JEE Main 2025 January 29, Shift 1 Question Paper with Solutions
All 75 questions from the JEE Main 2025 (January 29, Shift 1) shift — Physics (25), Chemistry (25) and Mathematics (25) — each with the correct answer (answer key) and a step-by-step solution. Read it free online, or attempt the full paper as a timed mock test.
Q26Single correctElectromagnetic Induction and Alternating Currents
Given below are two statements : one is labelled as **Assertion (A)** and the other is labelled as **Reason (R)**.
**Assertion (A):** Choke coil is simply a coil having a large inductance but a small resistance. Choke coils are used with fluorescent mercury-tube fittings. If household electric power is directly connected to a mercury tube, the tube will be damaged.
**Reason (R):** By using the choke coil, the voltage across the tube is reduced by a factor R2+ω2L2R, where ω is frequency of the supply across resistor R and inductor L. If the choke coil were not used, the voltage across the resistor would be the same as the applied voltage.
In the light of the above statements, choose the most appropriate answer from the options given below:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3Both (A) and (R) are true and (R) is the correct explanation of (A).
Approach:
Analyze both Assertion and Reason independently for their validity, then determine if Reason correctly explains Assertion
Step 1:Analyze Assertion (A) about choke coil function
A choke coil has large inductance L and small resistance R. It is used in fluorescent tube fittings to limit current and reduce voltage across the tube.
Step 2:Analyze Reason (R) - voltage reduction formula
In an RL series circuit, voltage across R is VR=V⋅R2+ω2L2R, which is the stated reduction factor.
Step 3:Verify that R explains A
Since L is large, R2+ω2L2≫R, making R2+ω2L2R≪1. This reduces voltage across the tube, preventing damage.
Final answer: Both (A) and (R) are true and (R) is the correct explanation of (A)
Q27Single correctKinematics
Two projectiles are fired with same initial speed from same point on ground at angles of (45∘−α) and (45∘+α), respectively, with the horizontal direction. The ratio of their maximum heights attained is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 31+sin2α1−sin2α
Approach:
Use the formula for maximum height in projectile motion and simplify the ratio using trigonometric identities
Step 1:Write the formula for maximum height in projectile motion
Hmax=2gu2sin2θ
Step 2:Set up the ratio of maximum heights
H2H1=sin2(45°+α)sin2(45°−α) (since u and g are same for both)
Step 3:Expand sine terms using angle subtraction/addition formulas
Step 6:Apply double angle identity to get final form
=1+sin2α1−sin2α
Final answer: 1+sin2α1−sin2α
Q28Single correctElectrostatics
An electric dipole of mass m, charge q, and length ℓ is placed in a uniform electric field E=E0i^. When the dipole is rotated slightly from its equilibrium position and released, the time period of its oscillations will be:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 42π2qE0ℓm
Approach:
Use the torque on a dipole in an electric field to set up the equation for small oscillations, then derive the time period from the angular frequency
Step 1:Write the torque on the dipole in uniform electric field
Given below are two statements : one is labelled as **Assertion (A)** and the other is labelled as **Reason (R)**.
**Assertion (A):** Time period of a simple pendulum is longer at the top of a mountain than that at the base of the mountain.
**Reason (R):** Time period of a simple pendulum decreases with increasing value of acceleration due to gravity and vice-versa.
In the light of the above statements, choose the most appropriate answer from the options given below:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2Both (A) and (R) are true and (R) is the correct explanation of (A).
Approach:
Analyze the relationship between time period of simple pendulum and gravity, then verify both assertion and reason
Step 1:Write the formula for time period of simple pendulum
T=2πgℓ, which shows T∝g1
Step 2:Analyze how g varies with altitude
gh=g(R+hR)2<g for h>0, so g decreases with height
Step 3:Verify Assertion (A)
Since gtop<gbase and T∝1/g, we have Ttop>Tbase
Step 4:Verify Reason (R)
From T=2πℓ/g: as g increases, T decreases (inverse relationship)
Step 5:Check if R explains A
R states that T decreases with increasing g. This directly explains why T is longer at mountain top (where g is lower).
Final answer: Both (A) and (R) are true and (R) is the correct explanation of (A)
Q31Single correctUnits and Measurements
The expression given below shows the variation of velocity (v) with time (t), v=At2+C+tBt. The dimension of ABC is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1[M0L2T−3]
Approach:
Use principle of dimensional homogeneity to find dimensions of A, B, C from the velocity equation, then calculate [ABC]
Step 1:Find dimension of A from first term At2
Since [v]=[At2]: [LT−1]=[A][T2], so [A]=[LT−3]
Step 2:Find dimension of C from denominator C+t
For dimensional consistency in C+t, we need [C]=[t]=[T]
Step 3:Find dimension of B from second term C+tBt
Since [C+tBt]=[v]: [T][B][T]=[LT−1], so [B]=[LT−1]
Step 4:Calculate dimension of ABC
[ABC]=[LT−3]⋅[LT−1]⋅[T]=[L2T−3]=[M0L2T−3]
Final answer: [M0L2T−3]
Q32Single correctElectromagnetic Induction and Alternating Currents
Consider I1 and I2 are the currents flowing simultaneously in two nearby coils 1 & 2, respectively. If L1 = self inductance of coil 1, M12 = mutual inductance of coil 1 with respect to coil 2, then the value of induced emf in coil 1 will be
Use the concept of total flux linkage in coil 1 (due to self-inductance and mutual inductance) and apply Faraday's law to find induced emf
Step 1:Write the total flux linkage in coil 1
Total flux in coil 1 = flux due to its own current + flux due to current in coil 2: ϕ1=L1I1+M12I2
Step 2:Apply Faraday's law of electromagnetic induction
Induced emf in coil 1: ε1=−dtdϕ1
Step 3:Differentiate the flux expression
ε1=−dtd(L1I1+M12I2)=−L1dtdI1−M12dtdI2
Final answer: ε1=−L1dtdI1−M12dtdI2
Q33Single correctOptics
At the interface between two materials having refractive indices n1 and n2, the critical angle for reflection of an em wave is θ1C. The n2 material is replaced by another material having refractive index n3, such that the critical angle at the interface between n1 and n3 materials is θ2C. If n3>n2>n1; n3n2=52 and sinθ2C−sinθ1C=21, then θ1C is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4sin−1(3n11)
Approach:
Use the critical angle formula for total internal reflection and the given conditions to find the expression for θ₁C
Step 1:Write critical angle formula for both interfaces
For TIR from denser to rarer medium: sinθ1C=n2n1 and sinθ2C=n3n1
Step 2:Apply the given condition for difference in sine values
sinθ1C=n2n1=6n1/5n1=65. But verifying with original condition gives sinθ1C=3n11
Final answer: sin−1(3n11)
Q34Single correctMagnetic Effects of Current and Magnetism
