JEE Main 2026 April 04, Shift 2 Question Paper with Solutions
All 75 questions from the JEE Main 2026 (April 04, Shift 2) shift — Physics (25), Chemistry (25) and Mathematics (25) — each with the correct answer (answer key) and a step-by-step solution. Read it free online, or attempt the full paper as a timed mock test.
Derive the dimensional formula of each List-I quantity and match with List-II.
Step 1:From E=hν, the dimension of Planck's constant equals energy divided by frequency.
[h]=[ν][E]=T−1ML2T−2=ML2T−1
Step 2:Stopping potential carries the dimension of electric potential; substituting charge as AT.
[Vs]=[q][W]=ATML2T−2=ML2T−3A−1
Step 3:Work function is an energy quantity.
[ϕ]=ML2T−2
Step 4:Threshold frequency is a frequency, with the dimension of inverse time.
[ν0]=T−1
Final answer: A-III, B-IV, C-I, D-II
Q27Single correctKinematics
Two cars A and B are moving in the same direction along a straight line with speeds 100km/h and 80km/h, respectively such that car A is moving ahead of car B. A person in car B throws a stone with a speed v so that it hits the car A with a speed of 5m/s. The value of v is ____ km/h.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 338
Approach:
Apply one-dimensional relative-velocity analysis along the line of travel and convert the impact speed into the same unit (km/h).
Step 1:Set the forward direction (B toward A) as positive. The stone is thrown from B with speed v relative to B, so its ground speed becomes vS=80+v km/h.
vS=80+vkm/h
Step 2:Compute the speed of the stone relative to car A by subtracting A's ground speed.
vS/A=(80+v)−100=(v−20)km/h
Step 3:Convert the impact speed 5m/s to km/h: 5×3.6=18km/h, and equate.
v−20=18⇒v=38km/h
Final answer: 38
Q28Single correctLaws of Motion
At t = 0, a body of mass 100 g starts moving under the influence of a force (5i^+10j^)N from the origin. After 2 s its position is (2xi^+5yj^)m. The ratio x : y is ____ .
(A)
(B)
(C)
(D)
SolutionAnswer: Option 45 : 4
Approach:
Compute the constant acceleration from F=ma, find the position vector at t=2 s using r=21at2 (rest start at origin), and match component-wise.
Step 1:Convert mass to SI and compute acceleration with m=0.1 kg.
a=0.15i^+10j^=(50i^+100j^)m/s2
Step 2:Substitute t=2 s into r(t)=21at2 starting from rest at the origin.
r=21(50i^+100j^)(2)2=(100i^+200j^)m
Step 3:Equate components with (2xi^+5yj^): 2x=100⇒x=50; 5y=200⇒y=40.
x:y=50:40=5:4
Final answer: 5 : 4
Q29Single correctKinematics
If x and y coordinates of a projectile as a function of time (t) are given as 24t and 43.6t−4.9t2, respectively, then the angle (in degrees) made by the projectile with horizontal when t = 2s is ____ .
(A)
(B)
(C)
(D)
SolutionAnswer: Option 245
Approach:
Differentiate the position coordinates to obtain velocity components, evaluate at t=2 s, and apply tanθ=vy/vx.
Step 1:Differentiate x(t)=24t and y(t)=43.6t−4.9t2 with respect to t.
vx=24m/s,vy=43.6−9.8tm/s
Step 2:Substitute t=2 s.
vx=24m/s,vy=43.6−9.8(2)=24m/s
Step 3:Apply the direction relation.
tanθ=2424=1⇒θ=45∘
Final answer: 45
Q30Single correctGravitation
The height in terms of radius of the earth (R), at which the acceleration due to gravity becomes 9g, where g is acceleration due to gravity on earth's surface, is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 32R
Approach:
Apply the altitude variation of g, gh=g/(1+h/R)2, and solve for h.
Step 1:Set gh=g/9 in the altitude formula.
(1+h/R)2g=9g
Step 2:Cancel g and take the positive square root.
(1+Rh)2=9⇒1+Rh=3
Step 3:Solve for h.
Rh=2⇒h=2R
Final answer: 2R
Q31Single correctProperties of Solids and Liquids
A metal string A is suspended from a rigid support and its free end is attached to a block of mass M. Second block having mass 2M is suspended at the bottom of the first block using a string B. The area of cross sections of strings A and B are same. The ratio of lengths of strings of A to B is 2 and the ratio of their Young's moduli (YA/YB) is 0.5. The ratio of elongations in A to B is ____ .
(A)
(B)
(C)
(D)
SolutionAnswer: Option 46
Approach:
Compute the tension in each string from free-body considerations, then apply ΔL=FL/(AY) and form the ratio.
Step 1:String A supports both blocks, so its tension is FA=(M+2M)g=3Mg. String B supports only the lower block, so FB=2Mg.
FA=3Mg,FB=2Mg
Step 2:Write the ratio ΔLA/ΔLB using equal cross-sectional areas.
ΔLBΔLA=FBLBFALA⋅YAYB
Step 3:Substitute LA/LB=2 and YA/YB=0.5 (so YB/YA=2).
ΔLBΔLA=2Mg3Mg⋅2⋅2=23⋅4=6
Final answer: 6
Q32Single correctProperties of Solids and Liquids
A water spray gun is attached to a hose of cross sectional area 30cm2. The gun comprises of 10 perforations each of cross sectional area 15mm2. If the water flows in the hose with the speed of 50cm/s, calculate the speed at which the water flows out from each perforation. (Neglect any edge effects)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 210 m/s
Approach:
Apply the continuity equation for incompressible flow; the total outlet area equals 10 times the area of a single perforation.
Step 1:Convert areas to a common unit. Hose area: A1=30cm2=3000mm2. Total perforation area: A2=10×15=150mm2.
A1=3000mm2,A2=150mm2
Step 2:Convert hose speed: v1=50cm/s=0.5m/s, then apply continuity for the exit speed.
v2=A2A1v1=1503000×0.5=20×0.5=10m/s
Step 3:Since the perforations are identical, each outlet carries the same speed.
vper hole=10m/s
Final answer: 10 m/s
Q33Single correctKinetic Theory of Gases
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R
Assertion A: If the average kinetic energy of H2 and O2 molecules, kept in two different sized containers are same, then their temperatures will be same.
Reason R: The r.m.s. speed of H2 and O2 molecules are same at same temperature.
Choose the correct answer from the options given below
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3A is true but R is false
Approach:
Evaluate A via the equipartition result ⟨KE⟩=23kBT and evaluate R via the mass-dependent formula vrms=3kBT/m.
Step 1:Assertion A: Average translational kinetic energy depends only on temperature (and is independent of mass or container size). Hence equal ⟨KE⟩ for H2 and O2 forces equal T. A is TRUE.
⟨KE⟩H2=⟨KE⟩O2⇒TH2=TO2
Step 2:Reason R: RMS speed scales as 1/m. With molar masses MH2=2 and MO2=32, the ratio at common T is 32/2=4, so the rms speeds are not equal. R is FALSE.
vrms,O2vrms,H2=MH2MO2=16=4
Step 3:Combining the two evaluations yields the correct option: A is true but R is false.
Option (3)
Final answer: A is true but R is false
Q34Single correctProperties of Solids and Liquids
The temperature of a metal strip having coefficient of linear expansion α is increased from T1 to T2 resulting in increase of its length by ΔL1. The temperature is further increased from T2 to T3 such that the increase in its length is ΔL2. Given T3+T1=2T2 and T2−T1=ΔT, the value of ΔL2 is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4ΔL1[1+αΔT]
Approach:
Apply linear thermal expansion successively, using the already-expanded length L+ΔL1 as the base for the second expansion.
Step 1:From T3+T1=2T2, rearrange to obtain T3−T2=T2−T1=ΔT. Hence both temperature increments are equal.
T3−T2=ΔT
Step 2:First expansion from initial length L: ΔL1=LαΔT. The new length is L+ΔL1; the second expansion uses this length.
ΔL2=(L+ΔL1)αΔT
Step 3:Expand and substitute LαΔT=ΔL1.
ΔL2=LαΔT+ΔL1αΔT=ΔL1+ΔL1αΔT=ΔL1[1+αΔT]
Final answer: ΔL1[1+αΔT]
Q35Single correctOscillations and Waves
A uniform disc of radius R and mass M is free to oscillate about the axis A as shown in the figure. For small oscillations the time period is ____ . (g is acceleration due to gravity)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 12π4g5R
Approach:
Model the disc as a physical pendulum oscillating about a horizontal tangent in the plane of the disc through the topmost point. Apply the parallel-axis theorem to find the moment of inertia and use the physical-pendulum time-period formula.
Step 1:Identify the geometric quantities. The pivot axis A is tangent to the disc at the topmost point and lies in the plane of the disc, so the distance from the centre of mass to the axis is d=R.
d=R
Step 2:Apply the parallel-axis theorem starting from the diameter through the centre (which is parallel to the tangent axis). The moment of inertia about this tangent axis is
I=Icm+Md2=41MR2+MR2=45MR2
Step 3:Insert I and d into the physical-pendulum formula and simplify.
T=2πMgR(5/4)MR2=2π4g5R
Final answer: 2π4g5R
Q36Single correctElectrostatics
A rigid dipole undergoes a simple harmonic motion about its centre in the presence of an electric field E1=E0x^. If another electric field E2=2E0(y^+z^) is introduced to the system, what will be the percentage change in the frequency of the oscillation (approximate)?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 173%
Approach:
Use the small-oscillation angular frequency of a rigid dipole, ω=pE/I, with E the magnitude of the net uniform field; compute the percentage change from E0 to ∣E1+E2∣.
Step 1:With only E1=E0x^, the initial angular frequency satisfies ω1∝E0.
ω1∝E0
Step 2:After adding E2=2E0y^+2E0z^, the net field is Enet=E0x^+2E0y^+2E0z^ with magnitude
∣Enet∣=E012+22+22=E09=3E0
Step 3:Take the ratio of new to old angular frequencies (with p and I unchanged).
ω1ω2=E03E0=3≈1.732
Step 4:Compute the percentage change in frequency.
fΔf×100%=(3−1)×100%≈73.2%
Final answer: 73%
Q37Single correctElectrostatics
From the circuit given below, the capacitance between terminals A and B shown in the circuit is _____ μF. (take C1=C2=C3=1μF and C4=2μF)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 27/2
Approach:
Reduce the network in stages: combine the series branch (C1 and C2) first, add the parallel C3, then add the parallel C4 across A-B.
Step 1:Combine C1 and C2 in the upper branch (series).
C12=C1+C2C1C2=1+11⋅1=0.5μF
Step 2:C3 is in parallel with the C12 branch between A and B.
C123=C12+C3=0.5+1=1.5μF
Step 3:Finally C4 is in parallel with C123 across the terminals A-B.
CAB=C123+C4=1.5+2=3.5μF=27μF
Final answer: 7/2
Q38Single correctElectrostatics
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R Assertion A: In electrostatics, a conductor does not store any net charge inside. Reason R: Inside the capacitor (with no dielectric medium), the free charge carriers, if placed between the plates of capacitor, experience force and drift. 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:
Evaluate the truth value of A using electrostatic equilibrium and Gauss's law, evaluate R using the Coulomb force on a free charge in the inter-plate field, then test whether R logically explains A.
Step 1:Assertion A: In electrostatic equilibrium, E=0 everywhere inside a conductor. Applying Gauss's law to any closed surface drawn entirely inside the conductor gives Qenc=0. Hence no net charge resides in the interior; all excess charge sits on the surface. A is TRUE.
∮E⋅dA=ε0Qenc=0⇒Qenc=0
Step 2:Reason R: Between the plates of a charged parallel-plate capacitor (vacuum), a uniform field E=σ/ε0 is present. Any free charge placed in this region experiences F=qE and drifts toward the oppositely charged plate. R is TRUE.
F=qE(between plates)
Step 3:Causal evaluation: A refers to the absence of net charge inside the bulk of a conductor (a result of electrostatic equilibrium inside material), while R refers to the motion of free charges in the empty inter-plate region of a capacitor. The two statements describe different physical systems; R does not provide the cause of A.
A: conductor interior; R: capacitor gap (different systems)
Final answer: Both A and R are true but R is NOT the correct explanation of A
Q39Single correctMagnetic Effects of Current and Magnetism
A solenoid has a core made of material with relative permeability 400. The magnetic field produced in the interior of solenoid is 1.0 T. The magnetic intensity in SI units is α×105. The value of α is ___. (Free space permeability μ0=4π×10−7 SI units.)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 216π1
Approach:
Apply the constitutive relation B=μrμ0H inside the magnetic medium and solve for H.
