JEE Main 2021 February 24, Shift 2 Question Paper with Solutions
All 90 questions from the JEE Main 2021 (February 24, 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.
The period of oscillation of a simple pendulum is T=2πgL. Measured value of L is 1.0m from meter scale having a minimum division of 1mm and time of one complete oscillation is 1.95s measured from stopwatch of 0.01s resolution. The percentage error in the determination of g will be:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 41.13%
Approach:
Propagate the errors in L and T through g obtained from the pendulum period formula.
Step 1:
LΔL=1.00.001=0.001
Step 2:
TΔT=1.950.01=0.005128
Step 3:
gΔg=0.001+2(0.005128)=0.011256
Step 4:
gΔg×100=1.13%
Final answer: 1.13%
Q2Single correctWork, Energy and Power
A particle is projected with velocity v0 along x-axis. A damping force is acting on the particle which is proportional to the square of the distance from the origin i.e. ma=−αx2. The distance at which the particle stops:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2(2α3mv02)31
Approach:
Use a=v dv/dx and integrate from the launch speed to rest.
Step 1:
m∫v00vdv=−α∫0xx2dx
Step 2:
m(−2v02)=−3αx3
Step 3:
x3=2α3mv02
Step 4:
x=(2α3mv02)1/3
Final answer: (2α3mv02)31
Q3Single correctRotational Motion
A circular hole of radius (2a) is cut out of a circular disc of radius a as shown in figure. The centroid of the remaining circular portion with respect to point O will be:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 265a
Approach:
Treat the hole as negative mass and take moments of area about O.
Step 1:
A1=πa2,x1=a
Step 2:
A2=π(2a)2=4πa2,x2=23a
Step 3:
xcm=πa2−4πa2πa2(a)−4πa2⋅23a=43a−83a
Step 4:
xcm=3/45a/8=65a
Final answer: 65a
Q4Single correctGravitation
A body weighs 49N on a spring balance at the north pole. What will be its weight recorded on the same weighing machine, if it is shifted to the equator? [Use g=R2GM=9.8m s−2 and radius of earth, R=6400km.]
(A)
(B)
(C)
(D)
SolutionAnswer: Option 248.83N
Approach:
Find the mass from the polar reading, then apply the rotational reduction of effective gravity at the equator.
Step 1:
m=9.849=5kg
Step 2:
ω=864002π=7.272×10−5rad s−1
Step 3:
ω2R=(7.272×10−5)2(6.4×106)=0.0338m s−2
Step 4:
Weq=5(9.8−0.0338)=5(9.766)=48.83N
Final answer: 48.83N
Q5Single correctThermodynamics
If one mole of an ideal gas at (P1,V1) is allowed to expand reversibly and isothermally (A to B) its pressure is reduced to one-half of the original pressure (see figure). This is followed by a constant volume cooling till its pressure is reduced to one-fourth of the initial value (B→C). Then it is restored to its initial state by a reversible adiabatic compression (C to A). The net workdone by the gas is equal to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4RT[ln(2)−2(γ−1)1]
Approach:
Sum the work done in the three legs: isothermal expansion, isochoric cooling, and adiabatic compression.
Step 1:
A(P1,V1)→B(P1/2,2V1) isothermal, so WAB=RTlnV12V1=RTln2
Step 2:
B(P1/2,2V1)→C(P1/4,2V1) at constant volume, WBC=0
On the basis of kinetic theory of gases, the gas exerts pressure because its molecules:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2suffer change in momentum when impinge on the walls of container.
Approach:
Identify the microscopic origin of gas pressure in kinetic theory.
Step 1:
Molecules collide elastically with the walls and reverse their momentum component normal to the wall.
Step 2:
The rate of momentum transfer to the wall gives a force, and force per unit area gives pressure.
Final answer: suffer change in momentum when impinge on the walls of container.
Q7Single correctOscillations and Waves
In the given figure, a body of mass M is held between two massless springs, on a smooth inclined plane. The free ends of the springs are attached to firm supports. If each spring has spring constant k, the frequency of oscillation of given body is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 42π1M2k
Approach:
Combine the two springs acting on the block to get the effective constant, then apply the SHM frequency formula.
Step 1:
On displacement along the incline one spring stretches and the other compresses, both giving restoring force.
Step 2:
keff=k+k=2k
Step 3:
f=2π1M2k
Final answer: 2π1M2k
Q8Single correctOscillations and Waves
When a particle executes SHM, the nature of graphical representation of velocity as a function of displacement is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2elliptical
Approach:
Express SHM velocity in terms of displacement and identify the curve.
Step 1:
v2=ω2(A2−x2)
Step 2:
ω2A2v2+A2x2=1
Final answer: elliptical
Q9Single correctOscillations and Waves
Which of the following equations represents a travelling wave?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1y=Asin(15x−2t)
Approach:
A travelling wave has the form y=f(x ± vt), a single function of the combination x ± vt.
Step 1:
y=Asin(15x−2t) depends only on the combination (15x−2t)
Step 2:
The other forms factor space and time separately or grow unboundedly, so they are not travelling waves.
Final answer: y=Asin(15x−2t)
Q10Single correctElectrostatics
Two electrons each are fixed at a distance 2d. A third charge proton placed at the midpoint is displaced slightly by a distance x(x≪d) perpendicular to the line joining the two fixed charges. Proton will execute simple harmonic motion having angular frequency: (m= mass of charged particle)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3(2πε0md3q2)21
Approach:
Find the net restoring force on the displaced proton for small x and extract the angular frequency.
