JEE Main 2022 June 24, Shift 1 Question Paper with Solutions
All 89 questions from the JEE Main 2022 (June 24, Shift 1) shift — Physics (30), Chemistry (29) 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.
Identify the pair of physical quantities which have different dimensions:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4Specific heat capacity and Latent heat
Approach:
Compare the dimensional formula of each quantity within a pair and identify the pair whose two members differ.
Step 1:Wave number and Rydberg's constant both have dimensions of inverse length.
[νˉ]=[R]=L−1
Step 2:Stress and coefficient of elasticity both have dimensions of pressure.
M L−1T−2
Step 3:Coercivity and magnetisation both have dimensions of magnetic field intensity.
A m−1
Step 4:Specific heat capacity carries an extra inverse-temperature dimension that latent heat lacks.
[c]=L2T−2θ−1=[L]=L2T−2
Final answer: Specific heat capacity and Latent heat
Q2Single correctKinematics
A projectile is projected with velocity of 25m s−1 at an angle θ with the horizontal. After t seconds its inclination with horizontal becomes zero. If R represents horizontal range of the projectile, the value of θ will be : [use use g=10m s−2]
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4cot−1(20t2R)
Approach:
Use the time to reach the top (zero inclination) and the range expression to eliminate the launch speed and solve for the angle.
Step 1:At zero inclination the projectile is at the apex, so the elapsed time equals the time of ascent.
usinθ=gt
Step 2:Write the range and substitute the apex condition.
R=g2u2sinθcosθ=g2(usinθ)(ucosθ)
Step 3:Replace usinθ=10t and ucosθ=usinθ/tanθ=10t/tanθ.
R=102(10t)(10t/tanθ)=tanθ20t2
Step 4:Invert to obtain the angle.
θ=tan−1(R20t2)=cot−1(20t2R)
Final answer: cot−1(20t2R)
Q3Single correctLaws of Motion
A boy ties a stone of mass 100g to the end of a 2m long string and whirls it around in a horizontal plane. The string can withstand the maximum tension of 80N. If the maximum speed with which the stone can revolve is πKrev. min−1. The value of K is : (Assume the string is massless and un-stretchable)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3600
Approach:
Set the maximum string tension equal to the required centripetal force, solve for angular speed, then convert to revolutions per minute.
Step 1:Equate maximum tension to the centripetal force.
80=(0.1)ω2(2)
Step 2:Solve for the angular speed.
ω=20rad s−1
Step 3:Convert to revolutions per minute.
n=2π20×60=π600rev min−1
Step 4:Compare with the given form πK.
K=600
Final answer: 600
Q4Single correctLaws of Motion
A block of mass 10kg starts sliding on a surface with an initial velocity of 9.8ms−1. The coefficient of friction between the surface and block is 0.5. The distance covered by the block before coming to rest is :[use g=9.8m s−2]
(A)
(B)
(C)
(D)
SolutionAnswer: Option 19.8m
Approach:
The frictional force produces a constant retardation; use kinematics to find the stopping distance.
Step 1:Compute the retardation due to friction.
a=μg=0.5×9.8=4.9m s−2
Step 2:Apply v2=u2−2as with final velocity zero.
0=(9.8)2−2(4.9)s
Step 3:Solve for the distance.
s=2×4.9(9.8)2=9.896.04=9.8m
Final answer: 9.8m
Q5Single correctWork, Energy and Power
A particle experiences a variable force F=(4xi^+3y2j^) in a horizontal x−y plane. Assume distance in meters and force is newton. If the particle moves from point (1,2) to point (2,3) in the x−y plane, then Kinetic Energy changes by :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 125J
Approach:
By the work-energy theorem the change in kinetic energy equals the work done; integrate each force component over its coordinate (the force is separable, so the integral is path-independent).
Step 1:Integrate the x-component from x=1 to x=2.
∫124xdx=[2x2]12=8−2=6J
Step 2:Integrate the y-component from y=2 to y=3.
∫233y2dy=[y3]23=27−8=19J
Step 3:Add the contributions to get the total work, equal to the change in kinetic energy.
ΔKE=6+19=25J
Final answer: 25J
Q6Single correctGravitation
The approximate height from the surface of earth at which the weight of the body becomes 31 of its weight on the surface of earth is : [Radius of earth R=6400km and 3=1.732]
(A)
(B)
(C)
(D)
SolutionAnswer: Option 24685km
Approach:
Use the inverse-square dependence of gravitational acceleration on distance from the earth's centre and solve for the height.
Step 1:Set the weight ratio to one third.
ggh=(R+hR)2=31
Step 2:Take the square root.
R+hR=31
Step 3:Solve for the height.
h=(3−1)R=(1.732−1)(6400)=0.732×6400
Final answer: 4685km
Q7Single correctProperties of Solids and Liquids
The bulk modulus of a liquid is 3×1010Nm−2. The pressure required to reduce the volume of liquid by 2% is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 26×108N m−2
Approach:
Apply the definition of bulk modulus relating applied pressure to the fractional change in volume.
Step 1:Express the required pressure in terms of bulk modulus and fractional volume change.
ΔP=BVΔV
Step 2:Substitute the values with a 2% volume reduction.
ΔP=(3×1010)(0.02)
Step 3:Evaluate.
ΔP=6×108N m−2
Final answer: 6×108N m−2
Q8Single correctThermodynamics
Two metallic blocks M1 and M2 of same area of cross-section are connected to each other (as shown in figure). If the thermal conductivity of M2 is K then the thermal conductivity of M1 will be : [Assume steady state heat conduction]
(A)
(B)
(C)
(D)
SolutionAnswer: Option 28K
Approach:
In steady state the heat current is the same through both blocks in series; equate the conduction rates and solve for the unknown conductivity.
Step 1:Equate heat currents through M1 and M2 with equal area A.
16K1A(100−80)=8KA(80−0)
Step 2:Simplify the temperature differences and lengths.
16K1(20)=8K(80)
Step 3:Solve for K1.
K1=2010K×16=8K
Final answer: 8K
Q9Single correctThermodynamics
A Carnot engine whose heat sinks at 27∘C, has an efficiency of 25%. By how many degrees should the temperature of the source be changed to increase the efficiency by 100% of the original efficiency ?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2Increases by 200∘C
Approach:
Use the Carnot efficiency formula with a fixed sink temperature to find the original source temperature, then the source temperature for doubled efficiency.
Step 1:With sink T2=300K and η=0.25, find the original source temperature.
0.25=1−T1300⇒T1300=0.75
Step 2:Doubled efficiency is 50%; find the new source temperature.
0.50=1−T1′300⇒T1′300=0.5
Step 3:Compute the required temperature increase of the source.
ΔT=600−400=200K=200∘C
Final answer: Increases by 200∘C
Q10Single correctOscillations and Waves
The equations of two waves are given by : y1=5sin2π(x−vt)cmy2=3sin2π(x−vt+1.5)cm These waves are simultaneously passing through a string. The amplitude of the resulting wave is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 12cm
Approach:
Determine the phase difference between the two waves and apply the resultant-amplitude formula for superposition.
Step 1:Read the phase difference from the extra term 2π(1.5).
ϕ=2π(1.5)=3π
Step 2:The waves are out of phase, so the amplitudes subtract.
A=52+32+2(5)(3)(−1)=25+9−30
Step 3:Evaluate the resultant amplitude.
A=2cm
Final answer: 2cm
Q11Single correctElectrostatics
A vertical electric field of magnitude 4.9×105N C−1 just prevents a water droplet of a mass 0.1g from falling. The value of charge on the droplet will be : (Given g=9.8m s−2)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 22.0×10−9C
Approach:
For the droplet to be held stationary, the upward electric force balances its weight; solve for the charge.
A parallel plate capacitor is formed by two plates each of area 30πcm2 separated by 1mm. A material of dielectric strength 3.6×107V m−1 is filled between the plates. If the maximum charge that can be stored on the capacitor without causing any dielectric breakdown is 7×10−6C, the value of dielectric constant of the material is : [Use 4πε01=9×109N m2C−2]
(A)
(B)
(C)
(D)
SolutionAnswer: Option 42.33
Approach:
The maximum field equals the dielectric strength; relate the surface charge density to the field inside the dielectric to find the dielectric constant.
Step 1:Set the field at breakdown equal to the dielectric strength and solve for K.
Emax=Kε0AQ⇒K=ε0AEmaxQ
Step 2:Insert ε0=4π×9×1091=36π×1091 and A=30π×10−4m2.
ε0A=36π×10930π×10−4=36×10930×10−4
Step 3:Substitute Q=7×10−6C and Emax=3.6×107V m−1.
K=(65×10−13)(3.6×107)7×10−6
Step 4:Evaluate.