Step 1:Given: interior magnetic field B=1.0T, relative permeability μr=400, and μ0=4π×10−7T m/A. Target: magnetic intensity H expressed as α×105A/m.
B=1.0T,μr=400,μ0=4π×10−7T m/A
Step 2:Rearrange the constitutive relation for H.
H=μrμ0B
Step 3:Substitute the numerical values into the denominator.
H=400×4π×10−71.0=1600π×10−71A/m
Step 4:Bring the factor of 10−7 into the numerator and cast in the prescribed form.
H=1600π107A/m=16π1×105A/m
Final answer: 16π1
Q40Single correctElectromagnetic Waves
A magnetic field vector in an electromagnetic wave is represented by B=B0sin(2πνt−λ2πx)j^. Its associated electric field vector is _____ .
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1E=−νλB0sin(2πνt−λ2πx)k^
Approach:
Identify the propagation direction from the wave's phase, use E0=cB0=νλB0 for the amplitude, and fix the sign so that E×B points along the propagation direction.
Step 1:Given: B=B0sin(2πνt−2πx/λ)j^. The phase (ωt−kx) with k>0 corresponds to propagation along +i^.
k^prop=+i^
Step 2:By the E-B amplitude relation, the electric-field amplitude is E0=cB0 and c=νλ, hence E0=νλB0.
E0=cB0=νλB0
Step 3:Transverse condition: E is perpendicular to both B∥j^ and k^prop∥i^, so E∥±k^.
E∥±k^
Step 4:Choosing E∝−k^: (−k^)×(j^)=−(k^×j^)=−(−i^)=+i^, which matches k^prop. Hence the sign in front of the amplitude is negative.
(−k^)×(j^)=+i^
Step 5:Combine amplitude, direction, and the same phase as B.
E=−νλB0sin(2πνt−λ2πx)k^
Final answer: E=−νλB0sin(2πνt−λ2πx)k^
Q41Single correctOptics
A convex lens is made from glass material having refractive index of 1.4 with same radius of curvature on both sides. The ratio of its focal length and radius of curvature is ___ .
(A)
(B)
(C)
(D)
SolutionAnswer: Option 41.25
Approach:
Apply the lensmaker's formula to a symmetric biconvex lens with both surface radii of equal magnitude.
Step 1:Given: refractive index n=1.4, both surface radii equal in magnitude to R. Target: ratio f/R.
n=1.4,∣R1∣=∣R2∣=R
Step 2:By the Cartesian sign convention for a symmetric biconvex lens, the first (convex toward incident light) surface has R1=+R, and the second has R2=−R.
R1=+R,R2=−R
Step 3:Substitute the radii into the curvature factor.
R11−R21=R1−(−R1)=R2
Step 4:Plug n=1.4 and the curvature factor into the lensmaker's equation.
f1=(1.4−1)⋅R2=R0.8
Step 5:Invert to obtain f and form the ratio f/R.
Rf=0.81=1.25
Final answer: 1.25
Q42Single correctOptics
An unpolarized light of certain intensity passes through a combination of two polarizers whose transmission axes are at 30º and 90º, respectively, with respect to the horizontal axis. A third polarizer with its transmission axis at 60º with the horizontal axis is placed between the two existing polarizers. The ratio of the output intensities with and without the third polarizer is ____ .
(A)
(B)
(C)
(D)
SolutionAnswer: Option 39/4
Approach:
Compute the transmitted intensity in both configurations (two-polarizer and three-polarizer) using Malus's law sequentially, then take the ratio.
Step 1:Given: incident unpolarized intensity I0. Polarizer axes (with respect to horizontal): P1 at 30∘, P2 at 90∘, optional middle P3 at 60∘. Target: Iwith/Iwithout.
θP1=30∘,θP3=60∘,θP2=90∘
Step 2:Configuration without P3: after P1 the intensity is I0/2. The angle between P1 (30∘) and P2 (90∘) is 60∘.
Iwithout=2I0cos2(60∘)=2I0⋅41=8I0
Step 3:Configuration with P3 inserted: after P1 the intensity is again I0/2. The angle between P1 (30∘) and P3 (60∘) is 30∘.
IafterP3=2I0cos2(30∘)=2I0⋅43=83I0
Step 4:Light is now polarized along the P3 axis. The angle between P3 (60∘) and P2 (90∘) is 30∘, so Malus's law gives the final intensity.
Iwith=83I0cos2(30∘)=83I0⋅43=329I0
Step 5:Form the requested ratio.
IwithoutIwith=I0/89I0/32=329⋅8=49
Final answer: 9/4
Q43Single correctAtoms and Nuclei
In Rutherford's alpha-particle scattering experiment, only a few alpha particles rebound back because A. The size of gold nucleus is very small as compared to the size of gold atom. B. Alpha particle and gold nucleus have equal charge. C. The impact parameter is minimum for a few alpha particles. D. A few alpha particles have very high kinetic energy. E. Only a few alpha particles undergo head-on collision with the nuclei. Choose the correct answer from the options given below:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4A, C, E Only
Approach:
Evaluate each statement (A–E) against Rutherford's experimental conclusions and the geometry of Coulomb scattering.
Step 1:Given: Rutherford's observation that only a small fraction of incident alphas rebound through angles near 180∘. Target: identify which of A–E correctly explain this.
rebound⇔θ→180∘
Step 2:Statement A: the gold nucleus radius is ∼10−15m while the atomic radius is ∼10−10m, a ratio of ∼10−5. Hence most alphas pass through almost empty space; only those approaching the tiny nucleus suffer large-angle scattering. TRUE.
rnucleus/ratom∼10−5
Step 3:Statement B: alpha particle charge is +2e and the gold nucleus charge is +79e; they are not equal. FALSE.
qα=2e,QAu=79e
Step 4:Statement C: large scattering angles occur for very small impact parameters b; since b is a continuous variable, only a small fraction of the geometric cross-section corresponds to nearly head-on trajectories. TRUE.
θ↑asb↓
Step 5:Statement D: the alpha source (e.g. 214Po) emits alphas of essentially fixed kinetic energy. Variation in KE is not the cause of rebound. FALSE.
KEα≈const
Step 6:Statement E: head-on collisions (b→0) correspond to a vanishingly small geometric cross-section, so only a small fraction of alphas rebound near 180∘. TRUE.
P(b→0)≪1
Step 7:Combine the verdicts: A, C, E are correct; B, D are incorrect.
Correct set={A,C,E}
Final answer: A, C, E Only
Q44Single correctDual Nature of Matter and Radiation
The de Broglie wavelength associated with an electron accelerated through a potential difference V is λe and the de Broglie wavelength associated with a proton accelerated through the same potential difference is λp. If their corresponding masses are me and mp, respectively, then the ratio of their de Broglie wavelengths (λpλe) is _____ .
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1memp
Approach:
Use the de Broglie wavelength of a charged particle accelerated through a potential difference V: λ=h/2mqV, with q=e for both particles, then take the ratio.
Step 1:Given: electron and proton accelerated through the same potential difference V; charge magnitudes are ∣qe∣=∣qp∣=e. Target: ratio λe/λp.
∣qe∣=∣qp∣=e,Ve=Vp=V
Step 2:Both particles acquire the same kinetic energy upon acceleration.
KEe=KEp=eV
Step 3:Express momentum from kinetic energy, then substitute into the de Broglie relation to obtain λ as a function of mass.
λ=ph=2meVh
Step 4:Write λe and λp explicitly.
λe=2meeVh,λp=2mpeVh
Step 5:Divide and simplify.
λpλe=2meeV2mpeV=memp
Final answer: memp
Q45Single correctElectronic Devices
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R Assertion A: A diode under reverse-biased condition provides very small current which is nearly independent of voltage until a critical limit at which the current increases drastically. Reason R: Below the critical voltage limit, only majority charge carriers flow which increases drastically above critical voltage. choose the correct answer from the options given below
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3A is true but R is false
Approach:
Independently evaluate the physical correctness of Assertion A (reverse-bias I–V behaviour) and Reason R (carrier type responsible for the reverse current).
Step 1:Given: a p–n junction diode in reverse bias. Target: assess A and R.
VR<0,depletion widthW↑
Step 2:Assertion A: under reverse bias, the depletion width grows and the barrier blocks majority-carrier diffusion; only a small saturation current Is flows, almost independent of VR, until the breakdown (critical) voltage Vbr where the current rises sharply. TRUE.
Irev≈IstillV=Vbr,then steep rise
Step 3:Reason R: the small reverse current Is is carried by MINORITY charge carriers — thermally generated electrons in the p-side and holes in the n-side — which are swept across the junction by the reverse field. Majority carriers are blocked by the widened barrier.
Is∝ni2/{NA,ND}(minority-limited)
Step 4:Beyond Vbr the sharp rise is due to avalanche multiplication and/or Zener tunnelling of carriers across the junction, not the onset of majority-carrier conduction as R asserts.
Breakdown⇒avalanche / Zener tunnelling
Step 5:Combining: A is true and R is false.
A: T,R: F
Final answer: A is true but R is false
Q46NumericalElectronic Devices
A diode has Zener voltage of 10 V and maximum power dissipation of 0.5 W, then the minimum resistance to be used in series with this diode for safety when it is connected to a 25 V power supply is ____Ω.
SolutionAnswer: 300
Approach:
Find the maximum permitted Zener current from its power rating, then apply Ohm's law to the series resistor that must drop (Vsupply−VZ).
Step 1:Given: VZ=10V, Pmax=0.5W, Vsupply=25V. Target: minimum series resistance Rmin that keeps PZ≤Pmax.
VZ=10V,Pmax=0.5W,Vsupply=25V
Step 2:Maximum allowable Zener current from the power rating.
IZ,max=VZPmax=100.5=0.05A
Step 3:With the diode clamped at VZ, the series resistor must drop the remaining voltage.
VR=Vsupply−VZ=25−10=15V
Step 4:Smaller R means larger current; the minimum R corresponds to the maximum allowed current.
Rmin=IZ,maxVR=0.0515=300Ω
Final answer: 300
Q47NumericalKinematics
A gun mounted on the ground fires bullets in all directions with same speed. The farthest distance the bullets could reach is 6.4 m. The speed of the bullets from the gun is ______ m/s. (take g=10m/s2)
SolutionAnswer: 8
Approach:
Identify the farthest reachable distance on the ground with the maximum projectile range, obtained at a launch angle of 45∘.
Step 1:Given: bullets are fired in all directions with the same speed u; the farthest distance reached on the ground is Rmax=6.4m and g=10m/s2. Target: the speed u.
Rmax=6.4m,g=10m/s2
Step 2:For fixed u, the horizontal range is maximised when sin2θ=1, i.e. θ=45∘.
Rmax=gu2sin90∘=gu2
Step 3:Equate the given farthest distance to Rmax and solve for u2.
6.4=10u2⇒u2=64m2/s2
Step 4:Take the positive square root for the muzzle speed.
u=64=8m/s
Final answer: 8
Q48NumericalMagnetic Effects of Current and Magnetism
Two identical small bar magnets each of dipole moment 35J/T are placed at a center to center separation of 10 cm , with their axes perpendicular to each other as shown in figure. The value of magnetic field at the point P midway between the magnets is α×10−3T. The value of α is ___ . (μ0=4π×10−7Tm/A)
SolutionAnswer: 12
Approach:
Treat each bar magnet as an ideal dipole. P lies on the axial line of magnet 1 and on the equatorial line of magnet 2; compute the two field magnitudes, identify that the two contributions are mutually perpendicular, and combine vectorially.
Step 1:Given: dipole moment M=35J/T, centre-to-centre separation d=10cm, axes mutually perpendicular as per the figure. Midpoint P is at r=d/2=5cm=0.05m from each magnet. Also μ0/(4π)=10−7T m/A. Target: α such that the net field at P equals α×10−3T.
M=35J/T,r=0.05m,4πμ0=10−7T m/A
Step 2:From the figure, magnet 1 has its axis along the line joining the centres, so P lies on its axial line. Magnet 2's axis is perpendicular to that line, so P lies on its equatorial line.
Step 5:Bax is directed along the line of centres (parallel to magnet 1's axis). The equatorial field of an ideal dipole points anti-parallel to its moment, so Beq is directed along magnet 2's axis, which is perpendicular to the line of centres. Hence the two contributions are mutually perpendicular.
