JEE Main 2026 April 05, Shift 2 Question Paper with Solutions
All 74 questions from the JEE Main 2026 (April 05, Shift 2) shift — Physics (25), Chemistry (25) and Mathematics (24) — 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.
Match List - I with List - II. where h (Planck's constant), G (gravitational constant) and c (speed of light in vacuum) as fundamental units. Choose the correct answer from the options given below:
List-I
List-II
A. Meter (L)
I.Ghc
B. Second (S)
II.c5Gh
C. Kilogram (M)
III.GhK2L2c3
D. Kelvin (K)
IV.c3Gh
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2A-IV, B-II, C-I, D-III
Approach:
Given fundamental units h (Planck constant), G (gravitational constant), c (speed of light) plus Boltzmann constant K. Target: identify the dimensional combination representing length (Meter), time (Second), mass (Kilogram) and temperature (Kelvin). Principle: dimensional analysis using Planck units.
Step 1:Dimensional check for length using G,h,c gives Gh/c3 since [Gh/c3]=[m2]. Hence Meter (A) corresponds to List-II entry IV.
L=c3Gh
Step 2:For time, divide Planck length by c, equivalent to Gh/c5 since [Gh/c5]=[s2]. Therefore Second (B) corresponds to entry II.
T=c5Gh
Step 3:For mass, the combination hc/G has dimensions of [kg2]. Therefore Kilogram (C) corresponds to entry I.
M=Ghc
Step 4:Temperature requires Boltzmann constant K. Treating L as the unit of length, the combination K2L2c3/(Gh) reduces to [K2] after substituting L2=Gh/c3, giving the Planck temperature squared. Therefore Kelvin (D) corresponds to entry III.
Θ=GhK2L2c3
Final answer: A-IV,B-II,C-I,D-III
Q27Single correctUnits and Measurements
In an experiment to determine the resistance of a given wire using Ohm's law, the voltmeter and ammeter readings are noted as 10 V and 5 A, respectively. The least counts of voltmeter and ammeter are 500 mV and 200 mA, respectively. The estimated error in the resistance measurement is _____ Ω.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 40.18
Approach:
Given voltmeter reading V=10V, ammeter reading I=5A, least counts ΔV=0.5V and ΔI=0.2A. Target: estimated absolute error ΔR. Principle: error propagation for the quotient R=V/I.
Step 1:Compute the measured resistance.
R=5A10V=2Ω
Step 2:Compute the relative errors from the least counts.
VΔV=100.5=0.05,IΔI=50.2=0.04
Step 3:Add the fractional errors.
RΔR=0.05+0.04=0.09
Step 4:Multiply by R to obtain the absolute error.
ΔR=0.09×2Ω=0.18Ω
Final answer: 0.18Ω
Q28Single correctWork, Energy and Power
A mass of 1 kg is kept on an inclined plane with 30∘ inclination with respect to the horizontal plane and it is at rest initially. Then the whole assembly is moved up with constant velocity of 4 m/s. The work done by the frictional force in time 2 s is _____ J. (Take g=10m/s2)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 120
Approach:
Given mass m=1kg, incline angle θ=30∘, vertical assembly speed v=4m/s, time t=2s, g=10m/s2, block at rest on the incline. Target: work done by friction on the block. Principle: static friction equals gravity component along incline; work is dot product with the displacement vector.
Step 1:Since the block is at rest on the incline while the assembly moves uniformly, the net force on the block is zero. Friction up the incline balances the gravity component mgsinθ down the incline.
f=mgsin30∘=1×10×0.5=5N
Step 2:Compute the displacement of the block; the assembly moves vertically upward at constant velocity.
d=vt=4m/s×2s=8m vertically
Step 3:Friction on the block points up along the incline, which itself makes θ=30∘ with the horizontal; therefore friction makes 30∘ with the horizontal. The block's displacement is purely vertical (upward) because the assembly is translated vertically at 4m/s. The angle ϕ between the friction vector and the vertical displacement equals 90∘−θ=60∘.
ϕ=90∘−30∘=60∘⇒cosϕ=21
Step 4:Apply the work formula.
Wf=fdcosϕ=5×8×0.5=20J
Final answer: 20J
Q29Single correctKinematics
The velocity (v) versus time (t) plot of a particle is shown in the figure, for a time interval of 40 s. The total distance travelled by the particle and the average velocity during this period are, respectively _____.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3100 m and zero
Approach:
Given a velocity-time plot with a triangular positive pulse rising from 0 to +5m/s at t=10s and back to zero at t=20s, then a symmetric negative pulse reaching −5m/s at t=30s and returning to zero at t=40s. Target: total distance and average velocity over 40s. Principle: distance equals area between ∣v∣ and the time axis; displacement equals signed area.
Step 1:Compute the area of the positive triangular pulse spanning 0 to 20s with peak height +5m/s.
A+=21×20s×5m/s=50m
Step 2:Compute the magnitude of the negative triangular pulse spanning 20 to 40s with depth −5m/s.
∣A−∣=21×20s×5m/s=50m
Step 3:Total distance is the sum of magnitudes; displacement is the signed sum.
s=A++∣A−∣=100m,Δx=A+−∣A−∣=0
Step 4:Average velocity from displacement and total time.
vˉ=40s0=0m/s
Final answer: 100mand0m/s
Q30Single correctRotational Motion
A wheel initially at rest is subjected to a uniform angular acceleration about its axis. In the first 2 s it rotates through an angle θ1 and in the next 2 s it rotates through an angle θ2. The ratio θ1θ2 is _____.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 23
Approach:
Given a wheel starting from rest under uniform angular acceleration α, rotating through θ1 in the first 2s and θ2 in the next 2s. Target: ratio θ2/θ1. Principle: rotational kinematics with constant angular acceleration from rest.
Step 1:Angle traversed in the first 2s.
θ1=21α(2)2=2α
Step 2:Total angle traversed in 4s.
θtot=21α(4)2=8α
Step 3:Angle in the next 2s obtained by subtraction.
θ2=θtot−θ1=8α−2α=6α
Step 4:Ratio of the two angles.
θ1θ2=2α6α=3
Final answer: 3
Q31Single correctRotational Motion
An object of uniform density rolls up the curved path with the initial velocity v0 as shown in the figure. If the maximum height attained by the object is 10g7v02 (g = acceleration due to gravity), the object is a _____.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4solid sphere
Approach:
Given an object rolling without slipping up a curved track with initial speed v0, reaching maximum height h=7v02/(10g). Target: identify the object via its moment-of-inertia coefficient k=I/(mR2). Principle: energy conservation between the bottom (translational + rotational kinetic energy) and the top (gravitational potential energy), with the rolling constraint v0=Rω.
Step 1:Apply energy conservation from the bottom to the maximum height h with surface assumed smooth enough for energy conservation under rolling.
21mv02(1+k)=mgh
Step 2:Substitute the given height.
21v02(1+k)=g⋅10g7v02=107v02
Step 3:Solve for k.
1+k=57⇒k=52
Step 4:The coefficient k=2/5 identifies the object as a solid sphere.
I=52mR2⇒solid sphere
Final answer: solid sphere
Q32Single correctGravitation
A body of mass m is taken from the surface of earth to a height equal to twice the radius of earth (Re). The increase in potential energy will be _____. (g is acceleration due to gravity at the surface of earth)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 432mgRe
Approach:
Given a body of mass m raised from Earth's surface (r=Re) to a height 2Re above the surface (so r=3Re). Target: increase in gravitational potential energy expressed in terms of g at the surface. Principle: exact gravitational potential energy formula combined with g=GM/Re2.
Step 1:Identify initial and final distances from Earth's centre.
ri=Re,rf=Re+2Re=3Re
Step 2:Compute the change in potential energy.
ΔU=−3ReGMm−(−ReGMm)=ReGMm(1−31)=3Re2GMm
Step 3:Substitute GM=gRe2 to eliminate GM.
ΔU=3Re2(gRe2)m=32mgRe
Final answer: 32mgRe
Q33Single correctProperties of Solids and Liquids
Eight mercury drops, each of radius r, coalesce to form a bigger drop. The surface energy released in this process is _____ (S is the surface tension of mercury).
(A)
(B)
(C)
(D)
SolutionAnswer: Option 216πr2S
Approach:
Given 8 mercury drops each of radius r coalescing into one larger drop, surface tension S. Target: surface energy released. Principle: volume conservation determines the radius of the merged drop; the released energy equals S times the decrease in total surface area.
Step 1:Equate total volume before and after merging.
8⋅34πr3=34πR3⇒R3=8r3⇒R=2r
Step 2:Compute initial total surface area of the 8 small drops.
Ai=8⋅4πr2=32πr2
Step 3:Compute final surface area of the merged drop.
Af=4π(2r)2=16πr2
Step 4:Released surface energy equals surface tension times the decrease in area.
ΔE=S(Ai−Af)=S(32πr2−16πr2)=16πr2S
Final answer: 16πr2S
Q34Single correctKinetic Theory of Gases
An ideal gas at pressure P and temperature T is expanding such that PT3= constant. The coefficient of volume expansion of the gas is _____.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3T4
Approach:
Given an ideal gas obeying PV=nRT undergoing a process with PT3=constant. Target: coefficient of volume expansion γ=(1/V)(dV/dT). Principle: eliminate P between the two relations to express V as a function of T alone, then differentiate.
Step 1:From the process constraint, express P as a function of T.
PT3=k⇒P=T3k
Step 2:Substitute into the ideal gas law to express V as a function of T.
V=PnRT=knRT⋅T3=knRT4
Step 3:Differentiate V with respect to T.
dTdV=k4nRT3
Step 4:Compute the coefficient of volume expansion.
γ=V1dTdV=(nR/k)T4(4nR/k)T3=T4
Final answer: T4
Q35Single correctOscillations and Waves
Match List - I with List - II.
List - I
List - II
A.sin2(ωt)
I. Periodic with time period T=ωπ but not simple harmonic motion (SHM)
B.sin3(2ωt)
II. Periodic with time period T=ω2π but Not SHM
C.sin(ωt)+cos(πωt)
III. Periodic with time period T=ωπ and SHM
D.cos(ωt)+cos(2ωt)
IV. Non-periodic
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1A-III, B-I, C-IV, D-II
Approach:
Givens: four time functions in List-I. Target: identify periodicity and SHM nature, then pair with List-II descriptors. Principle: a function is SHM when its second derivative is proportional to a sinusoidal function about a fixed mean with a single angular frequency; a sum of two harmonics is periodic only if the ratio of their periods is rational.
Step 1:Reduce A using the power-reduction identity to expose its harmonic content.
sin2(ωt)=21−21cos(2ωt)
Step 2:Apply the relevant identity to B using the triple-angle identity; the result is a superposition of two SHMs.
sin3(2ωt)=43sin(2ωt)−sin(6ωt)
Step 3:Examine C using the rational-ratio test.