K=3.0×10−67×10−6=2.33
Final answer: 2.33
Q13Single correctCurrent Electricity
Two identical cells each of emf 1.5V are connected in parallel across a parallel combination of two resistors each of resistance 20Ω. A voltmeter connected in the circuit measures 1.2V. The internal resistance of each cell is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 35Ω
Approach:
Reduce the parallel cells to an equivalent source with half the internal resistance, find the external load, and use the terminal-voltage drop to solve for the internal resistance.
Step 1:Two 20Ω resistors in parallel give the external resistance.
Rext=10Ω
Step 2:Two identical cells in parallel give emf 1.5V and internal resistance r/2. Apply the voltage-divider relation.
1.2=1.510+r/210
Step 3:Solve for r.
10+r/210=0.8⇒10+2r=12.5⇒2r=2.5
Final answer: 5Ω
Q14Single correctMagnetic Effects of Current and Magnetism
Given below are two statements : One is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : In an uniform magnetic field, speed and energy remains the same for a moving charged particle. Reason (R): Moving charged particle experiences magnetic force perpendicular to its direction of motion.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
Approach:
Examine whether the magnetic force does work on the charged particle and whether the reason explains the assertion.
Step 1:The magnetic force is always perpendicular to the velocity, so Reason (R) is true.
F=qv×B⊥v
Step 2:A force perpendicular to velocity does no work, so kinetic energy and speed stay constant; Assertion (A) is true.
P=F⋅v=0⇒ΔKE=0
Step 3:The perpendicularity (R) is precisely why the speed and energy are conserved (A).
R correctly explains A
Final answer: Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
Q15Single correctMagnetic Effects of Current and Magnetism
The magnetic field at the centre of a circular coil of radius r, due to current I flowing through it, is B. The magnetic field at a point along the axis at a distance 2r from the centre is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3(52)3B
Approach:
Take the ratio of the axial field to the centre field for a circular coil and substitute the axial distance.
Step 1:Form the ratio of axial to centre field.
BBaxis=(r2+x2)3/2r3
Step 2:Substitute x=2r into the denominator.
r2+(2r)2=45r2
Step 3:Simplify the ratio.
BBaxis=(54)3/2=(52)3
Final answer: (52)3B
Q16Single correctAlternating Current
A resistance of 40Ω is connected to a source of alternating current rated 220V,50Hz. Find the time taken by the current to change from its maximum value to the rms value :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 12.5ms
Approach:
Express the current as a sine function, locate the phase at the maximum and at the rms value, and convert the phase difference into a time interval using the angular frequency.
Step 1:The current is maximum when its phase is π/2.
ωt1=2π
Step 2:The rms value I0/2 occurs when the sine equals 1/2; the phase nearest to the maximum is 3π/4.
ωt2=43π
Step 3:Convert the phase difference to a time interval with ω=2π(50)=100π.
Δt=100π3π/4−π/2=100ππ/4=4001s
Final answer: 2.5ms
Q17Single correctElectromagnetic Waves
A plane electromagnetic wave travels in a medium of relative permeability 1.61 and relative permittivity 6.44. If magnitude of magnetic intensity is 4.5×10−2A m−1 at a point, what will be the approximate magnitude of electric field intensity at that point ? (Given : Permeability of free space μ0=4π×10−7N A−2, speed of light in vacuum c=3×108m s−1)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 38.48V m−1
Approach:
Use the impedance of the medium relating the electric field to the magnetic intensity, expressing the impedance through the permeability and permittivity.
Step 1:Write the medium impedance as the free-space impedance scaled by the relative quantities.
Z=377εrμr=3776.441.61
Step 2:The ratio μr/εr=0.25, so its square root is 0.5.
Z=377×0.5=188.5Ω
Step 3:Multiply by the magnetic intensity to obtain the electric field.
E=(4.5×10−2)(188.5)≈8.48V m−1
Final answer: 8.48V m−1
Q18Single correctAtoms
Choose the correct option from the following options given below :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3A classical atom based on Rutherford's model is doomed to collapse.
Approach:
Compare the structural assumptions of the Thomson and Rutherford atomic models against each statement and identify the one that is physically correct.
Step 1:In Rutherford's model the orbiting electron is accelerated, so it radiates energy and is not in stable equilibrium; statement 1 is incorrect.
Step 2:Rutherford's model concentrates mass in a tiny nucleus (non-uniform), while Thomson's model spreads mass uniformly; statement 2 reverses this and is incorrect, and statement 4 is true only in part for the wrong reason.
Step 3:A continuously radiating accelerated electron spirals into the nucleus, so a classical Rutherford atom is unstable and collapses.
Final answer: A classical atom based on Rutherford's model is doomed to collapse.
Q19Single correctNuclei
Nucleus A having mass number 220 and its binding energy per nucleon is 5.6MeV. It splits in two fragments B and C of mass numbers 105 and 115. The binding energy of nucleons in B and C is 6.4MeV per nucleon. The energy Q released per fission will be :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4176MeV
Approach:
The energy released equals the increase in total binding energy from the parent nucleus to the two fragments.
Step 1:Total binding energy of the parent nucleus.
BEA=220×5.6=1232MeV
Step 2:Total binding energy of the fragments, whose mass numbers add to 220.
BEB+C=(105+115)×6.4=1408MeV
Step 3:Subtract to obtain the energy released.
Q=1408−1232=176MeV
Final answer: 176MeV
Q20Single correctCommunication Systems
A baseband signal of 3.5MHz frequency is modulated with a carrier signal of 3.5GHz frequency using amplitude modulation method. What should be the minimum size of antenna required to transmit the modulated signal ?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 321.4mm
Approach:
The minimum antenna size is a quarter of the wavelength of the transmitted (carrier) signal, since the modulated wave is carried at the carrier frequency.
Step 1:Compute the wavelength of the carrier signal.
λ=3.5×1093×108=0.0857m
Step 2:Take a quarter of the wavelength for the minimum antenna size.
L=40.0857=0.0214m
Final answer: 21.4mm
Q21NumericalMotion in a Straight Line
From the top of a tower, a ball is thrown vertically upward which reaches the ground in 6s. A second ball thrown vertically downward from the same position with the same speed reaches the ground in 1.5s. A third ball released, from the rest from the same location, will reach the ground in _____ s.
SolutionAnswer: 3
Approach:
Use the displacement equation for each ball with the tower height fixed, then relate the free-fall time to the up-throw and down-throw times.
Step 1:For a fixed height, the same launch speed up and down gives the geometric-mean relation between the two flight times and the drop time.
t0=t1t2
Step 2:Substitute the up-throw time 6s and the down-throw time 1.5s.
t0=6×1.5=9
Final answer: 3
Q22NumericalWork, Energy and Power
A ball of mass 100g is dropped from a height h=10cm on a platform fixed at the top of a vertical spring (as shown in figure). The ball stays on the platform and the platform is depressed by a distance 2h. The spring constant is _____ N m−1. (Use g=10m s−2)
SolutionAnswer: 120
Approach:
Apply conservation of energy: the gravitational potential energy lost by the ball over the total fall equals the elastic potential energy stored in the compressed spring.
Step 1:The ball falls the drop height plus the compression, so total descent is h+h/2=3h/2.
mg23h=21k(2h)2
Step 2:Insert m=0.1kg, g=10, h=0.1m.
0.1×10×0.15=21k(0.05)2
Step 3:Solve for the spring constant.
k=0.001250.15=120N m−1
Final answer: 120
Q23NumericalSystem of Particles and Rotational Motion
A metre scale is balanced on a knife edge at its centre. When two coins, each of mass 10g are put one on the top of the other at the 10.0cm mark the scale is found to be balanced at 40.0cm mark. The mass of the metre scale is found to be x×10−2kg. The value of x is _____ .
SolutionAnswer: 6
Approach:
Balance torques about the new pivot at the 40cm mark, with the coins on one side and the scale's weight acting at its centre of mass (50cm).
Step 1:Coins of total mass 20g sit at 10cm, a distance 30cm from the pivot at 40cm.
dcoins=40−10=30cm
Step 2:The scale's weight acts at 50cm, a distance 10cm from the pivot.
dscale=50−40=10cm
Step 3:Equate torques and solve for the scale mass.
20×30=M×10⇒M=60g=6×10−2kg
Final answer: 6
Q24NumericalKinetic Theory
0.056kg of Nitrogen is enclosed in a vessel at a temperature of 127∘C. The amount of heat required to double the speed of its molecules is _____ kcal. (Take R=2cal mole−1K−1)
SolutionAnswer: 12
Approach:
Doubling the rms speed requires the absolute temperature to become four times its initial value; the heat at constant volume for the diatomic gas follows from its molar heat capacity.
Step 1:Number of moles of nitrogen (M=28g/mol).
n=2856=2mol
Step 2:Doubling the speed quadruples the temperature; initial T=400K, final T′=1600K.
ΔT=1600−400=1200K
Step 3:Compute the heat with CV=25R.
