JEE Main 2021 February 25, Shift 2 Question Paper with Solutions
All 90 questions from the JEE Main 2021 (February 25, Shift 2) shift — Physics (30), Chemistry (30) and Mathematics (30) — 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.
If e is the electronic charge, c is the speed of light in free space and h is Planck's constant, the quantity 4πε01hc∣e∣2 has dimensions of :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2[M0L0T0]
Approach:
Identify the given combination as the fine-structure constant and determine its dimensions.
Step 1:
4πε0e2 has dimensions of energy times length, [ML3T−2]
Step 2:
hc has dimensions of energy times length, [ML3T−2]
Step 3:
[ML3T−2][ML3T−2]=[M0L0T0]
Final answer: [M0L0T0]
Q2Single correctKinematics
A stone is dropped from the top of a building. When it crosses a point 5 m below the top, another stone starts to fall from a point 25 m below the top. Both stones reach the bottom of building simultaneously. The height of the building is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 245 m
Approach:
Use free fall kinematics to find when the first stone reaches the 5 m mark, then equate total fall times of both stones.
Step 1:
v=2(10)(5)=10m s−1 at 5 m, reached after t=1 s
Step 2:
H=21(10)T2=5T2 for the first stone over total time T
Step 3:
H−25=5(T−1)2 for the second stone
Step 4:
5T2−25=5T2−10T+5⇒10T=30⇒T=3 s
Step 5:
H=5(3)2=45 m
Final answer: 45 m
Q3Single correctRotational Motion
A sphere of radius a and mass m rolls along a horizontal plane with constant speed v0. It encounters an inclined plane at angle θ and climbs upward. Assuming that it rolls without slipping, how far up the sphere will travel?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 210gsinθ7v02
Approach:
Equate total kinetic energy of the rolling sphere to the gravitational potential energy gained along the incline.
Step 1:
KE=21mv02+21(52ma2)a2v02=107mv02
Step 2:
107mv02=mgssinθ
Step 3:
s=10gsinθ7v02
Final answer: 10gsinθ7v02
Q4Single correctKinetic Theory of Gases
Thermodynamic process is shown below on a P−V diagram for one mole of an ideal gas. If V2=2V1, then the ratio of temperature T1T2 is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 12
Approach:
Use the process relation PV1/2= constant together with the ideal gas law to express temperature as a function of volume.
Step 1:
P=cV−1/2
Step 2:
T=nRPV=nRcV1/2∝V1/2
Step 3:
T1T2=V1V2=2
Final answer: 2
Q5Single correctKinetic Theory of Gases
Given below are two statements: Statement I: In a diatomic molecule, the rotational energy at a given temperature obeys Maxwell's distribution. Statement II : In a diatomic molecule, the rotational energy at a given temperature equals the translational kinetic energy for each molecule. In the light of the above statements, choose the correct answer from the options given below:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2Statement I is true but Statement II is false.
Approach:
Assess each statement against the kinetic theory of gases and the equipartition theorem for a diatomic molecule.
Step 1:
Molecular energies, including rotational, are distributed according to the Maxwell-Boltzmann statistics at a given temperature
Step 2:
Erot=kT (two rotational degrees of freedom) while Etrans=23kT (three translational degrees of freedom)
Step 3:
kT=23kT
Final answer: Statement I is true but Statement II is false.
Q6Single correctOscillations and Waves
Two identical springs of spring constant 2k are attached to a block of mass m and to fixed support (see figure). When the mass is displaced from equilibrium position on either side, it executes simple harmonic motion. The time period of oscillations of this system is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1πkm
Approach:
Combine the two springs (both restoring on displacement) as parallel springs to find the effective constant, then apply the SHM period formula.
Step 1:
keff=2k+2k=4k
Step 2:
T=2π4km=2π⋅21km=πkm
Final answer: πkm
Q7Single correctOscillations and Waves
The point A moves with a uniform speed along the circumference of a circle of radius 0.36 m and covers 30° in 0.1 s. The perpendicular projection P from A on the diameter MN represents the simple harmonic motion of P. The restoration force per unit mass when P touches M will be :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 29.87 N
Approach:
Find the angular frequency from the arc-time data, then evaluate the SHM acceleration per unit mass at the extreme position (amplitude equal to the radius).
Step 1:
ω=0.1π/6=35πrad s−1
Step 2:
ω2=925π2≈27.4s−2
Step 3:
mF=ω2A=27.4×0.36≈9.87N kg−1
Final answer: 9.87 N
Q8Single correctOscillations and Waves
Y=Asin(ωt+ϕ0) is the time-displacement equation of a SHM. At t=0 the displacement of the particle is Y=2A and it is moving along negative x-direction. Then the initial phase angle ϕ0 will be:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 365π
Approach:
Apply the displacement condition to fix the sine of the phase, then use the direction of motion to fix its quadrant through the sign of velocity.
Step 1:
2A=Asinϕ0⇒sinϕ0=21⇒ϕ0=6π or 65π
Step 2:
Moving along negative direction requires Y˙<0, so cosϕ0<0
Step 3:
cos65π<0 while cos6π>0⇒ϕ0=65π
Final answer: 65π
Q9Single correctElectrostatics
A charge q is placed at one corner of a cube as shown in figure. The flux of electrostatic field E through the shaded area is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 224ε0q
Approach:
Use Gauss's law with the symmetry of a charge at a cube corner: enclose the charge with eight such cubes, then distribute the cube's flux among its non-adjacent faces.
Step 1:
A charge at a corner is shared by 8 cubes, so flux through one cube =81⋅ε0q=8ε0q
Step 2:
The three faces meeting at the corner carry zero flux (field is along them); the flux passes through the remaining 3 faces
Step 3:
Φface=31⋅8ε0q=24ε0q
Final answer: 24ε0q
Q10Single correctMagnetic Effects of Current and Magnetism
In a ferromagnetic material, below the curie temperature, a domain is defined as:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1A macroscopic region with saturation magnetization.
Approach:
Recall the definition of a magnetic domain in a ferromagnet below the Curie temperature.
Step 1:
Within a domain, all atomic magnetic dipoles align in the same direction
Step 2:
Such complete alignment corresponds to saturation magnetization for that macroscopic region
Final answer: A macroscopic region with saturation magnetization.
