JEE Main 2026 April 08, Shift 2 Question Paper with Solutions
All 72 questions from the JEE Main 2026 (April 08, Shift 2) shift — Physics (24), Chemistry (24) and Mathematics (24) — each with the correct answer (answer key) and a step-by-step solution. Read it free online, or attempt the full paper as a timed mock test.
A new unit (α) of length is chosen such that it is equal to the speed of light in vacuum. What is the distance between Venus and Earth in terms of α units if light takes 6 min. 40s to cover this distance?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2400α
Approach:
The chosen unit of length satisfies 1α=c in magnitude (the numerical value of the speed of light in vacuum). Since distance equals speed multiplied by time, the light-path distance in units of α reduces to the elapsed time in seconds.
Step 1:Given: speed of light c (m/s), travel time t=6min40s, and the new unit 1α=c. Target: distance d expressed in units of α.
c=c(m/s),1α=c,t=6min40s
Step 2:Convert the time into seconds.
t=6×60+40=360+40=400s
Step 3:Compute the path distance and convert to the new unit by dividing by α=c.
d=ct=c×400s;αd=c400cs=400
Step 4:Dimensional check: α carries the dimension of c (length/time), and d/α=(ct)/c=t leaves a pure number equal to the seconds elapsed, giving 400. Result: the distance is 400α.
αd=[c][c][T]=[T](numeric)=400
Final answer: d=400α
Q27Single correctUnits and Measurements
Consider the equation H=tsxpϵqEr Where H = magnetic field; E = electric field, ϵ = permittivity, x = distance, t = time. The values of p, q, r and s respectively are:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 11,1,1,1
Approach:
Apply the principle of dimensional homogeneity. Express each quantity in base dimensions (M, L, T, A), substitute into H=xpϵqErt−s, and equate the exponent of each base dimension on both sides to obtain a solvable linear system. Here H is the magnetic field strength (units A/m), consistent with the option set.
Step 1:Base dimensions of each quantity. Target: exponents p,q,r,s.
Step 2:Substitute into H=xpϵqErt−s and equate exponents of M, L, T, A.
M:0=−q+r;A:1=2q−r;T:0=4q−3r−s;L:−1=p−3q+r
Step 3:Solve the system. From M, r=q. Substituting into A: 1=2q−q=q, so q=1 and r=1. From T: s=4q−3r=4−3=1. From L: −1=p−3(1)+1, giving p=1.
q=1,r=1,s=1,p=1
Step 4:Back-substitution of the full product xϵEt−1. Result: the exponents are 1,1,1,1.
L⋅(M−1L−3T4A2)⋅(MLT−3A−1)⋅T−1=M0L−1T0A=[H]
Final answer: p=1,q=1,r=1,s=1
Q28Single correctLaws of Motion
A car moving with a speed of 54 km/h takes a turn of radius 20 m. A simple pendulum is suspended from the ceiling of the car. Determine the angle made by the string of the pendulum with the vertical during the turning. (Take g = 10 m/s2)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3tan−1(1.125)
Approach:
In the non-inertial frame of the turning car, the bob experiences gravity mg downward and a horizontal centrifugal pseudo-force mv2/r outward. The string aligns with the resultant, so the angle with the vertical satisfies tanθ=(v2/r)/g.
Step 1:Given: speed v=54km/h, turn radius r=20m, g=10m/s2. Target: angle θ of the string with the vertical.
Step 4:Dimensional check: v2/(rg) has dimensions (m2s−2)/(m⋅ms−2), which is dimensionless, suitable for a tangent. Result: the angle follows.
θ=tan−1(1.125)
Final answer: θ=tan−1(1.125)
Q29Single correctKinematics
A gas balloon is going up with a constant velocity of 10 m/s. When this balloon reached a height of 75 m, a stone is dropped from it and balloon keeps moving up with the same velocity. The height of the balloon when the stone hits the ground is ...... m. (Take g = 10 m/s2)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4125
Approach:
At release the stone shares the balloon's upward velocity of 10 m/s and then moves under gravity from a height of 75 m. Solve for the time to reach the ground, then add the distance the balloon rises in that time (at 10 m/s) to its release height of 75 m.
Step 1:Quantities (upward positive): stone's initial velocity u=+10m/s, ground displacement s=−75m, acceleration a=−10m/s2; balloon rises at v=10m/s. Target: balloon height when the stone lands.
u=+10m/s,s=−75m,a=−10m/s2
Step 2:Apply the stone's displacement equation.
−75=10t−21(10)t2⇒−75=10t−5t2⇒5t2−10t−75=0
Step 3:Solve the quadratic and keep the positive root for time of flight.
t2−2t−15=(t−5)(t+3)=0⇒t=5s
Step 4:Add the balloon's rise during this time to its release height. Check: substituting t=5 into the stone equation gives 10(5)−5(25)=50−125=−75m, matching the ground level.
hballoon=75+10×5=75+50=125m
Final answer: hballoon=125 m
Q30Single correctOptics
A thin biconvex lens is prepared from the glass (μ=1.5) both curved surfaces of which have equal radii of 20 cm each. Left side surface of the lens is silvered from outside to make it reflecting. To have the position of image and object at the same place, the object should be placed, from the lens at a distance of ......... cm.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 110
Approach:
A lens with one face silvered acts as an equivalent mirror whose power is Peq=2Plens+Pmirror (light crosses the lens twice and reflects once). The object coincides with its own image when placed at the centre of curvature of the equivalent mirror, i.e. at 2feq.
Step 3:Combine into the equivalent mirror power and its focal length.
Peq=2×201+101=101+101=51cm−1⇒feq=5cm
Step 4:For a self-coincident image the object sits at the centre of curvature of the equivalent concave mirror, a distance 2feq. At u=Req=2feq, mirror rays retrace and form the image on the object.
u=2feq=2×5=10cm
Final answer: u=10 cm
Q31Single correctKinematics
Two identical bodies, projected with the same speed at two different angles cover the same horizontal range R. If the time of flight of these bodies are 5 s and 10 s, respectively, then the value of R is ________ m. (Take g = 10 m/s2)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1250
Approach:
Equal range from the same speed at two different angles requires complementary projection angles θ and 90∘−θ. The product of their times of flight relates to the common range through R=21gT1T2.
Step 1:Given: complementary angles θ and 90∘−θ, common speed u, flight times T1=5s, T2=10s, g=10m/s2. Target: range R.
T1=g2usinθ,T2=g2ucosθ
Step 2:Form the product of the two flight times.
T1T2=g24u2sinθcosθ=g22u2sin2θ=g2R
Step 3:Rearrange and substitute the numerical values.
R=21gT1T2=21×10×5×10=250m
Step 4:Dimensional check: gT1T2 has dimensions (m/s2)(s)(s)=m, correct for a range.
[gT1T2]=s2m⋅s2=m
Final answer: R=250 m
Q32Single correctRotational Motion
A solid cylinder having radius R and length L is slipping on a rough horizontal plane. At time t = 0 the cylinder has a translational velocity v0=49 m/s, perpendicular to its axis and a rotational velocity v0/4R about the centre. The time taken by the cylinder to start rolling is ____ seconds. (coefficient of kinetic friction μK=0.25 and g = 9.8 m/s2)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 25
Approach:
The translational velocity exceeds the surface speed of rotation (v0>ω0R=v0/4), so the contact point slips forward and kinetic friction acts backward. Friction decelerates the centre of mass while its torque about the centre increases the angular velocity. Pure rolling begins when v(t)=ω(t)R.
Step 1:Given: v0=49m/s, ω0=v0/(4R) so ω0R=v0/4=12.25m/s, μK=0.25, g=9.8m/s2. Target: time t to begin rolling.
μKg=0.25×9.8=2.45m/s2,ω0R=449=12.25m/s
Step 2:Write the translational velocity, decelerating under friction.
v(t)=v0−μKgt=49−2.45t
Step 3:Surface rotational speed. Torque τ=μKMgR=Iα with I=21MR2 gives α=2μKg/R, so ω(t)R=ω0R+2μKgt.
ω(t)R=12.25+2(2.45)t=12.25+4.9t
Step 4:Apply the rolling condition v(t)=ω(t)R and solve. Check: at t=5, v=49−12.25=36.75m/s and ωR=12.25+24.5=36.75m/s, equal.
49−2.45t=12.25+4.9t⇒36.75=7.35t⇒t=5s
Final answer: t=5 s
Q33Single correctProperties of Solids and Liquids
A liquid of density 600 kg/m3 flowing steadily in a tube of varying cross-section. The cross-section at a point A is 1.0 cm2 and that at B is 20 mm2. Both the points A and B are in same horizontal plane, the speed of the liquid at A is 10 cm/s. The difference in pressures at A and B points is ______ Pa.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 472
Approach:
Find the speed at B from the equation of continuity, then apply Bernoulli's equation at the same horizontal level (no height term) to obtain the pressure difference PA−PB=21ρ(vB2−vA2).
Step 4:Dimensional check: ρv2 gives (kg/m3)(m2/s2)=kgm−1s−2=Pa. The smaller area at B yields higher speed and lower pressure, consistent with a positive PA−PB.
[ρv2]=kgm−1s−2=Pa
Final answer: ΔP=72 Pa
Q34Single correctProperties of Solids and Liquids
A spherical liquid drop of radius R acquires the terminal velocity v1 when falls through a gas of viscosity η. Now the drop is broken into 64 identical droplets and each droplet acquires terminal velocity v2 falling through the same gas. The ratio of terminal velocities v1/v2 is ____ .
(A)
(B)
(C)
(D)
SolutionAnswer: Option 416
Approach:
By Stokes' law the terminal velocity of a sphere is proportional to the square of its radius. Conserving total volume when the drop splits into 64 identical droplets fixes the small-droplet radius, after which the velocity ratio follows from the radius ratio squared.
Step 1:Given: original radius R with terminal velocity v1; 64 identical droplets of radius r with terminal velocity v2, same ρ, σ, η, g. Target: v1/v2.
v1∝R2,v2∝r2
Step 2:Apply volume conservation to find the droplet radius.
34πR3=64×34πr3⇒R3=64r3⇒r=4R
Step 3:Form the velocity ratio using vT∝r2.
v2v1=r2R2=(R/4R)2=42=16
Step 4:Consistency check: the larger drop falls faster, so v1/v2>1, and a fourfold larger radius scales velocity by 42=16.
v2v1=16>1
Final answer: v2v1=16
Q36Single correctThermodynamics
Initial pressure and volume of a monoatomic ideal gas are P and V. The change in internal energy of this gas in adiabatic expansion to volume Vfinal=27V is ____ J.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3−34PV
Approach:
Applying the adiabatic relation PVγ=const for a monoatomic gas (γ=5/3) gives the final pressure, and ΔU=γ−1P2V2−P1V1 then gives the change in internal energy, since the internal energy of an ideal gas depends only on the state quantity PV.
Step 1:Given the initial state P1=P,V1=V, final volume V2=27V, and a monoatomic gas with γ=5/3 and γ−1=2/3. The target is ΔU in joules.
P1=P,V1=V,V2=27V,γ=35
Step 2:The final pressure follows from the adiabatic condition.
P2=P(V2V1)γ=P(271)5/3=P⋅(27)5/31=35P=243P
Step 3:The product P2V2 goes into the internal-energy expression.
Step 4:An adiabatic expansion does positive work with Q=0, so ΔU=−W<0 and the negative sign is consistent. The change in internal energy is −34PV.
ΔU=−34PV<0
Final answer: ΔU=−34PV J
Q37Single correctOscillations and Waves
The frequency of oscillation of a mass m suspended by a spring is υ1. If the length of the spring is cut to half, the same mass oscillates with frequency υ2. The value of υ2/υ1 is ______ .
(A)
(B)
(C)
(D)
SolutionAnswer: Option 32
Approach:
The force constant of a uniform spring is inversely proportional to its natural length, so halving the length doubles the constant. The relation υ=2π1k/m with the same mass gives the ratio.
Step 1:The original spring of length L has constant k, and the same mass m hangs from it. The target is the ratio υ2/υ1 after the spring length is reduced to L/2.
υ1=2π1mk
Step 2:Since k∝1/L, cutting the length in half doubles the force constant.
k2=k⋅L/2L=2k
Step 3:The new frequency forms a ratio with the original, and m cancels.
υ2=2π1m2k;υ1υ2=k/m2k/m=2
Step 4:A stiffer spring with larger k raises the frequency; doubling k scales the frequency by 2, consistent with the result.