Step 1:
Distance from each fixed charge to the displaced proton: r=d2+x2
Step 2:
Net restoring component: F=2⋅4πε01(d2+x2)q2⋅d2+x2x≈2πε0d3q2x
Step 3:
keff=2πε0d3q2
Step 4:
ω=mkeff=(2πε0md3q2)1/2
Final answer: (2πε0md3q2)21
Q11Single correctMagnetic Effects of Current and Magnetism
A soft ferromagnetic material is placed in an external magnetic field. The magnetic domains:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4may increase or decrease in size and change its orientation.
Approach:
Recall how domains respond to an applied field in a soft ferromagnet.
Step 1:
Domains favourably aligned with the field grow while unfavourable ones shrink.
Step 2:
Domain magnetization also rotates toward the field direction.
Final answer: may increase or decrease in size and change its orientation.
Q12Single correctElectromagnetic Induction and Alternating Currents
The figure shows a circuit that contains four identical resistors with resistance R=2.0Ω, two identical inductors with inductance L=2.0mH and an ideal battery with E.M.F. E=9V. The current i just after the switch S is closed will be:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 32.25A
Approach:
Just after closing, inductors carry zero current and behave as open branches; reduce the circuit to resistors only.
Step 1:
At t=0+ the two inductor branches act as open circuits and carry no current.
Step 2:
Current flows through the series resistor and the single middle resistor: Rtotal=R+R=2R=4Ω
Step 3:
i=RtotalE=49=2.25A
Final answer: 2.25A
Q13Single correctDual Nature of Matter and Radiation
An X-ray tube is operated at 1.24 million volt. The shortest wavelength of the produced photon will be:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 110−3nm
Approach:
The shortest wavelength corresponds to the entire electron energy eV converted to one photon.
Match each electromagnetic source to its physical origin.
Step 1:
Microwaves are generated by a magnetron: (a)-(ii)
Step 2:
Infrared arises from vibration of atoms and molecules: (b)-(iv)
Step 3:
Gamma rays come from radioactive decay of nucleus: (c)-(i)
Step 4:
X-rays come from transitions of inner shell electrons: (d)-(iii)
Final answer: (a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)
Q15Single correctOptics
If the source of light used in a Young's double slit experiment is changed from red to violet:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1consecutive fringe lines will come closer.
Approach:
Relate fringe width to wavelength and compare red versus violet.
Step 1:
Violet light has a shorter wavelength than red light.
Step 2:
β=dλD decreases as λ decreases
Final answer: consecutive fringe lines will come closer.
Q16Single correctDual Nature of Matter and Radiation
The de Broglie wavelength of a proton and α-particle are equal. The ratio of their velocities is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 14:1
Approach:
Equate the de Broglie wavelengths of the two particles and relate their velocities through their masses.
Step 1:
mpvph=mαvαh
Step 2:
mpvp=mαvα
Step 3:
vαvp=mpmα=14
Final answer: 4:1
Q17Single correctAtoms and Nuclei
According to Bohr atom model, in which of the following transitions will the frequency be maximum?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4n=2 to n=1
Approach:
Compute the emitted frequency for each transition, which is proportional to the difference of the inverse squares of the principal quantum numbers.
Step 1:
3→2:41−91=0.139
Step 2:
5→4:161−251=0.0225
Step 3:
4→3:91−161=0.0486
Step 4:
2→1:11−41=0.75
Final answer: n=2 to n=1
Q18Single correctElectronic Devices
The logic circuit shown above is equivalent to :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3C=Aˉ⋅B (input A inverted then AND with B)
Approach:
Reduce the given gate network to a Boolean expression and match it to the option whose gate combination yields the same expression.
Step 1:
OR output=A+Bˉ
Step 2:
C=A+Bˉ=Aˉ⋅B
Step 3:
Option 3:C=Aˉ⋅B
Final answer: C=Aˉ⋅B (Option 3)
Q19Single correctElectronic Devices
Zener breakdown occurs in a p−n junction having p and n both:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4heavily doped and have narrow depletion layer.
Approach:
Relate Zener breakdown conditions to doping level and depletion-layer width.
Step 1:
Heavy doping⇒narrow depletion layer
Step 2:
Narrow d⇒very high field E
Step 3:
High field⇒Zener breakdown
Final answer: heavily doped and have narrow depletion layer.
Q20Single correctElectronic Devices
Given below are two statements: Statement I : p−n junction diodes can be used to function as a transistor, simply by connecting two diodes, back to back, which acts as the base terminal Statement II: In the study of transistors, the amplification factor β indicates ratio of the collector current to the base current. 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 false but Statement II is true
Approach:
Evaluate each statement against the physics of transistors.
Step 1:
Two back-to-back diodes do not form a transistor
Step 2:
β=IBIC
Final answer: Statement I is false but Statement II is true
Q21NumericalWork, Energy and Power
Two solids A and B of mass 1 kg and 2 kg respectively are moving with equal linear momentum. The ratio of their kinetic energies (K.E.)A:(K.E.)B will be 1A, so the value of A will be __________.
SolutionAnswer: 2
Approach:
Express kinetic energy in terms of momentum and mass for equal momenta and form the ratio.
Step 1:
(K.E.)B(K.E.)A=p2/2mBp2/2mA=mAmB
Step 2:
mAmB=12=2
Final answer: 2
Q22NumericalRotational Motion
A uniform thin bar of mass 6 kg and length 2.4 meter is bent to make an equilateral hexagon. The moment of inertia about an axis passing through the centre of mass and perpendicular to the plane of hexagon is __________ ×10−1kg m2.
SolutionAnswer: 8
Approach:
Treat each of the six sides as a rod and apply the parallel-axis theorem about the hexagon centre, then sum.
Step 1:
L=62.4=0.4m,m=66=1kg
Step 2:
d=2L3=0.23,d2=0.12
Step 3:
Ione=121(0.4)2+1(0.12)=0.01333+0.12=0.1333
Step 4:
I=6×0.1333=0.8=8×10−1kg m2
Final answer: 8
Q23NumericalProperties of Solids and Liquids
A uniform metallic wire is elongated by 0.04 m when subjected to a linear force F. The elongation, if its length and diameter is doubled and subjected to the same force will be __________ cm.