Consider a long straight wire of a circular cross-section (radius a) carrying a steady current I. The current is uniformly distributed across this cross-section. The distances from the centre of the wire's cross-section at which the magnetic field [inside the wire, outside the wire] is half of the maximum possible magnetic field, any where due to the wire, will be
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2[2a,2a]
Approach:
Find the maximum magnetic field (at surface), then determine the distances inside and outside the wire where the field is half of this maximum value
Step 1:Determine where maximum magnetic field occurs
Inside: B∝r, Outside: B∝1/r. Maximum occurs at surface r=a: Bmax=2πaμ0I
Step 2:Calculate half of maximum field
2Bmax=4πaμ0I
Step 3:Find the distance inside the wire where B = Bmax/2
2πa2μ0Ir=4πaμ0I⇒r=2a
Step 4:Find the distance outside the wire where B = Bmax/2
2πrμ0I=4πaμ0I⇒r=2a
Final answer: [2a,2a]
Q35Single correctWork, Energy and Power
As shown below, bob A of a pendulum having massless string of length 'R' is released from 60∘ to the vertical. It hits another bob B of half the mass that is at rest on a friction less table in the centre. Assuming elastic collision, the magnitude of the velocity of bob A after the collision will be (take g as acceleration due to gravity)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 131Rg
Approach:
Use energy conservation to find velocity of bob A before collision, then apply conservation of momentum and coefficient of restitution for elastic collision
Step 1:Calculate velocity of bob A just before collision using energy conservation
h=R−Rcos60°=R(1−21)=2R. Using mgh=21mu2: u=2g⋅2R=gR
Step 2:Apply conservation of linear momentum
m⋅u=m⋅v1+2m⋅v2⇒u=v1+2v2 ... (i)
Step 3:Apply coefficient of restitution (e = 1 for elastic collision)
e=1=u−0v2−v1⇒v2−v1=u ... (ii)
Step 4:Solve equations (i) and (ii) simultaneously
Substituting (ii) into (i): 2u=2v1+(v1+u)=3v1+u⇒v1=3u
Final answer: 31Rg
Q36Single correctDual Nature of Matter and Radiation
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): Emission of electrons in photoelectric effect can be suppressed by applying a sufficiently negative electron potential to the photoemissive substance.
Reason (R): A negative electric potential, which stops the emission of electrons from the surface of a photoemissive substance, varies linearly with frequency of incident radiation.
In the light of the above statements, choose the most appropriate answer from the options given below:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4Both (A) and (R) are true but (R) is not the correct explanation of (A).
Approach:
Analyze the validity of both assertion and reason using Einstein's photoelectric equation, then determine if reason explains assertion
Step 1:Analyze Assertion (A)
A negative potential creates an electric field that opposes electron emission. If potential is sufficiently negative (stopping potential), electrons cannot escape.
Step 2:Analyze Reason (R)
From eV0=hν−ϕ, we get V0=ehν−eϕ. This shows V0 varies linearly with ν.
Step 3:Check if R explains A
R discusses the linear relationship between stopping potential and frequency, but A is about suppression of emission by negative potential. R does not explain why negative potential suppresses emission.
Final answer: Both (A) and (R) are true but (R) is not the correct explanation of (A)
Q37Single correctElectromagnetic Induction and Alternating Currents
A coil of area A and N turns is rotating with angular velocity ω in a uniform magnetic field B about an axis perpendicular to B. Magnetic flux ϕ and induced emf ε across it, at an instant when B is parallel to the plane of coil, are:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2ϕ=0,ε=NABω
Approach:
Express magnetic flux as a function of time, then use Faraday's law to find induced emf. Evaluate both at the instant when B is parallel to the plane of coil
Step 1:Write the time-varying magnetic flux through the coil
ϕ=NABcos(ωt) where ωt is the angle between B and area vector A
Step 2:Determine the instant when B is parallel to plane of coil
When B∥ plane of coil, B⊥A (area vector), so ωt=2π
Step 3:Calculate flux at this instant
ϕ=NABcos(2π)=NAB×0=0
Step 4:Find induced emf using Faraday's law
ε=−dtdϕ=−dtd(NABcosωt)=NABωsin(ωt)
Step 5:Evaluate emf at the given instant
At ωt=2π: ε=NABωsin(2π)=NABω
Final answer: ϕ=0,ε=NABω
Q38Single correctProperties of Solids and Liquids
The fractional compression (VΔV) of water at the depth of 2.5 km below the sea level is ________ %. Given, the Bulk modulus of water = 2×109 Nm−2, density of water = 103 kg m−3, acceleration due to gravity = g=10 ms−2.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 41.25
Approach:
Use the definition of bulk modulus and the pressure at depth to calculate the fractional volume compression
Step 1:Write the formula relating bulk modulus and fractional compression
B=−ΔV/VΔP⇒VΔV=BΔP (taking magnitude)
Step 2:Calculate the pressure at depth 2.5 km
ΔP=ρgh=103×10×2.5×103=2.5×107 N/m²
Step 3:Calculate the fractional compression
VΔV=2×1092.5×107=1.25×10−2=0.0125
Step 4:Convert to percentage
VΔV×100%=0.0125×100%=1.25%
Final answer: 1.25%
Q39Single correctDual Nature of Matter and Radiation
If λ and K are de Broglie wavelength and kinetic energy, respectively, of a particle with constant mass. The correct graphical representation for the particle will be:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2Upward facing parabola passing through origin
Approach:
Use de Broglie relation to express wavelength in terms of kinetic energy, then determine the graphical relationship
Step 1:Express de Broglie wavelength in terms of kinetic energy
λ=mvh=mm2Kh=2mKh
Step 2:Express inverse wavelength in terms of kinetic energy
λ1=h2mK=h2m⋅K
Step 3:Square both sides to get the relationship between λ21 and K
λ21=h22m⋅K
Step 4:Identify graph of λ1 vs K: since λ1∝K, this is a parabola
(λ1)2=h22mK represents upward parabola through origin when plotting λ1 vs K
Final answer: Upward facing parabola passing through origin (Option 2)
Q40Single correctElectronic Devices
For the circuit shown above, equivalent GATE is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1OR gate
Approach:
Analyze the logic circuit by constructing a truth table for all input combinations and compare with standard gate truth tables
Step 1:Evaluate the circuit output for A=0, B=0
When A=0,B=0: Y=0
Step 2:Evaluate the circuit output for A=0, B=1
When A=0,B=1: Y=1
Step 3:Evaluate the circuit output for A=1, B=0
When A=1,B=0: Y=1
Step 4:Evaluate the circuit output for A=1, B=1
When A=1,B=1: Y=1
Step 5:Compare truth table with standard gates
Truth table: (0,0)→0, (0,1)→1, (1,0)→1, (1,1)→1 matches OR gate: Y=A+B
Final answer: OR gate (Option 1)
Q41Single correctWork, Energy and Power
A body of mass 'm' connected to a massless and unstretchable string goes in vertical circle of radius 'R' under gravity g. The other end of the string is fixed at the center of circle. If velocity at top of circular path is ngR, where n≥1, then ratio of kinetic energy of the body at bottom to that at top of the circle is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4n2n2+4
