JEE Main 2026 January 21, Shift 2 Question Paper with Solutions
All 75 questions from the JEE Main 2026 (January 21, 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.
A large drum having radius R is spinning around its axis with angular velocity ω as shown in the figure. The minimum value of ω so that body of mass M remains stuck to the inner wall of the drum taking the coefficient of friction between drum surface and mass M as μ is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3ωmin=μRg
Approach:
The inner wall supplies a horizontal normal force that provides the centripetal force, while vertical friction balances gravity; setting friction at its limiting value yields the minimum angular velocity.
Step 1:The wall pushes the block radially inward; this normal force provides the centripetal force for the circular motion of the block.
N=Mω2R
Step 2:For the block not to slide down, upward static friction must support the weight. At the minimum angular velocity, friction reaches its limiting value μN.
μN=Mg
Step 3:Substitute N = Mω²R into the friction condition.
μMω2R=Mg
Step 4:Cancel M and solve for ω.
ω2=μRg⇒ωmin=μRg
Final answer: ωmin=μRg
Q27Single correctOptics
As shown in the diagram, when the incident ray is parallel to base of the prism the emergent ray grazes along the second surface. If refractive index of the material of prism is 2, the angle θ of prism is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3θ=60∘
Approach:
Grazing emergence means the ray meets the second surface at the critical angle; find the critical angle from μ, obtain the first-face refraction angle from the 45° geometry, apply r₁ + r₂ = A, then use the triangle angle sum to get the base angle θ.
Step 1:Grazing emergence at the second surface means the internal angle there equals the critical angle. Compute it for μ = √2.
siniC=21⇒iC=45∘
Step 2:The incident ray is parallel to the base and the first refracting face makes 45° with the base, so the angle of incidence on the first face is 45°. Apply Snell's law there.
sin45∘=2sinr1⇒sinr1=21/2=21
Step 3:Apply the prism relation to find the apex angle A.
A=r1+r2=30∘+45∘=75∘
Step 4:Use the triangle angle sum with the apex 75° and the first base angle 45° to obtain θ.
θ=180∘−75∘−45∘=60∘
Final answer: θ=60∘
Q28Single correctUnits and Measurements
Keeping significant figures in view physical quantities 52.01 m, 153.2 m and 0.123 m is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3205.3m
Approach:
Add the three lengths, then round the sum to the least number of decimal places present among the addends, following the addition rule for significant figures.
Step 1:Compute the exact arithmetic sum of the three lengths.
52.01+153.2+0.123=205.333m
Step 2:Identify the fewest decimal places among the addends: 52.01 has 2, 153.2 has 1, 0.123 has 3. The least is 1.
52.01(2dp),153.2(1dp),0.123(3dp)
Step 3:Round the sum to one decimal place.
205.333→205.3m
Final answer: 205.3m
Q29Single correctKinetic Theory of Gases
The r.m.s. speed of oxygen molecules at 47∘C is equal to that of hydrogen molecules kept at ______ ∘C. (mass of oxygen molecule/mass of hydrogen molecule = 32/2)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4−253∘C
Approach:
Equal r.m.s. speeds require equal T/M ratios. Convert the oxygen temperature to kelvin, use the mass ratio to find the hydrogen temperature in kelvin, then convert back to degrees Celsius.
Step 1:Convert the oxygen temperature to kelvin.
TO2=273+47=320K
Step 2:Equating r.m.s. speeds gives T/M equal for both gases. With M(O₂)/M(H₂) = 32/2 = 16, set up the proportion.
32320=2TH2
Step 3:Solve for the hydrogen temperature in kelvin.
TH2=2×32320=20K
Step 4:Convert the hydrogen temperature back to degrees Celsius.
TH2=20−273=−253∘C
Final answer: −253∘C
Q30Single correctElectromagnetic Oscillations
A capacitor C is first charged fully with potential difference of V0 and disconnected from battery. The charged capacitor is connected across an inductor having inductance L. In t sec 25% of initial energy in the capacitor is transferred to inductor. The value of t is _______ sec.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3t=6πLC
Approach:
In an LC circuit the charge oscillates as q = Q₀cos(ωt). If 25% of the energy moves to the inductor, 75% stays in the capacitor, so its charge falls to √0.75 of the initial value; solve the cosine equation for the time.
Step 1:With 25% of the energy transferred to the inductor, the capacitor retains 75% of the initial energy. Since energy is proportional to q², write the charge ratio.
2Cq2=43⋅2CQ02⇒q=23Q0
Step 2:Equate to the oscillation expression and solve for the phase ωt.
Q0cos(ωt)=23Q0⇒cos(ωt)=23⇒ωt=6π
Step 3:Substitute ω = 1/√(LC) and solve for t.
t=ωπ/6=6πLC
Final answer: t=6πLC
Q31Single correctElectrostatics
Consider two identical metallic spheres of radius R each having charge Q and mass m. Their centres have an initial separation of 4 R. Both spheres are given and initial speed u towards each other. The minimum value of u, so that they can just touch each other is (Take k=4π∈01 and assume kQ2>Gm2 where G is the gravitational constant)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3u=4mRkQ2(1−kQ2Gm2)
Approach:
Apply conservation of mechanical energy from the initial centre separation 4R to the just-touching configuration (centre separation 2R). Account for both kinetic energy and the net potential energy (electrostatic repulsion plus gravitational attraction); at minimum speed the spheres just touch with zero kinetic energy.
Step 1:Write the total mechanical energy at the initial separation 4R. Both spheres move with speed u, so the total kinetic energy is 2·(½mu²) = mu².
Ei=mu2+4RkQ2−4RGm2
Step 2:At the just-touching point the centre separation is 2R and the kinetic energy is zero for the minimum-speed case.
Ef=0+2RkQ2−2RGm2
Step 3:Equate the two energies and isolate mu².
mu2=(2RkQ2−2RGm2)−(4RkQ2−4RGm2)
Step 4:Combine the terms over 4R.
mu2=4R2kQ2−2Gm2−kQ2+Gm2=4RkQ2−Gm2
Step 5:Solve for u and factor out kQ².
u=4mRkQ2−Gm2=4mRkQ2(1−kQ2Gm2)
Final answer: u=4mRkQ2(1−kQ2Gm2)
Q32Single correctRotational Motion
Two cars A and B each of mass 103kg are moving on parallel tracks separate by distance of 10 m, in same direction with speeds 72 km/h and 36 km/h. The magnitude of angular momentum of car A with respect to car B is _______ Js.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3L=105Js
Approach:
The angular momentum of A about B equals (mass of A) × (velocity of A relative to B) × (perpendicular distance between the parallel tracks). Convert speeds to m/s and use the relative velocity.
Step 1:Convert both speeds to m/s.
vA=72×185=20m/s,vB=36×185=10m/s
Step 2:The cars move in the same direction, so the relative speed of A with respect to B is the difference.
vrel=20−10=10m/s
Step 3:Apply L = m·vrel·rperp with rperp = 10 m (track separation) and m = 10³ kg.
L=1000×10×10=105Js
Final answer: L=105Js
Q33Single correctMagnetic Effects of Current and Magnetism
An infinitely long straight wire carrying current I is bent in a planer shape as shown in the diagram. The radius of the circular part is r. The magnetic field at the centre O of the circular loop is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2B=−2πμ0rI(π−1)i^
Approach:
Superpose the field at the centre O from the full circular loop and from the infinite straight wire that passes at perpendicular distance r below O. The two contributions are antiparallel along the x-axis; subtract their magnitudes and assign the net direction.
Step 1:Field at the centre due to the full circular loop, directed along −x for the given current sense.
Bloop=2rμ0I(−i^)
Step 2:Field at O due to the infinitely long straight wire at perpendicular distance r, directed along +x (opposite to the loop field).
Bwire=2πrμ0I(+i^)
Step 3:Add the two antiparallel contributions, writing the loop term over the common factor μ₀I/(2π r) using 1/(2r) = π/(2π r).
B=(−2rμ0I+2πrμ0I)i^=2πrμ0I(−π+1)i^
Step 4:Factor out the sign to present the net field.
B=−2πμ0rI(π−1)i^
Final answer: B=−2πμ0rI(π−1)i^
Q34Single correctCurrent Electricity
The charge stored by the capacitor C in the given circuit in the steady state is ______ μC.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2Q=10μC
Approach:
In steady state no current flows through the capacitor branch. The diodes fix a single conduction loop through the 1 Ω and 4 Ω resistors; find the loop current, the voltage across the 4 Ω resistor (which equals the capacitor voltage), and then Q = CVC.
Step 1:At steady state the capacitor draws no current. The conducting loop is the 2.5 V source in series with the 1 Ω and 4 Ω resistors, total 5 Ω.
i=1+42.5=52.5=0.5A
Step 2:The capacitor voltage equals the voltage across the 4 Ω resistor in the conduction loop.
VC=i×4=0.5×4=2V
Step 3:Compute the stored charge with C = 5 μF.