υ1υ2=2≈1.414
Final answer: υ1υ2=2
Q38Single correctDual Nature of Matter and Radiation
A monochromatic source of light operating at 15 kW emits 2.5×1022 photons/s. The region of an electromagnetic spectrum to which the emitted electromagnetic radiation belongs to ____. (Take h=6.6×10−34 J.s and c=3×108 m/s).
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4Ultraviolet
Approach:
The energy per photon equals the emitted power divided by the photon emission rate. This converts to a wavelength via λ=hc/E, which is classified against the visible band (400-700 nm).
Step 1:Power P=15kW=15×103W, photon rate n=2.5×1022s−1, h=6.6×10−34J⋅s, and c=3×108m/s. The target is the spectral region.
P=15×103W,n=2.5×1022s−1
Step 2:Energy of a single photon.
E=nP=2.5×102215×103=6×10−19J
Step 3:The photon energy converts to a wavelength.
Step 4:Since 330nm<400nm, the violet edge of the visible band, the radiation lies in the ultraviolet region.
λ=330nm<400nm
Final answer: Ultraviolet
Q39Single correctMagnetic Effects of Current and Magnetism
A current carrying circular loop of radius 2 cm with unit normal n^=2k^+i^ is placed in a magnetic field, B=Bo(3i^+2k^). If Bo=4×10−3 T and current I=1002 A, the torque experienced by the loop is ____ Wb.A. (π=3.14).
(A)
(B)
(C)
(D)
SolutionAnswer: Option 45024×10−7j^
Approach:
The magnetic dipole moment is m=IAn^ with A=πr2, and the torque is the cross product τ=m×B.
Step 1:Here r=2cm=0.02m, I=1002A, n^=2i^+k^, and B=4×10−3(3i^+2k^)T. The area follows.
A=πr2=3.14×(0.02)2=1.256×10−3m2
Step 2:Magnetic moment vector; the 2 in I cancels the 2 in n^.
Step 4:Both m and B lie in the x-z plane, so their cross product must be along ±j^; the net coefficient (0.012−0.008)=0.004 is positive, giving +j^.
τ=5024×10−7j^
Final answer: τ=5024×10−7j^ Wb⋅A
Q40Single correctElectromagnetic Induction and Alternating Currents
A 30 cm long solenoid has 10 turns per cm and area of 5 cm2. The current through the solenoid coil varies from 2 A to 4 A in 3.14 s. The e.m.f. induced in the coil is α×10−5 V. The value of α is ____ .
(A)
(B)
(C)
(D)
SolutionAnswer: Option 212
Approach:
The self-inductance is L=μ0n2Al with n the turns per unit length, and the induced emf follows from ε=LdI/dt.
Step 1:Here n=10turns/cm=1000m−1, A=5cm2=5×10−4m2, l=0.30m, dI=4−2=2A, dt=3.14s, and μ0=4π×10−7T⋅m/A.
n=1000m−1,A=5×10−4m2,l=0.30m
Step 2:Self-inductance of the solenoid.
L=μ0n2Al=4π×10−7×(1000)2×5×10−4×0.30=4π×10−7×150
Step 3:Induced emf from the rate of current change.
ε=LdtdI=1.884×10−4×3.142=1.2×10−4V
Step 4:From ε=α×10−5V, α=12. The factor π in μ0 and the dt=3.14s nearly cancel, leaving a clean numeric value.
α=10−51.2×10−4=12
Final answer: α=12
Q41Single correctElectrostatics
Two point charges q1=3μC and q2=−4μC are placed at points (2i^+3j^+3k^) and (i^+j^+k^) respectively. Force on charge q2 is ____ N. (Take 4πϵ01=9×109 SI Units).
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2(4i^+8j^+8k^)×10−3
Approach:
The displacement vector from q1 to q2 and its magnitude feed Coulomb's law in vector form F=4πϵ01r3q1q2r12 for the force on q2.
Step 1:Here q1=3×10−6C at (2,3,3), q2=−4×10−6C at (1,1,1), and k=9×109. The target is the force on q2.
Step 4:Multiplying the prefactor by r12 gives the force. Opposite charges attract, so the force on q2 points toward q1, i.e. along +(i^+2j^+2k^), consistent with the positive components.
F=−4×10−3(−i^−2j^−2k^)=(4i^+8j^+8k^)×10−3N
Final answer: (4i^+8j^+8k^)×10−3 N
Q42Single correctOptics
Light ray incident along a vector AO(AO=2i^−3j^) emerges out along vector OB(OB=Ci^−4j^) as shown in the figure below. The value of C is ____ . (Medium 1 above the interface has μ1=1, medium 2 below has μ2=1.5.)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 11.6
Approach:
The interface is horizontal, so the normal is the y-direction. The angle each ray makes with the normal has a sine equal to (horizontal component)/(vector magnitude). Snell's law μ1sinα=μ2sinβ then gives C.
Step 1:The incident ray AO=2i^−3j^ is in medium 1 (μ1=1) and the refracted ray OB=Ci^−4j^ is in medium 2 (μ2=1.5), with the normal along j^. The sine of incidence:
∣AO∣=22+32=13;sinα=132
Step 2:Sine of the angle of refraction from the components of OB.
Step 4:Substituting C=1.6 gives sinβ=1.6/18.56=0.3714, so μ2sinβ=1.5×0.3714=0.557 and μ1sinα=2/13=0.5547. Both sides agree, and β<α (bending toward the normal in the denser medium) as shown.
μ1sinα=0.555≈μ2sinβ=0.557
Final answer: C=1.6
Q43Single correctDual Nature of Matter and Radiation
K1 and K2 be the maximum kinetic energies of photoelectrons emitted from a surface of a given material for the light of wavelength λ1 and λ2, respectively. If λ1=2λ2 then the work function of material is given by :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4K2−2K1
Approach:
Einstein's photoelectric equation holds for both wavelengths. Since λ1=2λ2, the photon energy at λ2 is twice that at λ1. Eliminating the photon energies expresses the work function in K1,K2.
Step 1:With E1=hc/λ1 and E2=hc/λ2 as the photon energies, the photoelectric equation for each wavelength is:
E1=ϕ+K1,E2=ϕ+K2
Step 2:Photon energy is inversely proportional to wavelength, and λ1=2λ2 gives E2=2E1.
E2=λ2hc=λ1/2hc=2E1
Step 3:Substituting E1=ϕ+K1 and E2=ϕ+K2 into E2=2E1 and solving for ϕ.
ϕ+K2=2(ϕ+K1)=2ϕ+2K1⇒ϕ=K2−2K1
Step 4:The shorter wavelength λ2 delivers higher photon energy, giving the larger kinetic energy K2; back-substitution reproduces E2=2E1, consistent with the result.
ϕ=K2−2K1
Final answer: ϕ=K2−2K1
Q44Single correctAtoms and Nuclei
Two radioactive substances A and B of mass numbers 200 and 212 respectively, shows spontaneous α-decay with same Q value of 1 MeV. The ratio of energies of α-rays produced by A and B is ____ .
(A)
(B)
(C)
(D)
SolutionAnswer: Option 326002597
Approach:
In two-body alpha decay, momentum conservation gives the alpha kinetic energy as Kα=Q⋅AA−4, with A the parent mass number. The ratio for the two parents removes the common Q.
Step 1:Parent A has A1=200 and parent B has A2=212, both with Q=1MeV, and the daughter mass number is A−4 in each case. The target is Kα,A/Kα,B.
Step 4:Dividing numerator and denominator by their common factor 16.
4160041552=41600/1641552/16=26002597
Final answer: 26002597
Q45Single correctElectronic Devices
The output Y for the given inputs A and B to the circuit is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2Waveform 2
Approach:
The network has two AND gates feeding a final OR gate. The gate outputs are evaluated interval by interval from the input waveforms for A and B, and the output waveform Y is then assembled.
Step 1:From the input waveforms, B = 1 during 0-1 s (A = 0); A = 1 during 1-2 s (B = 0); both A and B = 0 during 2-3 s. The two AND gates feed an OR gate, so the effective output is high when either A or B is high.
0−1s:A=0,B=1;1−2s:A=1,B=0;2−3s:A=0,B=0
Step 2:Interval 0-1 s: A = 0, B = 1. With one input high, the OR combination of the gate outputs drives Y high.
Y(0−1)=A+B=0+1=1
Step 3:Interval 1-2 s: A = 1, B = 0. One input high again keeps Y high.
Y(1−2)=A+B=1+0=1
Step 4:Interval 2-3 s: A = 0, B = 0, so Y = 0. Thus Y is a single continuous high pulse from 0 to 2 s and low afterward, which is Waveform 2.
Y(2−3)=0+0=0;Y=1for0≤t≤2s,Y=0fort>2s
Final answer: Y is high over 0-2 s and low afterward, corresponding to Waveform 2.
Q46NumericalElectrostatics
A parallel plate capacitor is having separation between plates 0.885mm. It has a capacitance of 1μF when the space between the plates is filled with an insulating material of resistivity 1×1013Ωm and resistance 17.7×1014Ω. Relative permittivity of the insulating material is α×107. The value of α is ________. (Take permittivity of free space =8.85×10−12F/m)
SolutionAnswer: 2
Approach:
The same slab of dielectric fixes both the resistance R=ρd/A and the capacitance C=ε0εrA/d. The geometric ratio d/A is extracted from the resistance relation and substituted into the capacitance relation to solve for the relative permittivity εr.
Step 1:The givens with units are: resistivity ρ=1×1013Ωm, resistance R=17.7×1014Ω, capacitance C=1×10−6F, ε0=8.85×10−12F/m; the target is εr=α×107, with α to be found.
ρ=1013Ωm,R=17.7×1014Ω,C=10−6F
Step 2:From the resistance relation the geometric factor d/A is extracted.
Ad=ρR=1×101317.7×1014=177m−1
Step 3:Rearranging the capacitance relation gives εr in terms of C, ε0 and d/A.
εr=ε0ACd=ε0C⋅Ad=8.85×10−121×10−6×177
Step 4:Evaluating the product yields εr.
εr=1.130×105×177=2.0×107
Step 5:Dimensional consistency: ε0C⋅Ad=F/mF⋅m−1=m⋅m−1 is dimensionless, consistent with a relative permittivity.
[ε0C][Ad]=F/mF⋅m−1=1
Step 6:Comparing εr=α×107 with the computed value gives α.
α×107=2.0×107⇒α=2
Final answer: α=2
Q47NumericalOptics
Some distant star is to be observed by some telescope of diameter of objective lens a, at an angular resolution of 3.0×10−7 radian. If the wavelength of light from the star reaching the telescope is 500nm, the minimum diameter of the objective lens of the telescope is ______ cm. (nearest integer)
SolutionAnswer: 203
Approach:
The angular resolution of a telescope objective is set by the Rayleigh criterion for a circular aperture, θ=1.22λ/a. Rearranging for the aperture diameter a and substituting the given wavelength and angular resolution yields the minimum lens diameter.
Step 1:The givens with units are: angular resolution θ=3.0×10−7rad, wavelength λ=500nm=5×10−7m; the target is the aperture diameter a in cm.
θ=3.0×10−7rad,λ=5×10−7m
Step 2:The Rayleigh resolution criterion for the circular objective is written.
θ=a1.22λ
Step 3:Rearranging for the aperture diameter a.
a=θ1.22λ=3.0×10−71.22×5×10−7
Step 4:Evaluating the quotient gives a in metres.
a=2.033m
Step 5:Converting to centimetres and rounding to the nearest integer.
a=2.033m×100=203.3cm≈203cm
Step 6:The minimum objective diameter follows.
amin=203cm
Final answer: amin=203cm
Q48NumericalMagnetic Effects of Current and Magnetism
A 5mg particle carrying a charge of 5π×10−6C is moving with velocity of (3i^+2k^)×10−2m/s in a region having magnetic field B=0.1k^Wb/m2. It moves a distance of α meter along k^ when it completes 5 revolutions. The value of α is ______.
SolutionAnswer: 2
Approach:
The velocity component perpendicular to B (the i^ part) drives uniform circular motion, while the component along B (the k^ part) is unaffected and produces a steady drift, making the path a helix. The displacement along k^ over 5 revolutions equals the parallel velocity times the time for 5 cyclotron periods.
Step 1:The givens with units are: mass m=5mg=5×10−6kg, charge q=5π×10−6C, field B=0.1Wb/m2 along k^; the target is the axial distance α over n=5 revolutions.
m=5×10−6kg,q=5π×10−6C,B=0.1T
Step 2:The velocity is resolved relative to B (along k^): the perpendicular component along i^ and the parallel component along k^.
v⊥=3×10−2m/s,v∥=2×10−2m/s
Step 3:The cyclotron period is computed from mass, charge and field.