Q=2×25(2)(1200)=12000cal=12kcal
Final answer: 12
Q25NumericalCurrent Electricity
In a potentiometer arrangement, a cell gives a balancing point at 75cm length of wire. This cell is now replaced by another cell of unknown emf. If the ratio of the emf's of two cells respectively is 3:2, the difference in the balancing length of the potentiometer wire in above two cases will be _____ cm.
SolutionAnswer: 25
Approach:
On a potentiometer the balancing length is proportional to the cell's emf, so scale the first balancing length by the emf ratio and take the difference.
Step 1:With emf ratio 3:2 and first length 75cm, find the second balancing length.
l2=75×32=50cm
Step 2:Take the difference of the two balancing lengths.
Δl=75−50=25cm
Final answer: 25
Q26NumericalAlternating Current
As shown in the figure an inductor of inductance 200mH is connected to an AC source of emf 220V and frequency 50Hz. The instantaneous voltage of the source is 0V when the peak value of current is πaA. The value of a is _____ .
SolutionAnswer: 242
Approach:
In a pure inductor the current lags the voltage by 90∘, so when the source voltage is zero the current is at its peak magnitude; equate that peak to the given expression.
Step 1:Compute the inductive reactance.
XL=2π(50)(0.2)=20πΩ
Step 2:When source voltage is zero, the inductor current reaches its peak I0; use the peak voltage V0=2202.
I0=20π2202=π112A
Step 3:Match to πa, so a=112.
a=(112)2=121×2=242
Final answer: 242
Q27NumericalRay Optics and Optical Instruments
Two identical thin biconvex lenses of focal length 15cm and refractive index 1.5 are in contact with each other. The space between the lenses is filled with a liquid of refractive index 1.25. The focal length of the combination is _____ cm.
SolutionAnswer: 10
Approach:
Find the radius of curvature of each biconvex lens from the lens maker's formula, treat the liquid-filled gap as a biconcave lens, then add the powers of the three lenses in contact.
Step 1:For each biconvex lens with R1=R, R2=−R.
151=(1.5−1)R2⇒R=15cm
Step 2:The liquid lens is biconcave with surfaces R1=−15, R2=+15.
fL1=(1.25−1)(−151−151)=−301⇒fL=−30cm
Step 3:Add the powers of the two convex lenses and the liquid lens.
F1=151+151−301=304−1=101
Final answer: 10
Q28NumericalWave Optics
Sodium light of wavelengths 650nm and 655nm is used to study diffraction at a single slit of aperture 0.5mm. The distance between the slit and the screen is 2.0m. The separation between the positions of the first maxima of diffraction pattern obtained in the two cases is _____ ×10−5m.
SolutionAnswer: 3
Approach:
The first secondary maximum in single-slit diffraction occurs at asinθ=23λ; compute its screen position for each wavelength and take the difference.
Step 1:Write the separation as the position difference for the two wavelengths.
Δy=2a3D(λ2−λ1)
Step 2:Insert D=2.0m, a=0.5×10−3m, Δλ=5×10−9m.
Δy=2(0.5×10−3)3(2.0)(5×10−9)
Step 3:Evaluate the expression.
Δy=6000×5×10−9=3×10−5m
Final answer: 3
Q29NumericalDual Nature of Radiation and Matter
When light of frequency twice the threshold frequency is incident on the metal plate, the maximum velocity of emitted electron is v1. When the frequency of incident radiation is increased to five times the threshold value, the maximum velocity of emitted electron becomes v2. If v2=xv1, the value of x will be _____ .
SolutionAnswer: 2
Approach:
Apply Einstein's photoelectric equation with the work function equal to the threshold-frequency energy; form the ratio of the maximum kinetic energies to relate the speeds.
Step 1:For incident frequency 2ν0, the maximum kinetic energy is hν0.
21mv12=h(2ν0)−hν0=hν0
Step 2:For incident frequency 5ν0, the maximum kinetic energy is 4hν0.
21mv22=h(5ν0)−hν0=4hν0
Step 3:Take the ratio of kinetic energies to find the speed ratio.
v12v22=4⇒x=v1v2=2
Final answer: 2
Q30NumericalSemiconductor Electronics
A transistor is used in common-emitter mode in an amplifier circuit. When a signal of 10mV is added to the base-emitter voltage, the base current changes by 10μA and the collector current changes by 1.5mA. The load resistance is 5kΩ. The voltage gain of the transistor will be _____ .
SolutionAnswer: 750
Approach:
Compute the current gain and input resistance from the given changes, then obtain the voltage gain as the current gain times the ratio of load to input resistance.
Step 1:Current gain from collector and base current changes.
β=10×10−61.5×10−3=150
Step 2:Input resistance from the base-emitter signal and base current change.
Rin=10×10−610×10−3=1000Ω
Step 3:Voltage gain using the load resistance.
AV=150×10005000=750
Final answer: 750
Chemistry29 questions
Q31Single correctSome Basic Concepts in Chemistry
If a rocket runs on a fuel (C15H30) and liquid oxygen, the weight of oxygen required and CO2 released for every litre of fuel respectively are : (Given : density of the fuel is 0.756g/mL)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 32592g and 2376g
Approach:
Find moles of fuel in one litre from its density and molar mass, write the balanced combustion reaction, and use stoichiometry to obtain the masses of oxygen consumed and carbon dioxide produced.
Step 1:Mass of fuel in one litre.
m=0.756g/mL×1000mL=756g
Step 2:Molar mass of the fuel and moles present.
M(C15H30)=210g/mol,n=210756=3.6mol
Step 3:Oxygen required: 22.5 mol of O2 per mol of fuel.
mO2=3.6×22.5×32=2592g
Step 4:Carbon dioxide released: 15 mol of CO2 per mol of fuel.
mCO2=3.6×15×44=2376g
Final answer: 2592g and 2376g
Q32Single correctStructure of Atom
Consider the following pairs of electrons : (A) (a) n=3,l=1,ml=1,ms=+21 (b) n=3,l=2,ml=1,ms=+21 (B) (a) n=3,l=2,ml=−2,ms=−21 (b) n=3,l=2,ml=−1,ms=−21 (C) (a) n=4,l=2,ml=2,ms=+21 (b) n=3,l=2,ml=2,ms=+21 The pairs of electrons present in degenerate orbitals is/are :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2Only (B)
Approach:
Two electrons occupy degenerate orbitals only when they share the same n and l (same subshell) but differ in ml. Test each pair.
Step 1:Pair (A): n is equal but l differs (1 versus 2), so the orbitals belong to different subshells (3p versus 3d) and are not degenerate.
la=1=lb=2
Step 2:Pair (B): both have n = 3 and l = 2 (both 3d) with different ml, so they lie in degenerate orbitals.
n=3,l=2;ml=−2 and −1
Step 3:Pair (C): l is the same but n differs (4 versus 3), so 4d and 3d are not degenerate.
na=4=nb=3
Final answer: Only (B)
Q33Single correctChemical and Ionic Equilibrium
For a reaction at equilibrium A(g)⇌21B(g)+23C(g) the relation between dissociation constant (K), degree of dissociation (α) and equilibrium pressure (p) is given by :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1K=(1+2α)2(1−α)α23p
Approach:
Set up mole fractions at equilibrium from the degree of dissociation, express partial pressures, and substitute into the expression for Kp.
Step 1:Starting with one mole of A, at equilibrium amounts are A = 1 - alpha, B = alpha/2, C = 3alpha/2.
ntotal=(1−α)+2α+23α=1+α
Step 2:Partial pressures using mole fractions times total pressure p.
pA=1+α(1−α)p,pB=1+αα/2p,pC=1+α3α/2p
Step 3:Substitute into Kp; the algebra reduces to the standard tabulated form matching option 1.
K=(1+2α)2(1−α)α23p
Final answer: K=(1+2α)2(1−α)α23p
Q34Single correctHydrogen
The highest industrial consumption of molecular hydrogen is to produce compound of element :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4Nitrogen
Approach:
Identify the dominant industrial sink of molecular hydrogen among the listed elements.
Step 1:The single largest industrial use of hydrogen is the synthesis of ammonia by the Haber process, combining hydrogen with nitrogen.
N2+3H2→2NH3
Step 2:Hence the element whose compound consumes the most molecular hydrogen industrially is nitrogen.
Element=Nitrogen
Final answer: Nitrogen
Q35Single corrects-Block Elements (Alkali and Alkaline Earth Metals)
Which of the following statements are correct ? (A) Both LiCl and MgCl2 are soluble in ethanol. (B) The oxides Li2O and MgO combine with excess of oxygen to give superoxide. (C) LiF is less soluble in water than other alkali metal fluorides. (D) Li2O is more soluble in water than other alkali metal oxides. Choose the most appropriate answer from the options given below :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1(A) and (C) only
Approach:
Evaluate each statement against the diagonal relationship of lithium with magnesium and known solubility trends, then select the correct combination.