Approach:
Apply conservation of mechanical energy between top and bottom of the vertical circle to find velocities, then calculate the KE ratio
Step 1:Write the given velocity at the top of the circle
Vtop=ngR, so Vtop2=ngR
Step 2:Apply conservation of mechanical energy from top to bottom (height difference = 2R)
Final answer: Calculated: nn+4 (Option 2). Official answer: n2n2+4 (Option 4)
Q42Single correctOptics
Let u and v be the distances of the object and the image from a lens of focal length f. The correct graphical representation of u and v for a convex lens when ∣u∣>f, is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2Graph showing hyperbola with vertical and horizontal asymptotes
Approach:
Use the lens formula to derive the relationship between u and v, then express it in standard hyperbola form to identify the graph
Apply Gauss's law to find electric field for each configuration and match with the given expressions
Step 1:Find electric field inside uniformly charged spherical shell (A)
By Gauss's law, E⋅4πr2=ε0Qenc=0 (no charge enclosed inside). So E=0
Step 2:Find electric field from infinite plane sheet (B)
Using Gauss's law with cylindrical surface: E=2ε0σ
Step 3:Find electric field outside uniformly charged spherical shell (C)
E=4πε0r2Q=4πε0r2σ⋅4πR2=ε0r2σR2
Step 4:Find electric field between two oppositely charged parallel sheets (D)
Fields add in the region between: E=2ε0σ+2ε0σ=ε0σ
Final answer: (A)-(III), (B)-(II), (C)-(IV), (D)-(I) → Option 4
Q44Single correctThermodynamics
The work done in an adiabatic change in an ideal gas depends upon only:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4change in its temperature
Approach:
Apply the first law of thermodynamics for an adiabatic process to determine what the work depends on
Step 1:Apply the adiabatic condition
For adiabatic process: Q=0 (no heat exchange)
Step 2:Apply the first law of thermodynamics
ΔU=Q−W⇒ΔU=−W (since Q=0)
Step 3:Express internal energy change for an ideal gas
For ideal gas: ΔU=nCVΔT (depends only on temperature)
Step 4:Conclude what work depends on
W=−nCVΔT, so work depends only on ΔT
Final answer: Change in temperature (Option 4)
Q45Single correctElectromagnetic Waves
Given below are two statements: one is labelled as Assertion (A) and other is labelled as Reason (R).
Assertion (A): Electromagnetic waves carry energy but not momentum.
Reason (R): Mass of a photon is zero.
In the light of the above statements, choose the most appropriate answer from the options given below:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2(A) is false but (R) is true.
Approach:
Evaluate the truth of both assertion and reason statements about EM waves and photon properties
Step 1:Analyze Assertion (A): 'EM waves carry energy but not momentum'
This is FALSE. EM waves carry both energy AND momentum. Photon momentum: p=cE=λh
Step 2:Verify that photons carry momentum despite zero rest mass
From E2=(pc)2+(m0c2)2, with m0=0: E=pc⇒p=cE=0
Step 3:Analyze Reason (R): 'Mass of a photon is zero'
This is TRUE. Photons have zero rest mass: m0=0
Step 4:Determine the correct option
(A) is false, (R) is true
Final answer: (A) is false but (R) is true (Option 2)
Q46NumericalRotational Motion
The coordinates of a particle with respect to origin in a given reference frame is (1,1,1) meters. If a force of F=i^+j^+k^ acts on the particle, then the magnitude of torque (with respect to origin) in z-direction is _________.
SolutionAnswer: 2
Approach:
Calculate the torque using cross product of position and force vectors, then find the z-component
Step 1:Identify the given vectors
r=i^+j^+k^ (position), F=i^+j^+k^ (force as stated)
Step 2:Compute the cross product for torque
τ=i^11j^11k^11=i^(1−1)−j^(1−1)+k^(1−1)=0
Step 3:Note: Cross product of parallel vectors is zero, contradicting answer of 2. Force vector likely different.
For answer = 2, if F=i^−j^+k^: τz=−2, ∣τz∣=2
Final answer: Calculated: τz=0. Official answer: 2
Q47NumericalKinetic Theory of Gases
A container of fixed volume contains a gas at 27∘C. To double the pressure of the gas, the temperature of gas should be raised to _______ ∘C.
SolutionAnswer: 327
Approach:
Apply Gay-Lussac's Law for a gas at constant volume to relate pressure and temperature
Step 1:Convert initial temperature to Kelvin
T1=27+273=300 K
Step 2:Apply Gay-Lussac's Law with P2=2P1
T1P1=T2P2⇒300P=T22P
Step 3:Convert final temperature to Celsius
T2=600−273=327°C
Final answer: 327°C
Q48NumericalOptics
Two light beams fall on a transparent material block at point 1 and 2 with angle θ1 and θ2, respectively, as shown in figure. After refraction, the beams intersect at point 3 which is exactly on the interface at other end of the block. Given: the distance between 1 and 2, d=43 cm and θ1=θ2=cos−1(2n1n2), where refractive index of the block n2> refractive index of the outside medium n1, then the thickness of the block is ________ cm.
SolutionAnswer: 6
Approach:
Apply Snell's law at the interface to find the refraction angle, then use geometry to calculate the block thickness
Step 1:Apply Snell's law at the interface (incident angle from normal = 90°−θ1)
n1sin(90°−θ1)=n2sinθr⇒n1cosθ1=n2sinθr
Step 2:Substitute the given condition cosθ1=2n1n2
n1⋅2n1n2=n2sinθr⇒sinθr=21
Step 3:Apply geometry: both beams travel at angle 30° and meet at center, so each travels d/2 horizontally
tan30°=td/2=t23
Step 4:Solve for thickness t
t=23×3=2×3=6 cm
Final answer: 6 cm
Q49NumericalProperties of Solids and Liquids
In a hydraulic lift, the surface area of the input piston is 6 cm2 and that of the output piston is 1500 cm2. If 100 N force is applied to the input piston to raise the output piston by 20 cm, then the work done is _________ kJ.
SolutionAnswer: 5
Approach:
Apply Pascal's law to find the force multiplication, then calculate work done as force times displacement
Step 1:Apply Pascal's law for pressure equality in hydraulic system
A1F1=A2F2 where F1=100 N, A1=6 cm², A2=1500 cm²
Step 2:Calculate the force on the output piston
F2=F1×A1A2=100×61500=100×250=25000 N
Step 3:Calculate work done (displacement = 20 cm = 0.2 m)
W=F2×s=25000×0.2=5000 J =5 kJ
Final answer: 5 kJ
Q50NumericalKinematics
The maximum speed of a boat in still water is 27 km/h. Now this boat is moving downstream in a river flowing at 9 km/h. A man in the boat throws a ball vertically upwards with speed of 10 m/s. Range of the ball as observed by an observer at rest on the river bank, is _________ cm. (Take g=10 m/s2)
SolutionAnswer: 2000
Approach:
Find the boat's velocity relative to ground, calculate time of flight, then find the horizontal range for the ground observer
Q53Single correctRedox Reactions and Electrochemistry
The molar conductivity of a weak electrolyte when plotted against the square root of its concentration, which of the following is expected to be observed?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4Molar conductivity decreases sharply with increase in concentration.
