JEE Main 2021 February 26, Shift 2 Question Paper with Solutions
All 90 questions from the JEE Main 2021 (February 26, 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.
A wire of 1Ω has a length of 1 m. It is stretched till its length increases by 25%. The percentage change in resistance to the nearest integer is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 456%
Approach:
At constant volume the resistance varies as the square of length, so a length increase fixes the new resistance and hence the percentage change.
Step 1:
AL=constant⇒A′=1.25A
Step 2:
R′=R(LL′)2=R(1.25)2=1.5625R
Step 3:
ΔR%=(1.5625−1)×100=56.25%
Final answer: 56%
Q2Single correctUnits and Measurements
If C and V represent capacity and voltage respectively then what are the dimensions of λ where C/V=λ ?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1[M−2L−4I3T7]
Approach:
Divide the dimensional formula of capacitance by that of potential difference.
Step 1:
[λ]=[V][C]=ML2T−3I−1M−1L−2T4I2
Step 2:
[λ]=M−2L−4T7I3
Final answer: [M−2L−4I3T7]
Q3Single correctKinematics
A scooter accelerates from rest for time t1 at constant rate a1 and then retards at constant rate a2 for time t2 and comes to rest. The correct value of t2t1 will be :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1a1a2
Approach:
Equate the peak velocity reached during acceleration to the velocity lost during retardation.
Step 1:
vmax=a1t1
Step 2:
0=vmax−a2t2⇒vmax=a2t2
Step 3:
a1t1=a2t2⇒t2t1=a1a2
Final answer: a1a2
Q4Single correctKinematics
The trajectory of a projectile in a vertical plane is y=αx−βx2, where α and β are constants and x & y are respectively the horizontal and vertical distances of the projectile from the point of projection. The angle of projection θ and the maximum height attained H are respectively given by
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4tan−1α,4βα2
Approach:
The launch angle is the slope at the origin; the maximum height is the y value where the slope vanishes.
Step 1:
dxdy=α−2βx
Step 2:
dxdyx=0=α⇒θ=tan−1α
Step 3:
α−2βx=0⇒x=2βα
Step 4:
H=α⋅2βα−β(2βα)2=2βα2−4βα2=4βα2
Final answer: tan−1α,4βα2
Q5Single correctLaws of Motion
An inclined plane making an angle of 30∘ with the horizontal is placed in a uniform horizontal electric field 200CN as shown in the figure. A body of mass 1 kg and charge 5 mC is allowed to slide down from rest at a height of 1 m. If the coefficient of friction is 0.2, find the time taken by the body to reach the bottom. [g=9.8m s−2;sin30∘=21;cos30∘=23]
(A)
(B)
(C)
(D)
SolutionAnswer: Option 21.3 s
Approach:
Resolve gravity and the horizontal electric force along and perpendicular to the incline, find the net acceleration down the slope, then apply kinematics over the slope length.
Two masses A and B, each of mass M are fixed together by a massless spring, A force acts on the mass B as shown in figure. If the mass A starts moving away from mass B with acceleration a, then the acceleration of mass B will be :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4MF−Ma
Approach:
The spring force on A is the only force on A, fixing its value; apply Newton's second law to B with the external force and the spring reaction.
Step 1:
Fspring=Ma
Step 2:
F−Ma=MaB
Step 3:
aB=MF−Ma
Final answer: MF−Ma
Q7Single correctRotational Motion
A cord is wound round the circumference of wheel of radius r, The axis of the wheel is horizontal and the moment of inertia about it is I. A weight mg is attached to the cord at the end. The weight falls from rest. After falling through a distance h, the square of angular velocity of wheel will be
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1I+mr22mgh
Approach:
Apply energy conservation: the loss in gravitational potential energy equals the combined translational kinetic energy of the weight and rotational kinetic energy of the wheel, with the constraint v equals omega r.
Step 1:
mgh=21m(ωr)2+21Iω2
Step 2:
mgh=21ω2(I+mr2)
Step 3:
ω2=I+mr22mgh
Final answer: I+mr22mgh
Q8Single correctProperties of Solids and Liquids
The length of metallic wire is l1 when tension in it is T1. It is l2 when the tension is T2. The original length of the wire will be :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3T2−T1T2l1−T1l2
Approach:
Express each stretched length using Hooke's law about the natural length, then eliminate the elastic constant between the two equations.
Step 1:
l1=l0(1+AYT1),l2=l0(1+AYT2)
Step 2:
T2l1−T1l2=l0(T2−T1)
Step 3:
l0=T2−T1T2l1−T1l2
Final answer: T2−T1T2l1−T1l2
Q9Single correctKinetic Theory of Gases
The internal energy (U), pressure (P) and volume (V) of an ideal gas are related as U=3PV+4. The gas is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2polyatomic only
Approach:
Compare the given relation with the kinetic-theory expression for internal energy to read off the degrees of freedom.
Step 1:
U=3PV+4⇒2fPV=3PV
Step 2:
2f=3⇒f=6
Step 3:
f=6⇒polyatomic (3 translational + 3 rotational)
Final answer: polyatomic only
Q10Single correctOscillations and Waves
Given below are two statements: Statement I: A second's pendulum has a time period of 1 second. Statement II: It takes precisely one second to move between the two extreme positions. In the light of the above statements, choose the correct answer from the options given below
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1Statement I is false but Statement II is true
Approach:
Recall the definition of a second's pendulum and the meaning of motion between extreme positions.
Step 1:
T=2s=1s
Step 2:
extreme to extreme=2T=22=1s
Final answer: Statement I is false but Statement II is true
Q11Single correctOscillations and Waves
A particle executes S.H.M., the graph of velocity as a function of displacement is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4an ellipse
Approach:
Write the velocity-displacement relation for simple harmonic motion and recognise its conic form.
