JEE Main 2023 January 25, Shift 1 Question Paper with Solutions
All 90 questions from the JEE Main 2023 (January 25, Shift 1) shift — Physics (30), Chemistry (30) and Mathematics (30) — each with the correct answer (answer key) and a step-by-step solution. Read it free online, or attempt the full paper as a timed mock test.
A message signal of frequency 5 kHz is used to modulate a carrier signal of frequency 2 MHz. The bandwidth for amplitude modulation is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 310kHz
Approach:
The bandwidth required for an amplitude modulated signal equals twice the frequency of the modulating signal.
Step 1:Identify the modulating (message) signal frequency.
fm=5kHz
Step 2:Apply the amplitude modulation bandwidth relation.
BW=2×5kHz
Final answer: 10kHz
Q2Single correctDual Nature of Matter and Radiation
Electron beam used in an electron microscope, when accelerated by a voltage of 20 kV, has a de-Broglie wavelength of λ0. If the voltage is increased to 40 kV, then the de-Broglie wavelength associated with the electron beam would be :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 32λ0
Approach:
The de-Broglie wavelength of an accelerated electron varies inversely with the square root of the accelerating voltage.
Step 1:Express the proportionality between wavelength and voltage.
λ∝V1
Step 2:Form the ratio for the doubled voltage.
λ0λnew=40kV20kV=21
Step 3:Solve for the new wavelength.
λnew=2λ0
Final answer: 2λ0
Q3Single correctKinetic Theory of Gases
The root mean square velocity of molecules of gas is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1Proportional to square root of temperature (T)
Approach:
The root mean square speed of gas molecules follows from the kinetic theory expression relating speed to absolute temperature.
Step 1:State the kinetic theory expression for rms speed.
vrms=M3RT
Step 2:Isolate the temperature dependence with all other quantities fixed.
vrms∝T
Final answer: Proportional to square root of temperature (T)
Q4Single correctWave Optics
In Young's double slits experiment, the position of 5th bright fringe from the central maximum is 5 cm. The distance between slits and screen is 1 m and wavelength of used monochromatic light is 600 nm. The separation between the slits is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 260μm
Approach:
The position of the nth bright fringe relates the slit separation to the fringe distance, screen distance and wavelength.
Step 1:List the given quantities.
n=5,y5=5cm,D=1m,λ=600nm
Step 2:Rearrange the fringe position formula to obtain the slit separation.
d=y5nλD=5×10−25×600×10−9×1
Step 3:Convert to micrometres.
d=60μm
Final answer: 60μm
Q5Single correctUnits and Measurements
Choose the correct answer from the options given below :
List I
List II
A.. Surface tension
I..kg m−1s−1
B.. Pressure
II..kg m−1s−2
C.. Viscosity
III..kg s−2
D.. Impulse
IV..kg m s−1
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3A-III, B-II, C-I, D-IV
Approach:
Each physical quantity is reduced to its SI dimensional units and matched against the listed combinations.
Step 1:Surface tension has units of force per length.
N m−1=kg s−2
Step 2:Pressure has units of force per area.
N m−2=kg m−1s−2
Step 3:Coefficient of viscosity has units of pascal second.
Pa s=kg m−1s−1
Step 4:Impulse equals force times time, the unit of momentum.
N s=kg m s−1
Final answer: A-III, B-II, C-I, D-IV
Q6Single correctThermal Properties of Matter
A bowl filled with very hot soup cools from 98∘C to 86∘C in 2 minutes when the room temperature is 22∘C. How long it will take to cool from 75∘C to 69∘C?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 21.4 minutes
Approach:
Newton's law of cooling relates the rate of temperature drop to the difference between the average body temperature and the surroundings.
Step 1:Apply the law to the first interval using the mean temperature.
212=−k[298+86−22]
Step 2:Apply the law to the second interval, where the temperature falls by 6∘C.
dt6=−k[275+69−22]
Step 3:Divide the two relations to eliminate the cooling constant.
26⋅6dt=5070
Final answer: 1.4 minutes
Q7Single correctOscillations
T is the time period of simple pendulum on the earth's surface. Its time period becomes xT when taken to a height R (equal to earth's radius) above the earth's surface. Then, the value of x will be:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 32
Approach:
The pendulum period varies inversely with the square root of gravitational acceleration, which weakens with altitude.
Step 1:Evaluate gravity at a height equal to the earth's radius.
gR=(2R)2Gm=4g
Step 2:Write the period at this height.
TR=2πgRl=2πg/4l
Step 3:Compare with the surface period.
TR=2T
Final answer: 2
Q8Single correctThermodynamics
A Carnot engine with efficiency 50% takes heat from a source at 600 K. In order to increase the efficiency to 70%, keeping the temperature of sink same, the new temperature of the source will be :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 41000K
Approach:
Carnot efficiency fixes the ratio of sink to source temperature; the fixed sink temperature found from the first case gives the new source temperature.
Step 1:Use the 50% efficiency at the source of 600 K to find the sink temperature.
21=1−600Tsink
Step 2:Apply the 70% efficiency with the same sink temperature.
107=1−Tsource300
Step 3:Solve for the new source temperature.
Tsource=1000K
Final answer: 1000K
Q9Single correctNuclei
The ratio of the density of oxygen nucleus (816O) and helium nucleus (24He) is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 31:1
Approach:
Nuclear radius scales with the cube root of the mass number, so the density of nuclear matter is independent of the nucleus.
Step 1:Express the nuclear volume using the radius relation.
V=34πR03A
Step 2:Form the density as mass over volume.
ρ=34πR03AmA
Step 3:Take the ratio for the two nuclei.
ρheliumρoxygen=1
Final answer: 1:1
Q10Single correctLaws of Motion
A car is moving with a constant speed of 20 m/s in a circular horizontal track of radius 40 m. A bob is suspended from the roof of the car by a massless string. The angle made by the string with the vertical will be : (Take g=10 m/s2)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 44π
Approach:
In the rotating frame of the car the bob hangs under gravity and the outward pseudo (centrifugal) force; the string angle balances these.
Step 1:Equate the horizontal and vertical components acting on the bob.
tanθ=Rgv2
Step 2:Substitute the given values.
tanθ=40×10(20)2=400400
Step 3:Solve for the angle.
θ=tan−1(1)=45∘=4π
Final answer: 4π
Q11Single correctLaws of Motion
An object of mass 8 kg is hanging from one end of a uniform rod CD of mass 2 kg and length 1 m pivoted at its end C on a vertical wall as shown in figure. It is supported by a cable AB such that the system is in equilibrium. The tension in the cable is : (Take g=10 m/s2)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3300N
Approach:
Taking torques about the pivot C balances the cable tension against the rod's weight and the suspended load.
Step 1:Set the net torque about the pivot to zero, with the cable making 30 degrees and acting at 0.5 m.
2T×0.6=20×0.5+80×1
Step 2:Evaluate the right-hand side.
T×0.3=10+80=90
Step 3:Solve for the tension.
T=3900=300N
Final answer: 300N
Q12Single correctMotion in a Straight Line
A car travels a distance of 'x' with speed v1 and then same distance 'x' with speed v2 in the same direction. The average speed of the car is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3v1+v22v1v2
Approach:
Average speed equals total distance divided by total time for the two equal-distance segments.
Step 1:Express the total distance and total time.
vavg=v1x+v2x2x
Step 2:Cancel the common distance and simplify.
vavg=v1+v22v1v2
Final answer: v1+v22v1v2
Q13Single correctElectromagnetic Waves
An electromagnetic wave is transporting energy in the negative z direction. At a certain point and certain time the direction of electric field of the wave is along positive y direction. What will be the direction of the magnetic field of the wave at that point and instant?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4Positive direction of x
Approach:
For an electromagnetic wave the direction of energy flow is along the cross product of the electric and magnetic fields.
Step 1:Identify the propagation and electric field directions.
c^=−z^,E^=+y^
Step 2:Require the electric, magnetic and propagation directions to satisfy the cross product.
E^×B^=c^⇒y^×B^=−z^
Step 3:State the magnetic field direction.
B^=+x^
Final answer: Positive direction of x
Q14Single correctCurrent Electricity
A uniform metallic wire carries a current 2 A, when 3.4 V battery is connected across it. The mass of uniform metallic wire is 8.92×10−3 kg, density is 8.92×103 kg/m3 and resistivity is 1.7×10−8Ω−m. The length of wire is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3l=10m
Approach:
The resistance from Ohm's law combines with the resistivity formula, in which the area is replaced using mass and density.