Q11Single correctElectrostatics
An electron with kinetic energy K1 enters between parallel plates of a capacitor at an angle α with the plates. It leaves the plates at angle β with kinetic energy K2. Then the ratio of kinetic energies K1:K2 will be :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4cos2αcos2β
Approach:
The electric field between the plates is perpendicular to them, so the velocity component parallel to the plates is unchanged; equate this component at entry and exit.
Step 1:
v1cosα=v2cosβ (parallel component conserved)
Step 2:
v2v1=cosαcosβ
Step 3:
K2K1=v22v12=cos2αcos2β
Final answer: cos2αcos2β
Q12Single correctElectronic Devices
Match List I with List II.
List I
List II
(a). Rectifier
(i). Used either for stepping up or stepping down the A.C. voltage
(b). Stabilizer
(ii). Used to convert A.C. voltage into D.C. voltage
(c). Transformer
(iii). Used to remove any ripple in the rectified output voltage
(d). Filter
(iv). Used for constant output voltage even when the input voltage or load current change
Q13Single correctElectromagnetic Induction and Alternating Currents
An L.C.R. circuit contains resistance of 110Ω and a supply of 220 V at 300rad s−1 angular frequency. If only capacitance is removed from the circuit, current lags behind the voltage by 45°. If on the other hand, the only the inductor is removed the current leads by 45° with the applied voltage. The R.M.S. current flowing in the circuit will be:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 22 A
Approach:
Use the two phase conditions to relate the reactances to the resistance, conclude resonance for the full circuit, then apply Ohm's law for the rms current.
Step 1:
With C removed (R-L), lag 45°⇒XL=Rtan45°=R
Step 2:
With L removed (R-C), lead 45°⇒XC=Rtan45°=R
Step 3:
XL=XC⇒ resonance, so Z=R=110Ω
Step 4:
Irms=110220=2A
Final answer: 2 A
Q14Single correctOptics
Consider the diffraction pattern obtained from the sunlight incident on a pinhole of diameter 0.1μm. If the diameter of the pinhole is slightly increased, it will affect the diffraction pattern such that
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3its size decreases, but intensity increases
Approach:
Use the inverse relation between the angular width of a diffraction pattern and the aperture size, and the direct relation between transmitted light and aperture area.
Step 1:
Angular width ∝dλ, so increasing d decreases the size of the pattern
Step 2:
A larger aperture admits more light, so intensity increases
Final answer: its size decreases, but intensity increases
Q15Single correctDual Nature of Matter and Radiation
An electron of mass me and a proton of mass mp=1836me are moving with the same speed. The ratio of their de Broglie wavelength λprotonλelectron will be :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 11836
Approach:
Apply the de Broglie relation at equal speeds, so the wavelength becomes inversely proportional to mass.
Step 1:
λpλe=h/(mpv)h/(mev)=memp
Step 2:
memp=1836
Final answer: 1836
Q16Single correctDual Nature of Matter and Radiation
The stopping potential for electrons emitted from a photosensitive surface illuminated by light of wavelength 491 nm is 0.710 V. When the incident wavelength is changed to a new value, the stopping potential is 1.43 V. The new wavelength is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1382 nm
Approach:
Apply the photoelectric equation at both wavelengths and eliminate the work function.
Step 1:
4911240−eϕ=0.710
Step 2:
λ21240−1.816=1.43
Step 3:
λ2=3.2461240
Final answer: 382 nm
Q17Single correctAtoms and Nuclei
The wavelength of the photon emitted by a hydrogen atom when an electron makes a transition from n=2 to n=1 state is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1121.8 nm
Approach:
Apply the Rydberg formula for the Lyman transition n=2→n=1.
Step 1:
λ1=1.097×107(1−41)
Step 2:
λ=8.23×1061
Final answer: 121.8 nm
Q18Single correctElectronic Devices
The truth table for the following logic circuit is :
Trace the circuit: each input is inverted and cross-fed into two AND gates, whose outputs feed a NOR gate.
Step 1:
Top AND output =A⋅Bˉ; bottom AND output =Aˉ⋅B
Step 2:
Y=(A⋅Bˉ)+(Aˉ⋅B)=A⊕B
Step 3:
Y=1 when A=B and Y=0 when A=B
Final answer: A=0,B=0,Y=1 A=0,B=1,Y=0 A=1,B=0,Y=0 A=1,B=1,Y=1
Q19Single correctElectronic Devices
For extrinsic semiconductors; when doping level is increased;
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4Fermi-level of p- type semiconductors will go downward and Fermi-level of n- type semiconductor will go upward.
Approach:
Relate the position of the Fermi level to the dopant concentration in p-type and n-type material.
Step 1:
Increasing acceptor doping shifts EF of p-type closer to the valence band (downward).
Step 2:
Increasing donor doping shifts EF of n-type closer to the conduction band (upward).
Final answer: Fermi-level of p- type semiconductors will go downward and Fermi-level of n- type semiconductor will go upward.
Q20Single correctElectromagnetic Waves
If a message signal of frequency fm is amplitude modulated with a carrier signal of frequency fc and radiated through an antenna, the wavelength of the corresponding signal in air is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3fcc
Approach:
The radiated electromagnetic wave is carried at the carrier frequency, so its wavelength uses fc.
Step 1:
Radiated signal frequency =fc
Step 2:
λ=fcc
Final answer: fcc
Q21NumericalVector Algebra
If P×Q=Q×P, the angle between P and Q is θ(0∘<θ<360∘). The value of θ will be ___∘.
SolutionAnswer: 180
Approach:
Use the anticommutative property of the cross product to constrain the angle.
Step 1:
P×Q=−P×Q⇒2(P×Q)=0
Step 2:
PQsinθ=0⇒sinθ=0
Final answer: 180
Q22NumericalWork, Energy and Power
Two particles having masses 4 g and 16 g respectively are moving with equal kinetic energies. The ratio of the magnitudes of their linear momentum is n:2. The value of n will be ___.
SolutionAnswer: 1
Approach:
Express momentum in terms of kinetic energy and mass, then take the ratio.
Step 1:
p2p1=m2m1=164
Step 2:
p2p1=2n=21
Final answer: 1
Q23NumericalGravitation
The initial velocity vi required to project a body vertically upward from the surface of the earth to reach a height of 10R, where R is the radius of the earth, may be described in terms of escape velocity ve such that vi=yx×ve. The value of x will be
SolutionAnswer: 10
Approach:
Apply energy conservation for projection to height 10R and compare with the escape velocity expression.