Step 1:
v2=ω2(A2−x2)
Step 2:
A2x2+ω2A2v2=1
Final answer: an ellipse
Q12Single correctOscillations and Waves
A tuning fork A of unknown frequency produces 5 beats s−1 with a fork of known frequency 340 Hz. When fork A is filed, the beat frequency decreases to 2 beats s−1. What is the frequency of fork A ?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1335 Hz
Approach:
Use the two possible frequencies from the initial beat count and decide using the effect of filing, which raises a fork's frequency.
Step 1:
fA=345Hz or 335Hz
Step 2:
filing raises fA;beats drop 5→2
Step 3:
fA=335→338⇒∣338−340∣=2
Final answer: 335 Hz
Q13Single correctElectrostatics
Given below are two statements Statement I : An electric dipole is placed at the centre of a hollow sphere. The flux of electric field through the sphere is zero, but the electric field is not zero anywhere in the sphere. Statement II : If R is the radius of a solid metallic sphere and Q be the total charge on it. The electric field at any point on the spherical surface of radius r(<R) is zero but the electric flux passing through this closed spherical surface of radius r is not 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:
Apply Gauss's law to each described situation and compare the enclosed charge with the claims about field and flux.
Step 1:
qenc=0 (dipole)⇒Φ=0,E=0
Step 2:
charge resides on surface R;qenc(r<R)=0⇒Φ=0
Step 3:
Statement II claims flux =0⇒false
Final answer: Statement I is true but Statement II is false
Q14Single correctElectromagnetic Induction and Alternating Currents
An aeroplane, with its wings spread 10 m, is flying at a speed of 180 km h−1 in a horizontal direction. The total intensity of earth's field at that part is 2.5×10−4 Wb m−2 and the angle of dip is 60∘. The emf induced between the tips of the plane wings will be
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1108.25 mV
Approach:
Only the vertical component of the earth's field contributes to the motional emf across the horizontal wings; compute it from the dip angle.
Step 1:
v=180km h−1=50m s−1
Step 2:
BV=2.5×10−4sin60∘=2.165×10−4Wb m−2
Step 3:
ε=(2.165×10−4)(10)(50)=0.10825V
Step 4:
ε=108.25mV
Final answer: 108.25 mV
Q15Single correctElectromagnetic Induction and Alternating Currents
Find the peak current and resonant frequency of the following circuit (as shown in figure).
(A)
(B)
(C)
(D)
SolutionAnswer: Option 40.2 A and 50 Hz
Approach:
Compute the inductive and capacitive reactances at the driving angular frequency, find the impedance and hence the peak current, then evaluate the resonant frequency from L and C.
Step 1:
XL=100×0.1=10Ω,XC=100×100×10−61=100Ω
Step 2:
Z=1202+(10−100)2=14400+8100=150Ω
Step 3:
I0=15030=0.2A
Step 4:
f0=2π0.1×100×10−61≈50Hz
Final answer: 0.2 A and 50 Hz
Q16Single correctOptics
Given below are two statements : one is labeled as Assertion A and the other is labeled as Reason R. Assertion A: For a simple microscope, the angular size of the object equals the angular size of the image. Reason R: Magnification is achieved as the small object can be kept much closer to the eye than 25 cm and hence it subtends a large angle. In the light of the above statements, choose the most appropriate answer from the options given below:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4Both A and R are true and R is the correct explanation of A
Approach:
Evaluate the assertion and reason about angular magnification of a simple microscope.
Step 1:
A simple microscope produces a virtual, erect, magnified image; the eye sees the image and the object subtends the same angle at the eye as the image.
Step 2:
The object is placed within the focal length, closer than the least distance of distinct vision 25cm, so it subtends a larger angle, giving angular magnification.
Final answer: Both A and R are true and R is the correct explanation of A
Q17Single correctOptics
The incident ray, reflected ray and the outward drawn normal are denoted by the unitvectors a,b and c respectively. Then choose the correct relation for these vectors.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3b=a−2(a⋅c)c
Approach:
Apply the vector law of reflection about the surface normal.
Step 1:
Reflection reverses the normal component of the incident unit vector while keeping the tangential component unchanged.
Step 2:
b=a−2(a⋅c)c with c the outward normal.
Final answer: b=a−2(a⋅c)c
Q18Single correctAtoms and Nuclei
The recoil speed of a hydrogen atom after it emits a photon in going from n=5 state to n=5 state will be
(A)
(B)
(C)
(D)
SolutionAnswer: Option 14.17m s−1
Approach:
Use conservation of momentum: the photon momentum equals the recoil momentum of the atom for the n=5→n=1 transition (stem prints 5→5, a paper typo).
Step 1:
E=13.6(1−251)=13.056eV=2.09×10−18J
Step 2:
v=mcE=(1.67×10−27)(3×108)2.09×10−18
Final answer: 4.17m s−1
Q19Single correctAtoms and Nuclei
A radioactive sample is undergoing α decay, At any time t1, its activity is A and another time t2, the activity is 5A. What is the average life time for the sample ?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3ln5t2−t1
Approach:
Relate activities at two times by exponential decay and express mean life as the reciprocal of the decay constant.
Step 1:
A/5A=eλ(t2−t1)⇒5=eλ(t2−t1)
Step 2:
λ=t2−t1ln5⇒τ=λ1=ln5t2−t1
Final answer: ln5t2−t1
Q20Single correctElectronic Devices
Draw the output signal Y in the given combination of gates.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1(see figure)
Approach:
Read input waveforms A and B, pass A through the NAND gate, OR with B, then invert to get Y.
Step 1:
A is high during 0 to 2s and 3 to 4s; B is high during 1 to 2s and 3 to 4s.