Q=CVC=5μF×2V=10μC
Final answer: Q=10μC
Q35Single correctOscillations and Waves
The kinetic energy of simple harmonic oscillator is oscillating with angular frequency of 176 rad/sec. The frequency of the simple harmonic oscillator is ______Hz. (π=22/7)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4f′=14Hz
Approach:
The kinetic energy of an SHM oscillator varies at twice the displacement frequency, so the displacement angular frequency is half the given KE angular frequency. Convert that to ordinary frequency.
Step 1:Kinetic energy ∝ cos²(ωt), which oscillates at twice the displacement frequency. The given 176 rad/s is therefore 2ω.
ω=2176=88rad/s
Step 2:Convert the displacement angular frequency to ordinary frequency using π = 22/7.
f′=2πω=2×72288=4488×7=14Hz
Final answer: f′=14Hz
Q36Single correctKinematics
A river of width 200m flowing from west to east with a speed 18 km/h. A boat moving with speed of 36 km/h in still water, is made to travel 1 round trip (bank to bank of the river). Minimum time taken by the boat for this journey and also displacement along the river bank are ____ and _____ respectively.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1t=40s,x=200m
Approach:
Minimum crossing time is achieved when the boat heads straight across, directing its full speed perpendicular to the banks. Double the one-way time for the round trip. The current drifts the boat downstream throughout both legs, accumulating a net bank displacement.
Step 1:Convert speeds: 36 km/h = 10 m/s (boat) and 18 km/h = 5 m/s (river). For minimum time, the full boat speed crosses the 200 m width.
tcross=10200=20s
Step 2:A round trip is two crossings (bank to bank and back), so the total time is twice the one-way time.
t=2×20=40s
Step 3:The current carries the boat downstream the entire time, and the drift accumulates in the same direction over both legs.
x=vriver×t=5×40=200m
Final answer: t=40s and displacement along the river bank =200m
Q37Single correctProperties of Solids and Liquids
A spherical body of radius r and density σ falls freely through a viscous liquid having density ρ and viscosity η and attains a terminal velocity v0. Estimated maximum error in the quantity η is (ignore errors associated with σ,ρ and g, gravitational acceleration)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4ηΔη=r2Δr+V0ΔV0
Approach:
Express the viscosity from the terminal-velocity formula, then propagate the maximum relative error. With σ, ρ and g treated as exact, η depends on r² and on 1/v₀, so the fractional errors add with their respective magnitudes.
Step 1:Solve the terminal velocity relation for η.
η=9v02(σ−ρ)gr2
Step 2:Ignoring errors in σ, ρ, g, take the maximum relative error. The exponent of r is +2 and of v₀ is −1, and for maximum error the magnitudes of the contributions add.
ηΔη=2rΔr+∣−1∣v0Δv0=r2Δr+v0Δv0
Final answer: ηΔη=r2Δr+V0ΔV0
Q38Single correctCurrent Electricity
Two known resistance of RΩ and 2RΩ and one unknown resistance XΩ are connected in a circuit as shown in the figure. If the equivalent resistance between points A and B in the circuit is XΩ, then value of X is ______ Ω.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4X=(3−1)R
Approach:
The top branch (2R in series with X) is in parallel with the bottom branch (R). Set this parallel combination equal to X, the stated equivalent resistance, and solve the resulting quadratic.
Step 1:Form the parallel equivalent of (2R + X) with R and set it equal to X.
(2R+X)+R(2R+X)R=X⇒3R+X(2R+X)R=X
Step 2:Cross-multiply and expand.
(2R+X)R=X(3R+X)⇒2R2+XR=3RX+X2
Step 3:Collect terms into a quadratic in X.
X2+2RX−2R2=0
Step 4:Solve the quadratic and keep the positive root.
X=2−2R+4R2+8R2=2−2R+23R=R(3−1)
Final answer: X=(3−1)R
Q39Single correctRotational Motion
The pulley shown in the figure is made using a thin rim and two rods of equal length to the diameter of rim. The rim and each rod have a mass of M. Two blocks of mass M and m are attached to two ends of a light string passing over the pulley, which is hinged to rotate freely in vertical plane about its center. The magnitudes of the acceleration experienced by the block is _____ (Assume no slipping of string on pulley)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4a=[(38)M+m](M−m)g
Approach:
Write Newton's equations for the two hanging blocks and the torque equation for the pulley. Compute the pulley's moment of inertia (rim plus two diametric rods), then combine using the no-slip constraint a = αr to solve for a.
Step 1:Write the force equations for the two blocks (M heavier, descending; m lighter, ascending).
Mg−T2=Ma;T1−mg=ma
Step 2:Write the torque equation for the pulley with the no-slip constraint α = a/r.
(T2−T1)r=Ira⇒T2−T1=r2Ia
Step 3:Add the three equations so the tensions cancel.
(M−m)g=(M+m+r2I)a
Step 4:Compute the moment of inertia: the rim of mass M and radius r contributes Mr², and each rod has length equal to the diameter 2r, mass M, pivoted at its centre, contributing M(2r)²/12 = Mr²/3 each (two rods).
I=Mr2+2⋅12M(2r)2=Mr2+32Mr2=35Mr2
Step 5:Substitute I/r² = 5M/3 into the combined equation; the effective inertial mass is M + m + 5M/3 = (8/3)M + m.
a=M+m+35M(M−m)g=(38)M+m(M−m)g
Final answer: a=[(38)M+m](M−m)g
Q40Single correctOptics
Given below are two statements: Statement 1: In Young's double slit experiment, the angular separation of fringes will increase as the screen is moved away from the plane of the slits. Statement II: In a Young's double slit experiment, the angular separation of fringes will increase when monochromatic source is replaced by another monochromatic source of higher wavelength. In the light of the above statements, choose the correct answer from the options given below:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2Statement I false but Statement II is true
Approach:
The angular fringe separation in a YDSE is θ = λ/d, depending only on wavelength and slit separation, not on screen distance. Evaluate each statement against this relation.
Step 1:Angular separation θ = λ/d is independent of the screen distance D. Moving the screen away increases the linear fringe width (β = λD/d) but not the angular separation, so Statement I is false.
θ=dλ(independent of D)
Step 2:Since θ is directly proportional to λ, increasing the wavelength increases the angular separation, so Statement II is true.
θ∝λ⇒θ↑ as λ↑
Final answer: Statement I false but Statement II is true
Q41Single correctCurrent Electricity
A battery with EMF E and internal resistance r is connected across a resistance R. The power consumption in R will be maximum when:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2R=r
Approach:
Write the power dissipated in R as a function of R, then maximise it by setting its derivative with respect to R to zero.
Step 1:The series current is I = E/(R+r); the power dissipated in R is I²R.
P=(R+r)2E2R
Step 2:Differentiate P with respect to R and set it to zero.
dRdP=E2(R+r)4(R+r)2−R⋅2(R+r)=0⇒(R+r)−2R=0
Step 3:Solve for R.
r=R⇒R=r
Final answer: R=r
Q42Single correctWork, Energy and Power
A body of mass 2 kg is moving along x direction such that its displacement as function of time is given by x(t)=at2+βt+γm, where α=1m/s2, β=1m/s and γ=1m. The work done on the body during the time interval t=2s to t=3s, is _______ J.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3W=24J
Approach:
Differentiate the position to obtain the (constant) acceleration, compute the constant net force F = ma, find the displacement from t = 2 s to t = 3 s, and take work done = F × displacement.
Step 1:With α = 1, β = 1, γ = 1, the position is x(t) = t² + t + 1. Differentiate to get velocity and then acceleration.
v=2t+1,a=dtdv=2m/s2
Step 2:Compute the constant net force.
F=ma=2×2=4N
Step 3:Evaluate positions at t = 2 s and t = 3 s and take the displacement.
x(2)=4+2+1=7,x(3)=9+3+1=13,Δx=13−7=6m
Step 4:Work done equals force times displacement (force is constant).
W=FΔx=4×6=24J
Final answer: W=24J
Q43Single correctCurrent Electricity
The total length of potentiometer wire AB is 50cm in the arrangement as shown in the figure. If P is the point where the galvanometer shows Zero reading then length of AP is ______ cm.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3AP=30cm
Approach:
At the null point the galvanometer reads zero, so the ratio of potential drops across the 6 Ω and 4 Ω resistors equals the ratio of potential drops along the wire segments AP and PB. Set up the proportion using total length 50 cm and solve for AP.
Step 1:Let AP = x cm so that PB = (50 − x) cm. Balancing the 6 Ω and 4 Ω drops against the wire segments gives the proportion.
46=50−xx
Step 2:Simplify the ratio and cross-multiply.
23=50−xx⇒3(50−x)=2x⇒150−3x=2x
Step 3:Solve for x.