Periods: T1=ω2π for sin(ωt), T2=πω2π=ω2 for cos(πωt); T2T1=π∈/Q.
Step 4:Examine D using LCM of periods.
Periods: ω2π and ωπ; LCM =ω2π; superposition of two SHMs of different angular frequencies is periodic but not SHM.
Final answer: A-III, B-I, C-IV, D-II
Q36Single correctElectromagnetic Induction and Alternating Currents
A metal rod of length L rotates about one end at origin with a uniform angular velocity ω. The magnetic field radially falls off as B(r)=B0e−λr; λ being a positive constant. The emf induced (neglecting the centripetal force on electrons in the rod) is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1B0ω[λ21−e−λL(λ21+λL)]
Approach:
Givens: rod of length L rotating with angular velocity ω about one end at origin in a radial magnetic field B(r)=B0e−λr. Target: emf across the rod. Principle: motional emf dε=B(r)vdr with v=ωr integrated along the rod.
Step 1:Set up the elemental emf at radial position r with element speed v=ωr.
dε=B0e−λr(ωr)dr
Step 2:Integrate from r=0 to r=L using integration by parts.
Under steady state condition the potential difference across the capacitor in the circuit is _____ V.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 10.5
Approach:
Givens: 2 V battery in series with 6 Ω; across the source terminals, two parallel branches: a 2 Ω resistor and a series combination of 4 Ω and 2 μF. Target: steady-state voltage across the capacitor. Principle: in DC steady state the capacitor draws zero current, so the branch containing it carries no current and the 4 Ω resistor has zero drop.
Step 1:Apply the steady-state condition to the capacitor branch.
IC=0⇒I4Ω=0, so the entire current from the battery flows through the 6 Ω source resistor and the 2 Ω parallel resistor.
Step 2:Compute the loop current using Kirchhoff's voltage law on the active loop (2 V source, 6 Ω, 2 Ω).
I=6Ω+2Ω2V=0.25A
Step 3:Determine the capacitor voltage. The capacitor sits in parallel with the 2 Ω resistor through the (zero-drop) 4 Ω resistor, so VC equals the drop across the 2 Ω resistor.
VC=IR2Ω=0.25A×2Ω=0.5V
Final answer: 0.5V
Q38Single correctMagnetic Effects of Current and Magnetism
A particle of charge q and mass m is projected from origin with an initial velocity v=(2v0x^+2v0y^). There exists a uniform magnetic field B=B0z^ and a space varying electric field E=E0e−λxx^ within the region 0≤x≤L. After travelling a distance such that x-coordinate has changed from x=0 to x=L, the change in the kinetic energy is _____.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1λqE0(1−e−λL)
Approach:
Givens: charge q and mass m, initial velocity v=(v0/2)x^+(v0/2)y^, uniform B=B0z^, position-dependent E=E0e−λxx^ confined to 0≤x≤L. Target: change in kinetic energy between x=0 and x=L. Principle: by the work-energy theorem only the net work matters; the magnetic force is perpendicular to v and contributes zero work.
Step 1:Identify zero work from the magnetic force since q(v×B)⊥v at every instant.
WB=0
Step 2:Express the electric work along the x-direction; the y^ component of displacement does not couple to E=E0e−λxx^.
WE=∫x=0x=LqE0e−λxdx
Step 3:Evaluate the integral.
WE=qE0[−λe−λx]0L=λqE0(1−e−λL)
Final answer: ΔKE=λqE0(1−e−λL)
Q39Single correctElectromagnetic Waves
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): The electromagnetic wave exerts pressure on the surface on which they are allowed to fall. Reason (R): There is no mass associated with the electromagnetic waves. In the light of the above statements, choose the correct answer from the options given below:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2Both (A) and (R) are true but (R) is not the correct explanation of (A)
Approach:
Givens: Assertion that EM waves exert pressure on absorbing/reflecting surfaces; Reason that EM waves carry no mass. Target: evaluate truth values and the causal link. Principle: radiation pressure arises from momentum flux p=E/c carried by EM waves, independent of any rest-mass consideration.
Step 1:Evaluate Assertion (A).
EM waves transport linear momentum density S/c2 (with S the Poynting vector); striking a surface, this momentum is transferred at rate Prad=I/c for absorption (and 2I/c for perfect reflection).
Step 2:Evaluate Reason (R).
Photons, the quanta of EM radiation, have zero rest mass: m0,γ=0. Thus no rest mass is associated with EM waves.
Step 3:Examine the causal link.
Radiation pressure is produced by momentum transfer p=E/c, which exists for massless quanta because relativistic momentum requires only nonzero energy. Hence the absence of mass is not the cause of pressure; pressure follows from energy/momentum flux.
Final answer: Both (A) and (R) are true but (R) is not the correct explanation of (A)
Q40Single correctOptics
A thin convex lens and a thin concave lens are kept in contact and are co-axial. Which of the following statements is correct for this combination of two lenses?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1behaves as concave lens if ∣fconvex∣>∣fconcave∣
Approach:
Givens: thin convex lens (focal length +∣fconvex∣) in contact and coaxial with thin concave lens (focal length −∣fconcave∣). Target: identify the correct statement about the combination. Principle: thin lenses in contact combine through power addition; the sign of 1/f determines whether the system is converging or diverging.
Step 1:Substitute the signed focal lengths with the standard sign convention.
Step 2:Analyse the sign of 1/f under the condition ∣fconvex∣>∣fconcave∣.
Since ∣fconvex∣>∣fconcave∣ implies ∣fconvex∣1<∣fconcave∣1, the difference f1<0.
Step 3:Inspect the remaining option about interchanging positions.
The combination formula f1=f11+f21 is symmetric in f1↔f2, so interchanging the lenses leaves f unchanged.
Final answer: Behaves as a concave (diverging) lens when ∣fconvex∣>∣fconcave∣.
Q41Single correctOptics
An object AB is placed 15 cm on the left of a convex lens P of focal length 10 cm. Another convex lens Q is now placed 15 cm right of lens P. If the focal length of lens Q is 15 cm, the final image is _____.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2real, formed at 7.5 cm right of lens Q, with a size same as that of AB
Approach:
Givens: object AB at 15 cm left of convex lens P (fP=+10 cm); convex lens Q (fQ=+15 cm) at 15 cm right of P. Target: location, nature, and size of the final image. Principle: apply the thin-lens equation successively, treating the image from P as the object for Q with appropriate sign of object distance. Sign convention used throughout: light travels left to right; distances measured from the lens are positive to the right and negative to the left; a real object lies on the incident side (u<0), while a real image and a virtual object lie on the transmitted side (v,u>0).
Step 1:Compute the image formed by lens P with uP=−15 cm (real object on incident side) and fP=+10 cm (converging lens).
Step 2:Locate I1 with respect to lens Q, which sits 15 cm to the right of P. Because 30>15, I1 lies on the transmitted side of Q — the converging rays from P would have met I1 past Q. Q intercepts the rays before they converge, so I1 acts as a virtual object whose distance from Q is taken as positive.
Distance of I1 from Q =30−15=15cm on the transmitted side ⇒uQ=+15cm.
Step 3:Apply the lens equation for Q with fQ=+15 cm.
Final answer: Real image, 7.5 cm to the right of lens Q, with size equal to that of AB.
Q42Single correctOptics
The maximum intensity in a Young's double slit experiment is I0. Distance between the slits (d) is 5λ, where λ is the wavelength of light used. The intensity of the fringe, exactly opposite to one of the slits on the screen, placed at D=10d is _____.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 22I0
Approach:
Givens: YDSE with maximum intensity I0, slit separation d=5λ, screen distance D=10d=50λ. Target: intensity at the point directly opposite one of the slits, at y=d/2 from the central axis. Principle: small-angle path difference Δx=yd/D converts to phase difference ϕ=(2π/λ)Δx; intensity follows I=I0cos2(ϕ/2).
Step 1:Place the screen point opposite one slit at vertical distance y=d/2 from the central axis.
Δx=D(d/2)d=2Dd2
Step 2:Substitute d=5λ and D=10d=50λ.
Δx=2(50λ)(5λ)2=100λ25λ2=4λ
Step 3:Convert the path difference to phase and evaluate the intensity.
ϕ=λ2π⋅4λ=2π; I=I0cos2(4π)=I0⋅21
Final answer: I=2I0
Q43Single correctDual Nature of Matter and Radiation
An electron is travelling with a velocity v in free space and when it enters a medium, its velocity is reduced by 20%. The de Broglie wavelength of electron in the medium is αλ0, where λ0 is its de Broglie wavelength in free space. The value of α is _____ .
(A)
(B)
(C)
(D)
SolutionAnswer: Option 31.25
Approach:
Given: free-space electron speed v with de Broglie wavelength λ0; in the medium the speed is reduced by 20% so v′=0.8v. Target: the ratio α=λ/λ0 where λ is the wavelength in the medium. Principle: de Broglie relation λ=h/(mv) with non-relativistic electron mass.
Step 1:Express the free-space wavelength using the de Broglie relation.
λ0=mvh
Step 2:Express the wavelength in the medium with the reduced speed v′=0.8v.
λ=m(0.8v)h
Step 3:Form the ratio λ/λ0 to identify α.
λ0λ=h/(mv)h/(0.8mv)=0.81
Final answer: 1.25
Q44Single correctAtoms and Nuclei
Assuming the experimental mass of 612C as 12 u, the mass defect of 612C atom is _____ MeV/c2. (Mass of proton =1.00727u, mass of neutron =1.00866u, 1u=931.5MeV/c2 and c is the speed of the light in vacuum).
(A)
(B)
(C)
(D)
SolutionAnswer: Option 289.03
Approach:
Given: experimental atomic mass of 612C is 12u; mp=1.00727u; mn=1.00866u; conversion 1u=931.5MeV/c2. Target: mass defect Δm of 612C expressed in MeV/c2. Principle: mass defect equals total mass of constituent nucleons minus the nuclear (here, atomic) mass.
Step 1:Sum the masses of six protons and six neutrons.
6×1.00727+6×1.00866
Step 2:Subtract the experimental atomic mass to obtain the mass defect in atomic mass units.
Δm=12.09558−12=0.09558u
Step 3:Convert the mass defect to MeV/c2 using 1u=931.5MeV/c2.
0.09558×931.5
Final answer: 89.03MeV/c2
Q45Single correctElectronic Devices
In a semiconductor p-n diode, the doping concentrations on p-side and n-side are 1015atoms/cm3 and 1018atoms/cm3, respectively. Which one of the following statements is true?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2The depletion region width is more on p -side compared to that in n-side
Approach:
Given: doping concentrations NA=1015cm−3 on the p-side and ND=1018cm−3 on the n-side. Target: identify the correct statement about the relative widths of the depletion region. Principle: in an abrupt p-n junction the total ionized charge on each side of the metallurgical interface must be equal in magnitude.