Step 5:Axial displacement equals the parallel velocity times the elapsed time.
α=v∥t=(2×10−2)(100)=2m
Step 6:By units, (m/s)(s)=m, and the period is independent of speed (cyclotron property), confirming consistency. The axial distance follows.
[α]=m/s⋅s=m
Final answer: α=2m
Q49NumericalCurrent Electricity
The stored charge in the capacitor in steady state of the following circuit is ______ μC.
SolutionAnswer: 200
Approach:
In steady state no current flows through the capacitor branch, so that branch is treated as open and current flows only through the resistive ladder. Node potentials are obtained by Kirchhoff's current law applied to the reduced network; the potential difference appearing across the open capacitor branch is then used in Q=CV.
Step 1:Source E=12V, capacitance C=100μF=100×10−6F. In steady state the capacitor current is zero, so its branch is an open circuit and carries no current.
IC=0⇒capacitor branch open
Step 2:With the capacitor branch open, no current passes through the top 10Ω resistor leading to the capacitor node, so that node sits at the potential of the junction before it. The four active interior nodes (top-mid B, top-right C, bottom-mid E, bottom-right F) are referenced to the source negative at 0V and the positive at 12V.
V+=12V,V−=0V
Step 3:Applying KCL at the interior nodes for the current-carrying ladder (5Ω and 4Ω top, 10Ω and 4Ω vertical, 2Ω and 2Ω bottom). Solving the linear node equations gives the node potentials.
VB=7V,VC=5V,VE=2V,VF=3V
Step 4:The capacitor sits between the top-right node (at VC=5V, carried unchanged across the current-free 10Ω) and the bottom-right node F at 3V. The steady-state voltage across the capacitor is the difference.
VCcap=5−3=2V
Step 5:Applying Q=CV with C=100μF and V=2V.
Q=CV=(100μF)(2V)=200μC
Step 6:The assumption IC=0 is self-consistent because the computed node potentials satisfy KCL with no current in the capacitor branch. The stored charge follows.
Q=200μC
Final answer: Q=200μC
Q50NumericalKinematics
Two masses of 3.4kg and 2.5kg are accelerated from an initial speed of 5m/s and 12m/s, respectively. The distances traversed by the masses in the 5th second are 104m and 129m, respectively. The ratio of their momenta after 10s is 8x. The value of x is ______.
SolutionAnswer: 9
Approach:
The distance covered in the n-th second, sn=u+2a(2n−1), fixes each acceleration. The velocity after 10s follows from v=u+at, the momentum from p=mv, and the ratio p1/p2 is matched to x/8.
Step 1:The givens with units are: mass 1 m1=3.4kg, u1=5m/s, s5(1)=104m; mass 2 m2=2.5kg, u2=12m/s, s5(2)=129m; t=10s. The target is x where p1/p2=x/8.
m1=3.4,u1=5;m2=2.5,u2=12;n=5,t=10
Step 2:The n-th-second relation for mass 1 (n=5, so 2n−1=9) is solved for a1.
104=5+2a1(9)⇒99=4.5a1⇒a1=22m/s2
Step 3:The same relation for mass 2 is solved for a2.
129=12+2a2(9)⇒117=4.5a2⇒a2=26m/s2
Step 4:The velocities after 10s follow from v=u+at.
v1=5+22×10=225m/s;v2=12+26×10=272m/s
Step 5:The momenta p=mv are computed and the ratio is formed.
p2p1=2.5×2723.4×225=680765=89
Step 6:Since 765/680=1.125=9/8, a pure dimensionless ratio, matching to x/8 gives x.
8x=89⇒x=9
Final answer: x=9
Chemistry24 questions
Q51Single correctSome Basic Concepts in Chemistry
Match List - I with List - II.
List - I Mass of substance
List - II Number of atoms
A. 1.8 mg water
I.2×10−4×NA
B. 9.8 mg sulphuric acid
II.1.5×10−4×NA
C. 1.8 mg carbon
III.3×10−4×NA
D. 5.85 mg salt (NaCl)
IV.7×10−4×NA
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3A-III, B-IV, C-II, D-I
Approach:
For each mass, the moles follow from the molar mass, multiplied by the number of atoms in one formula unit, giving the total number of atoms as a multiple of Avogadro's number NA, which fixes the matching List-II value.
Step 1:Each List-I mass converts to moles and then to total atoms. Water H2O has M=18 and 3 atoms per molecule; H2SO4 has M=98 and 7 atoms; carbon has M=12 and 1 atom; NaCl has M=58.5 and 2 atoms per formula unit.
Step 4:Each mass gives an integral or half-integral multiple of 10−4 mol, and every List-II value is used exactly once, giving a one-to-one assignment.
A→III,B→IV,C→II,D→I
Final answer: A-III, B-IV, C-II, D-I
Q52Single correctSolutions
Given below are two statements: Given: Molar mass of C, H, O, Cl are 12, 1, 16 and 35.5 gmol−1, respectively. Statement I: In 30%(w/w) solution of methanol in CCl4 (at T K), the mole fraction of CCl4 is equal to 0.33. Statement II: Mixture of methanol and CCl4 shows positive deviation from Raoult's law. In the light of the above statements, choose the correct answer from the options given below:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1Both Statement I and Statement II are true
Approach:
A 100 g basis of the 30% (w/w) solution gives the masses of methanol and CCl4; converting to moles yields the mole fraction of CCl4 for Statement I, while the intermolecular interactions fix the deviation type in Statement II.
Step 1:In 100 g of a 30% (w/w) methanol solution there are 30 g methanol (CH3OH, M=32) and 70 g CCl4 (M=12+4×35.5=154).
mCH3OH=30g,mCCl4=70g
Step 2:The masses convert to moles.
nCH3OH=3230=0.9375,nCCl4=15470=0.4545
Step 3:The mole fraction of CCl4 follows from the moles.
xCCl4=0.4545+0.93750.4545=1.3920.4545≈0.33
Step 4:Pure methanol is strongly hydrogen bonded; adding inert CCl4 disrupts this H-bonding, raising the escaping tendency so the observed vapour pressure exceeds the Raoult prediction, that is, positive deviation. Statement II is true, so both statements are true.
pobserved>pRaoult
Final answer: Both Statement I and Statement II are true
Q53Single correctChemical Bonding and Molecular Structure
Bromine trifluoride autoionizes to form BrF2⊕ and BrF4⊖. The shapes of the cation and anion are respectively ______, and ______.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1bent, square planar
Approach:
For each ion, the valence electrons on the central Br atom (adjusted for charge) give the bond pairs and lone pairs, the steric number follows, and VSEPR gives the actual shape.
Step 1:Br has 7 valence electrons. In BrF2+ one electron is removed for the positive charge; in BrF4− one electron is added for the negative charge. Each Br–F bond uses one electron from Br.
Br:7valence electrons
Step 2:BrF2+ has 7−1=6 electrons on Br; two are used in the two Br–F bonds, leaving 4 electrons = 2 lone pairs. So 2 bond pairs and 2 lone pairs give steric number 4 (sp3).
BrF2+:2bp+2lp,SN=4
Step 3:BrF4− has 7+1=8 electrons on Br; four are used in the four Br–F bonds, leaving 4 electrons = 2 lone pairs. So 4 bond pairs and 2 lone pairs give steric number 6 (sp3d2); the two lone pairs occupy axial positions.
BrF4−:4bp+2lp,SN=6
Step 4:SN=4 with 2 lone pairs gives bent geometry, and SN=6 with 2 axial lone pairs gives square planar, matching the known interhalogen ion shapes.
cation: bent,anion: square planar
Final answer: bent, square planar
Q54Single correctSolutions
Which of the following statements are not correct? A. For water, magnitude of Kb is more than the magnitude of Kf. B. The elevation in boiling point of water when a non-volatile solute is added to it is larger in magnitude than its depression in freezing point. C. Osmotic pressure measurement is preferred over any other colligative property to determine molar mass of proteins and polymers. D. The dimerised form of benzoic acid in benzene is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3A, B and D only
Approach:
The question asks which statements are NOT correct. Each statement is tested against the known values of the ebullioscopic and cryoscopic constants of water, the relative magnitudes of the colligative effects, the suitability of osmotic pressure for macromolecules, and the actual cyclic structure of the benzoic acid dimer.
Step 1:For water Kb=0.52Kkgmol−1 and Kf=1.86Kkgmol−1, hence Kb<Kf. Statement A claims Kb>Kf, which is false.
Kb(0.52)<Kf(1.86)
Step 2:For the same molality, ΔTb=Kbm and ΔTf=Kfm; since Kb<Kf, the freezing point depression exceeds the boiling point elevation. Statement B claims the elevation is larger, which is false.
ΔTf>ΔTb
Step 3:Osmotic pressure is large and measurable at room temperature even for very dilute macromolecular solutions, so it is the preferred colligative property for proteins and polymers. Statement C is correct, so it is not among the wrong statements.
π=CRT
Step 4:Benzoic acid in benzene forms a cyclic dimer held by TWO O–H⋯O hydrogen bonds (each carbonyl O of one molecule binds the O–H of the other). The depicted linkage with a single hydrogen bond is not this cyclic dimer, so Statement D is incorrect. Therefore the incorrect statements are A, B and D.
cyclic dimer: two O-H⋯O bonds
Final answer: A, B and D only
Q55Single correctEquilibrium
Consider the following reactions in which all the reactants and products are present in gaseous state: 2xy⇌x2+y2 ; K1=2.5×105 xy+21z2⇌xyz ; K2=5×10−3 The value of K3 for the equilibrium 21x2+21y2+21z2⇌xyz is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 31.0×10−5
Approach:
The two given equilibria are manipulated (reversed, halved, then added) so their sum reproduces the target equilibrium, and the equilibrium constants combine accordingly: reciprocal on reversal, power 1/2 on halving, and product on addition.
Step 1:The target is 21x2+21y2+21z2⇌xyz. Reaction 1 makes x2+y2 from 2xy; it must be reversed and halved to produce xy from 21x2+21y2. Reaction 2 already converts xy+21z2 to xyz.
21x2+21y2⇌xy
Step 2:Reversing and halving reaction 1 gives the constant (1/K1)1/2.
Ka=(K11)1/2=(2.5×1051)1/2=(4×10−6)1/2=2×10−3
Step 3:Reaction 2 is used as given.
Kb=K2=5×10−3
Step 4:Adding the two manipulated reactions reproduces the target stoichiometry exactly, so the constants multiply.
K3=Ka×Kb=(2×10−3)(5×10−3)=1.0×10−5
Final answer: K3=1.0×10−5
Q56Single correctRedox Reactions and Electrochemistry
Given at 298 K: EFe2+/Fe⊖=X Volt EFe3+/Fe⊖=Y Volt The EFe3+/Fe2+⊖ in Volt at 298 K is given by:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 23Y−2X
Approach:
Standard electrode potentials are not directly additive when the electron counts differ; instead the standard Gibbs energies ΔG⊖=−nFE⊖ of the half reactions combine and convert back to a potential for the target single-electron couple.
Step 1:The two given half reactions carry their Gibbs energies. Fe3++3e−→Fe has n=3 and ΔG1⊖=−3FY. Fe2++2e−→Fe has n=2 and ΔG2⊖=−2FX.
ΔG1⊖=−3FY,ΔG2⊖=−2FX
Step 2:The target half reaction Fe3++e−→Fe2+ is obtained as (reaction 1) minus (reaction 2), so the Gibbs energies subtract.
ΔG3⊖=ΔG1⊖−ΔG2⊖=−3FY−(−2FX)=−F(3Y−2X)
Step 3:For the target couple n=1, so ΔG3⊖=−1⋅FEFe3+/Fe2+⊖. Equating the two expressions for ΔG3⊖ gives the potential.
−FE⊖=−F(3Y−2X)⇒E⊖=3Y−2X
Step 4:The electron balance holds (3=2+1) and ΔG is additive, giving the target potential as 3Y−2X.
EFe3+/Fe2+⊖=3Y−2X
Final answer: 3Y−2X
Q57Single correctChemical Kinetics
Given below are two statements: R=8.314JK−1mol−1 and 1cal=4.2J Statement I: When Ea=12.6kcal/mol, the room temperature rate constant is doubled by a 10∘C increase in temperature (298 K to 308 K). Statement II: For a first order reaction A→B, the plot of t1/2/s versus [A]0/mol L−1 is a straight line passing through the origin with positive slope. Here [A]0 is the initial concentration of A and t1/2 is the half life of reaction. In the light of the above statements, choose the correct answer from the options given below:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3Statement I is true but Statement II is false
Approach:
The two-temperature Arrhenius equation tests whether the rate constant doubles for Ea=12.6 kcal/mol over 298 K to 308 K, and the first-order half-life expression settles the dependence of t1/2 on initial concentration in Statement II.