Step 1:Find the resistance of the wire.
R=23.4=1.7Ω
Step 2:Substitute the area in terms of mass and density into the resistance relation.
1.7=(m/d)ρl2=10−61.7×10−8×l2
Step 3:Solve for the length.
l=10m
Final answer: l=10m
Q15Single correctOscillations
In an LC oscillator, if values of inductance and capacitance become twice and eight times, respectively, then the resonant frequency of oscillator becomes x times its initial resonant frequency ω0. The value of x is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 241
Approach:
The resonant frequency of an LC circuit varies inversely with the square root of the inductance-capacitance product.
Step 1:Replace the inductance with twice its value and the capacitance with eight times its value.
ω=2L⋅8C1=4LC1
Step 2:Express the new frequency as a multiple of the original.
ω=4ω0
Final answer: 41
Q16Single correctMagnetic Effects of Current and Magnetism
A solenoid of 1200 turns is wound uniformly in a single layer on a glass tube 2 m long and 0.2 m in diameter. The magnetic intensity at the center of the solenoid when a current of 2 A flows through it is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 31.2×103A m−1
Approach:
The magnetic intensity inside a long solenoid is the product of the number of turns per unit length and the current.
Step 1:Compute the turns per unit length from the total turns and the length of the tube.
n=21200=600m−1
Step 2:Multiply by the current to find the magnetic intensity at the center.
H=nI=600×2=1200A m−1
Final answer: 1.2×103A m−1
Q17Single correctSemiconductor Electronics
Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R Assertion A : Photodiodes are used in forward bias usually for measuring the light intensity. Reason R : For a p-n junction diode, at applied voltage V the current in the forward bias is more than the current in the reverse bias for ∣Vz∣>x≥∣V0∣ where V0 is the threshold voltage and Vz is the breakdown voltage. In the light of the above statements, choose the correct answer from the options given below
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1A is false but R is true
Approach:
Evaluate the operating mode of a photodiode against the stated assertion and the diode current relation in the reason.
Step 1:A photodiode used to measure light intensity is operated in reverse bias, where the reverse current varies linearly with illumination, so the assertion is false.
Step 2:The reason correctly describes that for a p-n junction the forward-bias current exceeds the reverse-bias current in the stated voltage range, so the reason is true.
Final answer: A is false but R is true
Q18Single correctOscillations
Assume that the earth is a solid sphere of uniform density and a tunnel is dug along its diameter throughout the earth. It is found that when a particle is released in this tunnel, it executes a simple harmonic motion. The mass of the particle is 100 g. The time period of the motion of the particle will be (approximately) (Take g=10m s−2, radius of earth =6400km)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 41 hour 24 minutes
Approach:
A particle in a tunnel through the earth experiences a restoring force proportional to displacement, giving simple harmonic motion with period set by the surface gravity and earth radius.
Step 1:Substitute the earth radius and surface gravity into the period expression.
T=2π106400000
Step 2:Evaluate the period in seconds.
T=2π×800=5024s
Step 3:Convert the period to hours and minutes.
5024s≈1h24min
Final answer: 1 hour 24 minutes
Q19Single correctElectrostatics
A parallel plate capacitor has plate area 40cm2 and plates separation 2 mm. The space between the plates is filled with a dielectric medium of a thickness 1 mm and dielectric constant 5. The capacitance of the system is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2310ε0F
Approach:
Treat the partially filled capacitor as a series combination of a vacuum gap and a dielectric slab, using the standard partial-fill capacitance formula.
Step 1:Insert the plate area, total separation, slab thickness and dielectric constant into the partial-fill formula.
C=(2×10−3−1×10−3)+51×10−3ε0(40×10−4)
Step 2:Simplify the numerator and denominator to obtain the capacitance.
C=620ε0=310ε0F
Final answer: 310ε0F
Q20Single correctMagnetic Effects of Current and Magnetism
Choose the correct answer from the options given below :
List I (Current configuration)
List II (Magnitude of Magnetic Field at point P)
A..
I..B0=4πrμ0I[π+2]
B..
II..B0=2rμ0I
C..
III..B0=4πrμ0I[π−1]
D..
IV..B0=4πrμ0I[π+1]
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3A-III, B-I, C-IV, D-II
Approach:
Compute the net magnetic field at the marked point for each current configuration by superposing the contributions of the straight segments and circular arcs, then match to List II.
Step 1:For configuration A, the arc and straight portions combine to give the field proportional to the bracket pi minus one.
B0=4πrμ0I[π−1]
Step 2:For configuration B, the contributions add to the bracket pi plus two.
B0=4πrμ0I[π+2]
Step 3:For configuration C, the field reduces to the bracket pi plus one.
B0=4πrμ0I[π+1]
Step 4:For configuration D, a complete circular loop gives the field of a full circle.
B0=2rμ0I
Final answer: A-III, B-I, C-IV, D-II
Q21NumericalWork, Energy and Power
An object of mass 'm' initially at rest on a smooth horizontal plane starts moving under the action of force F=2N. In the process of its linear motion, the angle θ (as shown in figure) between the direction of force and horizontal varies as θ=kx, where k is a constant and x is the distance covered by the object from its initial position. The expression of kinetic energy of the object will be E=knsinθ, the value of n is ______.
SolutionAnswer: 2
Approach:
Use the work-energy theorem, integrating the horizontal component of the force over the displacement with the angle varying as a linear function of distance.
Step 1:Express the kinetic energy as the integral of the horizontal force component, with the angle equal to k times distance.
E=∫0x2cos(kx)dx
Step 2:Evaluate the integral to obtain the kinetic energy in terms of the angle.
E=k2sinkx=k2sinθ
Step 3:Compare with the given form to read off the numerator constant.
knsinθ=k2sinθ⇒n=2
Final answer: 2
Q22NumericalProperties of Solids
As shown in the figure, in an experiment to determine Young's modulus of a wire, the extension-load curve is plotted. The curve is a straight line passing through the origin and makes an angle of 45∘ with the load axis. The length of wire is 62.8cm and its diameter is 4 mm. The Young's modulus is found to be x×104N m−2. The value of x is ______.
SolutionAnswer: 5
Approach:
Relate Young's modulus to the slope of the extension-load line through the wire geometry, then evaluate with the slope equal to one for a forty-five degree line.
Step 1:Write Young's modulus as the reciprocal slope times the geometric factor of length over cross-sectional area.
Y=slope1⋅πr2l=(slope)π(2×10−3)262.8×10−2
Step 2:Substitute the length and radius with unit slope and evaluate.
Y=(1)×5×104N m−2
Step 3:Compare with the given form to read off the coefficient.
x×104=5×104⇒x=5
Final answer: 5
Q23NumericalElectrostatics
A uniform electric field of 10 N/C is created between two parallel charged plates (as shown in figure). An electron enters the field symmetrically between the plates with a kinetic energy 0.5 eV. The length of each plate is 10 cm. The angle (θ) of deviation of the path of electron as it comes out of the field is ______ (in degree).
SolutionAnswer: 45
Approach:
Find the transverse velocity gained from the electric force over the time of flight inside the plates and compare it with the entry velocity to obtain the deflection angle.
Step 1:Express the tangent of the deflection angle as the transverse velocity from the electric acceleration over the entry velocity, using the time of flight equal to the plate length over the entry velocity.
tanθ=v(meE)T=mv2eER
Step 2:Substitute the field, plate length and kinetic energy, with the electronic charge cancelling against the kinetic energy in electron volts.
tanθ=2×(0.5eV)(e)(10)×(10×10−2)=1
Step 3:Take the inverse tangent to obtain the deflection angle.
θ=45∘
Final answer: 45
Q24NumericalAlternating Current
An LCR series circuit of capacitance 62.5 nF and resistance of 50Ω, is connected to an A.C. source of frequency 2.0 kHz. For maximum value of amplitude of current in circuit, the value of inductance is ______ mH. (Take π2=10)
SolutionAnswer: 100
Approach:
Maximum current amplitude occurs at resonance, where the inductive and capacitive reactances are equal; solve for the inductance.
Step 1:At resonance the inductive reactance equals the capacitive reactance, giving the inductance in terms of angular frequency and capacitance.