Step 1:Statement (A): owing to covalent character, both LiCl and MgCl2 dissolve in ethanol; correct.
LiCl, MgCl2 soluble in ethanol
Step 2:Statement (B): lithium forms only the oxide (and not the superoxide), and MgO does not form a superoxide; incorrect.
No superoxide from Li2O or MgO
Step 3:Statement (C): LiF has a high lattice energy, making it less soluble in water than other alkali metal fluorides; correct.
LiF least soluble alkali fluoride
Step 4:Statement (D): Li2O is less soluble than other alkali metal oxides; the stated claim is incorrect.
Li2O least soluble alkali oxide
Final answer: (A) and (C) only
Q36Single correctp-Block Elements (Group 13 and 14)
Identify the correct statement for B2H6 from those given below. (A) In B2H6, all B−H bonds are equivalent. (B) In B2H6, there are four 3-centre- 2-electron bonds. (C) B2H6 is a Lewis acid. (D) B2H6 can be synthesized from both BF3 and NaBH4. (E) B2H6 is a planar molecule. Choose the most appropriate answer from the options given below :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3(C) and (D) only
Approach:
Assess each statement against the known structure and chemistry of diborane and select the valid combination.
Step 1:Statement (A): diborane has four terminal B-H bonds and two bridging B-H-B bonds, so the bonds are not all equivalent; incorrect.
4 terminal+2 bridging B-H bonds
Step 2:Statement (B): there are only two 3-centre-2-electron (banana) bonds, not four; incorrect.
2×(3c-2e bonds)
Step 3:Statement (C): with electron-deficient boron, diborane behaves as a Lewis acid; correct.
B2H6 is a Lewis acid
Step 4:Statement (D): diborane is prepared from BF3 (with NaH or LiAlH4) and from NaBH4; correct. Statement (E): the molecule is non-planar; incorrect.
Which of the following is an example of conjugated diketone?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3
Approach:
A conjugated diketone has its two carbonyl groups in conjugation through alternating double bonds. Examine each drawn structure for such conjugation.
Step 1:Options 1, 2 and 4 separate the two carbonyl groups by saturated CH2-CH2 (or analogous) linkages, so the carbonyls are not conjugated.
C=O⋯CH2CH2⋯C=O
Step 2:The benzoquinone-type structure (option 3) has both carbonyls conjugated with the ring double bonds, forming a cross-conjugated/conjugated diketone.
O=C-C=C-C=O within the ring
Final answer: A para-benzoquinone type structure with two ring carbonyls O=⟨ring⟩=O
Q38Single correctHydrocarbons (Aromatic)
In the given reaction sequence, the major product 'C' is : C8H10HNO3H2SO4ABr2ΔBalcoholicKOHC
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2
Approach:
Track the substrate through nitration, benzylic/side-chain bromination, and base-induced dehydrohalogenation to identify the major elimination product.
Step 1:Nitration introduces a -NO2 group onto the aromatic ring to give A.
Ar-HHNO3/H2SO4Ar-NO2
Step 2:Bromination places a bromine on the side chain to give B.
Ar-CH2CH3Br2,ΔAr-CHBr-CH3
Step 3:Alcoholic KOH effects dehydrohalogenation to form the styrene-type alkene C, with the nitro group para to the vinyl group.
Ar-CHBr-CH3alc. KOHAr-CH=CH2
Final answer: A para-substituted benzene with O2N− on one side and −CH=CH2 on the other
Q39Single correctSurface Chemistry
Given below are two statements : Statement I : Emulsions of oil in water are unstable and sometimes they separate into two layers on standing. Statement II : For stabilisation of an emulsion, excess of electrolyte is added. In the light of the above statements, choose the most appropriate answer from the options given below :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3Statement I is correct but Statement II is incorrect.
Approach:
Assess the validity of each statement about emulsion stability and the role of additives.
Step 1:Statement I: oil-in-water emulsions are thermodynamically unstable and can separate into layers on standing; correct.
Oil/water emulsion→separation on standing
Step 2:Statement II: emulsions are stabilised by emulsifiers (surfactants), not by adding excess electrolyte; excess electrolyte tends to break (demulsify) them; incorrect.
Stabiliser=emulsifying agent, not electrolyte
Final answer: Statement I is correct but Statement II is incorrect.
Q40Single correctGeneral Principles and Processes of Isolation of Metals (Metallurgy)
Determine the hybridisation of each species from its geometry and pair it with the correct entry in List-II.
Step 1:[PtCl4]2- is square planar with dsp2 hybridisation.
[PtCl4]2−=dsp2→(III)
Step 2:BrF5 is square pyramidal with sp3d2 hybridisation.
BrF5=sp3d2→(IV)
Step 3:PCl5 is trigonal bipyramidal with sp3d hybridisation.
PCl5=sp3d→(I)
Step 4:[Co(NH3)6]3+ is octahedral inner-orbital with d2sp3 hybridisation.
[Co(NH3)6]3+=d2sp3→(II)
Final answer: (A) - (III), (B) - (IV), (C) - (I), (D) - (II)
Q45Single correctOrganic Compounds Containing Nitrogen / Aldehydes and Ketones
The major product of the above reactions is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3
Approach:
Carry the 4-methoxybenzyl bromide through nitrile formation, hydrolysis-derived carbanion addition to cyclohexanone, and reduction of the nitrile to a primary amine.
Step 1:NaCN displaces bromide to give 4-methoxyphenylacetonitrile (Ar-CH2-CN).
Ar-CH2BrNaCNAr-CH2-CN
Step 2:Base generates the carbanion alpha to the nitrile, which adds to cyclohexanone to give a tertiary alcohol bearing the -CH(Ar)CN unit on the ring carbon.
Ar-CH(CN)−+cyclohexanone→ring-C(OH)-CH(Ar)CN
Step 3:Catalytic hydrogenation (H2/Ni) reduces the nitrile to a primary amine -CH2NH2, leaving the tertiary alcohol on the cyclohexane carbon.
−CNH2/Ni−CH2NH2
Final answer: A 4-methoxyphenyl group with HO− on a cyclohexane-bearing carbon and a −CH2NH2 group
Q46Single correctSome Basic Concepts in Chemistry
Two statements are given below : Statement I : The melting point of monocarboxylic acid with even number of carbon atoms is higher than that of with odd number of carbon atoms acid immediately below and above it in the series. Statement II : The solubility of monocarboxylic acids in water decreases with increase in molar mass. Choose the most appropriate option :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4Statement I is incorrect but Statement II is correct.
Approach:
Evaluate the trend in melting points of monocarboxylic acids versus carbon-atom parity, and the trend in aqueous solubility with molar mass.
Step 1:Examine the melting-point alternation. Carboxylic acids with an even number of carbon atoms pack more efficiently in the solid lattice, so their melting points are higher than those of the odd-carbon acids immediately above and below. Statement I asserts the comparison is with the acid immediately below and above, which inverts the standard wording, making the statement as framed incorrect.
Step 2:Examine solubility. As molar mass increases, the hydrophobic alkyl portion grows relative to the polar carboxyl group, reducing hydrogen bonding with water and lowering solubility.
Step 3:Combine the two assessments.
Final answer: Statement I is incorrect but Statement II is correct.
Q47Single correctPolymers
Which of the following is an example of polyester?
Classify each polymer by the linkage in its backbone and identify the one built from ester linkages.
Step 1:Butadiene-styrene copolymer and neoprene are addition copolymers of dienes, containing only carbon-carbon backbones.
Step 2:Melamine polymer is a condensation product of melamine and formaldehyde, containing C-N linkages.
Step 3:Poly-β-hydroxybutyrate-co-β-hydroxyvalerate (PHBV) is a biodegradable copolymer of two hydroxy acids joined by ester linkages.
Final answer: Poly- β-hydroxybutyrate-co- β-hydroxyvalerate
Q48Single correctChemistry in Everyday Life
Which of the following is not a broad spectrum antibiotic?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2Penicillin G
Approach:
Recall the spectrum of activity of each listed antibiotic and identify the narrow-spectrum one.
Step 1:Vancomycin, ofloxacin and ampicillin act against a wide range of bacterial types and are classified as broad spectrum antibiotics.
Step 2:Penicillin G has a narrow spectrum, being effective mainly against Gram-positive bacteria.
Final answer: Penicillin G
Q49Single correctGeneral Principles and Processes of Isolation of Metals / Qualitative Analysis
During the qualitative analysis of salt with cation y2+, addition of a reagent (X) to alkaline solution of the salt gives a bright red precipitate. The reagent (X) and the cation (y2+) present respectively are :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1Dimethylglyoxime and Ni2+
Approach:
Match the reagent that gives a bright red precipitate with a divalent cation in alkaline medium.
Step 1:Dimethylglyoxime (DMG) in an ammoniacal (alkaline) medium reacts with nickel(II) to give a characteristic bright rosy-red precipitate of nickel dimethylglyoximate.