Step 2:
The two-input NAND with both inputs A acts as NOT A; OR with B then inverted by the final NOT gives Y=A+B=A⋅B.
Step 3:
A=1,B=0 during 0 to 1s and 2 to 3s.
Final answer: Output waveform of option (1)
Q21NumericalGravitation
In the reported figure of earth, the value of acceleration due to gravity is same at point A and C but it is smaller than that of its value at point B (surface of the earth). The value of OA : AB will be x:5. The value of x is ______
SolutionAnswer: 4
Approach:
Set the gravitational acceleration inside the earth at A equal to that above the surface at C, with B on the surface, using R=6400 km and C at 3200 km above the surface.
Step 1:
gROA=g(R+h)2R2 with R=6400 km, h=3200 km, R+h=9600 km.
1 mole of rigid diatomic gas performs a work of 5Q when heat Q is supplied to it. The molar heat capacity of the gas during this transformation is 8xR. The value of x is [R universal gas constant ]
SolutionAnswer: 25
Approach:
Apply the first law with W=Q/5 for one mole of a rigid diatomic gas and express the molar heat capacity.
Step 1:
ΔU=Q−W=Q−5Q=54Q
Step 2:
54CΔT=25RΔT⇒C=825R
Step 3:
8xR=825R⇒x=25
Final answer: 25
Q23NumericalThermodynamics
The volume V of a given mass of monoatomic gas changes with temperature T according to the relation V=KT32. The workdone when temperature changes by 90 K will be xR. The value of x is [R universal gas constant ]
SolutionAnswer: 60
Approach:
Express pressure from the ideal gas law using V=KT2/3 and integrate P\,dV over the temperature change.
Step 1:
P=VnRT=KT2/3nRT=KnRT1/3; dV=32KT−1/3dT
Step 2:
W=∫KnRT1/3⋅32KT−1/3dT=32nRΔT
Step 3:
W=32(1)R(90)=60R⇒x=60
Final answer: 60
Q24NumericalOscillations and Waves
A particle executes S.H.M. with amplitude A and time period T. The displacement of the particle when its speed is half of maximum speed is 2xA. The value of x is
SolutionAnswer: 3
Approach:
Use the SHM velocity relation with speed equal to half the maximum speed to find the displacement.
Step 1:
ωA2−y2=2ωA⇒A2−y2=4A2
Step 2:
y2=43A2⇒y=23A
Step 3:
2xA=23A⇒x=3
Final answer: 3
Q25NumericalOscillations and Waves
Time period of a simple pendulum is T. The time taken to complete 85 oscillations starting from mean position is 12αT. The value of α is ______
SolutionAnswer: 7
Approach:
Track the pendulum bob from the mean position through successive extreme and mean crossings to cover five-eighths of an oscillation.
Step 1:
From the mean position, a half oscillation (mean to mean) takes 2T, reaching mean again after passing one extreme.
Step 2:
The remaining one-eighth of an oscillation from the mean covers a quarter swing toward the next extreme; the bob travels to amplitude and the total covers 85 of the motion in time 127T.
Step 3:
12αT=127T⇒α=7
Final answer: 7
Q26NumericalElectrostatics
27 similar drops of mercury are maintained at 10 V each. All these spherical drops combine into a single big drop. The potential energy of the bigger drop is ______ times that of a smaller drop.
SolutionAnswer: 243
Approach:
Relate the big drop radius and charge to a small drop, then compare electrostatic self-energies.
Step 1:
n=27⇒R=271/3r=3r; total charge Q=27q.
Step 2:
Ubig=8πε0(3r)(27q)2=3729⋅8πε0rq2=243Usmall
Final answer: 243
Q27NumericalOptics
A point source of light S, placed at a distance 60 cm infront of the centre of a plane mirror of width 50 cm, hangs vertically on a wall. A man walks infront of the mirror along a line parallel to the mirror at a distance 1.2 m from it (see in the figure). The distance between the extreme points where he can see the image of the light source in the mirror is ______ cm
SolutionAnswer: 150
Approach:
Locate the image behind the mirror and use similar triangles through the mirror edges projected onto the man's line of motion.
Step 1:
Image S' lies 60cm behind the mirror; the man is 120cm in front, so S' to man's line =60+120=180cm.
Step 2:
Each mirror edge is 25cm from the axis at 60cm from S'; spread on the man's line =25×60180=75cm each side.
Step 3:
Total span =2×75=150cm
Final answer: 150
Q28NumericalDual Nature of Matter and Radiation
Two stream of photons, possessing energies equal to twice and ten times the work function of metal are incident on the metal surface successively. The value of ratio of maximum velocities of the photoelectrons emitted in the two respective cases is x:3. The value of x is
SolutionAnswer: 1
Approach:
Apply the photoelectric equation to each photon energy and take the ratio of resulting maximum velocities.
Step 1:
KE1=2ϕ−ϕ=ϕ; KE2=10ϕ−ϕ=9ϕ
Step 2:
v2v1=9ϕϕ=31
Step 3:
x:3=1:3⇒x=1
Final answer: 1
Q29NumericalElectronic Devices
The zener diode has a Vz=30 V. The current passing through the diode for the following circuit is ______ mA.
SolutionAnswer: 9
Approach:
Hold the load voltage at the zener voltage, find the series and load currents, and take their difference for the diode current.
Step 1:
Load (5 kΩ) voltage =30V, so Iload=500030=6mA
Step 2:
Iseries=400090−30=400060=15mA
Step 3:
Idiode=15−6=9mA
Final answer: 9
Q30NumericalElectromagnetic Waves
If the highest frequency modulating a carrier is 5 kHz, then the number of AM broadcast stations accommodated in a 90 kHz bandwidth are
SolutionAnswer: 9
Approach:
Find the bandwidth needed per AM station (twice the highest modulating frequency) and divide the total bandwidth by it.
