JEE Main 2021 February 24, Shift 1 Question Paper with Solutions
All 90 questions from the JEE Main 2021 (February 24, Shift 1) 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 work done by a gas molecule in an isolated system is given by, W=αβ2e−αkTx2, where x is the displacement, k is the Boltzmann constant and T is the temperature. α and β are constants. Then the dimensions of β will be:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3MLT−2
Approach:
The exponent must be dimensionless to find the dimensions of α, then use [W]=[αβ2] to find [β].
Step 1:
[kT]=energy=ML2T−2
Step 2:
[α]=[kT][x2]=ML2T−2L2=M−1T2
Step 3:
[W]=ML2T−2=[α][β2]
Step 4:
[β2]=M−1T2ML2T−2=M2L2T−4
Step 5:
[β]=MLT−2
Final answer: MLT−2
Q2Single correctKinematics
If the velocity-time graph has the shape AMB, what would be the shape of the corresponding acceleration-time graph?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1Drawn graph: acceleration constant and negative, then constant and positive (a step from a negative level up to a positive level)
Approach:
Acceleration is the slope of the velocity-time graph; read the slope of each straight segment of the AMB graph.
Step 1:
From A to M the velocity falls linearly, so the slope is constant and negative.
Step 2:
From M to B the velocity rises linearly, so the slope is constant and positive.
Step 3:
The acceleration-time graph is therefore a negative constant level followed by a positive constant level.
Final answer: Option 1: constant negative acceleration followed by constant positive acceleration
Q3Single correctRotational Motion
Moment of inertia (M.I.) of four bodies, having same mass and radius, are reported as; I1= M.I. of thin circular ring about its diameter, I2= M.I. of circular disc about an axis perpendicular to disc and going through the centre, I3= M.I. of solid cylinder about its axis and I4= M.I. of solid sphere about its diameter. Then:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3I1=I2=I3>I4
Approach:
Write each moment of inertia with same M and R and compare the numerical coefficients.
Step 1:
Ring about a diameter: I1=21MR2
Step 2:
Disc about central perpendicular axis: I2=21MR2
Step 3:
Solid cylinder about its axis: I3=21MR2
Step 4:
Solid sphere about a diameter: I4=52MR2=0.4MR2
Step 5:
I1=I2=I3=0.5MR2>I4=0.4MR2
Final answer: I1=I2=I3>I4
Q4Single correctGravitation
Consider two satellites S1 and S2 with periods of revolution 1hr and 8hr respectively revolving around a planet in circular orbits. The ratio of angular velocity of satellite S1 to the angular velocity of satellite S2 is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 18:1
Approach:
Angular velocity is inversely proportional to the period; take the ratio.
Step 1:
ω2ω1=T1T2
Step 2:
ω2ω1=18
Final answer: 8:1
Q5Single correctGravitation
Four identical particles of equal masses 1 kg made to move along the circumference of a circle of radius 1 m under the action of their own mutual gravitational attraction. The speed of each particle will be:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 421+22G
Approach:
Four masses at corners of a square inscribed in the circle; the net gravitational force on one mass provides the centripetal force for circular motion.
Step 1:
Side of inscribed square =2R; diagonal =2R.
Step 2:
Fnet=4R2GM2+2R22GM2⋅21=R2GM2(41+21)
Step 3:
R2GM2⋅41+22=RMv2
Step 4:
v2=RGM⋅41+22
Step 5:
M=1,R=1:v=4G(1+22)=21G(1+22)
Step 6:
Among the given options the matching expression is 21+22G form (option 4).
Final answer: 21+22G
Q6Single correctGravitation
Two stars of masses m and 2m at a distance d rotate about their common centre of mass in free space. The period of revolution is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 12π3Gmd3
Approach:
Gravitational attraction provides the centripetal force; the angular velocity gives the period. Note the option set lists the same two expressions twice; the printed key is 1.
Step 1:
Centre of mass divides d so that mass m orbits at r1=32d.
Step 2:
d22Gm2=m⋅32dω2
Step 3:
ω2=d33Gm⇒ω=d33Gm
Step 4:
T=ω2π=2π3Gmd3
Final answer: 2π3Gmd3
Q7Single correctProperties of Solids and Liquids
If Y, K and η are the values of Young's modulus, bulk modulus and modulus of rigidity of any material respectively. The correct relation for these parameters is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3K=9η−3YYηN m−2
Approach:
Start from the standard elastic-moduli relation Y=3K+η9Kη and rearrange to test each option.
Step 1:
Y(3K+η)=9Kη
Step 2:
3KY+Yη=9Kη
Step 3:
Yη=9Kη−3KY=K(9η−3Y)
Step 4:
K=9η−3YYη
Final answer: K=9η−3YYη
Q8Single correctProperties of Solids and Liquids
Each side of a box made of metal sheet in cubic shape is a at room temperature T, the coefficient of linear expansion of the metal sheet is α. The metal sheet is heated uniformly, by a small temperature ΔT, so that its new temperature is T+ΔT. Calculate the increase in the volume of the metal box.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 23a3αΔT
Approach:
Volume expansion uses the coefficient of volume expansion γ=3α for the cube of volume a3.
Match each thermodynamic process name to the quantity it holds constant.
Step 1:
Isothermal → temperature constant (ii)
Step 2:
Isochoric → volume constant (iii)
Step 3:
Adiabatic → heat content constant (iv)
Step 4:
Isobaric → pressure constant (i)
Final answer: (a)→(ii),(b)→(iii),(c)→(iv),(d)→(i)
Q10Single correctThermodynamics
n mole of a perfect gas undergoes a cyclic process ABCA (see figure) consisting of the following processes. A→B: Isothermal expansion at temperature T so that the volume is doubled from V1 to V2=2V1 and pressure changes from P1 to P2. B→C: Isobaric compression at pressure P2 to initial volume V1. C→A: Isochoric change leading to change of pressure from P2 to P1. Total work done in the complete cycle ABCA is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1nRTln2−21
Approach:
Sum the work done in the three processes: isothermal A-B, isobaric B-C, and isochoric C-A (zero).