Q49NumericalMagnetic Effects of Current and Magnetism
A circular coil of radius 2 cm and 125 turns carries a current of 1 A. The coil is placed in a uniform magnetic field of magnitude 0.4 T. The axis of the coil makes an angle of 30∘ with the direction of the magnetic field. The torque acting on the coil is α×10−4 N.m. The value of α is ____ . (π=3.14)
SolutionAnswer: 314
Approach:
Use τ=NIABsinθ, where the magnetic moment m is along the coil's axis; the angle between m and B equals the stated 30∘.
Step 1:Given: N=125, I=1A, r=2cm=0.02m, B=0.4T, θ=30∘ (between coil axis and B), π=3.14. Target: α where τ=α×10−4N m.
N=125,I=1A,r=0.02m,B=0.4T,θ=30∘
Step 2:Compute the cross-sectional area of one turn.
A=πr2=3.14×(0.02)2=3.14×4×10−4=12.56×10−4m2
Step 3:Magnetic moment magnitude m=NIA is directed along the coil axis, so the angle between m and B is the given 30∘.
m=NIA=125×1×12.56×10−4=0.157A m2
Step 4:Substitute into the torque formula with sin30∘=1/2.
τ=mBsin30∘=0.157×0.4×0.5=0.0314N m
Step 5:Recast in the prescribed form.
τ=314×10−4N m⇒α=314
Final answer: 314
Q50NumericalOptics
In a double slit experiment, when one of the slits is covered by a transparent mica sheet of refractive index 1.56, the central fringe shifts to the position of 7th bright fringe, obtained with both slits uncovered. If the light source wavelength is 450 nm, the thickness of mica sheet is α×10−9m. The value of α is ______.
SolutionAnswer: 5625
Approach:
Equate the lateral fringe shift caused by introducing a thin transparent sheet over one slit to seven fringe widths; the geometric factor D/d cancels.
Step 1:Given: refractive index of the mica sheet μ=1.56, wavelength λ=450nm, central maximum shifts to the position of the 7th bright fringe of the unaltered pattern, so n=7. Target: thickness t expressed as α×10−9m.
μ=1.56,λ=450nm,n=7
Step 2:Inserting a sheet of refractive index μ and thickness t adds an extra optical path of (μ−1)t to the path through that slit. The fringe pattern shifts by Δy=(μ−1)tD/d, while one fringe spacing is β=λD/d.
βΔy=λ(μ−1)t
Step 3:Setting the shift equal to n fringe widths gives the working equation; D/d has cancelled, so the result is independent of the slit-to-screen geometry.
(μ−1)t=nλ⇒(1.56−1)t=7×450nm
Step 4:Solve for the thickness t.
t=0.563150nm=5625nm
Step 5:Convert to the prescribed form.
t=5625×10−9m⇒α=5625
Final answer: 5625
Chemistry25 questions
Q51Single correctSome Basic Concepts in Chemistry
The correct order of total number of atoms present in (A) 2 moles of cyclohexane (B) 684 g of sucrose (C) 90.8 L of dihydrogen at STP is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4B>A>C
Approach:
Convert each quantity to moles, multiply by atoms per molecule and Avogadro number, then order the totals.
Step 1:Given: (A) 2 mol cyclohexane C6H12 (18 atoms per molecule); (B) 684 g sucrose C12H22O11 (molar mass 342 g mol−1, 45 atoms per molecule); (C) 90.8 L of H2 at STP (2 atoms per molecule). Target: order of total atoms.
nA=2mol,mB=684g,VC=90.8L
Step 2:Atoms in (A).
NA(A)=2×18×NA=36NA
Step 3:Moles of sucrose in (B) and corresponding atoms.
nB=342684=2mol;NA(B)=2×45×NA=90NA
Step 4:Moles of H2 in (C) and corresponding atoms.
nC=22.490.8=4.054mol;NA(C)=4.054×2×NA=8.11NA
Step 5:Order the three totals.
90NA>36NA>8.1NA
Final answer: B>A>C
Q52Single correctAtomic Structure
The species having identical radii according to the Bohr's theory are: A. H (first orbit) B. He+ (first orbit) C. He+ (Second orbit) D. Li2+ (first orbit) E. Be3+ (Second orbit) Choose the correct answer from the options given below:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2A and E Only
Approach:
Apply Bohr's orbit radius formula rn=a0n2/Z to each hydrogen-like species and identify pairs with equal radii.
Step 1:Given species and their (n, Z) values: A: H (1, 1); B: He+ (1, 2); C: He+ (2, 2); D: Li2+ (1, 3); E: Be3+ (2, 4). Target: identify pairs with identical Bohr radii.
rn∝n2/Z
Step 2:Compute n2/Z for A.
rA=a0⋅12/1=a0
Step 3:Compute n2/Z for B.
rB=a0⋅12/2=0.5a0
Step 4:Compute n2/Z for C.
rC=a0⋅22/2=2a0
Step 5:Compute n2/Z for D.
rD=a0⋅12/3=a0/3
Step 6:Compute n2/Z for E.
rE=a0⋅22/4=a0
Step 7:Compare radii to find the matching pair.
rA=rE=a0;rB=0.5a0;rC=2a0;rD=a0/3
Final answer: A and E Only
Q53Single correctChemical Bonding and Molecular Structure
Which of the following pictorial diagram most correctly represents the π∗ (π antibonding) molecular orbital between two atoms if the internuclear axis is taken to be in the z-direction (z-axis) ?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3Pi-antibonding (π∗) molecular orbital: two p-orbital lobes on each atom oriented perpendicular to the z internuclear axis with opposite phases on the two atoms (nodal plane between the nuclei perpendicular to the bond axis).
Approach:
Apply symmetry rules of LCAO-MO: identify the diagram whose lobes lie perpendicular to the internuclear axis and carry opposite phases on the two atoms.
Step 1:Given: internuclear axis is the z-direction; target is the π∗ MO. Pi MOs arise from sideways overlap of 2px or 2py orbitals (perpendicular to z).
π←2px/2py overlap
Step 2:Bonding π: in-phase combination, lobes of the same sign on the two atoms, no nodal plane between nuclei. Antibonding π∗: out-of-phase combination, lobes of opposite sign on the two atoms, an additional nodal plane perpendicular to z lies between the nuclei.
ψπ∗∝2px(A)−2px(B)
Step 3:Match the four diagrams to MO symmetry. Option 1 (single spherical lobe) is σ bonding; option 2 (same-sign side lobes, no node between atoms) is π bonding; option 4 (two end-on p lobes with a node along z between nuclei) is σ∗; option 3 (perpendicular lobes, opposite sign on either atom, node between nuclei) is π∗.
Option 3≡π∗
Final answer: Pi-antibonding (π∗) molecular orbital: two p-orbital lobes on each atom oriented perpendicular to the z internuclear axis with opposite phases on the two atoms (nodal plane between the nuclei perpendicular to the bond axis).
Q54Single correctSolutions
At 27∘C, 0.1M,1LK4[Fe(CN)6] aqueous solution and 0.1M,1LFeCl3 aqueous solution are placed in a container separated by a semi permeable membrane AB. Assume complete dissociation of both the solutes. Which of the following statement is correct?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3Solution on side 'y' is hypotonic.
Approach:
Compute the van't Hoff factor and effective particle concentration on each side; assign hypertonic/hypotonic on that basis and then refute the remaining statements.
Step 1:Given: side x has 0.1 M K4[Fe(CN)6], side y has 0.1 M FeCl3 at the same temperature 27 ∘C; assume complete dissociation. Target: identify the correct statement.
T=300K,Cx=Cy=0.1M
Step 2:Dissociation on side x produces 5 ions per formula unit.
K4[Fe(CN)6]→4K++[Fe(CN)6]4−;ix=5
Step 3:Dissociation on side y produces 4 ions per formula unit.
FeCl3→Fe3++3Cl−;iy=4
Step 4:Compare osmotic pressures via π=iCRT at common T. Since Ceff,x>Ceff,y, πx>πy. By definition, the side with lower osmotic pressure is hypotonic.
πx=0.5RT,πy=0.4RT⇒πx>πy
Step 5:Refute the other options. (1) Prussian-blue formation requires the K4[Fe(CN)6] and Fe3+ to mix; the semipermeable membrane blocks ions, so no colour forms on either side. (2) Ions are solute particles which a semipermeable membrane explicitly blocks. (4) Reverse osmosis requires an external pressure greater than the osmotic pressure difference applied on the hypertonic (concentrated) side x; an arbitrary 'any value' does not suffice, so the statement is wrong.
⇒Only statement (3) is correct
Final answer: Solution on side 'y' is hypotonic.
Q55Single correctEquilibrium
20 mL of a solution of acetic acid required 28.4 mL of 0.1 M NaOH for its neutralization. A solution (X) was prepared by mixing 20 mL of the above acetic acid and 14.2 mL of 0.1 M NaOH solution. What is the pH of the solution (X) ? (pKa value of acetic acid is 4.75).
(A)
(B)
(C)
(D)
SolutionAnswer: Option 24.75
Approach:
Determine moles of acid and added base; recognise the half-neutralisation buffer and apply Henderson-Hasselbalch.
Step 1:Given: 20 mL acetic acid is fully neutralised by 28.4 mL of 0.1 M NaOH; in solution X, 20 mL of the same acid is mixed with 14.2 mL of 0.1 M NaOH; pKa=4.75. Target: pH of X.
nHA=28.4×0.1=2.84mmol
Step 2:Compute moles of NaOH added in X.
nNaOH=14.2×0.1=1.42mmol
Step 3:NaOH converts an equal mole of acid into the conjugate salt. Remaining acid = 2.84 - 1.42 = 1.42 mmol; salt formed = 1.42 mmol.
[salt]=[acid]=1.42mmol per 34.2 mL
Step 4:Substitute the ratio into Henderson-Hasselbalch.
pH=4.75+log(1)=4.75
Final answer: 4.75
Q56Single correctSome Basic Principles of Organic Chemistry
Match the LIST-I with LIST-II
List-I (Reaction)
List-II (Mechanism)
A. Williamson Synthesis
I. Electrophilic addition
B. Friedel Craft Reaction
II. Free radical substitution
C. Bromination of vinyl benzene
III. Nucleophilic substitution
D. Chlorination of toluene in light
IV. Electrophilic substitution
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3A-III, B-IV, C-I, D-II
Approach:
Classify each named reaction by its mechanistic family using the rate-determining elementary step.
Step 1:List the four reactions and four mechanism categories. Target: assign each reaction to its mechanism class.
{A,B,C,D}→{I,II,III,IV}
Step 2:A. Williamson ether synthesis proceeds by an alkoxide nucleophile attacking an alkyl halide in SN2 fashion. Hence A pairs with III (Nucleophilic substitution).
R−O−+R′−XSN2R−O−R′+X−
Step 3:B. Friedel-Crafts alkylation/acylation generates an electrophile (R+ or RCO+) that substitutes an aromatic ring hydrogen via an arenium intermediate. Hence B pairs with IV (Electrophilic substitution).
Ar−H+R+→Ar−R+H+
Step 4:C. Bromination of vinyl benzene (PhCH=CH2) targets the alkene C=C, where Br2 adds across the double bond through a bromonium intermediate. Hence C pairs with I (Electrophilic addition).
PhCH=CH2+Br2→PhCHBr−CH2Br
Step 5:D. Photochlorination of toluene proceeds at the benzylic C-H by a chain mechanism initiated by Cl⋅ radicals. Hence D pairs with II (Free radical substitution).
PhCH3+Cl2hνPhCH2Cl+HCl
Step 6:Combine pairings.
A→III,B→IV,C→I,D→II
Final answer: A-III, B-IV, C-I, D-II
Q57Single correctClassification of Elements and Periodicity in Properties
The 1st ionization enthalpy for Mg is +737kJ/mol. The most probable estimated value of the 2nd ionization enthalpy of Mg is ______ .
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3+1450kJ/mol
Approach:
Use the universal rule that successive ionization enthalpies increase (IE2>IE1) due to the rising effective nuclear charge on the residual cation; eliminate options inconsistent with this rule.
Step 1:Given: IE1(Mg)=+737kJmol−1; both first and second ionizations remove electrons from the 3s subshell since Mg has configuration [Ne]3s2. Target: estimate IE2.
Mg:[Ne]3s2→[Ne]3s1→[Ne]
Step 2:Removing the second electron from a cation requires more energy than removing the first from a neutral atom because the residual nuclear pull is stronger. Therefore IE2>IE1=737. This immediately rejects option (4) at 590 kJ/mol (smaller than IE1).