Step 1:
BWstation=2×5kHz=10kHz
Step 2:
N=10kHz90kHz=9
Final answer: 9
Chemistry30 questions
Q31Single correctClassification of Elements and Periodicity in Properties
The correct order of electron gain enthalpy is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3S>Se>Te>O
Approach:
Rank the group-16 elements by the magnitude (most negative) of their electron gain enthalpy, accounting for the anomalously low value of oxygen.
Step 1:
Oxygen is small, so added-electron repulsion lowers its electron gain enthalpy below the lower congeners.
Step 2:
Down the group from S: S>Se>Te
Step 3:
S>Se>Te>O
Final answer: S>Se>Te>O
Q32Single correctp-Block Elements
Which pair of oxides is acidic in nature?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4B2O3,SiO2
Approach:
Classify each listed oxide as acidic, basic or amphoteric and find the pair where both members are acidic.
Step 1:
CaO and BaO are basic oxides of alkaline earth metals.
Step 2:
B2O3,SiO2,N2O5 are acidic non-metal oxides.
Step 3:
Both members acidic only in B2O3,SiO2
Final answer: B2O3,SiO2
Q33Single correctp-Block Elements
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A : In TlI3, isomorphous to CsI3, the metal is present in +1 oxidation state. Reason R : Tl metal has fourteen f electrons in its electronic configuration. In the light of the above statements, choose the most appropriate answer from the options given below:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4Both A and R are correct but R is NOT the correct explanation of A
Approach:
Evaluate the truth of the assertion and the reason independently, then test whether the reason explains the assertion.
Step 1:
TlI3 is actually Tl+(I3)−, a triiodide; thallium is +1.
Step 2:
Tl:[Xe]4f145d106s26p1
Step 3:
The +1 state arises from the triiodide ion / inert pair effect, not from the f electrons.
Final answer: Both A and R are correct but R is NOT the correct explanation of A
Q34Single correctChemical Bonding and Molecular Structure
SolutionAnswer: Option 2DRAWN: tetralin bearing OH on the aromatic ring (top), Cl on the saturated-ring carbon (bottom-left) and CH2Cl on the aromatic ring (bottom)
Approach:
Apply the selective action of SOCl2: it converts alcoholic (aliphatic and benzylic) hydroxyl groups to chlorides but leaves phenolic hydroxyl groups unchanged.
Step 1:
The phenolic -OH on the aromatic ring is unreactive toward SOCl2.
Step 2:
The benzylic -CH2OH is converted to -CH2Cl.
Step 3:
The secondary aliphatic -OH on the saturated ring is converted to -Cl.
Final answer: DRAWN: tetralin bearing OH on the aromatic ring (top), Cl on the saturated-ring carbon (bottom-left) and CH2Cl on the aromatic ring (bottom)
Q43Single correctHydrocarbons
Considering the above reaction, the major product among the following is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4DRAWN: benzene ring bearing a −CH2CH3 group (ethylbenzene)
Approach:
Step 1 (Zn/HCl, Clemmensen) reduces the ketone carbonyl to a methylene; step 2 (Cr2O3, 773 K, 10-20 atm) is catalytic reductive aromatization (dehydrocyclization) of the resulting alkane.
Step 1:
The ketone (C2H5)2CH-CO-CH2CH3 is 3-ethylhexan-... a C8 ketone.
Step 2:
Clemmensen: C=O→CH2⇒3-ethylhexane (C8H18).
Step 3:
Cr2O3,773K: the six-carbon main chain cyclizes and aromatizes; the ethyl branch remains as a substituent.
Final answer: DRAWN: benzene ring bearing a −CH2CH3 group (ethylbenzene)
SolutionAnswer: Option 2DRAWN: benzene fused to a seven-membered carbocyclic ring carrying a −CHO on a ring double bond (a benzo-fused cycloheptene carbaldehyde)
Approach:
The substrate has two ortho -CH2CH2CHO chains; dilute NaOH promotes an intramolecular aldol condensation between the two aldehyde chains.
Step 1:
The α-carbon of one -CH2CH2CHO chain attacks the carbonyl of the other chain.
Step 2:
Dehydration of the aldol gives an α,β-unsaturated aldehyde within the new ring.
Step 3:
Counting the two aromatic junction carbons plus the five chain carbons closes a seven-membered ring fused to benzene, bearing a -CHO on the double bond.
Final answer: DRAWN: benzene fused to a seven-membered carbocyclic ring carrying a −CHO on a ring double bond (a benzo-fused cycloheptene carbaldehyde)
Track each functional group transformation: crossed Cannizzaro on the aldehyde, Williamson etherification of the benzylic alcohol, then HI cleavage of both ethers.
Step 1:
Ar-CHO+HCHO+NaOH→Ar-CH2OH+HCOONa
Step 2:
Ar-CH2OHNaH, CH3CH2BrAr-CH2OCH2CH3
Step 3:
CH3O-ArHI,ΔHO-Ar+CH3I
Step 4:
Ar-CH2OCH2CH3HI,ΔAr-CH2I
Final answer: 4-hydroxybenzyl iodide (HO-C6H4-CH2I, para)
Identify the monosaccharide units released on hydrolysis of each disaccharide.
Step 1:
Sucrose hydrolyses to α−D−Glucose and β−D−Fructose.
Step 2:
Lactose hydrolyses to β−D−Galactose and β−D−Glucose.
Step 3:
Maltose hydrolyses to two α−D−Glucose units.
Final answer: (a)→(ii),(b)→(i),(c)→(iii)
Q50Single correctBiomolecules
Seliwanoff test and Xanthoproteic test are used for the identification of ___ and ___ respectively.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2ketoses, proteins
Approach:
Match each named test to the species it detects.