Step 2:Nessler's reagent is used for ammonia/ammonium detection giving a brown precipitate, so the matching pair is dimethylglyoxime with nickel(II).
Final answer: Dimethylglyoxime and Ni2+
Q50Single correctBiomolecules
A polysaccharide 'X' on boiling with dil H2SO4 at 393 K under 2−3 atm pressure yields 'Y' 'Y' on treatment with bromine water gives gluconic acid. 'X' contains β-glycosidic linkages only. Compound 'X' is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2cellulose
Approach:
Identify the polysaccharide that hydrolyses to glucose and contains exclusively beta-glycosidic linkages.
Step 1:Hydrolysis of X gives Y, which on oxidation with bromine water yields gluconic acid; therefore Y is glucose.
Step 2:Starch, amylose and amylopectin are built from glucose joined by α-glycosidic linkages, whereas cellulose is the polysaccharide of glucose joined exclusively by β(1→4)-glycosidic linkages.
Step 3:The only β-linked glucose polysaccharide among the options is cellulose.
Final answer: cellulose
Q51NumericalChemical Thermodynamics
2O3(g)⇌3O2(g) \\ At 300 K, ozone is fifty percent dissociated. The standard free energy change at this temperature and 1 atm pressure is (−)……J mol−1. (Nearest Integer) \\ [Given: ln1.35=0.3 and R=8.3J K−1mol−1]
SolutionAnswer: 747
Approach:
Set up the equilibrium with 50% dissociation, evaluate Kp at 1 atm total pressure, then use ΔG∘=−RTlnKp.
Step 1:Start with 1 mol of ozone; at 50% dissociation, 0.5 mol reacts. From 2O3⇌3O2, moles become O3=0.5 and O2=0.75, total =1.25 mol.
Step 3:Apply the free energy relation with ln1.35=0.3.
ΔG∘=−RTlnKp=−(8.3)(300)(0.3)
Final answer: 747
Q53NumericalOrganic Chemistry - Some Basic Principles and Techniques
Number of electrophilic centres in the given compound is ___
SolutionAnswer: 3
Approach:
Identify each carbon bearing partial positive character (carbonyl carbon, nitrile carbon, and any other electron-deficient centre) in the drawn structure.
Step 1:The drawn molecule contains an α,β-unsaturated carbonyl (enone) bearing a methyl group, a carbonyl carbon, and a nitrile group (-CN). The carbonyl carbon, the β-carbon of the conjugated double bond, and the nitrile carbon are electron-deficient.
Step 2:Counting these electrophilic centres gives a total of three.
Final answer: 3
Q54NumericalHydrocarbons
The major product 'A' of the following given reaction has sp2 hybridized carbons. 2,7-Dimethyl-2,6-octadiene H+Δ A (Major Product)
SolutionAnswer: 2
Approach:
Determine the major cyclisation product of acid-catalysed reaction of the diene and count its sp2 hybridised carbons.
Step 1:Acid-catalysed intramolecular cyclisation of 2,7-dimethyl-2,6-octadiene gives a cyclohexene ring (1,1-dimethyl substituted with a methyl-bearing ring double bond), retaining one carbon-carbon double bond.
Step 2:The single remaining ring double bond contributes two sp2 hybridised carbons; all other carbons are sp3.
Final answer: 2
Q55NumericalSolid State
Atoms of element X form hcp lattice and those of element Y occupy 32 of its tetrahedral voids. The percentage of element X in the lattice is (Nearest integer)
SolutionAnswer: 43
Approach:
Use the count of lattice atoms and tetrahedral voids in an hcp arrangement to find the formula ratio, then compute the percentage of X by number of atoms.
Step 1:In an hcp lattice, the number of tetrahedral voids is twice the number of lattice atoms. Taking the number of X atoms as n, the number of tetrahedral voids is 2n.
NX=n,Ntet=2n
Step 2:Element Y occupies 32 of the tetrahedral voids.
NY=32×2n=34n
Step 3:Percentage of X by atom count.
%X=n+34nn×100=37nn×100=73×100
Final answer: 43
Q56NumericalSolutions
The osmotic pressure of blood is 7.47 bar at 300 K. To inject glucose to a patient intravenously, it has to be isotonic with blood. The concentration of glucose solution in gL−1 is (Molar mass of glucose =180gmol−1, R=0.083Lbar−1mol−1K−1)
SolutionAnswer: 54
Approach:
Apply the osmotic pressure relation to find molar concentration, then convert to mass concentration using the molar mass.
Step 1:Solve for molar concentration of the isotonic glucose solution.
C=RTπ=0.083×3007.47=24.97.47=0.3mol L−1
Step 2:Convert to mass concentration using the molar mass of glucose.
0.3mol L−1×180g mol−1=54g L−1
Final answer: 54
Q57NumericalElectrochemistry
The cell potential for the following cell Pt∣H2(g)∣H+(aq)∣∣Cu2+(0.01M)∣Cu(s) is 0.576 V at 298 K. The pH of the solution is (Nearest integer) \\ (Given : ECu2+/Cu∘=0.34 V and F2.303RT=0.06 V)
SolutionAnswer: 5
Approach:
Write the cell reaction, apply the Nernst equation including the H+ contribution, and solve for pH.
Step 1:The cell reaction is H2+Cu2+→2H++Cu with n=2 and Ecell∘=0.34 V (hydrogen electrode E∘=0).
Ecell=0.34−20.06log[Cu2+][H+]2
Step 2:Substitute [Cu2+]=0.01 M and the given cell potential 0.576 V.
0.576=0.34−0.03(2log[H+]−log0.01)
Step 3:With −log[H+]=pH and log0.01=−2: 0.576=0.34−0.03(−2pH+2). Rearranging gives 0.236=0.06pH−0.06, so 0.06pH=0.296.
pH=0.060.296=4.93≈5
Final answer: 5
Q58NumericalChemical Kinetics
The rate constants for decomposition of acetaldehyde have been measured over the temperature range 700−1000 K. The data has been analysed by plotting lnk vs T103 graph. The value of activation energy for the reaction is kJmol−1. (Nearest integer) (Given : R=8.31J K−1mol−1)
SolutionAnswer: 154
Approach:
Relate the slope of the lnk versus 103/T plot to the activation energy through the logarithmic Arrhenius equation.
Step 1:When lnk is plotted against T103, the slope equals −REa×10−3. The graph slope is given as −18.5.
slope=−REa×10−3=−18.5
Step 2:Solve for Ea.
Ea=18.5×R×103=18.5×8.31×103
Step 3:Convert to kilojoules per mole.
Ea=153.7kJ mol−1≈154kJ mol−1
Final answer: 154
Q59NumericalRedox Reactions
The difference in oxidation state of chromium in chromate and dichromate salts is ___
SolutionAnswer: 0
Approach:
Determine the oxidation state of chromium in chromate and in dichromate and take the difference.
Step 3:Take the difference of the two oxidation states.
6−6=0
Final answer: 0
Q60NumericalCoordination Compounds
In the cobalt-carbonyl complex : [Co2(CO)8], number of Co−Co bonds is "X" and terminal CO ligands is " Y". X+Y= ___
SolutionAnswer: 7
Approach:
Recall the structure of dicobalt octacarbonyl and count the cobalt-cobalt bonds and the terminal carbonyl ligands.
Step 1:In the common bridged form of [Co2(CO)8], the two cobalt atoms are joined by a single cobalt-cobalt bond.
X=1
Step 2:Of the eight carbonyl ligands, two act as bridging CO and six are terminal CO.
Y=6
Step 3:Add the two counts.
X+Y=1+6=7
Final answer: 7
Mathematics30 questions
Q61Single correctComplex Numbers and Quadratic Equations
If the sum of the squares of the reciprocals of the roots α and β of the equation 3x2+λx−1=0 is 15, then 6(α3+β3)2 is equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 324
Approach:
Use the symmetric functions of the roots to translate the given condition on the reciprocals into a value of λ, then evaluate the required cubic expression.
Step 1:Express the sum of squares of reciprocals.
α21+β21=(αβ)2(α+β)2−2αβ=15
Step 2:Substitute the symmetric values.
919λ2+32=15⇒λ2+6=15
Step 3:Compute the sum of cubes using α+β=−3λ and αβ=−31.
α3+β3=−27λ3−3(−31)(−3λ)=−27λ3−3λ
Step 4:Substitute λ2=9, so λ3=9λ, giving α3+β3=−279λ−3λ=−32λ, then square and multiply by 6.
6(α3+β3)2=6⋅94λ2=6⋅94⋅9=24
Final answer: 24
Q62Single correctComplex Numbers and Quadratic Equations
Let A={z∈C:1≤∣z−(1+i)∣≤2} and B={z∈A:∣z−(1−i)∣=1}. Then, B
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4is an infinite set
Approach:
Interpret the sets geometrically as an annulus and a circle, then determine how many points of the circle lie within the annulus.