Step 1:Apply charge neutrality of the depletion region: the negative acceptor charge on the p-side equals the positive donor charge on the n-side per unit junction area.
NAWp=NDWn
Step 2:Substitute the given concentrations to compute the ratio of depletion widths.
WnWp=NAND=10151018
Step 3:Interpret the result: the depletion width extends much further into the lightly doped p-region than into the heavily doped n-region.
Wp≫Wn
Final answer: The depletion region width is more on the p-side compared to that in n-side
Q46NumericalProperties of Solids and Liquids
A copper wire of length 3 m is stretched by 3 mm by applying an external force. The volume of the wire is 600×10−6m3. The elastic potential energy stored in the wire in stretched condition would be _____ J. (Given Young modulus of copper =1.1×1011N/m2)
SolutionAnswer: 33
Approach:
Given: copper wire of original length L=3m, extension ΔL=3mm=3×10−3m, volume V=600×10−6m3, Young's modulus Y=1.1×1011N/m2. Target: elastic potential energy U stored in the stretched wire (in joules). Principle: energy density of an elastically strained solid is 21Yε2 for longitudinal strain ε.
Step 1:Compute the longitudinal strain produced in the wire.
ε=33×10−3=10−3
Step 2:Form the elastic energy density using Young's modulus and the strain.
u=21(1.1×1011)(10−3)2=21(1.1×1011)(10−6)
Step 3:Multiply the energy density by the volume of the wire to obtain the total stored elastic energy.
U=(5.5×104)(600×10−6)=5.5×6×100
Final answer: 33J
Q47NumericalProperties of Solids and Liquids
The heat extracted out of x gram of water initially at 50∘C to cool it down to 0∘C is sufficient to evaporate (1000−x) gram of water also initially at 50∘C. The value of x (closest integer) is _____ . (Take latent heat of water 2256kJ/kg.K, specific heat capacity of water 4200J/kg.K)
SolutionAnswer: 922
Approach:
Given: xg of water at 50∘C is cooled to 0∘C; the heat removed is supplied to (1000−x)g of water initially at 50∘C to heat it to 100∘C and then vaporise it completely; specific heat capacity c=4200J/(kg⋅K); latent heat of vaporisation L=2256kJ/kg. Target: closest integer value of x. Principle: conservation of energy between heat lost by the cooling water and heat absorbed by the heating–then–vaporising water.
Step 1:Express the heat extracted from the cooling mass (working with mass in grams, c=4.2J/(g⋅K)).
Qcool=x×4.2×50=210xJ
Step 2:Express the heat required to raise the remaining (1000−x)g from 50∘C to 100∘C and then vaporise it. The latent heat per gram is L=2256J/g.
Qabs=(1000−x)[4.2×50+2256]=(1000−x)(2466)J
Step 3:Equate the two heats and solve for x.
210x=2466(1000−x)⇒210x+2466x=2.466×106
Step 4:Divide to obtain the value of x and round to the nearest integer.
x=26762.466×106≈921.52
Final answer: 922g
Q48NumericalElectromagnetic Induction and Alternating Currents
A series LCR circuit with R=20Ω, L=1.6H and C=40μF is connected to a variable frequency a.c. source. The inductive reactance at resonant frequency is _____ Ω.
SolutionAnswer: 200
Approach:
Given: series LCR circuit with R=20Ω, L=1.6H, C=40μF=40×10−6F, driven at variable angular frequency ω. Target: inductive reactance XL at the resonant frequency. Principle: at series resonance XL=XC, with ω0=1/LC, leading to XL=L/C.
Step 1:Combine XL=ω0L with ω0=1/LC to remove the explicit frequency.
XL=LCL=CL
Step 2:Substitute the numerical values of L and C.
CL=40×10−61.6=4×104
Step 3:Take the square root to obtain the reactance at resonance.
XL=4×104=2×102Ω
Final answer: 200Ω
Q49NumericalCurrent Electricity
When an external resistance of 5Ω is connected across terminals of a cell, a current of 0.25A flows through it. When the 5Ω resistor is replaced by a 2Ω resistor, a current of 0.5A flows through it. The internal resistance of the cell is _____ Ω.
SolutionAnswer: 1
Approach:
Given: an external resistance R1=5Ω produces a current I1=0.25A; replacing it with R2=2Ω yields I2=0.5A. Target: internal resistance r of the cell. Principle: a cell of EMF ε and internal resistance r obeys ε=I(R+r) for any external load R.
Step 1:Write the EMF equation for each load using the same EMF ε.
ε=0.25(5+r)andε=0.5(2+r)
Step 2:Equate the two expressions for ε and expand.
0.25(5+r)=0.5(2+r)⇒1.25+0.25r=1.0+0.5r
Step 3:Solve the linear equation for the internal resistance.
0.25=0.25r
Final answer: 1Ω
Q50NumericalElectromagnetic Induction and Alternating Currents
A circular loop of radius 20 cm and resistance 2Ω is placed in a time varying magnetic field B=(2t2+2t+3)T. At t=0, for the plane of the loop being perpendicular to the magnetic field, the induced current in the loop at t=3s is 50αA. The value of α is _____ . (Take π=22/7)
SolutionAnswer: 44
Approach:
Given: circular loop of radius r=20cm=0.20m and resistance R=2Ω placed perpendicular to a time-varying field B=2t2+2t+3T; the loop's plane is normal to B, so the field is along the area vector. Induced current at t=3s is expressed as α/50A; π=22/7. Target: integer α. Principle: Faraday's law of electromagnetic induction with Φ=BA.
Step 1:Compute the loop area and differentiate the field with respect to time.
A=πr2=π(0.20)2=0.04πm2;dtdB=4t+2
Step 2:Evaluate dB/dt at t=3s and compute the magnitude of the induced EMF.
dtdBt=3=4(3)+2=14T/s;∣ε∣=AdtdB=0.04π×14
Step 3:Apply Ohm's law to obtain the induced current.
I=R∣ε∣=20.56π=0.28πA
Step 4:Substitute π=22/7 and match with the given form I=α/50A.
I=0.28×722=76.16=0.88A=5044A
Final answer: α=44
Chemistry25 questions
Q51Single correctSome Basic Concepts in Chemistry
What volume of hydrogen gas at STP would be liberated by action of 50 mL of H2SO4 of 50% purity (density = 1.3 g.mL−1) on 20 g of zinc? Given: Molar mass of H, O, S, Zn are 1, 16, 32, 65 g mol−1 respectively.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 36.892 L
Approach:
Given 50 mL of H2SO4 solution of density 1.3 g/mL and 50% purity reacting with 20 g of Zn. The principle is stoichiometric limiting reagent analysis for Zn+H2SO4→ZnSO4+H2; required is the volume of H2 liberated at STP.
Step 1:Compute total mass of acid solution and then mass of pure H2SO4 available.
msoln=50×1.3=65g;mH2SO4=0.5×65=32.5g
Step 2:Convert reactant masses to moles using molar masses: M(H2SO4)=98, M(Zn)=65.
nH2SO4=9832.5=0.3316;nZn=6520=0.3077
Step 3:Identify the limiting reagent. The 1:1 stoichiometry of Zn+H2SO4→ZnSO4+H2 makes Zn limiting since 0.3077<0.3316. Therefore moles of H2 produced equal moles of Zn consumed.
nH2=nZn=0.3077
Step 4:Convert moles of H2 to volume at STP using 22.4L/mol.
V=0.3077×22.4=6.892L
Final answer: 6.892L
Q52Single correctAtomic Structure
Which of the following statement(s) is/are true? A. If two orbitals have the same value of (n + l), then the orbital with lower value of n will have lower energy. B. Energies of the orbitals in the same subshell increase with increase in atomic number. C. The size of 2pz orbital is less than the size of 3pz orbital. D. Among 5f, 6s, 4d, 5p and 5d, no one of the orbitals have 2 radial nodes. Choose the correct answer from the options given below:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2A and C only
Approach:
Each of statements A through D concerns orbital energies, sizes, or radial nodes. Each is to be evaluated using Madelung's (n+l) rule, hydrogenic size scaling, screening effects on subshell energy, and the radial-node count n−l−1.
Step 1:Evaluate A. For two orbitals with the same (n+l), Madelung rule states the orbital with smaller n has lower energy (e.g., 2p with n+l=3 lies below 3s). Therefore A is TRUE.
A: TRUE
Step 2:Evaluate B. Increase in atomic number Z enhances effective nuclear charge on each subshell, lowering (making more negative) the orbital energy rather than raising it. Therefore B is FALSE.
Enl∝−Zeff2⇒E decreases as Z rises
Step 3:Evaluate C. Mean radial size of a hydrogenic orbital scales with n2; hence the 3pz orbital is larger than the 2pz orbital. Therefore C is TRUE.
⟨r⟩∝n2⇒r(3p)>r(2p)
Step 4:Evaluate D. Apply n−l−1 to each orbital: 5f has 5−3−1=1 node; 6s has 6−0−1=5 nodes; 4d has 4−2−1=1 node; 5p has 5−1−1=3 nodes; 5d has 5−2−1=2 nodes. Since 5d possesses exactly 2 radial nodes, the claim that none has 2 nodes is FALSE.
5d radial nodes=5−2−1=2
Step 5:Combine the truth values: only A and C are true.
True set={A,C}
Final answer: A and C only
Q53Single correctChemical Bonding and Molecular Structure
The covalent radii of atoms A and B are rA and rB respectively. The covalent bond length and total length of AB molecule are respectively:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1(rA+rB),2(rA+rB)
Approach:
Given a diatomic molecule AB with covalent radii rA and rB. Bond length between bonded atoms is the sum of covalent radii. Total molecular length spans from the outer edge of atom A to the outer edge of atom B, requiring inclusion of both radii on the ends in addition to the internal bond.
Step 1:Identify the covalent bond length. For an A-B covalent bond, the internuclear distance equals the sum of the covalent radii of the two atoms.
dA−B=rA+rB
Step 2:Construct the total length of the AB molecule. Starting at the far edge of atom A, traverse radius rA to reach nucleus A, then bond length (rA+rB) to reach nucleus B, then radius rB to reach the far edge of atom B.
L=rA+(rA+rB)+rB=2rA+2rB
Step 3:Pair the bond length with the total length to match the option format.
(dA−B,L)=((rA+rB),2(rA+rB))
Final answer: (rA+rB),2(rA+rB)
Q54Single correctChemical Thermodynamics
Consider the following data for the reaction X2(g)+Y2(g)⇌2XY(g) at 600 K. The ΔrG∘ (in kJ mol−1) for the reaction is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2−10
Approach:
Given table values ΔfH∘ (kJ/mol) and Sm∘ (J/mol K) for XY, X2, Y2 at T=600K and reaction X2(g)+Y2(g)⇌2XY(g). The principle: compute ΔrH∘, ΔrS∘, then ΔrG∘=ΔrH∘−TΔrS∘. Given ΔrG∘=−X kJ/mol, find X.