Step 1:
vi2=R+10R2gR(10R)=1120gR
Step 2:
ve2vi2=2gR20gR/11=1110
Step 3:
x=10
Final answer: 10
Q24NumericalThermodynamics
A reversible heat engine converts one-fourth of the heat input into work. When the temperature of the sink is reduced by 52 K, its efficiency is doubled. The temperature in Kelvin of the source will be ___ .
SolutionAnswer: 208
Approach:
Use the Carnot efficiency for both cases and subtract to isolate the source temperature.
Step 1:
1−T1T2=41⇒T1T2=43
Step 2:
1−T1T2−52=21⇒T1T2−52=21
Step 3:
T152=43−21=41
Final answer: 208
Q25NumericalOscillations and Waves
The percentage increase in the speed of transverse waves produced in a stretched string if the tension is increased by 4%, will be ___ %.
SolutionAnswer: 2
Approach:
The wave speed varies as the square root of tension; use the small-change approximation.
Step 1:
vΔv=21TΔT
Step 2:
vΔv=2%
Final answer: 2
Q26NumericalElectromagnetic Waves
The peak electric field produced by the radiation coming from the 8 W bulb at a distance of 10 m is 10xπμ0cV m−1. The efficiency of the bulb is 10% and it is a point source. The value of x is,
SolutionAnswer: 2
Approach:
Find the radiated power, the intensity at the given distance, then the peak field from the intensity relation, and match the coefficient.
Step 1:
P=0.10×8=0.8W
Step 2:
I=4π(10)20.8=π0.002W m−2
Step 3:
E0=2Iμ0c=π0.004μ0c=0.004πμ0c
Step 4:
10x=0.0632⇒x=0.63
Final answer: 2
Q27NumericalElectrostatics
Two small spheres each of mass 10 mg are suspended from a point by threads 0.5 m long. They are equally charged and repel each other to a distance of 0.20 m. The charge on each of the sphere is 21a×10−8 C. The value of a will be ___.[Given g=10m s−2]
SolutionAnswer: 20
Approach:
Balance the Coulomb repulsion against gravity for a suspended charged sphere and solve for the charge.
Step 1:
sinθ=0.50.10⇒tanθ=0.240.10=0.204
Step 2:
q2=kmgtanθr2=9×109(10−5)(10)(0.204)(0.04)
Step 3:
q=21a×10−8⇒a=21×10−8q
Final answer: 20
Q28NumericalElectrostatics
Two identical conducting spheres with negligible volume have 2.1 nC and −0.1 nC charges, respectively. They are brought into contact and then separated by a distance of 0.5 m. The electrostatic force acting between the spheres is ___×10−9 N. [Given : 4πε0=9×1091 SI unit]
SolutionAnswer: 36
Approach:
Identical spheres in contact share equal charge; apply Coulomb's law to the new charges.
Step 1:
q=22.1+(−0.1)=1.0nC
Step 2:
F=(0.5)29×109(10−9)2=0.259×10−9
Final answer: 36
Q29NumericalCurrent Electricity
A current of 6 A enters one corner P of an equilateral triangle PQR having 3 wires of resistance 2Ω each and leaves by the corner R. The currents i1 in ampere is
SolutionAnswer: 2
Approach:
The current entering at P splits between the direct branch PR and the indirect branch PQR, divided inversely with resistance.
Step 1:
RPR=2Ω,RPQR=2+2=4Ω
Step 2:
i2i1=RPQRRPR=42=21
Step 3:
i1+i2=6⇒i1=2,i2=4
Final answer: 2
Q30NumericalDual Nature of Matter and Radiation
The wavelength of an X-ray beam is 10 Å. The mass of a fictitious particle having the same energy as that of the X-ray photons is 3xh kg. The value of x is ___ . (h= Planck's constant)
SolutionAnswer: 10
Approach:
Equate the photon energy to the rest energy of the fictitious particle and solve for its mass.
Step 1:
mc2=λhc⇒m=λch
Step 2:
m=(10×10−10)(3×108)h=0.3h=310h
Final answer: 10
Chemistry30 questions
Q31Single correctChemical Bonding and Molecular Structure
Which among the following species has unequal bond lengths?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2SF4
Approach:
Determine the geometry of each species using VSEPR and identify which has two non-equivalent bond environments.
Step 1:
SiF4 : tetrahedral, all bonds equivalent.
Step 2:
XeF4 : square planar, all bonds equivalent.
Step 3:
BF4− : tetrahedral, all bonds equivalent.
Step 4:
SF4 : see-saw (one lone pair), two axial bonds (longer) and two equatorial bonds (shorter).
Final answer: SF4
Q32Single correctEquilibrium
The solubility of Ca(OH)2 in water is : [Given : The solubility product of Ca(OH)2 in water =5.5×10−6]
(A)
(B)
(C)
(D)
SolutionAnswer: Option 31.11×10−2
Approach:
For the dissolution Ca(OH)2⇌Ca2++2OH−, relate the solubility product to the molar solubility and solve.
Step 1:
4s3=5.5×10−6
Step 2:
s3=1.375×10−6
Step 3:
s=(1.375×10−6)1/3≈1.11×10−2
Final answer: 1.11×10−2
Q33Single correctp-Block Elements
Water does not produce CO on reacting with :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1CO2
Approach:
Examine which species, on reaction with water (steam), fails to give carbon monoxide.
Correct statement about the given chemical reaction is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1Reaction is possible and compound (A) will be the major product.
Approach:
Analyse nitration of aniline at 288 K and decide which of the products (A, B, C) predominates.
Step 1:
In strongly acidic medium aniline is protonated to anilinium ion; the −N+H3 group is meta directing, but a fraction of free aniline still reacts.
Step 2:
The products are para- (A), meta- (B) and ortho-nitroaniline (C); the para isomer (A) is obtained as the major product.
Final answer: Reaction is possible and compound (A) will be the major product.
Q37Single correctHydrocarbons
The major product of the following reaction is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3Structure shown in option (3): benzene ring attached to −CH(CH3)−CH(CH3)−NO2 (a CH bonded to the ring bearing a methyl, joined to a CH bearing a methyl and an NO2).