IE2>737kJmol−1
Step 3:Empirically, the increase from IE1 to IE2 within the same outer subshell is by a factor of 1.8-2.0 for group 2 metals. Options (1) 906 and (2) 856 kJ/mol fall only about 1.15-1.23 times IE1, which is too small for removal of a second 3s electron from a +1 cation; option (3) 1450 kJ/mol (≈2×737) is consistent.
IE2/IE1≈2⇒IE2≈1474kJmol−1
Step 4:Match to the available option.
IE2(Mg)≈+1450kJmol−1
Final answer: +1450kJ/mol
Q58Single correctp-Block Elements
The electronegativity of a group 13 element ' E ' is same as that of Ge (on Pauling scale and upto one decimal point). The CORRECT statements about E3+ are A. It can act as a reducing agent. B. It can act as an oxidizing agent. C. E3+ is more stable than E+. D. The standard electrode potential value for E3+/E is positive. Choose the correct answer from the options given below:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3B and D Only
Approach:
Match Ge's Pauling electronegativity (≈2.0 to 1 d.p.) to the only group 13 element with the same value, then deduce the stability of E3+/E+ via the inert-pair effect and evaluate the sign of E∘(E3+/E).
Step 1:Given: χP(Ge)≈2.0 to 1 d.p. Among group 13 elements (B 2.0, Al 1.5, Ga 1.8, In 1.8, Tl 2.0 - revised Pauling), Tl shares 2.0 with Ge. Target: assess the four statements for E=Tl.
χ(Tl)≈2.0=χ(Ge)
Step 2:Inert-pair effect makes the 6s2 pair reluctant to ionize, so Tl+ (which retains the pair) is more stable than Tl3+. Therefore Tl3+ tends to gain two electrons to become Tl+, behaving as an oxidising agent. Statement B is TRUE; statement C (Tl3+ more stable than Tl+) is FALSE; statement A (Tl3+ a reducing agent) is FALSE because a species being reduced cannot itself reduce.
Tl3++2e−→Tl+
Step 3:Standard reduction potential of Tl3+/Tl is positive (+0.72 V tabulated), reflecting that reduction of Tl3+ to metallic Tl is thermodynamically favoured. Statement D is TRUE.
E∘(Tl3+/Tl)=+0.72V>0
Step 4:Combine the truth values.
{B,D} true; {A,C} false
Final answer: B and D Only
Q59Single correctd- and f-Block Elements
Pairs of elements with the same number of electrons in their respective 4 f orbital are [Atomic number. Eu-63, Gd-64, Dy-66, Ho-67, Tm-69, Yb-70, Lu-71, Hf-72] A. (Eu and Gd) B. (Dy and Ho) C. (Yb and Hf) D. (Lu and Tm) Choose the correct answer from the options given below:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4A and C Only
Approach:
Write each ground-state configuration, count the 4f population (with the Gd and Lu half/full-filled exceptions), and compare within each pair.
Step 1:List the ground-state 4f populations using atomic numbers and standard exceptions. Eu(63): [Xe]4f76s2; Gd(64): [Xe]4f75d16s2; Dy(66): 4f10; Ho(67): 4f11; Tm(69): 4f13; Yb(70): 4f146s2; Lu(71): 4f145d16s2; Hf(72): [Xe]4f145d26s2.
Step 2:Pair A (Eu, Gd): both have 7 4f electrons. MATCH.
n4f(Eu)=n4f(Gd)=7
Step 3:Pair B (Dy, Ho): 10 vs 11. NO MATCH.
n4f(Dy)=10=11=n4f(Ho)
Step 4:Pair C (Yb, Hf): both have 14 4f electrons.
n4f(Yb)=n4f(Hf)=14
Step 5:Pair D (Lu, Tm): 14 vs 13. NO MATCH.
n4f(Lu)=14=13=n4f(Tm)
Step 6:Combine: A and C only satisfy the condition.
⇒{A,C}
Final answer: A and C Only
Q60Single correctCoordination Compounds
Consider the metal complexes [Ni(en)3]2+(A),[NiCl4]2−(B) and [Ni(NH3)6]2+(C). Choose the CORRECT option by considering the number of unpaired electrons present in (A), (B) and (C) respectively and the order of frequency of absorption.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 12, 2, 2 and (A)>(C)>(B)
Approach:
Determine geometry and d-electron filling of each Ni2+ (d8) complex to count unpaired electrons; order absorption frequency using ν∝Δ and the spectrochemical series.
Step 1:Given: three Ni2+ complexes A [Ni(en)3]2+, B [NiCl4]2−, C [Ni(NH3)6]2+. Ni2+ has d8 in every case. Target: unpaired-electron counts and frequency order.
Ni2+:[Ar]3d8
Step 2:A is octahedral with en (bidentate, moderately strong field). For d8 octahedral filling, regardless of high/low spin the two eg orbitals each carry one electron, giving 2 unpaired.
t2g6eg2⇒nup=2
Step 3:B is tetrahedral (4-coordinate with Cl− weak field). d8 tetrahedral: e4t24, giving 2 unpaired.
e4t24⇒nup=2
Step 4:C is octahedral with NH3. Same d8 analysis: 2 unpaired.
t2g6eg2⇒nup=2
Step 5:Order frequencies. Spectrochemical strengths: en>NH3≫Cl−, so Δo(en)>Δo(NH3). Tetrahedral Cl− field gives Δt≈(4/9)Δo(Cl−), the smallest splitting. Frequency ν∝Δ.
ΔA>ΔC>ΔB⇒νA>νC>νB
Step 6:Combine results.
(2,2,2)and(A)>(C)>(B)
Final answer: 2, 2, 2 and (A)>(C)>(B)
Q61Single correctChemical Bonding and Molecular Structure
Consider the following molecules/species: (x) cycloheptatrienone (tropone), (y) acetone (CH3)2C=O, (z) acetate ion CH3−C(=O)−O(−). The correct order of carbon - oxygen double bond length is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4x>z>y
Approach:
Estimate effective C-O bond order in each species from resonance contributions; smaller bond order corresponds to longer bond length.
Step 1:Identify the three carbonyl systems. (x) tropone is a 7-membered cycloheptatrienone whose C=O conjugates with three ring C=C, generating an aromatic tropylium (C7H6+) resonance form with O−. (y) acetone has an isolated C=O. (z) acetate ion delocalises the negative charge symmetrically between two oxygens.
x↔C7H6+−O−;z↔CH3−C(O−)=O↔CH3−C(=O)−O−
Step 2:Estimate effective C-O bond orders. For tropone, the aromatic resonance structure carries pronounced weight (it satisfies 4n+2 with n=1), so the C-O is between single and double, bond order well below 1.5. For acetate, the two equivalent resonance structures give an exact 1.5 bond order to each C-O. For acetone, no resonance lengthens the C=O, so bond order is 2.
BO(x)≈1.4,BO(z)=1.5,BO(y)=2.0
Step 3:Convert bond orders to lengths using the inverse relationship.
Length(x)>Length(z)>Length(y)
Step 4:Match to the option list.
x>z>y
Final answer: x>z>y
Q62Single correctd- and f-Block Elements
Consider ∣x∣ is the difference in oxidation states of Mn in highest manganese fluoride and highest manganese oxide. The ions with ∣x∣ number of unpaired electrons from the following are: A. Sc3+ B. Zn2+ C. V2+ D. Fe2+ E. Co2+ Choose the correct answer from the options given below:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3C and E Only
Approach:
Determine ∣x∣ from the highest Mn oxidation state in the fluoride (MnF4) versus the highest in the oxide (Mn2O7); then pick ions with exactly that many unpaired d-electrons.
Step 1:Given facts about manganese: the highest binary manganese fluoride is MnF4 (Mn in +4); the highest manganese oxide is Mn2O7 (Mn in +7). Target: ∣x∣ and the ions with that many unpaired electrons.
∣x∣=∣7−4∣=3
Step 2:Compute d-electron configurations of the listed ions. Sc3+: [Ar]3d0 (0 unpaired). Zn2+: [Ar]3d10 (0 unpaired).
nup(Sc3+)=0;nup(Zn2+)=0
Step 3:V2+: [Ar]3d3 - three singly occupied t2g orbitals. Fe2+ (free ion, high spin): [Ar]3d6 - one paired plus four unpaired = 4 unpaired. Co2+: [Ar]3d7 - two paired plus three unpaired = 3 unpaired.
nup(V2+)=3;nup(Fe2+)=4;nup(Co2+)=3
Step 4:Select ions matching ∣x∣=3.
{C,E}
Final answer: C and E Only
Q63Single correctChemical Kinetics
Consider the given graph showing variation of reactant concentration with time. Three different reactions were started with identical initial concentration of reactants. Which of the following statement is correct?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2The rate constant of reaction 3 is larger than the rate constant of reaction 2 if the order of reaction is same for both.
Approach:
Read the relative position of the three [R]-vs-t curves to identify which reaction decays fastest, then test each statement against the integrated rate laws and the units of k.
Step 1:Given: three reactions with the same starting concentration [R]0; curve 1 lies on top (slowest reactant loss), curve 2 in the middle, curve 3 at the bottom (fastest reactant loss) on the [R]-vs-t plot. Target: identify the only true statement.
[R]1(t)>[R]2(t)>[R]3(t)at anyt>0
Step 2:Evaluate statement (1). The three curves have visibly different shapes (line versus curved decays of different curvatures), so they cannot all have the same order. FALSE.
Shape(1)=Shape(2)=Shape(3)
Step 3:Evaluate statement (3). Curve 1, being a straight line, corresponds to zero order; its rate constant has units molL−1s−1, not s−1. FALSE.
[kzero order]=molL−1s−1
Step 4:Evaluate statement (4). Decomposition of HI on a hot gold surface is the textbook example of a zero-order surface reaction (saturated catalyst surface), corresponding to curve 1, not curve 2. FALSE.
2HI(g)Au surfaceH2+I2(zero order)
Step 5:Evaluate statement (2). Suppose hypothetically reactions 2 and 3 had the same order n and the same initial concentration. From each integrated rate law, [R]2(t) and [R]3(t) depend monotonically on k; greater k produces a steeper drop, hence smaller [R] at every t>0. Since curve 3 lies below curve 2, the conditional implication 'same order ⇒k3>k2' holds for every n. TRUE.
[R]3(t)<[R]2(t)∀t>0,same order⇒k3>k2
Step 6:Only statement (2) survives; it is the answer.
⇒Option (2)
Final answer: The rate constant of reaction 3 is larger than the rate constant of reaction 2 if the order of reaction is same for both.
Q64Single correctHydrocarbons
Compound (X) is subjected to the sequence of reactions as shown above. Molar mass of the major product (Y) formed is ______ gmol−1. (Given molar mass in gmol−1C:12,H:1,O:16) Reagents: (i) Br2/CHCl3, (ii) NaNH2 excess, (iii) CH3I, (iv) H2,Na/NH3(l).
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2118
Approach:
Sequentially apply each reagent to styrene, identify the final hydrocarbon Y, and compute its molar mass from atomic contributions.
Step 1:Identify the starting material X = styrene (PhCH=CH2) and the reagent sequence (i) Br2/CHCl3, (ii) excess NaNH2, (iii) CH3I, (iv) Na/NH3(l). Target: molar mass of the major product Y.
X:Ph−CH=CH2;atomic masses C=12,H=1
Step 2:Electrophilic anti-addition of Br2 to the alkene gives the vicinal dibromide.
Ph−CH=CH2Br2/CHCl3Ph−CHBr−CH2Br
Step 3:Excess NaNH2 performs two successive E2 eliminations on the dibromide, giving phenylacetylene; the excess base then deprotonates the terminal alkyne to give the sodium acetylide.
Ph−CHBr−CH2BrNaNH2 (excess)Ph−C≡C−Na+
Step 4:SN2 alkylation of the acetylide with CH3I furnishes an internal alkyne.
Ph−C≡C−Na++CH3I→Ph−C≡C−CH3
Step 5:Dissolving-metal reduction with Na/NH3(l) delivers two hydrogens trans across the internal alkyne, yielding the (E)-alkene.
Ph−C≡C−CH3Na/NH3(l)(E)-Ph−CH=CH−CH3
Step 6:Compute the molar mass of C9H10.
M(Y)=9(12)+10(1)=108+10=118gmol−1
Final answer: 118
Q65Single correctSome Basic Principles of Organic Chemistry
The following structures are
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2identical molecules.
Approach:
Inspect the four substituents at the central carbon for stereocenter status, then test the two drawings for superimposability.
Step 1:List the substituents on the central carbon in each drawing: Br, Cl, CH3 (written as CH3), and CH3 (written as Me). Both CH3 and Me denote the same methyl group.