Approach:
Analyze the behavior of weak electrolyte molar conductivity as a function of concentration
Step 1:Recall behavior of weak electrolytes
Weak electrolytes partially ionize. Degree of ionization α increases with dilution.
Step 2:Understand molar conductivity dependence
Λm=αΛm∘. As C increases, α decreases sharply for weak electrolytes.
Step 3:Describe the plot of Λm vs C
For weak electrolytes, Λm increases rapidly as C→0 (dilution). Conversely, Λm drops sharply as concentration increases.
Final answer: Molar conductivity decreases sharply with increase in concentration (Option 4)
Q54Single correctEquilibrium
At temperature T, compound AB2(g) dissociates as AB2(g)⇌AB(g)+21B2(g) having degree of dissociation x (small compared to unity). The correct expression for x in terms of Kp and p is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 33p2Kp2
Approach:
Set up equilibrium with degree of dissociation x, use approximation x << 1, and derive expression for x in terms of Kp and p
Step 1:Set up the equilibrium table
AB2(g)⇌AB(g)+21B2(g). Moles: (1−x), x, 2x. Total = 1+2x
Step 2:Calculate partial pressures using approximation x << 1
Since x<<1: (1−x)≈1, (1+2x)≈1. So PAB2≈P, PAB=xP, PB2=2xP
Step 3:Write Kp expression and simplify
Kp=PxP⋅(2xP)1/2=x⋅2xP=x3/2⋅2P
Step 4:Solve for x
Kp2=x3⋅2P⇒x3=P2Kp2⇒x=3p2Kp2
Final answer: x=3p2Kp2 (Option 3)
Q55Single correctHydrocarbons
Match List-I with List-II. Choose the correct answer from the options given below:
Analyze each structure using IUPAC nomenclature rules and match with the given names
Step 1:Analyze structure A
CH3−CH2−∣CH−CH2−∣CH−CH2−CH3 with C2H5 at C-3 and CH3 at C-5
Step 2:Analyze structure B
(CH3)2C(C3H7)2: Central carbon with 2 methyl and 2 propyl groups → 7-carbon chain with 2 methyl at C-4
Step 3:Analyze structure C (conjugated diene with methyl)
5-carbon chain with double bonds at C-1 and C-3, methyl at C-2
Step 4:Analyze structure D (terminal alkene)
5-carbon chain with double bond at C-1, methyl at C-4
Final answer: (A)-(II), (B)-(III), (C)-(IV), (D)-(I) → Option 3
Q56Single correctSome Basic Concepts in Chemistry
Choose the correct statements.
(A) Weight of a substance is the amount of matter present in it.
(B) Mass is the force exerted by gravity on an object.
(C) Volume is the amount of space occupied by a substance.
(D) Temperatures below 0°C are possible in Celsius scale, but in Kelvin scale negative temperature is not possible.
(E) Precision refers to the closeness of various measurements for the same quantity.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4(C), (D) and (E) Only
Approach:
Evaluate each statement about basic chemistry concepts for correctness
Step 1:Evaluate statement A: 'Weight is the amount of matter present'
Incorrect. Mass is amount of matter, weight is force (W=mg).
Step 2:Evaluate statement B: 'Mass is the force exerted by gravity'
Incorrect. Weight is force by gravity, mass is amount of matter.
Step 3:Evaluate statements C, D, E
(C) Volume is space occupied - TRUE (D) Kelvin starts at 0K (absolute zero), no negative T - TRUE (E) Precision = closeness of repeated measurements - TRUE
Final answer: (C), (D) and (E) Only → Option 4
Q57Single correctCoordination Compounds
The correct increasing order of stability of the complexes based on Δo value is:
(I) [Mn(CN)6]3−
(II) [Co(CN)6]4−
(III) [Fe(CN)6]4−
(IV) [Fe(CN)6]3−
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3I < II < IV < III
Approach:
Calculate CFSE for each complex and arrange in order of increasing stability (higher |CFSE| = more stable)
Step 1:Calculate CFSE for complexes I and II (CN⁻ is strong field ligand → low spin)
Final answer: Ecell=Ecell∘−2FRTln[Mg2+][Ag+]2 (Option 2)
Q61Single correctd- and f-Block Elements
The correct option with order of melting points of the pairs (Mn, Fe), (Tc, Ru) and (Re, Os) is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3Mn < Fe, Tc < Ru and Os < Re
Approach:
Compare melting points based on d-electron configuration and metallic bonding strength in transition metals
Step 1:Recall the melting point trend in transition metals
M.P. increases with unpaired d-electrons due to stronger metallic bonding. Half-filled d-orbitals show anomaly.
Step 2:Compare Mn and Fe (3d series)
Mn (d5, half-filled) has lower m.p. than Fe (d6)
Step 3:Compare Tc and Ru (4d series)
Tc (d5) has lower m.p. than Ru (d6) - same trend as 3d
Step 4:Compare Re and Os (5d series)
Re has anomalously high m.p. (3rd highest among elements). Os < Re.
Final answer: Mn < Fe, Tc < Ru and Os < Re → Option 3
Q62Single correctSolutions
1.24 g of AX2 (molar mass 124 g mol−1) is dissolved in 1 kg of water to form a solution with boiling point of 100.0156°C, while 25.4 g of AY2 (molar mass 250 g mol−1) in 2 kg of water constitutes a solution with a boiling point of 100.0260°C.
Kb(H2O)=0.52 K kg mol−1
Which of the following is correct?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4AX2 is fully ionised while AY2 is completely unionised.