ωL=ωC1⇒L=ω2C1
Step 2:Substitute the angular frequency and capacitance, using pi squared equal to ten.
L=(2π×2×103)2×62.5×10−91
Final answer: 100
Q25NumericalCurrent Electricity
In the given circuit, the equivalent resistance between the terminal A and B is ______ Ω.
SolutionAnswer: 10
Approach:
Redraw the network to identify series and parallel combinations, then reduce stepwise to a single equivalent resistance between the terminals.
Step 1:Redraw the circuit so that the three-ohm resistor is in series with the parallel grouping of the remaining resistors.
Step 2:Combine the parallel and series branches to obtain the equivalent resistance between A and B.
RAB=10Ω
Final answer: 10
Q26NumericalWaves
The distance between two consecutive points with phase difference of 60∘ in a wave of frequency 500 Hz is 6.0 m. The velocity with which wave is traveling is ______ km/s.
SolutionAnswer: 18
Approach:
Relate the given path difference and phase difference to the wavelength, then use the wave speed equation with the frequency.
Step 1:Convert the sixty degree phase difference to a fraction of the wavelength and equate to the given separation.
Δx=2πλ(3π)=6λ=6m
Step 2:Multiply the wavelength by the frequency to find the wave speed.
U=fλ=500×36=18000m/s
Final answer: 18
Q27NumericalSystem of Particles and Rotational Motion
ICM is the moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of disc. IAB is its moment of inertia about an axis AB perpendicular to plane and parallel to axis CM at a distance 32R from center. Where R is the radius of the disc. The ratio of IAB and ICM is x:9. The value of x is ______.
SolutionAnswer: 17
Approach:
Apply the parallel axis theorem to shift the perpendicular axis from the center to a point two-thirds of the radius away, then form the requested ratio.
Step 1:Add the shift term for a distance of two-thirds the radius to the central moment of inertia.
IAB=21MR2+M(32R)2=21MR2+94MR2
Step 2:Form the ratio of the shifted moment of inertia to the central moment of inertia.
ICMIAB=1/217/18=917
Step 3:Compare the ratio with x to nine to read off the value of x.
x:9=17:9⇒x=17
Final answer: 17
Q28NumericalRay Optics
A ray of light is incident from air on a glass plate having thickness 3 cm and refractive index 2. The angle of incidence of a ray is equal to the critical angle for glass-air interface. The lateral displacement of the ray when it passes through the plate is ______ ×10−2 cm. (given sin15∘=0.26)
SolutionAnswer: 52
Approach:
Find the critical angle as the angle of incidence, determine the refraction angle inside the plate, then apply the lateral displacement formula for a parallel plate.
Step 1:Set the angle of incidence equal to the critical angle for the glass-air interface.
sini=21⇒i=45∘
Step 2:Apply Snell's law at the air-to-glass entry to find the refraction angle.
sinr=μsini=21⇒r=30∘
Step 3:Substitute into the lateral displacement formula with the given sine of fifteen degrees.
d=cosr3sin(15∘)=233×0.26
Step 4:Express the displacement in the requested units.
d=0.52cm=52×10−2cm
Final answer: 52
Q29NumericalVectors
If P=3i^+3j^+2k^ and Q=4i^+3j^+2.5k^ then, the unit vector in the direction of P×Q is x1(3i^+j^−23k^). The value of x is ______.
SolutionAnswer: 4
Approach:
Evaluate the cross product of the two vectors, take its magnitude, and identify the normalizing factor that produces the stated unit vector.
Step 1:Expand the determinant to obtain the cross product vector.
P×Q=23i^+21j^−3k^
Step 2:Compute the magnitude of the cross product.
∣P×Q∣=43+41+3=2
Step 3:Divide by the magnitude and factor out one quarter to match the given form.
n^=41(3i^+j^−23k^)⇒x=4
Final answer: 4
Q30NumericalAtoms and Nuclei
The wavelength of the radiation emitted is λ0 when an electron jumps from the second excited state to the first excited state of hydrogen atom. If the electron jumps from the third excited state to the second excited state of hydrogen atom, the wavelength of the radiation emitted will be x20λ0. The value of x is ______.
SolutionAnswer: 27
Approach:
Use the Rydberg relation for each transition, form the ratio of the two wavelengths, and equate to the given expression to find x.
Step 1:Write the reciprocal wavelength for the transition from the second excited state to the first excited state, that is from level three to level two.
λ01=R(41−91)=365R
Step 2:Write the reciprocal wavelength for the transition from the third excited state to the second excited state, that is from level four to level three.
λ1=R(91−161)=1447R
Step 3:Take the ratio of the wavelengths and simplify to match the given form.
SolutionAnswer: Option 2A = 2-bromo-4-nitrotoluene; E = 2-bromobenzoic acid
Approach:
Track the functional group transformations of p-nitrotoluene through bromination, reduction, diazotisation and subsequent steps to identify intermediates A and final product E.
Step 1:Bromination of the toluene ring directs ortho to methyl giving the brominated nitrotoluene as A.
Ar-CH3Br2A
Step 2:Sn/HCl reduces the nitro group; diazotisation, deamination and oxidation of the methyl convert the chain to the carboxylic acid bearing bromine, giving E.
KMnO4/KOH
Final answer: A = 2-bromo-4-nitrotoluene; E = 2-bromobenzoic acid
Q32Single correctSolid State
A cubic solid is made up of two elements X and Y. Atoms of X are present on every alternate corner and one at the center of cube. Y is at 31rd of the total faces. The empirical formula of the compound is
(A)
(B)
(C)
(D)
SolutionAnswer: Option
Approach:
Count the net contribution of X from alternate corners plus the body centre and of Y from one third of the faces, then reduce to the empirical formula.
Step 1:Four alternate corners contribute one eighth each and the body centre contributes one, giving the X count.
4×81+1=1.5
Step 2:One third of the six faces, that is two faces, each contribute one half, giving the Y count.
6×31×21=1
Step 3:Combining the counts yields the empirical ratio, which matches none of the printed options.
X1.5Y1=X3Y2
Final answer: No answer is correct
Q33Single correctp-Block Elements
Reaction of thionyl chloride with white phosphorus forms a compound [A], which on hydrolysis gives [B], a dibasic acid. [A] and [B] are respectively
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2PCl3 and H3PO3
Approach:
Determine the product of white phosphorus with thionyl chloride and identify the dibasic acid produced on its hydrolysis.
Step 1:White phosphorus reacts with thionyl chloride to form phosphorus trichloride as compound A.
P4+8SOCl2→4PCl3+4SO2+2S2Cl2
Step 2:Hydrolysis of phosphorus trichloride yields phosphorous acid, a dibasic acid B.
PCl3HydrolysisH3PO3
Final answer: PCl3 and H3PO3
Q34Single correctSome Basic Concepts in Chemistry
'25 volume' hydrogen peroxide means
(A)
(B)
(C)
(D)
SolutionAnswer: Option 41 L marketed solution contains 75 g of H2O2.
Approach:
Convert the volume strength to molarity using the standard relation, then find the mass of hydrogen peroxide in one litre.
Step 1:Divide the volume strength by 11.2 to obtain the molarity.
M=11.225=2.23M
Step 2:Multiply molarity by the molar mass 34 to obtain the mass of hydrogen peroxide in one litre.
2.23×34=75g
Final answer: 1 L marketed solution contains 75 g of H2O2.
Q35Single correctAlcohols, Phenols and Ethers
In the cumene to phenol preparation in presence of air, the intermediate is
Identify the species formed when cumene reacts with air before acid-catalysed rearrangement to phenol and acetone.
Step 1:Aerial oxidation of the benzylic carbon of cumene gives cumene hydroperoxide as the intermediate.
C(CH3)2-O-O-H
Step 2:Acid treatment of this hydroperoxide rearranges it to phenol and acetone.
H+phenol+acetone
Final answer: Cumene hydroperoxide, Ph-C(CH3)2-O-O-H
Q36Single correctAtomic Structure
The radius of the 2nd orbit of Li2+ is x. The expected radius of the 3rd orbit of Be3+ is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 31627x
Approach:
Apply the Bohr radius dependence on principal quantum number squared over nuclear charge to both ions and take their ratio.