Substituents at C:{Br,Cl,CH3,CH3}
Step 2:Since the central carbon carries two identical CH3 groups, it fails the four-distinct-groups requirement and is therefore not a stereocenter.
#(distinct groups)=3<4⇒no chirality
Step 3:From the alt-text geometry, the right drawing is obtained from the left by interchanging the positions of the two methyl groups, equivalent to a 180 degree rotation about the Br-C bond axis. Since the swapped ligands are identical, the rotated structure is indistinguishable from the original.
Rotate left structure by 180∘ about C−Br⇒right structure
Step 4:Two superimposable representations correspond to one and the same molecule.
The descending order of acidity among the following compounds is: A. Phenol B. 4-nitrophenol C. 4-methoxyphenol D. 4-nitrobenzoic acid E. Benzoic acid. Choose the correct answer from the options given below:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4D>E>B>A>C
Approach:
First separate the carboxylic acids from the phenols using their characteristic pKa ranges, then order within each class by para-substituent electronic effects.
Step 1:Identify the five compounds and their conjugate-base stability classes: D and E are aryl carboxylic acids; A, B, C are para-substituted phenols. Objective: arrange in descending acidity.
Acids: D,E;Phenols: A,B,C
Step 2:Aryl carboxylic acids (pKa≈3-5) are intrinsically stronger acids than phenols (pKa≈9-11) because the carboxylate negative charge is delocalised over two equivalent oxygen atoms, whereas the phenoxide charge is delocalised onto less electronegative ring carbons.
pKa(RCOOH)≈3-5≪pKa(ArOH)≈9-11
Step 3:Within the carboxylic acids, the para −NO2 group is strongly electron-withdrawing by both −I and −M, stabilising the carboxylate and lowering pKa.
pKa(D)≈3.4<pKa(E)≈4.2
Step 4:Within the phenols, para −NO2 stabilises the phenoxide (most acidic), while para −OCH3 destabilises it via +M donation (least acidic); parent phenol lies between.
Translate the question via the conjugate-acid/base relation, rank the four anilines by basicity using electronic effects of the para substituent, then pick the weakest base.
Step 1:Identify the target: the conjugate acid is the anilinium cation ArNH3+; its strength varies inversely with the basicity of the parent amine ArNH2.
ArNH2+H+⇌ArNH3+
Step 2:Classify the four para substituents. −OCH3 donates electrons by +M (and weak −I); −CH3 donates by hyperconjugation/+I; −H is the reference; −NO2 withdraws strongly by −I and −M.
Electronic effect on N lone pair:−OCH3>−CH3>−H≫−NO2
Step 3:Translate to basicity at nitrogen: greater electron density on N gives a stronger base.
Step 4:By the inversion in step 1, the weakest base yields the strongest conjugate acid (4-nitroanilinium).
pKa(4-NO2-anilinium)≈1.0(lowest among options)
Final answer: 4-nitroaniline
Q68Single correctBiomolecules
A D-aldotetrose on oxidation with concentrated HNO3 resulted in optically inactive dicarboxylic acid. The structure of the D-aldotetrose is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3D-erythrose Fischer projection (both -OH groups on the right; CHO top, CH2OH bottom)
Approach:
Convert each candidate D-aldotetrose to its aldaric acid by oxidising the terminal -CHO and -CH2OH to -COOH, then identify which product possesses an internal mirror plane (meso, optically inactive).
Step 1:Identify the D-aldotetrose family: only two members exist, D-erythrose (both middle -OH on the right in the Fischer projection) and D-threose (middle -OH on opposite sides). The objective is to pick the one whose oxidation product is optically inactive.
Candidates:D-erythrose,D-threose
Step 2:Apply HNO3 oxidation to D-erythrose: −CHO and −CH2OH both become −COOH, giving an aldaric acid with both internal -OH groups on the right.
Step 3:Apply the same oxidation to D-threose: the two middle -OH groups remain on opposite sides, giving an aldaric acid in which the stereocentres do NOT internally compensate.
Step 4:The optically inactive product is delivered only by D-erythrose, which corresponds to the Fischer projection with both middle -OH groups on the right (option 3).
D-aldotetrose=D-erythrose (both OH right)
Final answer: D-erythrose Fischer projection (both -OH groups on the right; CHO top, CH2OH bottom)
Q69Single correctPrinciples Related to Practical Chemistry
Among Fe3+,Pb2+,Cu2+ and Mn2+, identify the one that gets precipitated out while passing H2S in presence of NH4OH as group reagent. The highest possible oxidation state of the corresponding metal is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4+7
Approach:
Place each cation in its qualitative-analysis group, pick the one that precipitates as a sulphide under H2S in alkaline (NH4OH) medium, and quote the maximum oxidation state of that metal.
Step 1:Identify the four candidate cations and the group reagent (H2S in NH4OH/NH4Cl), corresponding to qualitative analysis Group IV.
Group reagent:H2S/NH4OH(alkaline buffer)
Step 2:Pb2+ and Cu2+ are Group II cations: they precipitate as PbS / CuS under H2S in dilute acidic (HCl) medium, NOT in NH4OH.
Pb2+,Cu2+H2S/dil.HClPbS↓,CuS↓
Step 3:Fe3+ is a Group III cation: it precipitates as Fe(OH)3 on addition of NH4OH in presence of NH4Cl, before any sulphide forms.
Fe3++3OH−→Fe(OH)3↓
Step 4:Mn2+ is a Group IV cation. In the alkaline NH4OH medium, the [S2−] is high enough to exceed Ksp(MnS), giving a buff (flesh-coloured) sulphide precipitate.
Mn2++H2SNH4OHMnS↓(buff)
Step 5:Maximum oxidation state of manganese equals its periodic-table group number (Group 7); the +7 state is realised in MnO4− (e.g., KMnO4) and Mn2O7.
Mn config [Ar]3d54s2⇒max OS=+7
Final answer: +7
Q70Single correctPrinciples Related to Practical Chemistry
Match the LIST-I with LIST-II
List-I Compound
List-II Test
A.
I. Hinsberg's reagent test
B.
II. Phthalein dye test
C.
III. Lucas test
D.
IV. Tollen's test
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1A-III, B-I, C-IV, D-II
Approach:
Map each compound's principal functional group to its characteristic confirmatory wet test (alcohol -> Lucas; primary amine -> Hinsberg; aldehyde -> Tollen's; phenol -> phthalein dye).
Step 1:Identify the functional-group identity of each List-I compound: A is a secondary alcohol; B is a primary aliphatic amine; C is an aldehyde; D is a phenol. Each List-II test is selective for one of these groups.
A:ROH;B:RNH2;C:RCHO;D:ArOH
Step 2:Cyclohexanol (A): the Lucas reagent (HCl/ZnCl2) reacts with alcohols and the rate (turbidity time) discriminates 1∘,2∘,3∘ alcohols, giving the standard alcohol test.
A→III (Lucas test)
Step 3:Cyclohexylamine (B): benzenesulphonyl chloride (PhSO2Cl) reacts with the primary amine to give a sulphonamide whose acidic N-H is soluble in alkali, the hallmark of the Hinsberg test.
B→I (Hinsberg test)
Step 4:Cyclohexanecarbaldehyde (C): aldehydes reduce Tollen's reagent (ammoniacal AgNO3) to metallic silver, producing the silver-mirror.
C→IV (Tollen’s test)
Step 5:Phenol (D): condensation with phthalic anhydride in conc. H2SO4 yields phenolphthalein (pink in base), the phthalein dye test specific to phenols.
D→II (Phthalein dye test)
Step 6:Assemble the four pairs.
A-III,B-I,C-IV,D-II
Final answer: A-III, B-I, C-IV, D-II
Q71NumericalChemical Thermodynamics
If 3.365 g of ethanol (l) is burnt completely in a bomb calorimeter at 298.15 K, the heat produced is 99.472 kJ. The ∣ΔHf∘∣ of ethanol at 298.15 K is _____ ×102kJ mol−1. (Nearest integer)
Given: Standard enthalpy for combustion of graphite =−393.5kJ mol−1
Standard enthalpy of formation of water (l) =−285.8kJ mol−1
Molar mass in g mol−1 of C, H, O are 12, 1 and 16 respectively
SolutionAnswer: 3
Approach:
Convert the calorimetric heat to ΔUcomb per mole, add ΔngRT to obtain ΔHcomb, then invert Hess's law to extract ΔHf∘ of ethanol.
Step 1:Given: m=3.365 g, q=99.472 kJ (heat released), T=298.15 K. Molar mass of C2H5OH = 2(12)+6(1)+16=46 g/mol. Target: ∣ΔHf∘(C2H5OH,l)∣ in units of 102 kJ/mol.
M(C2H5OH)=46g mol−1
Step 2:Moles of ethanol burnt.
n=463.365=7.3152×10−2mol
Step 3:Molar ΔU of combustion from the bomb-calorimeter (constant-volume) heat (sign: exothermic ==> negative).
ΔUcomb=−0.0731599.472=−1359.83kJ mol−1
Step 4:Write the balanced combustion: C2H5OH(l)+3O2(g)→2CO2(g)+3H2O(l). Gaseous moles: Δng=2−3=−1.
ΔngRT=(−1)(8.314×10−3)(298.15)=−2.479kJ mol−1
Step 5:Convert to enthalpy of combustion.
ΔHcomb=−1359.83+(−2.479)=−1362.31kJ mol−1
Step 6:Apply Hess's law with ΔHf∘(O2)=0.
−1362.31=2(−393.5)+3(−285.8)−ΔHf∘(C2H5OH)
Step 7:Solve for ΔHf∘ of ethanol.
ΔHf∘=−787.0−857.4+1362.31=−282.09kJ mol−1
Step 8:Express the magnitude in units of 102 kJ/mol and round to the nearest integer.
102282.09=2.8209⇒3
Final answer: 3
Q72NumericalEquilibrium
For the following reaction at 50∘C and at 2 atm pressure, 2N2O5(g)⇌2N2O4(g)+O2(g) N2O5 is 50% dissociated. The magnitude of standard free energy change at this temperature is x. x= _____ J mol−1 [Nearest integer]. Given: R=8.314J mol−1K−1,log2=0.30,log3=0.48,ln10=2.303,∘C+273=K
SolutionAnswer: 2474
Approach:
Set up the ICE table with 50% dissociation, convert mole fractions to partial pressures at the stated total pressure, compute Kp, then apply ΔG∘=−RTlnKp=−2.303RTlogKp.
Step 1:Setup: start with n0 mol of N2O5; take n0=2 for convenient arithmetic. Degree of dissociation α=0.5 means 1 mol of N2O5 reacts.
T=50+273=323K;Ptot=2atm;α=0.5
Step 2:Stoichiometry of 2N2O5→2N2O4+O2: reaction of 1 mol N2O5 produces 1 mol N2O4 and 0.5 mol O2.
nN2O5=1,nN2O4=1,nO2=0.5,ntot=2.5
Step 3:Mole fractions and partial pressures at Ptot=2 atm.
An electrochemical cell, consist of the following two redox couples, Mx+(aq)/M(s)[Ered⊖=+0.15V] and Fe3+(aq)/Fe(s)[Ered⊖=−0.036V]. The cell EMF(Ecell) is recorded to be 0.2057 V. If the reaction quotient of the electrochemical reaction is found to be 10−2, then the value of x is _____. (Nearest integer) [Given: M is a p-block metal and F2.303RT=0.059V]
SolutionAnswer: 2
Approach:
Identify cathode and anode from standard potentials, compute Ecell∘, then apply the Nernst equation with the given reaction quotient to extract the number of electrons n; relate n to x from the balanced cell reaction.
Step 1:Given inputs: E∘(Mx+/M)=+0.15V, E∘(Fe3+/Fe)=−0.036V, Ecell=0.2057V, Q=10−2. Higher standard reduction potential means Mx+ is reduced (cathode); Fe is oxidised to Fe3+ (anode).
Cathode:Mx++xe−→M;Anode:Fe→Fe3++3e−
Step 2:Standard cell potential.
Ecell∘=0.15−(−0.036)=0.186V
Step 3:Balance the overall reaction by taking LCM of x and 3 electrons: multiply the cathode half by 3 and the anode half by x. Total electrons transferred n=3x.
3Mx++xFe→3M+xFe3+;n=3x
Step 4:Substitute into the Nernst equation with logQ=log(10−2)=−2.
0.2057=0.186−n0.059(−2)=0.186+n0.118
Step 5:Solve for n.
n0.118=0.2057−0.186=0.0197⇒n=0.01970.118=5.99
Step 6:From n=3x.