Step 1:Compute ΔrH∘ using product XY (coefficient 2) and reactants X2, Y2 (each coefficient 1).
ΔrH∘=2(42)−[8+60]=84−68=16kJ/mol
Step 2:Compute ΔrS∘ analogously.
ΔrS∘=2(200)−[140+250]=400−390=10J/(molK)
Step 3:Apply Gibbs equation at T=600 K, converting entropy to kJ/(mol K) for consistency.
ΔrG∘=16−600×0.010=16−6=10kJ/mol
Step 4:Equate ΔrG∘=−XkJ/mol from the problem statement to solve for X.
−X=10⇒X=−10
Final answer: −10
Q55Single correctChemical Thermodynamics
The correct order of molar heat capacities measured at 298 K and 1 bar is:
Required: rank molar heat capacities of Cu(s), Br2(l) and He(g) at 298 K and 1 bar. Principle: monoatomic ideal gas has only 3 translational degrees of freedom giving Cp,m≈25R≈20.8J/(mol K); a metallic solid follows Dulong-Petit giving ≈3R≈25J/(mol K); a diatomic liquid with strong intermolecular interactions and additional rotational/vibrational contributions stores far more energy per Kelvin per mole.
Step 1:Estimate Cp,m for He(g) using monoatomic ideal-gas relation.
Cp,m(He)≈25(8.314)≈20.8J/(mol K)
Step 2:Estimate Cp,m for Cu(s) using Dulong-Petit law at room temperature (Cu sits close to the classical limit).
Cp,m(Cu)≈3R≈24.5J/(mol K)(tabulated≈24.4)
Step 3:Estimate Cp,m for Br2(l). The liquid stores translational, rotational, librational and vibrational energy plus the energy required to weaken intermolecular interactions; tabulated value is ≈75.7 J/(mol K), far above Cu and He.
Cp,m(Br2,l)≈75.7J/(mol K)
Step 4:Rank the three values in descending order.
75.7>24.4>20.8⇒Br2(l)>Cu(s)>He(g)
Final answer: Bromine(l) > Copper(s) > Helium(g)
Q56Single correctEquilibrium
The reaction A(g)⇌B(g)+C(g) was initiated with the amount 'a' of A(g). At equilibrium it is found that the amount of A(g) remaining is (a−x) at a total pressure of p. The equilibrium constant Kp of the reaction can be calculated from the expression:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2a2−x2x2⋅p
Approach:
Given A(g)⇌B(g)+C(g) with initial moles a of A, equilibrium moles a−x of A, x each of B and C, and total pressure p. Required: closed-form Kp in terms of a, x, p. Principle: use mole fractions to obtain partial pressures, then apply law of mass action.
Step 1:Sum equilibrium moles. With A: a−x, B: x, C: x the total is ntot=(a−x)+x+x=a+x.
ntot=a+x
Step 2:Write the partial pressure of each species via its mole fraction and total pressure p.
pA=a+xa−xp,pB=pC=a+xxp
Step 3:Insert into the mass-action expression for Kp.
Kp=pApBpC=a+xa−xp(a+xxp)2
Step 4:Simplify. Multiply numerator and denominator: (a−x)p/(a+x)x2p2/(a+x)2=(a+x)(a−x)x2p=a2−x2x2p.
Kp=a2−x2x2p
Final answer: a2−x2x2p
Q57Single correctRedox Reactions and Electrochemistry
One half cell in a voltaic cell is constructed from a silver rod dipped in silver nitrate solution of unknown concentration. The other half cell consists of a zinc rod dipped in 1 molar solution of ZnSO4. A voltage of 1.60 V is measured at 298 K for this cell. What is the concentration of Ag+ ions used in terms of logx, where x=[Ag+]? Given: EZn2+/Zn∘=−0.76 V,EAg+/Ag∘=+0.80 V,F2.303RT=0.059 V.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 25.94
Approach:
Given a galvanic cell Zn|Zn2+(1 M) ‖ Ag+(x M)|Ag at 298 K with Ecell=1.60 V, EZn2+/Zn∘=−0.76 V, EAg+/Ag∘=+0.80 V, F2.303RT=0.059 V. Principle: apply Nernst equation to the spontaneous net reaction Zn+2Ag+→Zn2++2Ag (n=2) and solve for logx.
Step 1:Compute Ecell∘ taking Ag+/Ag as cathode (reduction) and Zn2+/Zn as anode (oxidation).
Ecell∘=0.80−(−0.76)=1.56V
Step 2:Identify n=2 from the balanced net reaction and write the reaction quotient: Q=[Zn2+]/[Ag+]2=1/x2.
Q=x21,n=2
Step 3:Apply the Nernst equation with Ecell=1.60 V.
Given below are two statements. Statement I: The number of pairs among {Al2O3,Cr2O3}, {Cl2O7,Mn2O7}, {Na2O,V2O3} and {CO,N2O} that contain oxides of same nature (acidic, basic, neutral or amphoteric) is 4. Statement II: Among Na2O, Al2O3, CO and Cl2O7, the most basic and acidic oxides are Na2O and Cl2O7, respectively. 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 are true
Approach:
Statement I lists four pairs of oxides; for each pair, identify the acid-base character of both members and count how many pairs share a common nature. Statement II compares the strengths of Na2O, Al2O3, CO and Cl2O7 to verify that the most basic is Na2O and the most acidic is Cl2O7. Principles: nature of metal/non-metal oxides, periodic-table trends, and dependence on oxidation state.
Step 1:Pair 1 [Al2O3,Cr2O3]: Al2O3 dissolves in both acids and bases (amphoteric); Cr2O3 likewise reacts with acids and with fused alkali (amphoteric). Both amphoteric; same nature.
Al2O3,Cr2O3:amphoteric
Step 2:Pair 2 [Cl2O7,Mn2O7]: Cl2O7 is the anhydride of HClO4 (strongly acidic); Mn2O7 is the anhydride of HMnO4 (strongly acidic). Same nature.
Cl2O7,Mn2O7:acidic
Step 3:Pair 3 [Na2O,V2O3]: Na2O is a typical alkali-metal oxide (basic). V2O3 contains V in its low oxidation state +3, where the oxide is basic (lower oxidation states of d-block oxides are basic; higher oxidation states such as V2O5 are amphoteric/acidic). Same nature.
Na2O,V2O3:basic
Step 4:Pair 4 [CO,N2O]: CO is a classic neutral oxide; N2O is likewise a neutral oxide. Same nature.
CO,N2O:neutral
Step 5:Count: all four pairs share the same character within the pair. Therefore the count is 4 and Statement I is TRUE.
count=4
Step 6:Evaluate Statement II among Na2O,Al2O3,CO,Cl2O7: Na2O is the most basic (group-1 oxide); Cl2O7 is the most acidic (highest oxidation state of Cl, anhydride of HClO4). Statement II is TRUE.
Most basic: Na2O;Most acidic: Cl2O7
Step 7:Both statements true; select the corresponding option.
I true∧II true
Final answer: Both Statement I and Statement II are true
Q59Single correctp-Block Elements
Given below are two statements: Statement I: Aluminium upon reaction with NaOH forms [Al(OH)6]3− ion. Statement II: The geometry of ICl4−, ClO3− and IBr2− is square planar, pyramidal and linear respectively. 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 but Statement II is true
Approach:
Evaluate Statement I by recalling the product of the Al/NaOH reaction and Statement II by applying VSEPR to ICl4-, ClO3- and IBr2-.
Step 1:Aluminium with aqueous sodium hydroxide produces sodium tetrahydroxoaluminate, not the hexahydroxo species.
2Al+2NaOH+6H2O⟶2Na[Al(OH)4]+3H2
Step 2:VSEPR for ICl4−: central I has 7 valence electrons plus 1 from the charge, minus 4 used in σ-bonds to Cl, giving 4 non-bonding electrons (2 lone pairs).
AX4E2,SN=6⇒square planar
Step 3:VSEPR for ClO3− and IBr2−.
ClO3−:AX3E⇒pyramidal;IBr2−:AX2E3⇒linear
Step 4:Combine evaluations of the two statements.
I: false,II: true
Final answer: Statement I is false but Statement II is true
Q60Single correctd- and f-Block Elements
Given below are two statements: Statement I: Presence of large number of unpaired electrons in transition metal atoms results in higher enthalpies of their atomisation. Statement II: dxy=dxz=dyz<dx2−y2=dz2 and dx2−y2=dz2<dxy=dxz=dyz are the d-orbital splittings in [Fe(H2O)6]3+ and [Ni(CO)4]2− complex ions respectively. 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 are true
Approach:
Link atomisation enthalpy to unpaired d-electron count, then identify the geometry of each complex and apply the corresponding crystal-field splitting pattern.
Step 1:Atomisation enthalpy in transition metals scales with metal-metal bonding from unpaired d-electrons.
ΔHatom∝nunpaired d-electrons
Step 2:[Fe(H2O)6]3+ is six-coordinate octahedral, so the octahedral CFT pattern applies.
dxy=dxz=dyz<dx2−y2=dz2
Step 3:[Ni(CO)4] is four-coordinate tetrahedral, so the tetrahedral CFT pattern applies.
dx2−y2=dz2<dxy=dxz=dyz
Step 4:Combine both evaluations.
I: true,II: true
Final answer: Both Statement I and Statement II are true
Q61Single correctCoordination Compounds
Identify the correct statements from the following: A. [Fe(C2O4)3]3− is the most stable complex among [Fe(OH)6]3−, [Fe(C2O4)3]3− and [Fe(SCN)6]3−. B. The stability of [Cu(NH3)4]2+ is greater than that of [Cu(en)2]2+. C. The hybridization of Fe in K4[Fe(CN)6] is d2sp3. D. [Fe(NO2)3Cl3]3− exhibits linkage isomerism. E. NO2− and SCN− ligands are NOT ambidentate ligands. Choose the correct answer from the options given below:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3A, C and D only
Approach:
Test each statement against the chelate effect, hybridization of low-spin Fe(II) hexacyano complex, linkage isomerism criteria and ambidentate behaviour of NO2−/SCN−.
Step 1:Statement A: among [Fe(OH)6]3−, [Fe(C2O4)3]3− and [Fe(SCN)6]3−, the oxalato complex uses a chelating bidentate ligand and is therefore the most stable.
C2O42−bidentate chelate⇒highest stability
Step 2:Statement B: ethylenediamine is bidentate while ammonia is monodentate, so the en complex is more stable than the ammine complex.
