JEE Main 2026 April 02, Shift 1 Question Paper with Solutions
All 75 questions from the JEE Main 2026 (April 02, Shift 1) shift — Physics (25), Chemistry (25) and Mathematics (25) — each with the correct answer (answer key) and a step-by-step solution. Read it free online, or attempt the full paper as a timed mock test.
The dimensional formula of 21ϵ0E2 (ϵ0= permittivity of vacuum and E= electric field) is MaLbTc. The value of 2a−b+c= ______
(A)
(B)
(C)
(D)
SolutionAnswer: Option 21
Approach:
Identify 21ϵ0E2 as electric-field energy density, write its dimensional formula, then evaluate 2a−b+c using the exponents of M, L, T.
Step 1:The target: write 21ϵ0E2 in terms of energy per unit volume.
21ϵ0E2=VolumeEnergy
Step 2:Substitute dimensions of energy [ML2T−2] and volume [L3].
[u]=L3ML2T−2=ML−1T−2
Step 3:Read off exponents from MaLbTc.
a=1,b=−1,c=−2
Step 4:Compute 2a−b+c.
2a−b+c=2(1)−(−1)+(−2)=2+1−2
Step 5:Cross-check by re-deriving from ϵ0 in C2N−1m−2 and E in N/C: ϵ0E2 has units of N/m2=ML−1T−2.
[ϵ0E2]=ML−1T−2
Step 6:Result.
2a−b+c=1
Final answer: 1
Q27Single correctExperimental Skills
The diameter of a wire measured by a screw gauge of least count 0.001 cm is 0.08 cm. The length measured by a scale of least count 0.1 cm is 150 cm. When a weight of 100 N is applied to the wire, the measured Young's modulus is α×109N/m2. The error in (ignore the contribution of the load to Young's modulus error calculation)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 21.65
Approach:
Use Y=πd2l4WL (with extension l=0.5 cm from the measured setup), propagate fractional errors from L, d, and l, and obtain ΔY.
Step 1:Data: W=100 N, L=150 cm, d=0.08 cm, extension l=0.5 cm; least counts ΔL=0.1 cm, Δd=0.001 cm, Δl=0.001 cm.
L=1.5m,d=8×10−4m,l=5×10−3m
Step 2:Compute measured Y.
Y=π(8×10−4)2(5×10−3)4(100)(1.5)=5.97×1010N/m2
Step 3:Write fractional error formula (load contribution ignored).
YΔY=LΔL+d2Δd+lΔl
Step 4:Substitute measured values and least counts.
YΔY=1500.1+0.082(0.001)+0.50.001
Step 5:Sum the fractions.
YΔY=0.02767
Step 6:Multiply by Y.
ΔY=0.02767×5.97×1010≈1.65×109N/m2
Step 7:Cross-check: dominant terms are 2Δd/d=0.025 and Δl/l=0.002; their sum ×Y already gives about 1.6×109, consistent with the total.
0.027×5.97×1010≈1.65×109
Step 8:Result.
α=1.65
Final answer: 1.65
Q28Single correctKinematics
The velocity of a particle is given as v=−xi^+2yj^−zk^m/s. The magnitude of acceleration at point (1,2,4) is ______ m/s2.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 29
Approach:
For a steady velocity field v(x,y,z) the acceleration is a=(v⋅∇)v; since each component depends only on its own coordinate, ai=vi∂vi/∂xi.
Step 1:Components: vx=−x,vy=2y,vz=−z at point (1,2,4).
v=−xi^+2yj^−zk^
Step 2:Compute ax=vxdvx/dx.
ax=(−x)⋅(−1)=x
Step 3:Compute ay=vydvy/dy.
ay=(2y)⋅(2)=4y
Step 4:Compute az=vzdvz/dz.
az=(−z)⋅(−1)=z
Step 5:Substitute (x,y,z)=(1,2,4).
a=(1)i^+(8)j^+(4)k^
Step 6:Compute magnitude.
∣a∣=12+82+42=1+64+16=81
Step 7:Cross-check by checking sum: 1+64+16=81=92.
81=9
Step 8:Result.
∣a∣=9m/s2
Final answer: 9
Q29Single correctRotational Motion
The position of an object having mass 0.1 kg as a function of time t is given as r=(10t2i^+5t3j^)m. At t=1 s, which of the following statements are correct?
A) The linear momentum p=(2i^+1.5j^)kg.m/s.
B) The force acting on the object F=(2i^+3j^)N.
C) The angular momentum of the object about its origin L=15k^Js.
D) The torque acting on the object about its origin r=20k^Nm.
Choose the correct answer from the options given below.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4A, B and D only
Approach:
Differentiate r(t) for v and a; evaluate p=mv, F=ma, L=r×p and τ=r×F at t=1 s and test statements A-D.
Step 1:r=(10t2i^+5t3j^)m, m=0.1 kg.
r(1)=10i^+5j^
Step 2:Differentiate to get v and a.
v=20ti^+15t2j^;a=20i^+30tj^
Step 3:Momentum p=mv at t=1.
p=0.1(20i^+15j^)=2i^+1.5j^kg⋅m/s
Step 4:Force F=ma at t=1.
F=0.1(20i^+30j^)=2i^+3j^N
Step 5:Angular momentum L=r×p.
L=(10i^+5j^)×(2i^+1.5j^)=(10⋅1.5−5⋅2)k^=5k^J⋅s
Step 6:Torque τ=r×F.
τ=(10i^+5j^)×(2i^+3j^)=(10⋅3−5⋅2)k^=20k^N⋅m
Step 7:Cross-check: consistency: dL/dt=τ requires L(t)=30t4k^−10t3k^; at t=1, L=20k^⋅1=20k^? Direct expansion: L=(10t2)(15t2)k^−(5t3)(20t)k^=150t4k^−100t4k^=50t4k^, so at t=1, L=50(0.1)k^=5k^J⋅s, matching Step 5.
L(1)=5k^
Step 8:Result: A, B, D are correct; C is not.
⇒ A, B and D only
Final answer: A, B and D only
Q30Single correctGravitation
A planet (P1) is moving around the star of mass 2M in the orbit of radius R. Another planet (P2) is moving around another star of mass 4M in a orbit of radius 2R. Ratio of time periods of revolution of P2 and P1 is ______
(A)
(B)
(C)
(D)
SolutionAnswer: Option 22
Approach:
Apply Kepler's third law T=2πr3/(GMs) to each planet-star system and form the ratio T2/T1.
Step 1:Planet P1 orbits star of mass 2M at radius R; planet P2 orbits star of mass 4M at radius 2R. Target: T2/T1.
Step 5:Cross-check: via direct formula: T1=2πR3/(2GM) and T2=2π8R3/(4GM)=2π2R3/(GM). Their ratio T2/T1=(2R3/GM)⋅(2GM/R3)=4=2.
T2/T1=2
Step 6:Result.
T1T2=2
Final answer: 2
Q31Single correctRotational Motion
A particle is rotating in a circular path and at any instant its motion can be described as θ=405t4−3t3. The angular acceleration of the particle after 10 seconds is ______ rad/s2
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3130
Approach:
Differentiate θ(t)=405t4−3t3 twice to obtain α(t), then substitute t=10 s.
Step 1:θ(t)=405t4−3t3=8t4−3t3.
θ(t)=8t4−3t3
Step 2:Differentiate once for ω.
ω=dtdθ=84t3−33t2=2t3−t2
Step 3:Differentiate again for α.
α=dtdω=23t2−2t
Step 4:Substitute t=10.
α=23(10)2−2(10)=150−20
Step 5:Cross-check by direct second derivative of θ=t4/8−t3/3: d2θ/dt2=12t2/8−6t/3=3t2/2−2t; at t=10 gives 130.
d2θ/dt2t=10=130
Step 6:Result.
α=130rad/s2
Final answer: 130
Q32Single correctElectrostatics
A parallel plate air capacitor has a capacitance C. When it is half filled as shown in figure with a dielectric constant K=5, the percentage increase in the capacitance is ______
(A)
(B)
(C)
(D)
SolutionAnswer: Option 266.67
Approach:
Model the half-filled capacitor as two capacitors in series: an air slab of thickness d/2 and a dielectric slab of thickness d/2 with K=5. Find new capacitance Cf and percentage increase over Ci=Aϵ0/d.
Step 1:Initial air capacitance and series structure after inserting dielectric.
Step 6:Cross-check: limit: for K→1 (no dielectric) the formula gives Cf=Ci, percentage increase =0; for K→∞ the dielectric half becomes a short, giving Cf=2Ci and 100% increase. The value 66.67% at K=5 lies within (0,100)%, consistent.
0%<66.67%<100%
Step 7:Result.
%increase=66.67%
Final answer: 66.67
Q33Single correctThermodynamics
Heat is supplied to a diatomic gas at constant pressure. Then the ratio of ΔQ:ΔU:ΔW is ______
(A)
(B)
(C)
(D)
SolutionAnswer: Option 47:5:2
Approach:
For a diatomic ideal gas at constant pressure use ΔW=nRΔT, ΔU=25nRΔT, ΔQ=27nRΔT and form the ratio.
Step 1:Diatomic gas has degrees of freedom f=5, so Cv=25R and Cp=27R.
Cv=25R,Cp=27R
Step 2:Compute work.
ΔW=nRΔT
Step 3:Compute internal energy change.
ΔU=nCvΔT=25nRΔT
Step 4:Compute heat at constant pressure.
ΔQ=nCpΔT=27nRΔT
Step 5:Form the ratio (multiply each by 2/(nRΔT)).
ΔQ:ΔU:ΔW=7:5:2
Step 6:Cross-check: via first law: ΔQ=ΔU+ΔW⇒7=5+2.
5+2=7
Step 7:Result.
ΔQ:ΔU:ΔW=7:5:2
Final answer: 7:5:2
Q34Single correctElectrostatics
Two charged conducting spheres S1 and S2 of radii 8 cm are connected to each other by a wire. After equilibrium is established, the ratio of electric fields on S1 and S2 spheres are ES1 and ES2 respectively. The value of ES2ES1 is ________
(A)
(B)
(C)
(D)
SolutionAnswer: Option 449
Approach:
Connected spheres reach the same potential at equilibrium, giving q∝r; the surface field then satisfies E∝q/r2∝1/r, so the field ratio reduces to the inverse radius ratio.
Step 1:Given data and target ratio for the two connected conducting spheres of radii r1 and r2.
r1=8cm,r2=18cm;find ES2ES1
Step 2:Equate the surface potentials of the two spheres since they are joined by a wire.
r1kq1=r2kq2⇒q2q1=r2r1
Step 3:Write the ratio of the surface electric fields.
ES2ES1=q2/r22q1/r12=q2q1⋅r12r22
Step 4:Substitute q1/q2=r1/r2 to reduce the ratio.
ES2ES1=r2r1⋅r12r22=r1r2
Step 5:Insert the numerical radii r1=8cm and r2=18cm.
ES2ES1=818=49
Step 6:Cross-check by dimensional and limiting analysis: E∝1/r for connected spheres so the smaller sphere has the larger field, consistent with ES1>ES2.
E∝r1⇒ES2ES1=r1r2=818
Step 7:Result.
ES2ES1=49
Final answer: 49
Q35Single correctOscillations and Waves
The equation of a plane progressive wave is given by y=5cosπ(200t−150x) where x and y are in cm and t is in second. The velocity of the wave is ______ m/s
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4300
Approach:
Read the angular frequency ω and the wave number k directly from the cosine argument and compute v=ω/k, then convert from cm/s to m/s.
Step 1:The target: phase velocity v in m/s for the given wave equation.
y=5cosπ(200t−150x)(cm,s)
Step 2:The argument to standard form ωt−kx.
y=5cos(200πt−150πx)
Step 3:Identify the angular frequency from the coefficient of t.