Step 1:For the second orbit of lithium ion the radius equals the base radius times four over three, fixing the base radius.
rLi2+=r0×322=x⇒r0=43x
Step 2:For the third orbit of beryllium ion the radius equals the base radius times nine over four.
rBe3+=r0×432
Step 3:Substituting the base radius gives the required radius in terms of x.
rBe3+=43x×432=1627x
Final answer: 1627x
Q37Single correctGeneral Principles of Metallurgy
Which one of the following reactions does not occur during extraction of copper?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1CaO+SiO2→CaSiO3
Approach:
Recall the slag-forming and roasting reactions of the copper extraction process and identify the reaction not part of it.
Step 1:During copper extraction iron(II) oxide is removed as ferrous silicate slag, and copper sulphide and iron sulphide undergo roasting.
FeO+SiO2→FeSiO3
Step 2:Calcium oxide combining with silica to form calcium silicate belongs to other metallurgies and does not occur here.
CaO+SiO2→CaSiO3
Final answer: CaO+SiO2→CaSiO3
Q38Single correctBiomolecules
Correct match is
Row I
Row II
A.
(i).α-D-(-)-Fructofuranose
B.
(ii).β-D-(-)-Fructofuranose
C.
(iii).α-D-(-) Glucopyranose,
D.
(iv).β-D-(-)-Glucopyranose
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1A→iii, B→iv, C→i, D→ii
Approach:
Identify each Haworth structure in Row I as a pyranose or furanose with its anomeric configuration and match to the names in Row II.
Step 1:Structures A and B are six-membered glucopyranose rings differing in anomeric hydroxyl orientation, matching the alpha and beta glucopyranose names.
A→iii, B→iv
Step 2:Structures C and D are five-membered fructofuranose rings, matching the alpha and beta fructofuranose names.
C→i, D→ii
Final answer: A→iii, B→iv, C→i, D→ii
Q39Single correctChemistry in Everyday Life
Which of the following statements is incorrect for antibiotics?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1An antibiotic should promote the growth or survival of microorganisms.
Approach:
Recall the defining properties of antibiotics and identify the statement contradicting them.
Step 1:Antibiotics inhibit the growth or survival of microorganisms rather than promote it, so the first statement is incorrect.
Step 2:The remaining statements correctly describe antibiotics as low-concentration metabolic or synthetic-analogue agents.
Final answer: An antibiotic should promote the growth or survival of microorganisms.
Q40Single corrects-Block Elements
Choose the correct answer from the options given below :
LIST I (Elements)
LIST II (Colour imparted to the flame)
A.K
I.Brick Red
B.Ca
II.Violet
C.Sr
III.Apple Green
D.Ba
IV.Crimson Red
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1A-II, B-I, C-IV, D-III
Approach:
Assign the characteristic flame colour of each metal and match to the list.
Step 1:Potassium gives a violet flame and calcium gives a brick red flame.
Step 2:Strontium gives a crimson red flame and barium gives an apple green flame.
Final answer: A-II, B-I, C-IV, D-III
Q41Single correctHaloalkanes and Haloarenes
The compound which will have the lowest rate towards nucleophilic aromatic substitution on treatment with OH− is
Compare the position of the electron-withdrawing nitro group relative to the halogen in each structure to rank the rate of nucleophilic aromatic substitution.
Step 1:Aryl halides bearing the nitro group at the ortho or para position to the halogen react faster than the meta isomer.
Step 2:The structure with the nitro group meta to the chlorine has the lowest rate, identifying option 2.
Final answer: 1-chloro-3-nitrobenzene (meta isomer)
Q42Single correctHydrocarbons
The correct sequence of reagents for the preparation of Q and R is:
Trace the conversion of the n-alkane P through aromatisation and side-chain oxidation to give benzoic acid Q and benzyl alcohol R, selecting the matching reagent sequence.
Step 1:Aromatisation of the alkane over chromium(III) oxide at high temperature and pressure forms the toluene ring system.
Cr2O3,770K, 20 atm
Step 2:Chromyl chloride oxidation gives an aldehyde, which through base-mediated Cannizzaro-type steps and acidification yields benzoic acid and benzyl alcohol.
CrO2Cl2,H3O+NaOHH3O+
Final answer: (i) Cr2O3,770K, 20 atm; (ii) CrO2Cl2,H3O+;(iii) NaOH; (iv) H3O+
Q43Single correctAldehydes, Ketones and Carboxylic Acids
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R: Assertion A: Acetal/Ketal is stable in basic medium. Reason R: The high leaving tendency of alkoxide ion gives the stability to acetal/ketal in basic medium. In the light of the above statements, choose the correct answer from the options given below:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4A is true but R is false
Approach:
Assess the truth of the assertion about acetal stability and the reason about alkoxide leaving tendency.
Step 1:Acetals and ketals are stable in basic conditions but hydrolyse readily under acidic conditions, so the assertion is true.
Step 2:Alkoxide is a poor leaving group, so the high leaving tendency stated in the reason is false.
Final answer: A is true but R is false
Q44Single correctSome Basic Principles of Organic Chemistry
Which of the following conformations will be the most stable?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1Anti (staggered) Newman projection of n-butane
Approach:
Rank the Newman projection conformations of butane by torsional and steric strain to find the most stable one.
Step 1:The stability order of butane conformers places the anti form highest, followed by gauche, partially eclipsed and fully eclipsed.
Anti>Gauche>Partially eclipsed>Fully eclipsed
Step 2:The anti conformation with the two methyl groups opposite corresponds to option 1.
Final answer: Anti (staggered) Newman projection of n-butane
The correct order in aqueous medium of basic strength in case of methyl substituted amines is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1Me2NH>MeNH2>Me3N>NH3
Approach:
Balance inductive electron donation, steric hindrance and solvation of the conjugate acid to order the basic strength of methylamines in water.
Step 1:In aqueous medium the combined effect of inductive donation and solvation makes the secondary amine the strongest base.
Me2NH>MeNH2
Step 2:The tertiary amine suffers reduced solvation and the unsubstituted ammonia is weakest, completing the order.
Me2NH>MeNH2>Me3N>NH3
Final answer: Me2NH>MeNH2>Me3N>NH3
Q46Single corrects-Block Elements
Compound A reacts with NH4Cl and forms a compound B. Compound B reacts with H2O and excess of CO2 to form compound C which on passing through or reaction with saturated NaCl solution forms sodium hydrogen carbonate. Compound A, B and C, are respectively
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2Ca(OH)2,NH3,NH4HCO3
Approach:
Identify the species in the Solvay (ammonia-soda) process. Slaked lime reacts with ammonium chloride to liberate ammonia, which with water and carbon dioxide forms ammonium bicarbonate, the precursor to sodium bicarbonate.
Step 1:Compound A, slaked lime, treated with ammonium chloride releases ammonia gas.
Ca(OH)2+NH4Cl→CaCl2+NH3+H2O
Step 2:Compound B, ammonia, with water and excess carbon dioxide forms ammonium bicarbonate.
NH3+H2O+CO2→NH4HCO3
Step 3:Compound C reacts with saturated brine to precipitate sodium bicarbonate, confirming the assignment.
NH4HCO3+NaCl→NH4Cl+NaHCO3
Final answer: Ca(OH)2,NH3,NH4HCO3
Q47Single correctQualitative Analysis
Correct match is
List-I (Cations)
List-II (Group reagents)
A.Pb2+,Cu2+
(i).H2S gas in presence of dilute HCl
B.Al3+,Fe3+
(ii).(NH4)2CO3 in presence of NH4OH
C.Co2+,Ni2+
(iii).NH4OH in presence of NH4Cl
D.Ba2+,Ca2+
(iv).H2S in presence of NH4OH
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1A→i;B→iii;C→iv;D→ii
Approach:
Assign each cation pair to its analytical group reagent in the classical scheme of qualitative inorganic analysis.
Step 1:Group II cations precipitate as sulphides with hydrogen sulphide in dilute acid.
Pb2+,Cu2+→H2S/dilute HCl
Step 2:Group III cations precipitate as hydroxides with ammonia in presence of ammonium chloride.
Al3+,Fe3+→NH4OH/NH4Cl
Step 3:Group IV cations precipitate as sulphides with hydrogen sulphide in ammoniacal medium.
Co2+,Ni2+→H2S/NH4OH
Step 4:Group V cations precipitate as carbonates with ammonium carbonate in ammoniacal medium.