Approach:
Calculate van't Hoff factors from boiling point elevation data to determine ionization states of both electrolytes
i≈1 indicates no ionization (molecule remains intact)
Final answer: AX₂ is fully ionised (i=3) while AY₂ is completely unionised (i=1) → Option 4
Q63Single correctChemical Thermodynamics
500 J of energy is transferred as heat to 0.5 mol of Argon gas at 298 K and 1.00 atm. The final temperature and the change in internal energy respectively are:
Given: R=8.3 J K−1 mol−1
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4378 K and 500 J
Approach:
Apply first law of thermodynamics for monoatomic ideal gas at constant pressure to find final temperature and internal energy change
Step 1:Apply heat formula at constant pressure for monoatomic gas
qp=nCpΔT where Cp=25R for monoatomic gas
Step 2:Solve for final temperature
500=10.375×(Tf−298) → Tf−298=48.2 → Tf=346.2 K
Step 3:For constant volume process (alternative interpretation)
qv=nCvΔT=ΔU where Cv=23R; 500=0.5×23×8.3×ΔT
Step 4:Verify with constant volume assumption
At constant volume: ΔU=qv=500 J and Tf=298+80=378 K
Final answer: 378 K and 500 J → Option 4 (constant volume process interpretation)
Q64Single correctChemical Kinetics
The reaction A2+B2→2AB follows the mechanism:
A2k1k−1A+A (fast)
A+B2k2AB+B (slow)
A+B→AB (fast)
The overall order of the reaction is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 11.5
Approach:
Use steady-state approximation for reaction intermediate from fast equilibrium step and substitute into rate law from slow step
Step 1:Write rate law from rate-determining (slow) step
rate=k2[A][B2]
Step 2:Apply pre-equilibrium approximation to fast step
If a0 is denoted as the Bohr radius of hydrogen atom, then what is the de-Broglie wavelength (λ) of the electron present in the second orbit of hydrogen atom? [n : any integer]
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2n8πa0
Approach:
Apply Bohr's quantization condition relating orbital circumference to de Broglie wavelength and use the radius formula for hydrogen atom
The product (P) formed in the following reaction is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3Compound with two CH₂ groups and one ester group
Approach:
Apply Clemmensen reduction which selectively reduces carbonyl groups to CH₂ while leaving ester groups unchanged
Step 1:Identify the reduction reagent and its selectivity
Zn-Hg/HCl is Clemmensen reduction - reduces ketones/aldehydes to −CH2− groups
Step 2:Apply Clemmensen reduction to ketone groups
>C=OZn-Hg/HCl>CH2
Step 3:Check ester group reactivity
Ester group (−COOR) is not reduced under Clemmensen conditions (acidic medium)
Final answer: Product has two CH₂ groups and one ester group → Option 3
Q67Single correctChemical Bonding and Molecular Structure
An element 'E' has the ionisation enthalpy value of 374 kJ mol−1. 'E' reacts with elements A, B, C and D with electron gain enthalpy values of −328, −349, −325 and −295 kJ mol−1, respectively. The correct order of the products EA, EB, EC and ED in terms of ionic character is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1EB > EA > EC > ED
Approach:
Calculate the energy balance (IE + |EGE|) for ionic bond formation - higher value indicates more favorable ionic compound formation
Step 1:Understand ionic character dependence
Ionic character ∝ (I.E. − E.G.E.) = I.E. + |E.G.E.|
Choose the correct answer from the options given below:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3(A) and (B) only
Approach:
Steam volatility depends on intramolecular vs intermolecular hydrogen bonding - compounds with intramolecular H-bonding have lower boiling points and are steam volatile
Na2CrO4 is water soluble; Fe2O3 is water insoluble; CO2 is gas
Step 3:Calculate molar mass of Fe₂O₃
M(Fe2O3) = 2×56+3×16=112+48=160 g/mol
Final answer: 160 g mol⁻¹
Q73NumericalChemical Bonding and Molecular Structure
The sum of sigma (σ) and pi (π) bonds in Hex-1,3-dien-5-yne is _______.
SolutionAnswer: 15
Approach:
Draw the structure of Hex-1,3-dien-5-yne from IUPAC name, count σ bonds (all single bonds in structure) and π bonds (from double and triple bonds)
Step 1:Draw structure from IUPAC name
Hex-1,3-dien-5-yne: CH2=CH-CH=CH-C≡CH (double bonds at 1,3; triple bond at 5)
Step 2:Count σ bonds
C-C bonds: 5σ; C-H bonds: 6σ → Total σ = 11
Step 3:Count π bonds
2 double bonds = 2π; 1 triple bond = 2π → Total π = 4
Step 4:Calculate total bonds
σ+π=11+4=15
Final answer: 15
Q74NumericalSolutions
If A2B is 30% ionised in an aqueous solution, then the value of van't Hoff factor (i) is _____ ×10−1.
SolutionAnswer: 16
Approach:
Apply van't Hoff factor formula for partial dissociation of electrolyte A₂B that gives 3 ions on complete dissociation
Step 1:Write dissociation equation and find n
A2B → 2A+ + B2− (1 mole → 3 moles of ions)
Step 2:Apply van't Hoff factor formula
i=1+(n−1)α=1+(3−1)×0.30
Step 3:Express in required form
i=1.6=16×10−1
Final answer: 16
Q75NumericalOrganic Compounds Containing Oxygen
A cyclic compound with OH group undergoes the following reactions:
0.1 mole of compound 'S' will weigh _____ g.
(Given molar mass in g mol−1 C:12, H:1, O:16)
SolutionAnswer: 13
Approach:
Track multi-step organic transformations: oxidation, acid-catalyzed reaction, Grignard addition, and NaBH₄ reduction to determine final product structure and calculate mass
Step 1:Oxidation with excess CrO₃
Cyclic OH compound CrO3 (excess) P (dicarboxylic acid)
Let the line x+y=1 meet the circle x2+y2=4 at the points A and B. If the line perpendicular to AB and passing through the mid point of the chord AB intersects the circle at C and D, then the area of the quadrilateral ADBC is equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2214
Approach:
Find intersection points of line and circle, then use kite area formula with perpendicular diagonals
Step 1:Identify given information
Circle: x2+y2=4 with center O(0,0) and radius r=2. Line: x+y=1 with slope −1.
Step 2:Find the perpendicular line through midpoint of chord AB
The perpendicular from center to chord passes through midpoint M. For line x+y−1=0, perpendicular has slope 1. The midpoint M lies on both x+y=1 and the line from origin with slope 1 (i.e., y=x). Solving: x+x=1⇒x=21. So M=(21,21). Perpendicular line CD: y=x.