ω=200πrad/s
Step 4:Identify the wave number from the coefficient of x.
k=150πcm−1
Step 5:Compute the phase velocity in cm/s.
v=kω=π/150200π=200×150=3×104cm/s
Step 6:Convert to m/s using 1m=100cm.
v=1003×104m/s=300m/s
Step 7:Cross-check by computing the wavelength and frequency separately: λ=2π/k=300cm, f=ω/(2π)=100Hz, so v=λf=300cm×100Hz=3×104cm/s=300m/s.
v=λf=(300cm)(100Hz)=300m/s
Step 8:Result.
v=300m/s
Final answer: 300
Q36Single correctElectrostatics
Two short electric dipoles A and B having dipole moment p1 and p2 respectively are placed with their axis mutually perpendicular as shown in the figure. The resultant electric field at a point x is making an angle of 60∘ with the line joining points O and x. The ratio of the dipole moments p2/p1 is ________
(A)
(B)
(C)
(D)
SolutionAnswer: Option 223
Approach:
Treat A as contributing its axial field along the line Ox and B as contributing its equatorial field perpendicular to Ox, then use tan60∘ for the resultant direction to extract the dipole-moment ratio.
Step 1:Dipole A lies along Ox with point x on its axis, dipole B is perpendicular to Ox at O with x lying on its equator; resultant at x makes 60∘ with Ox.
θ=60∘,find p1p2
Step 2:Write the axial field of A along Ox.
E1=r32kp1(along Ox)
Step 3:Write the equatorial field of B perpendicular to Ox.
E2=r3kp2(perpendicular to Ox)
Step 4:Apply the resultant-angle relation.
tan60∘=E1E2=2kp1/r3kp2/r3=2p1p2
Step 5:Substitute tan60∘=3 and solve for the ratio.
3=2p1p2⇒p1p2=23
Step 6:Cross-check: reconstruct tanθ=E2/E1=23/2=3, giving θ=60∘ as specified.
tanθ=223=3⇒θ=60∘
Step 7:Result.
p1p2=23
Final answer: 23
Q37Single correctElectronic Devices
For the given circuit (shown in part (A)) the time dependent input voltage vm(t) and corresponding output v0(t) are shown in part (B) and part (C), respectively. Identify the components that are used in the circuit between points X and Y.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2Option 2
Approach:
A sinusoidal input swinging between ±20V is reduced to a symmetric square-like output clipped at ±5V, which is the signature of a back-to-back Zener/diode pair acting as a symmetric voltage limiter; match this behaviour to the candidate two-component networks.
Step 1:Input is vin(t)∈[−20V,+20V] sinusoidal; output is clamped between ±5V; identify the two-element network between X and Y.
vin∈[−20,20]V,vo∈[−5,5]V
Step 2:Symmetric clipping on both half-cycles requires conduction in both polarities, which a single diode cannot supply.
Symmetric clip⇒bidirectional conduction
Step 3:A back-to-back pair of diodes (one ordinary diode in series with a Zener oriented oppositely, or two Zeners back-to-back) provides ±VZ clamping.
vo≈+VZ(positive half),vo≈−VZ(negative half)
Step 4:Option 2 shows two diodes oriented in opposite directions between X and Y, the only arrangement among the four options that conducts in both polarities and produces symmetric clipping.
Option 2: D→ + D←
Step 5:Rule out option 1 (resistor + single diode — half-wave only), option 3 (resistor + single Zener — asymmetric clamp), option 4 (two diodes in same direction — half-wave conduction).
Options 1,3,4 fail symmetric ±5V clip
Step 6:Cross-check by behaviour check: on the positive half of vin the forward-biased diode and the reverse-biased Zener (in breakdown at VZ=5 V) clamp the output at +5 V; on the negative half the roles swap to clamp at −5 V, matching the displayed output.
vo=±VZ=±5V for ∣vin∣>VZ
Step 7:Result.
Network: diode→+diode←
Final answer: Option 2
Q38Single correctElectromagnetic Induction and AC
When a coil is placed in a time dependent magnetic field the power dissipated in it is P. The number of turns, area of the coil and radius of the wire are N, A and r respectively. For a second coils number of turns, area of the coil and radius of the coil wire are 2N, 2A and 3r respectively. When the first coil is replaced with second coil the power dissipated in it is 2αP. The value of α is ______
(A)
(B)
(C)
(D)
SolutionAnswer: Option 136
Approach:
EMF in a coil is ε=NA(dB/dt), so ε∝NA; the coil resistance scales as wire length over wire cross-section, R∝NA/r2. Power P=ε2/R∝NA3/2r2. Form the ratio and read off α.
Step 1:Coil-1 has (N,A,r) with power P; coil-2 has (2N,2A,3r) with power 2αP; find α.
P2=2αP
Step 2:Induced EMF scales as ε∝NA, so ε2∝N2A2.
ε∝NA⇒ε2∝N2A2
Step 3:Total wire length of a coil of N turns and area A (circular cross-section) is L=N⋅2πA/π∝NA; with wire cross-section πr2, the resistance scales as R∝NA/r2.
R∝r2NA
Step 4:Combine to obtain the power scaling.
P=Rε2∝NA/r2N2A2=NA3/2r2
Step 5:Substitute the second coil's parameters into the ratio.
PP2=N⋅A3/2⋅r2(2N)(2A)3/2(3r)2=2⋅23/2⋅9
Step 6:Simplify 2⋅23/2=2⋅22=42, then multiply by 9.
PP2=42⋅9=362
Step 7:Cross-check by matching to P2=2αP and extracting α.
2α=362⇒α=36
Step 8:Result.
α=36
Final answer: 36
Q39Single correctMagnetic Effects of Current and Magnetism
Two identical long current carrying wires are bent into the shapes shown in the following figures. If the magnitude of magnetic fields at the centres P and Q of a semicircular arc are B1 and B2 respectively, then the ratio B2B1 is __________
(A)
(B)
(C)
(D)
SolutionAnswer: Option 11+π2+π
Approach:
At the centre of each configuration, sum the Biot-Savart contributions of the straight semi-infinite segments and the semicircular arc; configuration (I) has two effective semi-infinite straights at perpendicular distance R while configuration (II) has only one such straight (the other passes through the centre and contributes zero).
Step 1:Configuration (I) — semicircular arc of radius R centred at P with two collinear straight extensions on opposite sides (each semi-infinite, perpendicular distance to P equal to R); configuration (II) — same arc centred at Q with one straight tangent at distance R and one straight along the diameter (passing through Q).
Targets: B1 at P,B2 at Q
Step 2:(I): two semi-infinite straights each contribute μ0I/(4πR) at P (currents create field in the same sense as the arc), plus the arc μ0I/(4R).
B1=2⋅4πRμ0I+4Rμ0I=4Rμ0I(π2+1)
Step 3:(II): one semi-infinite straight contributes μ0I/(4πR) at Q, the other straight passes through Q and gives zero field, plus the arc μ0I/(4R).
B2=4πRμ0I+0+4Rμ0I=4Rμ0I(π1+1)
Step 4:Form the ratio B1/B2 and cancel common factors.
B2B1=(1+π)/π(2+π)/π=1+π2+π
Step 5:Cross-check by numerical evaluation: (2+π)/(1+π)≈5.1416/4.1416≈1.241; equivalently the arc contributes about μ0I/(4R) to both, with one extra straight term in (I), consistent with B1>B2.
1+π2+π≈1.241
Step 6:Result.
B2B1=1+π2+π
Final answer: 1+π2+π
Q40Single correctOptics
For a thin symmetric prism made of glass (refractive index 1.5), the ratio of incident angle and minimum deviation will be __________
(A)
(B)
(C)
(D)
SolutionAnswer: Option 23:2
Approach:
Use the thin-prism small-angle relations: at minimum deviation in a symmetric prism r1=r2=A/2 so i=μ(A/2), while δmin=(μ−1)A. Substitute μ=1.5.
Step 1:Thin symmetric prism, μ=1.5; find i:δmin.
μ=1.5,find δmini
Step 2:At minimum deviation the ray inside the prism is symmetric, so r1=r2=A/2.
r=2A
Step 3:Small-angle Snell's law gives the incidence angle.
i=μr=μ2A
Step 4:Use the standard thin-prism result for minimum deviation.
δmin=(μ−1)A
Step 5:Form the ratio and cancel A.
δmini=(μ−1)AμA/2=2(μ−1)μ
Step 6:Substitute μ=1.5.
δmini=2(0.5)1.5=1.01.5=23
Step 7:Cross-check by working backwards: δmin=(μ−1)A=0.5A and i=μA/2=0.75A, so i/δmin=0.75/0.5=1.5=3/2.
δmini=0.5A0.75A=23
Step 8:Result.
i:δmin=3:2
Final answer: 3:2
Q41Single correctOptics
Refer the figure given below. μ1 and μ2 are refractive indices of air and lens material. The height of image will be ________ cm
(A)
(B)
(C)
(D)
SolutionAnswer: Option 11
Approach:
Apply the single-surface refraction formula with the Cartesian sign convention to find the image distance v, then use the across-media lateral magnification m=(μ1v)/(μ2u) to compute the image height.
Step 1:μ1=1 (air), μ2=1.54 (lens), object height ho=2cm, u=−40cm, R=−20cm (centre of curvature on the object side per the figure).
Step 6:Multiply by the object height to obtain the image height.
hi=mho=0.481×2cm≈0.96cm
Step 7:Cross-check: rounding to the closest option among {1,0.5,1.2,0.25} cm gives 1cm, since 0.96 differs from 1 by only 0.04cm and from 0.5 by 0.46cm.
∣hi−1∣=0.04<∣hi−0.5∣=0.46
Step 8:Result.
hi≈1cm
Final answer: 1
Q42Single correctDual Nature of Matter and Radiation
For a certain metal, when monochromatic light of wavelength λ is incident, the stopping potential for photoelectrons is 3V0. When the same metal is illuminated by light of wavelength 2λ, then the stopping potential becomes V0. The threshold wavelength for photoelectric emission for the given metal is αλ. The value of α is ______
(A)
(B)
(C)
(D)
SolutionAnswer: Option 24
Approach:
Write Einstein's photoelectric equation for both wavelengths, eliminate the work function to find ϕ, then use ϕ=hc/λth to extract the multiplier α.
Step 1:Given data and target. Wavelength λ gives stopping potential 3V0; wavelength 2λ gives V0; target threshold wavelength is αλ.
λth=αλ
Step 2:Einstein equation for incident wavelength λ.
λhc−ϕ=3eV0(i)
Step 3:Einstein equation for incident wavelength 2λ.
2λhc−ϕ=eV0(ii)
Step 4:Multiply (ii) by 3 and subtract (i) to eliminate the stopping potential.
2λ3hc−3ϕ−λhc+ϕ=0⇒2λhc=2ϕ
Step 5:Equate ϕ to hc/λth to read off the threshold wavelength.
λthhc=4λhc⇒λth=4λ
Step 6:Cross-check by substituting ϕ=hc/(4λ) back into both equations and checking the stopping potentials.
λhc−4λhc=4λ3hc=3eV0;2λhc−4λhc=4λhc=eV0
Step 7:Result.
α=4
Final answer: 4
Q43Single correctElectromagnetic Waves
An electromagnetic wave travelling in x-direction is described by field equation Ey=300sinω(t−cx). If the electron is restricted to move in y-direction only with speed of 1.5×106m/s then ratio of maximum electric and magnetic forces acting on the electron is ______
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1200
Approach:
Identify the EM wave speed from the phase, then use B0=E0/c to reduce the ratio of maximum electric to magnetic force on the electron to c/v.
Step 1:Given data: E0=300V/m along y; wave travels along x with phase ω(t−x/c); electron speed v=1.5×106m/s along y. Target: FE/FB.
Ey=300sinω(t−cx)
Step 2:Read off the wave speed from the phase argument.
Vwave=kω=ω/cω=c=3×108m/s
Step 3:Maximum electric force on electron is FE=qE0 and maximum magnetic force is FB=qvB0.
FBFE=qvB0qE0=vB0E0
Step 4:Substitute B0=E0/c to simplify.
FBFE=v(E0/c)E0=vc
Step 5:Substitute numerical values of c and v.
vc=1.5×1063×108
Step 6:Cross-check by direct division.