SolutionAnswer: 2
Approach:
Use the elongation formula and scale length and area for doubled length and doubled diameter.
Step 1:
L→2L,A=4πd2→4A
Step 2:
Δl′=(4A)YF(2L)=21AYFL=2Δl
Step 3:
Δl′=20.04=0.02m=2cm
Final answer: 2
Q24NumericalKinetic Theory of Gases
The root-mean-square speed of molecules of a given mass of a gas at 27∘C and 1 atmosphere pressure is 200m s−1. The root-mean-square speed of molecules of the gas at 127∘C and 2 atmosphere pressure is 3xm s−1. The value of x will be __________.
SolutionAnswer: 400
Approach:
Use that rms speed depends only on temperature, scaling with the square root of absolute temperature.
Step 1:
T1=300K,T2=400K
Step 2:
v1v2=T1T2=300400=32
Step 3:
v2=200⋅32=3400m s−1
Final answer: 400
Q25NumericalOscillations and Waves
Two cars are approaching each other at an equal speed of 7.2km hr−1. When they see each other, both blow horns having a frequency of 676 Hz. The beat frequency heard by each driver will be __________ Hz. [Velocity of sound in air is 340m s−1].
SolutionAnswer: 8
Approach:
Convert the car speed, apply the Doppler formula for source and observer moving toward each other, then take the beat with the own horn frequency.
Step 1:
7.2km hr−1=2m s−1
Step 2:
f′=676⋅340−2340+2=676⋅338342=684Hz
Step 3:
fbeat=684−676=8Hz
Final answer: 8
Q26NumericalElectrostatics
A point charge of +12μC is at a distance 6 cm vertically above the centre of a square of side 12 cm as shown in figure. The magnitude of the electric flux through the square will be __________ ×103N m2C−1.
SolutionAnswer: 226
Approach:
Place the charge at the centre of an imaginary cube of side equal to the square, so the square is one face, and apply Gauss law with the flux split equally among six faces.
Step 1:
Charge at centre of cube of side 12cm
Step 2:
ϕ=6ε0q=6×8.854×10−1212×10−6
Step 3:
ϕ=2.259×105=226×103N m2C−1
Final answer: 226
Q27NumericalCurrent Electricity
A cylindrical wire of radius 0.5 mm and conductivity 5×107S m−1 is subjected to an electric field of 10 mV m−1. The expected value of current in the wire will be x3π mA. The value of x is __________.
SolutionAnswer: 5
Approach:
Use the microscopic form of Ohm law to get current density, multiply by cross-sectional area, and match to the given form.
Step 1:
r=0.5×10−3m,E=10−2V m−1
Step 2:
I=5×107⋅10−2⋅π(0.5×10−3)2
Step 3:
I=0.125πA=125πmA
Step 4:
x3=125⇒x=5
Final answer: 5
Q28NumericalElectromagnetic Induction and Alternating Currents
A series LCR circuit is designed to resonate at an angular frequency ω0=105rad s−1. The circuit draws 16 W power from 120 V source at resonance. The value of resistance R in the circuit is __________ Ω.
SolutionAnswer: 900
Approach:
At resonance the impedance is purely resistive, so the average power equals the square of the source voltage divided by the resistance.
Step 1:
R=PV2=161202
Step 2:
R=1614400=900Ω
Final answer: 900
Q29NumericalElectromagnetic Waves
An electromagnetic wave of frequency 3 GHz enters a dielectric medium of relative electric permittivity 2.25 from vacuum. The wavelength of this wave in that medium will be __________ ×10−2 cm.
SolutionAnswer: 667
Approach:
Find the vacuum wavelength from the frequency, then divide by the refractive index, which is the square root of the relative permittivity.
Step 1:
λ0=3×1093×108=0.1m
Step 2:
n=2.25=1.5
Step 3:
λ=1.50.1=0.0667m=6.667cm=667×10−2cm
Final answer: 667
Q30NumericalElectromagnetic Waves
A signal of 0.1 kW is transmitted in a cable. The attenuation of cable is −5 dB per km and cable length is 20 km. The power received at the receiver is 10−x W. The value of x is __________. [Gain in dB =10log10(PiPO)]
SolutionAnswer: 8
Approach:
Compute the total attenuation over the cable length, apply the dB gain definition, and solve for the received power.
Step 1:
Total gain=−5×20=−100dB
Step 2:
−100=10log10100PO⇒log10100PO=−10
Step 3:
PO=100×10−10=10−8W
Final answer: 8
Chemistry30 questions
Q31Single correctAtomic Structure
According to Bohr's atomic theory: (A) Kinetic energy of electron is ∝n2Z2. (B) The product of velocity (v) of electron and principal quantum number (n), vn∝Z2. (C) Frequency of revolution of electron in an orbit is ∝n3Z3. (D) Coulombic force of attraction on the electron is ∝n4Z3. Choose the most appropriate answer from the options given below:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2(A) and (D) only
Approach:
Apply Bohr model relations for radius, velocity, energy, frequency and force and test each statement.
Step 1:
KE∝n2Z2
Step 2:
v⋅n∝nZ⋅n=Z
Step 3:
f=2πrv∝n2/ZZ/n=n3Z2
Step 4:
F∝r2Z∝Z⋅(n2Z)2=n4Z3
Final answer: (A) and (D) only
Q32Single correctChemical Bonding and Molecular Structure
The correct set from the following in which both pairs are in correct order of melting point is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4LiF > LiCl; MgO > NaCl
Approach:
Compare lattice energies; higher charge and smaller anion give higher melting point.