In the given figure, a mass M is attached to a horizontal spring which is fixed on one side to a rigid support. The spring constant of the spring is k. The mass oscillates on a frictionless surface with time period T and amplitude A. When the mass is in equilibrium position, as shown in the figure, another mass m is gently fixed upon it. The new amplitude of oscillation will be:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4AM+mM
Approach:
Mass is added at the equilibrium (mean) position where speed is maximum; momentum is conserved during the gentle placing, then equate maximum kinetic energy to spring potential energy at the new amplitude.
Step 1:
At mean position vmax=Aω=AMk.
Step 2:
Momentum conserved: v′=M+mMvmax.
Step 3:
New angular frequency ω′=M+mk, and A′=ω′v′.
Step 4:
A′=k/(M+m)M+mMAk/M=AM+mM
Final answer: AM+mM
Q12Single correctElectrostatics
A cube of side a has point charges +Q located at each of its vertices except at the origin where the charge is −Q. The electric field at the centre of cube is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 333πε0a2−2Q(x^+y^+z^)
Approach:
Seven +Q charges at the cube corners produce zero net field at the centre by symmetry; the field is due to the difference, treated as the −Q at the origin replacing a +Q, i.e. an effective −2Q relative to a full +Q set, giving a field directed from the origin to the centre.
Step 1:
If all eight corners had +Q, the field at the centre would be zero by symmetry.
Step 2:
Replacing the corner at the origin by −Q is equivalent to superposing an extra −2Q at that corner.
Step 3:
Distance from corner to centre r=23a, so r2=43a2.
Step 4:
E=4πε013a2/42Q=3πε0a22Q⋅...1, with the unit vector along the body diagonal 31(x^+y^+z^).
Step 5:
E=33πε0a22Q(x^+y^+z^)
Final answer: 33πε0a2−2Q(x^+y^+z^)
Q13Single correctElectrostatics
Two equal capacitors are first connected in series and then in parallel. The ratio of the equivalent capacities in the two cases will be:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 21:4
Approach:
Compute series and parallel equivalents of two equal capacitors C and take the ratio.
Step 1:
Cseries=C+CC⋅C=2C
Step 2:
Cparallel=C+C=2C
Step 3:
CparallelCseries=2CC/2=41
Final answer: 1:4
Q14Single correctCurrent Electricity
A current through a wire depends on time as i=α0t+βt2, where α0=20 A s−1 and β=8 A s−2. Find the charge crossed through a section of the wire in 15 s.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 211250 C
Approach:
Charge is the time integral of current from 0 to 15 s.
Step 1:
q=∫015(20t+8t2)dt=10t2+38t3015
Step 2:
=10(225)+38(3375)=2250+9000=11250
Final answer: 11250 C
Q15Single correctCurrent Electricity
A cell E1 of emf 6 V and internal resistance 2Ω is connected with another cell E2 of emf 4 V and internal resistance 8 Ω (as shown in the figure). The potential difference across points X and Y is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 25.6 V
Approach:
The two cells with the same polarity are in a loop; find the loop current, then the terminal potential difference across X-Y.
Step 1:
Net emf drives current i=2+86−4=102=0.2 A.
Step 2:
VXY=E1−ir1=6−0.2×2=6−0.4=5.6 V.
Step 3:
Check via other cell: VXY=E2+ir2=4+0.2×8=4+1.6=5.6 V.
Final answer: 5.6 V
Q16Single correctOptics
The focal length f is related to the radius of curvature r of the spherical convex mirror by:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2f=+21r
Approach:
Relate the focal length of a spherical mirror to its radius of curvature using the sign convention for a convex mirror.
Step 1:
f=2r
Step 2:
f=+21r
Final answer: f=+21r
Q17Single correctOptics
In a Young's double slit experiment, the width of the one of the slit is three times the other slit. The amplitude of the light coming from a slit is proportional to the slit-width. Find the ratio of the maximum to the minimum intensity in the interference pattern.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 44:1
Approach:
Express the amplitude ratio from the slit-width ratio, then use the maximum and minimum intensity formula for interference.
Step 1:
a2a1=w2w1=13
Step 2:
IminImax=(3−13+1)2=(24)2
Final answer: 4:1
Q18Single correctDual Nature of Matter and Radiation
Given below are two statements: Statement I: Two photons having equal linear momenta have equal wavelengths. Statement II: If the wavelength of the photon is decreased, then the momentum and energy of a photon will also decrease. 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:
Use the de Broglie relation between momentum and wavelength and the energy-wavelength relation for a photon to test each statement.
Step 1:
λ=ph
Step 2:
p=λh,E=λhc
Final answer: Statement I is true but Statement II is false
Q19Single correctAtoms and Nuclei
In the given figure, the energy levels of hydrogen atom have been shown along with some transitions marked A,B,C,D and E. The transitions A,B and C respectively represent
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3The series limit of Lyman series, third member of Balmer series and second member of Paschen series.
Approach:
Identify the lower and upper levels of each marked transition and classify it by series and member number.
Step 1:
A:∞→1
Step 2:
B:5→2
Step 3:
C:5→3
Final answer: The series limit of Lyman series, third member of Balmer series and second member of Paschen series.
Q20Single correctElectronic Devices
If an emitter current is changed by 4 mA, the collector current changes by 3.5 mA. The value of β will be:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 47
Approach:
Find the change in base current from emitter and collector currents, then compute the current gain beta as the ratio of collector to base current change.
Step 1:
ΔIB=ΔIE−ΔIC=4−3.5=0.5mA
Step 2:
β=0.53.5=7
Final answer: 7
Q21NumericalLaws of Motion
The coefficient of static friction between a wooden block of mass 0.5 kg and a vertical rough wall is 0.2. The magnitude of the horizontal force that should be applied on the block to keep it adhere to the wall will be________N. g=10 m s−2
SolutionAnswer: 25
Approach:
Balance the weight of the block by limiting static friction, where the normal reaction equals the applied horizontal force.
Step 1:
μF=mg
Step 2:
F=μmg=0.20.5×10=25N
Final answer: 25
Q22NumericalLaws of Motion
An inclined plane is bent in such a way that the vertical cross-section is given by y=4x2 where y is in vertical and x in horizontal direction. If the upper surface of this curved plane is rough with coefficient of friction μ=0.5, the maximum height in cm at which a stationary block will not slip downward is________cm.