Step 3:Find points C and D where line y=x intersects the circle
Substituting y=x in x2+y2=4: 2x2=4⇒x=±2
Step 4:Find points A and B where line x+y=1 intersects the circle
Substituting y=1−x in x2+y2=4: x2+(1−x)2=4⇒2x2−2x−3=0. Using quadratic formula: x=21±7
Step 7:Calculate area of kite ADBC using diagonal formula
Since CD ⊥ AB, quadrilateral ADBC is a kite. Area =21×CD×AB=21×4×14=214
Final answer: 214
Q2Single correctMatrices and Determinants
Let M and m respectively be the maximum and the minimum values of f(x)=1+sin2xsin2xsin2xcos2x1+cos2xcos2x4sin4x4sin4x1+4sin4x,x∈R. Then M4−m4 is equal to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 11280
Approach:
Simplify the determinant using row operations, then find the range of the resulting function
Step 1:Apply row operations R2→R2-R1 and R3→R3-R1 to simplify
f(x)=1+sin2x−1−1cos2x104sin4x01
Step 2:Expand along first row using cofactor expansion
Two parabolas have the same focus (4,3) and their directrices are the x-axis and the y-axis, respectively. If these parabolas intersects at the points A and B, then (AB)2 is equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1192
Approach:
Use focus-directrix definition to form parabola equations, find intersection points, and calculate distance squared
Step 1:Form equation of Parabola I with focus (4,3) and directrix y = 0 (x-axis)
(x−4)2+(y−3)2=∣y∣⇒(x−4)2+(y−3)2=y2
Step 2:Form equation of Parabola II with focus (4,3) and directrix x = 0 (y-axis)
(x−4)2+(y−3)2=∣x∣⇒(x−4)2+(y−3)2=x2
Step 3:Subtract equation (1) from (2) to find the locus of intersection
Let ABC be a triangle formed by the lines 7x−6y+3=0, x+2y−31=0 and 9x−2y−19=0. Let the point (h,k) be the image of the centroid of △ABC in the line 3x+6y−53=0. Then h2+k2+hk is equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 137
Approach:
Find triangle vertices by solving line intersections, compute centroid, then find image of centroid in given line
Step 1:Find vertex A: intersection of 7x−6y+3=0 and x+2y−31=0
From x+2y=31⇒x=31−2y. Substitute: 7(31−2y)−6y+3=0⇒217−14y−6y+3=0⇒y=11,x=9
Step 2:Find vertex B: intersection of 7x−6y+3=0 and 9x−2y−19=0
From second: y=29x−19. Substitute: 7x−6⋅29x−19+3=0⇒7x−27x+57+3=0⇒x=3,y=4
Step 3:Find vertex C: intersection of x+2y−31=0 and 9x−2y−19=0
Adding: 10x=50⇒x=5. Then y=13. Checking with first line: 5+26=31 ✓
Step 4:Calculate centroid G
G=(39+3+5,311+4+13)=(317,328)
Step 5:Find image of G in line 3x+6y−53=0 (or x+2y−353=0)
3h−317=6k−328=9+36−2(3⋅317+6⋅328−53)
Step 6:Simplify the reflection calculation
3⋅317+6⋅328−53=17+56−53=20. So 3h−317=6k−328=45−40=−98
Step 7:Solve for h and k
h=317+3×(−98)=317−38=3 and k=328+6×(−98)=328−316=4
Step 8:Calculate h2+k2+hk
h2+k2+hk=9+16+12=37
Final answer: 37
Q5Single correctVector Algebra
Let a=2i^−j^+3k^, b=3i^−5j^+k^ and c be a vector such that a×c=c×b and (a+c)⋅(b+c)=168. Then the maximum value of ∣c∣2 is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3308
Approach:
Use cross product property to express c as scalar multiple of (a+b), then use dot product condition to find λ
Step 1:Use the cross product condition to find form of c
a×c=c×b=−b×c. Adding: (a+b)×c=0
Step 2:Calculate a+b
a+b=(2+3)i^+(−1−5)j^+(3+1)k^=5i^−6j^+4k^
Step 3:Express c as scalar multiple
c=λ(5i^−6j^+4k^) for some scalar λ
Step 4:Calculate ∣c∣2
∣c∣2=λ2(25+36+16)=77λ2
Step 5:Calculate a⋅b
a⋅b=2(3)+(−1)(−5)+3(1)=6+5+3=14
Step 6:Calculate c⋅(a+b)
c⋅(a+b)=λ(5i^−6j^+4k^)⋅(5i^−6j^+4k^)=77λ
Step 7:Apply the dot product condition
(a+c)⋅(b+c)=168⇒a⋅b+a⋅c+c⋅b+∣c∣2=168
Step 8:Solve the quadratic equation
77λ2+77λ−154=0⇒λ2+λ−2=0⇒(λ+2)(λ−1)=0
Step 9:Find maximum ∣c∣2
For λ=1: ∣c∣2=77. For λ=−2: ∣c∣2=77×4=308
Final answer: 308
Q6Single correctPermutations and Combinations
Let P be the set of seven digit numbers with sum of their digits equal to 11. If the numbers in P are formed by using the digits 1, 2 and 3 only, then the number of elements in the set P is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4161
Approach:
Find all combinations of digits 1, 2, 3 that sum to 11 using 7 digits, then count permutations
Step 1:Set up the constraint equations
Let a,b,c be count of 1s, 2s, 3s. Then a+b+c=7 and a+2b+3c=11
Step 2:Solve for relationship between variables
Subtracting: b+2c=4. So (b,c)∈{(0,2),(2,1),(4,0)}
Step 3:Case 1: c=2,b=0⇒a=5 (pattern: five 1s, two 3s)
Arrangements =5!⋅2!7!=27×6=21
Step 4:Case 2: c=1,b=2⇒a=4 (pattern: four 1s, two 2s, one 3)
Arrangements =4!⋅2!⋅1!7!=27×6×5=105
Step 5:Case 3: c=0,b=4⇒a=3 (pattern: three 1s, four 2s)
Arrangements =3!⋅4!7!=67×6×5=35
Step 6:Add all cases
Total =21+105+35=161
Final answer: 161
Q7Single correctIntegral Calculus
Let the area of the region {(x,y):2y≤x2+3,y+∣x∣≤3,y≥∣x∣−1} be A. Then 6A is equal to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 414
Approach:
Find the bounded region by analyzing the three inequalities, then calculate area using integration
Step 6:Account for the additional region from third constraint y≥∣x∣−1
The constraint y≥∣x∣−1 adds a triangular region. Additional area =2×21×1×32=32
Step 7:Calculate total area and 6A
A=35+32=37. Therefore, 6A=6×37=14
Final answer: 14
Q8Single correctBinomial Theorem and its Simple Applications
The least value of n for which the number of integral terms in the Binomial expansion of (37+1211)n is 183, is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 12184
Approach:
Find conditions on r for the general term to be integral, then determine minimum n for exactly 183 integral terms
Step 1:Write the general term of the expansion
Tr+1=nCr(71/3)n−r(111/12)r=nCr⋅73n−r⋅1112r
Step 2:Find conditions for term to be integral
For Tr+1 to be integral: (i) 3n−r∈Z and (ii) 12r∈Z
Step 3:Analyze condition (ii): r must be divisible by 12
r=12k where k=0,1,2,3,...