Step 1:
F−<Cl−⇒LiF lattice energy > LiCl
Step 2:
MgO: 2+/2− vs NaCl: 1+/1−
Final answer: LiF > LiCl; MgO > NaCl
Q33Single correctChemical Bonding and Molecular Structure
The correct shape and I−I−I bond angles respectively in I3− ion are:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4Linear; 180∘
Approach:
Determine hybridisation and geometry of the central iodine in triiodide using VSEPR.
Step 1:
Central I: 2 bond pairs + 3 lone pairs, sp3d
Step 2:
3 lone pairs occupy equatorial positions
Step 3:
Bond angle=180∘
Final answer: Linear; 180∘
Q34Single correctClassification of Elements and Periodicity in Properties
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: Hydrogen is the most abundant element in the Universe, but it is not the most abundant gas in the troposphere. Reason R: Hydrogen is the lightest element. In the light of the above statements, choose the correct answer from the options given below:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2Both A and R are true and R is the correct explanation of A
Approach:
Evaluate the truth of the assertion and reason and whether the reason explains the assertion.
Step 1:
Hydrogen abundant in stars but escapes Earth gravity
The correct order of the following compounds showing increasing tendency towards nucleophilic substitution reaction is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2(i) < (ii) < (iii) < (iv)
Approach:
More electron-withdrawing nitro groups (especially ortho/para to Cl) increase reactivity in aromatic nucleophilic substitution.
Step 1:
(i) chlorobenzene: no NO2
Step 2:
(ii) p-nitrochlorobenzene: one NO2
Step 3:
(iii) 2,4-dinitrochlorobenzene: two NO2
Step 4:
(iv) 2,4,6-trinitrochlorobenzene: three NO2
Final answer: (i) < (ii) < (iii) < (iv)
Q39Single correctPrinciples Related to Practical Chemistry
Given below are two statements: Statement I: The value of the parameter "Biochemical Oxygen Demand (BOD)" is important for survival of aquatic life. Statement II: The optimum value of BOD is 6.5ppm. In the light of the above statements, choose the most appropriate answer from the options given below:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1Statement I is true but Statement II is false
Approach:
Judge each statement against standard water-quality criteria for BOD.
Step 1:
BOD measures oxygen demand and indicates water quality
Step 2:
Clean water BOD<5ppm; polluted water BOD≈17ppm
Final answer: Statement I is true but Statement II is false
Q40Single correctSome Basic Principles of Organic Chemistry
Most suitable salt which can be used for efficient clotting of blood will be?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1FeCl3
Approach:
Identify the salt providing high-charge cations that coagulate the negatively charged blood colloid most effectively.
Which of the following reagent is suitable for the preparation of the product in the above reaction?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3NH2−NH2/C2H5O⊖Na⊕
Approach:
The carbonyl group is reduced to a methylene while the carbon-carbon double bonds are retained, indicating Wolff-Kishner reduction.
Step 1:
C=O converted to CH2 without touching C=C
Step 2:
NH2NH2 with base (Wolff-Kishner)
Final answer: NH2−NH2/C2H5O⊖Na⊕
Q46Single correctHydrocarbons
Which one of the following carbonyl compounds cannot be prepared by addition of water on an alkyne in the presence of HgSO4 and H2SO4?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1CH3−CH2−CHO
Approach:
Acid-catalysed hydration of an alkyne (oxymercuration) follows Markovnikov addition, so terminal alkynes other than ethyne give methyl ketones and only ethyne gives an aldehyde; identify the carbonyl that cannot arise this way.
The diazonium salt of which of the following compounds will form a coloured dye on reaction with β-Naphthol in NaOH?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2C6H5NH2
Approach:
Azo dye formation by coupling with beta-naphthol requires a benzene diazonium salt, which forms only from a primary aromatic amine bearing the amino group directly on the ring.
Match each medicinal compound in List-I with its pharmacological role in List-II.
Step 1:
Valium
Step 2:
Morphine
Step 3:
Norethindrone
Step 4:
Vitamin B12
Final answer: (a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)
Q51NumericalSome Basic Concepts in Chemistry
The formula of a gaseous hydrocarbon which requires 6 times of its own volume of O2 for complete oxidation and produces 4 times its own volume of CO2 is CxHy. The value of y is __________.
SolutionAnswer: 8
Approach:
Use Gay-Lussac's law of combining volumes for the combustion stoichiometry of the hydrocarbon to find the number of hydrogen atoms.
Step 1:
x=4
Step 2:
x+4y=6
Step 3:
4+4y=6⇒4y=2⇒y=8
Final answer: 8
Q52NumericalPrinciples Related to Practical Chemistry
1.86 g of aniline completely reacts to form acetanilide. 10% of the product is lost during purification. Amount of acetanilide obtained after purification (in g) is __________ ×10−2.
SolutionAnswer: 243
Approach:
Convert the mass of aniline to moles, find the theoretical mass of acetanilide formed in a 1:1 conversion, then apply the 10% loss.
Step 1:
n=931.86=0.02mol
Step 2:
m=0.02×135=2.70g
Step 3:
2.70×0.90=2.43g=243×10−2g
Final answer: 243
Q53NumericalSome Basic Concepts in Chemistry
The volume occupied by 4.75 g of acetylene gas at 50∘C and 740 mm Hg pressure is __________ L. (Rounded off to the nearest integer) [Given R=0.0826L atm K−1mol−1]
SolutionAnswer: 5
Approach:
Find moles of acetylene from its mass, then apply the ideal gas law with temperature in kelvin and pressure in atm.