[Cu(en)2]2+>[Cu(NH3)4]2+
Step 3:Statement C: in K4[Fe(CN)6] the metal centre is Fe2+ (d6), and CN− is a strong-field ligand giving a low-spin octahedral complex.
d6low spin, octahedral⇒d2sp3
Step 4:Statement D: NO2− is ambidentate (binds through N as nitro or O as nitrito), so a complex containing NO2− ligands exhibits linkage isomerism.
[Fe(NO2)3Cl3]3−⇒nitro/nitrito linkage isomers
Step 5:Statement E: NO2− (nitro/nitrito) and SCN− (thiocyanato-S/isothiocyanato-N) are textbook ambidentate ligands; the assertion that they are NOT ambidentate is false.
NO2−,SCN−are ambidentate
Step 6:Compile the correct statements.
Correct: A, C, D
Final answer: A, C and D only
Q62Single correctPurification and Characterisation of Organic Compounds
Match List - I with List - II.
List - I (Purification technique)
List - II (Used to separate)
A. Simple distillation
I. Steam volatile compound
B. Fractional distillation
II. Two liquids with large difference in boiling points
C. Steam distillation
III. Liquid decomposing at its boiling point
D. Distillation under reduced pressure
IV. Two liquids with close boiling points
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2A-II, B-IV, C-I, D-III
Approach:
Match each distillation technique to the separation scenario it is designed to address.
Step 1:Simple distillation is suitable when the components boil at temperatures differing by more than about 25∘C.
ΔTbp>25∘C
Step 2:Fractional distillation uses a fractionating column for liquids whose boiling points are close.
ΔTbpsmall⇒fractional column
Step 3:Steam distillation co-distils volatile, water-immiscible compounds with steam below their normal boiling points.
ptotal=pwater+pcompound=1atm
Step 4:Reduced-pressure distillation lowers the boiling point so heat-sensitive liquids distil before decomposing.
p↓⇒Tbp↓<Tdecomp
Step 5:Assemble the matching.
A-II,B-IV,C-I,D-III
Final answer: A-II, B-IV, C-I, D-III
Q63Single correctHydrocarbons
IUPAC name of the some alkenes are given below. Find out the correct stability order. A. 2-Methylbut-2-ene B. cis-But-2-ene C. 2,3-Dimethylbut-2-ene D. Prop-1-ene Choose the correct answer from the options given below:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1C>A>B>D
Approach:
Classify each alkene by the number of alkyl substituents on the doubly bonded carbons; greater substitution gives greater hyperconjugative stabilization.
Step 1:Determine the substitution pattern at the C=C for each alkene.
SolutionAnswer: Option 4Statement I is false but Statement II is true
Approach:
Generate the true semicarbazone connectivity from acetaldehyde and semicarbazide for Statement I, and apply the acid-catalysed hydrolysis of a hemiacetal for Statement II.
Step 1:The terminal -NH2 of semicarbazide is the nucleophile; it attacks the carbonyl carbon of acetaldehyde and water is eliminated to give the semicarbazone.
Step 2:Compare the cited structure CH3-CH=N-C(H)(=O)-N-NH2 with the genuine semicarbazone: the carbon bearing the C=O in semicarbazide sits between two nitrogens, whereas the cited formula places a CH=O group bonded only to one nitrogen.
Given below are two statements: Statement I: Heating benzamide with bromine in an ethanolic solution of sodium hydroxide will give benzylamine. Statement II: Nitration of aniline with HNO3/H2SO4 at 288 K produces m-nitroaniline in higher amount than o-nitroaniline (pH adjusted). 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 but Statement II is true
Approach:
Evaluate each statement on the basis of Hofmann bromamide degradation and electrophilic nitration of aniline at low temperature.
Step 1:Hofmann bromamide degradation of benzamide produces the amine with one carbon less than the starting amide, i.e., aniline, not benzylamine.
C6H5CONH2Br2/NaOH-EtOHC6H5NH2
Step 2:In HNO3/H2SO4 at 288 K, aniline is largely protonated to the anilinium ion, −NH3+, which is strongly deactivating and meta-directing.
C6H5NH2H+C6H5NH3+HNO3m-nitroanilinium
Step 3:Under these conditions m-nitroaniline (~47%) exceeds o-nitroaniline (~2%), confirming Statement II.
%m-NO2>%o-NO2
Final answer: Statement I is false but Statement II is true
Q68Single correctBiomolecules
Identify the incorrect statement about tertiary structure of proteins.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3The structure remains intact when exposed to pH changes.
Approach:
Identify which statement contradicts the known properties of protein tertiary structure.
Step 1:Tertiary structure of proteins arises from interactions between R-groups: hydrogen bonds, disulphide (-S-S-) links, van der Waals and electrostatic attractions, supporting Option 2.
Step 2:Tertiary structures fall in fibrous and globular classes (Option 1) and arise from the folding of a linear polypeptide that already has secondary structure (Option 4).
Step 3:Therefore both molecules share three similar chiral carbons at C3, C4, C5, confirming Statement II.
C3≡C3,C4≡C4,C5≡C5
Final answer: Both Statement I and Statement II are true
Q70Single correctPrinciples Related to Practical Chemistry
A paper dipped in a dil. H2SO4 solution of 'X' upon treatment with SO2 gas turns into green. The compound 'X' is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4K2Cr2O7
Approach:
Identify the substance whose acidified solution undergoes a colour change from orange to green upon reduction by SO2.
Step 1:In acidic medium K2Cr2O7 exists as orange dichromate, Cr2O72−, with Cr in +6 oxidation state.
K2Cr2O7/H2SO4→orange solution of Cr(VI)
Step 2:SO2 acts as a reducing agent and reduces Cr(VI) to Cr(III), which is green in aqueous acidic solution.
Cr2O72−+3SO2+2H+→2Cr3++3SO42−+H2O
Step 3:Other options fail this specific test: KI-starch turns blue (with oxidisers, but SO2 does not), KMnO4 would decolourise (purple to colourless), and Pb(CH3COO)2 gives a white/black precipitate with H2S not green with SO2.
Only K2Cr2O7shows orange→green with SO2
Final answer: K2Cr2O7
Q71NumericalCoordination Compounds
The total number of unpaired electrons present in the d3,d4 (low spin), d5 (high spin), d6 (high spin) and d7 (low spin) octahedral complex systems is ____.
SolutionAnswer: 15
Approach:
Apply crystal field splitting in an octahedral field and count unpaired electrons for each specified configuration.
RMgI when treated with ice cold water liberated a gas which occupied 1.4dm3/g at STP. The gas produced is further reacted with iodine in presence of HIO3 to give compound (X). Compound (X) in presence of Na and dry ether produced compound (Y). Molar mass of compound (Y) is ____ g mol−1. (Nearest integer)
SolutionAnswer: 30
Approach:
Identify the alkane gas from its molar volume, then trace iodination by I2/HIO3 followed by Wurtz coupling with Na/dry ether to find compound (Y).
Step 1:Gas occupies 1.4dm3/g at STP. Using Vm=22.4dm3mol−1, molar mass of gas =22.4/1.4=16g mol−1, identifying it as CH4. Hence R = CH3 and the Grignard is CH3MgI.
Mgas=1.422.4=16g mol−1⇒CH4
Step 2:Methane upon iodination in presence of HIO3 (which re-oxidises HI back to I2, driving the equilibrium forward) gives methyl iodide as compound (X).
CH4+I2HIO3CH3I+HI;5HI+HIO3→3I2+3H2O
Step 3:Wurtz reaction of CH3I with Na in dry ether couples two methyl groups giving ethane as compound (Y), with molar mass 2(12)+6(1)=30g mol−1.
2CH3I+2Nadry etherCH3-CH3+2NaI;MY=30
Final answer: 30
Q73NumericalSolutions
20 g hemoglobin in a 1 L aqueous solution (A) at 300 K is separated from pure water by semi permeable membrane. At equilibrium the height of solution in a tube dipped in a solution (A) is found to be 80.0 mm higher than the tube dipped in water. The molar mass of hemoglobin is ____ kg mol−1. (Nearest integer) (Given: g=10m s−2, R=8.3kPa dm3K−1mol−1, density of solution =1000kg m−3)
SolutionAnswer: 62
Approach:
Compute osmotic pressure from the height of liquid column using π=ρgh, then apply the van't Hoff equation to obtain moles and hence molar mass of hemoglobin.
Step 1:Convert the height into osmotic pressure: h=80.0mm=0.080m, ρ=1000kg m−3, g=10m s−2.
π=ρgh=1000×10×0.080=800Pa=0.800kPa
Step 2:Apply van't Hoff equation with V=1dm3, R=8.3kPa dm3K−1mol−1, T=300K to obtain moles of hemoglobin.
n=RTπV=8.3×3000.800×1=3.213×10−4mol
Step 3:Divide given mass 20 g by moles to obtain molar mass and convert to kg/mol.
M=3.213×10−420=6.224×104g mol−1=62.24kg mol−1
Final answer: 62
Q74NumericalElectrochemistry
At 298 K, the molar conductivity of x%(w/w) MX solution (aqueous) is 123.5S cm2mol−1. The conductance of same solution is 1.9×10−3S. The value of x is ____ ×10−2. (Given: cell constant =1.3cm−1; molar mass of MX is 75g mol−1, density of aqueous solution of MX at 298 K is 1.0g mL−1)
SolutionAnswer: 15
Approach:
Determine conductivity from conductance and cell constant, derive molar concentration from molar conductivity, then convert to percentage (w/w).
Step 2:Apply Λm=κ×1000/C to find molarity of MX in mol/L.
C=Λmκ×1000=123.52.47×10−3×1000=0.02mol L−1
Step 3:Mass of MX per litre =0.02×75=1.5g. With density 1.0g mL−1, 1 L of solution weighs 1000 g, hence %(w/w)=1.5/1000×100=0.15=15×10−2.
%(w/w)=10001.5×100=0.15=15×10−2⇒x=15
Final answer: 15
Q75NumericalChemical Kinetics
For a reaction A→P at T\,K, the half life (t1/2) is plotted as a function of initial concentration [A]0 of A as given below.
SolutionAnswer: 90
Approach:
Use the order test t1/2∝[A]01−n. The plot is a straight line through the origin (linear increase of t1/2 with [A]0), which identifies the reaction as zero order (n=0); the corresponding half-life relation is t1/2=[A]0/(2k), hence the ratio t1/2/[A]0 is constant across the line. Use this constant to solve for x from the two highlighted points.
Step 1:Read the two highlighted data points from the plot: at [A]0=1.5×10−3mol L−1 the half-life is xs, and at [A]0=4×10−3mol L−1 the half-life is 240s.