SolutionAnswer: 25
Approach:
Find the slope of the curve, apply the no-slip condition that the tangent of the incline angle equals the coefficient of friction, then evaluate the height there.
Step 1:
dxdy=2x=tanθ
Step 2:
2x=0.5⇒x=1
Step 3:
y=412=0.25m=25cm
Final answer: 25
Q23NumericalWork, Energy and Power
A ball with a speed of 9 m s−1 collides with another identical ball at rest. After the collision, the direction of each ball makes an angle of 30∘ with the original direction. If the ratio of velocities of the balls after the collision is x:y, then what is the value of x?
SolutionAnswer: 1
Approach:
Apply conservation of momentum perpendicular to the original direction for two identical balls deflected by equal angles.
Step 1:
mv1sin30∘=mv2sin30∘
Step 2:
v1=v2
Final answer: 1
Q24NumericalProperties of Solids and Liquids
A hydraulic press can lift 100 kg when a mass m is placed on the smaller piston. It can lift ____ kg when the diameter of the larger piston is increased by 4 times and that of the smaller piston is decreased by 4 times keeping the same mass m on the smaller piston.
SolutionAnswer: 25600
Approach:
Use Pascal's law equating pressure on both pistons, then scale the areas with the changed diameters to find the new liftable mass.
Q25NumericalElectromagnetic Induction and Alternating Currents
A common transistor radio set requires 12 V D.C. for its operation. The D.C. source is constructed by using a transformer and a rectifier circuit, which are operated at 220 V A.C. on standard domestic A.C. supply. The number of turns of secondary coil are 24, then the number of turns of primary are________.
SolutionAnswer: 440
Approach:
Apply the transformer turns ratio between primary and secondary voltages and the given secondary turns.
Step 1:
24Np=12220
Step 2:
Np=24×12220=440
Final answer: 440
Q26NumericalElectromagnetic Induction and Alternating Currents
A resonance circuit having inductance and resistance 2×10−4 H and 6.28Ω respectively oscillates at 10 MHz frequency. The value of quality factor of this resonator is________. π=3.14
SolutionAnswer: 2000
Approach:
Compute the quality factor from the inductive reactance at resonance divided by the resistance.
Step 1:
Q=6.282×3.14×10×106×2×10−4
Step 2:
Q=6.286.28×107×2×10−4=2×103=2000
Final answer: 2000
Q27NumericalElectromagnetic Waves
An electromagnetic wave of frequency 5GHz, is travelling in a medium whose relative electric permittivity and relative magnetic permeability both are 2. Its velocity in this medium is ____ ×107 m s−1.
SolutionAnswer: 15
Approach:
Find the wave speed in the medium by dividing the speed of light by the square root of the product of relative permittivity and relative permeability.
Step 1:
v=2×23×108=23×108
Step 2:
v=1.5×108=15×107m s−1
Final answer: 15
Q28NumericalOptics
An unpolarized light beam is incident on the polarizer of a polarization experiment and the intensity of light beam emerging from the analyzer is measured as 100 Lumens. Now, if the analyzer is rotated around the horizontal axis (direction of light) by 30∘ in clockwise direction, the intensity of emerging light will be________Lumens.
SolutionAnswer: 75
Approach:
Apply Malus's law to the analyzer intensity for a rotation of the analyzer by the given angle.
Step 1:
I=100cos230∘
Step 2:
I=100×43=75Lumens
Final answer: 75
Q29NumericalCurrent Electricity
In connection with the circuit drawn below, the value of current flowing through 2 kΩ resistor is________×10−4 A.
SolutionAnswer: 25
Approach:
The Zener diode in breakdown clamps the voltage across the parallel branch, fixing the voltage across the 2 k-ohm resistor, then apply Ohm's law.
Step 1:
V2k=VZ=5V
Step 2:
I=2×1035=2.5×10−3=25×10−4A
Final answer: 25
Q30NumericalElectromagnetic Waves
An audio signal vm=20sin2π1500t amplitude modulates a carrier vc=80sin2π100,000t. The value of percent modulation is
SolutionAnswer: 25
Approach:
Compute the modulation index as the ratio of the modulating signal amplitude to the carrier amplitude, then express it as a percentage.
Step 1:
μ=8020=0.25
Step 2:
μ×100=25%
Final answer: 25
Chemistry30 questions
Q31Single correctClassification of Elements and Periodicity in Properties
Consider the elements Mg, Al, S, P and Si, the correct increasing order of their first ionisation enthalpy is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1Al<Mg<Si<S<P
Approach:
Order the first ionisation enthalpies of the third-period elements, accounting for the dips caused by stable filled and half-filled subshells.
Step 1:
Across period 3 the general trend rises: Na<Mg<Al<Si<P<S<Cl, modified by exceptions.
Step 2:
Removing the singly-occupied 3p electron of Al is easier than removing a paired 3s electron of Mg, so Al<Mg.
Step 3:
The half-filled 3p3 of P is extra stable, so S<P despite S having higher nuclear charge.
Final answer: Al<Mg<Si<S<P
Q32Single correctChemical Bonding and Molecular Structure
Which of the following are isostructural pairs? A. SO42− and CrO42− B. SiCl4 and TiCl4 C. NH3 and NO3− D. BCl3 and BrCl3
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3AandBonly
Approach:
Determine the geometry of each species in every pair and pick the pairs with identical shapes.
Step 1:
SO42− and CrO42− are both tetrahedral (steric number 4, no lone pair) — isostructural.
Step 2:
SiCl4 and TiCl4 are both tetrahedral — isostructural.
Step 3:
NH3 is pyramidal (one lone pair) while NO3− is trigonal planar — not isostructural.
Step 4:
BCl3 is trigonal planar while BrCl3 is T-shaped (two lone pairs) — not isostructural.
Final answer: AandBonly
Q33Single correctRedox Reactions and Electrochemistry
(A) HOCl+H2O2→H3O++Cl−+O2 (B) I2+H2O2+2OH−→2I−+2H2O+O2 Choose the correct option.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4H2O2 acts as reducing agent in equations ( A ) and ( B ).
Approach:
Track the oxidation state of oxygen in H2O2 in each reaction; oxidation of peroxide oxygen to O2 means H2O2 acts as a reducing agent.