Step 1:The activation energy in joules is Ea=12.6kcal/mol=12600cal/mol×4.2=52920J/mol, with T1=298K and T2=308K.
Ea=52920Jmol−1
Step 2:Substituting into the Arrhenius equation, 2981−3081=298×30810=1.0897×10−4 and REa=8.31452920=6365.
lnk1k2=6365×1.0897×10−4≈0.693
Step 3:Exponentiating gives the ratio. Since ln2=0.693, the rate constant doubles. Statement I is true.
k1k2=e0.693≈2
Step 4:For a first-order reaction t1/2=0.693/k is independent of [A]0, so the plot of t1/2 versus [A]0 is a horizontal line, not a line through the origin with positive slope. Statement II is false; hence Statement I true and Statement II false.
t1/2=f([A]0)(first order)
Final answer: Statement I is true but Statement II is false
Q58Single correctClassification of Elements and Periodicity in Properties
Match List - I with List - II.
List - I Electronic configuration of neutral atom (where n = 2)
List - II 1st Ionization Energy (kJmol−1)
A.ns2
I. 2080
B.ns2np1
II. 899
C.ns2np3
III. 800
D.ns2np6
IV. 1402
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1A-II, B-III, C-IV, D-I
Approach:
Each electronic configuration identifies the second-period (n=2) element, whose first ionization energy follows from periodic trends, including the dip at boron (easier removal of a 2p electron versus the filled 2s of beryllium) and the elevated value for nitrogen (half-filled 2p stability) and neon (noble gas).
Step 1:ns2 with n=2 is 2s2 = beryllium (Be). The first ionization energy of Be is about 899 kJmol−1.
Be(2s2):IE1≈899
Step 2:ns2np1 is 2s22p1 = boron (B). Removing the higher-energy, less-penetrating 2p electron is easier than removing from Be's filled 2s, so IE1 is lower, about 800 kJmol−1.
B(2s22p1):IE1≈800
Step 3:ns2np3 is 2s22p3 = nitrogen (N). The half-filled 2p3 subshell is extra stable, raising IE1 to about 1402 kJmol−1.
N(2s22p3):IE1≈1402
Step 4:ns2np6 is 2s22p6 = neon (Ne), a noble gas with the highest IE1 of the period, about 2080 kJmol−1. The ordering 800<899<1402<2080 matches B < Be < N < Ne, confirming the assignment.
Ne(2s22p6):IE1≈2080
Final answer: A-II, B-III, C-IV, D-I
Q59Single correctp-Block Elements
Find the correct statements related to group 15 hydrides. A. Reducing nature increases from NH3 to BiH3. B. Tendency to donate lone pair of electrons decreases from NH3 to BiH3. C. The stability of hydrides decreases from NH3 to BiH3. D. HEH bond angle decreases from NH3 to SbH3 (E = Elements of group 15). Choose the correct answer from the options given below:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3A, B, C and D
Approach:
The established down-the-group trends for group 15 hydrides (from NH3 to BiH3) apply: E–H bond strength falls, which sets the reducing power, basicity, thermal stability and bond angle; each statement is tested against these trends.
Step 1:Down the group the E–H bond weakens, so the hydride releases hydrogen more readily and acts as a stronger reducing agent. Hence reducing nature increases from NH3 to BiH3. Statement A is correct.
reducing power:NH3<BiH3
Step 2:The lone pair occupies an increasingly large and diffuse orbital down the group, lowering its donating tendency, so basicity decreases from NH3 to BiH3. Statement B is correct.
basicity:NH3>BiH3
Step 3:Weaker E–H bonds down the group make the hydrides thermally less stable, so stability decreases from NH3 to BiH3. Statement C is correct.
thermal stability:NH3>BiH3
Step 4:As the electronegativity of the central atom falls, the bonding pairs lie farther out and the bond angle contracts: NH3107∘>PH393.6∘>AsH391.8∘>SbH391.3∘. Statement D is correct, so all four statements are correct.
107∘>93.6∘>91.8∘>91.3∘
Final answer: A, B, C and D
Q60Single correctd- and f-Block Elements
Given below are two statements: Statement I: The number of pairs among [Ti4+,V2+], [V2+,Mn2+], [Mn2+,Fe3+] and [V2+,Cr2+] in which both ions are coloured is 3. Statement II: The number of pairs among [La3+,Yb2+], [Lu3+,Ce4+] and [Ac3+,Lr3+] ions in which both are diamagnetic is 3. In the light of the above statements, choose the correct from the options given below:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1Both Statement I and Statement II are correct
Approach:
For Statement I, an ion is coloured when it has partially filled d-orbitals allowing d–d transitions; the count is the pairs in which BOTH ions are coloured. For Statement II, an f-block ion is diamagnetic when its f-subshell is empty (f0) or completely filled (f14); the count is the pairs in which BOTH ions are diamagnetic.
Step 2:For Statement I: [Ti4+,V2+] has Ti4+ colourless (fails); [V2+,Mn2+] both coloured (counts); [Mn2+,Fe3+] both coloured (counts); [V2+,Cr2+] both coloured (counts). Total = 3, so Statement I is correct.
both-coloured pairs=3
Step 3:For Statement II, the f-configurations are La3+=f0, Yb2+=f14, Lu3+=f14, Ce4+=f0, Ac3+=f0 (5f0), Lr3+=f14. Every ion is either f0 or f14, hence all are diamagnetic.
f0orf14⇒diamagnetic
Step 4:Each of the three pairs in Statement II has both ions diamagnetic, so the count is 3 and Statement II is correct. Therefore both statements are correct.
both-diamagnetic pairs=3
Final answer: Both Statement I and Statement II are correct
Q61Single correctd- and f-Block Elements
Given below are two statements for catalytic properties of transition metals. Statement I: First row transition metals which act as catalyst utilise their 3d electrons only for formation of bonds between reactant molecules and atoms on the surface of catalyst. Statement II: There is increase in the concentration of reactants on the surface of catalyst which strengthens the bonds in reacting molecules. In the light of the above statements, choose the correct answer from the options given below :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2Both Statement I and Statement II are incorrect
Approach:
Evaluate the validity of each statement against the accepted mechanism of heterogeneous catalysis by first-row transition metals.
Step 1:Given: two statements about how first-row transition metals function as heterogeneous catalysts. Principle: catalytic activity arises from the availability of unpaired d electrons and partially filled d orbitals together with variable oxidation states; both 3d and 4s electrons participate in surface bonding.
3d and 4s electrons participate
Step 2:Statement I claims the metals use their 3d electrons ONLY. Bond formation with adsorbed reactants involves both 3d and 4s electrons (and the variable valencies they provide), so restricting it to 3d electrons alone is wrong.
not3donly
Step 3:Statement II claims that increased reactant concentration on the surface strengthens the bonds in the reacting molecules. Adsorption indeed raises local reactant concentration, but the surface interaction WEAKENS the bonds within reactant molecules, lowering the activation energy; it does not strengthen them.
adsorptionweakensreactantbonds
Step 4:Both statements are wrong, so the correct choice is the option stating both are incorrect.
Bothincorrect
Final answer: Both Statement I and Statement II are incorrect
Q62Single correctPurification and Characterisation of Organic Compounds
Given below are two statements : Statement I: Vapours of the liquid with higher boiling point condense before vapours of the liquid with lower boiling points in fractional distillation. Statement II: The vapours rising up in the fractionating column become richer in high boiling component of the mixture. In the light of the above statements, choose the correct answer from the options given below :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3Statement I is true but Statement II is false
Approach:
Analyse the behaviour of the two components inside a fractionating column with respect to condensation order and vapour composition as height increases.
Step 1:Given: a mixture undergoing fractional distillation. Principle: the component with the higher boiling point is less volatile, while the lower boiling component is more volatile and remains in the vapour phase longer.
lessvolatile=higherb.p.
Step 2:Statement I: as the mixed vapours ascend and cool, the higher boiling (less volatile) component reaches its condensation point first and condenses earlier, falling back down the column. This is correct.
higherb.p.→condensesfirst
Step 3:Statement II: because the high boiling component repeatedly condenses and drains back, the vapours that continue rising become progressively enriched in the LOW boiling (more volatile) component, not the high boiling one.
risingvapourenrichedinlowb.p.component
Step 4:Statement I true and Statement II false gives the option stating I is true but II is false.
Itrue,IIfalse
Final answer: Statement I is true but Statement II is false
Q63Single correctSome Basic Principles of Organic Chemistry
The major product of which of the following reaction is not obtained by rearrangement reaction?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2tert-butyl alcohol dehydration to 2-methylpropene (no rearrangement)
Approach:
For each option determine whether the carbocation (or skeletal) intermediate must rearrange to reach the stated major product; the required reaction is the one whose product forms without any rearrangement.
Step 1:Option 1: Friedel-Crafts alkylation of benzene with 1-chloropropane generates a primary n-propyl cation, which undergoes a 1,2-hydride shift to the more stable secondary isopropyl cation, giving cumene. The product arises through rearrangement.
CH3CH2CH2+→(CH3)2CH+
Step 2:Option 3: n-hexane treated with anhydrous AlCl3/HCl undergoes acid-catalysed isomerisation to the branched 2-methylpentane, a skeletal rearrangement. Option 4: the branched primary/secondary alcohol forms a less stable cation that undergoes a hydride/methyl shift before eliminating to the more substituted alkene; rearrangement is involved.
skeletal/1,2-shift
Step 3:Option 2: protonation and loss of water from tert-butyl alcohol gives the tertiary tert-butyl cation, which is already the most stable possible cation. It directly loses a beta proton to form 2-methylprop-1-ene, with no shift.
(CH3)3C+→(CH3)2C=CH2
Step 4:Only option 2 yields its major product without any rearrangement; the other three proceed via hydride/methyl/skeletal shifts.
answer=tert-butanoldehydration
Final answer: Dehydration of tert-butyl alcohol to 2-methylprop-1-ene proceeds without rearrangement
Q65Single correctHydrocarbons
n-Butane on monochlorination under photochemical condition gives an optically active compound "P". "P" on further chlorination gives dichloro compounds. The number of dichloro compounds obtained (ignore stereoisomers) is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 24
Approach:
Identify the optically active monochloro product P, then enumerate every distinct constitutional dichloro product obtained by replacing a second hydrogen of P, ignoring stereoisomers.
Step 1:Monochlorination of n-butane (CH3-CH2-CH2-CH3) gives 1-chlorobutane and 2-chlorobutane. 1-Chlorobutane has no stereocentre, whereas 2-chlorobutane, CH3-CHCl-CH2-CH3, has C2 bonded to four different groups and is optically active. Hence P = 2-chlorobutane.
P=CH3-CHCl-CH2-CH3
Step 2:Label the carbons of P: C1 (CH3), C2 (CHCl), C3 (CH2), C4 (CH3). Replacing one further hydrogen at each position gives the dichloro products. Substitution at C1 gives 1,2-dichlorobutane; at C2 gives 2,2-dichlorobutane; at C3 gives 2,3-dichlorobutane.
C1→1,2;C2→2,2;C3→2,3
Step 3:Substitution at C4 places the two chlorines on carbons 2 and 4 of the original chain; renumbering from the C4 end gives the lowest locants 1,3, i.e. 1,3-dichlorobutane, which is constitutionally distinct from the previous three.
C4→1,3-dichlorobutane
Step 4:The distinct constitutional dichloro compounds are 1,2-dichlorobutane, 2,2-dichlorobutane, 2,3-dichlorobutane and 1,3-dichlorobutane, giving a total of 4 (stereoisomers ignored).
total=4
Final answer: 4 dichloro compounds (1,2-, 1,3-, 2,2- and 2,3-dichlorobutane)
Given below are two statements : Statement I: Due to increase in van der Waals forces, the order of boiling points is CH3CH2CH2I>CH3CH2I>CH3I.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1Both Statement I and Statement II are true
Approach:
Test the boiling-point trend of the homologous alkyl iodides (Statement I) and the melting/boiling-point comparison of the dichlorobenzene isomers based on symmetry and dipole moment (Statement II).
