0} is ee2+8e+1, then the value of a is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 35
Approach:
Set up and evaluate the double integral for the area, then solve for a.
The roots of the quadratic equation 3x2 - px + q = 0 are 10th and 11th terms of an arithmetic progression with common difference 23. If the sum of the first 11 terms of this arithmetic progression is 88, then q - 2p is equal to
SolutionAnswer: 474
Approach:
Use AP formulas to find the roots, then apply Vieta's formulas.
Step 1:Set up AP with common difference d = 3/2
Let first term be a. Then a10=a+9⋅23=a+227, a11=a+10⋅23=a+15
The number of ways 5 boys and 4 girls can sit in a row so that either all the boys sit together or no two boys sit together, is
SolutionAnswer: 17280
Approach:
Use complementary cases: (1) all boys together, (2) no two boys together.
Step 1:Case 1: All 5 boys sit together
Treat boys as single unit. Total units = 1 (boys block) + 4 (girls) = 5
Step 2:Count arrangements for Case 1
5 units can be arranged in 5! ways. Boys within block: 5! ways. Total = 5!×5!=120×120=14400
Step 3:Case 2: No two boys sit together
First arrange 4 girls in a row: 4! ways. This creates 5 gaps: □G□G□G□G□
Step 4:Place boys in gaps
5 boys in 5 gaps: 5! ways. Total = 4!×5!=24×120=2880
Step 5:Total arrangements
Total = 14400+2880=17280
Final answer: [object Object]
Q24NumericalCo-ordinate Geometry
The focus of the parabola y2=4x+16 is the centre of the circle C of radius 5. If the values of λ, for which C passes through the point of intersection of the lines 3x−y=0 and x+λy=4, are λ1 and λ2, λ1<λ2, then 12λ1+29λ2 is equal to
SolutionAnswer: 15
Approach:
Find focus of parabola (circle center), find line intersection point, use circle equation.
Vertex at (−4,0), 4a=4⇒a=1. Focus at (−4+1,0)=(−3,0)
Step 2:Find intersection of lines 3x - y = 0 and x + λy = 4
From y=3x: x+3λx=4⇒x=1+3λ4, y=1+3λ12
Step 3:Apply circle equation (x+3)² + y² = 25
(1+3λ4+3)2+(1+3λ12)2=25
Step 4:Simplify and form quadratic
(7+9λ)2+144=25(1+3λ)2⇒81λ2+126λ+193=225λ2+150λ+25
Step 5:Solve quadratic
λ=12−1±1+168=12−1±13
Step 6:Calculate 12λ₁ + 29λ₂
12⋅(−67)+29⋅1=−14+29=15
Final answer: [object Object]
Q25NumericalComplex Numbers and Quadratic Equations
Let α,β be the roots of the equation x2−ax−b=0 with Im(α)<Im(β). Let Pn=αn−βn. If P3=−57i, P4=−37i, P5=117i and P6=457i, then ∣α4+β4∣ is equal to
SolutionAnswer: 31
Approach:
Use recurrence relation for Pₙ = αⁿ - βⁿ and given values to find α⁴ + β⁴.
Step 1:Use recurrence relation Pₙ = aPₙ₋₁ + bPₙ₋₂
P5=aP4+bP3: 117i=a(−37i)+b(−57i)
Step 2:Use another recurrence equation
P6=aP5+bP4: 457i=a(117i)+b(−37i)
Step 3:Solve system for a and b
From (i): 3a+5b=−11. From (ii): 11a−3b=45. Solving: a=3,b=−4
How many questions are in the JEE Main 2025 January 23, Shift 2 paper?
The JEE Main 2025 January 23, 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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