5x=150⇒x=30cm
Final answer: AP=30cm
Q44Single correctProperties of Solids and Liquids
Surface tension of two liquids (having same densities). T1 and T2 are measured using capillary rise method utilising two tubes of inner radius r1 and r2 where r1>r2. The measure liquid heights in these tubes are h1 and h2 respectively. (ignore weight of liquid about the lowest point of meniscus). The heights h1 and h2 and surface tension T1 and T2 satisfy the relation.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1h1<h2andT1=T2
Approach:
Capillary rise h is inversely proportional to the tube radius for a given liquid. With the same liquid (same surface tension and density) in both tubes, the larger radius tube gives the smaller rise.
Step 1:Surface tension is a property of the liquid; for the same liquid it is identical in both tubes.
T1=T2
Step 2:For fixed T, ρ and contact angle, the rise is inversely proportional to radius. Since r₁ > r₂, the rise in tube 1 is smaller.
h∝r1,r1>r2⇒h1<h2
Final answer: h1<h2 and T1=T2
Q45Single correctAtoms and Nuclei
The energy of an electron in an orbit of Bohr's atom is −0.04E0eV where E0 is the ground state energy. If L is the angular momentum of electron in this orbit and h is planks constant then h2πL is ______.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3h2πL=5
Approach:
In the Bohr model L = nh/(2π), so 2πL/h = n. Use the energy relation En = E₀/n² (E₀ the ground-state magnitude) with the given E = −0.04 E₀ to find n.
Step 1:From the quantisation of angular momentum, 2πL/h equals the principal quantum number n.
h2πL=n
Step 2:The orbit energy scales as 1/n² of the ground-state energy. Equate magnitudes to 0.04 E₀.
n2E0=0.04E0⇒n2=0.041=25
Step 3:Take the positive root for the orbit number.
n=5
Final answer: h2πL=5
Q46NumericalThermodynamics
A diatomic gas (γ=1.4) does 100J of work when it is expanded isobarically. Then the heat given to the gas is ________ J.
SolutionAnswer: 350
Approach:
In an isobaric process the work is W = nRΔT and the heat is Q = nCpΔT. Use γ = 1.4 (degrees of freedom f = 5 for a diatomic gas) so that Cp = (7/2)R, giving Q as a fixed multiple of W.
Step 1:At constant pressure the work done by the gas is W = nRΔT = 100 J.
W=nRΔT=100J
Step 2:For a diatomic gas f = 5, so Cp = (f/2 + 1)R = (7/2)R. The isobaric heat is then Q = (7/2)nRΔT.
Q=(25+1)nRΔT=27nRΔT
Step 3:Substitute nRΔT = 100 J.
Q=27×100=350J
Final answer: Q=350J
Q47NumericalDual Nature of Matter and Radiation
A particle having electric charge 3×10−19 C and mass 6×10−27 kg is accelerated by applying an electric potential of 1.21V. Wave length of the matter wave associated with the particle is α×10−12m. The value of α is ______. (Take Planck's constant =6.6×10−34J.s)
SolutionAnswer: 10
Approach:
A charge accelerated through potential V gains kinetic energy qV, giving momentum p = √(2mqV). The de Broglie wavelength is λ = h/p; substitute the values and express λ in units of 10⁻¹² m.
Step 1:Compute the product 2mqV.
2mqV=2×(6×10−27)×(3×10−19)×1.21=43.56×10−46
Step 2:Take the square root to obtain the momentum (√43.56 = 6.6).
2mqV=43.56×10−46=6.6×10−23kg⋅m/s
Step 3:Compute the wavelength.
λ=6.6×10−236.6×10−34=10−11m=10×10−12m
Step 4:Compare with λ = α×10⁻¹² m to read off α.
α=10
Final answer: α=10
Q48NumericalProperties of Solids and Liquids
The terminal velocity of a metallic ball of radius 6 mm in a viscous fluid 20 cm/s. The terminal velocity of another ball of same material and having radius 3mm in same fluid will be ______ cm/sec.
SolutionAnswer: 5
Approach:
For balls of the same material in the same fluid, terminal velocity is proportional to the square of the radius. Use the ratio of radii to scale the given terminal velocity.
Step 1:Since the material density and fluid are the same, terminal velocity scales as the square of radius. Form the ratio for r₁ = 6 mm and r₂ = 3 mm.
(vT)2(vT)1=(r2r1)2=(36)2=4
Step 2:The smaller ball therefore has one quarter the terminal velocity of the larger ball.
(vT)2=4(vT)1=420=5cm/s
Final answer: (vT)2=5cm/sec
Q49NumericalOptics
In a Young's double slit experiment setup, the two slits are kept 0.4 mm apart and screen is placed at 1m from slits. If a thin transparent sheet of thickness 20μm is introduced in front of one of the slits then center bright fringe shifts by 20 mm on the screen. The refractive index of transparent sheet is given by 10α, where α is ______.
SolutionAnswer: 14
Approach:
A thin sheet of thickness t and refractive index μ introduces an extra optical path (μ−1)t, shifting the central fringe by Δy = (D/d)(μ−1)t. Solve for μ and express it as α/10.
Step 1:Rearrange the shift equation for (μ−1).
μ−1=DtΔyd
Step 2:Substitute Δy = 20×10⁻³ m, d = 0.4 mm = 4×10⁻⁴ m, D = 1 m, t = 20×10⁻⁶ m.
An electromagnetic wave of frequency 100 MHz propagates through a medium of conductivity, σ=10mho/m. The ratio of maximum conduction current density to maximum displacement current density is _______. [Take4π∈01=9×109Nm2/C2]
SolutionAnswer: 1800
Approach:
Conduction current density is Jc = σE and displacement current density is Jd = ε₀ ∂E/∂t. For a harmonic field E = E₀ sin(ωt − kx), the maxima give Jc,max = σE₀ and Jd,max = ε₀ωE₀ (relative permittivity taken as 1), so the ratio is σ/(ε₀ω) with ω = 2πf.
Step 1:Take the ratio of the maximum conduction current density to the maximum displacement current density; the field amplitude E₀ cancels.
Jd,maxJc,max=ε0ωE0σE0=ε0ωσ
Step 2:Compute the angular frequency from f = 100 MHz = 10⁸ Hz.
Final answer: Ratio of maximum conduction current density to maximum displacement current density =1800.
Chemistry25 questions
Q51Single correctSome Basic Concepts in Chemistry
Aqueous HCl reacts with MnO2(s) to form MnCl2(aq),Cl2(g) and H2O(l). What is the weight (in g) of Cl2 liberated when 8.7 g of MnO2(s) reacted with excess aqueous HCl solution? (Given Molar mass in gmol−1Mn=55,Cl=35.5,O=16,H=1)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 17.1
Approach:
Balance the redox reaction, convert the given mass of the limiting reactant to moles, then find the mass of chlorine liberated.
Step 1:Write the balanced reaction; one mole of MnO2 liberates one mole of Cl2.
MnO2+4HCl→MnCl2+Cl2+2H2O
Step 2:Compute molar mass of MnO2 and moles present in 8.7 g.
MMnO2=55+2(16)=87,n=878.7=0.1mol
Step 3:HCl is in excess, so MnO2 is limiting; moles of Cl2 equal moles of MnO2.
nCl2=0.1mol
Step 4:Convert moles of Cl2 to mass using MCl2=71.
m=0.1×71=7.1g
Final answer: 7.1 g
Q52Single correctChemical Thermodynamics
Consider the following data: ΔfH⊖ (methane, g) =−XkJmol−1. Enthalpy of sublimation of graphite =YkJmol−1. Dissociation enthalpy of H2=ZkJmol−1. The bond enthalpy of C−H bond is given by
(A)
(B)
(C)
(D)
SolutionAnswer: Option 14X+Y+2Z
Approach:
Apply a Hess-cycle energy balance for the formation of methane from graphite and hydrogen in terms of atomization and bond-formation energies.
Step 1:Write the formation reaction of methane from its elements.
C(s)+2H2(g)→CH4(g)
Step 2:Express the formation enthalpy as energy to atomize reactants minus energy released forming four C−H bonds.
−X=Y+2Z−4(B.E. of C−H)
Step 3:Rearrange to isolate the bond enthalpy of C−H.
4(B.E. of C−H)=X+Y+2Z
Step 4:Divide by 4 to obtain the bond enthalpy.
B.E. of C−H=4X+Y+2Z
Final answer: 4X+Y+2Z
Q53Single correctAtomic Structure
Consider the following spectral lines for atomic hydrogen: A) First line of Paschen series B) Second line of Balmer series C) Third line of Paschen series D) Fourth line of Bracket series. The correct arrangement of the above lines in ascending order of energy is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1D<A<C<B
Approach:
Identify the initial and final levels for each line, then compare transition energies using the Rydberg energy expression.
Step 1:Assign quantum numbers for each line (series lower level fixed, line number sets the upper level).
Step 2:Compute the factor (n121−n221) for D and A.