Step 1:Given: comparison of boiling points of CH3I, CH3CH2I and CH3CH2CH2I, and of melting/boiling points of para- and ortho-dichlorobenzene. Principle: boiling point rises with molecular size (stronger van der Waals forces), while melting point depends strongly on crystal packing/symmetry and boiling point on net dipole moment.
principlesestablished
Step 2:Statement I: along the series methyl, ethyl, n-propyl iodide the molecular size and surface area increase, strengthening van der Waals forces, so the boiling point order CH3CH2CH2I > CH3CH2I > CH3I is correct.
CH3CH2CH2I>CH3CH2I>CH3I
Step 3:Statement II, melting point: the para isomer is highly symmetric and packs efficiently into the crystal lattice, so its melting point is higher than that of the ortho isomer. This part is correct.
m.p.(para)>m.p.(ortho)
Step 4:Statement II, boiling point: para-dichlorobenzene is symmetric with the two C-Cl dipoles cancelling (net dipole = 0), whereas ortho-dichlorobenzene has a finite net dipole moment and thus extra dipole-dipole attraction, giving it a higher boiling point. Therefore the para boiling point is lower than the ortho, so Statement II is fully true and both statements are true.
b.p.(para)<b.p.(ortho)
Final answer: Both Statement I and Statement II are true
SolutionAnswer: Option 3HOOC-(CH2)4-NH-CH(CH3)CH2OH (ring-opened amino acid with amino alcohol)
Approach:
Apply each reagent in sequence to the substrate, distinguishing the chemoselective behaviour of NaBH4 (reduces aldehyde, not amide) from base-promoted hydrolysis of the lactam ring, then the acidic work-up.
Step 1:Identify the substrate: a six-membered lactam (a ring amide, the carbonyl on a ring carbon adjacent to nitrogen) whose nitrogen bears a -CH(CH3)-CHO side chain. The two reactive carbonyls are the side-chain aldehyde and the ring amide.
lactamring+side-chainCHO
Step 2:Step (i) NaBH4/MeOH reduces the aldehyde selectively to a primary alcohol; NaBH4 is too weak to reduce the amide (lactam) carbonyl, which is therefore untouched.
−CH(CH3)CHO→−CH(CH3)CH2OH
Step 3:Step (ii) NaOH(aq.) with heat hydrolyses the cyclic amide, cleaving the C-N bond and opening the ring to give a pentanoate carboxylate at one end and a secondary amine (still carrying the -CH(CH3)CH2OH group) at the other.
ring→−OOC-(CH2)4-NH-CH(CH3)CH2OH
Step 4:Step (iii) H3O+ protonates the carboxylate to the free carboxylic acid, giving P = HOOC-(CH2)4-NH-CH(CH3)CH2OH, an open-chain amino acid bearing the amino-alcohol substituent.
HOOC-(CH2)4-NH-CH(CH3)CH2OH
Final answer: P = HOOC-(CH2)4-NH-CH(CH3)CH2OH, the ring-opened amino acid bearing an amino-alcohol substituent
Which statements are True? A. In Hoffmann bromamide degradation, 4 moles of NaOH and 2 moles of Br2 are consumed per mole of an amide B. Hoffmann bromamide reaction is not given by alkyl amides C. Primary amines can be synthesized by Hoffmann bromamide degradation. D. Secondary amide on reaction with Br2 and NaOH will give secondary amine. E. The by-products of Hoffmann degradation are Na2CO3 and H2O. Choose the correct answer from the options given below :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3C and E only
Approach:
Write the balanced equation for the Hoffmann bromamide degradation and use it, together with the requirement of a primary amide, to test each statement A-E.
Step 1:Write and balance the reaction: one mole of primary amide consumes one mole of Br2 and four moles of NaOH, producing the primary amine (with one fewer carbon), sodium carbonate, sodium bromide and water.
RCONH2+Br2+4NaOH→RNH2+Na2CO3+2NaBr+2H2O
Step 2:Statement A: the balanced equation uses 4 moles NaOH but only 1 mole Br2 per mole amide, so claiming 2 moles Br2 is wrong; A is false. Statement B: the reaction is given by both alkyl and aryl primary amides, so saying alkyl amides do not react is wrong; B is false.
1Br2,4NaOHperamide
Step 3:Statement C: the reaction converts a primary amide into a primary amine (with loss of one carbon), so primary amines can indeed be made this way; C is true. Statement D: the reaction requires a primary amide (-CONH2 with an N-H2); secondary amides lack the necessary N-H pattern and do not undergo the degradation to give secondary amines; D is false.
1∘amide→1∘amine
Step 4:Statement E: from the balanced equation the by-products include Na2CO3 and H2O (along with NaBr), so E is true. The true statements are C and E only.
by-products:Na2CO3,H2O,NaBr
Final answer: C and E only
Q69Single correctBiomolecules
The incorrect statement from the following with respect to carbohydrates is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2The monosaccharide units obtained from hydrolysis of oligosaccharides are always the same.
Approach:
Assess the truth of each statement about carbohydrate classification and behaviour and isolate the one that is false.
Step 1:Statement (1): every monosaccharide, whether aldose or ketose, carries a free aldehyde or alpha-hydroxy ketone group capable of reduction, so all monosaccharides are reducing sugars; this is true.
allmonosaccharidesreduce
Step 2:Statements (3) and (4): starch and cellulose are high molecular weight polysaccharides made of more than ten monosaccharide units (true), and D-(+)-glucose exists as an equilibrium of open-chain and cyclic forms responsible for mutarotation (true).
(3),(4)true
Step 3:Statement (2): hydrolysis of oligosaccharides need not give identical monosaccharides. Sucrose yields glucose + fructose and lactose yields glucose + galactose, both giving two different monosaccharides, so the claim that they are always the same is false.
sucrose→glucose+fructose
Step 4:The only false (incorrect) statement is statement (2).
incorrect=(2)
Final answer: The statement that the monosaccharide units from hydrolysis of oligosaccharides are always the same is incorrect
Q70Single correctBiomolecules
Which of the following amino acid will give violet coloured complex with neutral ferric chloride solution?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3Tyrosine
Approach:
Recognise that the violet colour with neutral ferric chloride is the characteristic phenol test, then identify which amino acid side chain contains a phenolic hydroxyl group.
Step 1:Neutral ferric chloride produces a violet/purple coloured complex specifically with phenols (compounds bearing an -OH directly on an aromatic ring), through formation of a coloured iron-phenoxide complex.
phenol+FeCl3→violet
Step 2:Examine the side chains: threonine and serine carry aliphatic (non-aromatic) -OH groups, and cysteine carries a thiol (-SH); none of these is phenolic, so none gives the violet colour.
Thr,Ser:−OH(aliphatic);Cys:−SH
Step 3:Tyrosine has a para-hydroxyphenyl side chain, i.e. a true phenolic -OH on the aromatic ring, which is the group required for the test.
tyrosine=p-hydroxyphenyl
Step 4:Only tyrosine satisfies the requirement, so tyrosine gives the violet coloured complex with neutral ferric chloride.
Tyrosine→violet
Final answer: Tyrosine
Q71NumericalCoordination Compounds
Number of paramagnetic complexes among the following is ______. [MnBr4]2−,[NiCl4]2−,[Ni(CN)4]2−,[Ni(CO)4],[CoF6]3−,[Fe(CN)6]4−,[Mn(CN)6]3−,[Ti(CN)6]3−,[Cu(H2O)6]2+,[Co(C2O4)3]3−
SolutionAnswer: 6
Approach:
For each complex the central metal oxidation state, the d-electron count, the coordination geometry, and the ligand field strength (strong/weak) are determined. By crystal field theory and valence bond theory, the unpaired electrons are counted; a complex is paramagnetic when it has at least one unpaired electron.
Step 1:The task is to count complexes with one or more unpaired electrons among the ten listed. The oxidation states and d-electron counts are: [MnBr4]2− Mn2+d5; [NiCl4]2− Ni2+d8; [Ni(CN)4]2− Ni2+d8; [Ni(CO)4] Ni0d10; [CoF6]3− Co3+d6; [Fe(CN)6]4− Fe2+d6; [Mn(CN)6]3− Mn3+d4; [Ti(CN)6]3− Ti3+d1; [Cu(H2O)6]2+ Cu2+d9; [Co(C2O4)3]3− Co3+d6.
Step 2:Consider the tetrahedral halide complexes (weak-field, high spin). [MnBr4]2−d5 tetrahedral fills e2t23 giving 5 unpaired electrons. [NiCl4]2−d8 tetrahedral fills e4t24 giving 2 unpaired electrons. Both are paramagnetic.
[MnBr4]2−:n=5;[NiCl4]2−:n=2
Step 3:Consider the strong-field and d10 Ni species. [Ni(CN)4]2−d8 with strong-field CN− is square planar, all electrons paired, n=0. [Ni(CO)4] is Ni0d10, completely filled, n=0. Both are diamagnetic.
[Ni(CN)4]2−:n=0;[Ni(CO)4]:n=0
Step 4:Consider the octahedral complexes. [CoF6]3−d6 with weak-field F− is high spin t2g4eg2, n=4 (paramagnetic). [Fe(CN)6]4−d6 with strong CN− is low spin t2g6, n=0 (diamagnetic). [Mn(CN)6]3−d4 low spin t2g4, n=2 (paramagnetic). [Ti(CN)6]3−d1, n=1 (paramagnetic). [Cu(H2O)6]2+d9, n=1 (paramagnetic). [Co(C2O4)3]3−d6 with chelating oxalate (strong field) is low spin t2g6, n=0 (diamagnetic).
Step 5:The paramagnetic complexes (n>=1) are [MnBr4]2−, [NiCl4]2−, [CoF6]3−, [Mn(CN)6]3−, [Ti(CN)6]3−, [Cu(H2O)6]2+. The diamagnetic complexes (n=0) are [Ni(CN)4]2−, [Ni(CO)4], [Fe(CN)6]4−, [Co(C2O4)3]3−. Total paramagnetic = 6.
2+4=6
Final answer: 6
Q72NumericalOrganic Compounds Containing Oxygen
'x' is the product which is obtained from benzene by reacting it with carbon monoxide and hydrogen chloride in the presence of cuprous chloride. 'y' is the major product obtained from the benzene by reacting it with ethanoyl chloride in the presence of anhydrous AlCl3. Product (major) obtained by heating x and y in the presence of alkali is z. Total number of π(pi) electrons in z is ______
SolutionAnswer: 16
Approach:
x is the Gattermann-Koch product of benzene and y is the Friedel-Crafts acylation product. The base-mediated crossed (Claisen-Schmidt) aldol condensation of x and y gives z, and every pi electron in z is counted.
Step 1:Benzene with CO and HCl over cuprous chloride (Gattermann-Koch reaction) introduces a formyl group, giving benzaldehyde as x.
C6H6+CO+HClCuClC6H5CHO
Step 2:Benzene with ethanoyl chloride (CH3COCl) and anhydrous AlCl3 undergoes Friedel-Crafts acylation to give acetophenone as y.
C6H6+CH3COClAlCl3C6H5COCH3
Step 3:Heating benzaldehyde and acetophenone in alkali drives a crossed Claisen-Schmidt aldol condensation: the α-hydrogen of acetophenone forms a carbanion that adds to the aldehyde carbonyl, and dehydration gives the α,β-unsaturated ketone (E)-chalcone.
C6H5CHO+CH3COC6H5OH−,ΔC6H5CH=CHCOC6H5
Step 4:Each benzene ring contributes 3 C=C, i.e. 6 pi electrons; with two rings that is 12. The conjugated alkene C=C contributes 2, and the carbonyl C=O contributes 2.
2×6+2+2
Step 5:Summing all pi electrons of chalcone gives the total.
12+2+2=16
Final answer: 16
Q73NumericalAtomic Structure
Consider two radiations of wavelengths 1. λ1=2000A˚ 2. λ2=6000A˚ The ratio of the energies of these two radiations (E2E1) is ______ (Nearest integer).
SolutionAnswer: 3
Approach:
Photon energy is inversely proportional to wavelength via E=hc/λ, so the energy ratio equals the inverse ratio of wavelengths.
Step 1:Two radiations have wavelengths λ1=2000A˚ and λ2=6000A˚; the target is the energy ratio E1/E2.
λ1=2000A˚,λ2=6000A˚
Step 2:By the Planck-Einstein relation, photon energy varies inversely with wavelength.
E1=λ1hc,E2=λ2hc
Step 3:Forming the ratio cancels the constant hc.
E2E1=hc/λ2hc/λ1=λ1λ2
Step 4:Substituting the wavelengths and evaluating.
E2E1=20006000=3
Step 5:The shorter wavelength (2000A˚) carries the higher energy; the ratio is the integer 3.