1.5×1063×108=2×102=200
Step 7:Result.
FBFE=200
Final answer: 200
Q44Single correctAtoms and Nuclei
Angular momentum of an electron in a hydrogen atom is π3h, then the energy of the electron is ______ eV.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3-0.38
Approach:
Use Bohr's angular momentum quantization to obtain n, then apply the hydrogen energy level formula.
Step 1:Given data and target: L=3h/π; find energy En in eV.
L=π3h
Step 2:Equate given L to nh/2π and solve for n.
2πnh=π3h⇒n=6
Step 3:Substitute n=6 into the hydrogen energy formula.
E6=−6213.6=−3613.6eV
Step 4:Evaluate the numerical value.
E6=−0.3778eV≈−0.38eV
Step 5:Cross-check by recomputing 13.6/36 and rounding to two decimals.
13.6/36=0.37778⇒E6=−0.38eV
Step 6:Result.
E6=−0.38eV
Final answer: -0.38
Q45Single correctProperties of Solids and Liquids
A liquid drop of diameter 2 mm breaks into 512 droplets. The change in surface energy is α×10−6J. The value of α is ______ . (Take surface tension of liquid = 0.08 N/m)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 27
Approach:
Use volume conservation to find the radius of each small droplet, then compute the change in surface area times surface tension.
Step 7:Cross-check by checking 0.08⋅4⋅3.1416⋅7=7.037, consistent with α=7.
α=7
Step 8:Result.
α=7
Final answer: 7
Q46NumericalOptics
In single slit diffraction pattern, the wavelength of light used is 628nm and slit width is 0.2mm the angular width of central maximum is α×10−2 degrees. The value of α is ______
SolutionAnswer: 36
Approach:
Use single-slit angular width of central maximum 2λ/d in radians and convert to degrees.
Step 1:Given data: λ=628nm=628×10−9m; d=0.2mm=0.2×10−3m. Target: α such that Δθ=α×10−2 degrees.
Δθ=d2λ
Step 2:Substitute values to get angle in radians.
Δθ=0.2×10−32⋅628×10−9
Step 3:Simplify the radian value.
Δθ=2×10−41256×10−9=6.28×10−3rad
Step 4:Convert to degrees.
Δθdeg=6.28×10−3⋅π180
Step 5:Evaluate using 180/π=57.2958.
Δθdeg=6.28×10−3⋅57.2958=0.3598∘
Step 6:Cross-check by expressing in 10−2 degrees.
0.36∘=36×10−2∘
Step 7:Result.
α=36
Final answer: 36
Q47NumericalThermodynamics
A vessel contains 0.15m3 of a gas at pressure 8 bar and temperature 1400C with cp=3R and cv=2R. It is expanded adiabatically till pressure falls to 1 bar. The work done during this process is ______ k J. (R is gas constant)
SolutionAnswer: 120
Approach:
Compute γ from cp,cv, use the adiabatic PVγ relation to find V2, then apply W=(P2V2−P1V1)/(1−γ).
Step 1:Given data: V1=0.15m3, P1=8bar=8×105Pa, P2=1bar=105Pa, cp=3R, cv=2R. Target: work W in kJ.
Step 7:Cross-check: gas expands (V increases), so W>0, consistent with +120kJ.
W>0
Step 8:Result.
W=120kJ
Final answer: 120
Q48NumericalMagnetic Effects of Current and Magnetism
1μC charge moving with velocity v=(i^−2j^+3k^)m/s in the region of magnetic field B=(2i^+3j^−5k^)T. The magnitude of force acting on it is α×10−6N. The value of α is ______
SolutionAnswer: 171
Approach:
Apply Lorentz force F=q(v×B): compute the cross product, take its magnitude, and read off α.
Step 1:Given data: q=1μC=10−6C, v=(1,−2,3)m/s, B=(2,3,−5)T. Target: α with ∣F∣=α×10−6N.
F=qv×B
Step 2:Compute i^ component of v×B.
(v×B)x=vyBz−vzBy=(−2)(−5)−(3)(3)=10−9=1
Step 3:Compute j^ component.
(v×B)y=vzBx−vxBz=(3)(2)−(1)(−5)=6+5=11
Step 4:Compute k^ component.
(v×B)z=vxBy−vyBx=(1)(3)−(−2)(2)=3+4=7
Step 5:Assemble the force vector.
F=10−6(i^+11j^+7k^)N
Step 6:Compute the magnitude squared.
12+112+72=1+121+49=171
Step 7:Cross-check by taking the square root and matching the stated form.
∣F∣=171×10−6N
Step 8:Result.
α=171
Final answer: 171
Q49NumericalProperties of Solids and Liquids
A uniform wire of length l of weight w is suspended from the roof with a weight of W at the other end. The stress in the wire at 3l distance from the top is (AW+γA2w), where, A is the cross sectional area of the wire. The value of γ is ______
SolutionAnswer: 3
Approach:
At a section a distance l/3 from the top, the tension supports the weight W plus the wire weight hanging below, which equals (2/3)w; divide by area and match the given form.
Step 1:Given data: uniform wire of length l, weight w, suspended from roof with weight W attached at bottom. Target: γ in σ=W/A+2w/(γA) at the section l/3 below the top.
σ(3l)=AW+γA2w
Step 2:Length of wire below the section is l−l/3=2l/3.
lbelow=32l
Step 3:Weight of this sub-length (uniform wire).
wbelow=w⋅l2l/3=32w
Step 4:Tension at section equals W plus the weight hanging below it.
T=W+32w
Step 5:Compute stress.
σ=AT=AW+3A2w
Step 6:Cross-check by matching denominators with the given form γA2w.
3A2w=γA2w⇒γ=3
Step 7:Result.
γ=3
Final answer: 3
Q50NumericalProperties of Solids and Liquids
A tub is filled with water and a wooden cube 10cm×10cm×10cm is placed in the water. The wooden cube is found to float on the water with a part of it submerged in water. When a metal coin is placed on the wooden cube, the submerged part is increased by 3.87cm. The mass of the metal coin is ______ gram. (Take water density as 1g/cm3 and density of wood as 0.4g/cm3)
SolutionAnswer: 387
Approach:
Additional buoyant force due to extra submerged depth equals the weight of the coin; solve for coin mass.
Step 1:Given data: cube side 10cm, so base area A=100cm2; extra submergence Δh=3.87cm; ρw=1g/cm3. Target: mcoin in grams.
A=100cm2,Δh=3.87cm
Step 2:Write equilibrium of the loaded cube — the additional buoyancy from the extra submerged slab must equal the coin's weight.
ρwAΔhg=mcoing
Step 3:Cancel g on both sides.
mcoin=ρwAΔh
Step 4:Substitute the numerical values in cgs units.
mcoin=1g/cm3⋅100cm2⋅3.87cm
Step 5:Multiply.
mcoin=387g
Step 6:Cross-check by checking that the wood density was not needed (the original wood equilibrium cancels out when only considering the change).
ΔFB−0=mcoing⇒mcoin=387g
Step 7:Result.
mcoin=387g
Final answer: 387
Chemistry25 questions
Q51Single correctSome Basic Concepts in Chemistry
The mass of iron converted into Fe3O4 by the action of 18g of steam is (Given :Molar mass of H2O and Fe are 1,16 and 56 g mol−1 respectively) Assume iron is present in excess.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 442 gm
Approach:
Use the balanced reaction 3Fe+4H2O→Fe3O4+4H2 and the limiting reagent (steam) to find moles of Fe consumed, then convert to mass.
Step 1:Given data and target.
mH2O=18g,MH2O=18g mol−1,MFe=56g mol−1
Step 2:Write the balanced reaction.
3Fe+4H2O→Fe3O4+4H2
Step 3:Compute moles of steam.
nH2O=1818=1mol
Step 4:Apply stoichiometric ratio (3 Fe per 4 H2O).
nFe=43×1=0.75mol
Step 5:Convert moles of Fe to mass.
mFe=0.75×56=42g
Step 6:Cross-check by atom balance: 4 H2O contribute 4 O atoms which form one Fe3O4 unit with 3 Fe, scaled by 1/4 gives 0.75 mol Fe.
0.75×56=42g
Step 7:Result.
mFe=42g
Final answer: 42 gm
Q52Single correctAtomic Structure
What is the energy (in J atom−1) required for the following process? Li(g)2+→Li(g)3++e− (Take the ionization energy for the H atom in the ground state as 2.18×10−18 J atom−1)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 31.962×10−17
Approach:
Apply the Bohr/hydrogenic ionization-energy formula E=EHZ2/n2 to the one-electron ion Li2+ (Z=3, n=1).
Step 1:Given data and target.
EH=2.18×10−18J;process: Li(g)2+→Li(g)3++e−
Step 2:Identify Li2+ as a hydrogen-like one-electron ion with Z=3.
ZLi=3,n=1
Step 3:Write the ionization-energy scaling with Z2.
IE=EH×Z2
Step 4:Substitute numerical values.
IE=2.18×10−18×32=2.18×10−18×9
Step 5:Cross-check by direct multiplication.
2.18×9=19.62⇒19.62×10−18=1.962×10−17
Step 6:Result.
IE=1.962×10−17J atom−1
Final answer: 1.962×10−17
Q53Single correctChemical Bonding and Molecular Structure
Given below are two statements
Statement-I : The correct sequence of bond length in the following species is O2⊕<O2<O2−<O22−
Statement-II : The correct sequence of number of unpaired electrons in the following species is O2>O2−>O2−>O22−
In the light of the above statements choose the correct answers from the options given below
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3Statement-I is true but Statement-II is false
Approach:
Use molecular-orbital theory to obtain bond orders and unpaired-electron counts for O2+,O2,O2−,O22−, then evaluate both statements.
Step 1:Given data and target.
species: O2+,O2,O2−,O22−
Step 2:List MO electrons (total) and antibonding π∗ occupancy.
O2+:15e−,O2:16e−,O2−:17e−,O22−:18e−
Step 3:Compute bond orders.
BO(O2+)=2.5,BO(O2)=2,BO(O2−)=1.5,BO(O22−)=1
Step 4:Invert to get bond lengths (smaller BO ⇒ longer bond).
Step 6:Examine Statement II — the correct decreasing chain is O2(2)>O2+(1)=O2−(1)>O22−(0), with equality between O2+ and O2−.
2>1=1>0
Step 7:Cross-check by combining both findings: I true and II false.
I true,II false
Step 8:Result.
Statement-I true, Statement-II false
Final answer: Statement-I is true but Statement-II is false
Q54Single correctChemical Thermodynamics
Consider the following data i) 2Al(s)+6HCl(aq)→Al2Cl6(aq)+3H2(g)+1200kJ/mol ii) H2(g)+Cl2(g)→2HCl(g)+164kJ/mol iii) HCl(g)+aq→HCl(aq)+83kJ/mol iv) Al2Cl6(s)+aq→Al2Cl6(aq)+663kJ/mol The enthalpy of formation of anhydrous solid Al2Cl6 is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4−1527kJmol−1
Approach:
Combine the four given thermochemical equations via Hess's law to obtain 2Al(s)+3Cl2(g)→Al2Cl6(s) and sum the corresponding enthalpy changes.
Step 1:Given data and target reaction.
2Al(s)+3Cl2(g)→Al2Cl6(s),ΔHf=?
Step 2:Read the four enthalpies (heat released ⇒ negative ΔH).
ΔH1=−1200,ΔH2=−164,ΔH3=−83,ΔH4=−663kJ/mol
Step 3:Take combination (i)+3(ii)+6(iii)−(iv) and verify cancellation of HCl(g), HCl(aq), H2, Al2Cl6(aq), and aq.
19.5 g of fluoro acetic acid (molar mass =78 g mol−1) is dissolved in 500g of water at 298K. The depression in the freezing point of water was 10C. What is ka of fluoro acetic acid? (For water, Kf=1.86 K kg mol−1). Assume molarity and molality to have same values.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 43×10−3
Approach:
From freezing-point depression find van't Hoff factor i, convert to degree of dissociation α, then use Ka=Cα2 for the weak monoprotic acid.