Ba2+,Ca2+→(NH4)2CO3/NH4OH
Final answer: A→i;B→iii;C→iv;D→ii
Q48Single correctChemical Kinetics
The variation of the rate of an enzyme catalyzed reaction with substrate concentration is correctly represented by graph
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4(c)
Approach:
Apply the saturation behaviour of enzyme catalysis: rate rises with substrate concentration and then levels off as the enzyme becomes saturated.
Step 1:At low substrate concentration the rate increases nearly linearly with substrate concentration.
v∝[S]for [S]≪Km
Step 2:At high substrate concentration every enzyme active site is occupied, so the rate becomes independent of substrate concentration and reaches a plateau.
v→Vmaxfor [S]≫Km
Step 3:The plot that rises and then flattens to a maximum is graph (c).
v-[S]saturation curve=graph (c)
Final answer: (c)
Q49Single correctEnvironmental Chemistry
Some reactions of NO2 relevant to photochemical smog formation are NO2sunlightX+Y X+O2→A A+NO2→B Identify A, B, X and Y.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2X=[O],Y=NO,A=O2,B=O3
Approach:
Trace the photochemical chain of nitrogen dioxide photolysis that generates ground-level ozone.
Step 1:Sunlight dissociates nitrogen dioxide into nitric oxide and atomic oxygen.
NO2hνNO+[O]
Step 2:Atomic oxygen combines with molecular oxygen of the air.
[O]+O2→O3
Step 3:The printed solution assigns A as molecular oxygen and B as ozone, consistent with option (2).
A=O2,B=O3
Final answer: X=[O],Y=NO,A=O2,B=O3
Q50Single correctPeriodic Properties
Inert gases have positive electron gain enthalpy. Its correct order is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1He<Xe<Kr<Ne
Approach:
Compare the magnitudes of the (positive) electron gain enthalpies of the noble gases from the tabulated values.
Step 1:List the positive electron gain enthalpy values in kilojoule per mole.
He=48,Ne=116,Ar=96,Kr=96,Xe=77
Step 2:Order the four listed gases by increasing magnitude of electron gain enthalpy.
48<77<96<116
Step 3:The increasing order corresponds to option (1).
He<Xe<Kr<Ne
Final answer: He<Xe<Kr<Ne
Q51NumericalSolutions
The osmotic pressure of solutions of PVC in cyclohexanone at 300 K are plotted on the graph. The molar mass of PVC is _______ g mol−1 (Nearest integer) (Given : R = 0.083 L atm K−1mol−1)
SolutionAnswer: 41500
Approach:
Relate the slope of the osmotic-pressure-versus-concentration plot to the molar mass through the van't Hoff equation.
Step 1:Express osmotic pressure in terms of mass concentration and molar mass.
π=volumemass⋅MRT
Step 2:Identify the slope of the plot with the quantity RT divided by molar mass.
slope=MRT=6×10−4
Step 3:Solve for the molar mass using R, T and the slope.
M=slopeRT=6×10−40.083×300
Final answer: 41500
Q52NumericalSome Basic Concepts of Chemistry
The density of a monobasic strong acid (Molar mass 24.2 g/mol) is 1.21 kg/L. The volume of its solution required for the complete neutralization of 25 mL of 0.24 M NaOH is _______ ×10−2 mL (Nearest integer)
SolutionAnswer: 12
Approach:
Equate the moles of monobasic acid to the moles of sodium hydroxide, expressing acid moles through its density and molar mass.
Step 1:Compute the molarity of the pure monobasic acid from its density and molar mass.
M=24.21.21×103=50M
Step 2:Apply the milliequivalence balance for a monobasic acid and a monoacidic base.
25×0.24×1=1×V×M
Step 3:Solve for the required volume of acid.
V=5025×0.24=0.12mL=12×10−2mL
Final answer: 12
Q53NumericalIonic Equilibrium
A litre of buffer solution contains 0.1 mole of each of NH3 and NH4Cl. On the addition of 0.02 mole of HCl by dissolving gaseous HCl, the pH of the solution is found to be _______ ×10−3 (Nearest integer) [Given : pKb(NH3)=4.745 log2=0.301 log3=0.477 T=298K]
SolutionAnswer: 9079
Approach:
Update the base and salt amounts after the added acid consumes ammonia, then apply the Henderson-Hasselbalch equation for the basic buffer and convert pOH to pH.
Step 1:Added hydrochloric acid converts an equal amount of ammonia into ammonium ion.
NH3+HCl→NH4Cl
Step 2:Apply the buffer equation for the new amounts.
pOH=4.745+log0.1−0.020.1+0.02=4.745+log(23)
Step 3:Convert the pOH to pH at 298 K.
pH=14−4.921=9.079=9079×10−3
Final answer: 9079
Q54NumericalCoordination Compounds
The number of paramagnetic species from the following is _______. [Ni(CN)4]2−,[Ni(CO)4],[NiCl4]2−, [Fe(CN)6]4−,[Cu(NH3)4]2+ [Fe(CN)6]3− and [Fe(H2O)6]2+
SolutionAnswer: 4
Approach:
Determine the number of unpaired electrons in each complex from the metal oxidation state and the ligand field strength, then count those with at least one unpaired electron.
Step 1:Evaluate the nickel complexes for unpaired electrons.
Q55NumericalPurification and Characterisation of Organic Compounds
In sulphur estimation, 0.471 g of an organic compound gave 1.4439 g of barium sulphate. The percentage of sulphur in the compound is _______ (Nearest Integer) (Given: Atomic mass Ba: 137 u, S: 32 u, O:16 u)
SolutionAnswer: 42
Approach:
In the Carius method all sulphur of the compound is precipitated as barium sulphate; the sulphur fraction of barium sulphate scaled by its mass gives the sulphur percentage.
Step 1:Compute the molar mass of barium sulphate.
M(BaSO4)=137+32+4×16=233
Step 2:Apply the sulphur percentage formula with the measured masses.
%S=23332×0.4711.4439×100
Step 3:Evaluate the expression.
%S=42%
Final answer: 42
Q56NumericalChemical Kinetics
For the first order reaction A→B, the half life is 30 min. The time taken for 75% completion of the reaction is_______ min. (Nearest integer) Given : log2=0.3010 log3=0.4771 log5=0.6989
SolutionAnswer: 60
Approach:
For a first order reaction the time for 75 percent completion equals two half lives, since three quarters reacted corresponds to one quarter remaining.
Step 1:After 75 percent completion the fraction remaining is one quarter, which equals two successive halvings.
[A]0[A]=41=(21)2
Step 2:Express the time as twice the half life.
t=2×t1/2=2×30
Step 3:State the result.
t=60min
Final answer: 60
Q57Numericald- and f-Block Elements
How many of the following metal ions have similar value of spin only magnetic moment in gaseous state? _______ (Given : Atomic number : V, 23; Cr, 24; Fe, 26; Ni, 28) V3+,Cr3+,Fe2+,Ni3+
SolutionAnswer: 2
Approach:
Find the number of unpaired d electrons for each ion and compare their spin only magnetic moments, counting how many share the same value.
Step 1:Determine unpaired electrons and magnetic moments for the d-configurations.
V3+(d2):8;Cr3+(d3):15;Fe2+(d6):24;Ni3+(d7):15
Step 2:Identify the ions sharing the same magnetic moment.
Cr3+andNi3+=15BM
Step 3:Count the matching ions.
n=2
Final answer: 2
Q58NumericalElectrochemistry
Consider the cell Pt(s)∣H2(g) (1 atm)∣H+(aq,[H+]=1)∥Fe3+(aq),Fe2+(aq)∣Pt(s) Given EFe3+/Fe2+o=0.771V and EH+/21H2o=0V, T=298K If the potential of the cell is 0.712 V, the ratio of concentration of Fe2+ to Fe3+ is _______ (Nearest integer)
SolutionAnswer: 10
Approach:
Write the Nernst equation for the single-electron cell, insert the standard and measured potentials, and solve for the concentration ratio.
Step 1:With unit hydrogen ion concentration and unit hydrogen pressure the reaction quotient reduces to the ratio of iron ion concentrations.
0.712=0.771−10.0591log[Fe3+][Fe2+]
Step 2:Isolate the logarithm of the concentration ratio.
−0.059=−0.0591log[Fe3+][Fe2+]
Step 3:Convert from the logarithm to the ratio.