E2E1=3
Final answer: 3
Q74NumericalChemical Thermodynamics
Consider the reaction: 2H2S(g)+3O2(g)→2H2O(l)+2SO2(g) The magnitude of enthalpy change for the reaction in kJmol−1 is ______. (Nearest integer) Given: ΔfH⊖(H2S)=−20.1kJmol−1, ΔfH⊖(H2O)=−286.0kJmol−1, ΔfH⊖(SO2)=−297.0kJmol−1
SolutionAnswer: 1126
Approach:
Hess's law in the form of standard formation enthalpies applies: the reaction enthalpy equals the sum over products minus the sum over reactants, with stoichiometric coefficients. Elemental O2 in its standard state has zero formation enthalpy. The magnitude is reported.
Step 1:Coefficients: 2H2S+3O2→2H2O+2SO2. Formation enthalpies (kJ/mol): H2S=−20.1, H2O=−286.0, SO2=−297.0, O2=0. The target is the magnitude of ΔrH.
ΔfH:H2S=−20.1,H2O=−286.0,SO2=−297.0,O2=0
Step 2:The Hess's law expression with stoichiometric coefficients is written out.
Step 4:Evaluating the reactant sum (O2 contributes zero).
2(−20.1)+3(0)=−40.2kJ
Step 5:Subtracting and taking the magnitude, rounded to the nearest integer.
ΔrH=−1166.0−(−40.2)=−1125.8kJmol−1,∣ΔrH∣≈1126
Final answer: 1126
Q75NumericalEquilibrium
Solid carbon, CaO and CaCO3 are mixed and allowed to attain equilibrium at T K. CaCO3(s)⇌CaO(s)+CO2(g)Kp1=0.08atm C(s)+CO2(g)⇌2CO(g)Kp2=2atm The partial pressure of CO is ____ ×10−1 atm
SolutionAnswer: 4
Approach:
Solids have unit activity, so the first equilibrium fixes the CO2 partial pressure at Kp1. Substituting this into the Kp2 expression of the second equilibrium gives the CO partial pressure.
Step 1:Kp1=0.08 atm for CaCO3(s)⇌CaO(s)+CO2(g); Kp2=2 atm for C(s)+CO2(g)⇌2CO(g). The target is p(CO) expressed in units of 10−1 atm.
Kp1=0.08atm,Kp2=2atm
Step 2:In the first equilibrium, CaCO3 and CaO are solids (activity 1), so Kp1 equals the CO2 partial pressure.
Kp1=pCO2=0.08atm
Step 3:In the second equilibrium, carbon is solid, so only CO2 and CO appear in Kp2.
Kp2=pCO2pCO2=2atm
Step 4:Solving for p(CO)2 by substituting p(CO2)=0.08 atm.
pCO2=Kp2×pCO2=2×0.08=0.16atm2
Step 5:Taking the square root gives p(CO), expressed in units of 10−1 atm.
pCO=0.16=0.4atm=4×10−1atm
Final answer: 4
Mathematics24 questions
Q1Single correctSets, Relations and Functions
Consider the relation R on the set {−2,−1,0,1,2} defined by (a,b)∈R if and only if 1+ab>0. Then, among the statements : I. The number of elements in R is 17 II. R is an equivalence relation
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1Only I is true
Approach:
Translate the defining condition into ab>−1, count the ordered pairs over the 5×5 grid by excluding those with ab≤−1, then test the equivalence-relation properties.
Step 1:Given set {−2,−1,0,1,2} with 25 possible ordered pairs. Target: count pairs with ab>−1 and decide whether R is reflexive, symmetric and transitive.
(a,b)∈R⇔ab>−1
Step 2:Identify the pairs that violate the condition, i.e. ab≤−1. Products equal to −1: (−1,1),(1,−1) give 2 pairs. Products equal to −2: (−1,2),(2,−1),(1,−2),(−2,1) give 4 pairs. Products equal to −4: (−2,2),(2,−2) give 2 pairs.
#{ab≤−1}=2+4+2=8
Step 3:All remaining pairs satisfy ab>−1 (pairs containing 0 give ab=0; equal-sign nonzero pairs give ab≥1). Hence ∣R∣=25−8=17, so statement I is true.
∣R∣=25−8=17
Step 4:Check equivalence: reflexive since a⋅a=a2≥0>−1; symmetric since ab=ba. Transitivity fails: (−2,0)∈R and (0,2)∈R, but (−2)(2)=−4>−1, so (−2,2)∈/R.
(−2)(2)=−4>−1
Step 5:Since transitivity fails, R is not an equivalence relation, so statement II is false. Only statement I holds.
I true,II false
Final answer: Statement I (|R| = 17) is true and statement II (R equivalence) is false, so only I is true.
Q2Single correctComplex Numbers and Quadratic Equations
The number of values of z∈C, satisfying the equations ∣z−(4+8i)∣=10 and ∣z−(3+5i)∣+∣z−(5+11i)∣=45, is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 22
Approach:
Read each equation as a locus in the plane: the first is a circle, the second is an ellipse with the given foci; then compare their axes to count common points.
Step 1:Target: count points z=(x,y) lying on both loci. Foci of the second equation are A=(3,5) and B=(5,11); the constant sum of distances is 45.
A=(3,5),B=(5,11),sum=45
Step 2:Compute the focal distance: AB=(5−3)2+(11−5)2=4+36=40=210, so c=10. The constant sum gives 2a=45, hence a=25 and a2=20.
c=10,a=25,a2=20
Step 3:Since 2a=45>AB=210, the second locus is a genuine ellipse with semi-minor axis b2=a2−c2=20−10=10, so b=10. Its centre is the midpoint of A,B, namely (4,8).
b2=20−10=10,centre=(4,8)
Step 4:The first equation is a circle centred at (4,8) with radius 10, concentric with the ellipse. Its radius equals the semi-minor axis b=10 and is less than a=25, so the circle meets the ellipse exactly at the two endpoints of the minor axis.
r=b=10<a=25
Step 5:Consistency check: on the minor axis the ellipse reaches distance b=10 from the centre, exactly the circle radius, while on the major axis it reaches a=25>10; thus contact occurs only at the two minor-axis endpoints.
10=b(minor),25=a(major)
Final answer: There are exactly 2 values of z.
Q3Single correctMatrices and Determinants
If the system of linear equations: x+y+z=6 x+2y+5z=10 2x+3y+λz=μ has infinitely many solutions, then the value of λ+μ equals:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 322
Approach:
For infinitely many solutions the augmented rank must equal the coefficient rank and be less than 3, so the third equation is a linear combination of the first two; match coefficients to find λ and μ.
Step 1:Given E1:x+y+z=6, E2:x+2y+5z=10, E3:2x+3y+λz=μ. Target: find λ,μ making E3 dependent, then compute λ+μ. Write E3=αE1+βE2 and match x and y coefficients.
α+β=2,α+2β=3
Step 2:Subtract the first matching equation from the second: (α+2β)−(α+β)=3−2, giving β=1, and then α=2−β=1.
β=1,α=1
Step 3:Match the z coefficient: λ=α(1)+β(5)=1+5=6.
λ=1+5=6
Step 4:Match the constant term: μ=α(6)+β(10)=6+10=16. Hence λ+μ=6+16=22.
μ=16,λ+μ=22
Step 5:Dependence check: with λ=6,μ=16, the third equation is exactly E1+E2 (x+y+z plus x+2y+5z=2x+3y+6z=16), so the coefficient and augmented matrices share rank 2<3, confirming infinitely many solutions.
E1+E2=2x+3y+6z=16
Final answer: λ+μ=22.
Q4Single correctMatrices and Determinants
Let A=α20134205 and B=1000−5α4α00−2α+adj(A). If det(B)=66, then det(adj(A)) equals :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3441
Approach:
Build adj(A) from cofactors, add the given matrix, impose det(B)=66 to solve for α, then apply det(adjA)=(detA)n−1.
Step 1:Given the 3×3 matrix A with parameter α and the matrix B formed by adding a known matrix to adj(A), with det(B)=66. Target: det(adjA). Evaluating det(A) along row 1: detA=α(3⋅5−0⋅4)−1(2⋅5−0⋅0)+2(2⋅4−3⋅0).
detA=15α−10+16=15α+6
Step 2:Form adj(A) from the cofactor transpose and add the given matrix to obtain B. Evaluating the determinant of the sum yields det(B)=30α+36.
det(B)=30α+36
Step 3:Impose det(B)=66: 30α+36=66⇒30α=30⇒α=1. Then detA=15(1)+6=21.
α=1,detA=21
Step 4:For a 3×3 matrix, det(adjA)=(detA)n−1=(detA)2=212=441.
det(adjA)=212=441
Step 5:Consistency check: detA=21=0, so A is invertible and the adjoint identity applies; substituting α=1 back gives detB=30+36=66, matching the data.
30(1)+36=66✓
Final answer: det(adjA)=441.
Q5Single correctTrigonometry
Let α=3+4+8+9+13+14+… upto 40 terms. If (tanβ)1020α is a root of the equation x2+x−2=0, β∈(0,2π), then sin2β+3cos2β is equal to :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 12
Approach:
Group the series into 20 pairs forming an AP to evaluate α, reduce the exponent, select the admissible root of the quadratic to find β, then evaluate the trigonometric expression.
Step 1:The 40-term series 3+4+8+9+13+14+… groups into consecutive pairs (3+4)+(8+9)+(13+14)+…, giving 20 pair-sums 7,17,27,…, an AP with first term a=7 and common difference d=10. Target: evaluate α, the exponent, β, and the expression.
7,17,27,…(20terms)
Step 2:Sum the AP: α=220[2(7)+(20−1)(10)]=10[14+190]=10⋅204=2040. Then the exponent is 1020α=10202040=2.
α=2040,1020α=2
Step 3:Factor the quadratic: x2+x−2=(x+2)(x−1)=0, with roots x=1 and x=−2. Since (tanβ)2≥0, the root must be 1, so tan2β=1. With β∈(0,π/2), tanβ=1 and β=π/4.
(tanβ)2=1⇒β=4π
Step 4:At β=π/4, cos2β=21, so sin2β+3cos2β=1+2cos2β=1+2⋅21=2.
1+2⋅21=2
Step 5:Consistency check: directly sin24π+3cos24π=21+3⋅21=21+23=2, agreeing with the identity-based value.
21+23=2
Final answer: sin2β+3cos2β=2.
Q6Single correctStatistics and Probability
A candidate has to go to the examination centre to appear in an examination. The candidate uses only one means of transportation for the entire distance out of bus, scooter and car. The probabilities of the candidate going by bus, scooter and car, respectively, are 52,51 and 52. The probabilities that the candidate reaches late at the examination centre are 51,31 and 41 if the candidate uses bus, scooter and car, respectively. Given that the candidate reached late at the examination centre, the probability that the candidate travelled by bus is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 23712
Approach:
Treat the three transport modes as mutually exclusive causes and 'reaching late' as the observed event, then apply Bayes' theorem.
Step 1:Given prior probabilities P(B)=52,P(S)=51,P(C)=52 and late probabilities P(L∣B)=51,P(L∣S)=31,P(L∣C)=41. Target: P(B∣L). Compute the bus joint probability P(B)P(L∣B)=52⋅51=252.
P(B)P(L∣B)=252
Step 2:Compute the other joint probabilities: scooter 51⋅31=151 and car 52⋅41=101.
151,101
Step 3:Total probability of being late, over LCD 150: P(L)=252+151+101=15012+15010+15015=15037.
Step 5:Consistency check: the three posterior numerators 12,10,15 over 37 sum to 3712+10+15=3737=1, confirming a valid probability distribution.
3712+10+15=1
Final answer: P(bus∣late)=3712.
Q7Single correctStatistics and Probability
A set of four observations has mean 1 and variance 13. Another set of six observations has mean 2 and variance 1. Then, the variance of all these 10 observations is equal to :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 36.04
Approach:
Recover each group's sum of squares from σ2=n∑x2−xˉ2, find the combined mean, then apply the variance formula to the pooled data.
Step 1:Group 1: n1=4,xˉ1=1,σ12=13. Group 2: n2=6,xˉ2=2,σ22=1. Target: pooled variance over all 10 observations. Recover group 1 sum of squares: ∑x12=n1(σ12+xˉ12)=4(13+1)=56.
∑x12=4(13+1)=56
Step 2:Recover group 2 sum of squares: ∑x22=n2(σ22+xˉ22)=6(1+4)=30.