Step 1:
n=264.75=0.1827mol
Step 2:
T=50+273=323K,P=760740=0.974atm
Step 3:
V=PnRT=0.9740.1827×0.0826×323=5.0L
Final answer: 5
Q54NumericalChemical Thermodynamics
Assuming ideal behaviour, the magnitude of log K for the following reaction at 25∘C is x×10−1. The value of x is __________. (Integer answer) [Given: ΔfG∘(HC≡CH)=−2.04×105J mol−1;ΔfG∘(C6H6)=−1.24×105J mol−1;R=8.314J K−1mol−1]
SolutionAnswer: 855
Approach:
Compute the standard Gibbs energy of the trimerisation of ethyne to benzene from formation energies, then convert to log K using the standard relation.
Step 1:
ΔG∘=(−1.24×105)−3(−2.04×105)=+4.88×105J
Step 2:
logK=2.303RT−ΔG∘=2.303×8.314×298−4.88×105
Step 3:
∣logK∣=85.5=855×10−1
Final answer: 855
Q55NumericalEquilibrium
The solubility product of PbI2 is 8.0×10−9. The solubility of lead iodide in 0.1 molar solution of lead nitrate is x×10−6 mol /L. The value of x is __________ (Rounded off to the nearest integer) [Given 2=1.41]
SolutionAnswer: 141
Approach:
Account for the common ion effect of lead nitrate, treating the lead-ion concentration as 0.1 M, then solve the solubility-product expression for the solubility of lead iodide.
Step 1:
[Pb2+]≈0.1M,[I−]=2s
Step 2:
8.0×10−9=0.1×(2s)2=0.4s2
Step 3:
s2=2×10−8⇒s=2×10−4=1.41×10−4=141×10−6
Final answer: 141
Q56NumericalSolutions
C6H6 freezes at 5.5∘C. The temperature at which a solution of 10 g of C4H10 in 200 g of C6H6 freeze is __________ ∘C. (nearest integer value), (The molal freezing point depression constant of C6H6 is 5.12∘C/m.)
SolutionAnswer: 1
Approach:
Compute the molality of the butane solution, apply the freezing point depression equation, and subtract the depression from the freezing point of pure benzene.
Step 1:
m=0.20010/58=0.862mol kg−1
Step 2:
ΔTf=5.12×0.862=4.41∘C
Step 3:
Tf=5.5−4.41=1.09∘C≈1∘C
Final answer: 1
Q57NumericalRedox Reactions and Electrochemistry
The magnitude of the change in oxidising power of the MnO4−/Mn2+ couple is x×10−4 V if the H+ concentration is decreased from 1 M to 10−4M at 25∘C. (Assume concentration of MnO4− and Mn2+ to be same on change in H+ concentration). The value of x is __________. (Rounded off to the nearest integer) [Given: F2.303RT=0.059]
SolutionAnswer: 3776
Approach:
Write the Nernst equation for the permanganate-manganese couple including the eight hydrogen ions, and evaluate the change in electrode potential caused only by the drop in hydrogen-ion concentration.
Step 1:
ΔE=50.059×8(log[H+]2−log[H+]1)
Step 2:
ΔE=50.059×8×(log10−4−log1)=50.059×8×(−4)
Step 3:
∣ΔE∣=0.3776V=3776×10−4V
Final answer: 3776
Q58NumericalChemical Kinetics
Sucrose hydrolyses in acid solution into glucose and fructose following first order rate law with a half-life of 3.33 h at 25∘C. After 9 h, the fraction of sucrose remaining is f. The value of log10(f1) is __________ ×10−2. (Rounded off to the nearest integer) [Assume: ln10 = 2.303, ln 2 = 0.693]
SolutionAnswer: 81
Approach:
Obtain the first-order rate constant from the half-life, then use the integrated first-order rate law to find the logarithm of the reciprocal of the remaining fraction at the given time.
Step 1:
k=3.330.693=0.208h−1
Step 2:
log10f1=2.3030.208×9=0.813
Step 3:
0.813=81.3×10−2≈81×10−2
Final answer: 81
Q59Numericalp-Block Elements
Among the following allotropic forms of sulphur, the number of allotropic forms, which will show paramagnetism is __________. (A) α-sulphur (B) β-sulphur (C) S2-form
SolutionAnswer: 1
Approach:
Determine the magnetic behaviour of each allotrope from its molecular structure and electron configuration.
Step 1:
alpha-sulphur and beta-sulphur are S8 rings
Step 2:
S2 form has two unpaired electrons in antibonding pi orbitals
Step 3:
count of paramagnetic forms = 1
Final answer: 1
Q60NumericalOrganic Compounds Containing Nitrogen
The total number of amines among the following which can be synthesized by Gabriel synthesis is __________.
SolutionAnswer: 3
Approach:
Gabriel phthalimide synthesis gives only primary amines from alkyl halides and fails for aromatic amines because aryl halides do not undergo the required nucleophilic substitution; count the structures that qualify.
Step 1:
(CH3)2CH-CH2-NH2 (isobutylamine)
Step 2:
CH3CH2NH2 (ethylamine)
Step 3:
C6H5-CH2-NH2 (benzylamine)
Step 4:
C6H5-NH2 (aniline)
Final answer: 3
Mathematics30 questions
Q61Single correctComplex Numbers and Quadratic Equations
Let a, b, c be in arithmetic progression. Let the centroid of the triangle with vertices (a, c), (2,b) and (a, b) be (310,37). If α,β are the roots of the equation ax2+bx+1=0, then the value of α2+β2−αβ is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1−25671
Approach:
Use the AP condition and the centroid coordinates to find a,b,c, then form the quadratic and evaluate the symmetric expression.
Step 1:
3a+2+a=310⇒2a+2=10⇒a=4
Step 2:
3c+b+b=37⇒c+2b=7 with 2b=a+c=4+c
Step 3:
c=23,b=411
Step 4:
α+β=−ab=−1611,αβ=a1=41
Step 5:
α2+β2−αβ=(α+β)2−3αβ=256121−43=−25671
Final answer: −25671
Q62Single correctPermutations and Combinations
If n≥2 is a positive integer, then the sum of the series n+1C2+2(2C2+3C2+4C2+…+nC2) is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 26n(n+1)(2n+1)
Approach:
Apply the hockey-stick identity to the bracketed sum, then combine with the leading term.