[Fe3+][Fe2+]=101=10
Final answer: 10
Q59NumericalChemical Bonding
The total number of lone pairs of electrons on oxygen atoms of ozone is_______
SolutionAnswer: 6
Approach:
Draw the resonance Lewis structure of ozone and count the lone pairs on the central and terminal oxygen atoms.
Step 1:The central oxygen carries a positive charge with one lone pair, bonded by a single and a double bond to the terminal atoms.
central O+:1lone pair
Step 2:The singly bonded terminal oxygen bears a negative charge with three lone pairs; the doubly bonded terminal oxygen has two lone pairs.
O−:3lone pairs;O(=):2lone pairs
Step 3:Sum the lone pairs over all three oxygen atoms.
1+3+2=6
Final answer: 6
Q60NumericalThermodynamics
An athlete is given 100 g of glucose (C6H12O6) for energy. This is equivalent to 1800kJ of energy. The 50% of this energy gained is utilized by the athlete for sports activities at the event. In order to avoid storage of energy, the weight of extra water he would need to perspire is_______ g (Nearest integer) Assume that there is no other way of consuming stored energy. Given : The enthalpy of evaporation of water is 45 kJ mol−1 Molar mass of C, H & O are 12, 1 and 16 g mol−1
SolutionAnswer: 360
Approach:
The unused half of the glucose energy must be dissipated by evaporating water; divide that energy by the molar enthalpy of evaporation and convert moles of water to mass.
Step 1:Half of the 1800 kJ supplied by glucose remains as the energy to be removed.
E=21800=900kJ
Step 2:Determine the moles of water whose evaporation absorbs this energy.
n=45900=20mol
Step 3:Convert moles of water to mass.
m=21800×4518=20×18=360g
Final answer: 360
Mathematics30 questions
Q61Single correctStatistics
The mean and variance of the marks obtained by the students in a test are 10 and 4 respectively. Later, the marks of one of the points P and Q revised is increased from 8 to 12. If the new mean of the marks of the students is increased from 8 to 12, then their new variance is equal to :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 33.96
Approach:
Find the number of students from the original mean and variance, then recompute mean and variance after one mark changes from 8 to 12.
Step 1:From the original mean and variance, determine the number of students.
xˉ=10,σ2=4⇒N∑xi2−100=4
Step 2:Changing one mark from 8 to 12 raises the mean to 10.2, fixing N.
N∑xi+4=10+N4=10.2⇒N=20
Step 3:Update the sum of squares and compute the new variance.
Construct the truth table for the compound statement over all truth-value assignments of p and q.
Step 1:Evaluate the antecedent and the consequent.
A=p∧(∼q),B=p⇒(∼q)
Step 2:If p∧(∼q) is true, then p is true and ∼q is true, so p⇒(∼q) is true.
A=T⇒B=T
Step 3:The full implication is true in every row of the truth table.
(p∧(∼q))⇒(p⇒(∼q))≡T
Final answer: a tautology
Q63Single correctCo-ordinate Geometry
The points of intersection of the line ax+by=0,(a=b) and the circle x2+y2−2x=0 are A(a,0) and B(1,β). The image of the circle with AB as a diameter in the line x+y+2=0 is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1x2+y2+5x+5y+12=0
Approach:
Find A and B on the circle, form the circle with AB as diameter, then reflect its centre in the given line.
Step 1:A and B lie on the line and circle.
α=0⇒A(0,0),β=1⇒B(1,1)
Step 2:Circle on AB as diameter has centre at the midpoint and radius half the length AB.
Centre(21,21),r=21
Step 3:Reflect the centre in x+y+2=0.
1x−21=1y−21=2−2(3)
Step 4:Write the reflected circle with the same radius.
(x+25)2+(y+25)2=21
Final answer: x2+y2+5x+5y+12=0
Q64Single correct3D Geometry
Consider the lines L1 and L2 given by L1:2x−1=1y−3=2z−2 L2:1x−2=2y−2=3z−3. A line L3 having direction ratios 1,−1,−2, intersects L1 and L2 at the points P and Q respectively. Then the length of line segment PQ is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 226
Approach:
Parametrize P on L1 and Q on L2, impose that PQ is parallel to (1,-1,-2), then compute the length.
Step 1:Take general points on each line.
P=(2λ+1,λ+3,2λ+2),Q=(μ+2,2μ+2,3μ+3)
Step 2:Direction ratios of PQ are proportional to 1,−1,−2.
12λ−μ−1=−1λ−2μ−1=−22λ−3μ−1
Step 3:Solve the resulting equations.
3λ−3μ=0,μ=3⇒λ=3,μ=3
Step 4:Compute the distance PQ.
∣PQ∣=22+22+42=24=26
Final answer: 26
Q65Single correctFunctions
Let f:(0,1)→R be a function defined by f(x)=1−e−x1, and g(x)=(f(−x)−f(x)). Consider (I) g is an increasing function in (0,1) (II) g is one-one in (0,1) Then,
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2Both (I) and (II) are true
Approach:
Simplify g(x), differentiate, and analyze its sign to determine monotonicity and injectivity.
Step 1:Form g(x) from f(-x) and f(x).
g(x)=1−ex1−1−e−x1=1−ex1+ex
Step 2:Differentiate g(x).
g′(x)=(1−ex)2(1−ex)ex−(1+ex)(−ex)=(1−ex)22ex>0
Step 3:A positive derivative makes g increasing, therefore one-one.
g′(x)>0⇒g increasing and one-one
Final answer: Both (I) and (II) are true
Q66Single correctMatrices and Determinants
Let S1 and S2 be respectively the sets of all a∈R−{0} for which the system of linear equations ax+2ay−3az=1 (2a+1)x+(2a+3)y+(a+1)z=2 (3a+5)x+(a+5)y+(a+2)z=3 has unique solution and infinitely many solutions. Then
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2S1=R−{0} and S2=Φ
Approach:
Evaluate the coefficient determinant; if it is nonzero for all admissible a, the system has a unique solution for every such a.
Step 1:Form the coefficient determinant.
Δ=a2a+13a+52a2a+3a+5−3aa+1a+2
Step 2:Expanding gives a nonzero quadratic in a after factoring out a.
Δ=a(15a2+31a+37)
Step 3:The quadratic 15a2+31a+37 has no real root, so for every a=0 the determinant is nonzero.
a=0⇒Δ=0
Step 4:Hence the unique-solution set is all admissible a and the infinite-solution set is empty.
S1=R−{0},S2=Φ
Final answer: S1=R−{0} and S2=Φ
Q67Single correctIntegral Calculus
The minimum value of the function f(x)=∫02e∣x−t∣dt is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 32(e−1)
Approach:
Split the integral over the cases x<0, 0<=x<=2 and x>2, evaluate f(x) in each region, and locate the minimum.
Step 1:For x>2 the integrand is ex−t.
f(x)=∫02ex−tdt=ex(1−e−2)
Step 2:For x<0 the integrand is et−x.
f(x)=∫02et−xdt=e−x(e2−1)
Step 3:For 0≤x≤2 split at t=x.
f(x)=∫0xex−tdt+∫x2et−xdt=ex−1+e2−x−1
Step 4:Minimize over [0,2]; the minimum occurs at x=1.
f(1)=e+e−2=2(e−1)
Final answer: 2(e−1)
Q68Single correctSequence and Series
Let y(x)=(1+x)(1+x2)(1+x4)(1+x8)(1+x16). Then y′−y′′ at x=−1 is equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 128
Approach:
Expand y(x) using the telescoping product, then differentiate and evaluate at x=-1.
Step 1:Multiply and divide by (1-x) to telescope the product.
y=1−x1−x32=1+x+x2+x3+⋯+x31
Step 2:Differentiate term by term and evaluate at x=-1.
y′=1+2x+3x2+⋯+31x30,y′(−1)=1−2+3−⋯+31=16
Step 3:Differentiate again and evaluate at x=-1.
y′′=2+6x+12x2+⋯+31⋅30x29,y′′(−1)=−12
Step 4:Combine the two values.
y′(−1)−y′′(−1)=16−(−12)=28
Final answer: 28
Q69Single correctDifferential Calculus
Let x=2 be a local minima of the function f(x)=2x4−18x2+8x+12,x∈(−4,4). If M is local maximum value of the function f in (−4,4), then M=
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1126−233
Approach:
Differentiate f, factor the cubic derivative using the given local minimum, locate the local maximum, and evaluate f there.