Step 1:
Seliwanoff test gives a rapid red colour with ketoses, distinguishing them from aldoses.
Step 2:
Xanthoproteic test gives a yellow colour with proteins bearing aromatic residues.
Final answer: ketoses, proteins
Q51NumericalSome Basic Concepts in Chemistry
The NaNO3 weighed out to make 50 mL of an aqueous solution containing 70.0 mg Na+ per mL isg. (Rounded off to the nearest integer) [Given : Atomic weight in gmol−1− Na : 23; N : 14; O : 16]
SolutionAnswer: 13
Approach:
Find total sodium ions, convert to moles, then to mass of sodium nitrate.
Step 1:
m(Na+)=70.0mg/mL×50mL=3500mg=3.5g
Step 2:
n(Na+)=233.5=0.1522mol
Step 3:
M(NaNO3)=23+14+3(16)=85g mol−1
Step 4:
m(NaNO3)=0.1522×85=12.93g
Final answer: 13
Q52NumericalAtomic Structure
A ball weighing 10 g is moving with a velocity of 90 m s−1. If the uncertainty in its velocity is 5%, then the uncertainty in its position is _ ×10−33 m. (Rounded off to the nearest integer) [Given: h=6.63×10−34 Js]
SolutionAnswer: 1
Approach:
Apply the Heisenberg uncertainty principle with the velocity uncertainty taken as 5% of the speed.
Step 1:
Δv=0.05×90=4.5m s−1,m=0.01kg
Step 2:
Δx=4πmΔvh=4π(0.01)(4.5)6.63×10−34
Step 3:
Δx=0.56556.63×10−34=1.17×10−33m
Final answer: 1
Q53NumericalChemical Thermodynamics
The average S−F bond energy in kJ mol−1 of SF6 is _. (Rounded off to the nearest integer) [Given : The values of standard enthalpy of formation of SF6(g), S(g) and F(g) are −1100, 275 and 80 kJ mol−1 respectively.]
SolutionAnswer: 309
Approach:
Form SF6 from gaseous atoms; the enthalpy of that step equals minus six S-F bond energies.
Step 1:
S(g)+6F(g)→SF6(g)
Step 2:
ΔH=−1100−[275+6(80)]=−1100−755=−1855kJ mol−1
Step 3:
ES-F=61855=309.2kJ mol−1
Final answer: 309
Q54NumericalEquilibrium
The pH of ammonium phosphate solution, if pka of phosphoric acid and pkb of ammonium hydroxide are 5.23 and 4.75 respectively, is
SolutionAnswer: 7
Approach:
Ammonium phosphate is a salt of a weak acid and a weak base; apply the salt-of-weak-acid-weak-base pH relation.
Step 1:
pH=7+21(5.23−4.75)
Step 2:
pH=7+21(0.48)=7+0.24=7.24
Final answer: 7
Q55NumericalRedox Reactions and Electrochemistry
In mildly alkaline medium, thiosulphate ion is oxidized by MnO4− to "A". The oxidation state of sulphur in "A" is_
SolutionAnswer: 6
Approach:
Identify the oxidation product of thiosulphate by permanganate in mildly alkaline medium and assign the oxidation state of sulphur in it.
Step 1:
S2O32−MnO4−,mildly alkalineSO42−
Step 2:
In SO42−: x+4(−2)=−2⇒x=+6
Final answer: 6
Q56NumericalSome Basic Concepts in Chemistry
The number of octahedral voids per lattice site in a lattice is_. (Rounded off to the nearest integer)
SolutionAnswer: 1
Approach:
Recall the standard count of octahedral voids relative to lattice points in close packing.
Step 1:
In a close-packed lattice the number of octahedral voids equals the number of lattice points.
Final answer: 1
Q57NumericalSolutions
When 12.2 g of benzoic acid is dissolved in 100 g of water, the freezing point of solution was found to be −0.93∘C (Kf(H2O)=1.86 K kg mol−1). The number (n) of benzoic acid molecules associated (assuming 100% association) is _.
SolutionAnswer: 2
Approach:
Find the van't Hoff factor from the freezing point depression, then deduce the association number for complete association.
Step 1:
m=0.112.2/122=1mol kg−1
Step 2:
i=KfmΔTf=1.86×10.93=0.5
Step 3:
0.5=1−1(1−n1)=n1⇒n=2
Final answer: 2
Q58NumericalRedox Reactions and Electrochemistry
Emf of the following cell at 298 K in V is x×10−2 Zn∣Zn2+(0.1M)∥Ag+(0.01M)∣Ag The value of x is _ (Rounded off to the nearest integer) [Given : EZn2+/Znθ=−0.76 V; EAg+/Agθ=+0.80 V; F2.303RT=0.059]
SolutionAnswer: 147
Approach:
Compute the standard cell potential, then apply the Nernst equation with n = 2 for the Zn-Ag cell.
If the activation energy of a reaction is 80.9 kJ mol−1, the fraction of molecules at 700 K, having enough energy to react to form products is e−x. The value of x is (Rounded off to the nearest integer) [Use R=8.31 J K−1mol−1]
SolutionAnswer: 14
Approach:
The fraction of molecules with energy above the activation energy is the Boltzmann factor; identify x as the exponent.
Step 1:
x=RTEa=8.31×70080900
Step 2:
x=581780900=13.9
Final answer: 14
Q60NumericalCoordination Compounds
The number of stereo isomers possible for [Co(ox)2(Br)(NH3)]2− is _. [ox = oxalate]
SolutionAnswer: 3
Approach:
Treat the complex as an octahedral M(AA)2bc type with two bidentate oxalates and two different monodentate ligands, and count geometrical and optical isomers.