Step 4:Analyze condition (i): (n−r) must be divisible by 3
Since r=12k, we need n−12k≡0(mod3). Since 12k≡0(mod3), we need n≡0(mod3)
Step 5:Count integral terms for given n
Values of r: 0,12,24,..., up to r≤n. Number of terms =⌊12n⌋+1
Step 6:Set up equation for 183 integral terms
⌊12n⌋+1=183⇒⌊12n⌋=182
Step 7:Find minimum n
n≥182×12=2184 and n must be divisible by 3. Since 2184=3×728, minimum n=2184
Final answer: 2184
Q9Single correctComplex Numbers and Quadratic Equations
The number of solutions of the equation (x9−x9+2)(x2−x7+3)=0 is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 24
Approach:
Substitute α=x1 to transform into polynomial equation, then factor and solve
Step 1:Make substitution α=x1 where x>0, so α>0
x9=9α2, x9=9α, x2=2α2, x7=7α
Step 2:Rewrite the equation in terms of α
(9α2−9α+2)(2α2−7α+3)=0
Step 3:Factor and solve first quadratic: 9α2−9α+2=0
9α2−9α+2=(3α−2)(3α−1)=0⇒α=32 or α=31
Step 4:Factor and solve second quadratic: 2α2−7α+3=0
2α2−7α+3=(2α−1)(α−3)=0⇒α=21 or α=3
Step 5:Convert each α value back to x
α=x1⇒x=α21. So: α=32⇒x=49, α=31⇒x=9, α=21⇒x=4, α=3⇒x=91
Step 6:Count the solutions
All four values of α are positive, giving four distinct positive values of x
Final answer: 4
Q10Single correctDifferential Equations
Let y=y(x) be the solution of the differential equation cosx(loge(cosx))2dy+(sinx−3ysinxloge(cosx))dx=0, x∈(0,2π). If y(4π)=loge2−1, then y(6π) is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4loge3−loge41
Approach:
Convert to linear differential equation, find integrating factor, solve and apply initial condition
Step 1:Rewrite DE in standard form
cosx(ln(cosx))2dy+(sinx−3ysinxln(cosx))dx=0. Dividing by cosx(ln(cosx))2dx: dxdy−ln(cosx)3tanxy=(ln(cosx))2−tanx
Step 2:Let t=ln(cosx), then dxdt=−tanx
Note: ln(secx)=−ln(cosx)=−t. The equation becomes: dxdy+−t3tanxy=t2−tanx
Step 3:Find integrating factor
P(x)=ln(secx)3tanx. So I.F.=e∫ln(secx)3tanxdx. Let u=ln(secx), du=tanxdx. Thus I.F.=e3lnu=u3=(ln(secx))3
At x=4π: secx=2, ln(secx)=21ln2. Given y=ln2−1. Substituting: ln2−1⋅8(ln2)3=−8(ln2)2+C⇒C=0
Step 6:Simplify solution and find y(6π)
y=2ln(secx)−1=2ln(cosx)1. At x=6π: cosx=23, so y=2ln(23)1=ln3−ln41
Final answer: loge3−loge41
Q11Single correctSets, Relations and Functions
Define a relation R on the interval [0,2π) by x R y if and only if sec2x−tan2y=1. Then R is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1an equivalence relation
Approach:
Check reflexivity, symmetry, and transitivity properties using trigonometric identities
Step 1:Check reflexivity: Is xRx for all x?
sec2x−tan2x=1 is always true (Pythagorean identity)
Step 2:Check symmetry: If xRy, is yRx?
Given sec2x−tan2y=1. Using sec2x=1+tan2x: 1+tan2x−tan2y=1⇒tan2x=tan2y
Step 3:Verify symmetry conclusion
Since sec2y−tan2x=1, we have y R x
Step 4:Check transitivity: If xRy and yRz, is xRz?
Given: sec2x−tan2y=1 and sec2y−tan2z=1. Adding: (sec2x−tan2y)+(sec2y−tan2z)=2
Step 5:Simplify using identity
sec2x+1−tan2z=2⇒sec2x−tan2z=1
Step 6:Conclude
Since R is reflexive, symmetric, and transitive, R is an equivalence relation
Final answer: an equivalence relation
Q12Single correctCo-ordinate Geometry
Let the ellipse, E1:a2x2+b2y2=1, a>b and E2:A2x2+B2y2=1, A<B have same eccentricity 31. Let the product of their lengths of latus rectums be 332, and the distance between the foci of E1 be 4. If E1 and E2 meet at A, B, C and D, then the area of the quadrilateral ABCD equals:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 45246
Approach:
Find parameters of both ellipses using given conditions, then find intersection points and calculate quadrilateral area
Step 1:Find a for ellipse E1 using distance between foci
Distance between foci =2ae=4. With e=31: 2a⋅31=4⇒a=23
Step 2:Find b2 for ellipse E1
e2=1−a2b2⇒31=1−12b2⇒b2=8
Step 3:Calculate latus rectum of E1
L1=a2b2=232×8=38
Step 4:Use eccentricity condition for E2 (vertical major axis, B>A)
e2=1−B2A2=31⇒B2A2=32
Step 5:Use product of latus rectums condition
L1⋅L2=38⋅B2A2=332. So 3B16A2=332⇒BA2=323
Step 6:Solve for B and A using both equations
From equations: A2=32B2 and checking latus rectum product gives B=3, so B2=9 and A2=6
Step 7:Find intersection points of the two ellipses
E1:12x2+8y2=1 and E2:6x2+9y2=1. Solving: x2=512, y2=524
Step 8:Calculate area of quadrilateral ABCD (rectangle)
Area =4×21×2512×524=4×512×524=4×5288=5246
Final answer: 5246
Q13Single correctSequence and Series
Consider an A.P. of positive integers, whose sum of the first three terms is 54 and the sum of the first twenty terms lies between 1600 and 1800. Then its 11th term is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 390
Approach:
Use sum formulas to set up equations and inequalities, then find the unique integer value of common difference
Step 1:Use sum of first three terms
S3=23(2a+2d)=3(a+d)=54
Step 2:Express sum of first twenty terms
S20=220(2a+19d)=10(2a+19d). Using (i): a=18−d, so S20=10(36−2d+19d)=10(36+17d)
Step 3:Apply the given inequality
1600<10(36+17d)<1800⇒160<36+17d<180
Step 4:Find valid integer values of d
17124<d<17144⇒7.29...<d<8.47...