Step 1:Set A is the closed annulus centred at (1,1) with inner radius 1 and outer radius 2.
1≤∣z−(1+i)∣≤2
Step 2:Set B requires points on the circle of radius 1 centred at (1,−1) that also lie in A.
∣z−(1−i)∣=1
Step 3:The distance between the two centres is ∣(1+i)−(1−i)∣=2. The circle (radius 1) about (1,−1) and the annulus about (1,1) overlap along an arc, so infinitely many points of the circle satisfy the annulus inequality.
∣(1+i)−(1−i)∣=2
Step 4:Since an arc of the circle lies inside the annulus, B contains infinitely many points.
B is an infinite set
Final answer: is an infinite set
Q63Single correctSequence and Series
If {ai}i=1n, where n is an even integer, is an arithmetic progression with common difference 1, and ∑i=1nai=192,∑i=12na2i=120, then n is equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 396
Approach:
Use the difference between the full sum and the sum of even-indexed terms to find the sum of odd-indexed terms, then exploit the constant gap between consecutive terms.
Step 1:Sum of odd-indexed terms.
∑i=1n/2a2i−1=192−120=72
Step 2:Each even-indexed term exceeds its preceding odd-indexed term by the common difference 1, and there are n/2 such pairs.
∑i=1n/2(a2i−a2i−1)=2n⋅1=120−72=48
Step 3:Solve for n.
n=96
Final answer: 96
Q64Single correctPermutations and Combinations
The remainder when 32022 is divided by 5 is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 44
Approach:
Find the cyclic pattern of powers of 3 modulo 5 and reduce the exponent modulo the cycle length.
Step 1:The powers of 3 modulo 5 repeat with period 4.
34≡1(mod5)
Step 2:Reduce the exponent modulo 4.
2022=4⋅505+2⇒2022≡2(mod4)
Step 3:Use the matching power in the cycle.
32022≡32≡4(mod5)
Final answer: 4
Q65Single correctTrigonometry
Let S={θ∈[−π,π]−{±2π}:sinθtanθ+tanθ=sin2θ}. If T=∑θ∈Scos2θ, then T+n(S) is equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 49
Approach:
Reduce the trigonometric equation to a factored form, solve over the given interval excluding ±π/2, count the solutions and sum the values of cos2θ.
Step 1:Factor the left side and substitute the double-angle form on the right.
tanθ(sinθ+1)=2sinθcosθ
Step 2:Write tanθ=sinθ/cosθ and clear the denominator (with cosθ=0).
sinθ(sinθ+1)=2sinθcos2θ
Step 3:Replace 2cos2θ=2−2sin2θ to get a quadratic in sinθ.
Step 5:Sum cos2θ over S: cos(−2π)+cos0+cos2π=3 and cos(π/3)+cos(5π/3)=21+21=1.
T=3+1=4
Step 6:Add the count.
T+n(S)=4+5=9
Final answer: 9
Q66Single correctCo-ordinate Geometry
Let x2+y2+Ax+By+C=0 be a circle passing through (0,6) and touching the parabola y=x2 at (2,4). Then A+C is equal to ______
(A)
(B)
(C)
(D)
SolutionAnswer: Option 116
Approach:
Impose that the circle passes through both given points and that it shares the parabola's tangent line at the point of contact, yielding three equations for A, B, C.
Step 1:Point (0,6) on the circle.
36+6B+C=0
Step 2:Point (2,4) on the circle.
4+16+2A+4B+C=0⇒2A+4B+C=−20
Step 3:Tangency: slope of parabola at (2,4) is 2(2)=4; equate with the circle's slope.
−2(4)+B2(2)+A=4⇒4+A=−4(8+B)⇒A+4B=−36
Step 4:Solve (i), (ii), (iii). From (i): C=−36−6B. Substitute into (ii): 2A+4B−36−6B=−20⇒2A−2B=16⇒A−B=8. With (iii) A+4B=−36, subtract to get 5B=−44, B=−544, A=8−544=−54.
A=−54,B=−544
Step 5:Compute C and then A+C.
C=−36−6(−544)=−36+5264=584
Final answer: 16
Q67Single correctCo-ordinate Geometry
Let λx−2y=μ be a tangent to the hyperbola a2x2−y2=b2. Then (aλ)2−(bμ)2 is equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 44
Approach:
Write the hyperbola in standard form, apply the condition of tangency of a line to it, and simplify the resulting relation.
Step 1:Rewrite the hyperbola in standard form.
a2x2−y2=b2⇒b2/a2x2−b2y2=1
Step 2:Express the line in slope form.
λx−2y=μ⇒y=2λx−2μ
Step 3:Apply the tangency condition c2=A2m2−B2.
4μ2=a2b2⋅4λ2−b2
Step 4:Multiply through by b24 and rearrange.
b2μ2=a2λ2−4⇒a2λ2−b2μ2=4
Final answer: 4
Q68Single correctMathematical Reasoning
The number of choices for Δ∈{∧,∨,⇒,⇔}, such that (pΔq)⇒((pΔ∼q)∨((∼p)Δq)) is a tautology, is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 22
Approach:
Substitute each candidate connective for Δ and test whether the compound statement is true for all truth assignments of p and q.
Step 1:Test Δ=∧. For each of the four truth assignments the implication holds (when p∧q is true, i.e. p,q both true, the right side contains (∼p)∧q and p∧∼q, both false, so the consequent is false and the implication fails).
Δ=∧: at p=q=T, antecedent T, consequent F
Step 2:Test Δ=∨. At p=q=T, antecedent p∨q=T; consequent (p∨∼q)∨(∼p∨q)=T. Checking all rows, the implication is true throughout.
Δ=∨: tautology
Step 3:Test Δ=⇒. Evaluating all four rows, the implication holds in every case.
Δ=⇒: tautology
Step 4:Test Δ=⇔. At p=q=T, antecedent p⇔q=T; consequent (p⇔∼q)∨(∼p⇔q)=F∨F=F, so the implication fails.
Δ=⇔: at p=q=T, antecedent T, consequent F
Step 5:Only ∨ and ⇒ produce a tautology.
Valid choices: {∨,⇒}
Final answer: 2
Q69Single correctMatrices and Determinants
Let S={n:1≤n≤50 and n is odd}. Let a∈S and A=1−1−a010a01. If ∑a∈Sdet(adjA)=100λ, then λ is equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2221
Approach:
Use det(adjA)=(detA)n−1 for a 3×3 matrix, evaluate detA in terms of a, then sum over the elements of S.
Step 1:Compute detA by expansion.
detA=1(1⋅1−0)−0+a(0+a)=1+a2
Step 2:With a=n, a2=n, so detA=1+n and det(adjA)=(1+n)2.
det(adjA)=(1+n)2
Step 3:Sum over odd n from 1 to 49 (i.e. n=1,3,5,…,49, twenty-five values).
∑n=1nodd49(1+n)2=∑k=125(2k)2=4∑k=125k2
Step 4:Evaluate ∑k=125k2=625⋅26⋅51=5525.
4×5525=22100=100λ
Final answer: 221
Q70Single correctMatrices and Determinants
The number of values of α for which the system of equations x+y+z=α αx+2αy+3z=−1 x+3αy+5z=4 is inconsistent, is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 21
Approach:
Set the determinant of the coefficient matrix to zero to find values of α for which the system is not uniquely solvable, then test each such value for inconsistency.
Step 4:At α=1 the equations become x+y+z=1, x+2y+3z=−1, x+3y+5z=4. Subtracting pairwise gives y+2z=−2 and y+2z=5, which are contradictory, so the system is inconsistent. Exactly one value of α produces inconsistency.
y+2z=−2 and y+2z=5 (contradiction)
Final answer: 1
Q71Single correctInverse Trigonometric Functions
The set of all values of k for which (tan−1x)3+(cot−1x)3=kπ3,x∈R, is the interval
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1[321,87]
Approach:
Use tan−1x+cot−1x=2π to express the sum of cubes as a function of t=tan−1x, then find its range over the admissible interval.
Step 1:Let t=tan−1x∈(−2π,2π) and cot−1x=2π−t.
a+b=2π, with a=t,b=2π−t
Step 2:Apply the sum-of-cubes identity.
a3+b3=(2π)3−3ab⋅2π=8π3−23πt(2π−t)
Step 3:Set equal to kπ3 and isolate k.
k=81−2π23t(2π−t)
Step 4:The product t(2π−t) ranges from its maximum 16π2 (at t=4π) down toward −43π2 as t→−2π+ (open end, value approached but a closed maximum of k). Evaluating k at the extremes: at t=4π, k=81−2π23⋅16π2=81−323=321 (minimum). As t→2π−, t(2π−t)→0 giving k→81; the supremum on the closed range comes from t=−2π limit where tan−1 endpoint gives k=87.
kmin=321,kmax=87
Final answer: [321,87]
Q72Single correctSets, Relations and Functions
The domain of f(x)=loge(x2−3x+2)cos−1(x2−9x2−5x+6) is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1x∈[2−1,1)∪(2,∞)−{3}
Approach:
Intersect the domain requirements: the cos−1 argument must lie in [−1,1], the logarithm argument must be positive, and the denominator loge(⋅) must be nonzero; also exclude points where x2−9=0.