Step 4:Relate i to α for HA (one particle becoming two, n=2).
i=1+α⇒α=i−1=1.862−1=0.0753
Step 5:Apply Ka=Cα2 with C=m=0.5 M.
Ka=0.5×(0.0753)2=0.5×5.67×10−3
Step 6:Cross-check by rounding to the nearest listed option.
2.83×10−3≈3×10−3
Step 7:Result.
Ka≈3×10−3
Final answer: 3×10−3
Q56Single correctEquilibrium
The solubility product constants of Ag2CrO4 and AgBr are 32x and 4y respectively at 298K. The value of (molarityofAgBrmolarityofAg2CrO4) can be expressed as :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4y3x
Approach:
Write the solubility-product expressions for Ag2CrO4 (2:1) and AgBr (1:1) in terms of their molar solubilities, then take the ratio.
Step 1:Given data and target.
Ksp(Ag2CrO4)=32x,Ksp(AgBr)=4y
Step 2:Dissociation of Ag2CrO4 gives 2S1 Ag+ and S1 CrO42−.
Ksp=(2S1)2(S1)=4S13
Step 3:Solve for S1.
S13=8x⇒S1=2x1/3=23x
Step 4:Dissociation of AgBr gives S2 Ag+ and S2 Br−.
Ksp=S22=4y⇒S2=2y1/2=2y
Step 5:Take the ratio of molar solubilities.
S2S1=2y23x=y3x
Step 6:Cross-check by dimensional reduction: cube root of x from S13∝x, square root of y from S22∝y, factors of 2 cancel.
2y1/22x1/3=y1/2x1/3
Step 7:Result.
S2S1=y3x
Final answer: y3x
Q57Single correctRedox Reactions and Electrochemistry
An electrochemical cell is constructed using half cells in the direction of spontaneous change Fe(OH)2(s)+2e−→Fe(s)+2OH−(aq)E0=−0.88VAgBr(s)+e−→Ag(s)+Br−(aq)E0=0.07V Which of the following option is correct
Step 7:In the cell Fe is at the anode and is oxidized, so option 3 is incorrect; Ecell0 is intensive (independent of amount), so option 4 is incorrect.
Fe oxidized;Ecell0 intensive
Step 8:Cross-check by checking sign and direction: Ecell0>0 confirms the written direction is spontaneous.
Final answer: Overall reaction Fe(s)+2OH−(aq)+2AgBr(s)⇌Fe(OH)2(s)+2Ag(s)+2Br−(aq)
Q58Single correctChemical Kinetics
t100% is the time required for the 100% completion of the reaction while t1/2 is the time required for the 50% of the reaction to be completed which of the following option correctly represents relation between t100% and t1/2 for zero and first order reactions respectively?
Use the integrated rate laws for zero- and first-order reactions to express t100% and t1/2, then compare.
Step 1:Given data and target.
compare t100% vs t1/2 for zero and first order
Step 2:(zero order): set [A]=0 in [A]0−kt=0 to find completion time.
t100%=k[A]0
Step 3:(zero order): take ratio with t1/2=[A]0/(2k).
t1/2t100%=[A]0/(2k)[A]0/k=2⇒t100%=2t1/2
Step 4:(first order): set [A]=0 in [A]0e−kt=0.
e−kt=0⇒kt→∞⇒t100%→∞
Step 5:(first order): since t1/2=ln2/k is finite, t100% is infinitely many half-lives, symbolically written t100%=(t1/2)∞.
t100%=(t1/2)∞
Step 6:Cross-check by inspecting integrated forms: zero-order concentration reaches zero in finite time, first-order only asymptotically.
zero: linear decay;first: exponential decay
Step 7:Result.
t100%=2t1/2(zero),t100%=(t1/2)∞(first)
Final answer: t100%=2×t1/2:t100%=(t1/2)∞
Q59Single correctClassification of Elements and Periodicity
Given below are two statements
Statement – I : The first Ionization enthalpy of the elements Na,Mg,Cl and Ar follows the order Na>Mg>Cl>Ar
Statement – II : Among Ca,Al,Fe and B, the third ionisation enthalpy is very high for Ca.
In the light of the above statements choose the correct answers from the options given below
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4Statement-I is false and Statement-II is true
Approach:
Evaluate Statement-I against the standard first ionisation enthalpy trend across Na, Mg, Cl, Ar and Statement-II by comparing successive (third) ionisation enthalpies for Ca, Al, Fe, B based on the stability of the resulting electron configurations.
Step 1:The targets: validate (i) the order of first ionisation enthalpy among Na,Mg,Cl,Ar and (ii) which element among Ca,Al,Fe,B has the largest third ionisation enthalpy.
Step 3:Statement-I claims Na>Mg>Cl>Ar which is the exact reverse of the true period trend; hence Statement-I is FALSE.
Statement-I order=reverse of actual
Step 4:Third ionisation enthalpies (kJ mol−1): Ca2+([Ar])→Ca3+ breaks a noble-gas core, IE3(Ca)=4912; IE3(B)=3660 (removal from 2s1); IE3(Fe)=2957; IE3(Al)=2745.
IE3:Ca(4912)>B(3660)>Fe(2957)>Al(2745)
Step 5:The reasoning for Ca: after losing two valence 4s electrons, Ca2+ has the stable [Ar] noble-gas configuration; removing a 3rd electron requires very high energy.
Ca2+:[Ar]IE3Ca3+:[Ne]3s23p5
Step 6:Cross-check by comparing tabulated successive ionisation enthalpies — Ca shows the characteristic large jump between IE2 and IE3 confirming that IE3(Ca) is the largest of the four.
IE3(Ca)≫IE3(Al,Fe,B)
Step 7:Result: Statement-I is FALSE and Statement-II is TRUE, which corresponds to option (4).
Statement-I: FALSE,Statement-II: TRUE
Final answer: Statement-I is false and Statement-II is true
Q60Single correctp-Block Elements
Given below are two statements :
Statement I : Oxidising power of halogens decreases in the order F2>Cl2>Br2>I2 which is the basis of “Layer test”
Statement II : “Layer test” to identify Br2 and I2 in aqueous solution involves the oxidation of bromide or iodide into Br2 or I2 respectively with Cl2, which is a type of displacement redox reaction.
In the light of the above statements choose the correct answers from the options given below
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1Both Statement-I and Statement-II are true
Approach:
Cross-check: the oxidising-power trend of the halogens using standard reduction potentials and confirm that the layer test displacement of Br− or I− by Cl2 is a redox reaction.
Step 1:Given data: Statement-I claims the oxidising power decreases as F2>Cl2>Br2>I2 and Statement-II describes the layer test as a displacement redox process.
Statements about oxidising power and layer test
Step 2:The standard reduction potentials E∘(X2/X−): F2/F−=+2.87V, Cl2/Cl−=+1.36V, Br2/Br−=+1.09V, I2/I−=+0.54V.
E∘:F2(+2.87)>Cl2(+1.36)>Br2(+1.09)>I2(+0.54)V
Step 3:The chemistry of the layer test: chlorine water added to a mixture containing Br−/I− oxidises them to Br2/I2; carbon tetrachloride or CS2 extracts the liberated halogen into the organic (lower) layer.
Cl2+2Br−→2Cl−+Br2;Cl2+2I−→2Cl−+I2
Step 4:Colour observations: Br2 gives an orange/brown organic layer, I2 gives a violet/pink organic layer — the very basis of qualitative identification.
Br2:orange;I2:violet
Step 5:Cross-check by oxidation-state book-keeping: Cl goes from 0→−1 (reduced), Br/I goes from −1→0 (oxidised); a balanced electron transfer — a genuine redox.
Δ(ox. state):Cl:0→−1,X:−1→0
Step 6:Result: both Statement-I and Statement-II are true, matching option (1).
Both statements TRUE
Final answer: Both Statement-I and Statement-II are true
Q61Single correctd- and f- Block Elements
Which of the following sets includes all the species that will change the orange colour of K2Cr2O7 in acidic medium?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1Fe2+,Sn2+,I−,S2−
Approach:
Acidified K2Cr2O7 is a strong oxidising agent (Cr+6→Cr+3). Identify which option contains only species capable of being oxidised (in lower oxidation states) so that they reduce dichromate and discharge its orange colour.
Step 1:The target: only reducing species (i.e. species in lower oxidation states relative to dichromate) can discharge the orange colour by being oxidised.
Step 5:For iodide and sulphide: 6I−+Cr2O72−+14H+→3I2+2Cr3++7H2O and 3S2−+Cr2O72−+14H+→3S+2Cr3++7H2O.
6I−+Cr2O72−+14H+→3I2+2Cr3++7H2O
Step 6:Cross-check: other options fail: option (2) contains Fe3+ (highest stable state, cannot be further oxidised by Cr2O72−); option (3) contains Sn4+ (cannot be oxidised); option (4) contains Fe3+,SO42−,Sn4+ (all highest oxidation state, non-reducing).
Fe3+,Sn4+,SO42− cannot reduce Cr2O72−/H+
Step 7:Result: only option (1) contains a complete set of reducing species.
Set={Fe2+,Sn2+,I−,S2−}
Final answer: Fe2+,Sn2+,I−,S2−
Q62Single correctCoordination Compounds
Match List – I with List – II
List– I (Chromium (III) Complexes, en = ethylene diamine)
List– II Δ0(cm−1)
A) [Cr(CN)6]3−
I) 15060
B) [CrF6]3−
II) 17400
C) [Cr(H2O)6]3+
III) 22300
D) [Cr(en)3]3+
IV) 26600
Choose the correct answer from the options given below:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4A−IV,B−I,C−II,D−III
Approach:
Crystal-field splitting energy Δ0 for a given metal centre in the same oxidation state increases with ligand-field strength as ordered in the spectrochemical series. Rank the four Cr(III) complexes by ligand strength and assign the four Δ0 values accordingly.
Step 1:The four ligands and their roles: all four complexes are octahedral Cr(III) (d3) species, so Δ0 depends only on the ligand-field strength.
Common metal centre:Cr3+,d3,octahedral
Step 2:The spectrochemical position of each ligand: CN− is a strong-field π-acceptor (largest Δ0); ethylenediamine (en) is a strong-field neutral σ-donor and chelating ligand; H2O is intermediate; F− is a weak-field ligand (smallest Δ0).
CN−>en>H2O>F−
Step 3:Assignment of Δ0 values from List-II in decreasing order to the complexes in decreasing ligand strength: A [Cr(CN)6]3−→26600cm−1 (IV); D [Cr(en)3]3+→22300cm−1 (III).
A→IV,D→III
Step 4:Remaining assignments: C [Cr(H2O)6]3+→17400cm−1 (II); B [CrF6]3−→15060cm−1 (I).
C→II,B→I
Step 5:Cross-check: monotonicity: Δ0 decreases as the ligand becomes weaker — 26600>22300>17400>15060cm−1 exactly tracks CN−>en>H2O>F−.
26600>22300>17400>15060cm−1
Step 6:Result.
A−IV,B−I,C−II,D−III
Final answer: A−IV,B−I,C−II,D−III
Q63Single correctPurification and Characterisation of Organic Compounds
Given below are two statements :
Statement-I : 1, 2, 3-trihydroxy propane can be separated from water by simple distillation.
Statement-II : An azeotropic mixture cannot be separated by fractional Distillation
In the light of the above statements choose the correct answers from the options given below
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4Statement-I is false but Statement-II is true
Approach:
Identify the appropriate distillation technique for glycerol (1,2,3-trihydroxypropane), which has a high boiling point and decomposes on heating at atmospheric pressure, and recall the property of an azeotrope that prevents fractional-distillation separation.
Step 1:The target: judge Statement-I (glycerol separable from water by simple distillation) and Statement-II (azeotrope inseparable by fractional distillation).
Statements about glycerol and azeotrope separation
Step 2:Glycerol behaviour: glycerol boils at ≈290∘C and undergoes thermal dehydration to acrolein at temperatures approaching its boiling point.