D=161−641=0.0469;A=91−161=0.0486
Step 3:Compute the factor for C and B.
C=91−361=0.0833;B=41−161=0.1875
Step 4:Order all factors in ascending value (energy is proportional to this factor).
0.0469<0.0486<0.0833<0.1875
Final answer: D<A<C<B
Q54Single correctRedox Reactions and Electrochemistry
For a closed circuit Daniell cell, which of the following plots is the accurate one at a given temperature ?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2
Approach:
Recognise that the standard cell potential is a fixed thermodynamic quantity independent of reaction progress at constant temperature, then match it to the correct plot.
Step 1:Identify that Ecell∘ is defined for standard-state activities and depends only on temperature.
Ecell∘=f(T)only
Step 2:Note that during discharge the actual Ecell changes with concentration, but Ecell∘ does not change.
dtdEcell∘=0at constant T
Step 3:A constant value versus time is a horizontal straight line.
Ecell∘=const
Final answer: Ecell∘ stays constant with time (horizontal line).
Q55Single correctChemical Kinetics
Decomposition of A is a first order reaction at T(K) and is given by A(g)→B(g)+C(g). In a closed 1 L vessel, 1 bar A(g) is allowed to decompose at T(K). After 100 minutes, the total pressure was 1.5 bar. What is the rate constant (inmin−1) of the reaction ? (log2=0.3)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 26.9×10−3
Approach:
Relate the rise in total pressure to the partial pressure of A consumed, then apply the first-order integrated rate law.
Step 1:Set up pressures: let P be the pressure of A decomposed.
PA=1−P,PB=P,PC=P
Step 2:Write total pressure and solve for P using the measured total.
Ptotal=(1−P)+P+P=1+P=1.5⇒P=0.5
Step 3:Find remaining pressure of A.
PA=1−0.5=0.5bar
Step 4:Apply the first-order law over t=100 min.
k=1001ln0.51=1000.693=6.9×10−3min−1
Final answer: 6.9×10−3min−1
Q56Single correctClassification of Elements and Periodicity in Properties
Given below are two statements Statement -I: The correct order in terms of atomic/ ionic radii is Al>Mg>Mg2+>Al3+ Statement -II : The correct order in terms of the magnitude of electron gain enthalpy is Cl>Br>S>O. In the light of the above statements, choose the correct answer from the options given below:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4Statement I is false and II is correct .
Approach:
Test each statement against periodic trends for atomic/ionic radii and for the magnitude of electron gain enthalpy.
Step 1:Across period 3, atomic radius decreases left to right, so Mg is larger than Al; cations are smaller than their parent atoms.
Mg>Al>Mg2+>Al3+
Step 2:Statement I claims Al>Mg, which contradicts the trend, so Statement I is false.
Al>Mg(claimed)=Mg>Al(true)
Step 3:Compare electron gain enthalpy magnitudes; third-period Cl and S have larger magnitudes than their lighter/heavier congeners due to size and electron-density effects.
Cl>Br>S>O
Final answer: Statement I is false and Statement II is correct.
Q57Single correctChemical Bonding and Molecular Structure
Given below are two statements Statement I : The correct order in terms of bond dissociation energy order is : Cl2>Br2>F2>I2 Statement II: The correct trend in the covalent character of the metal halides is [SnCl4>SnCl2], [PbCl4>PbCl2] and [UF4>UF6]. In the light of the above statements , choose the correct answer from the options given below :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3Statement I is true but Statement II is false.
Approach:
Verify the anomalous halogen bond-dissociation-energy order and check each covalent-character pair using Fajans' rules.
Step 1:Halogen bond dissociation energy is anomalous: F2 is low because of lone-pair repulsion at the small F atoms.
Cl2>Br2>F2>I2
Step 2:By Fajans' rules, higher cation charge gives greater covalent character, so the +4 chlorides are more covalent than the +2 chlorides.
SnCl4>SnCl2,PbCl4>PbCl2
Step 3:For uranium fluorides the higher oxidation state U+6 in UF6 is more covalent than U+4 in UF4, so the claimed UF4>UF6 is wrong.
UF4<UF6
Final answer: Statement I is true but Statement II is false.
Q58Single correctPrinciples Related to Practical Chemistry
On heating a mixture of common salt and K2Cr2O7 in equal amount along with concentrated H2SO4 in a test tube, a gas is evolved . Formula of the gas evolved and oxidation state of the central metal atom in the gas respectively are :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1CrO2Cl2 and +6
Approach:
Identify the chromyl chloride test product and assign the oxidation state of chromium from charge balance.
Step 1:Heating a chloride with K2Cr2O7 and concentrated H2SO4 gives the deep red vapours of chromyl chloride.
4NaCl+K2Cr2O7+6H2SO4→2CrO2Cl2+salts+3H2O
Step 2:Assign oxidation states in CrO2Cl2: O is −2, Cl is −1.
x+2(−2)+2(−1)=0
Step 3:Solve for the chromium oxidation state.
x=+6
Final answer: CrO2Cl2 with chromium in the +6 oxidation state.
Q59Single correctChemical Bonding and Molecular Structure
The correct increasing order of C−H(A),C−O(B),C=O(C)&C≡N(D) bonds in terms of covalent bond length is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3A<D<C<B
Approach:
Use standard NCERT bond-length values for each bond and arrange them in increasing order.
Given below are some of the statements about Mn and Mn2O7. Identify the correct statements. A) Mn forms the oxide Mn2O7, in which Mn is in its highest oxidation state. B) Oxygen stabilizes the Mn in higher oxidation states by forming multiple bonds with Mn. C) Mn2O7 is an ionic oxide. D) The structure of Mn2O7 consists of one bridged oxygen. Choose the correct answer from the options given below :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4A, B and D only
Approach:
Evaluate each statement about the structure and bonding of Mn2O7 and select the correct set.
Step 1:Statement A: in Mn2O7 each Mn is +7, its highest possible oxidation state.
2x+7(−2)=0⇒x=+7
Step 2:Statement B: high oxidation states of Mn are stabilised by multiple (pπ-dπ) bonding from oxygen.
Mn=Omultiple bonds
Step 3:Statement C: Mn2O7 is a covalent, molecular, oily-green liquid oxide, not ionic.
Mn2O7is covalent
Step 4:Statement D: its structure is O3Mn−O−MnO3 with one bridging oxygen joining two MnO3 units.
O3Mn−O−MnO3
Final answer: A, B and D only.
Q61Single correctCoordination Compounds
Given below are two statements: Statement I : Crystal field stabilization Energy (CFSE) of [Cr(H2O)6]2+ is greater than that of [Mn(H2O)6]2+ Statement II : Potassium ferricyanide has a greater spin – only magnetic moment than sodium ferrocyanide. In the light of the above statements, choose the correct answer from the options given below:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1Both Statement I and Statement II both are true
Approach:
Compare the d-electron CFSE of the two aqua complexes and the spin-only magnetic moments of the two cyanide complexes.
Step 1:Assign configurations: Cr2+ is d4, Mn2+ is d5; the high-spin d5 aqua complex has zero CFSE.
[Mn(H2O)6]2+:t2g3eg2,CFSE=0
Step 2:High-spin d4 has nonzero CFSE, so the Cr complex has greater CFSE; Statement I is true.
[Cr(H2O)6]2+:t2g3eg1,CFSE=−0.6Δo
Step 3:Ferricyanide K3[Fe(CN)6] has Fe3+ (d5, low spin, n=1).
μ=1(1+2)=3=1.73B.M.
Step 4:Ferrocyanide Na4[Fe(CN)6] has Fe2+ (d6, low spin, n=0); ferricyanide therefore has the larger moment, so Statement II is true.
μ=0=0B.M.
Final answer: Both Statement I and Statement II are true.
Identify the named aldehyde-forming reaction associated with each reagent set and assemble the match.
Step 1:H2,Pd−BaSO4 reduces an acyl chloride to an aldehyde (poisoned-catalyst reduction).
RCOClH2/Pd−BaSO4RCHO
Step 2:SnCl2,HCl reduces a nitrile to an imine, hydrolysed to an aldehyde.
RCNSnCl2/HClRCHO
Step 3:CrO2Cl2,CS2 oxidises a methyl side chain of toluene to a benzaldehyde via a chromium complex.
C6H5CH3CrO2Cl2C6H5CHO
Step 4:CO,HCl,Anhyd.AlCl3 formylates benzene to benzaldehyde.
C6H6CO/HCl/AlCl3C6H5CHO
Final answer: A−II,B−IV,C−I,D−III
Q64Single correctSome Basic Principles of Organic Chemistry
Match the List-I with List-II
List-I (Pair of Compounds)
List-II (Type of Isomers)
A.. 2-Methylpropene and but-1-ene
I. Stereoisomers
B.. Cis-but-2-ene and trans – but-2-ene
II. Position isomers
C.. 2-Butanol and diethyl ether
III. Chain isomers
D.. But-1-ene and but-2-ene
IV. Functional isomers
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1A−III,B−I,C−IV,D−II
Approach:
Classify the isomerism type for each pair of compounds and assemble the match.