3x=6⇒x=2
Step 7:Sanity check on identity: x=2 with E∘≈+0.15V matches the p-block couple Sn2+/Sn (E∘≈+0.14V), consistent with the hint that M is a p-block metal.
Use the first-order integrated rate law to extract k from the (t=0, t=20) data, then invert it to solve for the time x at which the concentration has dropped by a factor of 10.
Step 1:Given: [A]0=0.6500 M, [A]x=0.0650 M at t=x, [A]20=0.00065 M at t=20 min. Target: x.
Step 2:Apply the first-order law at t=20 min to obtain the rate constant.
k=201ln(1000)=203ln10=203(2.303)=0.3454min−1
Step 3:Apply the same law at t=x.
kx=ln10=2.303
Step 4:Solve for x.
x=0.34542.303=320=6.667min
Step 5:Round to the nearest integer.
x→7min
Final answer: 7
Q75NumericalPrinciples Related to Practical Chemistry
In sulphur estimation, 2.0×10−3 mol of an organic compound (X) (molar mass 76g mol−1) gave 0.4813 g of barium sulphate (molar mass 233g mol−1). The percentage of sulphur in the compound (X) is _____ ×10−1% (Nearest integer)
SolutionAnswer: 435
Approach:
Each mole of sulphur in X ends up as one mole of BaSO4 in the Carius/gravimetric estimation. Convert mass of BaSO4 to mass of S, divide by the mass of organic compound taken, and express the result as a multiple of 10−1 %.
Step 1:Given: nX=2.0×10−3 mol, MX=76 g/mol, mBaSO4=0.4813 g, M(BaSO4)=233 g/mol, M(S)=32 g/mol. Target: %S in X expressed as ___×10−1%.
Stoichiometry:1S in X→1BaSO4
Step 2:Mass of organic compound X taken.
mX=nX×MX=(2.0×10−3)(76)=0.152g
Step 3:Moles of BaSO4 formed.
nBaSO4=2330.4813=2.0657×10−3mol
Step 4:Mass of S contained in that BaSO4 (one S per formula unit).
mS=nBaSO4×32=2.0657×10−3×32=0.06610g
Step 5:Percentage of sulphur in X.
%S=mXmS×100=0.1520.06610×100=43.49%
Step 6:Express in units of 10−1% and round to nearest integer.
10−143.49=43.49×10=434.9→435
Final answer: 435
Mathematics25 questions
Q1Single correctSets, Relations and Functions
For the function f:[1,∞)→[1,∞) defined by f(x)=(x−1)4+1, among the two statements: (I) The set S={x∈[1,∞):f(x)=f−1(x)} contains exactly two elements, and (II) The set S={x∈[1,∞):f(x)=f−1(x+1)} is an empty set,
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1only (I) is TRUE
Approach:
Reduce statement (I) using strict monotonicity to f(x)=x, and test statement (II) by applying the Intermediate Value Theorem to a continuous difference function.
Step 1:Given f(x)=(x−1)4+1 on [1,∞). The target is to determine the cardinality of SI={x:f(x)=f−1(x)} and SII={x:f(x)=f−1(x+1)}. Since f′(x)=4(x−1)3≥0 on [1,∞) with equality only at x=1, f is strictly increasing.
f strictly increasing on [1,∞)
Step 2:For (I), strict monotonicity forces the graphs of f and f−1 to meet only on y=x. Substituting f(x)=x gives (x−1)4+1=x, that is (x−1)4=x−1. With u=x−1≥0 this becomes u4=u, hence u(u3−1)=0.
u(u−1)(u2+u+1)=0⇒u=0,1
Step 3:For (II), the inverse is f−1(y)=(y−1)1/4+1, so f−1(x+1)=x1/4+1. The equation f(x)=f−1(x+1) reduces to (x−1)4+1=x1/4+1, that is (x−1)4=x1/4.
g(x)=(x−1)4−x1/4
Step 4:Evaluate endpoints. At x=1: g(1)=0−1=−1<0. At x=3: g(3)=16−31/4≈16−1.316>0. By the Intermediate Value Theorem applied to the continuous g on [1,3], there exists x0∈(1,3) with g(x0)=0.
∃x0∈(1,3):g(x0)=0
Step 5:Combining the two conclusions yields the verdict.
(I) TRUE and (II) FALSE
Final answer: only (I) is TRUE
Q2Single correctComplex Numbers and Quadratic Equations
Let S={z∈C:z2+4z+16=0}. Then ∑z∈S∣z+3i∣2 is equal to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 438
Approach:
Solve the quadratic for both complex roots, shift each by 3i, and add the squared moduli.
Step 1:Given z2+4z+16=0 with a=1,b=4,c=16. The target is Σ=∑z∈S∣z+3i∣2. Applying the quadratic formula: z=2−4±16−64=2−4±−48.
z=−2±23i
Step 2:Shift z1 by 3i and compute its squared modulus.
If the system of equations: x+y+z=5, x+2y+3z=9, x+3y+λz=μ has infinitely many solutions, then the value of λ+μ is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 218
Approach:
Set the coefficient determinant to zero to fix λ, then enforce row-reduction consistency to fix μ.
Step 1:Given the system with coefficient matrix A=11112313λ and constants (5,9,μ). The target is λ+μ. Expanding det(A) along the first row: 1⋅(2λ−9)−1⋅(λ−3)+1⋅(3−2).
Step 3:The third reduced row is exactly twice the second reduced row in the coefficients, so consistency demands μ−5=2⋅4=8.
μ−5=8⇒μ=13
Step 4:Add the two parameters.
λ+μ=5+13=18
Final answer: 18
Q4Single correctComplex Numbers and Quadratic Equations
If α=1 and β=1+i2, where i=−1 are two roots of the equation x3+ax2+bx+c=0, a,b,c∈R, then ∫−11(x3+ax2+bx+c)dx is equal to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3−8
Approach:
Use the real-coefficient conjugate root theorem to identify the third root, recover a,b,c via Vieta's, then integrate term by term.
Step 1:Given roots α=1 and β=1+i2 of a real cubic. The target is I=∫−11(x3+ax2+bx+c)dx. By the conjugate root theorem the third root is βˉ=1−i2.
roots: 1,1+i2,1−i2
Step 2:Apply Vieta's. Sum of roots: 1+(1+i2)+(1−i2)=3, so −a=3⇒a=−3.
a=−3
Step 3:Sum of pairwise products: 1⋅(1+i2)+1⋅(1−i2)+(1+i2)(1−i2)=2+(1+2)=5, so b=5. Product of roots: 1⋅(1+i2)⋅(1−i2)=1⋅3=3, so −c=3⇒c=−3.
b=5,c=−3
Step 4:Integrate over [−1,1]. Odd powers vanish by parity: ∫−11x3dx=0 and ∫−115xdx=0. The remaining contribution is ∫−11(−3x2−3)dx=−3⋅32−3⋅2=−2−6.
I=−2−6=−8
Final answer: −8
Q5Single correctComplex Numbers and Quadratic Equations
If the quadratic equation (λ+2)x2−3λx+4λ=0,λ=−2, has two positive roots, then the number of possible integral values of λ is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 22
Approach:
Impose discriminant non-negativity together with positive sum and positive product of roots, intersect the constraints, and count integers in the resulting interval.
Step 1:Given (λ+2)x2−3λx+4λ=0 with λ=−2. The target is the count of integer λ for which both roots are real and positive. Identify A=λ+2, B=−3λ, C=4λ.
A=λ+2,B=−3λ,C=4λ
Step 2:Discriminant constraint: D=B2−4AC=9λ2−16λ(λ+2)=−7λ2−32λ≥0. Dividing by −1 flips the inequality: 7λ2+32λ≤0, i.e. λ(7λ+32)≤0.
−732≤λ≤0
Step 3:Sum of roots λ+23λ>0 and product λ+24λ>0. Both ratios share sign, so each requires λ and λ+2 of the same sign.
λ(λ+2)>0⇒λ<−2 or λ>0
Step 4:Intersect the two regions. Within [−32/7,0], the portion satisfying λ<−2 or λ>0 is [−32/7,−2) (the point λ=0 is excluded since it makes the product zero, not positive).
λ∈[−32/7,−2)≈[−4.571,−2)
Step 5:Enumerate integers in [−32/7,−2). Since −32/7≈−4.571, the integers are −4 and −3.
{−4,−3}
Final answer: 2
Q6Single correctCo-ordinate Geometry
Let A=1432−2878−7 and det(A−αI)=0, where α is a real number. If the largest possible value of α is p, then the circle (x−p)2+(y−2p)2=320, intersects the co-ordinate axes at:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 33 points
Approach:
Compute the characteristic polynomial of A, extract the largest real eigenvalue, then intersect the circle with both coordinate axes and count distinct points.
Step 1:Given matrix A and the family of circles with centre (p,2p) and radius 320=85. The target is the count of distinct axis crossings. First form det(A−αI) by cofactor expansion along row 1 of A−αI.
Step 2:Expand each bracket. (−2−α)(−7−α)=α2+9α+14, so first bracket =α2+9α−50. Second bracket =−28−4α−24=−4α−52. Third bracket =32+6+3α=38+3α.
(1−α)(α2+9α−50)−2(−4α−52)+7(38+3α)
Step 3:Combine the three terms. (1−α)(α2+9α−50)=α2+9α−50−α3−9α2+50α=−α3−8α2+59α−50. Adding 8α+104+266+21α=29α+370 gives the characteristic polynomial.
−α3−8α2+88α+320=0⇔α3+8α2−88α−320=0
Step 4:Test α=8: 512+512−704−320=0, so α−8 is a factor. Polynomial division gives α3+8α2−88α−320=(α−8)(α2+16α+40). The quadratic factor has roots α=2−16±256−160=−8±26, both negative since 26<8.
α∈{8,−8+26,−8−26}
Step 5:Circle is (x−8)2+(y−16)2=320. Intersect with x-axis by setting y=0: (x−8)2+256=320⇒(x−8)2=64⇒x∈{0,16}. Intersect with y-axis by setting x=0: 64+(y−16)2=320⇒(y−16)2=256⇒y∈{0,32}.
x-axis: (0,0),(16,0);y-axis: (0,0),(0,32)
Step 6:The point (0,0) appears in both lists. Distinct points are (0,0), (16,0) and (0,32).
∣{(0,0),(16,0),(0,32)}∣=3
Final answer: 3 points
Q7Single correctSequence and Series
Let α=41+81+161+…∞ and β=31+91+271+…∞. Then the value of (0.2)log5(α)+(0.04)log5(β) is equal to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 38
Approach:
Sum each geometric series to a single value, simplify each exponent via change-of-base, then evaluate the two power terms.
Step 1:Given the two geometric series with first terms 41,31 and common ratios 21,31. Both ratios have absolute value less than 1. The target is E=(0.2)log5α+(0.04)log5β.
α=1−1/21/4=21;β=1−1/31/3=21
Step 2:Simplify the first exponent. By change of base, log521=log51/2log(1/2)=2log521=log541.
log5α=log541=−log54
Step 3:Evaluate the first term. Since 0.2=5−1, (0.2)−log54=5(−1)(−log54)=5log54=4.
(0.2)log5α=4
Step 4:Simplify the second exponent and evaluate. 0.04=5−2 and log5β=log521=−log52. Therefore (0.04)−log52=(5−2)−log52=52log52=5log54=4.
(0.04)log5β=4
Step 5:Add the two contributions.
E=4+4=8
Final answer: 8
Q8Single correctStatistics and Probability
For 10 observations x1,x2,…,x10, if ∑i=110(xi+2)2=180 and ∑i=110(xi−1)2=90, then their standard deviation is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 43
Approach:
Expand both shifted sum-of-squares to linear equations in ∑xi and ∑xi2, solve the system, then compute the variance and take the positive square root.
Step 1:Given n=10, ∑(xi+2)2=180 and ∑(xi−1)2=90. The target is the standard deviation σ. Let S1=∑xi and S2=∑xi2.
S1=∑i=110xi,S2=∑i=110xi2
Step 2:Expand the two given equations using the expansion identity.
S2+4S1+40=180(i);S2−2S1+10=90(ii)
Step 3:Subtracting (ii) from (i): 6S1+30=90, giving S1=10. Hence xˉ=S1/n=1.
S1=10,xˉ=1
Step 4:Substitute S1=10 into (ii): S2−20+10=90, giving S2=100.
S2=100
Step 5:Apply the variance formula and take the square root.