Step 1:
The complex is of type [M(AA)2bc] with AA = oxalate, b = Br, c = NH3.
Step 2:
The trans arrangement of b and c is achiral, giving one isomer; the cis arrangement is chiral, giving a pair of enantiomers.
Step 3:
Total stereoisomers =1+2=3.
Final answer: 3
Mathematics30 questions
Q61Single correctPermutations and Combinations
A natural number has prime factorization given by n=2x3y5z, where y and z are such that y+z=5 and y−1+z−1=65,y>z. Then the number of odd divisors of n, including 1, is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 112
Approach:
Determine y and z from the two given relations, then count odd divisors from the odd part of the factorization.
Step 1:
yzy+z=65⇒yz5=65⇒yz=6
Step 2:
y+z=5,yz=6,y>z⇒y=3,z=2
Step 3:
n=2x3352;odd part=3352
Step 4:
odd divisors=(3+1)(2+1)=12
Final answer: 12
Q62Single correctSequence and Series
The sum of the series ∑n=1∞(2n+1)!n2+6n+10 is equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4841e−819e−1−10
Approach:
Express the numerator in terms of (2n+1) and (2n) so each term reduces to standard series for e and e−1.
If 0<a,b<1, and tan−1a+tan−1b=4π, then the value of (a+b)−(2a2+b2)+(3a3+b3)−(4a4+b4)+… is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4loge2
Approach:
Recognize each bracketed expansion as the logarithm series and use the inverse-tangent sum condition.
Step 1:
Series=ln(1+a)+ln(1+b)=ln((1+a)(1+b))
Step 2:
1−aba+b=tan4π=1⇒a+b=1−ab
Step 3:
(1+a)(1+b)=1+a+b+ab=1+1=2
Step 4:
ln((1+a)(1+b))=ln2
Final answer: loge2
Q64Single correctCo-ordinate Geometry
If the locus of the mid-point of the line segment from the point (3,2) to a point on the circle, x2+y2=1 is a circle of radius r, then r is equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 421
Approach:
Parametrize the midpoint and substitute the circle constraint to obtain the locus.
Step 1:
Let point on circle (α,β),α2+β2=1;midpoint (h,k)=(23+α,22+β)
Step 2:
(2h−3)2+(2k−2)2=1
Step 3:
(h−23)2+(k−1)2=41
Final answer: 21
Q65Single correctCo-ordinate Geometry
Let A(1,4) and B(1,−5) be two points. Let P be a point on the circle ((x−1))2+(y−1)2=1, such that (PA)2+(PB)2 have maximum value, then the points, P,\ A and B lie on
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2a straight line
Approach:
Write the sum of squared distances using the midpoint of AB and maximize over the circle to locate P.
Step 1:
M=(1,−21);PA2+PB2=2PM2+2AB2
Step 2:
PA2+PB2=2[(x−1)2+(y+21)2]+281
Step 3:
(x−1)2+(y−1)2=1⇒(x−1)2=1−(y−1)2; maximum of the expression occurs at x=1, giving P=(1,2)
Step 4:
A=(1,4),B=(1,−5),P=(1,2) all satisfy x=1
Final answer: a straight line
Q66Single correctLimit, Continuity and Differentiability
Let f(x) be a differentiable function at x=a with f′(a)=2 and f(a)=4. Then x→alimx−axf(a)−af(x) equals:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 34−2a
Approach:
The limit is a 0/0 form; rewrite to expose the derivative definition or apply L'Hopital.
Step 1:
Numerator at x=a:af(a)−af(a)=0,form 00
Step 2:
x−axf(a)−af(x)=f(a)−a⋅x−af(x)−f(a)
Step 3:
limx→a=f(a)−af′(a)=4−2a
Final answer: 4−2a
Q67Single correctSets, Relations and Functions
Let F1(A,B,C)=(A∧∼B)∨[∼C∧(A∨B)]∨∼A and F2(A,B)=(A∨B)∨(B→∼A) be two logical expressions. Then :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2F1 is not a tautology but F2 is a tautology
Approach:
Simplify each expression using De Morgan and the implication identity, then test for tautology.
Step 1:
F2=(A∨B)∨(∼B∨∼A)=A∨B∨∼B∨∼A
Step 2:
B∨∼B≡T⇒F2≡T
Step 3:
F1 at A=T,B=T,C=T:(T∧F)∨[F∧T]∨F=F
Final answer: F1 is not a tautology but F2 is a tautology
Q68Single correctMatrices and Determinants
Consider the following system of equations: x+2y−3z=a 2x+6y−11z=b x−2y+7z=c where a,b and c are real constants. Then the system of equations :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3has infinite number of solutions when 5a=2b+c
Approach:
Evaluate the coefficient determinant; since it vanishes, find the consistency condition on the constants.
Step 1:
12126−2−3−117=0
Step 2:
5(R1)=2(R2)+(R3) in coefficients
Step 3:
Consistency requires 5a=2b+c
Final answer: has infinite number of solutions when 5a=2b+c
Q69Single correctPermutations and Combinations
Let A={1,2,3,…,10} and f:A→A be defined as f(k)={k+1kif k is oddif k is even Then the number of possible functions g:A→A such that gof=f is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4105
Approach:
Determine the image of f, force g to fix those elements, and count free choices for the rest.
Q71Single correctLimit, Continuity and Differentiability
Let f:R→R be defined as f(x)=⎩⎨⎧2sin(−2πx),∣ax2+x+b∣,sin(πx),if x<−1if −1≤x≤1if x>1 If f(x) is continuous on R, then a+b equals :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4−1
Approach:
Match left and right limits to the middle branch at the junction points x=−1 and x=1.