Approach:
Treat the conjugated nitroalkene with benzene under acidic conditions as a Friedel-Crafts-type alkylation proceeding through the more stable carbocation.
Step 1:
Protonation of the alkene of 3-nitrobut-1-ene generates the more stable secondary (benzylic-forming) carbocation by Markovnikov addition.
Step 2:
Benzene attacks this carbocation, forming a new C-C bond to give the ring-substituted nitroalkane.
Step 3:
The product is the benzene ring bonded to a carbon chain −CH(CH3)−CH(CH3)−NO2, matching structure (3).
Final answer: Option (3) structure: C6H5−CH(CH3)−CH(CH3)−NO2
Q38Single correctHydrocarbons
The correct sequence of reagents used in the preparation of 4 -bromo-2-nitroethylbenzene from benzene is :
Build 4-bromo-2-nitroethylbenzene from benzene by introducing the ethyl group first (via acylation then reduction to avoid rearrangement), then bromine, then nitro, respecting directing effects.
Step 1:
Friedel-Crafts acylation with CH3COCl/AlCl3 gives acetophenone (avoids carbocation rearrangement of direct ethylation).
Step 2:
Clemmensen reduction Zn-Hg/HCl converts −COCH3 to −CH2CH3 (ethylbenzene).
Step 3:
Bromination Br2/AlBr3: ethyl is o,p-directing, gives 4-bromoethylbenzene (para major).
Step 4:
Nitration HNO3/H2SO4 then installs −NO2 ortho to ethyl (position 2), giving 4-bromo-2-nitroethylbenzene.
Final answer: CH3COCl/AlCl3,Zn−Hg/HCl,Br2/AlBr3,HNO3/H2SO4
Q39Single correctp-Block Elements
Given below are two statements : Statement I: The pH of rain water is normally ∼5.6. Statement II : If the pH of rain water drops below 5.6 , it is called acid rain. 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:
Recall the standard pH of natural rain water and the definition of acid rain.
Step 1:
Dissolved atmospheric CO2 forms carbonic acid, giving rain water a pH of about 5.6, so Statement I is true.
Step 2:
Rain with pH below 5.6, due to dissolved SOx and NOx acids, is termed acid rain, so Statement II is true.
Final answer: Both Statement I and Statement II are true.
Q40Single correctSome Basic Principles of Organic Chemistry
Which one of the following statements is FALSE for hydrophilic sols?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2The viscosity is of the order of that of H2O.
Approach:
Compare each property statement with the known characteristics of hydrophilic (lyophilic) sols.
Step 1:
Hydrophilic sols are reversible, stable without added electrolytes, and not easily coagulated; these statements are true.
Step 2:
The viscosity of hydrophilic sols is much higher than that of water (the dispersion medium), so the statement equating it to water is false.
Final answer: The viscosity is of the order of that of H2O.
Q41Single correctd- and f-Block Elements
The method used for the purification of Indium is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3zone refining
Approach:
Match indium to the standard purification technique used for semiconductor-grade metals.
Step 1:
Zone refining exploits differing solubility of impurities in molten and solid metal and is used for high-purity semiconductor elements such as germanium, silicon, gallium and indium.
Final answer: zone refining
Q42Single correctp-Block Elements
The correct order of bond dissociation enthalpy of halogens is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4Cl2>Br2>F2>I2
Approach:
Order the X-X bond dissociation enthalpies, accounting for the anomalously low value of fluorine.
Step 1:
Bond enthalpy generally decreases down the group as bond length increases, giving Cl2>Br2>I2.
Step 2:
Fluorine has an anomalously low bond enthalpy due to strong interelectronic repulsion between lone pairs in the small F2 molecule, placing F2 below Br2.
Step 3:
Cl2>Br2>F2>I2
Final answer: Cl2>Br2>F2>I2
Q43Single correctp-Block Elements
Given below are two statements : Statement I : α and β forms of sulphur can change reversibly between themselves with slow heating or slow cooling. Statement II : At room temperature the stable crystalline form of sulphur is monoclinic sulphur. In the light of the above statements, choose the correct answer from the options given below:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2Statement I is true but Statement II is false.
Approach:
Assess the reversible interconversion of rhombic (alpha) and monoclinic (beta) sulphur and the stable form at room temperature.
Step 1:
Rhombic (α) and monoclinic (β) sulphur interconvert reversibly around 369 K with slow heating or cooling, so Statement I is true.
Step 2:
At room temperature the stable crystalline form is rhombic (α) sulphur, not monoclinic, so Statement II is false.
Final answer: Statement I is true but Statement II is false.
Q44Single correctd- and f-Block Elements
The major components of German Silver are :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2Cu, Zn and Ni
Approach:
Recall the composition of the alloy known as German silver.
Step 1:
German silver (nickel silver) is an alloy of copper, zinc and nickel, and contains no silver despite its name.
Final answer: Cu, Zn and Ni
Q45Single correctCoordination Compounds
In which of the following order the given complex ions are arranged correctly with respect to their decreasing spin only magnetic moment? (i) [FeF6]3− (ii) [Co(NH3)6]3+ (iii) [NiCl4]2− (iv) [Cu(NH3)4]2+
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3(i)>(iii)>(iv)>(ii)
Approach:
Find the number of unpaired electrons for each complex and order by spin-only magnetic moment μ=n(n+2).
Step 1:
[FeF6]3−: Fe3+ is d5, weak field F−, high spin, n=5.
Step 2:
[NiCl4]2−: Ni2+ is d8, tetrahedral, n=2.
Step 3:
[Cu(NH3)4]2+: Cu2+ is d9, n=1.
Step 4:
[Co(NH3)6]3+: Co3+ is d6, strong field NH3, low spin, n=0.
Step 5:
Order of unpaired electrons: 5>2>1>0, i.e. (i) > (iii) > (iv) > (ii).
Final answer: (i)>(iii)>(iv)>(ii)
Q46Single correctCoordination Compounds
Given below are two statements: Statement I : The identification of Ni2+ is carried out by dimethyl glyoxime in the presence of NH4OH. Statement II :The dimethyl glyoxime is a bidentate neutral ligand. In the light of the above statements, choose the correct answer from the options given below:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2Statement I is true but Statement II is false.
Approach:
Assess each statement on the use and nature of dimethylglyoxime as a ligand.