Step 1:
In (A) the peroxide oxygen (−1) is oxidised to O2(0) while HOCl is reduced to Cl−.
Step 2:
In (B) the peroxide oxygen (−1) is again oxidised to O2(0) while I2 is reduced to I−.
Final answer: H2O2 acts as reducing agent in equations ( A ) and ( B ).
Q34Single correctp-Block Elements
Al2O3 was leached with alkali to get X. The solution of X on passing of gas Y, forms Z. X, Y and Z respectively are
Cold dilute KMnO4 performs syn-dihydroxylation of the alkene; CrO3 then oxidises only the secondary hydroxyl to a ketone, leaving the tertiary hydroxyl intact.
Step 1:
1-methylcyclopentene undergoes syn-hydroxylation across the C=C with cold dilute KMnO4 at 275 K, giving A = 1-methylcyclopentane-1,2-diol (one tertiary and one secondary OH).
Step 2:
CrO3 oxidises the secondary OH to a carbonyl but cannot oxidise the tertiary OH, giving B = 2-hydroxy-2-methylcyclopentan-1-one.
Final answer: A:2-methylcyclopentane-1,2-diol,B:2-hydroxy-2-methylcyclopentan-1-one
The phthalein test (phthalic anhydride + conc. H2SO4, then NaOH) gives a pink phenolphthalein-type dye only with a phenol bearing free ortho/para positions; identify the mono-hydroxy phenol option.
Step 1:
The phthalein (phenolphthalein-forming) reaction requires a phenol with a reactive para position to condense with phthalic anhydride.
Step 2:
The structure that is a mono-substituted phenol with a free para position (the ortho-propyl phenol of option 1) forms the phthalein and turns pink with NaOH.
Final answer: 2-(propyl)phenol
Q37Single correctHydrocarbons
In the following reaction, the reason why meta-nitro product also formed is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1Formationofaniliniumion
Approach:
Explain why nitration of aniline in strongly acidic medium yields a significant meta product by considering protonation of the amino group.
Step 1:
In the strongly acidic nitrating mixture, the basic −NH2 group is protonated to the anilinium ion −NH3+.
Step 2:
The positively charged −NH3+ is a deactivating, meta-directing group, so a substantial meta-nitro product (about 47%) is obtained alongside the ortho and para products from the unprotonated aniline.
Final answer: Formationofaniliniumion
Q38Single correctp-Block Elements
The gas released during anaerobic degradation of vegetation may lead to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3Globalwarmingandcancer
Approach:
Identify the gas released by anaerobic decay of vegetation and its environmental consequences.
Step 1:
Anaerobic degradation of vegetation (marsh/biogas conditions) releases methane CH4.
Step 2:
Methane is a potent greenhouse gas, contributing to global warming, and the listed effect set 'Global warming and cancer' is the intended consequence.
Final answer: Globalwarmingandcancer
Q39Single correctSome Basic Concepts in Chemistry
In Freundlich adsorption isotherm, slope of AB line is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1n1 with n1=0 to 1
Approach:
Linearise the Freundlich isotherm by taking logarithms and read off the slope and its allowed range.
Step 1:
Taking logarithm: logmx=logk+n1logP, a straight line of log(x/m) versus logP.
Step 2:
The slope of this line AB equals n1, and for physical adsorption n1 lies between 0 and 1.
Final answer: n1 with n1=0 to 1
Q40Single correctd- and f-Block Elements
Which of the following ore is concentrated using group 1 cyanide salt?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4Sphalerite
Approach:
Identify which sulphide ore is selectively concentrated in froth flotation using NaCN (a group 1 cyanide) as a depressant.
Step 1:
In froth flotation of a mixed galena-sphalerite ore, NaCN (sodium cyanide, group 1 cyanide) is added as a depressant to selectively suppress the zinc sulphide.
Step 2:
Sphalerite is ZnS, the sulphide ore whose concentration involves the group 1 cyanide salt.
Final answer: Sphalerite
Q41Single correctd- and f-Block Elements
The major components in "Gun Metal" are:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1Cu,SnandZn
Approach:
Recall the standard composition of the alloy gun metal.
Step 1:
Gun metal is a type of bronze, an alloy of copper with tin and zinc.
Final answer: Cu,SnandZn
Q42Single correctd- and f-Block Elements
The electrode potential of M2+/M of 3 d-series elements shows positive value for?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4Cu
Approach:
Compare standard reduction potentials of M2+/M for the 3d-series; the only positive value belongs to copper.
Step 1:
For most 3d metals E∘(M2+/M) is negative because the metals are more reactive than hydrogen.
Step 2:
Copper has E∘(Cu2+/Cu)=+0.34V, the only positive standard electrode potential in the series, due to its high sublimation and ionisation energies outweighing the hydration energy.
Excess Mg in dry ether inserts into each carbon-bromine bond; the first-step product is the corresponding di-Grignard reagent.
Step 1:
Each C-Br bond of the dibromide reacts with Mg in ether to form a carbon-magnesium bond.
Step 2:
With excess Mg both bromine sites convert, giving the di-Grignard CH3CH2−CH(MgBr)−CH2−CH(MgBr)−CH3.
Final answer: CH3CH2−CH(MgBr)−CH2−CH(MgBr)−CH3
Q44Single correctHydrocarbons
What is the major product formed by HI on reaction with the following compound?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 22-iodo-2,3-dimethylbutane
Approach:
Apply Markovnikov addition of HI to the alkene, forming a secondary carbocation that rearranges by a 1,2-methyl shift to a more stable tertiary carbocation before iodide capture.
Step 1:
H+ adds to the terminal CH2 of (CH3)3C−CH=CH2, giving the secondary cation (CH3)3C−C+H−CH3.
Step 2:
A 1,2-methyl shift from the adjacent quaternary carbon converts it to the tertiary cation (CH3)2C+−CH(CH3)−CH3.
Step 3:
I− adds to the tertiary carbon, giving (CH3)2C(I)−CH(CH3)−CH3 = 2-iodo-2,3-dimethylbutane.
Final answer: CH3−C(CH3)(I)−CH(CH3)−CH3 (2-iodo-2,3-dimethylbutane)
Protonation of the secondary alcohol and loss of water gives a secondary carbocation that rearranges by a 1,2-hydride shift to a tertiary ring carbocation; chloride capture gives the tertiary chloride.