Step 4:Pooled sum of squares ∑x2=56+30=86. Pooled variance =1086−(1.6)2=8.6−2.56=6.04.
8.6−2.56=6.04
Step 5:Consistency check: the pooled variance 6.04 lies between the group variances 1 and 13 and exceeds the weighted average of variances 104⋅13+6⋅1=5.8 by the between-group spread.
6.04>5.8=104⋅13+6⋅1
Final answer: The combined variance is 6.04.
Q8Single correctBinomial Theorem and its Simple Applications
If 26(323(12C2)+525(12C4)+727(12C6)+⋯+13213(12C12))=313−α, then α is equal to :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 351
Approach:
Recognize each term k+12k+1(k12) as the result of integrating (k12)xk over [0,2]; isolate the even-index terms by averaging (1+x)12 and (1−x)12.
Step 1:Each summand has the form k+12k+1(k12) for even k from 2 to 12. Target: α. Evaluate the two enabling integrals: ∫02(1+x)12dx=[13(1+x)13]02=13313−1 and ∫02(1−x)12dx=[−13(1−x)13]02=131−(−1)13=132.
∫02(1+x)12dx=13313−1,∫02(1−x)12dx=132
Step 2:Averaging the integrands isolates even powers, since 21[(1+x)12+(1−x)12]=∑keven(k12)xk. Integrating term by term over [0,2] gives ∑kevenk+12k+1(k12)=21[13313−1+132]=26313+1.
∑kevenk+12k+1(k12)=26313+1
Step 3:The given bracket starts at k=2, so it omits the k=0 term 121(012)=2. Hence the bracket equals 26313+1−2.
bracket=26313+1−2
Step 4:Multiply by 26: 26⋅bracket=(313+1)−52=313−51. Comparing with 313−α gives α=51.
313−51=313−α⇒α=51
Step 5:Numerical check using 313=1594323: the even-index bracket times 26 evaluates to 1594272=1594323−51, confirming α=51.
1594272=1594323−51
Final answer: α=51.
Q9Single correctPermutations and Combinations
A person has three different bags and four different books. The number of ways, in which he can put these books in the bags so that no bag is empty, is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 236
Approach:
Count onto (surjective) assignments of 4 distinct books to 3 distinct bags using the inclusion-exclusion principle.
Step 1:Each of the n=4 distinct books is placed into one of m=3 distinct bags, with the constraint that no bag is empty (surjection). Target: count such assignments. Total unrestricted assignments: 34=81.
34=81
Step 2:Subtract assignments leaving at least one chosen bag empty: (13)24=3⋅16=48.
(13)24=48
Step 3:Add back assignments where two chosen bags are empty (subtracted twice): (23)14=3⋅1=3.
(23)14=3
Step 4:Inclusion-exclusion gives the number of surjections: 81−48+3=36.
81−48+3=36
Step 5:Cross-check via Stirling numbers: the number of onto maps equals 3!S(4,3)=6⋅6=36, where S(4,3)=6 partitions 4 books into 3 nonempty unlabelled groups.
3!S(4,3)=6⋅6=36
Final answer: There are 36 such ways.
Q10Single correctCo-ordinate Geometry
If a straight line drawn through the point of intersection of the lines 4x+3y−1=0 and 3x+4y−1=0, meets the co-ordinate axes at the points P and Q, then the locus of the mid point of PQ is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2x+y-14xy=0
Approach:
Locate the fixed point of intersection, parametrize the variable line by its intercepts using the midpoint (h,k), then impose that the line passes through the fixed point to obtain the locus.
Step 1:Find the fixed point common to all such lines by solving 4x+3y=1 and 3x+4y=1. Subtracting: (4x+3y)−(3x+4y)=0⇒x−y=0⇒x=y. Substituting into 4x+3y=1: 7x=1, so x=y=71. Fixed point (71,71).
(71,71)
Step 2:Let the variable line meet the axes at P=(a,0) and Q=(0,b) with midpoint (h,k)=(2a,2b), so a=2h and b=2k. The line in intercept form is 2hx+2ky=1.
2hx+2ky=1
Step 3:Impose passage through (71,71): 2h1/7+2k1/7=1⇒14h1+14k1=1.
14h1+14k1=1
Step 4:Multiply through by 14hk: k+h=14hk⇒h+k−14hk=0. Replacing (h,k) by (x,y) gives the locus x+y−14xy=0.
x+y−14xy=0
Step 5:Consistency check: the fixed point (71,71) corresponds to the midpoint case h=k=71, giving 71+71−14⋅491=72−4914=72−72=0, satisfying the locus.
72−4914=0
Final answer: The locus of the midpoint of PQ is x+y−14xy=0.
Q11Single correctCo-ordinate Geometry
Let O be the vertex of the parabola y2=4x and its chords OP and OQ are perpendicular to each other. If the locus of the mid-point of the line segment PQ is a conic C, then the length of its latus rectum is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 22
Approach:
Parametrize P and Q on the parabola, impose perpendicularity of the chords from the vertex, then eliminate the parameters from the midpoint coordinates to obtain the locus and read off its latus rectum.
Step 1:Take P=(t12,2t1) and Q=(t22,2t2) on y2=4x, with vertex O=(0,0). Target: the locus of the midpoint M(h,k) of PQ. Slopes from O are mOP=t122t1=t12 and mOQ=t22.
mOP=t12,mOQ=t22
Step 2:Perpendicular chords give mOPmOQ=−1, so t12⋅t22=−1.
t1t24=−1
Step 3:Midpoint coordinates: h=2t12+t22 and k=22t1+2t2=t1+t2. Using t12+t22=(t1+t2)2−2t1t2=k2−2(−4)=k2+8 gives 2h=k2+8.
2h=k2+8
Step 4:Substitution check: replacing (h,k) by (x,y) yields y2=2x−8=2(x−4), of the standard form Y2=2X with 4a=2, so the latus rectum length is 2.
y2=2(x−4),4a=2
Final answer: The length of the latus rectum of the locus is 2.
Q12Single correctTrigonometry
Let α=3sin−1(116) and β=3cos−1(94), where inverse trigonometric functions take only the principal values. Given below are two statements : Statement I: cos(α+β)>0. Statement II: cos(α)<0. In the light of the above statements, choose the correct answer from the options given below :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1Both Statement I and Statement II are true
Approach:
Bound each inverse-trigonometric value to locate α, β and α+β in their quadrants, then determine the signs of the required cosines.
Step 1:Given α=3sin−1(6/11) and β=3cos−1(4/9). Target: signs of cosα and cos(α+β). With 6/11≈0.545, sin−1(0.545)≈0.5768 rad, so α=3(0.5768)≈1.731 rad.
α≈1.731 rad
Step 2:Since 2π≈1.571<1.731<π≈3.142, the angle α lies in the second quadrant, where cosine is negative.
α∈(2π,π)
Step 3:With 4/9≈0.444, cos−1(0.444)≈1.110 rad, so β=3(1.110)≈3.331 rad. Then α+β≈1.731+3.331=5.062 rad.
α+β≈5.062 rad
Step 4:Quadrant check on the sum: 23π≈4.712<5.062<2π≈6.283, so α+β lies in the fourth quadrant, where cosine is positive.
α+β∈(23π,2π)
Final answer: Both Statement I and Statement II are true.
Q13Single correctLimit, Continuity and Differentiability
For the function f(x)=esin∣x∣−∣x∣,x∈R, consider the following statements: Statement I : f is differentiable for all x∈R. Statement II: f is increasing in (−π,−2π). In the light of the above statements, choose the correct answer from the options given below :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1Both Statement I and Statement II are true
Approach:
Write f piecewise about x=0 (the only point where ∣x∣ is non-smooth), match one-sided derivatives there, then analyze the sign of f' on (−π,−2π).
Step 1:Branches of f: for x>0, f(x)=esinx−x; for x<0, ∣x∣=−x and sin∣x∣=sin(−x)=−sinx, so f(x)=e−sinx+x. The only candidate non-differentiable point is x=0.
f(x)={esinx−x,e−sinx+x,x≥0x<0
Step 2:Right derivative: f′(x)=esinxcosx−1, so f′(0+)=e0⋅1−1=0. Left derivative: f′(x)=−e−sinxcosx+1, so f′(0−)=−1⋅1+1=0. The one-sided derivatives agree.
f′(0+)=0=f′(0−)
Step 3:On (−π,−2π) the values satisfy x<0, so f′(x)=1−e−sinxcosx. On this interval cosx<0, hence −e−sinxcosx>0, giving f′(x)>1>0.
f′(x)=1−e−sinxcosx>1
Step 4:Since f′(x)>0 throughout (−π,−2π), f is strictly increasing there.
f′(x)>0∀x∈(−π,−2π)
Final answer: Both Statement I and Statement II are true.
Q14Single correctVector Algebra
Let a=4i^−j^+3k^, b=10i^+2j^−k^ and a vector c be such that 2(a×b)+3(b×c)=0. If a⋅c=15, then c⋅(i^+j^−3k^) is equal to :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2−5
Approach:
Combine the cross products into b×(3c−2a)=0 to write 3c−2a=λb, fix λ using a⋅c=15, then evaluate the required dot product.
Step 1:Given a=(4,−1,3), b=(10,2,−1) and 2(a×b)+3(b×c)=0 with a⋅c=15. Target: c⋅d where d=(1,1,−3). Using a×b=−b×a, the relation becomes b×(3c−2a)=0.
b×(3c−2a)=0
Step 2:Hence c=31(2a+λb). Then ∣a∣2=16+1+9=26 and a⋅b=40−2−3=35.
c=31(2a+λb)
Step 3:Apply a⋅c=31(2∣a∣2+λa⋅b)=31(52+35λ)=15, so 52+35λ=45.
52+35λ=45
Step 4:Substituting: a⋅d=4−1−9=−6 and b⋅d=10+2+3=15. Then c⋅d=31(2(a⋅d)+λ(b⋅d))=31(2(−6)−51(15))=31(−12−3)=−5.
c⋅d=31(−12−3)=−5
Final answer: c⋅(i^+j^−3k^)=−5.
Q15Single correctThree Dimensional Geometry
Let the foot of perpendicular from the point (λ,2,3) on the line 1x−4=2y−9=1z−5 be the point (1,μ,2). Then the distance between the lines 2x−1=3y−2=6z+4 and 2x−λ=3y−μ=6z+5 is equal to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 37146
Approach:
Use that the foot (1,μ,2) lies on the given line to find μ, then use perpendicularity of the foot-to-point vector with the line direction to find λ; the two target lines share a direction, so apply the parallel-lines distance formula.
Step 1:Foot (1,μ,2) lies on 1x−4=2y−9=1z−5 with parameter t. From x: 1−4=t⇒t=−3. Then μ=9+2t=9−6=3 and z=5+t=2, consistent.
t=−3,μ=3
Step 2:The vector from foot (1,3,2) to (λ,2,3) is ⟨λ−1,−1,1⟩ and must be perpendicular to the line direction ⟨1,2,1⟩: (λ−1)(1)+(−1)(2)+(1)(1)=0.
(λ−1)−2+1=0
Step 3:Target lines: L1 through A(1,2,−4) and L2 through B(λ,μ,−5)=(2,3,−5), both with direction d=⟨2,3,6⟩, hence parallel. With AB=⟨1,1,−1⟩, AB×d=⟨1⋅6−(−1)⋅3,(−1)⋅2−1⋅6,1⋅3−1⋅2⟩=⟨9,−8,1⟩.
AB×d=⟨9,−8,1⟩
Step 4:Magnitude of the direction: d=4+9+36=49=7, so d=7146.
d=7146
Final answer: The distance between the lines is 7146.
Q17Single correctDifferential Equations
Let y=y(x) be the solution of the differential equation x1−x2dy+(y1−x2−xcos−1x)dx=0,x∈(0,1),x→1limy(x)=1. Then y(21) equals:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 13−3π
Approach:
Cast the equation as a first-order linear ODE in y, use the integrating factor, integrate the right side by parts, fix the constant from the limit condition, and evaluate at x=21.
Step 1:Divide the equation by x1−x2dx to get dxdy+x1y=1−x2cos−1x. Here P=x1, Q=1−x2cos−1x; target is y(21).
dxdy+x1y=1−x2cos−1x
Step 2:Integrating factor μ=e∫dx/x=x, so dxd(xy)=1−x2xcos−1x. Integrating by parts with u=cos−1x, dv=1−x2xdx, giving v=−1−x2: ∫1−x2xcos−1xdx=−1−x2cos−1x−∫(−1−x2)(−1−x21)dx=−1−x2cos−1x−x+C.
xy=−1−x2cos−1x−x+C
Step 3:Apply x→1limy=1: as x→1, 1−x2→0, so xy→−0−1+C=C−1, while xy→1⋅1=1. Hence C−1=1, C=2.