Step 1:
At x=−1:2sin(2π)=2=∣a−1+b∣
Step 2:
At x=1:sin(π)=0=∣a+1+b∣⇒a+b+1=0
Step 3:
Check ∣a+b−1∣=∣−1−1∣=2
Final answer: −1
Q72Single correctLimit, Continuity and Differentiability
The triangle of maximum area that can be inscribed in a given circle of radius ′r′ is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1An equilateral triangle having each of its side of length 3r.
Approach:
The maximum-area inscribed triangle is equilateral; relate its side to the circumradius.
Step 1:
Maximum area inscribed triangle is equilateral
Step 2:
R=3a=r⇒a=3r
Final answer: An equilateral triangle having each of its side of length 3r.
Q73Single correctIntegral Calculus
For x>0, if f(x)=∫1x(1+t)logetdt, then f(e)+f(e1) is equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 221
Approach:
Substitute t→1/u in f(1/e) and combine with f(e) to obtain a clean integral.
Step 1:
f(e1)=∫11/e1+tlntdtt=1/u∫1eu(1+u)lnudu
Step 2:
f(e)+f(e1)=∫1elnt(1+t1+t(1+t)1)dt=∫1etlntdt
Step 3:
∫1etlntdt=[2(lnt)2]1e=21
Final answer: 21
Q74Single correctDifferential Equations
Let f(x)=∫0xetf(t)dt+ex be a differentiable function for all x∈R. Then f(x) equals :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 42e(ex−1)−1
Approach:
Differentiate the integral equation to get a first-order linear ODE, use the initial value at x=0.
Step 1:
f(0)=∫00(…)+e0=1
Step 2:
f′(x)=exf(x)+ex=ex(f(x)+1)
Step 3:
∫f+1df=∫exdx⇒ln(f+1)=ex+C
Step 4:
f(0)=1⇒ln2=1+C⇒C=ln2−1;f(x)+1=2eex−1
Final answer: 2e(ex−1)−1
Q75Single correctIntegral Calculus
Let A1 be the area of the region bounded by the curves y=sinx,y=cosx and y-axis in the first quadrant. Also, let A2 be the area of the region bounded by the curves y=sinx,y=cosx,x-axis and x=2π in the first quadrant. Then,
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2A1:A2=1:2 and A1+A2=1
Approach:
Use the intersection at x=π/4 to integrate the bounded regions for A1 and A2.
Step 1:
A1=∫0π/4(cosx−sinx)dx=2−1
Step 2:
A2=∫0π/4sinxdx+∫π/4π/2cosxdx=2−2
Step 3:
A1+A2=(2−1)+(2−2)=1
Step 4:
A2A1=2(2−1)2−1=21
Final answer: A1:A2=1:2 and A1+A2=1
Q76Single correctDifferential Equations
Let slope of the tangent line to a curve at any point P(x,y) be given by xxy2+y. If the curve intersects the line x+2y=4 at x=−2, then the value of y, for which the point (3,y) lies on the curve, is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3−1918
Approach:
Solve the Bernoulli differential equation from the given slope, apply the intersection condition to find the constant, then evaluate at the required abscissa.
Step 1:
dxdy−xy=y2
Step 2:
v=y1,dxdv+xv=−1
Step 3:
xv=−2x2+C⇒y1=−2x+xC
Step 4:
x=−2,y=3:31=1−2C⇒C=34
Step 5:
x=3:y1=−23+94=−1819
Final answer: −1918
Q77Single correctVector Algebra
If vectors a1=xi^−j^+k^ and a2=i^+yj^+zk^ are collinear, then a possible unit vector parallel to the vector xi^+yj^+zk^ is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 431(i^−j^+k^)
Approach:
Use the collinearity condition to express the components, identify the resultant vector, and normalise it.
Step 1:
x=λ,−1=λy,1=λz
Step 2:
λ=−1:x=−1,y=1,z=−1
Step 3:
xi^+yj^+zk^=−i^+j^−k^
Step 4:
u^=31(−i^+j^−k^)=−31(i^−j^+k^)
Final answer: 31(i^−j^+k^)
Q78Single correctThree Dimensional Geometry
Let L be a line obtained from the intersection of two planes x+2y+z=6 and y+2z=4. If point P(α,β,γ) is the foot of perpendicular from (3,2,1) on L, then the value of 21(α+β+γ) equals:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1102
Approach:
Parametrise the line of intersection, impose perpendicularity from the external point, solve for the parameter, and evaluate the foot.
Step 1:
z=t,y=4−2t,x=−2+3t
Step 2:
(3(−5+3t)−2(2−2t)+(t−1))=0
Step 3:
t=710
Step 4:
α=716,β=78,γ=710,α+β+γ=734
Step 5:
21⋅734=102
Final answer: 102
Q79Single correctThree Dimensional Geometry
If the mirror image of the point (1,3,5) with respect to the plane 4x−5y+2z=8 is (α,β,γ), then 5(α+β+γ) equals :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 247
Approach:
Apply the reflection formula across the plane using the normal vector and the signed value of the point.
Step 1:
4(1)−5(3)+2(5)−8=−9
Step 2:
a2+b2+c2=16+25+4=45
Step 3:
(α,β,γ)=(1,3,5)+4518(4,−5,2)=(513,1,529)
Step 4:
α+β+γ=547,5(α+β+γ)=47
Final answer: 47
Q80Single correctStatistics and Probability
A seven digit number is formed using digits 3,3,4,4,4,5,5. The probability, that number so formed is divisible by 2, is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 273
Approach:
Count total distinct arrangements of the multiset and favourable arrangements ending in an even digit.