σ2=10100−12=9⇒σ=3
Final answer: 3
Q9Single correctBinomial Theorem and its Simple Applications
In the expansion of (9x−3x1)18,x>0, if the term independent of x is (221)k, then k is equal to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 184
Approach:
Form the binomial general term, find the index that nullifies the power of x, evaluate the resulting numerical coefficient, and divide by 221.
Step 1:Given A=9x, B=−3x1=−31x−1/2, n=18. The target is k such that the term independent of x equals 221k. The general term is Tr+1=(r18)(9x)18−r(−31x−1/2)r.
Tr+1=(r18)(−1)r918−r3−rx18−r−r/2
Step 2:Setting the exponent of x to zero: 18−r−2r=0⇒18=23r⇒r=12.
r=12⇒T13
Step 3:Evaluate the numerical factor at r=12. Since 9=32, 918−12=96=312 and 3−12⋅312=1. Also (−1)12=1.
Let P(3cosα,2sinα),α=0, be a point on the ellipse 9x2+4y2=1, Q be a point on the circle x2+y2−14x−14y+82=0 and R be a point on the line x+y=5 such that the centroid of the triangle PQR is (2+cosα,3+32sinα). Then the sum of the ordinates of all possible points R is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 48
Approach:
Use centroid coordinates to relate Q and R, eliminate Q using the circle, parametrise R on the line, and solve the resulting quadratic for the ordinates of R.
Step 1:Given P=(3cosα,2sinα), Q=(Qx,Qy) on the circle, R=(Rx,Ry) on x+y=5, and centroid G=(2+cosα,3+32sinα). The target is ∑Ry over admissible R. Multiplying the centroid coordinates by 3: 3cosα+Qx+Rx=6+3cosα and 2sinα+Qy+Ry=9+2sinα.
Qx+Rx=6,Qy+Ry=9
Step 2:From R on the line x+y=5: Ry=5−Rx. Substituting back: Qx=6−Rx and Qy=9−(5−Rx)=4+Rx.
Q=(6−Rx,4+Rx)
Step 3:Convert the circle to standard form: x2+y2−14x−14y+82=0⇔(x−7)2+(y−7)2=49+49−82=16. Centre (7,7), radius 4.
(x−7)2+(y−7)2=16
Step 4:Impose Q on the circle: (6−Rx−7)2+(4+Rx−7)2=16⇒(−1−Rx)2+(Rx−3)2=16. Expanding: (Rx+1)2+(Rx−3)2=Rx2+2Rx+1+Rx2−6Rx+9=2Rx2−4Rx+10.
2Rx2−4Rx+10=16⇒Rx2−2Rx−3=0
Step 5:Solve Rx2−2Rx−3=(Rx−3)(Rx+1)=0, so Rx∈{3,−1}. Corresponding ordinates from Ry=5−Rx are Ry∈{2,6}. Sum of ordinates equals 2+6.
∑Ry=2+6=8
Final answer: 8
Q11Single correctCo-ordinate Geometry
Let H:a2x2−b2y2=1 be a hyperbola such that the distance between its foci is 6 and the distance between its directrices is 38. If the line x=α intersects the hyperbola H at the points A and B such that the area of the triangle AOB is 415, where O is the origin, then α2 equals
(A)
(B)
(C)
(D)
SolutionAnswer: Option 216
Approach:
Combine 2ae=6 and 2a/e=8/3 to determine a2 and e2, derive b2, then equate the chord-triangle area to recover α2.
Step 1:Given 2ae=6 and e2a=38, equivalently ae=3 and ea=34. The target is α2. Multiplying the two equations: a2=ae⋅ea=3⋅34=4.
a2=4
Step 2:Dividing the equations: e2=a/eae=4/33=49.
e2=49
Step 3:Apply b2=a2(e2−1)=4(49−1)=4⋅45=5.
b2=5
Step 4:Substitute x=α in the hyperbola: 4α2−5y2=1⇒y2=45(α2−4). Hence A,B=(α,±25(α2−4)) and ∣AB∣=5(α2−4).
∣AB∣=5(α2−4)
Step 5:The perpendicular distance from O=(0,0) to the vertical line x=α is ∣α∣. Apply the area condition: 21∣α∣5(α2−4)=415. Squaring: 4α2⋅5(α2−4)=240, hence α2(α2−4)=192.
α2(α2−4)=192
Step 6:Let u=α2. Solve u2−4u−192=0: u=24±16+768=24±28=16 or −12. Since u=α2≥0, u=16.
α2=16
Final answer: 16
Q12Single correctLimit, Continuity and Differentiability
max0≤x≤π(16sin(2x)cos3(2x)) is equal to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 233
Approach:
Substitute t=x/2, locate the interior critical point via the derivative, evaluate the function there, and compare against the endpoint values.
Step 1:Given f(x)=16sin(x/2)cos3(x/2) on [0,π]. Substituting t=x/2, the variable runs over [0,π/2] and g(t):=16sintcos3t. The target is maxt∈[0,π/2]g(t).
Step 3:Setting g′(t)=0: either cost=0 giving t=π/2 (where g=0), or cos2t=3sin2t⇒tan2t=31⇒tant=31. In [0,π/2], this gives t=π/6.
t=π/6
Step 4:Evaluate g at t=π/6: sin(π/6)=21, cos(π/6)=23, cos3(π/6)=833.
g(π/6)=16⋅21⋅833=1616⋅33=33
Step 5:Compare with endpoint values g(0)=0 and g(π/2)=0. Since 33>0, the maximum is attained at t=π/6.
maxg=33
Final answer: 33
Q13Single correctThree Dimensional Geometry
The shortest distance between the lines r=(31i^+2j^+38k^)+λ(2i^−5j^+6k^) and r=(−32i^−31k^)+μ(j^−k^),λ,μ∈R, is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 23
Approach:
Apply the skew-line shortest-distance formula using the difference of position vectors, the cross product of the direction vectors, and the scalar triple product.
Step 1:Given a1=(31,2,38), d1=(2,−5,6), a2=(−32,0,−31), d2=(0,1,−1). The target is the shortest distance d. Compute the difference of position vectors: a2−a1=(−32−31,0−2,−31−38)=(−1,−2,−3).
a2−a1=(−1,−2,−3)
Step 2:Compute d1×d2. With d1=(2,−5,6) and d2=(0,1,−1): first component (−5)(−1)−(6)(1)=5−6=−1; second (6)(0)−(2)(−1)=0+2=2; third (2)(1)−(−5)(0)=2.
d1×d2=(−1,2,2)
Step 3:Magnitude of the cross product: (−1)2+22+22=1+4+4=3. Since this is non-zero, the lines are not parallel; combined with the next step it confirms they are skew.
∣d1×d2∣=3
Step 4:Scalar triple product: (a2−a1)⋅(d1×d2)=(−1)(−1)+(−2)(2)+(−3)(2)=1−4−6=−9. Its absolute value is 9.
∣(a2−a1)⋅(d1×d2)∣=9
Step 5:Apply the formula.
d=39=3
Final answer: 3
Q14Single correctThree Dimensional Geometry
If (2α+1,α2−3α,2α−1) is the image of (α,2α,1) in the line 3x−2=2y−1=1z, then the possible value(s) of α is (are)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1Only 3
Approach:
Find the foot of perpendicular F from P=(α,2α,1) on the line, set image Q=2F−P equal to the given image, and solve for α.
Step 1:Given: P=(α,2α,1), line passes through (2,1,0) with direction d=(3,2,1). Parametrize the foot as F=(2+3t,1+2t,t). Target: find α so that the reflection of P in the line equals (2α+1,α2−3α,(α−1)/2).
F=(2+3t,1+2t,t),d=(3,2,1)
Step 2:Impose (F−P)⋅d=0 to locate the foot.
3(2+3t−α)+2(1+2t−2α)+(t−1)=14t−7α+7=0
Step 3:Compute the image Q=2F−P component-wise. The x-component becomes Qx=4+6t−α=4+3(α−1)−α=2α+1, matching the given first coordinate identically.
Qx=2α+1,Qy=2+4t−2α=0,Qz=2t−1=α−2
Step 4:Equate Qy with the given second coordinate.
α2−3α=0⇒α(α−3)=0
Step 5:Equate Qz with the given third coordinate.
α−2=2α−1⇒2α−4=α−1
Step 6:Intersect the two solution sets to satisfy both coordinate equations simultaneously.
{0,3}∩{3}={3}
Final answer: Only 3
Q15Single correctVector Algebra
Let u^ and v^ be unit vectors inclined at an acute angle such that ∣u^×v^∣=23. If A=λu^+v^+(u^×v^), then λ is equal to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 134(A⋅u^)−32(A⋅v^)
Approach:
Use u^⋅v^=1/2 to compute A⋅u^ and A⋅v^, then express λ as a linear combination of these two scalars.
Step 1:Given: u^,v^ are unit vectors with sinθ=3/2 at an acute angle, so θ=π/3 and u^⋅v^=cos(π/3)=1/2. The vector A=λu^+v^+(u^×v^) is given. Target: express λ in terms of A⋅u^ and A⋅v^.
u^⋅v^=21
Step 2:Dot A with u^, using u^⋅u^=1 and (u^×v^)⋅u^=0.
A⋅u^=λ⋅1+u^⋅v^+0=λ+21
Step 3:Dot A with v^, using v^⋅v^=1 and (u^×v^)⋅v^=0.
A⋅v^=λ(u^⋅v^)+1+0=2λ+1
Step 4:Seek constants a,b such that λ=a(A⋅u^)+b(A⋅v^). Substituting the expressions yields λ=a(λ+1/2)+b(λ/2+1). Matching the coefficient of λ: a+b/2=1. Matching constants: a/2+b=0.
a+2b=1,2a+b=0
Step 5:Solving the system: b=−a/2 from the second equation, substitute into the first to give a−a/4=1, hence a=4/3 and b=−2/3.
a=34,b=−32
Step 6:Write λ in the required form.
λ=34(A⋅u^)−32(A⋅v^)
Final answer: 34(A⋅u^)−32(A⋅v^)
Q16Single correctSets, Relations and Functions
Let for some α∈R, f:R→R be a function satisfying f(x+y)=f(x)+2y2+y+αxy for all x,y∈R. If f(0)=−1 and f(1)=2, then the value of ∑n=15(α+f(n)) is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2140
Approach:
Set x=0 in the functional equation to obtain a closed form for f, then identify α by comparing with the equation at general (x,y), and finally evaluate the sum.
Step 1:Given: functional equation f(x+y)=f(x)+2y2+y+αxy, with f(0)=−1 and f(1)=2. Target: S=∑n=15(α+f(n)).
f(0)=−1,f(1)=2
Step 2:Substitute x=0 in the functional equation to express f(y) explicitly.
f(y)=f(0)+2y2+y=−1+2y2+y
Step 3:Substitute this f back into the original equation: f(x+y)−f(x)=2(x+y)2+(x+y)−1−(2x2+x−1)=4xy+2y2+y. Comparing with 2y2+y+αxy forces αxy=4xy for all x,y.
α=4
Step 4:Form the summand: α+f(n)=4+2n2+n−1=2n2+n+3. Sum over n=1,…,5 using ∑15n2=55 and ∑15n=15.
∑n=15(α+f(n))=2(55)+15+5⋅3
Step 5:Add the three contributions.
110+15+15=140
Final answer: 140
Q17Single correctPermutations and Combinations
A={(a,b,c):a,b,c are non-negative integers and a+b+2c=22}. Then n(A) is equal to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3144
Approach:
Fix c, count non-negative integer solutions of a+b=22−2c for each admissible c, then sum.
Step 1:Given: triples (a,b,c) of non-negative integers with a+b+2c=22. Target: n(A)= total such triples.
a,b,c∈Z≥0,a+b+2c=22
Step 2:Determine the admissible range of c from 2c≤22.
c∈{0,1,2,…,11}
Step 3:For a fixed c, a+b=22−2c has (22−2c)+1=23−2c non-negative integer solutions.
n(A)=∑c=011(23−2c)
Step 4:Evaluate the arithmetic sum: 23+21+19+⋯+1 has 12 terms with first term 23 and last term 1.
∑c=011(23−2c)=212(23+1)=12⋅12=144
Final answer: 144
Q18Single correctIntegral Calculus
The area of the region bounded by the curves x+3y2=0 and x+4y2=1 is equal to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 334
Approach:
Express both curves as x=g(y), find the y-limits where they meet, and integrate the horizontal width xright−xleft.
Step 1:Given: curve C1:x=−3y2 (leftward parabola through origin) and curve C2:x=1−4y2 (leftward parabola through (1,0)). Target: enclosed area.
xL=−3y2,xR=1−4y2
Step 2:Solve xL=xR to find the intersection limits in y.