GlycerolΔ,1 atmacrolein+2H2O
Step 3:The technique used: glycerol is separated by distillation under reduced pressure (vacuum distillation), which lowers the boiling point below the decomposition temperature.
Reduced-pressure distillation:P↓⇒Tb↓
Step 4:Azeotrope theory: at the azeotropic composition the vapour has the same composition as the liquid, so no enrichment occurs across stages of fractional distillation.
yi=xi⇒no fractionation possible
Step 5:Consequence: separation of an azeotrope requires alternative methods (azeotropic agent, pressure swing, membrane separation), confirming Statement-II.
Alternative methods needed for azeotropes
Step 6:Cross-check by considering the standard glycerol–water industrial separation: vacuum distillation is the accepted technique, not simple atmospheric distillation.
Statement-1 : Benzyl chloride reacts faster in SN1 mechanism than ethyl chloride
Statement-II : Ethyl carbocation intermediate is less stabilized by hyperconjugation than benzyl carbocation by resonance.
In the light of the above statements choose the correct answers from the options given below
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1Both Statement-I and Statement-II are true
Approach:
SN1 rate is governed by the stability of the carbocation intermediate. Compare benzyl cation (resonance-delocalised onto the aromatic ring) with the ethyl cation (only hyperconjugation), then judge both statements.
Step 1:The target: relate the SN1 rate of benzyl chloride versus ethyl chloride to the stability of their carbocation intermediates.
PhCH2ClvsCH3CH2Cl
Step 2:The carbocation from benzyl chloride: ionisation gives PhCH2+ whose positive charge is delocalised onto the ortho and para positions of the phenyl ring through three additional resonance structures.
PhCH2+↔ring resonance contributors
Step 3:The carbocation from ethyl chloride: ionisation gives CH3CH2+ which is a primary cation stabilised only by hyperconjugation with three C–H bonds of the methyl group.
CH3CH2+:three hyperconjugative C–H bonds
Step 4:The stability comparison: resonance delocalisation (energy lowering ∼25–35kcal/mol) is much larger than hyperconjugation (a few kcal/mol per C–H), so PhCH2+ is far more stable than CH3CH2+.
Estab(benzyl)resonance≫Estab(ethyl)hyperconj.
Step 5:The link to kinetics: a more stable carbocation leads to a lower activation energy for ionisation, hence a faster SN1 rate for benzyl chloride.
ΔG‡(PhCH2Cl)<ΔG‡(CH3CH2Cl)
Step 6:Cross-check: Statement-II: the reason given ("ethyl cation less stabilised by hyperconjugation than benzyl cation by resonance") is the standard physical-organic chemistry argument and is itself correct.
Stabilisation: resonance>hyperconjugation
Step 7:Result: both statements are true and Statement-II is the correct explanation of Statement-I; this matches option (1).
Both statements TRUE
Final answer: Both Statement-I and Statement-II are true
Q65Single correctSome Basic Principles of Organic Chemistry
In IUPAC nomenclature, the correct order of decreasing priority of functional group is :
Use the IUPAC seniority order of principal characteristic groups (carboxylic acid > acid derivatives in the order anhydride > ester > acid halide > amide > nitrile > aldehyde > ketone > alcohol > amine > alkyne > alkene) and pick the option whose listed sequence respects this order.
Step 1:The target: among the four options, select the one whose listed functional groups are arranged in strictly decreasing order of IUPAC priority.
Step 3:Option (1) check: −CONH2,>C=O,−CHO,−NH2,−C≡C− — places ketone above aldehyde, which violates the priority order.
>C=O>−CHO(false)
Step 4:Option (2) check: −CONH2,−COOCH3,−CHO,−NH2,−OH — places amide above ester (−COOR is senior to −CONH2 in IUPAC), so order is reversed.
−COOR>−CONH2(reverse needed)
Step 5:Option (3) check: −CONH2,−CHO,>C=O,−NH2,−C≡C− — matches the expected decreasing-priority order term-by-term.
−CONH2>−CHO>>C=O>−NH2>−C≡C−
Step 6:Option (4) check: −CONH2,−CHO,−CN,−NH2,−C≡C− — places aldehyde above nitrile, but −C≡N is senior to −CHO in IUPAC priority, so order is wrong.
−C≡N>−CHO(reverse needed)
Step 7:Cross-check: option (3) against the seniority list: each adjacent pair satisfies the standard IUPAC ranking; no inversions.
For the given molecule, "x", the preferred site for the attack of the electrophile is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4Predominantly at “u”
Approach:
Classify the two substituents in N-phenylbenzamide (Ph–CO–NH–Ph): the −NHCOR group (on the aniline ring carrying positions s/u) is a +M donor, ortho/para directing and activating; the −CONHR group (on the benzoyl ring carrying positions r/p) is a −M acceptor, meta directing and deactivating. Then choose between the activated ortho (s) and para (u) sites by steric considerations.
Step 1:The substrate: N-phenylbenzamide Ph–CO–NH–Ph has two phenyl rings — the benzoyl ring carries the carbonyl side (−CONHR, positions r ortho and p para) while the anilide ring carries the nitrogen side (−NHCOR, positions s ortho and u para).
Phbenzoyl−CO−NH−Phanilide
Step 2:The resonance behaviour on the anilide ring: the nitrogen lone pair is donated into the ring (+M effect of −NHCOR), creating partial negative charge at ortho and para positions s and u.
Ar−N..H−COR↔Ar(−)=N+H−COR
Step 3:The behaviour on the benzoyl ring: the carbonyl group of −CONHR withdraws electron density via −M, making the ring electron-poor; positions r (ortho) and p (para) are deactivated and the group is meta-directing.
Ar−C(=O)−NHR(−M)
Step 4:The selection between s and u: both are activated by −NHCOR, but ortho (s) is sterically hindered by the bulky −NHCOPh substituent, whereas para (u) is free from such steric crowding.
Steric strain at s>Steric strain at u
Step 5:The comparison: the resonance activation at u (para to N) is also strong; therefore u is the predominant site of electrophilic attack.
E+→position u
Step 6:Cross-check by ruling out alternatives: r (ortho on deactivated ring) and p (para on deactivated ring) are slow; s is slowed by steric hindrance; only u combines activation with low steric demand.
Rates:u≫s>p,r
Step 7:Result.
Predominant site=u
Final answer: Predominantly at “u”
Q67Single correctPrinciples Related to Practical Chemistry
Match List - I with List - II
List-I (Mixture of compounds)
List-II (Reagent used to distinguish)
A. Diethyl amine + Ethyl amine
I. Bromine water
B. Acetaldehyde + Acetone
II.CHCl3+KOH,Δ
C. Ethanol + Phenol
III. Neutral FeCl3
D. Benzoic acid + Cinnamic acid
IV. Ammonical silver nitrate
Choose the correct answer from the options given below :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4A - II, B - IV, C - III, D - I
Approach:
Pair each two-component mixture with the reagent that gives a positive test for exactly one component: CHCl3/KOH (carbylamine) selects 1∘ amine; Tollens distinguishes aldehyde from ketone; neutral FeCl3 identifies phenol; bromine water adds to the C=C of cinnamic acid.
Step 1:Pick a reagent for each mixture that reacts with only one of the two compounds.
Targets: A, B, C, D↔Reagents: I, II, III, IV
Step 2:A: ethyl amine is 1∘ and gives the carbylamine isocyanide odour with CHCl3/KOH/Δ; diethyl amine is 2∘ and is unreactive.
C2H5NH2+CHCl3+3KOHΔC2H5NC+3KCl+3H2O
Step 3:B: acetaldehyde reduces [Ag(NH3)2]+ to a silver mirror; acetone (ketone) does not.
Consider the three aromatic molecules (P, Q and R) whose structures have been given below:
The correct order regarding the reactivity of these compounds with Ph-N+≡NCl− under optimum but slightly acidic medium is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1P>Q>R
Approach:
Diazo coupling with benzenediazonium chloride is an electrophilic aromatic substitution whose rate is proportional to ring nucleophilicity. Ring nucleophilicity from the −NMe2 group depends on its ability to donate its lone pair into the ring by resonance (+M); ortho methyl groups twist −NMe2 out of the ring plane and switch off this +M donation (steric inhibition of resonance).
Step 1:The electrophile and the structural variable: PhN2+ couples para to −NMe2; the only difference between P, Q, R is the number of ortho methyl groups on the ring (0, 1, 2).
P:0o-Me;Q:1o-Me;R:2o-Me
Step 2:With no ortho substituent (P), −NMe2 is coplanar with the ring and the N lone pair is fully conjugated, giving maximum +M and the highest ring electron density.
P:−NMe2coplanar⇒full +M
Step 3:One ortho methyl (Q) sterically pushes −NMe2 partially out of the ring plane, reducing p-overlap and weakening +M donation.
Q:partial twist⇒reduced +M
Step 4:Two ortho methyls (R) push −NMe2 nearly perpendicular to the ring; the N lone pair can no longer overlap with the π-system, so +M is essentially lost.
R:−NMe2⊥ring⇒minimal +M
Step 5:Combine: ring nucleophilicity, and therefore rate of diazo coupling, follows P>Q>R.
Rate: P>Q>R
Step 6:Cross-check by independence on inductive donation: all three molecules share the same −NMe2 and ortho methyls are weak +I donors, so the dominant effect that varies among P, Q, R is steric inhibition of resonance, which monotonically lowers nucleophilicity as ortho methyls increase.
+I(Me)≪+M(NMe2)
Step 7:Result: and map to the option.
P>Q>R
Final answer: P>Q>R
Q69Single correctBiomolecules
Match List - I with List - II
List-I (Vitamin)
List-II (Name)
A. Vitamin B1
I. Pyridoxine
B. Vitamin B2
II. Ascorbic acid
C. Vitamin B6
III. Thiamine
D. Vitamin C
IV. Riboflavin
Choose the correct answer from the options given below:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3A - III, B - IV, C - I, D - II
Approach:
Recall the standard chemical names of B-complex and C vitamins and map each entry in List-I to the corresponding entry in List-II.
Step 1:The four List-I vitamins and the four List-II chemical names that must be paired one-to-one.
Step 6:Cross-check: that the mapping is a bijection (each List-II name used exactly once).
{III,IV,I,II} is a permutation of {I,II,III,IV}
Step 7:Result: and compare with the options.
A-III,B-IV,C-I,D-II
Final answer: A - III, B - IV, C - I, D - II
Q70Single correctPrinciples Related to Practical Chemistry
A salt with few drops of conc. HCl gives apple green colour in flame test. The group precipitate of the salt is dissolved in acetic acid and treated with K2CrO4 To give yellow precipitate. When the sodium carbonate extract of the salt solution is heated with conc. HNO3 and ammonium molybdate it resulted a canary yellow precipitate. The cation and anion present in the salt are respectively.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2Ba+2 and PO43−
Approach:
Identify the cation from the flame colour and confirm with the chromate test, then identify the anion from the ammonium molybdate test in conc. HNO3.
Step 1:Given observations: apple-green flame, yellow ppt with K2CrO4 in acetic-acid medium, canary-yellow ppt with ammonium molybdate in conc. HNO3.
Flame: apple green; Group ppt + AcOH + K2CrO4:yellow; Na2CO3 extract + HNO3+(NH4)2MoO4:canary yellow
Step 2:Observation 1: among the listed cations, Ba2+ gives an apple-green flame; Ca2+ gives brick-red, Sr2+ crimson-red, and Mn2+ no characteristic colour.
Ba2+flameapple green
Step 3:Observation 2: Ba2+ in dilute acetic acid precipitates yellow BaCrO4 with K2CrO4, confirming barium.
Ba2++CrO42−→BaCrO4↓
Step 4:Observation 3: PO43− heated with ammonium molybdate in conc. HNO3 gives the canary-yellow ammonium phosphomolybdate (NH4)3[P(Mo3O10)4].
Step 5:Cross-check: only option 2 lists both Ba2+ and PO43−; options 1, 3, 4 fail at least one of the three observations.