Step 1:
2C2+3C2+…+nC2=n+1C3
Step 2:
Sum=n+1C2+2n+1C3
Step 3:
=2(n+1)n+2⋅6(n+1)n(n−1)=6n(n+1)(2n+1)
Final answer: 6n(n+1)(2n+1)
Q63Single correctLimit, Continuity and Differentiability
If P is a point on the parabola y=x2+4 which is closest to the straight line y=4x−1, then the co-ordinates of P are:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3(2,8)
Approach:
The closest point on the parabola has a tangent parallel to the given line, so equate the parabola's slope to the line's slope.
Step 1:
dxdy=2x=4⇒x=2
Step 2:
y=22+4=8
Final answer: (2,8)
Q64Single correctSets, Relations and Functions
The negation of the statement ∼p∧(p∨q) is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3p∨∼q
Approach:
Apply De Morgan's laws to negate the conjunction and simplify.
Step 1:
∼(∼p∧(p∨q))≡p∨∼(p∨q)
Step 2:
≡p∨(∼p∧∼q)≡(p∨∼p)∧(p∨∼q)≡p∨∼q
Final answer: p∨∼q
Q65Single correctSets, Relations and Functions
For the statements p and q, consider the following compound statements: (a)(∼q∧(p→q))→∼p (b)((p∨q)∧∼p)→q Then which of the following statements is correct?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2(a) and (b) both are tautologies.
Approach:
Evaluate each compound statement over all truth assignments of p and q.
Step 1:
(a): ∼q∧(p→q) is true only when q false and p false, where ∼p is true, so (a) holds for all assignments
Step 2:
(b): (p∨q)∧∼p is true only when p false and q true, where q is true, so (b) holds for all assignments
Final answer: (a) and (b) both are tautologies.
Q66Single correctTrigonometry
The angle of elevation of a jet plane from a point A on the ground is 60∘. After a flight of 20 seconds at the speed of 432 km / hour, the angle of elevation changes to 30∘. If the jet plane is flying at a constant height, then its height is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 112003m
Approach:
Convert speed to find horizontal distance covered, then use the two elevation angles to set up the height relation.
Step 1:
432km/h=120m/s,distance=120×20=2400m
Step 2:
x1=3h,x2=h3
Step 3:
h3−3h=32h=2400⇒h=12003m
Final answer: 12003m
Q67Single correctMatrices and Determinants
For the system of linear equations: x−2y=1,x−y+kz=−2,ky+4z=6,k∈R Consider the following statements: (A) The system has unique solution if k=2,k=−2. (B) The system has unique solution if k=−2. (C) The system has unique solution if k=2. (D) The system has no-solution if k=2. (E) The system has infinite number of solutions if k=−2. Which of the following statements are correct?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3(A) and (D) only
Approach:
Compute the coefficient determinant to locate non-unique cases, then test consistency at the critical values of k.
Step 1:
Δ=−(k−2)(k+2)=4−k2
Step 2:
Δ=0⇔k=2 and k=−2⇒ unique solution, so (A) holds
Step 3:
k=2: the equations become inconsistent ⇒ no solution, so (D) holds
Step 4:
k=−2: the system reduces to x=4z−5,y=2z−3⇒ infinitely many solutions, so (E) is false
Final answer: (A) and (D) only
Q68Single correctMatrices and Determinants
Let A and B be 3×3 real matrices such that A is a symmetric matrix and B is a skew-symmetric matrix. Then the system of linear equations (A2B2−B2A2)X=O, where X is a 3×1 column matrix of unknown variables and O is a 3×1 null matrix, has
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2infinitely many solutions
Approach:
Determine the symmetry type of the coefficient matrix to find its determinant, then classify the homogeneous system.
Step 1:
(A2B2)T=(B2)T(A2)T=B2A2 since B2T=B2,A2T=A2
Step 2:
M=A2B2−B2A2⇒MT=B2A2−A2B2=−M
Step 3:
M is a 3×3 skew-symmetric matrix ⇒detM=0
Step 4:
MX=O with detM=0⇒ infinitely many solutions
Final answer: infinitely many solutions
Q69Single correctTrigonometry
A possible value of tan(41sin−1863) is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 471
Approach:
Let the inner inverse-sine be 4θ, express sin4θ via the double-angle identities, and solve for tanθ.
Step 1:
sin4θ=863=837,cos4θ=81
Step 2:
tan4θ=37=1−6t2+t44t(1−t2)
Step 3:
t=71 satisfies the relation
Final answer: 71
Q70Single correctCo-ordinate Geometry
For which of the following curves, the line x+3y=23 is the tangent at the point (233,21)?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3x2+9y2=9
Approach:
Require the given point to lie on the curve and the implicit slope there to equal the line's slope.
Step 1:
(233)2+9(21)2=427+49=9, so the point lies on x2+9y2=9
Step 2:
2x+18yy′=0⇒y′=−9yx=−9/233/2=−31
Final answer: x2+9y2=9
Q71Single correctLimit, Continuity and Differentiability
Let f:R→R be defined as f(x)=⎩⎨⎧−55x,2x3−3x2−120x,2x3−3x2−36x−336,if x<−5if −5≤x≤4if x>4 Let A={x∈R:f is increasing}. Then A is equal to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2(−5,−4)∪(4,∞)
Approach:
Differentiate each piece and determine the sub-intervals where the derivative is positive within the respective domains.