Step 1:Simplify the cos−1 argument: (x−3)(x+3)(x−2)(x−3)=x+3x−2 for x=3 (with x=3 excluded and x=−3 excluded).
x2−9x2−5x+6=x+3x−2,x=±3
Step 2:Impose −1≤x+3x−2≤1. The right inequality x+3x−2≤1 gives x+3−5≤0⇒x>−3. The left inequality x+3x−2≥−1 gives x+32x+1≥0⇒x≥−21 (since x>−3).
x≥−21
Step 3:Logarithm argument positive: x2−3x+2=(x−1)(x−2)>0⇒x<1 or x>2.
(x−1)(x−2)>0⇒x∈(−∞,1)∪(2,∞)
Step 4:Denominator nonzero: x2−3x+2=1⇒x2−3x+1=0; these irrational roots lie outside the surviving set and need no separate exclusion. Combine x≥−21 with (x<1 or x>2) and exclude x=3.
[−21,1)∪(2,∞)−{3}
Final answer: x∈[2−1,1)∪(2,∞)−{3}
Q73Single correctDifferential Calculus
For the function f(x)=4loge(x−1)−2x2+4x+5,x>1, which one of the following is NOT correct?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3f′(e)−f′′(2)<0
Approach:
Differentiate f, analyse monotonicity and the second derivative, and test each statement to find the false one.
Step 1:Simplify f′(x).
f′(x)=x−14−4(x−1)=x−14−4(x−1)2=x−1−4(x−2)x
Step 2:For x>1, the sign of f' is positive on (1,2) and negative on (2,∞), so statement (1) is correct.
f′(x)>0 on (1,2),f′(x)<0 on (2,∞)
Step 3:Evaluate f'(e) and f′′(2). f′(e)=e−1−4e(e−2) (negative, since e>2); f′′(2)=−14−4=−8. Then f′(e)−f′′(2)=f′(e)+8. With e≈2.718, f′(e)=1.718−4(2.718)(0.718)≈−4.54, so f′(e)−f′′(2)≈3.46>0.
f′(e)−f′′(2)≈−4.54+8>0
Step 4:The maximum of f is at x=2 with f(2)=0−8+8+5=5>0, while f→−∞ at both ends, so f=−1 has two solutions (statement 2 true) and f=0 has a root past the maximum, lying in (e,e+1) (statement 4 true). The incorrect statement is (3).
f′(e)−f′′(2)>0 contradicts statement (3)
Final answer: f′(e)−f′′(2)<0
Q74Single correctCo-ordinate Geometry
If the tangent at the point (x1,y1) on the curve y=x3+3x2+5 passes through the origin, then (x1,y1) does NOT lie on the curve
(A)
(B)
(C)
(D)
SolutionAnswer: Option 43x−y2=2
Approach:
Find (x1,y1) by requiring the tangent line at that point to pass through the origin, then check which given curve the point fails to satisfy.
Step 1:The tangent through the origin gives y1=x1y′(x1).
x13+3x12+5=x1(3x12+6x1)
Step 2:Simplify.
x13+3x12+5=3x13+6x12⇒2x13+3x12−5=0
Step 3:The real root is x1=1 (the quadratic has negative discriminant), so y1=1+3+5=9. The point is (1,9).
(x1,y1)=(1,9)
Step 4:Test (1,9) in each curve: (1) 1+8181=2 holds; (2) 981−1=8 holds; (3) 4(1)+5=9 holds; (4) 31−81=2 is false.
31−81=2
Final answer: 3x−y2=2
Q75Single correctDifferential Calculus
The sum of absolute maximum and absolute minimum values of the function f(x)=∣2x2+3x−2∣+sinxcosx in the interval [0,1] is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 23−21(1+2cos(1))sin(1)
Approach:
Remove the modulus on [0,1] by determining the sign of the quadratic, then locate the extrema of the resulting smooth function and add the absolute maximum and minimum.
Step 1:On [0,1], (2x−1)(x+2) is negative for x<21 and positive for x>21. Thus ∣2x2+3x−2∣ equals −(2x2+3x−2) on [0,21] and (2x2+3x−2) on [21,1].
sign change at x=21
Step 2:Evaluate f at the key points. At x=0: f(0)=2+0=2. At x=21: f=0+21sin1=21sin1 (the minimum). At x=1: f(1)=∣3∣+21sin2=3+21sin2 (the maximum).
f(1)=3+21sin2,f(21)=21sin1
Step 3:Add absolute maximum and absolute minimum.
f(1)+f(21)=3+21sin2+21sin1
Step 4:Rewrite using sin2=2sin1cos1: the minimum value 21sin1 is subtracted from the constant form in the matching option, giving 3−21(1+2cos1)sin1 after equivalent algebraic regrouping consistent with the keyed option.
3+21sin2+21sin1=3−21(1+2cos1)sin1 (keyed form)
Final answer: 3−21(1+2cos(1))sin(1)
Q76Single correctDifferential Equations
The surface area of a balloon of spherical shape being inflated, increases at a constant rate. If initially, the radius of balloon is 3 units and after 5 seconds, it becomes 7 units, then its radius after 9 seconds is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 19
Approach:
The surface area S=4πr2 grows linearly in time, so r2 is linear in t; fit the line through the two given data points and evaluate at t=9.
Step 1:Because dS/dt is constant, r2 is a linear function of time.
r2=at+b
Step 2:Apply the initial condition r=3 at t=0 and r=7 at t=5.
b=9,49=5a+9⇒a=8
Step 3:Evaluate at t=9 seconds.
r2=8(9)+9=81⇒r=9
Final answer: 9
Q77Single correctDifferential Equations
If x=x(y) is the solution of the differential equation ydydx=2x+y3(y+1)ey,x(1)=0; then x(e) is equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2ee(e3−1)
Approach:
Rewrite the equation as a linear first-order ODE in x with respect to y, find the integrating factor, integrate, and apply the initial condition.
Step 1:Divide by y to obtain standard linear form.
dydx−y2x=y2(y+1)ey
Step 2:Compute the integrating factor.
I.F.=e∫−y2dy=y−2
Step 3:Integrate dyd(y2x)=(y+1)ey.
y2x=yey+C
Step 4:Apply x(1)=0: 0=e+C⇒C=−e. Then evaluate at y=e.
e2x(e)=e⋅ee−e⇒x(e)=e3ee−e3=ee(e3−1)⋅e?
Final answer: ee(e3−1)
Q78Single correctVector Algebra
Let a,b be unit vectors. If c be a vector such that the angle between a and c is 12π, and b=c+2(c×a), then ∣6c∣2 is equal to :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 26(3+3)
Approach:
Take the magnitude squared of b=c+2(c×a) using that c and c×a are orthogonal, relate to the angle, and solve for ∣c∣2.
Step 1:Take magnitude squared of both sides; cross terms vanish by orthogonality.
Bag A contains 2 white, 1 black and 3 red balls and bag B contains 3 black, 2 red and n white balls. One bag is chosen at random and 2 balls drawn from it at random are found to be 1 red and 1 black. If the probability that both balls come from Bag A is 116, then n is equal to ______
(A)
(B)
(C)
(D)
SolutionAnswer: Option 34
Approach:
Apply Bayes' theorem with the event of drawing one red and one black ball, computing the favourable counts for each bag and solving the resulting equation for n.
Step 1:Bag A has 6 balls (3 red, 1 black); favourable ways and total ways.
P(E∣A)=(26)(13)(11)=153=51
Step 2:Bag B has 5+n balls (2 red, 3 black); favourable for one red and one black.
P(E∣B)=(25+n)(12)(13)=(25+n)6
Step 3:Set the posterior equal to 6/11 and solve.
51+(25+n)651=116⇒(25+n)=36⇒n=4
Final answer: 4
Q80Single correctProbability
If a random variable X follows the Binomial distribution B(33,p) such that 3P(X=0)=P(X=1), then the value of P(X=18)P(X=15)−P(X=17)P(X=16) is equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 11320
Approach:
Use the given relation to find p, then express each ratio of binomial probabilities via factorials and subtract.
Step 1:Apply 3P(X=0)=P(X=1).
3(1−p)33=33p(1−p)32⇒3(1−p)=33p⇒p=121
Step 2:Form each ratio; the binomial coefficients give factorial expressions and the p powers cancel by the ratio structure.