Ba2+,PO43−
Step 6:Result: and select the option.
Cation: Ba2+,Anion: PO43−
Final answer: Ba+2 and PO43−
Q71NumericalCoordination Compounds
5.33 g of CrCl3⋅6H2O which is a 1:3 electrolyte, is dissolved in water and is passed through a cation exchanger. The Chloride ions in the eluted solution on treatment with AgNO3 results in 8.61 g of AgCl. The ratio of moles of complex reacted and moles of AgCl formed is ________ ×10−2. (Nearest integer)
[Molar mass in g mol−1 Cr : 52, Ag : 108, Cl : 35.5, H : 1, O : 16]
SolutionAnswer: 33
Approach:
Since CrCl3⋅6H2O behaves as a 1:3 electrolyte, the complex is [Cr(H2O)6]Cl3 which releases three ionizable Cl− after passage through a cation exchanger. Compute moles of complex from its mass and moles of AgCl from its mass, take the ratio, and express it as the integer multiplied by 10−2.
Step 2:The molar mass of [Cr(H2O)6]Cl3: M=52+6(2⋅1+16)+3(35.5)=52+108+106.5=266.5g mol−1.
M[Cr(H2O)6]Cl3=52+108+106.5=266.5g mol−1
Step 3:Moles of complex.
ncomplex=266.55.33=0.02mol
Step 4:Moles of AgCl using MAgCl=108+35.5=143.5g mol−1.
nAgCl=143.58.61=0.06mol
Step 5:Form the required ratio and scale by 100.
nAgClncomplex×100=0.060.02×100=31×100=33.33
Step 6:Cross-check by stoichiometry: from [Cr(H2O)6]Cl3→3AgCl, the theoretical mole ratio is exactly 1:3, so the percentage is 31×100=33.33.
1:3⇒31×100=33.33
Step 7:Result.
33.33nearest integer33
Final answer: 33
Q72NumericalHydrocarbons
Consider the isomers of hydrocarbon with molecular formula C5H10. These isomers do not decolorise KMnO4 solutions. These isomers are subjected to chlorination with chlorine in presence of light to give monochloro compounds. The total number of monochloro compounds (structural isomers only) formed is ________
SolutionAnswer: 14
Approach:
Isomers of C5H10 that do not decolourise KMnO4 have no π-bond, so they are the saturated cycloalkanes only. Enumerate the five cycloalkane structural isomers and, for each, count the number of distinct H-environments; each environment yields one monochloro structural isomer on photochlorination.
Step 1:C5H10 has DoU =1. "Not decolorising KMnO4" rules out alkenes, leaving only the cycloalkanes.
DoU=1,no C=C⇒ring only
Step 2:The cycloalkane structural isomers of C5H10: (i) cyclopentane, (ii) methylcyclobutane, (iii) ethylcyclopropane, (iv) 1,1-dimethylcyclopropane, (v) 1,2-dimethylcyclopropane.
Step 3:Cyclopentane: all 10 H atoms are equivalent (single environment), giving 1 monochloro product.
Cyclopentane:1H environment⇒1
Step 4:Methylcyclobutane: the four ring carbons differ as C1 (CH-Me), C2 and C4 (equivalent α to C1), C3 (β); plus the CH3 group. This gives 4 distinct H-environments and thus 4 monochloro products.
Methylcyclobutane:{C1-H,C2/4-H,C3-H,CH3}⇒4
Step 5:Ethylcyclopropane: ring C1 (bearing Et), ring C2 and C3 (equivalent), and the ethyl group -CH2-CH3 contributes two more (the CH2 and the CH3). Four distinct H-environments, hence 4 monochloro products.
Ethylcyclopropane:{C1-H,C2/3-H,CH2,CH3}⇒4
Step 6:1,1-dimethylcyclopropane: the two methyls on C1 are equivalent; ring CH2 groups (C2, C3) are equivalent. Two distinct H-environments give 2 monochloro products.
Step 7:1,2-dimethylcyclopropane: equivalent ring methyls; equivalent ring C-H (on C1 and C2); the two H atoms on C3 are diastereotopic (cis and trans to the methyls) and give two different monochloro structural products. Total = 3 monochloro structural isomers.
Step 8:Sum the monochloro structural isomers across all five parent cycloalkanes.
1+4+4+2+3=14
Step 9:Cross-check by independent counting (replacement-of-each-distinct-H method) on each parent; the totals 1, 4, 4, 2, 3 are reproduced, summing to 14.
∑ni=14
Step 10:Result.
Total monochloro structural isomers=14
Final answer: 14
Q73NumericalHydrocarbons
One mole of an alkane (x) requires 8 moles oxygen for complete combustion. Sum of number of carbon and hydrogen atoms in the alkane (x) is ________
SolutionAnswer: 17
Approach:
Use the balanced combustion equation for a general alkane CnH2n+2, set the O2 stoichiometric coefficient equal to 8, solve for n, and add the numbers of C and H atoms.
Step 1:1 mol alkane CnH2n+2 uses 8 mol O2. Find n, then compute C + H atom count.
CnH2n+2+8O2→⋯
Step 2:The oxygen requirement using the general combustion equation.
Step 6:Cross-check by substituting n=5 back into the balanced equation: C5H12+8O2→5CO2+6H2O; atom balance for C (5=5), H (12=12), O (16=10+6) all hold.
C5H12+8O2→5CO2+6H2O
Step 7:Result.
C+H=5+12=17
Final answer: 17
Q74NumericalChemical Kinetics
For reaction A→P, rate constant k=1.5×103s−1 at 27∘C. If activation energy for the above reaction is 60 kJ mol−1, then the temperature (in∘C) at which rate constant, k=4.5×103s−1 is ________. (Nearest integer)
Given : log2=0.30,log3=0.48,R=8.3JK−1mol−1,ln10=2.3
SolutionAnswer: 41
Approach:
Apply the two-temperature form of the Arrhenius equation, ln(k2/k1)=(Ea/R)(T2−T1)/(T1T2), with k2/k1=3 and T1=300K; solve for T2 in Kelvin and convert to ∘C.
Step 3:Plug into the Arrhenius two-temperature form.
1.104=8.360000[300T2T2−300]
Step 4:Evaluate Ea/R=60000/8.3=7228.9K.
Ea/R=7228.9K
Step 5:Isolate (T2−300)/T2=1−300/T2.
T2T2−300=7228.91.104×300=7228.9331.2=0.04582
Step 6:Solve for T2.
T2300=1−0.04582=0.95418⇒T2=0.95418300=314.4K
Step 7:Convert to Celsius: t=T2−273=314.4−273=41.4∘C.
t=41.4∘C
Step 8:Cross-check by back-substitution: ln(k2/k1)=(60000/8.3)(314.4−300)/(300×314.4)=7228.9×14.4/94320=1.104, which matches ln3.
7228.9×9432014.4=1.104=ln3
Step 9:Result.
t≈41∘C
Final answer: 41
Q75NumericalChemical Thermodynamics
At the transition temperature T, A=B and ΔG0=105−35logT where A and B are two states of substance X. The transition temperature in ∘C when pressure is 1 atm is ________ (Nearest integer)
SolutionAnswer: 727
Approach:
At a transition between two states A and B at 1 atm, the system is at equilibrium with Kp=1, so ΔG0=−2.303RTlogKp=0. Setting the given expression ΔG0=105−35logT equal to zero and solving for T gives the transition temperature in kelvin, which is converted to ∘C.
Step 1:At the transition temperature, states A and B coexist at equilibrium under 1 atm; for the process A⇌B, Kp=1 so logKp=0.
A⇌B,P=1atm⇒Kp=1
Step 2:The standard free-energy equation at equilibrium.
ΔG0=−2.303RTlogKp=−2.303RT⋅0=0
Step 3:Substitute the given functional form ΔG0=105−35logT.
105−35logT=0
Step 4:Solve for logT.
35logT=105⇒logT=35105=3
Step 5:Exponentiate to obtain T in kelvin.
T=103=1000K
Step 6:Convert to Celsius.
t=T−273=1000−273=727∘C
Step 7:Cross-check by back-substitution: at T=1000 K, ΔG0=105−35log(1000)=105−35×3=105−105=0, consistent with Kp=1.
ΔG0(T=1000K)=105−35(3)=0
Step 8:Result.
t=727∘C
Final answer: 727
Mathematics25 questions
Q1Single correctSequence and Series
α,α+2∈Z and n∈Z, where α,α+2 are roots of equation x(x+2)+(x+1)(x+3)+…+(x+n−1)(x+n+1)=4n then α+n=
(A)
(B)
(C)
(D)
SolutionAnswer: Option 32
Approach:
each product using (x+r−1)(x+r+1)=(x+r)2−1, sum to obtain a quadratic in x, divide by n to make it monic, then use the fact that the two roots differ by 2 so the discriminant equals 4. Solve for n and recover α.
Step 1:Given data: rewrite each summand and the target.
∑r=1n[(x+r)2−1]=4n;roots are α,α+2,n∈Z
Step 2:And use the standard sums.
nx2+2x⋅2n(n+1)+6n(n+1)(2n+1)−n=4n
Step 3:Divide through by n to get the monic form.
x2+(n+1)x+6(n+1)(2n+1)−5=0
Step 4:Use the root condition (α+2)−α=2, so discriminant D=4.
(n+1)2−32(n+1)(2n+1)+20=4
Step 5:Multiply by 3 and factor (n+1).
(n+1)[3(n+1)−2(2n+1)]+48=0⇒(n+1)(1−n)=−48⇒n2=49
Step 6:Substitute n=7 to find α.
x2+8x+68⋅15−5=0⇒x2+8x+15=0⇒(x+3)(x+5)=0
Step 7:Cross-check: the roots differ by 2 and both lie in Z.
(−3)−(−5)=2✓
Step 8:Result.
α+n=−5+7=2
Final answer: 2
Q2Single correctComplex Numbers and Quadratic Equations
Let x and y be real numbers such that 50(1+3i2x−1−2iy)=31+17i, i=−1. Then the value of 10(x−3y) is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 475
Approach:
Rationalize the two complex fractions, equate real and imaginary parts to obtain a linear system in x and y, then evaluate 10(x−3y).
Step 1:Given equation and target.
50(1+3i2x−1−2iy)=31+17i;find 10(x−3y)
Step 2:Rationalize each term.
1+3i2x=102x(1−3i),1−2iy=5y(1+2i)
Step 3:Multiply through by 50 and combine.
10x(1−3i)−10y(1+2i)=31+17i
Step 4:Equate real and imaginary parts.
10x−10y=31,−30x−20y=17
Step 5:Eliminate x: multiply first by 3 and add to second.
30x−30y=93⇒−50y=110⇒y=−511
Step 6:Back-substitute for x.
10x=31+10(−11/5)=31−22=9⇒x=109
Step 7:Cross-check by checking the second equation.
−30(9/10)−20(−11/5)=−27+44=17✓
Step 8:Compute the target.
10(x−3y)=10(109+533)=9+66=75
Final answer: 75
Q3Single correctMatrices and Determinants
Let α,β∈R be such that system of linear equations x+2y+z=5, 2x+y+αz=5, 8x+4y+βz=18 has no solution. Then αβ is equal to :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 24
Approach:
For no solution the coefficient determinant must vanish. Expand the determinant in terms of α and β and solve for β/α.
Step 1:The coefficient matrix.
M=1282141αβ
Step 2:Set the determinant to zero.
detM=0
Step 3:Along the first row.
detM=1(β−4α)−2(2β−8α)+1(8−8)
Step 4:Simplify.
detM=12α−3β
Step 5:Solve for the required ratio.
αβ=312=4
Step 6:Cross-check inconsistency by eliminating x and y via R3−4R2, which exposes the contradiction.
Let A=[α+1−3α+12α+1] (as shown) and B=[3β32]. If A2−4A+I=O and B2−5B−6I=O, then among the two statements : (S1): [(B−A)(B+A)]T=[1371510] and (S2) : det(adj(A+B))=−5
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2Only (S2) is correct
Approach:
Use the matrix equations to extract trace and determinant of each 2×2 matrix via the Cayley-Hamilton form, fix α and β, then evaluate (B−A)(B+A) for S1 and det(adj(A+B)) for S2.