Step 1:
x<−5: f′(x)=−55<0, decreasing
Step 2:
−5≤x≤4: f′(x)=6(x−5)(x+4)>0 for x<−4, so increasing on (−5,−4)
Step 3:
x>4: f′(x)=6(x−3)(x+2)>0 for all x>4, so increasing on (4,∞)
Final answer: (−5,−4)∪(4,∞)
Q72Single correctLimit, Continuity and Differentiability
If the curve y=ax2+bx+c,x∈R, passes through the point (1,2) and the tangent line to this curve at origin is y=x, then the possible values of a,\ b,\ c are:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3a=1,b=1,c=0
Approach:
Use passage through the given point, passage through the origin, and the tangent slope at the origin to solve for the coefficients.
Step 1:
Tangent y=x at origin ⇒c=0 and b=1
Step 2:
a+b+c=2⇒a=1
Final answer: a=1,b=1,c=0
Q73Single correctIntegral Calculus
The value of the integral, ∫13[x2−2x−2]dx, where [x] denotes the greatest integer less than or equal to x, is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4−2−3−1
Approach:
Write the integrand as ⌊(x−1)2−3⌋, find the sub-intervals where the floor is constant, and sum the contributions.
Step 1:
On [1,2): floor =−3; on [2,1+2): floor =−2
Step 2:
On [1+2,1+3): floor =−1; on [1+3,3): floor =0
Step 3:
−3(1)−2(2−1)−1(3−2)+0=−2−3−1
Final answer: −2−3−1
Q74Single correctIntegral Calculus
The area of the region: R={(x,y):5x2≤y≤2x2+9} is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2123 square units
Approach:
Find the intersection abscissae of the two parabolas, then integrate the vertical gap between them.
Step 1:
Area=∫−33(9−3x2)dx=2∫03(9−3x2)dx
Step 2:
=2[9x−x3]03=2(93−33)=123
Final answer: 123 square units
Q75Single correctLimit, Continuity and Differentiability
Let f be a twice differentiable function defined on R such that f(0)=1,f′(0)=2 and f′(x)=0 for all x∈R. If f(x)f′(x)f′(x)f′′(x)=0, for all x∈R, then the value of f(1) lies in the interval
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4(6,9)
Approach:
Expand the determinant into a differential relation, integrate to obtain f explicitly, then evaluate f(1).
Step 1:
ff′′−(f′)2=0⇒dxd(ff′)=0⇒ff′=f(0)f′(0)=2
Step 2:
ff′=2⇒f(x)=e2x using f(0)=1
Step 3:
f(1)=e2≈7.389∈(6,9)
Final answer: (6,9)
Q76Single correctDifferential Equations
If a curve y=f(x) passes through the point (1,2) and satisfies xdxdy+y=bx4, then for what value of b, ∫12f(x)dx=562 ?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 210
Approach:
Recognize the left side as the derivative of a product and integrate to find f(x), then impose the definite integral condition to solve for b.
Step 1:
dxd(xy)=bx4
Step 2:
xy=5bx5+C
Step 3:
2=5b+C
Step 4:
f(x)=5bx4+xC
Step 5:
∫12f(x)dx=2531b+Cln2
Step 6:
2531b−5(b−10)ln2=562
Step 7:
b=10
Final answer: 10
Q77Single correctIntegral Calculus
Let f(x) be a differentiable function defined on [0,2] such that f′(x)=f′(2−x) for all x∈(0,2), f(0)=1 and f(2)=e2. Then the value of ∫02f(x)dx is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 21+e2
Approach:
Use the symmetry condition to show f(x)+f(2-x) is constant, then apply the integral property.
Step 1:
dxd(f(x)+f(2−x))=f′(x)−f′(2−x)=0
Step 2:
f(x)+f(2−x)=f(0)+f(2)=1+e2
Step 3:
2I=∫02(f(x)+f(2−x))dx=(1+e2)⋅2
Step 4:
I=1+e2
Final answer: 1+e2
Q78Single correctThree Dimensional Geometry
The vector equation of the plane passing through the intersection of the planes r⋅(i^+j^+k^)=1 and r⋅(i^−2j^)=−2, and the point (1,0,2) is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1r⋅(i^+7j^+3k^)=7
Approach:
Form the family of planes through the line of intersection and fix the parameter using the given point.
Step 1:
r⋅[(1+λ)i^+(1−2λ)j^+k^]=1−2λ
Step 2:
(1+λ)+0+2=1−2λ
Step 3:
λ=−32
Step 4:
r⋅(31i^+37j^+k^)=37
Step 5:
r⋅(i^+7j^+3k^)=7
Final answer: r⋅(i^+7j^+3k^)=7
Q79Single correctThree Dimensional Geometry
Let a,b∈R. If the mirror image of the point P(a,6,9) with respect to the line 7x−3=5y−2=−9z−1 is (20,b,−a−9), then ∣a+b∣ is equal to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 488
Approach:
Impose that the midpoint of P and its image lies on the line and that PQ is perpendicular to the line direction.
Step 1:
d=(7,5,−9),point=(3,2,1)
Step 2:
(20−a,b−6,−a−18)⋅(7,5,−9)=0
Step 3:
a=−56,b=−32
Step 4:
∣a+b∣=∣−88∣=88
Final answer: 88
Q80Single correctStatistics and Probability
The probability that two randomly selected subsets of the set {1,2,3,4,5} have exactly two elements in their intersection, is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 429135
Approach:
Count ordered pairs of subsets and the favourable assignments where exactly two elements lie in both.
Step 1:
total pairs=25⋅25=45=1024
Step 2:
(25)⋅33=10⋅27=270
Step 3:
1024270=512135=29135
Final answer: 29135
Q81NumericalComplex Numbers and Quadratic Equations
The number of the real roots of the equation (x+1)2+∣x−5∣=427 is ________.
SolutionAnswer: 2
Approach:
Split the modulus into two cases and count valid real roots in each region.