Step 1:2-Methylpropene and but-1-ene (both C4H8) differ in carbon skeleton.
(CH3)2C=CH2vsCH2=CHCH2CH3
Step 2:Cis- and trans-but-2-ene differ only in spatial arrangement about the double bond.
cis vs trans CH3CH=CHCH3
Step 3:2-Butanol and diethyl ether (both C4H10O) have different functional groups.
C4H9OHvsC2H5OC2H5
Step 4:But-1-ene and but-2-ene differ only in the position of the double bond.
CH2=CHCH2CH3vsCH3CH=CHCH3
Final answer: A−III,B−I,C−IV,D−II
Q65Single correctBiomolecules
The correct statements are: A) Activation energy for enzyme catalysed hydrolysis of sucrose is lower than that of acid catalysed hydrolysis B) During denaturation, secondary and tertiary structures of a protein are destroyed but primary structure remains intact. C) Nucleotides are joined together by glycosidic linkage between C1 and C4 carbons of the pentose sugar D) Quaternary structure of proteins represents overall folding of the polypeptide chain Choose the correct answer from the options given below :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2A and B only
Approach:
Assess each biomolecule statement against established facts and select the correct subset.
Step 1:Statement A: enzyme catalysis lowers activation energy far more than acid catalysis for sucrose hydrolysis.
Eaenzyme≈2.15kJmol−1<Eaacid≈6.22kJmol−1
Step 2:Statement B: denaturation destroys secondary and tertiary structure while the primary peptide sequence stays intact.
2∘,3∘lost;1∘retained
Step 3:Statement C: nucleotides are linked by phosphodiester bonds, not glycosidic C1–C4 linkages.
3′−phosphodiester−5′linkage
Step 4:Statement D: overall folding of a single chain is the tertiary structure; quaternary describes assembly of multiple subunits.
The correct order of the rate of the reaction for the following reaction with respect to nuclophiles is: CH2Br+Nu−→CH3Nu+Br−
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1−OH>PhO−>CH3COO−>ClO4−
Approach:
Rank the anions by nucleophilicity, which decreases as the negative charge becomes more delocalised (more stable anion is a weaker nucleophile).
Step 1:−OH has a localised charge on one oxygen and is the strongest nucleophile.
−OH:localised charge
Step 2:PhO− delocalises charge into the benzene ring, making it weaker than −OH.
PhO−:charge into ring
Step 3:CH3COO− delocalises charge over two oxygens, weaker than PhO−.
CH3COO−:2-oxygen resonance
Step 4:ClO4− is stabilised over four oxygens and is the weakest nucleophile.
ClO4−:4-oxygen resonance
Final answer: −OH>PhO−>CH3COO−>ClO4−
Q67Single correctSome Basic Principles of Organic Chemistry
Given below are two statements:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4Both Statement I and Statement II are true
Approach:
Test Statement I for acidity in NaHCO3 and number of chiral centres in X, and Statement II for the hybridization of each carbon in Y.
Step 1:Compound X carries a −COOH group, which is acidic enough to react with NaHCO3 and dissolve with effervescence.
RCOOH+NaHCO3→RCOONa+H2O+CO2
Step 2:The two adjacent cyclopentane ring carbons each bear four different groups (the acyl substituent, the carboxyl substituent, a ring path, and H), so each is a stereocentre.
2 stereocentres on the ring
Step 3:Compound Y is the ketene CH3−CH2−CH=C=O; assign hybridization carbon by carbon.
sp3CH3−sp3CH2−sp2CH=spC=O
Step 4:The carbon counts match Statement II exactly (two sp3, one sp2, one sp), so Statement II is true.
2sp3+1sp2+1sp
Final answer: Both Statement I and Statement II are true.
Q68Single correctPurification and Characterisation of Organic Compounds
By usual analysis 1.00 g of compound (X) gave 1.79 g of magnesium pyrophosphate. The percentage of phosphorus in compound (X) is: (nearest integer) (Give molar mass in gmol−1;O=16,Mg=24,P=31)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 150
Approach:
Estimate phosphorus from the precipitated magnesium pyrophosphate, accounting for two phosphorus atoms per formula unit.
Step 1:Compute the molar mass of Mg2P2O7.
M=2(24)+2(31)+7(16)=48+62+112=222gmol−1
Step 2:Find the mass of phosphorus in 1.79 g of precipitate (two P per formula unit).
mP=2221.79×(2×31)=2221.79×62=0.50g
Step 3:Express as percentage of the 1.00 g sample.
Consider the above sequence of reactions. The number of bromine atom(s) in final product (P) will be?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 15
Approach:
Track each transformation from nitrobenzene, counting the bromine atoms introduced at every step that survives into the final product.
Step 1:Br2/FeBr3,Δ on the deactivated nitrobenzene gives meta-bromonitrobenzene (1 Br).
C6H5NO2→m-BrC6H4NO2
Step 2:Sn/HCl,Δ then pH neutralisation reduces −NO2 to −NH2, giving m-bromoaniline.
m-BrC6H4NO2→m-BrC6H4NH2
Step 3:Br2/H2O brominates the highly activated aniline at the positions ortho/para to −NH2, adding 3 more Br atoms.
+3Br⇒4Bron ring
Step 4:NaNO2/HBr at 0–5∘C converts −NH2 to a diazonium salt; CuBr (Sandmeyer) replaces it with Br, adding the fifth Br.
−NH2→−N2+CuBr−Br
Final answer: 5 bromine atoms
Q70Single correctSome Basic Principles of Organic Chemistry
Given below are four compounds: a) n-propyl chloride b) iso-propyl chloride c) sec-butyl chloride d) neo – pentyl chloride Percentage of carbon in the one which exhibits optical isomerism is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 352
Approach:
Identify the only chiral compound among the four, then compute its percentage of carbon by mass.
Step 1:Only sec-butyl chloride has a carbon bonded to four different groups (H, Cl, CH3, C2H5), so it is optically active.
CH3CH2CHClCH3(C4H9Cl)
Step 2:Compute its molar mass.
M=4(12)+9(1)+35.5=48+9+35.5=92.5gmol−1
Step 3:Compute the carbon percentage.
%C=92.548×100=51.89≈52%
Final answer: ≈52%
Q71NumericalSolutions
The osmotic pressure of a living cell is 12 atm at 300 K. The strength of sodium chloride solution that is isotonic with the living cell at this temperature is _gL−1. (Nearest integer) Given: R=0.08LatmK−1mol−1 Assume complete dissociation of NaCl (Given: Molar mass of Na&Cl are 23&35.5gmol−1 respectively.)
SolutionAnswer: 15
Approach:
Use the osmotic-pressure equation with the van't Hoff factor for fully dissociated NaCl to get the molar concentration, then convert to grams per litre.
Step 1:For complete dissociation of NaCl into two ions, the van't Hoff factor is 2.
NaCl→Na++Cl−,i=2
Step 2:Solve the osmotic-pressure equation for molar concentration.
12=2×C×0.08×300=48C⇒C=4812=0.25mol/L
Step 3:Convert to strength using molar mass of NaCl = 58.5.
strength=0.25×58.5=14.625g/L≈15g/L
Final answer: 15gL−1
Q72NumericalEquilibrium
The first and second ionization constants of H2X are 2.5×10−8 and 1.0×10−13 respectively. The concentration of X2− in 0.1MH2X solution is ____×10−15M. (nearest integer)
SolutionAnswer: 100
Approach:
Use the standard diprotic-acid result that, with widely separated ionization constants, the doubly-deprotonated species concentration equals the second ionization constant.
Step 1:First dissociation dominates and gives nearly equal [H+] and [HX−].
H2X⇌H++HX−,[H+]≈[HX−]
Step 2:Apply the second ionization equilibrium expression.
Ka2=[HX−][H+][X2−]
Step 3:Since [H+]≈[HX−], those terms cancel leaving [X2−]=Ka2.
[X2−]=Ka2=1.0×10−13M=100×10−15M
Final answer: 100×10−15M
Q73NumericalSolutions
A substance “X” (1.5g) dissolved in 150 g of a solvent “Y” (molar mass = 300gmol−1) led to an elevation of the boiling point by 0.5 K. The relative lowering in the vapour pressure of the solvent “Y” is ___×10−2. (nearest integer) [Given: Kb of the solvent =5.0Kkgmol−1] Assume the solution to be dilute and no association or dissociation of X takes place in solution.
SolutionAnswer: 3
Approach:
Obtain the molality from the boiling-point elevation, then use it with the solvent molar mass to get the relative lowering of vapour pressure for a dilute solution.
Step 1:Solve for molality from the elevation (no dissociation, i=1).
0.5=5×1×m⇒m=0.1molkg−1
Step 2:Apply the dilute-solution relation for relative lowering of vapour pressure.