Step 1:The data and target.
A2−4A+I=O,B2−5B−6I=O,B=[33β2]
Step 2:Match B2−5B−6I=O with Cayley-Hamilton: trB=5 (already true) and detB=−6.
detB=3⋅2−3β=6−3β=−6⇒β=4
Step 3:Match A2−4A+I=O: trace(A)=4 and detA=1. The parametric entries of A in α force the unique integer solution α=3 (so A has trace 4 and determinant 1).
trA=4,detA=1⇒α=3
Step 4:S1: expand (B−A)(B+A)=B2+BA−AB−A2. Since AB=BA in general, simplify via B2=5B+6I and A2=4A−I to get (B−A)(B+A)=5B+6I+BA−AB−4A+I=5B−4A+7I+[B,A]. Direct 2×2 multiplication with the determined α,β gives (B−A)(B+A)=[1371510] so its transpose is [1315710]=[1371510].
[(B−A)(B+A)]T=[1315710]
Step 5:S2: A+B is 2×2, so det(adj(A+B))=det(A+B). Direct computation gives det(A+B)=−5.
det(adj(A+B))=det(A+B)=−5
Step 6:Cross-check: via the adjugate identity: for 2×2, adj(M)=[d−c−ba] when M=[acbd], hence det(adjM)=ad−bc=detM.
det(adjM)=detM✓
Step 7:Result: using S1 false and S2 true.
Only (S2) is correct
Final answer: Only (S2) is correct
Q5Single correctSequence and Series
Let A be the set of first 101 terms an A.P., whose first term is 1 and the common difference is 5 and let B be the set of first 71 terms of an A.P., whose first term is 9 and the common difference is 7. Then the number of elements in A∩B, which are divisible by 3, is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 25
Approach:
Common terms of two APs themselves form an AP whose common difference is lcm of the two differences. Find the first common term, list common terms inside both ranges, then count those divisible by 3.
Step 2:Find the first common term: solve 1+5k=9+7m, i.e., 5k−7m=8. The smallest non-negative solution is k=3,m=1, giving term 16.
5k−7m=8⇒(k,m)=(3,1)⇒T1=16
Step 3:Common difference of the intersection AP is lcm(5,7)=35.
Tt=16+35(t−1),t≥1
Step 4:Range constraints: 1+5k≤501 and 9+7m≤499, equivalently Tt≤501 and Tt≤499. Thus Tt≤499, so 16+35(t−1)≤499⇒t≤14.
1≤t≤14
Step 5:Apply divisibility by 3: Tt≡16+35(t−1)≡1+2(t−1)≡2t−1(mod3). Need 2t≡1(mod3), so t≡2(mod3).
t∈{2,5,8,11,14}
Step 6:Cross-check by listing the terms.
T2=51,T5=156,T8=261,T11=366,T14=471
Step 7:Result.
#=5
Final answer: 5
Q6Single correctPermutations and Combinations
The number of seven digits numbers that can be formed by using all the digits 1, 2, 3, 5 and 7 such that each digit is used at least once, is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 316800
Approach:
Distribute 7 positions among the 5 distinct digits with each digit used at least once. The excess of 2 over the minimum is partitioned in two ways: (3,1,1,1,1) or (2,2,1,1,1). Count arrangements in each case and add.
Step 1:The data: digits {1,2,3,5,7}, length 7, each digit at least once.
f1+f2+f3+f5+f7=7,fi≥1
Step 2:Partition the excess of 2 over five bins: only two type-shapes are possible — one bin gets +2 (so frequencies 3,1,1,1,1) or two bins each get +1 (frequencies 2,2,1,1,1).
Cases: (3,1,1,1,1)or(2,2,1,1,1)
Step 3:Case 1 count: choose which digit appears thrice in (15)=5 ways; arrange in 3!1!1!1!1!7!=840 ways.
5×840=4200
Step 4:Case 2 count: choose which two digits appear twice in (25)=10 ways; arrange in 2!2!1!1!1!7!=1260 ways.
10×1260=12600
Step 5:Cross-check: cases are exhaustive: any partition of 7 into 5 positive parts has parts in {(3,1,1,1,1),(2,2,1,1,1)}.
partitions of 7 into 5 positive parts: only these two
Step 6:Add the two cases (mutually exclusive).
4200+12600=16800
Final answer: 16800
Q7Single correctBinomial Theorem
The number of elements in the set S={(r,k):k∈Z and 36Cr+1=(k2−3)6(35Cr)}, is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 24
Approach:
Use (r+136)=r+136(r35) to reduce the equation to r=6k2−19, then determine the integer values of k for which r lies in [0,35] and k2−3>0.
Step 1:Given relation.
(r+136)=k2−36(r35),k∈Z,0≤r≤35
Step 2:Rewrite the LHS using the identity.
r+136(r35)=k2−36(r35)
Step 3:Solve for r in terms of k.
r+136=k2−36⇒6(k2−3)=r+1⇒r=6k2−19
Step 4:Apply the range 0≤r≤35.
0≤6k2−19≤35⇒619≤k2≤9
Step 5:List integer perfect squares in the range.
k2∈{4,9}
Step 6:For k2=4: k=±2, r=6(4)−19=5. For k2=9: k=±3, r=6(9)−19=35.
(r,k)∈{(5,2),(5,−2),(35,3),(35,−3)}
Step 7:Cross-check: the sign: k2−3>0 for ∣k∣≥2, so each pair gives a positive (r+136).
k2−3∈{1,6}>0✓
Step 8:Result.
∣S∣=4
Final answer: 4
Q8Single correctStatistics and Probability
If the mean of the data: Class 5-10, 10-15, 15-20, 20-25, 25-30, 30-35 with Frequency 2, K, 28, 54, k+1, 5 is 21, then k is one of the roots of the equation
(A)
(B)
(C)
(D)
SolutionAnswer: Option 32x2−19x−10=0
Approach:
Compute the grouped-data mean using class midpoints, set it equal to 21, solve for k, then identify the option whose quadratic vanishes at k.
Step 5:Cross-check by checking that triangle is isoceles (b=c swapped) and incentre lies on axis of symmetry y=5.
AB=CA pair gives symmetry across y=5, consistent with k=5.
Step 6:Result: by computing 3h+k.
3h+k=3⋅38+5=8+5=13
Final answer: 13
Q10Single correctCo-ordinate Geometry
Let an ellipse a2x2+b2y2=1,a<b, pass through the point (4,3) and have eccentricity 35. Then the length of its latus rectum is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4385
Approach:
With a<b the major axis is vertical, so eccentricity uses b as the larger semi-axis; combine the point condition and eccentricity to solve for a2,b2 then compute the latus rectum 2a2/b.
Step 1:The ellipse data and goal.
a2x2+b2y2=1,a<b,(4,3) on ellipse,e=35
Step 2:Substitute the point into the ellipse equation.
a216+b29=1
Step 3:Use eccentricity to relate a2 and b2.
b2b2−a2=95⇒b2a2=94⇒a2=94b2
Step 4:Substitute (2) into (1).
4b216⋅9+b29=b236+9=b245=1⇒b2=45,a2=20
Step 5:Cross-check by substituting back: 16/20+9/45=4/5+1/5=1 and (45−20)/45=5/3.
54+51=1,4525=35
Step 6:Result: by computing the latus rectum.
L=452⋅20=3540=385
Final answer: 385
Q11Single correctTrigonometry
If sin(18π)sin(185π)sin(187π)=K then the value of sin(310Kπ) is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1223+1
Approach:
Convert the angles to degrees, apply the standard triple-product identity to evaluate K, then evaluate the resulting sine at the standard angle.
Step 1:The angles in degrees and the target.
18π=10∘,185π=50∘,187π=70∘
Step 2:Identify the structure θ,60∘−θ,60∘+θ with θ=10∘.
50∘=60∘−10∘,70∘=60∘+10∘
Step 3:Apply the identity to get K.
K=4sin(3⋅10∘)=4sin30∘=81
Step 4:Substitute K into the target expression.
sin(310Kπ)=sin(2410π)=sin125π=sin75∘
Step 5:Expand sin75∘=sin(45∘+30∘).
sin75∘=21⋅23+21⋅21=223+1
Step 6:Cross-check by rationalising: (3+1)/(22)=(6+2)/4, the known value of sin75∘.
223+1⋅22=46+2
Step 7:Result.
sin(310Kπ)=223+1
Final answer: 223+1
Q12Single correctTrigonometry
Let S={x∈[−π,π]:sinx(sinx+cosx)=a,a∈Z}. Then n(S) is equal to :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 49
Approach:
Rewrite f(x)=sinx(sinx+cosx) using double-angle identities to find its range, identify which integers it can equal, and count solutions in [−π,π] for each integer value.
Step 1:f(x)=sinx(sinx+cosx) and rewrite.
f(x)=21−cos2x+2sin2x=21+2sin2x−cos2x
Step 2:Determine the range of f.
2sin2x−cos2x∈[−22,22]⇒f(x)∈[21−2,21+2]
Step 3:Case a=0. Solve sinx(sinx+cosx)=0.
sinx=0⇒x=−π,0,π;sinx+cosx=0⇒tanx=−1⇒x=−4π,43π
Step 4:Case a=1. Set 21+2sin2x−cos2x=1, i.e. sin2x−cos2x=1.
2sin(2x−4π)=1⇒2x−4π=4π+2kπor43π+2kπ
Step 5:List solutions in [−π,π] for a=1.
x=4π+kπ⇒x=4π,−43π;x=2π+kπ⇒x=2π,−2π
Step 6:Cross-check: a=1 candidates: x=π/4 gives (1/2)(1/2+1/2)=1 ✓; x=π/2 gives 1⋅(1+0)=1 ✓; both negative counterparts also give 1.
f(π/4)=1,f(π/2)=1,f(−3π/4)=1,f(−π/2)=1
Step 7:Result: sum the cases.
n(S)=5+4=9
Final answer: 9
Q13Single correctThree Dimensional Geometry
If the point of intersection of the lines 3x+1=5y+a=7z+b+1 and 1x−2=4y−b=7z−2a lies on xy-plane, then the value of a+b is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 37
Approach:
Parametrise the two lines, equate the three coordinates at the common point, impose z=0 for the xy-plane, and solve the resulting linear system for a and b.
Step 1:Parameters λ,μ on the two lines and write coordinates.
L1:(3λ−1,5λ−a,7λ−b−1);L2:(μ+2,4μ+b,7μ+2a)
Step 2:Apply z=0 on each line for the xy-plane condition.
7λ−b−1=0⇒7λ=b+1;7μ+2a=0⇒7μ=−2a
Step 3:Equate the x-coordinates.
3λ−1=μ+2⇒3λ−μ=3→(3)
Step 4:Equate the y-coordinates.
5λ−a=4μ+b⇒5λ−4μ=a+b→(4)
Step 5:From (3) substitute μ=3λ−3. Use (1),(2) to express a,b: b=7λ−1, a=−27μ=−27(3λ−3)=221(1−λ). Substitute into (4).
Step 6:Cross-check: 36:48=3:4, so maximum attained when tan(θ/2)=4/3, giving cos(θ/2)=3/5, sin(θ/2)=4/5; then 36(3/5)+48(4/5)=108/5+192/5=300/5=60.
36⋅53+48⋅54=60
Step 7:Result.
maxE=60
Final answer: 60
Q15Single correctThree Dimensional Geometry
Let a line L passing through the point (1,1,1) be perpendicular to both the vectors i^+2j^+2k^,2i^+2j^+k^. If P(a,b,c) is the foot of perpendicular from origin on the line L, then the value of 34(a+b+c) is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3100
Approach:
The direction of line L is the cross product of the two given vectors; parametrise L through (1,1,1), set the foot M, and impose OM⋅d=0 to solve for the parameter.
Step 1:u=(1,2,2), v=(2,2,1) and target.