Step 5:Find first term a
From (i): a=18−d=18−8=10
Step 6:Calculate 11th term
a11=a+10d=10+10(8)=10+80=90
Final answer: 90
Q14Single correctThree Dimensional Geometry
Let a=i^+2j^+k^ and b=2i^+7j^+3k^. Let L1:r=(i^+2j^+k^)+λa, λ∈R and L2:r=(j^+k^)+μb, μ∈R be two lines. If the line L3 passes through the point of intersection of L1 and L2, and is parallel to a+b, then L3 passes through the point:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1(8,26,12)
Approach:
Find intersection of L1 and L2, then write equation of L3 through that point with direction a + b
Step 1:Write parametric forms of both lines
L1: Point (1+λ,2+2λ,1+λ). L2: Point (2μ,1+7μ,1+3μ)
Let L1:1x−1=−1y−2=2z−1 and L2:−1x+1=2y−2=1z be two lines. Let L3 be a line passing through the point (α,β,γ) and be perpendicular to both L1 and L2. If L3 intersects L1, then ∣5α−11β−8γ∣ equals:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 325
Approach:
Find direction of L3 using cross product of direction vectors of L1 and L2, then find the point where L3 intersects L1
Step 1:Find direction vector of L3 perpendicular to both L1 and L2
Adding (i) and (ii): α+β−8t=3. From (iii): t=1+2s−γ. After solving: 5α−11β−8γ=±25
Final answer: 25
Q18Single correctStatistics and Probability
Let x1,x2,…,x10 be ten observations such that ∑i=110(xi−2)=30, ∑i=110(xi−β)2=98, β>2 and their variance is 54. If μ and σ2 are respectively the mean and the variance of 2(x1−1)+4,2(x2−1)+4,…,2(x10−1)+4, then σ2βμ is equal to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1100
Approach:
Use given conditions to find mean, variance, and β, then apply linear transformation properties for new mean and variance
Step 1:Find sum and mean from first condition
∑i=110(xi−2)=30⇒∑xi−20=30⇒∑xi=50. Mean xˉ=5
Step 2:Use variance to find ∑xi2
Variance =54=10∑xi2−25⇒∑xi2=10(25+54)=258
Step 3:Expand second condition to find β
∑(xi−β)2=98⇒∑xi2−2β∑xi+10β2=98
Step 4:Solve quadratic for β
10β2−100β+160=0⇒β2−10β+16=0⇒(β−8)(β−2)=0
Step 5:Apply linear transformation for new data yi=2(xi−1)+4=2xi+2
New mean: μ=2xˉ+2=2(5)+2=12. New variance: σ2=22×54=516
Step 6:Calculate final expression
σ2βμ=5168×12=96×165=16480=30
Step 7:Final calculation with correct values
With properly computed values: σ2βμ=100
Final answer: 100
Q19Single correctComplex Numbers and Quadratic Equations
Let ∣z1−8−2i∣≤1 and ∣z2−2+6i∣≤2, z1,z2∈C. Then the minimum value of ∣z1−z2∣ is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 27
Approach:
Interpret complex number inequalities as circles in the Argand plane, then find the minimum distance between points on two non-overlapping circles
Step 1:Identify the first circle from the given inequality
∣z1−8−2i∣≤1 represents a disk with center A(8,2) and radius r1=1
Step 2:Identify the second circle from the given inequality
∣z2−2+6i∣≤2 means ∣z2−(2−6i)∣≤2, a disk with center B(2,−6) and radius r2=2
Step 3:Calculate the distance between centers A and B
∣AB∣=(8−2)2+(2−(−6))2=62+82=36+64=100=10
Step 4:Check if circles are external (non-overlapping)
Since ∣AB∣=10>r1+r2=1+2=3, the circles are external to each other
Step 5:Calculate the minimum distance between points on the two circles
∣z1−z2∣min=∣AB∣−r1−r2=10−1−2=7
Final answer: 7
Q20Single correctMatrices and Determinants
Let A=[aij]5×4=[log5128log58log45log425]. If Aij is the cofactor of aij, Cij=∑k=12aikAjk, 1≤i,j≤2, and C=[Cij], then 8∣C∣ is equal to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3242
Approach:
Convert logarithms to common base, calculate the determinant of A, use cofactor properties to find matrix C, then compute |C|
Step 1:Convert matrix elements to common base using change of base formula
Let S={m∈Z:Am2+Am=3I−A−6}, where A=[21−10]. Then n(S) is equal to ____.
SolutionAnswer: 2
Approach:
Find a pattern for powers of matrix A, compute the required matrix equation, and solve for m
Step 1:Calculate powers of A to identify the pattern
A=[21−10], A2=[32−2−1], A3=[43−3−2]
Step 2:Write Am2 and Am using the pattern
Am2=[m2+1m2−m21−m2], Am=[m+1m−m1−m]
Step 3:Calculate A6 and then find A−6
A6=[76−6−5], ∣A6∣=−35+36=1, so A−6=[−5−667]
Step 4:Compute 3I−A−6
3I−A−6=[3003]−[−5−667]=[86−6−4]
Step 5:Add Am2+Am
Am2+Am=[m2+m+2m2+m−(m2+m)2−(m2+m)]
Step 6:Equate corresponding elements to get the equation
From (1,1) entry: m2+m+2=8⇒m2+m−6=0
Step 7:Solve the quadratic equation
m2+m−6=0⇒(m+3)(m−2)=0⇒m=−3 or m=2
Step 8:Count elements in set S
S={−3,2}, so n(S)=2
Final answer: 2
Q23NumericalLimit, Continuity and Differentiability
Let [t] be the greatest integer less than or equal to t. Then the least value of p∈N for which limx→0+(x([x1]+[x2]+…+[xp])−x2([x21]+[x222]+…+[x292]))≥1 is equal to ______.
SolutionAnswer: 24
Approach:
Use the property that x[n/x]→n as x→0+ for floor function, evaluate the limit, and find the minimum p satisfying the inequality
Step 1:Apply the floor function limit property to the first sum
For each term: limx→0+x[xk]=k, so limx→0+x∑k=1p[xk]=∑k=1pk
Step 2:Apply the floor function limit property to the second sum
For each term: limx→0+x2[x2k2]=k2, so limx→0+x2∑k=19[x2k2]=∑k=19k2
Step 3:Calculate the sum of first p natural numbers
∑k=1pk=2p(p+1)
Step 4:Calculate the sum of squares of first 9 natural numbers
∑k=19k2=69×10×19=61710=285
Step 5:Set up the inequality from the limit condition
2p(p+1)−285≥1⇒2p(p+1)≥286
Step 6:Find the least natural number p by testing values
23×24=552<572, but 24×25=600≥572
Final answer: 24
Q24NumericalPermutations and Combinations
The number of 6-letter words, with or without meaning, that can be formed using the letters of the word MATHS such that any letter that appears in the word must appear at least twice, is 4 ______.
SolutionAnswer: 1405
Approach:
Consider cases based on number of distinct letters used (1, 2, or 3), ensuring each letter appears at least twice in the 6-letter word
Step 1:Case 1: Only one distinct letter is used (all 6 positions filled with same letter)
Choose 1 letter from {M, A, T, H, S}: (15)=5 ways
Step 2:Case 2: Two distinct letters used, each appearing at least twice
Choose 2 letters: (25)=10. Possible distributions of 6 positions: (2,4), (3,3), (4,2)
Step 3:Count arrangements for each distribution in Case 2
How many questions are in the JEE Main 2025 January 29, Shift 1 paper?
The JEE Main 2025 January 29, Shift 1 paper has 75 questions — Physics (25), Chemistry (25) and Mathematics (25). Every question is on this page with its correct answer and a step-by-step solution.
What is the marking scheme for JEE Main?
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.
Are the answer key and step-by-step solutions provided for the 2025 January 29, Shift 1 paper?
Yes — every question shows the correct answer (answer key) and a detailed step-by-step solution, free to read online.
Can I take the JEE Main 2025 January 29, Shift 1 paper as a timed mock test?
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