Step 2:Identify the order. A straight line of t1/2 vs [A]0 passing through the origin is consistent only with 1−n=1, i.e., n=0 (zero order). For zero-order kinetics, t1/2/[A]0 is constant along the line.
t1/2∝[A]0⇒[A]0t1/2=2k1=const
Step 3:Equate the ratio t1/2/[A]0 at the two points and solve for x.
Q1Single correctComplex Numbers and Quadratic Equations
Let α,β be the roots of the equation x2−x+p=0 and γ,δ be the roots of the equation x2−4x+q=0. If α,β,γ,δ are in G.P., then ∣p+q∣ equals:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 334
Approach:
Apply Vieta relations to both quadratics, parametrise the four roots as a four-term G.P. with first term α and common ratio r, then solve the resulting system for r and α to recover p and q.
Step 1:Identify the given data: α,β are roots of x2−x+p=0 and γ,δ are roots of x2−4x+q=0, with α,β,γ,δ in G.P. The target is ∣p+q∣.
α+β=1,αβ=p,γ+δ=4,γδ=q
Step 2:Parametrise the four-term G.P. as α,αr,αr2,αr3. Therefore α+β=α(1+r) and γ+δ=αr2(1+r).
α(1+r)=1,αr2(1+r)=4
Step 3:Divide the second equation by the first to eliminate α(1+r).
α(1+r)αr2(1+r)=14⇒r2=4
Step 4:Branch r=2: α(3)=1⇒α=1/3, giving β=2/3, γ=4/3, δ=8/3. Then p=αβ=2/9 and q=γδ=32/9, so ∣p+q∣=34/9, a non-integer value.
∣p+q∣=934
Step 5:Branch r=−2: α(1+(−2))=−α=1⇒α=−1. The four terms become −1,2,−4,8, producing the integer-valued root products required by the problem.
α=−1,β=2,γ=−4,δ=8
Step 6:Compute p and q from the products of roots.
p=αβ=(−1)(2)=−2,q=γδ=(−4)(8)=−32
Step 7:Evaluate ∣p+q∣.
∣p+q∣=∣−2−32∣=∣−34∣=34
Final answer: 34
Q2Single correctComplex Numbers and Quadratic Equations
Let z1,z2∈C be the distinct solutions of the equation z2+4z−(1+12i)=0. Then ∣z1∣2+∣z2∣2 is equal to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 434
Approach:
Solve the complex quadratic z2+4z−(1+12i)=0 by the quadratic formula, extract the principal square root of the resulting complex discriminant, and then compute the sum of squared moduli of the two roots.
Step 1:Apply the quadratic formula with a=1, b=4, c=−(1+12i). The discriminant is b2−4ac=16+4(1+12i)=20+48i.
z=2−4±20+48i=−2±5+12i
Step 2:Set 5+12i=a+ib with a,b∈R. Squaring gives a2−b2=5 and 2ab=12, hence ab=6.
a2−b2=5,ab=6
Step 3:Substitute b=6/a into a2−b2=5: a2−36/a2=5. Multiplying by a2 gives a4−5a2−36=0, a quadratic in a2.
a4−5a2−36=0⇒a2=25±25+144=25±13
Step 4:Therefore a=3 and b=6/3=2, giving 5+12i=3+2i.
5+12i=3+2i
Step 5:Substitute back to obtain the two roots.
z1=−2+(3+2i)=1+2i,z2=−2−(3+2i)=−5−2i
Step 6:Compute the squared moduli.
∣z1∣2=12+22=5,∣z2∣2=(−5)2+(−2)2=29
Step 7:Add the two squared moduli.
∣z1∣2+∣z2∣2=5+29=34
Final answer: 34
Q4Single correctMatrices and Determinants
Let M be a 3×3 matrix such that M120=120, M010=011 and M001=012. If Mxyz=3111, then x+y+z equals:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 14
Approach:
Recover the columns of M from the three given image equations using Mei as the i-th column, assemble M, then solve the linear system M(x,y,z)T=(3,1,11)T and add the components.
Step 1:Read the second image equation: M(0,1,0)T=Me2=(0,1,1)T. Hence the second column of M is (0,1,1)T.
col2(M)=(0,1,1)T
Step 2:Read the third image equation: M(0,0,1)T=Me3=(0,1,2)T. Hence the third column of M is (0,1,2)T.
col3(M)=(0,1,2)T
Step 3:From the first image equation, M(1,2,0)T=Me1+2Me2=(1,2,0)T. Substituting Me2=(0,1,1)T gives Me1=(1,2,0)T−2(0,1,1)T=(1,0,−2)T.
Me1=(1,0,−2)T
Step 4:Assemble the matrix from its three columns.
M=10−2011012
Step 5:Form the system M(x,y,z)T=(3,1,11)T by reading off rows.
x=3,y+z=1,−2x+y+2z=11
Step 6:Substitute x=3 into the third equation: −6+y+2z=11⇒y+2z=17. Subtracting y+z=1 from y+2z=17 gives z=16, hence y=1−z=−15.
x=3,y=−15,z=16
Step 7:Add the components.
x+y+z=3+(−15)+16=4
Final answer: 4
Q5Single correctSequence and Series
If the sum of the first 10 terms of the series 1+14⋅41+1+24⋅42+1+34⋅43+1+44⋅44+⋯ is nm, gcd(m,n)=1, then m+n is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3276
Approach:
Factor the denominator 1+4k4 via the Sophie Germain identity, split the general term into a difference of consecutive reciprocals through partial fractions, and exploit telescoping to sum the first ten terms.
Step 1:The k-th term is Tk=k/(1+k4⋅4)=k/(1+4k4). Apply the Sophie Germain factorisation to the denominator.
Tk=(2k2−2k+1)(2k2+2k+1)k
Step 2:The difference of the two factors equals (2k2+2k+1)−(2k2−2k+1)=4k. Therefore 4k appears as the numerator difference, enabling partial fractions.
2k2−2k+11−2k2+2k+11=(2k2−2k+1)(2k2+2k+1)4k
Step 3:Divide by 4 to obtain Tk as a telescoping difference.
Tk=41[2k2−2k+11−2k2+2k+11]
Step 4:Identify ak=1/(2k2−2k+1). Then ak+1=1/(2(k+1)2−2(k+1)+1)=1/(2k2+2k+1), so Tk=41(ak−ak+1).
Step 6:Check gcd(55,221). Factorisations: 55=5⋅11 and 221=13⋅17. The two prime sets are disjoint, so the fraction is already in lowest terms.
gcd(55,221)=1⇒m=55,n=221
Step 7:Add numerator and denominator.
m+n=55+221=276
Final answer: 276
Q6Single correctSequence and Series
Let A1,A2,A3,…,A39 be 39 arithmetic means between the numbers 109 and 159. Then the mean of A2,A4,A6,…,A38 is equal to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4134
Approach:
Form the arithmetic progression of 41 terms from 109 to 159 with 39 inserted means, write the general inserted term, identify the sub-AP of even-indexed means, and apply the mean-of-AP formula.
Step 1:Inserting 39 arithmetic means between 109 and 159 produces an AP whose first term is 109, last term is 159, and which has 41 terms in total.
d=39+1159−109=4050=45=1.25
Step 2:Write the k-th inserted mean.
Ak=109+kd=109+45k
Step 3:The selection A2,A4,…,A38 is itself an arithmetic progression with first term A2, common difference 2d=5/2, and 19 terms (since the indices 2,4,…,38 form an AP of length (38−2)/2+1=19).
Number of terms=19,common difference=5/2
Step 4:Compute the first and last terms of the sub-AP.
Step 5:Apply the mean-of-AP formula: the mean equals the average of the first and last terms.
Mean=2A2+A38=2111.5+156.5=2268
Step 6:Simplify.
Mean=134
Final answer: 134
Q7Single correctBinomial Theorem and its Simple Applications
The coefficient of x2 in the expansion of (2x2+x1)10, x=0, is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 23360
Approach:
Write the binomial general term of (2x2+1/x)10, combine the exponents of x, solve for the index r that gives exponent 2, then evaluate the numerical coefficient.
Step 1:Identify n=10, a=2x2, b=1/x and write the general term.
Tr+1=(r10)(2x2)10−r(x1)r
Step 2:Separate constants and powers of x.
Tr+1=(r10)210−rx2(10−r)x−r=(r10)210−rx20−3r
Step 3:Set the exponent equal to 2 and solve for r.
20−3r=2⇒3r=18⇒r=6
Step 4:Substitute r=6 to compute the coefficient.
coeff=(610)210−6=(610)24
Step 5:Evaluate (610)=(410)=4!10⋅9⋅8⋅7=245040=210 and 24=16.
(610)=210,24=16
Step 6:Multiply to obtain the coefficient of x2.
210×16=3360
Final answer: 3360
Q8Single correctStatistics and Probability
The probabilities that players A and B of a team are selected for the captaincy for a tournament are 0.6 and 0.4, respectively. If A is selected the captain, the probability that the team wins the tournament is 0.8 and if B is selected the captain, the probability that the team wins the tournament is 0.7. Then the probability, that the team wins the tournament, is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 20.76
Approach:
Model the captaincy as a partition into two mutually exclusive events A and B with P(A)+P(B)=1, then apply the law of total probability to the winning event W.
Step 1:Set up the events. Let A be the event that player A is chosen captain and B that player B is chosen. The two events partition the sample space, with P(A)=0.6 and P(B)=0.4 summing to 1.
P(A)=0.6,P(B)=0.4,P(A)+P(B)=1
Step 2:Record the conditional probabilities of winning given each captain choice.
P(W∣A)=0.8,P(W∣B)=0.7
Step 3:Apply the law of total probability.
P(W)=P(A)P(W∣A)+P(B)P(W∣B)=(0.6)(0.8)+(0.4)(0.7)
Step 4:Compute each product separately.
(0.6)(0.8)=0.48,(0.4)(0.7)=0.28
Step 5:Sum the contributions.
P(W)=0.48+0.28=0.76
Final answer: 0.76
Q9Single correctPermutations and Combinations
A box contains 5 blue, 6 yellow and 4 red balls. The number of ways of drawing 8 balls containing at least two balls of each colour is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 14100
Approach:
Reduce the problem to counting non-negative integer solutions of B′+Y′+R′=2 (after subtracting the mandatory 2 from each colour) with appropriate upper bounds, enumerate the resulting (B,Y,R) triples, and sum the products of combinations.
Step 1:Set up constraints: choose B blue, Y yellow, R red with B+Y+R=8, B≥2, Y≥2, R≥2, and supply bounds B≤5, Y≤6, R≤4.
B+Y+R=8,2≤B≤5,2≤Y≤6,2≤R≤4
Step 2:Substitute B=B′+2,Y=Y′+2,R=R′+2 with B′,Y′,R′≥0. The equation becomes B′+Y′+R′=2 with bounds B′≤3,Y′≤4,R′≤2. All non-negative triples summing to 2 trivially satisfy the upper bounds.