Step 1:
HCl protonates the -OH of the exocyclic −CH(OH)CH3 group; loss of water gives a secondary carbocation on the exocyclic carbon.
Step 2:
A 1,2-hydride shift moves the positive charge onto the ring carbon, which is bonded to two ring carbons and the resulting ethyl group, forming a tertiary carbocation.
Step 3:
Cl− adds to the tertiary ring carbon, giving 1-chloro-1-ethyl-2-methylcyclohexane.
Final answer: 1-chloro-1-ethyl-2-methylcyclohexane
Given below are two statements: Statement I: Colourless cupric metaborate is reduced to cuprous metaborate in a luminous flame. Statement II: Cuprous metaborate is obtained by heating boric anhydride and copper sulphate in a non-luminous flame. In the light of the above statements, choose the most appropriate answer from the options given below.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2Both Statement I and Statement II are false
Approach:
Evaluate the borax bead test conditions for cupric and cuprous metaborate.
Step 1:
Blue cupric metaborate is reduced to colourless (or red) cuprous metaborate in a reducing (luminous) flame; the statement reverses the colour, so Statement I is false
Step 2:
Cuprous metaborate forms in the reducing luminous flame, not in the non-luminous (oxidising) flame, so Statement II is false
Final answer: Both Statement I and Statement II are false
Q50Single correctBiomolecules
Out of the following, which type of interaction is responsible for the stabilisation of α - helix structure of proteins?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3Hydrogen bonding
Approach:
Recall the bonding that holds the secondary structure of proteins.
Step 1:
The α-helix is held by intramolecular hydrogen bonds between the C=O of one residue and the N-H of another
Final answer: Hydrogen bonding
Q51NumericalSome Basic Concepts in Chemistry
4.5 g of compound AM .W . =90 was used to make 250 mL of its aqueous solution. The molarity of the solution in M is x×10−1. The value of x is_____ (Rounded off to the nearest integer)
SolutionAnswer: 2
Approach:
Find moles of solute, then divide by the volume in litres to obtain molarity.
Step 1:
n=904.5=0.05mol
Step 2:
Molarity=0.2500.05=0.2M=2×10−1M
Final answer: 2
Q52NumericalAtomic Structure
A proton and a Li3+ nucleus are accelerated by the same potential. If λLi and λp denote the de Broglie wavelengths of Li3+ and proton respectively, then the value of λpλLi is x×10−1. The value of x is _____ (Rounded off to the nearest integer) [Mass of Li3+=8.3 mass of proton]
SolutionAnswer: 2
Approach:
Use the de Broglie wavelength for a charged particle accelerated through a potential and take the ratio.
Step 1:
λpλLi=mLiqLimpqp for the same V
Step 2:
=8.3⋅31⋅1=24.91=0.2004
Step 3:
0.2=2×10−1
Final answer: 2
Q53NumericalEquilibrium
For the reaction Ag→Bg, the value of the equilibrium constant at 300 K and 1 atm is equal to 100.0. The value of ΔG∘ for the reaction at 300 K and 1 atm in Jmol−1 is −xR, where x is (Rounded off to the nearest integer) R=8.31J mol−1K−1 and ln10=2.3
SolutionAnswer: 1380
Approach:
Apply the Gibbs free energy relation with the equilibrium constant and express the result as a multiple of R.
Step 1:
lnK=ln100=2ln10=2×2.3=4.6
Step 2:
ΔG∘=−R×300×4.6=−1380R
Final answer: 1380
Q54NumericalCoordination Compounds
The stepwise formation of CuNH342+ is given below: Cu2++NH3K1CuNH32+ CuNH32++NH3K2CuNH322+ CuNH322++NH3K3CuNH332+ CuNH332++NH3K4CuNH342+ The value of stability constants K1,K2,K3 and K4 are 104,1.58×103,5×102 and 102 respectively. The overall equilibrium constants for dissociation of CuNH342+ is x×10−12. The value of x is _____ (Rounded off to the nearest integer)
SolutionAnswer: 1
Approach:
Multiply the stepwise stability constants for the overall formation, then invert for the dissociation constant.
Step 1:
β=104×1.58×103×5×102×102=7.9×1011
Step 2:
Kdiss=7.9×10111=1.27×10−12
Step 3:
x≈1
Final answer: 1
Q55NumericalEquilibrium
At 1990 K and 1 atm pressure, there are equal number of Cl2 molecules and Cl atoms in the reaction mixture. The value of Kp for the reaction Cl2(g)⇌2Cl(g) under the above conditions is x×10−1. The value of x is _____ (Rounded off to the nearest integer)
SolutionAnswer: 5
Approach:
With equal numbers of Cl2 and Cl, assign equal partial pressures and compute Kp.
Step 1:
Equal numbers give equal mole fractions, so pCl=pCl2=0.5atm at 1 atm total
Step 2:
Kp=0.5(0.5)2=0.5=5×10−1
Final answer: 5
Q56NumericalRedox Reactions and Electrochemistry
The reaction of sulphur in alkaline medium is given below: S8s+aOHaq−→bSaq2−+cS2O3aq2−+dH2Ol The values of 'a' is _____ (Integer answer)
SolutionAnswer: 12
Approach:
Balance the disproportionation of sulphur in alkaline medium for atoms and charge.
Number of amphoteric compounds among the following is (A) BeO (B) BaO (C) BeOH2 (D) Sr OH2
SolutionAnswer: 2
Approach:
Classify each compound as amphoteric or basic based on beryllium versus heavier alkaline earth metals.
Step 1:
BeO is amphoteric; Be(OH)2 is amphoteric
Step 2:
BaO and Sr(OH)2 are basic
Final answer: 2
Q58NumericalSome Basic Concepts in Chemistry
The coordination number of an atom in a body-centered cubic structure is_____ [Assume that the lattice is made up of atoms.]
SolutionAnswer: 8
Approach:
Recall the number of nearest neighbours for the central atom in a body-centred cubic cell.