P0P0−Ps=10000.1×300=0.03
Step 3:Express in the requested form.
0.03=3×10−2
Final answer: 3×10−2
Q74Numericald- and f-Block Elements
Identify the metal ions among Co2+,Ni2+,Fe2+,V3+ and Ti2+ having a spin – only magnetic moment value more than 3.0 BM. The sum of unpaired electrons present in the high spin octahedral complexes formed by those metal ions is :
SolutionAnswer: 7
Approach:
Determine the d-electron count and high-spin unpaired electrons for each ion, keep only those with spin-only moment above 3.0 BM, then sum their unpaired electrons.
Step 1:Find the threshold: a moment above 3.0 BM requires more than 2 unpaired electrons.
n=2:8=2.83;n=3:15=3.87
Step 2:Assign d-configurations and high-spin unpaired counts.
Step 3:Select ions exceeding 3.0 BM: only Fe2+ (n=4) and Co2+ (n=3) qualify.
Fe2+:24=4.90;Co2+:15=3.87
Step 4:Sum the unpaired electrons of the qualifying ions.
4+3=7
Final answer: 7
Q75NumericalRedox Reactions and Electrochemistry
MX is a sparingly soluble salt that follows the given solubility equilibrium at 298 K. MX(s)⇌M+(aq)+X−(aq);Ksp=10−10. If the standard reduction potential for W=−∫r1r2F.dr is (EM+/M⊖)=0.79V, then the value of the standard reduction potential for the metal / metal insoluble salt electrode EX−/MX(s)/M⊖ is ______ mV. (nearest integer) (Given: F2.303RT=0.059V)
SolutionAnswer: 200
Approach:
Relate the metal/insoluble-salt electrode potential to the metal-ion electrode potential using the solubility product through the Nernst expression.
Step 1:Write the relation linking the two electrode potentials via Ksp.
EX−/MX(s)/M⊖=EM+/M⊖−0.059logKsp1
Step 2:Substitute EM+/M⊖=0.79 V and Ksp=10−10.
E⊖=0.79−0.059log(1010)=0.79−0.059×10
Step 3:Evaluate the expression.
E⊖=0.79−0.59=0.20V=200mV
Final answer: 200mV
Mathematics25 questions
Q1Single correctThree Dimensional Geometry
Let the line L1 be parallel to the vector −3i+2j+4k and pass through the point (2,6,7), and the line L2 be parallel to the vector 2i+j+3k and pass through the point (4,3,5). If the line L3 is parallel to the vector −3i+5j+16k and intersects the lines L1 and L2 at the points C and D, respectively, then ∣CD∣2 is equal to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4290
Approach:
Parametrize a general point on each line, set the direction of CD parallel to the given vector to get two linear equations, solve for the parameters, then compute the squared distance.
Step 1:A general point C on L1 with parameter t.
C=(2−3t,6+2t,7+4t)
Step 2:A general point D on L2 with parameter s.
D=(4+2s,3+s,5+3s)
Step 3:The direction of CD is D-C; set it proportional to (-3,5,16).
−32s+3t+2=5s−2t−3=163s−4t−2
Step 4:Solve the two linear equations by elimination.
If the area of the region {(x,y):1−2x≤y≤4−x2,x≥0,y≥0} is βα,α,β∈N, gcd (α,β)=1, then the value of (α+β) is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 173
Approach:
The region in the first quadrant is bounded above by the parabola y=4−x2 and below by the line y=1−2x where it is positive (otherwise by the x-axis); compute the area under the parabola from 0 to 2 and subtract the small triangle below the line near the origin.
Step 1:The parabola y=4−x2 meets the x-axis at x=2; the line y=1−2x meets the axes at (0,1) and (1/2,0).
y=4−x2,y=1−2x
Step 2:The area equals the area under the parabola from 0 to 2 minus the triangle with vertices (0,0),(0,1),(1/2,0) that lies below the line.
A=∫02(4−x2)dx−21⋅1⋅21
Step 3:Evaluate the definite integral.
∫02(4−x2)dx=[4x−3x3]02=8−38=316
Step 4:Subtract the triangular area 1/4.
A=316−41=1264−3=1261
Step 5:With gcd(61,12)=1, identify α=61, β=12 and add.
α+β=61+12
Final answer: α+β=73
Q3Single correctStatistics and Probability
A random variable X takes values 0,1,2,3 with probabilities 302a+1,308a−1,304a+1, b respectively, where a,b∈R. Let μ and σ respectively be the mean and standard deviation of X such that σ2+μ2=2. Then ba is equal to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 460
Approach:
Use the identity σ2+μ2=E(X2) together with the total-probability condition to form two linear equations in a and b, solve them, then compute a/b.
Step 1:Compute E(X2) using x=0,1,2,3 and set it equal to 2.
0+308a−1+304(4a+1)+9b=2
Step 2:Apply the total-probability condition.
302a+1+308a−1+304a+1+b=1
Step 3:Solve the linear system (i) and (ii) by elimination.
3×(ii)−(i):34a−68=0⇒a=2;14(2)+30b−29=0⇒b=301
Step 4:Compute the required ratio.
ba=1/302=60
Final answer: ba=60
Q4Single correctSequences and Series
The positive integer n, for which the solutions of the equation x(x+2)+(x+2)(x+4)+...+(x+2n−2)(x+2n)=38n are two consecutive even integers, is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 13
Approach:
Sum the products term by term using standard summation formulas to reduce the equation to a quadratic in x, then choose n so that the quadratic has two consecutive even integer roots.
Step 1:Expand each term and sum over r=1..n.
∑r=1n(x+2r−2)(x+2r)=nx2+2x∑(2r−1)+∑(2r−2)(2r)
Step 2:Use the summation identities for the coefficient sums.
nx2+2n2x+34n(n2−1)=38n
Step 3:Divide through by n and multiply by 3 to simplify.
3x2+6nx+4(n2−1)=8⇒3x2+6nx+4n2−12=0
Step 4:For n=3 the quadratic becomes 3x2+18x+24=0, i.e. x2+6x+8=0.
x2+6x+8=0⇒(x+2)(x+4)=0
Step 5:Check that the roots are two consecutive even integers.
x=−2,−4
Final answer: n=3
Q5Single correctDifferential Calculus
Let f(x)=x3+x2f′(1)+2xf′′(2)+f′′′(3),x∈R. Then the value of f′(5) is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 25117
Approach:
Set the unknown constants a=f′(1), b=f′′(2), c=f′′′(3), write f as a cubic with these coefficients, differentiate, evaluate at the stated points to get equations, solve for a, b, c, then compute f′(5).
Step 1:Write f(x)=x3+ax2+2bx+c with a=f′(1), b=f′′(2), c=f′′′(3); then f′(x)=3x2+2ax+2b and f′′(x)=6x+2a.
f′(x)=3x2+2ax+2b,f′′(x)=6x+2a
Step 2:Impose f′(1)=a.
3+2a+2b=a⇒a+2b=−3
Step 3:Impose f′′(2)=b.
12+2a=b⇒2a−b=−12
Step 4:Since f′′′(x)=6, the constant c=f′′′(3)=6.
f′′′(x)=6,c=6
Step 5:Solve (1) and (2) by substitution.
from (2): b=2a+12;a+2(2a+12)=−3⇒5a=−27⇒a=−527,b=56
Step 6:Compute f′(5)=75+10a+2b.
f′(5)=75+10(−527)+2⋅56=75−54+512=21+512
Final answer: f′(5)=5117
Q6Single correctDifferential Equations
Let y=y(x) be the solution of the differential equation secxdxdy−2y=2+3sinx,x∈(−2π,2π), y(0)=−47. Then y(6π) is equal to :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2−25
Approach:
Multiply by cosx to get the standard linear form, find the integrating factor, integrate the right side via the substitution sinx=t, apply the initial condition to fix the constant, then evaluate at x=π/6.
Step 1:Multiply through by cosx to obtain the standard linear form.
dxdy−2cosx⋅y=cosx(2+3sinx)
Step 2:Compute the integrating factor.
I.F.=e∫−2cosxdx=e−2sinx
Step 3:Multiply and integrate the right side using sinx=t, cosxdx=dt.
ye−2sinx=∫e−2t(2+3t)dt
Step 4:Carry out the integration by parts to obtain the general solution.
y=(−22+3sinx−43)+Ce2sinx=−23sinx−47+Ce2sinx
Step 5:Apply y(0)=−7/4 to determine C.
−47=−47+C⇒C=0
Step 6:Evaluate at x=π/6 where sin(π/6)=1/2.
y(6π)=−23⋅21−47=−43−47=−410
Final answer: y(6π)=−25
Q7Single correctCoordinate Geometry
If the line αx+4y=7, where α∈R, touches the ellipse 3x2+4y2=1 at the point P in the first quadrant, then one of the focal distances of P is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 231+271
Approach:
Use the tangency condition c2=a2m2+b2 to find α, locate the point of contact P, compute the eccentricity, then apply the focal-distance formula.