B′+Y′+R′=2,B′,Y′,R′∈{0,1,2}
Step 3:Enumerate the six triples (B',Y',R') summing to 2 and convert back to (B,Y,R): (0,0,2)→(2,2,4), (0,1,1)→(2,3,3), (0,2,0)→(2,4,2), (1,0,1)→(3,2,3), (1,1,0)→(3,3,2), (2,0,0)→(4,2,2).
(2,2,4),(2,3,3),(2,4,2),(3,2,3),(3,3,2),(4,2,2)
Step 4:Evaluate the multiplication-principle product for each case. Case (2,2,4): (25)(26)(44)=10⋅15⋅1=150. Case (2,3,3): (25)(36)(34)=10⋅20⋅4=800. Case (2,4,2): (25)(46)(24)=10⋅15⋅6=900.
150,800,900
Step 5:Continue: case (3,2,3): (35)(26)(34)=10⋅15⋅4=600. Case (3,3,2): (35)(36)(24)=10⋅20⋅6=1200. Case (4,2,2): (45)(26)(24)=5⋅15⋅6=450.
600,1200,450
Step 6:Sum across all six cases.
150+800+900+600+1200+450=4100
Final answer: 4100
Q10Single correctStatistics and Probability
A variable X takes values 0,0,2,6,12,20,…,n(n−1) with frequencies nC0,nC1,nC2,nC3,…,nCn, respectively. If the mean of this data is 60, then its median is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 156
Approach:
Given variable X with values xk=k(k−1) and frequencies fk=(kn) for k=0,1,…,n and mean 60; target is the median. Apply binomial identities ∑(kn)=2n and ∑k(k−1)(kn)=n(n−1)2n−2 to solve for n, then locate the median from cumulative frequencies.
Step 1:Compute the total frequency using the binomial sum identity.
∑k=0n(kn)=2n
Step 2:Compute the weighted sum ∑xkfk where xk=k(k−1).
∑k=0nk(k−1)(kn)=n(n−1)2n−2
Step 3:Form the mean by dividing the weighted sum by the total frequency.
Xˉ=2nn(n−1)2n−2=4n(n−1)
Step 4:Equate the mean to 60 and solve the resulting quadratic for n.
4n(n−1)=60⇒n(n−1)=240
Step 5:With n=16, total frequency is 216=65536; the median position is at N/2=32768.
N/2=215=32768
Step 6:Build cumulative frequencies (016)+(116)+⋯ to locate where the cumulative count first reaches 32768.
Step 7:Add (816)=12870 to the cumulative count to cross 32768.
26333+12870=39203≥32768
Step 8:The values xk=k(k−1) are sorted in non-decreasing order in k (since xk+1−xk=2k≥0), so cumulative frequencies built in the order k=0,1,2,… correspond to sorted data. The median position N/2=32768 lies in the half-open block (26333,39203] belonging to k=8, hence the median value is x8.
x8=8(8−1)=56
Final answer: 56
Q11Single correctCo-ordinate Geometry
Let the point P be the vertex of the parabola y=x2−6x+12. If a line passing through the point P intersects the circle x2+y2−2x−4y+3=0 at the points R and S, then the maximum value of (PR+PS)2 is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 220
Approach:
Find vertex P of y=x2−6x+12 and centre/radius of the circle x2+y2−2x−4y+3=0. For a chord through P meeting the circle at R,S, express PR+PS in terms of the perpendicular distance from the centre to the chord, then maximise.
Step 1:Complete the square in y=x2−6x+12=(x−3)2+3, giving the vertex P=(3,3).
y=(x−3)2+3
Step 2:Rewrite the circle in standard form: (x−1)2+(y−2)2=1+4−3=2, centre C=(1,2), radius r=2.
(x−1)2+(y−2)2=2
Step 3:Compute PC2=(3−1)2+(3−2)2=4+1=5; since PC2>r2, P lies outside the circle.
PC2=5,r2=2
Step 4:Let M be the foot of perpendicular from C to chord RS and h=CM. Then PM2=PC2−h2=5−h2 and MR=MS=r2−h2=2−h2.
PM=5−h2,MR=2−h2
Step 5:Since PM2=5−h2>2−h2=MR2, P lies beyond M outside the segment RS, so PR=PM−MR and PS=PM+MR both positive.
PR+PS=2PM=25−h2
Step 6:Square the sum: (PR+PS)2=4(5−h2), which is maximised when h=0 (chord through the centre).
(PR+PS)2=20−4h2
Final answer: 20
Q12Single correctCo-ordinate Geometry
Let the directrix of the parabola P:y2=8x cut the x-axis at the point A. Let B(α,β), α>1, be a point on P such that the slope of AB is 53. If BC is a focal chord of P, then six times the area of △ABC is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2160
Approach:
Identify A from the directrix of y2=8x, use the slope condition to pin down B via the parametric form, find C from the focal-chord relation, then compute the triangle's area by coordinates.
Step 1:Compare y2=8x with y2=4ax to get a=2; the directrix is x=−2, so A=(−2,0).
4a=8⇒a=2,directrix x=−2
Step 2:Parameterise B=(2t12,4t1) with α=2t12, β=4t1. The slope condition gives α+2β=53.
2t12+24t1=53
Step 3:Cross-multiply and rearrange to a quadratic in t1.
10t1=3(t12+1)⇒3t12−10t1+3=0
Step 4:Solve: t1=610±100−36=610±8, giving t1=3 or t1=31.
Step 6:For the focal chord through B, the other endpoint C has parameter t2=−1/t1=−31.
C=(2t22,4t2)=(92,−34)
Step 7:Apply the area-by-coordinates formula with A=(−2,0), B=(18,12), C=(92,−34).
Δ=21(−2)(12−(−34))+18((−34)−0)+92(0−12)
Step 8:Simplify: −380−38=−388; total inside absolute value =−388−24=−3160.
Δ=21⋅3160=380
Step 9:Multiply by 6 as required by the question.
6Δ=6⋅380=160
Final answer: 160
Q13Single correctCo-ordinate Geometry
Let the eccentricity e of a hyperbola satisfy the equation 6e2−11e+3=0. If the foci of the hyperbola are (3,5) and (3,−4), then the length of its latus rectum is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3215
Approach:
Solve the quadratic for e, retain the root exceeding 1 for a hyperbola, deduce a from the focal distance, find b2 via b2=a2(e2−1), and compute the latus rectum a2b2.
Step 1:Solve 6e2−11e+3=0 by the quadratic formula.
e=1211±121−72=1211±7
Step 2:Since a hyperbola requires e>1, retain e=23.
e=23
Step 3:Distance between foci (3,5) and (3,−4) is ∣5−(−4)∣=9; hence 2ae=9.
2a⋅23=9⇒a=3
Step 4:Compute b2 via b2=a2(e2−1)=9(49−1).
b2=9⋅45=445
Step 5:Apply the latus rectum formula L=a2b2.
L=32⋅445=345/2=215
Final answer: 215
Q14Single correctThree Dimensional Geometry
Let a triangle PQR be such that P and Q lie on the line 8x+3=2y−4=2z+1 and are at a distance of 6 units from R(1,2,3). If (α,β,γ) is the centroid of △PQR, then α+β+γ is equal to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 36
Approach:
Find the foot of perpendicular M from R(1,2,3) to the given line. Since P,Q on the line are equidistant from M and at distance 6 from R, P+Q=2M. The centroid is (P+Q+R)/3=(2M+R)/3, from which α+β+γ follows.
Step 1:Read off a point and direction of the line 8x+3=2y−4=2z+1: A=(−3,4,−1), d=(8,2,2), ∣d∣2=64+4+4=72.
A=(−3,4,−1),d=(8,2,2)
Step 2:Compute AR=R−A=(4,−2,4) and AR⋅d=4⋅8+(−2)⋅2+4⋅2=32−4+8=36.
AR⋅d=36
Step 3:Foot of perpendicular M=A+21d=(−3+4,4+1,−1+1)=(1,5,0).
M=(1,5,0)
Step 4:Compute RM2=(1−1)2+(5−2)2+(0−3)2=0+9+9=18.
RM=18=32
Step 5:With RP=RQ=6 and PM=QM=36−18=18=32, P and Q are reflections through M along the line, so P+Q=2M=(2,10,0).
If the distance of the point (α,2,5) from the image of the point (1,2,7) in the line 1x=1y−1=2z−2 is 211, then the sum of all possible values of α is equal to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 36
Approach:
Compute the image of (1,2,7) about the line by finding its foot of perpendicular N and using P′=2N−P. Equate the distance from (α,2,5) to P' to 11/2 and sum the roots.
Step 1:Read A=(0,1,2) and d=(1,1,2) from the line 1x=1y−1=2z−2; ∣d∣2=1+1+4=6.
A=(0,1,2),d=(1,1,2),∣d∣2=6
Step 2:Compute P−A=(1,1,5) and (P−A)⋅d=1+1+10=12. Parameter t=12/6=2.
t=2
Step 3:Foot of perpendicular N=A+2d=(2,3,6).
N=(2,3,6)
Step 4:Image P′=2N−P=(4−1,6−2,12−7)=(3,4,5).
P′=(3,4,5)
Step 5:Distance squared from (α,2,5) to P': (α−3)2+(2−4)2+(5−5)2=(α−3)2+4.
d2=(α−3)2+4
Step 6:Set d2=11/2 and isolate (α−3)2.
(α−3)2=211−4=23
Step 7:The two roots are α=3±3/2; their sum is 6.
α1+α2=(3+3/2)+(3−3/2)=6
Final answer: 6
Q16Single correctSets, Relations and Functions
Let A={1,4,7} and B={2,3,8}. Then the number of elements, in the relation R={((a1,a2),(a1′,a2′))∈(A×B,A×B):a1≤a1′anda2≥a2′}, is _____.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 236
Approach:
The condition factorises: choose (a1,a1′)∈A×A with a1≤a1′ independently of (a2,a2′)∈B×B with a2≥a2′. Count ordered pairs from a 3-element set with each non-strict inequality and multiply.
Step 1:For A={1,4,7} (size 3), count ordered pairs (a1,a1′) with a1≤a1′: pairs with a1<a1′ are (23)=3 and pairs with a1=a1′ are 3.
NA=3+3=6
Step 2:For B={2,3,8} (size 3), count ordered pairs (a2,a2′) with a2≥a2′ by the same enumeration: (23)+3=6.
NB=3+3=6
Step 3:The two conditions are independent (one constrains the first components, the other the second), so the total count multiplies.