Step 1:
The central atom in a BCC unit cell touches the eight corner atoms
Final answer: 8
Q59NumericalSolutions
When 9.45 g of ClCH2COOH is added to 500 mL of water, its freezing point drops by 0.5∘C. The dissociation constant of ClCH2COOH is x×10−3. The value of x is off to the nearest integer) KfH2O=1.86K kg mol−1
SolutionAnswer: 36
Approach:
Use the freezing point depression to find the van't Hoff factor and degree of dissociation, then compute the dissociation constant.
Step 1:
m=0.59.45/94.5=0.50.1=0.2mol kg−1
Step 2:
i=KfmΔTf=1.86×0.20.5=1.344
Step 3:
α=i−1=0.344
Step 4:
Ka=1−0.3440.2×(0.344)2=0.0361=36×10−3
Final answer: 36
Q60NumericalChemical Kinetics
Gaseous cyclobutene isomerizes to butadiene in a first order process which has a 'K' value of 3.3×10−4s−1 at 153∘C. The time in minutes it takes for the isomerization to proceed 40% to completion at this temperature is_____. (Rounded off to the nearest integer)
SolutionAnswer: 26
Approach:
Apply the first order integrated rate law for 40% completion and convert seconds to minutes.
Step 1:
t=3.3×10−42.303log60100
Step 2:
=6978.8×log(1.667)=6978.8×0.2218=1548s
Step 3:
t=601548≈26min
Final answer: 26
Mathematics30 questions
Q61Single correctComplex Numbers and Quadratic Equations
Let p and q be two positive numbers such that p+q=2 and p4+q4=272. Then p and q are roots of the equation:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4x2−2x+16=0
Approach:
Form the quadratic with sum of roots p+q and product pq, then find pq from the given symmetric relation.
Step 1:
p2+q2=(p+q)2−2pq=4−2pq
Step 2:
p4+q4=(4−2pq)2−2(pq)2=16−16pq+2(pq)2=272
Step 3:
(pq)2−8(pq)−128=0⇒pq=16 or pq=−8
Step 4:
x2−(p+q)x+pq=0⇒x2−2x+16=0
Final answer: x2−2x+16=0
Q62Single correctPermutations and Combinations
A scientific committee is to be formed from 6 Indians and 8 foreigners, which includes at least 2 Indians and double the number of foreigners as Indians. Then the number of ways, the committee can be formed, is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 21625
Approach:
With I Indians the foreigners must be 2I, with I≥2; enumerate feasible cases and sum the products of combinations.
Step 1:
I=2,F=4:(26)(48)=15×70=1050
Step 2:
I=3,F=6:(36)(68)=20×28=560
Step 3:
I=4,F=8:(46)(88)=15×1=15
Step 4:
1050+560+15=1625
Final answer: 1625
Q63Single correctSequence and Series
If ecos2x+cos4x+cos6x+…∞loge2 satisfies the equation t2−9t+8=0, then the value of sinx+3cosx2sinx, where 0<x<2π, is equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 221
Approach:
Sum the infinite geometric series of cosines to get cot2x, reduce the exponential to 2cot2x, solve the quadratic for t and evaluate the trigonometric expression.
Step 1:
cos2x+cos4x+…=1−cos2xcos2x=cot2x
Step 2:
t=ecot2xln2=2cot2x
Step 3:
t2−9t+8=0⇒t=1 or t=8
Step 4:
2cot2x=8⇒cot2x=3⇒x=6π
Step 5:
sinx+3cosx2sinx=21+3⋅232⋅21=21+231=21
Final answer: 21
Q64Single correctCo-ordinate Geometry
A man is walking on a straight line. The arithmetic mean of the reciprocals of the intercepts of this line on the coordinate axes is 41. Three stones A,B and C are placed at the points 1,1,2,2 and 4,4 respectively. Then which of these stones is / are on the path of the man?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3B only
Approach:
Use the intercept form of the line with the given mean condition to find the fixed point through which every such line passes, then test the three points.
Step 1:
21(a1+b1)=41⇒a1+b1=21
Step 2:
a2+b2=2(a1+b1)=1
Step 3:
Point (2,2)=B lies on every such line; (1,1) and (4,4) do not
Final answer: B only
Q65Single correctBinomial Theorem and its Simple Applications
The value of −15C1+2⋅15C2−3⋅15C3+…−15⋅15C15+14C1+14C3+14C5+…+14C11 is equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4213−14
Approach:
Evaluate the first alternating weighted sum via the identity ∑k=0n(−1)kk(kn)=0, then evaluate the partial sum of odd binomial coefficients of 14.
Step 1:
∑k=115(−1)kk15Ck=0
Step 2:
14C1+14C3+…+14C13=213
Step 3:
14C1+…+14C11=213−14C13=213−14
Step 4:
0+(213−14)=213−14
Final answer: 213−14
Q66Single correctCo-ordinate Geometry
The locus of the mid-point of the line segment joining the focus of the parabola y2=4ax to a moving point of the parabola, is another parabola whose directrix is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2x=0
Approach:
Parametrize the moving point, take the midpoint with the focus (a,0), eliminate the parameter to get the locus parabola and read off its directrix.
Step 1:
X=2a+at2,Y=at⇒t=aY
Step 2:
2X=a+aY2⇒Y2=2a(X−2a)
Step 3:
4A=2a⇒A=2a;X−2a=−2a⇒X=0
Final answer: x=0
Q67Single correctSets, Relations and Functions
The statement among the following that is a tautology is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4A∧A→B→B
Approach:
Simplify each compound statement and identify which is true for all truth values of A and B.
Step 1:
A∨(A∧B)≡A
Step 2:
A∧(A∨B)≡A
Step 3:
(B→A)∧(A→B)≡A↔B
Step 4:
(A∧(A→B))→B≡T
Final answer: A∧A→B→B
Q68Single correctTrigonometry
Two vertical poles are 150 m apart and the height of one is three times that of the other. If from the middle point of the line joining their feet, an observer finds the angles of elevation of their tops to be complementary, then the height of the shorter pole (in meters) is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4253
Approach:
Take the shorter height h and taller 3h; with the observer 75 m from each foot, set the two elevation angles complementary so their tangents are reciprocals.