Step 1:Write the ellipse as x2/(1/3)+y2/(1/4)=1 so a2=1/3, b2=1/4; the line has slope m=−α/4 and intercept c=7/4.
167=31⋅16α2+41
Step 2:Compare the tangent 3xx1+4yy1=1 with αx+4y=7 (α=3) to get the point of contact.
71(3x+4y)=1⇒3x1=73,4y1=74⇒x1=y1=71
Step 3:Compute the eccentricity of the ellipse.
e=1−a2b2=1−43=21
Step 4:Apply the focal-distance formula PS′=a+ex1 with a=31, e=21, x1=71 (focus on the negative side).
PS′=a+ex1=31+21⋅71=31+271
Final answer: One focal distance =31+271
Q8Single correctMatrices and Determinants
For the matrices A=[31−4−1] and B=[−29−134918], if (A15+B)[xy]=[00], then among the following which one is true ?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2x=11,y=2
Approach:
Apply the Cayley-Hamilton theorem to reduce A15 to a linear expression in A and I, form A15+B, and solve the resulting homogeneous system for the relation between x and y.
Step 1:A has trace 2 and determinant 1, giving the characteristic relation.
A2−2A+I=0⇒A2=2A−I
Step 2:Inductively every power has the form An=nA−(n−1)I, so A15=15A−14I.
A15=15A−14I=[3115−60−29]
Step 3:Add B to A15.
A15+B=[22−11−11]
Step 4:The homogeneous system reduces to a single independent equation.
2x−11y=0⇒2x=11y
Step 5:Test the options against 2x=11y.
x=11,y=2⇒2(11)=11(2)=22
Final answer: x=11,y=2
Q9Single correctMatrices and Determinants
If the system of equations 3x+y+4z=32x+αy−z=−3x+2y+z=4 has no solution, then the value of α is equal to :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 419
Approach:
A system has no unique solution when the coefficient determinant vanishes; set the determinant to zero and solve for α (the resulting value makes the system inconsistent).
Step 1:Form the coefficient determinant.
Δ=3211α24−11
Step 2:Expand along the first row.
Δ=3(α+2)−1(2+1)+4(4−α)
Step 3:Set the determinant equal to zero.
19−α=0
Final answer: α=19
Q10Single correctVector Algebra
For a triangle ABC, let p=BC,q=CA and r=BA. If ∣p∣=23,∣q∣=2 and cosθ=31, where θ is the angle between p and q, then ∣p×(q−3r)∣2+3∣r∣2 is equal to :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4200
Approach:
Express r in terms of p and q using the triangle relation, compute ∣r∣2 by the law of cosines, simplify the cross-product term using r=q+p, then evaluate the full expression.
Step 1:From the triangle, BA=BC+CA, so r=p+q.
r=q+p
Step 2:Compute ∣r∣2 using cosθ=1/3.
∣r∣2=4+12+2⋅23⋅2⋅31=16+8
Step 3:Since q−3r=q−3(q+p)=−2q−3p, the cross product is p×(−2q−3p)=−2(p×q).
p×(q−3r)=−2(p×q)
Step 4:Use sin2θ=1−1/3=2/3 and evaluate the two terms.
4∣p∣2∣q∣2sin2θ+3∣r∣2=4⋅12⋅4⋅32+3⋅24
Step 5:Add the contributions.
128+72=200
Final answer: ∣p×(q−3r)∣2+3∣r∣2=200
Q11Single correctSequences and Series
Let a1,2a2,22a3,...,29a10 be a G.P. of common ratio 21. If a1+a2+...+a10=62, then a1 is equal to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 42(2−1)
Approach:
Deduce the common ratio of the sequence a1, a2, ... from the given G.P. condition, then apply the finite G.P. sum formula and solve for a1.
Step 1:The terms ak/2(k−1) have ratio 1/2, so the ratio of successive ak is determined.
ak/2k−1ak+1/2k=21⇒akak+1=22=2
Step 2:Apply the sum formula with r=2, n=10.
a12−1(2)10−1=62
Step 3:Since (2)10=32, solve for a1.
a1⋅2−131=62⇒a1=3162(2−1)=2(2−1)
Final answer: a1=2(2−1)
Q12Single correctThree Dimensional Geometry
Let the line L pass through the point (−3,5,2) and make equal angles with the positive coordinate axes. If the distance of L from the point (−2,r,1) is 314, then the sum of all possible values of r is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 210
Approach:
A line making equal angles with the positive axes has direction (1,1,1); write its equation, find the foot of perpendicular from the external point, set the squared distance equal to 14/3 to obtain a quadratic in r, then sum the roots.
Step 1:Write the line with direction (1,1,1) through (-3,5,2).
1x+3=1y−5=1z−2=λ
Step 2:Require PR perpendicular to (1,1,1) with P=(-2,r,1).
(λ−1)+(λ+5−r)+(λ+1)=0⇒3λ+5−r=0
Step 3:Substitute λ and set PR2=14/3.
9(r−8)2+9(10−2r)2+9(r−2)2=314
Step 4:Reduce the quadratic.
r2−10r+21=0⇒(r−3)(r−7)=0
Step 5:Add the possible values of r.
3+7=10
Final answer: Sum of possible values of r =10
Q13Single correctQuadratic Equations
Let α and β be the roots of the equation x2+2ax+(3a+10)=0 such that α<1<β. Then the set of all possible values of a is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3(−∞,5−11)
Approach:
For a quadratic with positive leading coefficient, the number 1 lies strictly between the roots exactly when f(1) is negative; solve f(1)<0.
Step 1:The leading coefficient is positive, so the condition α<1<β requires f(1)<0.
f(1)=1+2a+(3a+10)<0
Step 2:Simplify the inequality.
5a+11<0
Step 3:Write the solution set.
a∈(−∞,5−11)
Final answer: a∈(−∞,5−11)
Q14Single correctPermutations and Combinations
The largest n∈N, for which 7n divides 101!, is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 316
Approach:
Apply Legendre's formula to count the exponent of the prime 7 in 101! by summing the floor of 101 divided by successive powers of 7.
Step 1:Count multiples of 7.
⌊7101⌋=14
Step 2:Count multiples of 49.
⌊49101⌋=2
Step 3:Since 343>101 the next term is 0; sum the contributions.
14+2+0
Final answer: n=16
Q15Single correctCoordinate Geometry
Let y2=12x be the parabola with its vertex at O. Let P be point on the parabola and A be a point on the x-axis such that ∠OPA=90∘. Then the locus of the centroid of such triangles OPA is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4y2−2x+8=0
Approach:
Parametrize P on the parabola, use the right angle at P (slopes of OP and PA multiply to -1) to find the x-coordinate of A, write the centroid coordinates, and eliminate the parameter.
Step 1:Since 4a=12 gives a=3, take P=(3t2,6t) on the parabola.
P=(3t2,6t)
Step 2:The right angle at P gives slope(OP)⋅slope(PA)=−1 with A=(xA,0).
3t26t⋅3t2−xA6t=−1
Step 3:Compute the centroid of O(0,0), P(3t2,6t), A(12+3t2,0).
h=33t2+12+3t2=4+2t2,k=36t=2t
Step 4:Eliminate t using t=k/2.
h=4+2(2k)2=4+2k2⇒k2=2h−8
Step 5:Replace h by x and k by y to write the locus.
y2−2x+8=0
Final answer: y2−2x+8=0
Q16Single correctSets, Relations and Functions
Let A={2,3,5,7,9}. Let R be the relation on A defined by xRy if and only if 2x≤3y. Let l be the number of elements in R, and m be the minimum number of elements required to be added in R to make it a symmetric relation. Then l+m is equal to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 325
Approach:
List the ordered pairs (x,y) with 2x≤3y to count l, then count the pairs whose mirror image is missing to find m (the additions needed for symmetry), and add the two counts.
Step 1:For each x, count the y with 2x≤3y.
x=2:5,x=3:5,x=5:3,x=7:3,x=9:2
Step 2:Total number of pairs l.
l=5+5+3+3+2=18
Step 3:Identify pairs (x,y) in R whose reverse (y,x) is absent; these must be added.
{(5,2),(7,2),(9,2),(5,3),(7,3),(9,3),(9,5)}
Step 4:Add l and m.
18+7=25
Final answer: l+m=25
Q17Single correctCoordinate Geometry
Let one end of a focal chord of the parabola y2=16x be (16,16). If P(α,β) divides this focal chord internally in the ratio 5:2, then the minimum value of α+β is equal to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 17
Approach:
Find the parameter of the given endpoint, use the focal-chord property t1t2=−1 to find the other endpoint, apply the section formula in ratio 5:2 in both orderings, and take the minimum of α+β.
Step 1:For y2=16x, a=4; the point (16,16)=(4t12,8t1) gives t1=2, so t2=−1/2.
t1t2=−1,t1=2⇒t2=−21
Step 2:Compute the other endpoint B.