Step 1:Differentiate f(x).
f′(x)=8x3−36x+8=4(2x3−9x+2)
Step 2:Factor using the known root x=2.
2x3−9x+2=(x−2)(2x2+4x−1)
Step 3:Solve the quadratic for the other critical points.
x=2−2±6
Step 4:Evaluate f at the local maximum.
f(x0)=126−233
Final answer: 126−233
Q70Single correctLimits
The value of n→∞lim2n4+4n+3−3n6+5n+41⋅2−3+2⋅3+5⋅6+⋯+(3n−2)⋅(3n−1)⋅3n is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 223(2+1)
Approach:
Express the numerator as a sum of products and the denominator via leading-order growth, then take the dominant-term limit.
Step 1:Write the numerator as a sum of triple products.
I=limn→∞2n4+4n+3−3n6+5n+4∑(3r−2)(3r−1)(3r)
Step 2:Leading order of numerator is the sum of products of order n4.
∑(3r−2)(3r−1)(3r)∼23n4
Step 3:Denominator leading order.
2n4+4n+3−3n6+5n+4∼n2(2−1)
Step 4:Divide leading terms and rationalize.
I=23⋅2−11=23(2+1)
Final answer: 23(2+1)
Q71Single correctCo-ordinate Geometry
The distance of the point (6,−22) from the common tangent y=mx+c,m>0, of the curves x=2y2 and x=1+y2 is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 45
Approach:
Write the tangent lines to both parabolas, equate them to find the common tangent, then apply the point-to-line distance formula.
Step 1:Tangent to y2=2x.
y=mx+8m1
Step 2:Tangent to y2=x−1.
y=mx−m+4m1
Step 3:Equate the intercepts and solve for m.
8m1=−m+4m1⇒m=221
Step 4:Common tangent and distance from the given point.
x−22y+1=0,d=1+8∣6−22(−22)+1∣=5
Final answer: 5
Q72Single correctMatrices and Determinants
Let x,y,z>1 and A=1logyxlogzxlogxy2logzylogxzlogyz3. Then ∣adj(adjA2)∣ is equal to :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 328
Approach:
Compute |A| using change-of-base for logarithms, then apply the adjugate-determinant identity twice.
Step 1:Express all entries with a common logarithm base.
Step 3:Apply the adjugate identity (n=3) twice to A2.
∣adj(adjA2)∣=∣A2∣(n−1)2=(∣A∣2)4
Step 4:Substitute |A| = 2.
∣A∣8=28
Final answer: 28
Q73Single correctIntegral Calculus
Let f(x)=∫(x2+1)(x2+3)2xdx. If f(3)=21(loge5−loge6), then f(4) is equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 321(loge17−loge19)
Approach:
Substitute t = x2, integrate by partial fractions, use the given value to fix the constant, then evaluate f(4).
Step 1:Substitute x2=t.
f(x)=∫(t+1)(t+3)dt
Step 2:Resolve into partial fractions and integrate.
f(x)=21loge(x2+3x2+1)+C
Step 3:Apply the given value f(3) to find C.
f(3)=21loge1210+C=21(loge5−loge6)⇒C=0
Step 4:Evaluate f(4).
f(4)=21loge(1917)=21(loge17−loge19)
Final answer: 21(loge17−loge19)
Q74Single correct3D Geometry
The distance of the point P(4,6,−2) from the line passing through the point (−3,2,3) and parallel to a line with direction ratios 3,3,−1 is equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 414
Approach:
Form the vector from the line's point A to P, project it onto the direction, and use the Pythagorean relation to get the perpendicular distance.
Step 1:Vector from A(-3,2,3) to P(4,6,-2).
AP=7i^+4j^−5k^,∣AP∣2=49+16+25=90
Step 2:Projection of AP onto the direction ⟨3,3,−1⟩.
projection=1921+12+5=1938
Step 3:Apply the Pythagorean relation.
PN2=90−19382=90−76=14
Step 4:Take the square root.
PN=14
Final answer: 14
Q75Single correctProbability
Let M be the maximum value of the product of two positive integers when their sum is 66. Let the sample space S={x∈Z:x(66−x)≥95M} and the event A={x∈S:x is a multiple of 3}. Then P(A) is equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 331
Approach:
Maximize the product under the fixed-sum constraint, determine the sample space from the inequality, count the favourable multiples of 3, and form the probability.
Step 1:Maximum product when the two integers are equal.
2x+y≥xy⇒33≥xy⇒M=332=1089
Step 2:Solve the defining inequality of S.
x(66−x)≥95⋅1089=605⇒x2−66x+605≤0
Step 3:Count integers in S and the multiples of 3.
n(S)=57,A={6,9,12,…,60},n(A)=19
Step 4:Form the probability.
P(A)=5719=31
Final answer: 31
Q76Single correctVector Algebra
Let a,b and c be three non zero vectors such that b⋅c=0 and a×(b×c)=2b−c. If d be a vector such that b⋅d=a⋅b, then (a×b)⋅(c×d) is equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 141
Approach:
Apply the vector triple product expansion to extract the dot products, then use the scalar quadruple product identity to evaluate the required expression.
Step 1:Expand the given triple product and compare with the right-hand side.
(a⋅c)b−(a⋅b)c=21b−21c
Step 2:Apply the scalar quadruple product identity with the given orthogonality b⋅c=0.
(a×b)⋅(c×d)=(a⋅c)(b⋅d)−(a⋅d)(b⋅c)
Step 3:Substitute b⋅d=a⋅b=21 and a⋅c=21.
=21⋅21=41
Final answer: 41
Q77Single correctBinomial Theorem
If ar is the coefficient of x10−r in the Binomial expansion of (1+x)10, then r=1∑10r3(ar−1ar)2 is equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 11210
Approach:
Express the ratio of consecutive binomial coefficients, square it, and evaluate the resulting summation using standard power sum formulas.
Step 1:Write the coefficient ratio and square it.
(ar−1ar)2=(r11−r)2
Step 2:Multiply by r3 and simplify the summand.
∑r=110r3⋅r2(11−r)2=∑r=110r(121−22r+r2)
Step 3:Apply the power-sum formulas with n=10.
121⋅210×11+(210×11)2−22⋅610×11×21
Final answer: 1210
Q78Single correctDifferential Equations
Let y=y(x) be the solution curve of the differential equation dxdy=xy(1+xy2(1+logex)),x>0,y(1)=3. Then 9y2(x) is equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 35−2x3(2+logex3)x2
Approach:
Reduce the equation to a Bernoulli form in y21, integrate, then apply the initial condition to fix the constant.
Step 1:Rearrange and substitute t=y2x2, giving a linear-type relation in t.
−y21t=∫x2(1+lnx)dx
Step 2:Apply y(1)=3 to evaluate the constant.
91=−2[31−91]+C
Step 3:Solve for 9y2(x) and simplify the logarithmic terms.
9y2=5−2x3(2+lnx3)x2
Final answer: 5−2x3(2+logex3)x2
Q79Single correctComplex Numbers and Quadratic Equations
Let z1=2+3i and z2=3+4i. The set S={z∈C:∣z−z1∣2−∣z−z2∣2=∣z1−z2∣2} represents a
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3straight line with the sum of its intercepts on the coordinate axes equals 14
Approach:
Write z=x+iy, expand the modulus condition into a linear equation, then find the axis intercepts.
Step 1:Substitute z=x+iy and the given points; expand both squared moduli.
(x−2)2+(y−3)2−(x−3)2−(y−4)2=1+1
Step 2:Simplify to the line equation.
2x+2y=9+⇒x+y=17
Step 3:Read off the intercepts: x-intercept =7, y-intercept =7.
7+7=14
Final answer: straight line with the sum of its intercepts on the coordinate axes equals 14
Q80Single correctVector Algebra
The vector a=−i^+2j^+k^ is rotated through a right angle, passing through the y-axis in its way and the resulting vector is b. Then the projection of 3a+2b on c=5i^+4j^+3k^ is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 432
Approach:
Use the rotation-through-90∘ via the y-axis condition to determine b, then compute the scalar projection of 3a+2b onto c.
Step 1:Pass through the y-axis: b=λj^+μa with b⋅a=0, giving b=λ(j^−(a⋅j^)∣a∣2a).
b=λ(i^−j^+k^)
Step 2:Choose b=2(i^−j^+k^) and form 3a+2b.