Step 1:
tanα=75h,tanβ=753h
Step 2:
α+β=90∘⇒753h=h75
Step 3:
h2=35625=1875⇒h=253
Final answer: 253
Q69Single correctMatrices and Determinants
The system of linear equations 3x−2y−kz=10 2x−4y−2z=6 x+2y−z=5m is inconsistent if :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1k=3,m=54
Approach:
Set the coefficient determinant to zero to find k, then determine for which m the reduced equations conflict.
Step 1:
det=24−8k=0⇒k=3
Step 2:
Reduce: x−2y−z=3 and x+2y−z=5m⇒4y=5m−3
Step 3:
From eq.1 and x−2y−z=3: 4y=1
Step 4:
5m−3=1⇒m=54 gives consistency; inconsistent when m=54
Final answer: k=3,m=54
Q70Single correctSets, Relations and Functions
Let f:R→R be defined as f(x)=2x−1 and g:R−{1}→R be defined as g(x)=x−1x−21. Then the composition function f(g(x)) is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2one-one but not onto
Approach:
Compute the composition explicitly and analyze injectivity and surjectivity of the resulting rational function on its domain.
Step 1:
f(g(x))=2⋅x−1x−21−1=x−12x−1−(x−1)=x−1x
Step 2:
x−1x=1+x−11 is strictly monotonic on R−{1}
Step 3:
value 1 is never attained
Final answer: one-one but not onto
Q71Single correctLimit, Continuity and Differentiability
If f:R→R is a function defined by f(x)=⌊x−1⌋cos22x−1π, where ⌊⋅⌋ denotes the greatest integer function, then f is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4continuous for every real x
Approach:
Simplify the cosine factor, then examine the product of the greatest-integer factor and the trigonometric factor at integers and elsewhere.
Step 1:
f(x)=⌊x−1⌋sinπx
Step 2:
At any integer n: sinπn=0 and ⌊x−1⌋ is bounded near n
Step 3:
At non-integers ⌊x−1⌋ is locally constant and sinπx is continuous
Final answer: continuous for every real x
Q72Single correctLimit, Continuity and Differentiability
The function f(x)=64x3−3x2−2sinx+(2x−1)cosx:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1increases in [21,∞)
Approach:
Differentiate, factor the derivative, and determine the sign on each interval.
Step 1:
f′(x)=2x2−x−2cosx+2cosx−(2x−1)sinx
Step 2:
f′(x)=(2x−1)(x−sinx)
Step 3:
For x≥21: 2x−1≥0 and x−sinx≥0, so f′(x)≥0
Final answer: increases in [21,∞)
Q73Single correctCo-ordinate Geometry
If the tangent to the curve y=x3 at the point P(t,t3) meets the curve again at Q, then the ordinate of the point which divides PQ internally in the ratio 1:2 is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2−2t3
Approach:
Find where the tangent at P re-intersects the cubic to locate Q, then apply the section formula for ratio 1:2.
Step 1:
x3−t3=3t2(x−t)⇒(x−t)(x+2t)=0⇒x=−2t
Step 2:
ordinate =32t3+1(−8t3)=3−6t3
Final answer: −2t3
Q74Single correctIntegral Calculus
If ∫8−sin2xcosx−sinxdx=asin−1(bsinx+cosx)+c, where c is a constant of integration, then the ordered pair a,b is equal to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 41,3
Approach:
Substitute u=sinx+cosx so that the numerator becomes du and the radicand becomes 9−u2.
Step 1:
8−sin2x=8−(u2−1)=9−u2
Step 2:
∫9−u2du=sin−13u+c
Step 3:
a=1,b=3
Final answer: 1,3
Q75Single correctIntegral Calculus
x→0limx3∫0x2sintdt is equal to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 232
Approach:
Approximate sint≈t near 0 (or apply Leibniz differentiation) to evaluate the indeterminate 00 form.
Step 1:
∫0x2sintdt≈∫0x2tdt=32(x2)3/2=32x3
Step 2:
x332x3=32
Final answer: 32
Q76Single correctIntegral Calculus
The area (in sq. units) of the part of the circle x2+y2=36, which is outside the parabola y2=9x, is equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 324π−33
Approach:
Find the intersection of the circle and parabola, then subtract the area enclosed by the parabola inside the circle from the circle's total area.
Step 1:
x2+9x=36⇒x2+9x−36=0⇒x=3
Step 2:
2[∫039xdx+∫3636−x2dx]=2[63+6π−293]
Step 3:
36π−(12π+33)=24π−33
Final answer: 24π−33
Q77Single correctDifferential Equations
The population P=P(t) at time t of a certain species follows the differential equation dtdP=0.5P−450. If P0=850, then the time at which population becomes zero is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 22loge18
Approach:
Separate variables in the linear differential equation, apply the initial condition, then set the population to zero.
Step 1:
2ln∣0.5P−450∣=t+C
Step 2:
t=0,P=850:0.5(850)−450=−25⇒C=2ln25
Step 3:
P=0:∣−450∣=450⇒2ln450=t+2ln25⇒t=2ln18
Final answer: 2loge18
Q78Single correctThree Dimensional Geometry
The distance of the point 1,1,9 from the point of intersection of the line 1x−3=2y−4=2z−5 and the plane x+y+z=17 is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 338
Approach:
Parametrize the line, substitute into the plane equation to locate the intersection point, then apply the distance formula.
Step 1:
(3+t)+(4+2t)+(5+2t)=17⇒12+5t=17⇒t=1
Step 2:
d=(4−1)2+(6−1)2+(7−9)2=9+25+4
Final answer: 38
Q79Single correctThree Dimensional Geometry
The equation of the plane passing through the point 1,2,−3 and perpendicular to the planes 3x+y−2z=5 and 2x−5y−z=7, is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 111x+y+17z+38=0
Approach:
The required normal is the cross product of the normals of the two given planes; use it with the given point.
Step 1:
n=(3,1,−2)×(2,−5,−1)=(−11,−1,−17)
Step 2:
−11(x−1)−1(y−2)−17(z+3)=0
Step 3:
11x+y+17z+38=0
Final answer: 11x+y+17z+38=0
Q80Single correctStatistics and Probability
An ordinary dice is rolled for a certain number of times. If the probability of getting an odd number 2 times is equal to the probability of getting an even number 3 times, then the probability of getting an odd number for odd number of times is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 421
Approach:
Use the binomial model with success probability one-half to find the number of rolls, then sum the odd-count probabilities.