Step 1:
x≥5:(x+1)2+(x−5)=427
Step 2:
x<5:(x+1)2+(5−x)=427
Step 3:
both <5
Final answer: 2
Q82NumericalComplex Numbers and Quadratic Equations
Let i=−1. If (1−i)24(−1+i3)21+(1+i)24(1+i3)21=k, and n=[∣k∣] be the greatest integral part of ∣k∣. Then ∑j=0n+5(j+5)2−∑j=0n+5(j+5) is equal to ________.
SolutionAnswer: 310
Approach:
Evaluate the complex expression to find k, take the integral part, then compute the difference of the two sums.
Step 1:
(1−i)24(−1+i3)21+(1+i)24(1+i3)21=0
Step 2:
n=[∣0∣]=0
Step 3:
∑j=05[(j+5)2−(j+5)]
Step 4:
20+30+42+56+72+90=310
Final answer: 310
Q83NumericalPermutations and Combinations
The students S1,S2,…,S10 are to be divided into 3 groups A,\ B and C such that each group has at least one student and the group C has at most 3 students. Then the total number of possibilities of forming such groups is ________.
SolutionAnswer: 31650
Approach:
Sum over the possible sizes of group C, choosing its members and distributing the rest into A and B with each non-empty.
Step 1:
c=1:(110)(29−2)=10⋅510=5100
Step 2:
c=2:(210)(28−2)=45⋅254=11430
Step 3:
c=3:(310)(27−2)=120⋅126=15120
Step 4:
5100+11430+15120=31650
Final answer: 31650
Q84NumericalSequence and Series
The sum of first four terms of a geometric progression (G. P. ) is 1265 and the sum of their respective reciprocals is 1865. If the product of first three terms of the G. P. is 1, and the third term is α, then 2α is ________.
SolutionAnswer: 3
Approach:
Use the ratio of the sum to the sum of reciprocals together with the product condition to find the third term.
Step 1:
65/1865/12=23=a2r3
Step 2:
a3r3=1⇒ar=1⇒a=r1
Step 3:
a2r3=r21⋅r3=r=23
Step 4:
α=ar2=r=23
Step 5:
2α=3
Final answer: 3
Q85NumericalBinomial Theorem and its Simple Applications
For integers n and r, let (rn)={nCr,0,if n≥r≥0otherwise. The maximum value of k for which the sum ∑i=0k(i10)(k−i15)+∑i=0k+1(i12)(k+1−i13) is maximum, is equal to ________.
SolutionAnswer: 12
Approach:
Apply Vandermonde's identity to each sum and combine using Pascal's rule, then locate the maximizing index.
Step 1:
∑i=0k(i10)(k−i15)=(k25)
Step 2:
∑i=0k+1(i12)(k+1−i13)=(k+125)
Step 3:
(k25)+(k+125)=(k+126)
Step 4:
(k+126) maximal at k+1=13
Final answer: 12
Q86NumericalCo-ordinate Geometry
Let a point P be such that its distance from the point (5,0) is thrice the distance of P from the point (−5,0). If the locus of the point P is a circle of radius r, then 4r2 (in the nearest integer) is equal to ________.
SolutionAnswer: 56
Approach:
Set up the distance condition, square it to obtain the circle equation, and extract the radius.
Step 1:
(x−5)2+y2=9[(x+5)2+y2]
Step 2:
x2+y2+225x+25=0
Step 3:
r2=(425)2−25=16225
Step 4:
4r2=4225=56.25≈56
Final answer: 56
Q87NumericalStatistics and Probability
If the variance of 10 natural numbers 1,1,1,…,1,k is less than 10, then the maximum possible value of k is ________.
SolutionAnswer: 11
Approach:
Express the variance of nine ones and k, impose the inequality, and find the largest natural number k.
Step 1:
xˉ=109+k
Step 2:
σ2=1009(k−1)2
Step 3:
1009(k−1)2<10⇒(k−1)2<91000
Step 4:
∣k−1∣<10.54⇒k≤11
Final answer: 11
Q88NumericalSets, Relations and Functions
If a+α=1,b+β=2 and af(x)+αf(x1)=bx+xβ,x=0, then the value of the expression x+x1f(x)+f(x1) is ________.
SolutionAnswer: 2
Approach:
Replace x by 1/x to obtain a second relation, solve the linear system for f(x) and f(1/x), then form the required ratio.
Step 1:
af(1/x)+αf(x)=xb+βx
Step 2:
f(x)+f(x1)=a+α(b+β)(x+x1)
Step 3:
x+1/xf(x)+f(1/x)=a+αb+β=12
Step 4:
=2
Final answer: 2
Q89NumericalCo-ordinate Geometry
If the area of the triangle formed by the x-axis, the normal and the tangent to the circle (x−2)2+(y−3)2=25 at the point (5,7) is A, then 24A is equal to ________.
SolutionAnswer: 1225
Approach:
Find the tangent and normal lines at the given point, locate their x-axis intercepts, and compute the triangle area.
Step 1:
radius slope=5−27−3=34
Step 2:
normal meets x-axis at (−41,0)
Step 3:
tangent slope=−43,meets x-axis at (343,0)
Step 4:
A=21⋅12175⋅7=241225
Step 5:
24A=1225
Final answer: 1225
Q90NumericalThree Dimensional Geometry
Let λ be an integer. If the shortest distance between the lines x−λ=2y−1=−2z and x=y+2λ=z−λ is 227, then the value of ∣λ∣ is ________.
SolutionAnswer: 1
Approach:
Write each line in point-direction form, apply the shortest-distance formula, set it equal to the given value, and select the integer solution.
How many questions are in the JEE Main 2021 February 24, Shift 2 paper?
The JEE Main 2021 February 24, 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 24, 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 24, 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.