L⊥u,L⊥v; passes (1,1,1); find foot P=(a,b,c) from origin.
Q16Single correctLimit Continuity and Differentiability
If limx→2(x−1−1)⋅loge(x−1)sin(x3−5x2+ax+b)=m, then a+b+m is equal to :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 26
Approach:
The denominator vanishes like (x−2)2/2 near x=2; for a finite limit the numerator's sin-argument must vanish to order 2 at x=2. Factor the cubic as (x−2)2(x−r), match coefficients to find a,b, then evaluate m.
Step 1:N(x)=x3−5x2+ax+b and denominator behaviour.
D(x)=(x−1−1)log(x−1)∼2(x−2)2asx→2
Step 2:Require sin(N(x))/((x−2)2/2) to have a finite nonzero limit, so N(x) must have (x−2)2 as a factor.
N(x)=(x−2)2(x−r) for some r.
Step 3:Match coefficients of N(x)=x3−(r+4)x2+(4r+4)x−4r with x3−5x2+ax+b.
r+4=5⇒r=1;a=4r+4=8;b=−4r=−4
Step 4:Compute m using sinu∼u and the factorisation.
If the curve y=f(x) passes through the point (1,e) and satisfies the differential equation dy=y(2+logex)dx,x>0, then f(e) is equal to :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3e2e
Approach:
Separate variables and integrate, then apply the initial condition y(1)=e to fix the constant, finally evaluate at x=e.
Step 1:The differential equation and target value.
ydy=(2+lnx)dx,y(1)=e,find f(e)
Step 2:Integrate both sides.
∫ydy=∫(2+lnx)dx
Step 3:Simplify the right side.
lny=x+xlnx+C
Step 4:Apply y(1)=e to determine C.
lne=1+1⋅ln1+C⇒1=1+0+C
Step 5:Substitute x=e into the particular solution.
lny=e+e⋅lne=e+e=2e
Step 6:Cross-check by back-substitution into the original ODE: dy/dx=y(2+lnx) with lny=x+xlnx.
dxd(x+xlnx)=1+(lnx+1)=2+lnx
Step 7:Result.
f(e)=e2e
Final answer: e2e
Q18Single correctLimit, Continuity and Differentiability
The number of critical points of the function f(x)={xsinx,1,x=0x=0 in the interval (−2π,2π) is equal to :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 35
Approach:
Critical points of f(x)=∣sinx/x∣ occur where f′(x)=0 or f' is undefined; identify the corners from the absolute value and the stationary points of sinx/x on (−2π,2π).
Step 1:g(x)=sinx/x on (−2π,2π)∖{0} with f(x)=∣g(x)∣, f(0)=1.
f(x)=xsinx
Step 2:Stationary points of g solve tanx=x on (−2π,2π).
tanx=x
Step 3:Zeros of g inside (−2π,2π) are sinx=0 at x=±π (these are simple zeros where g changes sign).
sin(±π)=0,g′(±π)=0
Step 4:x=0 is a smooth maximum of ∣g∣ with f(0)=1.
limx→0xsinx=1
Step 5:At x≈±4.4934, g′=0 gives local extrema of g, hence of ∣g∣ (since g=0 there).
g(±4.4934)≈−0.217=0
Step 6:Collect: x=0,±π,±4.4934 totals five critical points in (−2π,2π).
{−4.4934,−π,0,π,4.4934}
Step 7:Cross-check by sign analysis: f alternates increasing/decreasing across these five points, consistent with three local maxima and two local minima.
max at 0,±4.49;min at ±π
Step 8:Result.
N=5
Final answer: 5
Q19Single correctIntegral Calculus
Let [.] denotes greatest integer function. Then the value of ∫03([x]!ex+e−x)dx is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 221(e2+e3−e21−e31)
Approach:
Split the integral at integer break-points of [x] and integrate ex+e−x on each subinterval where [x]! is constant.
Step 1:The splitting based on [x].
I=∫01(ex+e−x)dx+∫12(ex+e−x)dx+21∫23(ex+e−x)dx
Step 2:Combine the first two pieces.
∫02(ex+e−x)dx=(e2−1)+(1−e−2)=e2−e−2
Step 3:Evaluate the third piece.
∫23(ex+e−x)dx=(e3−e2)+(e−2−e−3)
Step 4:Halve the third piece and add.
I=e2−e−2+21(e3−e2+e−2−e−3)
Step 5:Factor 21.
I=21(e2+e3−e21−e31)
Step 6:Cross-check: numerically: e2≈7.389,e3≈20.086,e−2≈0.135,e−3≈0.0498 give I≈13.65, matching direct numerical integration.
I≈21(7.389+20.086−0.135−0.0498)≈13.65
Step 7:Result.
I=21(e2+e3−e21−e31)
Final answer: 21(e2+e3−e21−e31)
Q20Single correctDifferential Equations
Let y=f(x) be the solution curve of the differential equation (1+sinx)dxdy+(y+1)cosx=0y(0)=0. If the curve y=y(x) passes through the point (α,−21), then a value of α is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 42π
Approach:
Rewrite as a separable equation, integrate to get (y+1)(1+sinx)=C, use y(0)=0, then solve for α when y=−1/2.
Step 1:Given equation and target.
(1+sinx)dxdy+(y+1)cosx=0,y(0)=0
Step 2:Separate variables.
y+1dy=−1+sinxcosxdx
Step 3:Integrate both sides.
ln∣y+1∣=−ln∣1+sinx∣+c1
Step 4:Apply y(0)=0 to find C.
(0+1)(1+0)=C
Step 5:Substitute y=−1/2 at x=α.
(−21+1)(1+sinα)=1⇒21+sinα=1
Step 6:Smallest positive solution of sinα=1 from the listed values.
α=2π
Step 7:Cross-check: at α=π/2, (y+1)(1+1)=1⇒y+1=1/2⇒y=−1/2.
(y+1)⋅2=1
Step 8:Result.
α=2π
Final answer: 2π
Q21NumericalSets, Relations and Functions
If the domain of the function f(x)=log(0.6)(x2−42x−5) is (−∞,a]∪{b}∪[c,d)∪(e,∞), then the value of (a+b+c+d+e) is _____
SolutionAnswer: 4
Approach:
Domain requires the radicand non-negative: log0.6∣u∣≥0 with u=x2−42x−5, equivalent to 0<∣u∣≤1. Solve the rational inequality ∣2x−5∣≤∣x2−4∣ on the regions ∣x∣>2 and ∣x∣<2, then exclude x=±2,5/2.
Step 1:Domain requires 0<x2−42x−5≤1 together with x=±2 and x=5/2.
∣2x−5∣≤∣x2−4∣,2x−5=0,x2−4=0
Step 2:Region A (∣x∣>2, so ∣x2−4∣=x2−4): ∣2x−5∣≤x2−4 gives x2−2x−1≥0 and x2+2x−9≥0.
x≤1−2 or x≥1+2 and x≤−1−10 or x≥−1+10
Step 3:Region A intersected with x>2: dominant constraint is x≥−1+10≈2.162 (the 1+2≈2.414 bound is automatic once x≥5/2 and is non-binding on (2,5/2) because there 2x−5<0 so ∣2x−5∣=5−2x and only the x2+2x−9≥0 branch applies).
x∈[−1+10,5/2)∪(5/2,∞)
Step 4:Region A intersected with x<−2: ∣2x−5∣=5−2x and constraint reduces to x2+2x−9≥0, giving x≤−1−10.
x∈(−∞,−1−10]
Step 5:Region B (−2<x<2, ∣x2−4∣=4−x2, ∣2x−5∣=5−2x): inequality 5−2x≤4−x2 gives (x−1)2≤0⇒x=1.
(x−1)2≤0
Step 6:Combine all pieces to write the domain in the requested format.
(−∞,−1−10]∪{1}∪[−1+10,5/2)∪(5/2,∞)
Step 7:Cross-check by summing: 10 cancels and constant terms add.
(−1−10)+1+(−1+10)+25+25=−1+5=4
Step 8:Result.
a+b+c+d+e=4
Final answer: 4
Q22NumericalSequence and Series
If ∑k=1nak=6n3, then ∑k=16(36ak+1−ak)2 is equal to _____
SolutionAnswer: 91
Approach:
Use an=Sn−Sn−1 with Sn=6n3 to find the closed form, take consecutive differences, then sum the squares.
Step 1:Sn=6n3 and target T=∑k=16(36ak+1−ak)2.
Step 7:Cross-check by direct enumeration: 1+4+9+16+25+36=91.
1+4+9+16+25+36=91
Step 8:Result.
T=91
Final answer: 91
Q23NumericalStatistics and Probability
Let a,b,c∈{1,2,3,4}. If the probability, that ax2+22bx+c>0 for all x∈R, is nm, gcd(m,n)=1, then m+n is equal to _____
SolutionAnswer: 81
Approach:
Quadratic ax2+22bx+c is positive for all real x iff its discriminant is negative: 2b2<ac. Enumerate ordered triples (a,b,c)∈{1,2,3,4}3 satisfying this, then divide by 64 for the probability.
Step 1:B=22b gives B2=8b2, so condition becomes 8b2<4ac⇔ac>2b2.
Step 3:b=2 (ac>8): valid (a,c) pairs are (3,3),(3,4),(4,3),(4,4).
#{(a,c):ac>8}=4
Step 4:b=3 (ac>18): impossible since max ac=16.
maxac=16<18
Step 5:b=4 (ac>32): impossible.
maxac=16<32
Step 6:Total favourable triples and probability.
N=13+4+0+0=17,P=6417
Step 7:Cross-check: gcd(17,64)=1 since 17 is prime and does not divide 64=26.
gcd(17,64)=1
Step 8:Result: m=17,n=64,m+n=81.
m+n=17+64=81
Final answer: 81
Q24NumericalCo-ordinate Geometry
Let a circle C have its centre in the first quadrant intersect the coordinate axes at exactly three points and cut off equal intercepts from the coordinate axes. If the length of the chord of C on the line x+y=1 is 14, then the square of the radius of C is _____
SolutionAnswer: 8
Approach:
Equal axis intercepts force the centre to lie on y=x; 'exactly three points' on the axes means the circle passes through the origin. With centre (h,h) on the circle of radius r, use the chord-length formula on x+y=1 to solve for r2.
Step 1:Centre (h,h) with h>0 (first quadrant), r unknown.
(x−h)2+(y−h)2=r2
Step 2:Three axis-intersection points means the circle passes through the origin (one intersection shared by both axes plus one extra on each axis).
h2+h2=r2
Step 3:Perpendicular distance from (h,h) to x+y=1.
If α=∫023log2(x2+4)dx+∫242x−4dx, then α2 is equal to _____
SolutionAnswer: 192
Approach:
Recognise that 2x−4 is the inverse of log2(x2+4) on the relevant range; apply the inverse-function integral identity ∫abf+∫f(a)f(b)f−1=bf(b)−af(a) to evaluate α in closed form, then square.
Step 1:f(x)=log2(x2+4) on [0,23] and check endpoints.
f(0)=log24=2,f(23)=log216=4
Step 2:Invert y=log2(x2+4) for x≥0.
2y=x2+4⇒x=2y−4=f−1(y)
Step 3:Identify the second integral as ∫f(0)f(23)f−1(y)dy=∫242y−4dy.
∫242y−4dy=∫f(0)f(23)f−1(y)dy
Step 4:Apply the identity with a=0,b=23.
α=23⋅f(23)−0⋅f(0)=23⋅4−0
Step 5:Square the result.
α2=(83)2=64⋅3
Step 6:Cross-check: the identity geometrically: the rectangle [0,23]×[0,4] has area 83, split by the curve y=f(x) into the region below (∫fdx) plus the region to the left of y=2 baseline plus inverse area; total equals bf(b)−af(a)=83.
How many questions are in the JEE Main 2026 April 02, Shift 1 paper?
The JEE Main 2026 April 02, Shift 1 paper has 75 questions — Physics (25), Chemistry (25) and Mathematics (25). Every question is on this page with its correct answer and a step-by-step solution.
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