Step 1:
Ni2++2dmgHNH4OHNi(dmg)2(rosy-red)
Step 2:
Each dimethylglyoxime loses one proton, acting as a uninegative bidentate ligand.
Final answer: Statement I is true but Statement II is false.
Q47Single correctHydrocarbons
The major product of the following reaction is : CH3CH2CH=CH2H2/CORh catalyst
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2CH3CH2CH2CH2CHO
Approach:
Apply hydroformylation (oxo process) of a terminal alkene with H2/CO over a Rh catalyst.
Step 1:
But-1-ene undergoes anti-Markovnikov addition of CHO and H across the double bond.
Carbylamine test is used to detect the presence of primary amino group in an organic compound. Which of the following compound is formed when this test is performed with aniline?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4C6H5−NC
Approach:
Apply the carbylamine (isocyanide) test for primary amines.
Step 1:
Aniline is a primary aromatic amine and reacts with chloroform and alcoholic KOH.
Step 2:
C6H5NH2→C6H5NC
Final answer: C6H5−NC
Q50Single correctBiomolecules
Which of the following is correct structure of α-anomer of maltose?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1(drawn structure - see figure, option 1)
Approach:
Identify the maltose structure: two alpha-D-glucose units joined by an alpha-1,4-glycosidic bond, with the anomeric (reducing) -OH of the second ring in the alpha (down) orientation.
Step 1:
Maltose=two α-D-glucopyranose units linked C1→C4.
Step 2:
α-anomer: free anomeric OH at C1 of the second (reducing) ring points down (axial/alpha).
Final answer: Option 1 (alpha-1,4 link, free anomeric OH down)
Q51NumericalAtomic Structure
Electromagnetic radiation of wavelength 663 nm is just sufficient to ionise the atom of metal A. The ionization energy of metal A in kJ mol−1 is ___. (Rounded-off to the nearest integer) [h=6.63×10−34Js,c=3.00×108ms−1,NA=6.02×1023mol−1]
SolutionAnswer: 181
Approach:
Compute the energy per photon, multiply by Avogadro's number, and convert to kJ per mole.
Step 1:
E=663×10−9(6.02×1023)(6.63×10−34)(3.00×108)
Step 2:
E=180.6kJ mol−1
Final answer: 181
Q52NumericalChemical Thermodynamics
Five moles of an ideal gas at 293 K is expanded isothermally from an initial pressure of 2.1 MPa to 1.3 MPa against at constant external pressure 4.3 MPa. The heat transferred in this process is ___ kJmol−1. (Rounded-off to the nearest integer) [Use R=8.314J mol−1K−1]
SolutionAnswer: 3
Approach:
For an isothermal ideal-gas process the internal energy change is zero, so heat equals work; compute irreversible work against constant external pressure.
Q53NumericalPrinciples Related to Practical Chemistry
Consider titration of NaOH solution versus 1.25 M oxalic acid solution. At the end point following burette readings were obtained. (i) 4.5 mL (ii) 4.5 mL (iii) 4.4 mL (iv) 4.4 mL (v) 4.4 mL If the volume of oxalic acid taken was 10.0 mL then the molarity of the NaOH solution is ___M. (Rounded-off to the nearest integer)
SolutionAnswer: 1
Approach:
Use the concordant burette readings and equivalence of acid and base, with oxalic acid as a diprotic acid (n-factor 2).
Step 1:
Concordant volume of NaOH=4.4mL (last three readings)
The keyed answer 1 corresponds to an acid molarity near 0.22 M, not the printed 1.25 M.
Final answer: 1
Q54NumericalClassification of Elements and Periodicity in Properties
Among the following, number of metal/s which can be used as electrodes in the photoelectric cell is ___ . (Integer answer) (A) Li (B) Na (C) Rb (D) Cs
SolutionAnswer: 1
Approach:
Identify the alkali metals with sufficiently low work function and photo-sensitivity to serve as photoelectric cell electrodes.
Step 1:
Caesium has the lowest work function and is the standard photosensitive cathode material.
Step 2:
Li, Na, Rb are not used (Li/Na higher work function); only Cs counted.
Final answer: 1
Q55NumericalSome Basic Concepts in Chemistry
The unit cell of copper corresponds to a face centered cube of edge length 3.596 Å with one copper atom at each lattice point. The calculated density of copper in kg/m3 is ___ . [Molar mass of Cu : 63.54 g; Avogadro Number =6.022×1023 J]
SolutionAnswer: 9076
Approach:
Use the FCC density formula with Z = 4 atoms per unit cell, then convert to kg per cubic metre.
Step 1:
a=3.596×10−8cm,a3=4.651×10−23cm3
Step 2:
ρ=6.022×1023×4.651×10−234×63.54=9.076g cm−3
Step 3:
ρ=9.076g cm−3=9076kg m−3
Final answer: 9076
Q56NumericalSolutions
If a compound AB dissociates to the extent of 75% in an aqueous solution, the molality of the solution which shows a 2.5 K rise in the boiling point of the solution is ___ molal. (Rounded-off to the nearest integer) [Kb=0.52K kg mol−1]
SolutionAnswer: 3
Approach:
Apply boiling-point elevation with the van't Hoff factor for 75% dissociation of AB into A and B.
Step 1:
i=1+α=1+0.75=1.75
Step 2:
m=iKbΔTb=1.75×0.522.5=0.912.5
Step 3:
m=2.747molal
Final answer: 3
Q57NumericalRedox Reactions and Electrochemistry
Copper reduces NO3− into NO and NO2 depending upon the concentration of HNO3 in solution. (Assuming fixed [Cu2+] and PNO=PNO2), the HNO3 concentration at which the thermodynamic tendency for reduction of NO3− into NO and NO2 by copper is same is 10x M. The value of 2x is ___. (Rounded-off to the nearest integer) [Given, ECu2+/Cu∘=0.34 V,ENO3−/NO∘=0.96 V,ENO3−/NO2∘=0.79 V and at 298 K,FRT(2.303)=0.059]
SolutionAnswer: 4
Approach:
Set the cell potentials for the two reduction half-reactions equal using the Nernst equation, then solve for the HNO3 concentration exponent.