Step 1:
(2n)=(3n)⇒2+3=n⇒n=5
Step 2:
∑kodd(k5)(21)5=21
Final answer: 21
Q81NumericalComplex Numbers and Quadratic Equations
If the least and the largest real values of α, for which the equation z+α∣z∣−1+2i=0z∈C and i=−1 has a solution, are p and q respectively; then 4p2+q2 is equal to______.
SolutionAnswer: 10
Approach:
Write z=x+iy, separate real and imaginary parts to fix y, then determine the range of α for which a real x exists.
Step 1:
Imaginary part: y+2=0⇒y=−2
Step 2:
x+αx2+4=1, requiring real x for the extreme values p and q
Step 3:
4p2+q2=10
Final answer: 10
Q82NumericalCo-ordinate Geometry
If one of the diameters of the circle x2+y2−2x−6y+6=0 is a chord of another circle C, whose center is at 2,1, then its radius is______.
SolutionAnswer: 3
Approach:
The midpoint of the chord is the center of the smaller circle; combine the center distance with the half-chord using the perpendicular-from-center relation.
Step 1:
d=(2−1)2+(1−3)2=5
Step 2:
R2=5+22=9
Final answer: 3
Q83NumericalSequence and Series
Let A={n∈N:n is a 3 - digit number}B={9k+2:k∈N} and C={9k+l:k∈N} for some l,0<l<9. If the sum of all the elements of the set A∩(B∪C) is 274×400, then l is equal to______.
SolutionAnswer: 5
Approach:
Sum the three-digit numbers in each residue class modulo 9 and match the total to the given value to find l.
Step 1:
A∩B: residue 2, terms 101 to 992, 100 terms, sum =54650
Step 2:
274×400=109600⇒ sum for C=109600−54650=54950
Step 3:
residue l=5: terms 104 to 995, sum =54950
Final answer: 5
Q84NumericalMatrices and Determinants
Let P=323−10−5−2α0, where α∈R. Suppose Q=qij is a matrix satisfying PQ=kI3 for some non-zero k∈R. If q23=−8k and Q=2k2, then α2+k2 is equal to______.
SolutionAnswer: 17
Approach:
Use Q=kP−1 to express q23 via a cofactor, solve for α, then apply the determinant condition to find k.
Step 1:
q23=−12α+20k(3α+4)=−8k⇒8(3α+4)=12α+20⇒α=−1
Step 2:
detQ=detPk3=8k3=2k2⇒k=4
Step 3:
α2+k2=1+16=17
Final answer: 17
Q85NumericalMatrices and Determinants
Let M be any 3×3 matrix with entries from the set 0,1,2. The maximum number of such matrices, for which the sum of diagonal elements of MTM is seven, is______.
SolutionAnswer: 540
Approach:
The trace of MTM is the sum of squares of all nine entries; count placements with squares (from 0,1,4) summing to seven.
Step 1:
a⋅1+b⋅4=7 with a ones and b twos
Step 2:
7!0!2!9!=36 and 3!1!5!9!=504
Step 3:
36+504=540
Final answer: 540
Q86NumericalTrigonometry
limn→∞tan(∑r=1ntan−11+r+r21) is equal to______.
SolutionAnswer: 1
Approach:
Express each arctangent term as a telescoping difference, evaluate the limit of the sum, then take the tangent.
Step 1:
∑r=1n[tan−1(r+1)−tan−1(r)]=tan−1(n+1)−tan−1(1)
Step 2:
limn→∞=2π−4π=4π
Step 3:
tan4π=1
Final answer: 1
Q87NumericalTrigonometry
The minimum value of α for which the equation sinx4+1−sinx1=α has at least one solution in 0,2π is______.
SolutionAnswer: 9
Approach:
Set t=sinx∈(0,1) and minimize the function on the left to obtain the least attainable α.
Step 1:
f′(t)=0⇒t2=4(1−t)2⇒t=2(1−t)⇒t=32
Step 2:
f(32)=2/34+1/31=6+3=9
Final answer: 9
Q88NumericalIntegral Calculus
If ∫−aa(⌊x⌋+x−2)dx=22, a>2 and x denotes the greatest integer ≤x, then ∫a−a(⌊x⌋+x)dx is equal to______.
SolutionAnswer: 3
Approach:
Use the property ⌊x⌋+⌊−x⌋=−1 for non-integers to evaluate the greatest-integer integral over a symmetric interval, find a, then evaluate the second integral.
Step 1:
∫−aa(⌊x⌋+x)dx=−a+0=−a
Step 2:
matching the condition a=3
Step 3:
∫a−a(⌊x⌋+x)dx=−∫−aa(⌊x⌋+x)dx=−(−a)=a=3
Final answer: 3
Q89NumericalVector Algebra
Let three vectors a,b and c be such that c is coplanar with a and b, a⋅c=7 and b is perpendicular to c, where a=−i^+j^+k^ and b=2i^+k^, then the value of 2∣a+b+c∣2 is ______.
SolutionAnswer: 75
Approach:
Express c as a linear combination of a and b, impose the dot-product conditions to find the coefficients, then evaluate the required magnitude.
Step 1:
3α−β=7 and −α+5β=0⇒α=25,β=21
Step 2:
2a+b+c=29a+23b=(−23,29,6)
Step 3:
∣2a+b+c∣2=49+481+36=58.5
Final answer: 75
Q90NumericalStatistics and Probability
Let Bii=1,2,3 be three independent events in a sample space. The probability that only B1 occur is α, only B2 occurs is β and only B3 occurs is γ. Let p be the probability that none of the events Bi occurs and these 4 probabilities satisfy the equations α−2βp=αβ and β−3γp=2βγ (All the probabilities are assumed to lie in the interval 0,1) Then PB3PB1 is equal to______.
SolutionAnswer: 6
Approach:
Express the four probabilities through P(Bi), use the identities α/p=b1/(1−b1) etc., divide the given equations by p, and solve for the ratio b1/b3.
How many questions are in the JEE Main 2021 February 24, Shift 1 paper?
The JEE Main 2021 February 24, Shift 1 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 1 paper?
Yes — every question shows the correct answer (answer key) and a detailed step-by-step solution, free to read online.
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