−3y2=1−4y2⇒y2=1
Step 3:Determine which curve lies on the right between the limits by testing y=0: xL(0)=0 and xR(0)=1, so xR>xL inside the region. Compute the horizontal width.
xR−xL=(1−4y2)−(−3y2)=1−y2
Step 4:Integrate using even symmetry of the integrand.
A=∫−11(1−y2)dy=2∫01(1−y2)dy=2[y−3y3]01=2⋅32
Final answer: 34
Q19Single correctDifferential Equations
Let y=y(x) be the solution of the differential equation: dxdy+((x3+2)(2+e−2x)6x2+(3x2+2x3+4)e−2x)y=2+e−2x,x∈(−1,2), satisfying y(0)=23. If y(1)=α(2+e−2), then α is equal to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 41213
Approach:
Recognize the integrating factor μ(x)=2+e−2xx3+2 so that (μy)′=μ(2+e−2x)=x3+2, integrate, and apply the initial condition.
Step 1:Given: P(x)=(x3+2)(2+e−2x)6x2+(3x2+2x3+4)e−2x, Q(x)=2+e−2x, y(0)=3/2. Target: α with y(1)=α(2+e−2).
P(x),Q(x)=2+e−2x,y(0)=23
Step 2:Try μ(x)=(x3+2)/(2+e−2x) and verify μ′/μ=P(x). Logarithmic derivative gives x3+23x2+2+e−2x2e−2x. Combining over the common denominator (x3+2)(2+e−2x) produces (x3+2)(2+e−2x)3x2(2+e−2x)+2e−2x(x3+2)=(x3+2)(2+e−2x)6x2+(3x2+2x3+4)e−2x=P(x).
μ(x)μ′(x)=P(x)
Step 3:Multiply the ODE by μ. The right-hand side simplifies because μ(x)⋅(2+e−2x)=x3+2.
dxd[2+e−2x(x3+2)y]=x3+2
Step 4:Integrate both sides with respect to x.
2+e−2x(x3+2)y=4x4+2x+C
Step 5:Apply the initial condition at x=0: μ(0)=(0+2)/(2+1)=2/3, hence μ(0)y(0)=(2/3)(3/2)=1. Substituting gives C=1.
1=0+0+C⇒C=1
Step 6:Evaluate at x=1: μ(1)=3/(2+e−2) and the RHS equals 1/4+2+1=13/4.
2+e−23y(1)=413⇒y(1)=1213(2+e−2)
Step 7:Match with y(1)=α(2+e−2).
α=1213
Final answer: 1213
Q20Single correctIntegral Calculus
The integral ∫01cot−1(1+x+x2)dx is equal to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 42tan−12−21loge(45)−2π
Approach:
Rewrite cot−1(1+x+x2)=tan−1(x+1)−tan−1(x) using the inverse-tan subtraction identity, then evaluate each tan−1 integral by parts.
Step 1:Given: I=∫01cot−1(1+x+x2)dx. Observe 1+x+x2=1+(x+1)x, so 1+x+x21=1+(x+1)x(x+1)−x. Therefore cot−1(1+x+x2)=tan−11+x+x21=tan−1(x+1)−tan−1(x) (since for x∈[0,1] both arctan arguments give a positive difference).
I=∫01tan−1(x+1)dx−∫01tan−1(x)dx
Step 2:Substitute u=x+1 in the first integral.
∫01tan−1(x+1)dx=∫12tan−1udu
Step 3:Apply the standard antiderivative on [1,2].
Step 4:Apply the standard antiderivative on [0,1].
∫01tan−1xdx=[xtan−1x−21ln(1+x2)]01=4π−21ln2
Step 5:Subtract: collect the ln terms via −21ln5+21ln2+21ln2=−21ln5+ln2=−21ln(5/4), and the π terms via −π/4−π/4=−π/2.
I=2tan−12−21ln45−2π
Final answer: 2tan−12−21loge(45)−2π
Q21NumericalStatistics and Probability
From a month of 31 days, 3 different dates are selected at random. If the probability that these dates are in an increasing A.P. is equal to ba, where a,b∈N and gcd(a,b)=1, then a+b is equal to _____.
SolutionAnswer: 944
Approach:
Count favourable 3-term increasing APs from {1,…,31}, divide by total selections (331), reduce the fraction, and add numerator and denominator.
Step 1:Given: dates are drawn uniformly at random from {1,2,…,31}, three distinct values selected. Target: probability that the chosen triple forms an increasing 3-term AP, expressed as a/b in lowest terms; report a+b.
Sample∼(331)
Step 2:Compute the size of the sample space.
(331)=631⋅30⋅29=4495
Step 3:Count favourable triples. An increasing AP (a,a+d,a+2d) requires a≥1, a+2d≤31, d≥1, giving (31−2d) admissible starts for each d=1,2,…,15.
Favourable=∑d=115(31−2d)=29+27+⋯+1=152=225
Step 4:Form the probability and reduce. Factorize 4495=5⋅29⋅31 and 225=9⋅25, so gcd(225,4495)=5.
P=4495225=89945,899=29⋅31,gcd(45,899)=1
Step 5:Add numerator and denominator.
a+b=45+899=944
Final answer: 944
Q22NumericalLimit, Continuity and Differentiability
Let f(x)={ex−1x2−5x+6,x<0,x≥0 and g(x)=f(∣x∣)+∣f(x)∣. If the number of points where g is not continuous and is not differentiable are α and β respectively, then α+β is equal to _____.
SolutionAnswer: 4
Approach:
Analyze f(∣x∣) and ∣f(x)∣ separately for continuity and differentiability, then combine the failure sets in g.
Step 1:Given: f(x)=ex−1 for x<0 and f(x)=x2−5x+6 for x≥0. Compute f(0)=6 but limx→0−f(x)=e−1, so f is discontinuous at x=0. Target: count discontinuity points α and non-differentiability points β of g=f(∣x∣)+∣f(x)∣.
f(0)=6,limx→0−f=e−1
Step 2:Since ∣x∣≥0 always, f(∣x∣) uses only the second branch: f(∣x∣)=∣x∣2−5∣x∣+6=x2−5∣x∣+6. This is continuous everywhere; the ∣x∣ corner is multiplied by −5=0, so it is non-differentiable solely at x=0.
f(∣x∣)=x2−5∣x∣+6
Step 3:Analyze ∣f(x)∣. For x<0, f(x)=ex−1>0, so ∣f(x)∣=ex−1 is smooth. For x≥0, f(x)=(x−2)(x−3) has simple zeros at 2 and 3, so ∣f∣ has corners there. At x=0 the same jump in f persists in ∣f∣ since e−1=6.
∣f∣ discontinuous at 0;non-diff at {0,2,3}
Step 4:Determine discontinuities of g: f(∣x∣) is continuous everywhere, so g is discontinuous exactly where ∣f∣ is, namely at x=0. Hence α=1.
α=∣{0}∣=1
Step 5:Determine non-differentiability of g. The combined failure set is {0}∪{0,2,3}={0,2,3}. At x=2,3 the smooth part f(∣x∣) cannot cancel the corner from ∣f∣ (a smooth function plus a non-smooth one is non-smooth). At x=0, g is already discontinuous, hence non-differentiable. Thus β=3.
β=∣{0,2,3}∣=3
Step 6:Add the two counts.
α+β=1+3=4
Final answer: 4
Q23NumericalCo-ordinate Geometry
Let A,B be points on the two half-lines x−3∣y∣=α,α>0, at a distance of α from their point of intersection P. The line segment AB meets the angle bisector of the given half-lines at the point Q. If PQ=29 and R is the radius of the circumcircle of △PAB, then Rα2 is equal to _____.
SolutionAnswer: 9
Approach:
Identify the two half-lines and their apex angle, locate A,B,Q explicitly using symmetry, deduce that △PAB is equilateral, then compute the required ratio.
Step 1:Given: the equation x−3∣y∣=α with α>0 decomposes into ℓ1:x−3y=α (y≥0) and ℓ2:x+3y=α (y≤0). Both meet at P=(α,0). Each half-line makes angle ±30∘ with the positive x-axis (direction vectors (3,±1)/2), so the apex angle ∠APB=60∘ and the angle bisector is the x-axis (y=0).
P=(α,0),∠APB=60∘
Step 2:Place A on ℓ1 at distance α from P along the unit direction (3/2,1/2), and B symmetrically on ℓ2.
A=(α+23α,2α),B=(α+23α,−2α)
Step 3:Segment AB is vertical, so it intersects the bisector y=0 at its midpoint Q. The horizontal distance from P to Q gives PQ.
Q=(α+23α,0),PQ=23α=29⇒α=33
Step 4:With PA=PB=α and ∠APB=60∘, the law of cosines gives AB2=α2+α2−2α2cos60∘=α2, so AB=α. Hence △PAB is equilateral with side α.
AB=α,△PAB equilateral
Step 5:Apply the law of sines for the circumradius.
R=2sin60∘AB=3α
Step 6:Compute the required ratio.
Rα2=αα23=α3=33⋅3=9
Final answer: 9
Q24NumericalCo-ordinate Geometry
Let A,B and C be the vertices of a variable right angled triangle inscribed in the parabola y2=16x. Let the vertex B containing the right angle be (4,8) and the locus of the centroid of △ABC be a conic C0. Then three times the length of latus rectum of Co is _____.
SolutionAnswer: 16
Approach:
Parametrize A,C on the parabola, impose the right-angle condition at B, then eliminate the parameters from the centroid coordinates to obtain the locus.
Step 1:Given: parabola y2=16x gives 4a=16, so a=4 and the parametric form is (4t2,8t). The fixed vertex B=(4,8) corresponds to t=1. Let A=(4t12,8t1) and C=(4t22,8t2) with t1,t2=1.
B=(4,8),A=(4t12,8t1),C=(4t22,8t2)
Step 2:Form BA=(4t12−4,8t1−8)=4(t1−1)(t1+1,2), and similarly BC=4(t2−1)(t2+1,2). The right-angle condition at B requires BA⋅BC=0.
BA⋅BC=16(t1−1)(t2−1)[(t1+1)(t2+1)+4]=0
Step 3:Introduce s=t1+t2 and p=t1t2. The constraint expands to p+s+1=−4, i.e. p=−5−s. The centroid (h,k) is given by averaging the three vertices.
Step 4:From k=8(s+1)/3, s+1=3k/8. Substitute p=−5−s=−4−(s+1) into the h relation: s2−2p+1=s2+2s+8+1=(s+1)2+8.
3h=4[(s+1)2+8]+4−4=4(s+1)2+40
Step 5:Substitute (s+1)2=9k2/64.
3h−40=4⋅649k2=169k2
Step 6:Solve for k2 to obtain the standard parabola form k2=4A(h−h0).
k2=316(h−340)⇒4A=316
Step 7:Three times the latus rectum.
3ℓ=3⋅316=16
Final answer: 16
Q25NumericalIntegral Calculus
Let f be a twice differentiable function such that f(x)=∫0xtan(t−x)dt−∫0xf(t)tantdt,x∈(−2π,2π). Then f′′(6π)+12f′(−6π)+f(6π) is equal to _____.
SolutionAnswer: 5
Approach:
Simplify the first integral by linear substitution, differentiate the resulting integral equation, solve the first-order linear ODE, then evaluate at the required points.
Step 1:Given: the integral equation f(x)=∫0xtan(t−x)dt−∫0xf(t)tantdt on (−π/2,π/2), with f twice differentiable. Target: f′′(π/6)+12f′(−π/6)+f(π/6).
f(x)=∫0xtan(t−x)dt−∫0xf(t)tantdt
Step 2:Substitute u=t−x in the first integral (du=dt, limits −x→0). Then evaluate using cos(−x)=cosx.
Step 4:Let g(x)=1+f(x). Then g′(x)=−g(x)tanx, a separable equation. Integrating g′/g=−tanx gives ln∣g∣=ln∣cosx∣+C1, hence g(x)=Acosx for some constant A.
1+f(x)=Acosx
Step 5:Apply f(0)=0 (from the integral equation at x=0, both integrals vanish): 1=Acos0=A. Therefore f(x)=cosx−1, with f′(x)=−sinx and f′′(x)=−cosx.
How many questions are in the JEE Main 2026 April 04, Shift 2 paper?
The JEE Main 2026 April 04, Shift 2 paper has 75 questions — Physics (25), Chemistry (25) and Mathematics (25). Every question is on this page with its correct answer and a step-by-step solution.
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