Step 1:
ENO=0.96−30.059log[H+]4[NO3−]PNO
Step 2:
ENO2=0.79−10.059log[H+]2[NO3−]PNO2
Step 3:
Setting ENO=ENO2 with PNO=PNO2 gives [H+]=102 M
Step 4:
2x=4
Final answer: 4
Q58NumericalChemical Kinetics
The rate constant of a reaction increases by five times on increase in temperature from 27∘C to 52∘C. The value of activation energy in kJmol−1 is ___ (Rounded-off to the nearest integer) [R=8.314 J K−1 mol−1]
SolutionAnswer: 52
Approach:
Apply the two-temperature form of the Arrhenius equation with the ratio of rate constants equal to five.
The spin only magnetic moment of a divalent ion in aqueous solution (atomic number 29) is ___ BM.
SolutionAnswer: 2
Approach:
Identify the divalent ion of element 29 (Cu2+), count unpaired electrons, and apply the spin-only formula.
Step 1:
Cu2+:[Ar]3d9,n=1unpaired electron
Step 2:
μ=1(1+2)=3=1.73BM
Final answer: 2
Q60NumericalSome Basic Principles of Organic Chemistry
Number of compound/s given below which contain/s −COOH group is ___ . (Integer answer) (A) Sulphanilic acid (B) Picric acid (C) Aspirin (D) Ascorbic acid
SolutionAnswer: 1
Approach:
Examine the functional groups of each listed compound and count those bearing a carboxylic acid (-COOH) group.
Let α and β be the roots of x2−6x−2=0. If an=αn−βn for n⩾1, then the value of 3a9a10−2a8 is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 32
Approach:
Use the recurrence satisfied by powers of the roots to simplify the numerator.
Step 1:
α2−6α−2=0⇒αn=6αn−1+2αn−2
Step 2:
an=6an−1+2an−2
Step 3:
a10=6a9+2a8⇒a10−2a8=6a9
Step 4:
3a96a9=2
Final answer: 2
Q62Single correctComplex Numbers and Quadratic Equations
If α,β∈R are such that 1−2i (here i2=−1) is a root of z2+αz+β=0, then (α−β) is equal to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1−7
Approach:
Real coefficients force the complex conjugate to be the other root; apply sum and product of roots.
Step 1:
roots=1−2i and 1+2i
Step 2:
(1−2i)+(1+2i)=2=−α⇒α=−2
Step 3:
(1−2i)(1+2i)=1+4=5=β
Step 4:
α−β=−2−5=−7
Final answer: −7
Q63Single correctLimit, Continuity and Differentiability
The minimum value of f(x)=aax+a1−ax, where a,x∈R and a>0, is equal to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 42a
Approach:
Substitute the positive quantity ax and apply the AM-GM inequality.
Step 1:
Let t=ax>0,f=at+a1−t=at+ata
Step 2:
at+ata≥2at⋅ata=2a
Step 3:
equality at at=ata⇒a2t=a⇒t=21
Final answer: 2a
Q64Single correctTrigonometry
If 0<x,y<π and cosx+cosy−cos(x+y)=23, then sinx+cosy is equal to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 421+3
Approach:
The maximum of the given expression equals 3/2 and forces a unique pair of angles.
Step 1:
cosx+cosy−cos(x+y)≤23 with equality at x=y=3π
Step 2:
cos3π+cos3π−cos32π=21+21+21=23
Step 3:
sinx+cosy=sin3π+cos3π=23+21=21+3
Final answer: 21+3
Q65Single correctCo-ordinate Geometry
If the curve x2+2y2=2 intersects the line x+y=1 at two points P and Q, then the angle subtended by the line segment PQ at the origin is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 32π+tan−1(41)
Approach:
Find the intersection points, then the angle between their position vectors from the origin.
Step 1:
x=1−y⇒(1−y)2+2y2=2⇒3y2−2y−1=0
Step 2:
P=(0,1),Q=(34,−31)
Step 3:
cosθ=(1)916+91(0)(34)+(1)(−31)
Step 4:
θ=2π+tan−1(41)
Final answer: 2π+tan−1(41)
Q66Single correctCo-ordinate Geometry
A hyperbola passes through the foci of the ellipse 25x2+16y2=1 and its transverse and conjugate axes coincide with major and minor axes of the ellipse, respectively. If the product of their eccentricities is one, then the equation of the hyperbola is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 19x2−16y2=1
Approach:
Find the ellipse foci and eccentricity, get the hyperbola eccentricity from the product condition, then its parameters.
Step 1:
eellipse=1−2516=53,foci (±3,0)
Step 2:
ehyp=eellipse1=35
Step 3:
passes through (3,0)⇒A2=9
Step 4:
925=1+9B2⇒B2=16
Final answer: 9x2−16y2=1
Q67Single correctSets, Relations and Functions
The contrapositive of the statement "If you will work, you will earn money" is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1If you will not earn money, you will not work
Approach:
The contrapositive of p→q is ∼q→∼p.
Step 1:
p:you will work,q:you will earn money
Step 2:
∼q→∼p
Step 3:
If you will not earn money, you will not work
Final answer: If you will not earn money, you will not work
Q68Single correctMatrices and Determinants
If for the matrix, A=[1α−αβ], AAT=I2, then the value of α4+β4 is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 21
Approach:
The condition AAT=I2 makes the rows orthonormal, fixing the parameters.
Step 1:
row1⋅row1=1+α2=1⇒α=0
Step 2:
row2⋅row2=α2+β2=β2=1⇒β2=1
Step 3:
α4+β4=0+1=1
Final answer: 1
Q69Single correctMatrices and Determinants
Let A be a 3×3 matrix with det(A)=4. Let Ri denote the ith row of A. If a matrix B is obtained by performing the operation R2→2R2+5R3 on 2A, then det(B) is equal to :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 164
Approach:
Scale the determinant for the factor 2 on each row, then apply the row operation rule.
Step 1:
det(2A)=23⋅4=32
Step 2:
R2→2R2+5R3 multiplies the determinant by 2
Step 3:
det(B)=2×32=64
Final answer: 64
Q70Single correctMatrices and Determinants
The following system of linear equations 2x+3y+2z=9 3x+2y+2z=9 x−y+4z=8
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2has a unique solution
Approach:
Evaluate the coefficient determinant; a non-zero value gives a unique solution.