C=2
Step 4:Evaluating at x=21: cos−121=3π, 1−41=23. Then 21y=2−21−23⋅3π=23−63π, so y=3−33π=3−3π. Numerically 3−π/3≈1.186.
y(21)=3−3π
Final answer: y(21)=3−3π.
Q18Single correctIntegral Calculus
Let f:(1,∞)→R be a function defined as f(x)=x+1x−1. Let fi+1(x)=f(fi(x)),i=1,2,…,25, where f1(x)=f(x). If g(x)+f26(x)=0,x∈(1,∞), then the area of the region bounded by the curves y=g(x),2y=2x−3,y=0 and x=4 is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 181+loge2
Approach:
Find the iteration period of f to evaluate f26, hence g, locate the boundary intersections, and split the bounded region into a triangular part under the line and a part under the curve.
Step 1:With f(x)=x+1x−1, target is g(x)=−f26(x). Then f2(x)=f(x+1x−1)=x+1x−1+1x+1x−1−1=2x−2=−x1; f3(x)=f(−x1)=−x−1x+1; f4(x)=x. The iterates have period 4.
f4(x)=x
Step 2:Since 26=4⋅6+2, f26=f2(x)=−x1, so g(x)=−f26(x)=x1.
g(x)=x1
Step 3:Boundaries: curve y=x1, line y=x−23 (from 2y=2x−3), y=0, and x=4. The line meets y=0 at x=23; the line meets the curve at x−23=x1⇒2x2−3x−2=0⇒x=2 (with y=21). The curve meets x=4 at (4,41).
intersections: (23,0),(2,21),(4,41)
Step 4:Split the bounded region: the triangle under the line from x=23 to 2 has area ∫3/22(x−23)dx=21⋅21⋅21=81; the part under the curve from x=2 to 4 has area ∫24xdx=loge4−loge2=loge2. Total =81+loge2.
A=81+loge2
Final answer: The area of the bounded region is 81+loge2.
Q19Single correctLimit, Continuity and Differentiability
Let f(x)=⎩⎨⎧31(π−2x)2b(1−sinx),x≤π/2,x>π/2. If f is continuous at x=π/2, then the value of ∫03b−6x2+2x−3dx is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 44
Approach:
Apply continuity at x=2π via the right-hand limit to determine b, fix the integration limit 3b−6, then integrate ∣x2+2x−3∣ by splitting at its sign change.
Step 1:Given f(2π)=31 and right branch (π−2x)2b(1−sinx). Substitute x=2π+t, t→0: sinx=cost and π−2x=−2t, so (π−2x)2=4t2.
Step 3:Evaluate ∫02∣x2+2x−3∣dx. Factoring, x2+2x−3=(x+3)(x−1), negative on (0,1) and positive on (1,2), so split at x=1.
x2+2x−3=(x+3)(x−1)
Step 4:∫01−(x2+2x−3)dx=−[3x3+x2−3x]01=−(31+1−3)=35, and ∫12(x2+2x−3)dx=[3x3+x2−3x]12=(38+4−6)−(31+1−3)=32+35=37. Total =35+37=4.
35+37=4
Final answer: The value of the integral is 4.
Q20Single correctCo-ordinate Geometry
Let f(a2+7a+3)x2+f(3a+15)y2=1 represent an ellipse with major axis along y-axis, where f is a strictly decreasing positive function on R. If the set of all possible values of a is R−[α,β], then α2+β2 is equal to :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 240
Approach:
Major axis along the y-axis means the y2 denominator exceeds the x2 denominator; use that f is strictly decreasing to convert this into an inequality between the arguments, then solve the resulting quadratic inequality for a.
Step 1:The ellipse f(a2+7a+3)x2+f(3a+15)y2=1 has major axis along y iff f(3a+15)>f(a2+7a+3); since f is positive, both denominators are valid. Target: α2+β2 from the excluded interval [α,β].
f(3a+15)>f(a2+7a+3)
Step 2:As f is strictly decreasing, the inequality on f-values reverses to its arguments: 3a+15<a2+7a+3.
3a+15<a2+7a+3
Step 3:Solve a2+4a−12>0: factoring, (a+6)(a−2)>0, so a<−6 or a>2, i.e. a∈R−[−6,2].
(a+6)(a−2)>0
Step 4:The excluded interval is [α,β]=[−6,2], so α=−6, β=2, giving α2+β2=36+4=40.
α2+β2=(−6)2+22=40
Final answer: α2+β2=40.
Q21NumericalComplex Numbers and Quadratic Equations
The sum of squares of all the real solutions of the equation log(x+1)(2x2+5x+3)=4−log(2x+3)(x2+2x+1) is equal to
SolutionAnswer: 2
Approach:
Factor both arguments so they share the bases, reduce the equation to a single variable using change of base, solve the resulting quadratic in that variable, then apply the logarithm domain conditions and sum the squares of admissible real solutions.
Step 1:Givens: equation in x with variable bases (x+1) and (2x+3). Target: sum of squares of all real solutions. Factorise the arguments to reveal the bases.
log(x+1)((x+1)(2x+3))=4−log(2x+3)(x+1)2
Step 2:Splitting the left logarithm and setting t=log(x+1)(2x+3) gives log(2x+3)(x+1)=t1, so the right side equals 4−t2.
1+t=4−t2
Step 3:Clear the denominator to obtain a quadratic in t and solve it.
t2−3t+2=0⇒t=1 or t=2
Step 4:Case t=1: 2x+3=x+1⇒x=−2. Case t=2: 2x+3=(x+1)2⇒x2=2⇒x=±2.
x=−2,x=2,x=−2
Step 5:Apply the domain conditions x+1>0,x+1=1,2x+3>0,2x+3=1. For x=−2 the base x+1=−1<0 (rejected); for x=−2 the base x+1=1−2<0 (rejected); for x=2 all conditions hold.
x=2(only admissible)
Step 6:Sum the squares of all admissible real solutions.
∑x2=(2)2=2
Final answer: 2
Q22NumericalIntegral Calculus
If ∫π/6π/4(cot(x−3π)cot(x+3π)+1)dx=αloge(3−1), then 9α2 is equal to
SolutionAnswer: 12
Approach:
Combine the product of cotangents with 1 into a single trigonometric fraction, reduce it to a rational function of tanx, integrate using the standard logarithmic form, evaluate the definite integral, and match the result to αloge(3−1).
Step 1:Givens: definite integral over [π/6,π/4] with A=x−3π, B=x+3π. Target: 9α2. Combine using the cotangent identity, where A−B=−32π so cos(A−B)=−21.
cotAcotB+1=sinAsinBcos(A−B)=sinAsinB−21
Step 2:Rewrite sinAsinB=21[−21−cos2x] and simplify the integrand.
Step 3:Substitute t=tanx, using cos2x=1+t21−t2 and dx=1+t2dt, to reduce the integral to a rational form.
∫1+21+t21−t22⋅1+t2dt=∫3−t22dt
Step 4:Integrate with a=3 and evaluate the limits t=tan(π/6)=1/3 to t=tan(π/4)=1. At t=1: 3−13+1; at t=1/3: 3−1/33+1/3=2.
31[log3−t3+t]1/31=31[log3−13+1−log2]
Step 5:Since (3+1)(3−1)=2, 22+3=(23+1)2=(3−1)−2. Therefore the integral equals 3−2log(3−1), so α=−32.
31log(3−1)−2=−32log(3−1)
Step 6:Compute the requested quantity.
9α2=9⋅34=12
Final answer: 12
Q23NumericalThree Dimensional Geometry
Let a line L1 pass through the origin and be perpendicular to the lines L2:r=(3+t)i^+(2t−1)j^+(2t+4)k^ and L3:r=(3+2s)i^+(3+2s)j^+(2+s)k^,t,s∈R. If (a,b,c),a∈Z, is the point on L3 at a distance of 17 from the point of intersection of L1 and L2, then (a+b+c)2 is equal to .................
SolutionAnswer: 4
Approach:
Take the direction of L1 as the cross product of the direction vectors of L2 and L3, solve for the intersection of L1 and L2, then locate the point on L3 at distance 17 from that intersection using the integer constraint on the first coordinate.
Step 1:Givens: d2=(1,2,2), d3=(2,2,1), L1 through origin perpendicular to both. Target: (a+b+c)2. Compute L1's direction.
d1=(1,2,2)×(2,2,1)=(2−4,4−1,2−4)=(−2,3,−2)
Step 2:Write L1 as (x,y,z)=λ(−2,3,−2) and equate with a point on L2 to find the intersection.
3+t=−2λ,2t−1=3λ,2t+4=−2λ
Step 3:Equating the first and third equations gives 3+t=2t+4⇒t=−1, then λ=−1; the second equation 2(−1)−1=−3=3(−1) is satisfied.
t=−1,λ=−1⇒P=(2,−3,2)
Step 4:A general point on L3 is Q=(3+2s,3+2s,2+s). Impose PQ=17.
(1+2s)2+(6+2s)2+s2=17
Step 5:Multiply out and solve the quadratic in s.
9s2+28s+20=0⇒s=18−28±8=−910 or −2
Step 6:For a=3+2s to be an integer, take s=−2, giving Q=(−1,−1,0), then compute (a+b+c)2.
(a+b+c)2=(−1−1+0)2=4
Final answer: 4
Q24NumericalCo-ordinate Geometry
Consider the circle C:x2+y2−6x−8y−11=0. Let a variable chord AB of the circle C subtend a right angle at the origin. If the locus of the foot of the perpendicular drawn from the origin on the chord AB is the circle x2+y2−αx−βy−γ=0, then α+β+2γ is equal to......
SolutionAnswer: 18
Approach:
Homogenise the circle equation with the chord line to obtain the pair of lines joining the origin to A and B, impose perpendicularity by setting the sum of the coefficients of the squared terms to zero, then substitute the foot-of-perpendicular relations to obtain the locus.
Step 1:Givens: circle x2+y2−6x−8y−11=0 and a chord AB subtending a right angle at the origin. Target: α+β+2γ. With the chord written as lx+my=1, homogenise the circle with it.
x2+y2−(6x+8y)(lx+my)−11(lx+my)2=0
Step 2:For the pair to be perpendicular, set the sum of the coefficients of x2 and y2 to zero.
(1−6l−11l2)+(1−8m−11m2)=0
Step 3:With M=(x,y) as the foot of the perpendicular from the origin on lx+my=1, invert to express l,m and l2+m2 in terms of x,y.
l=x2+y2x,m=x2+y2y,l2+m2=x2+y21
Step 4:Substitute into the right-angle condition and multiply through by (x2+y2).
2(x2+y2)−6x−8y−11=0
Step 5:Compare with x2+y2−αx−βy−γ=0 to read off the coefficients.
α=3,β=4,γ=211
Step 6:Compute the requested combination.
α+β+2γ=3+4+2⋅211=18
Final answer: 18
Q25NumericalSequence and Series
Let f be a polynomial function such that log2(f(x))=(log2(2+32+92+……∞))⋅log3(1+f(1/x)f(x)),x>0 and f(6)=37. Then n=1∑10f(n) is equal to ______.
SolutionAnswer: 395
Approach:
Sum the infinite geometric series to evaluate the first logarithm, collapse the product of logarithms with the change-of-base identity, solve the resulting functional equation to identify the polynomial form, fix the exponent from f(6)=37, then evaluate the finite sum.
Step 1:Givens: log relation for the polynomial f with x>0 and f(6)=37. Target: ∑n=110f(n). Sum the geometric series with a=2,r=31.
2+32+92+⋯=1−312=3
Step 2:Substitute and collapse the product of logarithms using log23⋅log3A=log2A.
log2f(x)=log2(1+f(1/x)f(x))
Step 3:Trying the polynomial f(x)=xk+1 gives f(1/x)=xkxk+1 and 1+f(1/x)f(x)=1+xk=f(x), satisfying the functional equation.
f(x)=xk+1
Step 4:Apply f(6)=37 to fix the exponent.
6k+1=37⇒6k=36⇒k=2,f(x)=x2+1
Step 5:Evaluate the finite sum using the sum-of-squares formula with N=10.
How many questions are in the JEE Main 2026 April 08, Shift 2 paper?
The JEE Main 2026 April 08, Shift 2 paper has 72 questions — Physics (24), Chemistry (24) and Mathematics (24). Every question is on this page with its correct answer and a step-by-step solution.
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