∣R∣=NA⋅NB=6⋅6
Final answer: 36
Q17Single correctLimit, Continuity and Differentiability
Let f(x)=y→0limy3(1−cos(xy))tan(xy). Then the number of solutions of the equation f(x)=sinx, x∈R is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 33
Approach:
Evaluate the limit defining f(x) by Taylor-expanding cos(xy) and tan(xy) to leading order in y, then count the real solutions of f(x)=sinx by analysing the difference function on the regions where sinx is bounded.
Step 1:Substitute the leading-order expansions of cos(xy) and tan(xy) as y→0.
1−cos(xy)=2x2y2+O(y4),tan(xy)=xy+O(y3)
Step 2:Multiply the two expansions and divide by y3 to obtain f(x).
f(x)=limy→0y3(2x2y2+O(y4))(xy+O(y3))=2x3
Step 3:The equation reduces to 2x3=sinx. Since ∣sinx∣≤1, any real solution satisfies ∣x3/2∣≤1, i.e. ∣x∣≤21/3≈1.26.
2x3=sinx,∣x∣≤21/3
Step 4:Let g(x)=sinx−2x3. Both summands are odd, so g is odd; non-zero solutions come in ± pairs and x=0 is one solution since g(0)=0.
g(0)=0,g(−x)=−g(x)
Step 5:Examine x>0. Near x=0, sinx≈x exceeds x3/2, so g(0+)>0. At x=π/2≈1.57>21/3, sinx=1 while x3/2≈1.94, giving g(π/2)<0. By continuity there is exactly one positive root in (0,21/3); for x>21/3 the right side x3/2 exceeds 1≥sinx, ruling out further positive roots.
g(0+)>0,g(21/3)=sin(21/3)−1<0
Step 6:By oddness, exactly one negative solution exists, mirroring the positive one. Adding the x=0 solution gives three real roots in total.
positive roots=1,negative roots=1,plus x=0
Final answer: 3
Q18Single correctIntegral Calculus
Let (21−a+21+a), f(a), (3a+3−a) be in A.P. and α be the minimum value of f(a). Then the value of the integral ∫loge(α−1)loge(α)e2x−e−2xdx is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 241loge(34)
Approach:
Apply the A.P. middle-term condition to express f(a), minimise via AM-GM at a=0 to obtain α, then evaluate the definite integral by the substitution u=e2x.
Step 1:From the A.P. condition, isolate f(a) as the average of the outer terms.
Let f:[1,∞)→R be a differentiable function defined as f(x)=∫1xf(t)dt+(1−x)(logex−1)+e. Then the value of f(f(1)) is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 11+ee
Approach:
Evaluate f(1) from the integral equation, differentiate to derive a linear ODE in f, integrate using factor e−x, fix the constant, then compute f(f(1)).
Step 1:Substituting x=1 collapses both the integral and the (1−x) factor to zero.
f(1)=0+(1−1)(loge1−1)+e=e
Step 2:Differentiating the given relation by Leibniz and the product rule on (1−x)(logex−1).
f′(x)=f(x)−(logex−1)+(1−x)⋅x1=f(x)−logex+x1
Step 3:Multiplying by integrating factor e−x and recognising the RHS as a perfect derivative.
dxd(e−xf(x))=e−x(x1−logex)=dxd(e−xlogex)
Step 4:Applying f(1)=e fixes the integration constant.
e−1⋅e=e−1⋅0+C⇒C=1⇒f(x)=logex+ex
Step 5:Computing f(f(1))=f(e) from the explicit formula.
f(e)=logee+ee=1+ee
Final answer: 1+ee
Q20Single correctLimit, Continuity and Differentiability
Let f(x) and g(x) be twice differentiable functions satisfying f′′(x)=g′′(x) for all x∈R, f′(1)=2g′(1)=4 and g(2)=3f(2)=9. Then f(25)−g(25) is equal to :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 240
Approach:
Since f′′=g′′ identically, the difference f−g is a linear polynomial; determine its coefficients from the data at x=1 and x=2, then evaluate at x=25.
Step 1:Unpacking the chained equalities gives the four boundary values.
f′(1)=4,g′(1)=2,f(2)=3,g(2)=9
Step 2:Since f′′(x)−g′′(x)=0 on R, integrating twice produces a linear difference.
f′(x)−g′(x)=A⇒f(x)−g(x)=Ax+B
Step 3:Substituting x=1 for the slope and x=2 for the intercept.
A=f′(1)−g′(1)=4−2=2;f(2)−g(2)=3−9=−6=2(2)+B⇒B=−10
Step 4:Evaluating at x=25.
f(25)−g(25)=2(25)−10=40
Final answer: 40
Q21NumericalSets, Relations and Functions
Let A={1,4,7} and B={2,3,8}. Then the number of elements in the relation R={((a1,b1),(a2,b2))∈(A×B)×(A×B):a1+b2 divides a2+b1} is ____.
SolutionAnswer: 18
Approach:
Enumerate the nine elements of A×B, iterate over all 81 ordered pairs, and tally those for which (a1+b2) divides (a2+b1).
Step 4:Exhaustive enumeration of the off-diagonal hits identifies nine ordered pairs satisfying D∣N: ((1,2),(4,2)):3∣6; ((1,2),(7,2)):3∣9; ((1,2),(7,8)):9∣9; ((1,8),(1,2)):3∣9; ((1,8),(4,2)):3∣12; ((1,8),(4,3)):4∣12; ((1,8),(7,2)):3∣15; ((4,8),(4,2)):6∣12; ((7,8),(1,2)):9∣9.
Off-diagonal contribution=9
Step 5:Sum diagonal and off-diagonal contributions.
∣R∣=9+9=18
Final answer: 18
Q22NumericalCo-ordinate Geometry
From the point (−1,−1), two rays are sent making angles of 45∘ with the line x+y=0. These rays get reflected from the mirror x+2y=1. If the equations of the reflected rays are ax+by=9 and cx+dy=7, a,b,c,d∈Z, then the value of ad+bc is ____.
SolutionAnswer: 7
Approach:
Identify the two incident rays by adjusting the slope of x+y=0 by ±45∘, locate their intersections with the mirror, reflect the source point across the mirror, then write each reflected ray through its incidence point and the image.
Step 1:Slope of x+y=0 is −1 (inclination 135∘). Adjusting by ±45∘ produces inclinations of 90∘ and 180∘.
Ray 1: x=−1;\ Ray 2: y=−1
Step 2:Intersecting each ray with x+2y=1.
x=−1⇒y=1⇒P1=(−1,1);\ y=−1⇒x=3⇒P2=(3,−1)
Step 3:Reflecting source (−1,−1) across x+2y−1=0: here ax0+by0+c=−1−2−1=−4, a2+b2=5.
Step 4:Reflected ray 1 joins P1=(−1,1) and the image; direction (8/5,6/5)∥(4,3), slope 3/4.
y−1=43(x+1)⇒3x−4y+7=0⇒−3x+4y=7
Step 5:Reflected ray 2 joins P2=(3,−1) and the image; direction (−12/5,16/5)∥(−3,4), slope −4/3.
y+1=−34(x−3)⇒4x+3y=9
Step 6:Computing the required combination.
ad+bc=(4)(4)+(3)(−3)=16−9=7
Final answer: 7
Q23NumericalTrigonometry
If S={θ∈[−π,π]:cosθcos25θ=cos7θcos27θ}, then n(S) is equal to ____.
SolutionAnswer: 19
Approach:
Convert both products to sums, reduce to cos(3θ/2)=cos(21θ/2), solve the two resulting families on [−π,π], then subtract overlapping solutions via inclusion-exclusion.
Step 1:Expanding both sides by the product-to-sum identity.
Step 2:Applying the cosine equality general solution.
221θ=2kπ±23θ⇒9θ=2kπor12θ=2kπ
Step 3:Counting solutions of each family inside [−π,π].
Family 1: ∣2k/9∣≤1⇒k∈{−4,…,4}, giving 9 values. Family 2: ∣k/6∣≤1⇒k∈{−6,…,6}, giving 13 values.
Step 4:Identifying overlap. Common solutions satisfy 92k1=6k2⇔4k1=3k2 with ∣k1∣≤4,∣k2∣≤6, giving (k1,k2)∈{(0,0),(3,4),(−3,−4)} and θ∈{0,±2π/3}.
Common count =3
Step 5:Applying inclusion-exclusion.
n(S)=9+13−3=19
Final answer: 19
Q24NumericalIntegral Calculus
Let f:R→R be a function such that f(x)+3f(2π−x)=sinx, x∈R. Let the maximum value of f on R be α. If the area of the region bounded by the curves g(x)=x2 and h(x)=βx3,β>0, is α2, then 30β3 is equal to ____.
SolutionAnswer: 16
Approach:
Generate a second linear relation by replacing x with π/2−x, solve the pair for f(x), extract amplitude α, equate the area between y=x2 and y=βx3 to α2, and compute 30β3.
Step 1:Substituting x→π/2−x in the functional relation.
f(2π−x)+3f(x)=cosx
Step 2:Solving the linear pair by eliminating f(π/2−x): multiply the new equation by 3 and subtract the original.
Step 4:Curves meet at x=0 and x=1/β; on (0,1/β) the parabola lies above the cubic since βx<1.
Area=∫01/β(x2−βx3)dx=3β31−4β31=12β31
Step 5:Equating area to α2 and solving.
12β31=325⇒12β3=532⇒β3=158⇒30β3=16
Final answer: 16
Q25NumericalDifferential Equations
Let y=y(x) be the solution of the differential equation (tanx)1/2dy=(sec3x−(tanx)3/2y)dx,0<x<2π, y(4π)=562. If y(3π)=54α, then α4 equals ____.
SolutionAnswer: 48
Approach:
Rewrite the equation in standard linear form dy/dx+tanx⋅y=sec3x/tanx, multiply by integrating factor secx, substitute u=tanx, fix the constant from y(π/4), evaluate y(π/3), then compute α4.
Step 1:Dividing the given relation by (tanx)1/2dx produces the standard linear form.
dxdy+tanxy=tanxsec3x
Step 2:Multiplying by secx collapses the LHS into an exact derivative.
dxd(ysecx)=tanxsec4x=tanx(1+tan2x)sec2x
Step 3:Substituting u=tanx, du=sec2xdx, and integrating term by term.
∫u1+u2du=∫(u−1/2+u3/2)du=2u+52u5/2+C
Step 4:Imposing y(π/4)=62/5 with tan(π/4)=1,sec(π/4)=2.
562⋅2=2+52+C⇒512=512+C⇒C=0
Step 5:Evaluating at x=π/3 with tan(π/3)=3,sec(π/3)=2.
How many questions are in the JEE Main 2026 April 05, Shift 2 paper?
The JEE Main 2026 April 05, Shift 2 paper has 74 questions — Physics (25), Chemistry (25) and Mathematics (24). 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.
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Yes — every question shows the correct answer (answer key) and a detailed step-by-step solution, free to read online.
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