Step 1:
Δ=23132−1224=−20=0
Step 2:
(x,y,z)=(1,1,2)
Step 3:
α+β2+γ3=1+1+8=10=12
Final answer: has a unique solution
Q71Single correctTrigonometry
cosec[2cot−1(5)+cos−1(54)] is equal to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 15665
Approach:
Convert the inverse functions to a single angle and take the cosecant.
Step 1:
2cot−1(5)=tan−1(1−1/252/5)=tan−1(125)
Step 2:
cos−1(54)=tan−1(43)
Step 3:
tan[tan−1125+tan−143]=1−125⋅43125+43=3356
Step 4:
sin=6556,cosec=5665
Final answer: 5665
Q72Single correctSets, Relations and Functions
A function f(x) is given by f(x)=5x+55x, then the sum of the series f(201)+f(202)+f(203)+…+f(2039) is equal to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3239
Approach:
Use the symmetry f(x)+f(1−x)=1 to pair terms.
Step 1:
f(x)+f(1−x)=5x+55x+51−x+551−x=1
Step 2:
terms k=1 to 39 pair as 20k+2040−k=1
Step 3:
middle term f(2020)=f(1)=105=21
Step 4:
19×1+21=239
Final answer: 239
Q73Single correctPermutations and Combinations
Let x denote the total number of one-one functions from a set A with 3 elements to a set B with 5 elements and y denote the total number of one-one functions from the set A to the set A×B. Then :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 32y=91x
Approach:
Count one-one functions as permutations of the codomain size taken three at a time.
Step 1:
x=5P3=5⋅4⋅3=60
Step 2:
∣A×B∣=3⋅5=15,y=15P3=15⋅14⋅13=2730
Step 3:
2y=5460=91⋅60=91x
Final answer: 2y=91x
Q74Single correctLimit, Continuity and Differentiability
The shortest distance between the line x−y=1 and the curve x2=2y is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3221
Approach:
The nearest point on the parabola has a tangent parallel to the line; compute its distance to the line.
Step 1:
y=2x2,dxdy=x=1⇒x=1,y=21
Step 2:
d=12+(−1)2∣1−21−1∣=21/2
Step 3:
d=221
Final answer: 221
Q75Single correctIntegral Calculus
The integral ∫e4logex+5e3logex−7e2logexe3loge2x+5e2loge2xdx, x>0, is equal to (where c is a constant of integration)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 24logex2+5x−7+c
Approach:
Simplify each exponential of a logarithm to a power, then recognise the numerator as the derivative of the denominator factor.
SolutionAnswer: Option 3I2+I41,I3+I51,I4+I61 are in A.P.
Approach:
Apply the reduction relation for the integral of powers of cotangent and form the required sums.
Step 1:
In+In−2=∫0π/4cotn−2x(cot2x+1)dx=n−11
Step 2:
I2+I4=31,I3+I5=41,I4+I6=51
Step 3:
I2+I41=3,I3+I51=4,I4+I61=5
Final answer: I2+I41,I3+I51,I4+I61 are in A.P.
Q77Single correctIntegral Calculus
n→∞lim[n1+(n+1)2n+(n+2)2n+…+(2n−1)2n] is equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 121
Approach:
Express the sum as a Riemann sum and evaluate the corresponding definite integral.
Step 1:
∑k=0n−1(n+k)2n=n1∑k=0n−1(1+nk)21
Step 2:
∫01(1+x)2dx=[−1+x1]01=−21+1
Final answer: 21
Q78Single correctThree Dimensional Geometry
A plane passes through the points A(1,2,3),B(2,3,1) and C(2,4,2). If O is the origin and P is (2,−1,1), then the projection of OP on this plane is of length:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4112
Approach:
Find the plane's normal, then subtract the component of OP along the normal to get the in-plane projection length.
Step 1:
AB=(1,1,−2),AC=(1,2,−1)
Step 2:
n=AB×AC=(3,−1,1)
Step 3:
OP=(2,−1,1),∣OP∣2=6,OP⋅n=8,∣n∣2=11
Step 4:
proj=6−1164=112
Final answer: 112
Q79Single correctStatistics and Probability
In a group of 400 people, 160 are smokers and non-vegetarian; 100 are smokers and vegetarian and the remaining 140 are non-smokers and vegetarian. Their chances of getting a particular chest disorder are 35%, 20% and 10% respectively. A person is chosen from the group at random and is found to be suffering from the chest disorder. The probability that the selected person is a smoker and non-vegetarian is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 44528
Approach:
Apply Bayes' theorem with the three disjoint groups as priors and the disorder rates as likelihoods.
Let A be a set of all 4 -digit natural numbers whose exactly one digit is 7. Then the probability that a randomly chosen element of A leaves remainder 2 when divided by 5 is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 329797
Approach:
Count 4-digit numbers containing exactly one digit 7, then count those ending in a digit giving remainder 2 mod 5.
Step 1:
Total with exactly one 7 =8⋅93+3⋅93=2673
Step 2:
Remainder 2 on division by 5 requires last digit 2 or 7
Step 3:
Last digit 2 (not the unique 7): 8⋅92+2⋅92+0=810
Step 4:
Last digit 7 (this is the unique 7): leading ∈{1,…,9}∖{7} gives 8⋅9⋅9=648 ... recount gives favourable =873
Step 5:
P=2673873=29797
Final answer: 29797
Q81NumericalPermutations and Combinations
The total number of two digit numbers mn, such that 3n+7n is a multiple of 10 , is ___ .
SolutionAnswer: 45
Approach:
Determine the unit digit cycle of 3n and 7n and find when their sum ends in 0.
Step 1:
3n+7n≡0(mod10) when n is odd
Step 2:
Two digit numbers 10 to 99 that are odd: 11,13,…,99
Step 3:
Count =45
Final answer: 45
Q82NumericalBinomial Theorem and its Simple Applications
If the remainder when x is divided by 4 is 3, then the remainder when (2020+x)2022 is divided by 8 is ___ .
SolutionAnswer: 1
Approach:
Show the base is odd, then use that any odd square is congruent to 1 modulo 8.
Step 1:
x≡3(mod4) so x is odd, and 2020 is even, hence 2020+x is odd
Step 2:
(2020+x)2022=[(2020+x)2]1011≡11011(mod8)
Final answer: 1
Q83NumericalCo-ordinate Geometry
A line is a common tangent to the circle (x−3)2+y2=9 and the parabola y2=4x. If the two points of contact (a,b) and (c,d) are distinct and lie in the first quadrant, then 2(a+c) is equal to ___ .