3a+2b=3(−i^+2j^+k^)+2(i^−j^+k^)
Step 3:Project onto c=5i^+4j^+3k^, with ∣c∣=50.
50(−1)(5)+(4)(4)+(5)(3)=50−5+16+15=5026
Final answer: 32
Q81NumericalThree Dimensional Geometry
Let the equation of the plane passing through the line x−2y−z−5=0=x+y+3z−5 and parallel to the line x+y+2z−7=0=2x+3y+z−2 be ax+by+cz=65. Then the distance of the point (a,b,c) from the plane 2x+2y−z+16=0 is _____.
SolutionAnswer: 9
Approach:
Form the family of planes through the given line, impose parallelism with the second line to fix the parameter, then compute the point-plane distance.
Step 1:Take the plane (x−2y−z−5)+λ(x+y+3z−5)=0 and require its normal to be perpendicular to the second line's direction; solve for λ.
λ=12
Step 2:Identify the coefficients.
a=13,b=10,c=35
Step 3:Apply the distance formula to the plane 2x+2y−z+16=0.
d=9∣26+20−35+16∣=327
Final answer: 9
Q82NumericalConic Sections
The vertices of a hyperbola H are (±6,0) and its eccentricity is 25. Let N be the normal to H at a point in the first quadrant and parallel to the line 2x+y=22. If d is the length of the line segment of N between H and the y-axis then d2 is equal to _____.
SolutionAnswer: 216
Approach:
Determine the hyperbola from its vertices and eccentricity, find the first-quadrant point where the normal has the given slope, then compute the squared segment length to the y-axis.
Step 1:With a=6 and e=25, obtain b2.
b2=36(45−1)=9
Step 2:Impose the normal slope equal to that of 2x+y=22 (slope −2) to locate the first-quadrant point.
N:(62,3)
Step 3:Compute the segment of the normal from this point to the y-axis and square it.
d2=72+144=216
Final answer: 216
Q83NumericalComplex Numbers and Quadratic Equations
Let S={α:log2(92α−4+13)−log2(25⋅32α−4+1)=2}. Then the maximum value of β for which the equation x2−2(α∈s∑α)2x+α∈s∑(α+1)2β=0 has real roots, is _____.
SolutionAnswer: 25
Approach:
Solve the logarithmic equation for α via a substitution, evaluate the two required summations over S, then impose the non-negative discriminant condition to bound β.
Step 1:Set t=32α−4, reduce the log equation to a linear relation and solve.
25⋅32α−4+192α−4+13=4⇒t2−10t+9=0
Step 2:Evaluate the summations over S={2,3}.
∑α=5,∑(α+1)2=9+16=25
Step 3:Impose D≥0 on x2−50x+25β=0.
(50)2−4(25β)≥0⇒β≤25
Final answer: 25
Q84NumericalRelations and Functions
For some a,b,c∈N, let f(x)=ax−3 and g(x)=xb+c,x∈R. If (fog)−1(x)=(2x−7)31, then (fog)(ac)+(gof)(b) is equal to _____.
SolutionAnswer: 2039
Approach:
Invert fog symbolically, match it to the given inverse to read off a,b,c, then evaluate the two composite values.
Step 1:Match (fog)−1(x)=(ax+3−ca)1/b with (2x−7)1/3.
a=2,b=3,c=5
Step 2:Evaluate (fog)(ac)=(fog)(10).
f(g(10))=2(103+5)−3=2010−3=2007
Step 3:Evaluate (gof)(b)=(gof)(3) and add.
g(f(3))=(2⋅3−3)3+5=27+5=32
Final answer: 2039
Q85NumericalPermutations and Combinations
Let x and y be distinct integers where 1≤x≤25 and 1≤y≤25. Then, the number of ways of choosing x and y, such that x+y is divisible by 5, is _____.
SolutionAnswer: 120
Approach:
Classify the integers 1 to 25 by their residue modulo 5, count residue pairs whose sum is divisible by 5, then subtract the equal x=y cases.
Step 1:Each residue class {5k},{5k+1},…,{5k+4} contains 5 numbers from 1 to 25.
5 numbers per residue class
Step 2:Pairs summing to a multiple of 5: (0,0),(1,4),(2,3) residue pairs; count ordered choices of distinct integers.
5×4+2(5×5)×2=20+100
Step 3:Confirm exclusion of x=y cases (already removed in distinctness).
Total=120
Final answer: 120
Q86NumericalBinomial Theorem
The constant term in the expansion of (2x+x71+3x2)5 is _____.
SolutionAnswer: 1080
Approach:
Factor out x351 to convert the trinomial into a polynomial, then find the coefficient of the power of x that cancels the factored term.
Step 1:Write the expression as x351(2x8+1+3x9)5.
x351(1+x8(3x+2))5
Step 2:The term with x35 uses one factor of x8(3x+2) raised so that 8×4+9×?; matching gives coefficient from (45)(34)(2)1(3)3.
5C4×4C3(2)1(3)3
Step 3:Multiply out.
5×4×2×27=1080
Final answer: 1080
Q87NumericalPermutations and Combinations
Let S={1,2,3,5,7,10,11}. The number of non-empty subsets of S that have the sum of all elements a multiple of 3, is _____.
SolutionAnswer: 43
Approach:
Group the elements by residue modulo 3, then count subsets in each residue group whose combined residue sums to 0, multiplying independent choices.
Step 1:Classify: one element is of 3k type, three are of 3k+1 type, and three are of 3k+2 type.
∣3k∣=1,∣3k+1∣=3,∣3k+2∣=3
Step 2:Count subset combinations whose total residue is 0(mod3) across the groups (the 3k element is free in 2 ways).
combinations summing to 0(mod3)
Step 3:Remove the empty subset.
44−1=43
Final answer: 43
Q88NumericalInverse Trigonometric Functions
If the sum of all the solutions of tan−1(1−x22x)+cot−1(2x1−x2)=3π,−1<x<1,x=0, is α−34, then α is equal to _____.
SolutionAnswer: 2
Approach:
Split the domain by the sign of x, reduce each inverse term using the double-angle substitution x=tanθ, solve for x on each branch, then sum the solutions.
Step 1:Case −1<x<0: the identity gives 2tan−1x+π+2tan−1x=3π.
tan−1x=−6π⇒x=−31
Step 2:Case 0<x<1: 2tan−1x+2tan−1x=3π leads to a second root.
x=6π-branch root
Step 3:Sum the solutions and match to α−34.
Sum=−31+2−3=2−34
Final answer: 2
Q89NumericalSequences and Series
Let A1,A2,A3 be the three A.P. with the same common difference d and having their first terms as A,A+1,A+2, respectively. Let a,b,c be the 7th,9th,17th terms of A1,A2,A3, respectively such that a2bc71717111+70=0. If a=29, then the sum of first 20 terms of an AP whose first term is c−a−b and common difference is 12d, is equal to _____.
SolutionAnswer: 495
Approach:
Express a,b,c in terms of A and d, use a=29 and the determinant condition to solve for d and the required quantities, then apply the AP sum formula.
Step 1:Write a=A+6d, b=A+1+8d, c=A+2+16d; use a=29 and the determinant condition to solve.
a=A+6d=29,d=6,A=−7
Step 2:Form the AP with first term c−a−b and common difference 12d=21.
c−a−b=91−29−42=20,12d=126=21
Step 3:Apply the sum formula for 20 terms.
S20=220[2×20+19×126]=10[40+219]
Final answer: 495
Q90NumericalIntegral Calculus
In the area enclosed by the parabolas P1:2y=5x2 and P2:x2−y+6=0 is equal to the area enclosed by P1 and y=ax,a>0, then a3 is equal to _____.
SolutionAnswer: 600
Approach:
Find the intersection of the two parabolas, compute the enclosed area A1, then set the area between P1 and the line y=ax equal to it and solve for a3.
Step 1:Intersect P1 and P2: x2+6=25x2 gives x=±2. Compute A1.
A1=∫−22(x2+6−25x2)dx=16
Step 2:Area between P1:y=25x2 and the line y=ax from 0 to 52a.
How many questions are in the JEE Main 2023 January 25, Shift 1 paper?
The JEE Main 2023 January 25, Shift 1 paper has 90 questions — Physics (30), Chemistry (30) and Mathematics (30). Every question is on this page with its correct answer and a step-by-step solution.
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