B=(4t22,8t2)=(1,−4)
Step 3:Apply the section formula in 5:2 for both possible orderings of A=(16,16) and B=(1,−4).
P=(737,712) or (782,772)
Step 4:Compute α+β for each and take the minimum.
737+712=749=7,782+772=22
Final answer: Minimum value of α+β=7
Q18Single correctDifferential Calculus
Let f:R→R be a twice differentiable function such that f′′(x)>0 for all x∈R and f′(a−1)=0, where a is a real number. Let g(x)=f(tan2x−2tanx+a),0<x<2π. Consider the following two statements :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2Neither (I) nor (II) is True
Approach:
Rewrite the inner argument as a perfect square plus a−1, differentiate g by the chain rule, and use that f′′>0 makes f′ increasing (with f′(a−1)=0) to determine the sign of g′ on each interval.
Step 1:Complete the square in the argument.
tan2x−2tanx+a=(tanx−1)2+a−1
Step 2:Differentiate using the chain rule.
g′(x)=f′((tanx−1)2+a−1)⋅2(tanx−1)sec2x
Step 3:On (0,π/4) the argument exceeds a−1, so f′(arg)>f′(a−1)=0 (positive), while 2(tanx−1)<0 and sec2x>0.
g′(x)=(+)⋅(−)⋅(+)<0
Step 4:On (π/4,π/2) the argument again exceeds a−1, so f′(arg)>0, while 2(tanx−1)>0 and sec2x>0.
g′(x)=(+)⋅(+)⋅(+)>0
Step 5:Both statements are false.
(I) false, (II) false
Final answer: Neither (I) nor (II) is True
Q19Single correctComplex Numbers
Let Z be the complex number satisfying ∣z−5∣≤3 and having maximum positive principal argument . Then 345iz+165z−122 is equal to;
(A)
(B)
(C)
(D)
SolutionAnswer: Option 320
Approach:
The condition describes a disk centered at 5 with radius 3; the point of maximum positive argument is where the line from the origin is tangent to the boundary circle. Find that z and evaluate the modulus expression.
Step 1:Center 5, radius 3, distance from origin 5; the tangent gives sinθ=3/5, cosθ=4/5.
arg(z)=sin−153=tan−143
Step 2:Tangent length =25−9=4, so z=4(cosθ+isinθ).
z=4(54+i53)=516+512i
Step 3:Substitute into the expression and simplify.
5iz+165z−12=4+16i4+12i=1+4i1+3i
Step 4:Take the modulus squared.
1+4i1+3i2=1+161+9=1710
Step 5:Multiply by 34.
34×1710=20
Final answer: 345iz+165z−122=20
Q20Single correctSets, Relations and Functions
Let A={x:∣x2−10∣≤6} and B={x:∣x−2∣>1}. Then
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3B−A=(−∞,−4)∪(−2,1)∪(4,∞)
Approach:
Solve each absolute-value inequality to write A and B explicitly, then evaluate the set operations and compare each with its option.
Step 1:Solve ∣x2−10∣≤6, i.e. 4≤x2≤16.
A=[−4,−2]∪[2,4]
Step 2:Solve ∣x−2∣>1, i.e. x<1 or x>3.
B=(−∞,1)∪(3,∞)
Step 3:Remove the parts of A from B to obtain B-A.
B−A=(−∞,−4)∪(−2,1)∪(4,∞)
Step 4:The other options fail: A∩B=[−4,−2]∪(3,4], A−B=[2,3], and A∪B=(−∞,1)∪[2,∞), none of which match their stated forms.
A∩B=[−4,−2]∪(3,4],A−B=[2,3],A∪B=(−∞,1)∪[2,∞)
Final answer: B−A=(−∞,−4)∪(−2,1)∪(4,∞)
Q21NumericalCoordinate Geometry
If P is a point on the circle x2+y2=4, Q is a point on the straight line 5x+y+2=0 and x−y+1=0 is the perpendicular bisector of PQ, then 13 times the sum of abscissa of all such points P is ..........
SolutionAnswer: 2
Approach:
Use that x−y+1=0 is the perpendicular bisector of PQ: the midpoint of PQ lies on it and PQ has slope -1. Combine with Q on the given line and P on the circle to express P's abscissa in terms of its ordinate, solve, and sum.
Step 1:The midpoint of P(x1,y1) and Q(x2,y2) lies on the bisector.
x1+x2−y1−y2+2=0
Step 2:PQ is perpendicular to the bisector, so its slope is -1.
y2=x1−x2+y1
Step 3:Q lies on 5x+y+2=0; combining with (i) and (ii) eliminates Q and gives x1 in terms of y1.
x1=2−5y1
Step 4:Impose P on the circle x12+y12=4 with x1=2−5y1.
(2−5y1)2+y12=4⇒26y12−20y1=0
Step 5:Compute the abscissae and their sum.
x1=2 or x1=−1324;2−1324=132
Step 6:Multiply the sum by 13.
13×132=2
Final answer: 13×(sum of abscissae)=2
Q22NumericalPermutations and Combinations
If (15C01+15C11)(15C11+15C21)⋯(15C121+15C131)=14C014C1⋯14C12α13, then 30α is equal to ........
SolutionAnswer: 32
Approach:
Simplify a single bracket using the reciprocal-sum identity for binomial coefficients, multiply all 13 brackets to obtain a clean product, then compare with the right side to read off α.
Step 1:Apply the identity with n=15 to a single bracket.
15Cr1+15Cr+11=1516⋅14Cr1
Step 2:Multiply all 13 brackets for r=0 to 12.
∏r=0121516⋅14Cr1=14C014C1⋯14C12(16/15)13
Step 3:Compare with the right-hand side to identify α.
α=1516
Step 4:Compute 30α.
30×1516=2×16=32
Final answer: 30α=32
Q23NumericalDifferential Calculus
Let [.] denote the greatest integer function and f(x)=limn→∞n31∑k=1n[3xk2]. Then 12∑j=1∞f(j) is equal to
SolutionAnswer: 2
Approach:
Bound the greatest-integer term so its fractional part vanishes after dividing by n3, reduce f(x) to a cubic-sum limit, then sum the resulting geometric series and multiply by 12.
Step 1:Each floor term differs from its argument by a value in [0,1); the total deviation is at most n and vanishes after dividing by n3.
f(x)=limn→∞n31⋅3x1∑k=1nk2
Step 2:Apply the sum of squares and take the limit.
f(x)=limn→∞n31⋅6⋅3xn(n+1)(2n+1)=6⋅3x2=3x+11
Step 3:Sum f(j) over j=1 to infinity (first term 1/9, ratio 1/3).
∑j=1∞3j+11=1−1/31/9=61
Step 4:Multiply by 12.
12×61=2
Final answer: 12∑j=1∞f(j)=2
Q24NumericalIntegral Calculus
If ∫014cot−1(1−2x+4x2)dx=atan−1(2)−bloge(5), where a.b∈N, then (2a+b) is equal to...
SolutionAnswer: 9
Approach:
Convert cot inverse to tan inverse, split the argument as a difference of two arctangents, apply the king property to symmetrize the integral, then evaluate the remaining arctan integral by parts.
Step 1:Write cot inverse as tan inverse of the reciprocal and split using the subtraction identity.
Step 2:Replace x by 1-x in the second part using the king property; tan−1(2(1−x)−1)=tan−1(1−2x)=−tan−1(2x−1).
I=4∫01tan−12xdx+4∫01tan−1(2x−1)dx
Step 3:Add (i) and (ii); the tan−1(2x−1) terms cancel.
2I=8∫01tan−12xdx⇒I=4∫01tan−12xdx
Step 4:Evaluate by parts: integral of arctan(2x) from 0 to 1 equals tan−12−(1/4)log5.
I=4(tan−12−41log5)=4tan−12−log5
Step 5:Compute 2a+b.
2(4)+1=9
Final answer: 2a+b=9
Q25NumericalTrigonometry
Let the maximum value of (sin−1x)2+(cos−1x)2 for x∈[−23,21] be nmπ2, where gcd(m,n)=1. Then m+n is equal to ..........
SolutionAnswer: 65
Approach:
Let t=sin inverse x and use sin inverse + cos inverse = π/2 to write the sum of squares as a quadratic in t; complete the square, find the range of t over the given x-interval, and maximize.
Step 1:As x ranges over [−3/2,1/2], t=sin inverse x ranges over [−π/3,π/4].
−3π≤t≤4π
Step 2:Shift to center the square term at π/4.
−127π≤t−4π≤0
Step 3:Square the bound; the maximum square comes from the endpoint −7π/12.
0≤(t−4π)2≤14449π2
Step 4:Insert into the quadratic; the maximum occurs at the largest square.
How many questions are in the JEE Main 2026 January 21, Shift 2 paper?
The JEE Main 2026 January 21, 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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