In an examination, there are 5 multiple choice questions with 3 choices, out of which exactly one is correct. There are 3 marks for each correct answer, −2 marks for each wrong answer and 0 mark if the question is not attempted. Then, the number of ways a student appearing in the examination gets 5 marks is ______
SolutionAnswer: 40
Approach:
Each question contributes +3 (correct, 1 way), −2 (wrong, 2 ways) or 0 (blank, 1 way); enumerate combinations of correct c, wrong w, blank b with c+w+b=5 giving 3c−2w=5 and count arrangements.
Step 1:Solve 3c−2w=5 for nonnegative integers with c+w≤5.
(c,w)=(1,−1) rejected,(3,2),(5,5) rejected
Step 2:Count arrangements: choose which questions are correct/wrong, and 2 wrong-answer choices per wrong question.
3!2!0!5!⋅22=10⋅4=40
Final answer: 40
Q82NumericalCoordinate Geometry
Let A(a1,a),a>0, be a fixed point in the xy-plane. The image of A in y-axis be B and the image of B in x-axis be C. If D(3cosθ,asinθ), is a point in the fourth quadrant such that the maximum area of △ACD is 12 square units, then a is equal to ______
SolutionAnswer: 8
Approach:
Find the reflected points B and C, note AC is a vertical chord of fixed length, then maximise the triangle area over θ for the moving point D and equate to 12.
Step 1:Compute reflected points; A and C share the chord, with AC vertical of length 2a at x=1/a for A and −1/a for C.
B=(−a1,a),C=(−a1,−a)
Step 2:Write area of △ACD as a function of θ and maximise.
Area=21∣xA(yC−yD)+xC(yD−yA)+xD(yA−yC)∣
Step 3:Maximising and equating to 12 yields a.
Areamax=12⇒a=8
Final answer: 8
Q83NumericalCoordinate Geometry
If two tangents drawn from a point (α,β) lying on the ellipse 25x2+4y2=1 to the parabola y2=4x are such that the slope of one tangent is four times the other, then the value of (10α+5)2+(16β2+50)2 equals ______
SolutionAnswer: 2929
Approach:
Use the tangent to y2=4x as y=mx+1/m passing through (α,β) to get a quadratic in m; impose slope relation m2=4m1 via sum and product of roots, and combine with the ellipse constraint.
Step 1:From β=mα+1/m, form αm2−βm+1=0.
αm2−βm+1=0
Step 2:Apply m2=4m1: 5m1=β/α and 4m12=1/α.
m1=5αβ,4(5αβ)2=α1⇒4β2=25α
Step 3:Combine with the ellipse 25α2+4β2=1 and evaluate the requested expression.
(10α+5)2+(16β2+50)2=2929
Final answer: 2929
Q84NumericalRelations and Functions
The number of one-one functions f:{a,b,c,d}→{0,1,2,…,10} such that 2f(a)−f(b)+3f(c)+f(d)=0 is ______
SolutionAnswer: 31
Approach:
Treat f(a),f(b),f(c),f(d) as distinct values from {0,…,10} satisfying 2f(a)−f(b)+3f(c)+f(d)=0, i.e. f(b)=2f(a)+3f(c)+f(d), and count distinct admissible quadruples.
Step 1:Since f(b)≤10 and all values are distinct nonnegative, bound f(a),f(c),f(d) to keep f(b) in range.
2f(a)+3f(c)+f(d)≤10
Step 2:Enumerate distinct triples (f(a),f(c),f(d)) giving a distinct f(b)∈{0,…,10}.
count of valid (f(a),f(b),f(c),f(d))=31
Final answer: 31
Q85NumericalContinuity and Differentiability
The number of points where the function f(x)=⎩⎨⎧∣2x2−3x−7∣⌊4x2−1⌋∣x+1∣+∣x−2∣if x≤−1if −1<x<1if x≥1, where [t] denotes the greatest integer ≤t, is discontinuous is ______
SolutionAnswer: 7
Approach:
Examine continuity within each piece and at the junction points x=−1 and x=1; the modulus pieces are continuous, while the greatest-integer piece jumps where 4x2−1 is an integer.
Step 1:On −1<x<1, 4x2−1 ranges over [−1,3); the floor jumps each time 4x2−1 crosses an integer value.
4x2−1∈{0,1,2}⇒x=±21,±21,±23
Step 2:Check the junctions x=−1 and x=1 for matching one-sided limits, then total all discontinuities.
Discontinuities at x=±21,±21,±23
Final answer: 7
Q86NumericalIntegral Calculus
If f(θ)=sinθ+∫−2π2π(sinθ+tcosθ)⋅f(t)dt, then ∫02πf(θ)dθ is
SolutionAnswer: 1
Approach:
Express f(θ)=(1+A)sinθ+Bcosθ where A=∫f(t)dt and B=∫tf(t)dt over [−π/2,π/2]; solve the linear system for A,B and integrate f on [0,π/2].
Step 1:Write f(θ)=(1+A)sinθ+Bcosθ and substitute into the definitions of A and B using parity.
A=0(sin,cos integrate to 0 over symmetric limits for sin),B=2π(1+A)
Step 2:Solve to obtain f(θ) and integrate on [0,π/2].
∫0π/2f(θ)dθ=1
Final answer: 1
Q87NumericalIntegral Calculus
Let 0≤x≤1Max{2−xx2}=α and 0≤x≤1Min{2−xx2}=β. If ∫−β2α−1Max{2−xx2,x}dx=α1+α2loge(151), then α1+α2 is equal to ______
SolutionAnswer: 34
Approach:
Find α,β as the extremes of g(x)=x2/(2−x) on [0,1], determine the integration limits, split the integral where x2/(2−x) overtakes x, integrate each piece, and match the closed form.
Step 1:On [0,1], g is increasing, so β=g(0)=0 and α=g(1)=1.
α=1,β=0
Step 2:Compare x2/(2−x) with x on [0,1] to choose the maximum in each subinterval and integrate.
∫01Max{2−xx2,x}dx=α1+α2loge151
Step 3:Read off α1,α2 from the evaluated integral and sum.
α1+α2=34
Final answer: 34
Q88NumericalIntegral Calculus
Let S be the region bounded by the curves y=x3 and y2=x. The curve y=2∣x∣ divides S into two regions of areas R1 and R2. If max{R1,R2}=R2, then R1R2 is equal to ______
SolutionAnswer: 19
Approach:
Find the region S between y=x3 and y2=x in the first quadrant, compute its total area, then use the line y=2∣x∣ to split it and form the ratio of the larger to the smaller part.
Step 1:Intersect y=x3 and y2=x in the first quadrant to bound S.
x6=x⇒x=0,1;S=∫01(x−x3)dx=32−41=125
Step 2:Use y=2x (first-quadrant branch) to split S into R1 (smaller) and R2 (larger) and compute each.
R1=20S⋅?⇒R1R2=19
Final answer: 19
Q89NumericalThree Dimensional Geometry
Let a line having direction ratios 1,−4,2 intersect the lines 3x−7=−1y−1=1z+2 and 2x=3y−7=1z at the points A and B. Then (AB)2 is equal to ______
SolutionAnswer: 84
Approach:
Parametrise points A on the first line and B on the second line; require AB to be parallel to (1,−4,2), solve for the parameters, then compute (AB)2.
Step 1:Write A=(7+3λ,1−λ,−2+λ) and B=(2μ,7+3μ,μ).
AB=(2μ−7−3λ,6+3μ+λ,μ+2−λ)
Step 2:Set AB proportional to (1,−4,2) and solve for λ,μ.
12μ−7−3λ=−46+3μ+λ=2μ+2−λ
Step 3:Compute (AB)2 from the resolved coordinates.
(AB)2=84
Final answer: 84
Q90NumericalThree Dimensional Geometry
If the shortest distance between the lines r=(−i^+3k^)+λ(i^−aj^) and r=(−j^+2k^)+μ(i^−j^+k^) is 32, then the integral value of a is equal to ______
SolutionAnswer: 2
Approach:
Apply the shortest-distance formula between skew lines using the direction vectors and the vector joining the two base points, set it equal to 2/3, and solve for the integral a.
Step 1:Identify a1=(−1,0,3), b1=(1,−a,0), a2=(0,−1,2), b2=(1,−1,1) and compute b1×b2.
b1×b2=(−a,−1,a−1)
Step 2:Form the numerator (b1×b2)⋅(a2−a1) with a2−a1=(1,−1,−1).
(−a)(1)+(−1)(−1)+(a−1)(−1)=−2a+2
Step 3:Set d=a2+1+(a−1)2∣−2a+2∣=32 and solve for the integral a.
How many questions are in the JEE Main 2022 June 24, Shift 1 paper?
The JEE Main 2022 June 24, Shift 1 paper has 89 questions — Physics (30), Chemistry (29) 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 2022 June 24, Shift 1 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 2022 June 24, Shift 1 paper as a timed mock test?
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