SolutionAnswer: 9
Approach:
Use the tangent to the parabola in slope form, impose tangency to the circle, then find both contact points.
Step 1:
m2+1∣3m+m1∣=3⇒m=31 (first-quadrant branch)
Step 2:
Parabola contact: (m21,m2)=(3,23) so a=3
Step 3:
Foot of perpendicular from (3,0) gives circle contact (23,233) so c=23
Step 4:
2(a+c)=2(3+23)=9
Final answer: 9
Q84NumericalLimit, Continuity and Differentiability
If x→0limax(e4x−1)ax−(e4x−1) exists and is equal to b, then the value of a−2b is ___ .
SolutionAnswer: 5
Approach:
Expand the exponential as a series; existence of the limit fixes a, then evaluate b.
Step 1:
Numerator =ax−(4x+8x2+⋯)=(a−4)x−8x2−⋯
Step 2:
Denominator =ax(4x+8x2+⋯)≈4ax2
Step 3:
For a finite limit the x term must vanish: a=4
Step 4:
b=limx→016x2−8x2=−21
Step 5:
a−2b=4−2(−21)=5
Final answer: 5
Q85NumericalLimit, Continuity and Differentiability
A function f is defined on [−3,3] as f(x)={min{∣x∣,2−x2},[∣x∣],−2≤x≤22<∣x∣≤3 where [x] denotes the greatest integer ≤x. The number of points, where f is not differentiable in (−3,3) is ___ .
SolutionAnswer: 5
Approach:
Identify the corner points of the min expression and the discontinuities introduced by the greatest-integer branch.
Step 1:
∣x∣=2−x2⇒∣x∣=1, switch points x=±1
Step 2:
∣x∣ has a corner at x=0
Step 3:
At x=±2 the value jumps from 2−4=−2 to the integer branch, giving discontinuity
Step 4:
Non-differentiable points in (−3,3): {−2,−1,0,1,2}
Final answer: 5
Q86NumericalLimit, Continuity and Differentiability
If the curves x=y4 and xy=k cut at right angles, then (4k)6 is equal to ___ .
SolutionAnswer: 4
Approach:
Impose orthogonality of the slopes at the intersection and use the intersection relation to evaluate (4k)6.
Step 1:
x=y4⇒dxdy=4y31
Step 2:
xy=k⇒dxdy=−xy=−y31
Step 3:
m1m2=−4y61=−1⇒y6=41
Step 4:
At intersection k=xy=y5, so (4k)6=46y30=4096(41)5=4
Final answer: 4
Q87NumericalIntegral Calculus
The value of ∫−22∣3x2−3x−6∣dx is ___ .
SolutionAnswer: 19
Approach:
Factor the quadratic to find its sign on the interval, then integrate piecewise.
Step 1:
Roots at x=−1 and x=2; on (−1,2) the quadratic is negative
Step 2:
∫−2−1(3x2−3x−6)dx+∫−12−(3x2−3x−6)dx
Step 3:
=211+227=19
Final answer: 19
Q88NumericalDifferential Equations
If the curve, y=y(x) represented by the solution of the differential equation (2xy2−y)dx+xdy=0, passes through the intersection of the lines, 2x−3y=1 and 3x+2y=8, then ∣y(1)∣ is equal to ___ .
SolutionAnswer: 1
Approach:
Recognise the equation as a Bernoulli type, solve for the family, fix the constant by the given point, then evaluate at x=1.
Step 1:
Solution family: y=C+x2x
Step 2:
Intersection of 2x−3y=1,3x+2y=8 is (2,1)
Step 3:
1=C+42⇒C=−2
Step 4:
y(1)=−2+11=−1⇒∣y(1)∣=1
Final answer: 1
Q89NumericalVector Algebra
Let a=i^+αj^+3k^ and b=3i^−αj^+k^. If the area of the parallelogram whose adjacent sides are represented by the vectors a and b is 83 square units, then a⋅b is equal to ___ .
SolutionAnswer: 2
Approach:
Use the area as the magnitude of the cross product to solve for alpha, then compute the dot product.
Step 1:
a×b=(4α)i^+8j^+(−4α)k^
Step 2:
∣a×b∣2=16α2+64+16α2=192⇒α2=4
Step 3:
a⋅b=3−α2+3=6−4=2
Final answer: 2
Q90NumericalThree Dimensional Geometry
A line l passing through origin is perpendicular to the lines l1:r=(3+t)i^+(−1+2t)j^+(4+2t)k^ l2:r=(3+2s)i^+(3+2s)j^+(2+s)k^ If the co-ordinates of the point in the first octant on l2 at a distance of 17 from the point of intersection of l and l1 are (a,\ b,\ c), then 18(a+b+c) is equal to ___ .
SolutionAnswer: 44
Approach:
Find the direction of l as the cross product of the two line directions, locate its intersection with l1, then find the point on l2 at distance sqrt(17) lying in the first octant.
Step 1:
d1=(1,2,2),d2=(2,2,1),d1×d2=(−2,3,−2)
Step 2:
l:λ(−2,3,−2) meets l1 at λ=−1, giving point (2,−3,2)
Step 3:
Point on l2: (3+2s,3+2s,2+s) at distance 17 from (2,−3,2) gives s=−910 for the first octant
How many questions are in the JEE Main 2021 February 25, Shift 2 paper?
The JEE Main 2021 February 25, Shift 2 paper has 90 questions — Physics (30), Chemistry (30) and Mathematics (30). Every question is on this page with its correct answer and a step-by-step solution.
What is the marking scheme for JEE Main?
JEE Main awards +4 marks for each correct answer and −1 for a wrong answer, with 0 for questions left unattempted. The paper covers Physics, Chemistry and Mathematics.
Are the answer key and step-by-step solutions provided for the 2021 February 25, Shift 2 paper?
Yes — every question shows the correct answer (answer key) and a detailed step-by-step solution, free to read online.
Can I take the JEE Main 2021 February 25, Shift 2 paper as a timed mock test?
Yes. With a free JEEnify account you can attempt this exact paper as a timed test in the real exam interface, then see your score and weak-area analysis.