Step 1:
2!3!2!7!=210
Step 2:
2!2!2!6!=90
Step 3:
P=21090=73
Final answer: 73
Q81NumericalSequence and Series
Let α and β be two real numbers such that α+β=1 and αβ=−1. Let pn=(α)n+(β)n, pn−1=11 and pn+1=29 for some integer n≥1. Then, the value of pn2 is ______.
SolutionAnswer: 324
Approach:
Use the linear recurrence satisfied by power sums of the roots of x2−x−1=0.
Step 1:
α,β are roots of x2−x−1=0
Step 2:
29=pn+11
Step 3:
pn2=182=324
Final answer: 324
Q82NumericalComplex Numbers and Quadratic Equations
Let z be those complex numbers which satisfy ∣z+5∣≤4 and z(1+i)+zˉ(1−i)≥−10, i=−1. If the maximum value of ∣z+1∣2 is α+β2, then the value of (α+β) is
SolutionAnswer: 48
Approach:
Translate the constraints into a disk and a half-plane in the Cartesian plane, then maximise the squared distance from (−1,0).
Step 1:
(x+5)2+y2≤16,x−y≥−5
Step 2:
maximise (x+1)2+y2
Step 3:
circle ∩ line y=x+5: 2(x+5)2=16⇒x=−5−22,y=−22
Step 4:
(−4−22)2+(22)2=32+162
Step 5:
α+β=48
Final answer: 48
Q83NumericalPermutations and Combinations
The total number of 4-digit numbers whose greatest common divisor with 18 is 3 is ______.
SolutionAnswer: 1000
Approach:
Numbers with gcd=3 with 18=2⋅32 are odd multiples of 3 that are not divisible by 9; count such 4-digit numbers.
Step 1:
count n∈[1000,9999] with 3∣n,n odd, 9∤n
Step 2:
direct enumeration yields 1000
Final answer: 1000
Q84NumericalSequence and Series
If the arithmetic mean and the geometric mean of the pth and qth terms of the sequence −16,8,−4,2,… satisfy the equation 4x2−9x+5=0, then p+q is equal to ______.
SolutionAnswer: 10
Approach:
Identify the GP, take the roots of the quadratic as the AM and GM values, and find the term indices that produce them.
Step 1:
roots x=45,x=1
Step 2:
t4=2,t6=21
Step 3:
p+q=4+6=10
Final answer: 10
Q85NumericalCo-ordinate Geometry
Let L be a common tangent line to the curves 4x2+9y2=36 and (2x)2+(2y)2=31. Then the square of the slope of the line L is ______.
SolutionAnswer: 3
Approach:
Write the tangent-line condition for the ellipse and for the circle, then equate the two expressions for the intercept squared.
Step 1:
9x2+4y2=1,c2=9m2+4
Step 2:
x2+y2=431,c2=431(1+m2)
Step 3:
9m2+4=431(1+m2)⇒5m2=15
Final answer: 3
Q86NumericalStatistics and Probability
Let X1,X2,…,X18 be eighteen observations such that ∑i=118(Xi−α)=36 and ∑i=118(Xi−β)2=90, where α and β are distinct real numbers. If the standard deviation of these observations is 1, then the value of ∣α−β∣ is ______.
SolutionAnswer: 4
Approach:
Express the mean via the first sum, expand the second sum about the mean, and use the variance to solve for the offset.
Step 1:
Xˉ=α+2
Step 2:
∑(Xi−Xˉ)2=18⋅1=18
Step 3:
90=18+18(Xˉ−β)2⇒(Xˉ−β)2=4
Step 4:
α+2−β=±2⇒α−β=−4
Final answer: 4
Q87NumericalMatrices and Determinants
If the matrix A=10302000−1 satisfies the equation A20+αA19+βA=100040001 for some real numbers α and β, then β−α is equal to ______.
SolutionAnswer: 4
Approach:
Compute the relevant powers of the matrix entrywise and match against the right-hand side to solve for the scalars.
Step 1:
A20+αA19+βA=diag-liketarget
Step 2:
solving the entry equations gives α=−2,β=2
Step 3:
β−α=2−(−2)=4
Final answer: 4
Q88NumericalDifferential Equations
Let the normals at all the points on a given curve pass through a fixed point (a,b). If the curve passes through (3,−3) and (4,−22), given that a−22b=3, then (a2+b2+ab) is equal to ______.
SolutionAnswer: 9
Approach:
A curve all of whose normals pass through one point is a circle centred at that point; impose equal distances to the two given points and the linear condition.
Step 1:
equidistance from center (a,b)
Step 2:
solving with a−22b=3 gives a=3,b=0
Step 3:
a2+b2+ab=9+0+0=9
Final answer: 9
Q89NumericalLimit, Continuity and Differentiability
Let a be an integer such that all the real roots of the polynomial 2x5+5x4+10x3+10x2+10x+10 lie in the interval (a,a+1). Then, ∣a∣ is equal to ______.
SolutionAnswer: 2
Approach:
Locate the real root by sign analysis of the polynomial and identify the unit interval containing it.
Step 1:
the polynomial has a single real root
Step 2:
root ≈−1.232∈(−2,−1)
Step 3:
∣a∣=2
Final answer: 2
Q90NumericalIntegral Calculus
If Im,n=∫01xm−1(1−x)n−1dx, for m,n≥1, and ∫01(1+x)m+nxm−1+xn−1dx=αIm,n, α∈R, then α equals ______.
SolutionAnswer: 1
Approach:
Recognise Im,n as the Beta function and apply the standard identity reducing the given integral to it.
Step 1:
the substitution x→t1−t maps the integral to B(m,n)
How many questions are in the JEE Main 2021 February 26, Shift 2 paper?
The JEE Main 2021 February 26, 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.
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Yes — every question shows the correct answer (answer key) and a detailed step-by-step solution, free to read online.
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