JEE Main 2024 April 05, Shift 1 Question Paper with Solutions
All 88 questions from the JEE Main 2024 (April 05, Shift 1) shift — Physics (29), 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.
The angle between vector Q and the resultant of (2Q+2P) and (2Q−2P) is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 20∘
Approach:
Add the two given vectors to find the resultant, then compare its direction with Q.
Step 1:Add the two vectors.
(2Q+2P)+(2Q−2P)=4Q
Step 2:The resultant 4Q points along Q.
R=4Q
Step 3:The angle between Q and a vector parallel to it is zero.
θ=0∘
Final answer: 0∘
Q2Single correctPhysics and Measurement
Time periods of oscillation of the same simple pendulum measured using four different measuring clocks were recorded as 4.62 s, 4.632 s, 4.6 s and 4.64 s. The arithmetic mean of these readings in correct significant figure is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 34.6 s
Approach:
Compute the arithmetic mean, then round to the least number of decimal places among the readings as dictated by significant-figure rules.
Step 1:Sum the four readings.
4.62+4.632+4.6+4.64=18.492
Step 2:Divide by the number of readings.
Tˉ=418.492=4.623
Step 3:The least precise reading (4.6 s) has one decimal place, so the mean is rounded to one decimal place.
Tˉ≈4.6 s
Final answer: 4.6 s
Q3Single correctGravitation
If G be the gravitational constant and u be the energy density then which of the following quantity have the dimensions as that of the uG :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4Force per unit mass
Approach:
Write the dimensions of G and energy density u, form uG, and compare with each option.
Step 1:Form the product uG.
[uG]=[ML−1T−2][M−1L3T−2]=[L2T−4]
Step 2:Take the square root.
[uG]=[LT−2]
Step 3:Force per unit mass equals acceleration, with dimension [LT−2].
[mF]=[LT−2]
Final answer: Force per unit mass
Q4Single correctLaws of Motion
A wooden block mass 5 kg rests on a soft horizontal floor. When an iron cylinder of mass 25 kg is placed on the top of the block, the floor yields and the block and the cylinder together go down with an acceleration of 0.1 ms−2. The action force of the system on the floor is equal to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2291 N
Approach:
Apply Newton's second law to the combined block-cylinder system moving downward, find the normal reaction from the floor, then use the third law for the action on the floor.
Step 1:Total mass of the system.
M=5+25=30 kg
Step 2:Apply Newton's second law for downward acceleration with g=9.8 ms−2.
N=M(g−a)=30(9.8−0.1)
Step 3:Evaluate the reaction.
N=291 N
Step 4:By Newton's third law, the system pushes on the floor with this force.
F=N=291 N
Final answer: 291 N
Q5Single correctWork, Energy and Power
A body of mass 50 kg is lifted to a height of 20 m from the ground in the two different ways as shown in the figures. The ratio of work done against the gravity in both the respective cases, will be :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 41:1
Approach:
Recognize that gravity is a conservative force, so work done against gravity depends only on the change in height, not on the path taken.
Step 1:Case 1, pulled straight up to height h.
W1=mgh
Step 2:Case 2, raised along the ramp to the same height h.
W2=mgh
Step 3:Form the ratio.
W2W1=mghmgh=1
Final answer: 1:1
Q6Single correctRotational Motion
Ratio of radius of gyration of a hollow sphere to that of a solid cylinder of equal mass, for moment of Inertia about their diameter axis AB as shown in figure is 8/x. The value of x is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 467
Approach:
Find the radius of gyration of the hollow sphere about its diameter and of the solid cylinder (radius R, length 4R) about the transverse axis AB through one end, then form their ratio and equate to 8/x.
Step 1:Radius of gyration of the hollow sphere about its diameter.
k12=32R2
Step 2:For the cylinder, radius R and length L=4R, about the transverse axis AB at the end face.
k22=4R2+3(4R)2=4R2+316R2=1267R2
Step 3:Form the ratio of radii of gyration squared.
k22k12=67/122/3=678
Step 4:Equate to 8/x squared.
x8=678⇒x=67
Final answer: 67
Q7Single correctGravitation
A simple pendulum doing small oscillations at a place R height above earth surface has time period of T1=4 s. T2 would be it's time period if it is brought to a point which is at a height 2R from earth surface. Choose the correct relation [R= radius of earth] :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 43T1=2T2
Approach:
Express g at heights R and 2R using the inverse-square law, then use T∝1/g to relate the two time periods.
Step 1:At height R the distance from centre is 2R.
g1=(2R)2gR2=4g
Step 2:At height 2R the distance from centre is 3R.
g2=(3R)2gR2=9g
Step 3:Since T∝1/g, form the ratio.
T1T2=g2g1=g/9g/4=23
Step 4:Rearrange the ratio.
2T2=3T1
Final answer: 3T1=2T2
Q8Single correctGravitation
In hydrogen like system the ratio of coulombian force and gravitational force between an electron and a proton is in the order of :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 11039
Approach:
Both forces scale as inverse square of separation, so the ratio is independent of distance and reduces to a ratio of constants and charges/masses.
Step 1:Form the ratio, cancelling the r2 dependence.
FgFe=4πε0Gmempe2
Step 2:Substitute 4πε01=9×109, e=1.6×10−19 C.
4πε01e2=9×109×(1.6×10−19)2≈2.3×10−28
Step 3:Substitute G=6.67×10−11, me=9.1×10−31 kg, mp=1.67×10−27 kg.
Gmemp≈6.67×10−11×9.1×10−31×1.67×10−27≈1.0×10−67
Step 4:Divide to obtain the order of magnitude.
FgFe≈1.0×10−672.3×10−28≈2.3×1039
Final answer: 1039
Q9Single correctGravitation
Match List I with List II :
List I
List II
A. Kinetic energy of planet
I.−aGMm
B. Gravitation Potential energy of sun-planet system
II.2aGMm
C. Total mechanical energy of planet
III.rGm
D. Escape energy at the surface of planet for unit mass object
Use the standard expressions for kinetic, potential and total energy of a planet in a circular orbit of radius a, and the escape energy per unit mass at the planet surface.
Step 1:Kinetic energy of the orbiting planet matches (II).
Step 2:Sun-planet potential energy matches (I).
Step 3:Total mechanical energy matches (IV).
Step 4:Escape energy per unit mass at the planet surface matches (III).
Final answer: (A)-(II), (B)-(I), (C)-(IV), (D)-(III)
Q10Single correctProperties of Solids and Liquids
Given below are two statements : Statement I : When a capillary tube is dipped into a liquid, the liquid neither rises nor falls in the capillary. The contact angle may be 0∘. Statement II : The contact angle between a solid and a liquid is a property of the material of the solid and liquid as well. In the light of the above statement, choose the correct answer from the options given below :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3Statement I is false but Statement II is true
Approach:
Examine each statement against the physics of capillary action and contact angle.
Step 1:For a contact angle of 0∘, cosθ=1, giving the maximum rise, not zero rise; so Statement I is false.
h=ρgr2Tcos0∘=ρgr2T>0
Step 2:The contact angle is determined by the surface tensions of the solid-liquid, solid-gas and liquid-gas interfaces, hence depends on both materials; Statement II is true.
cosθ=γLGγSG−γSL
Step 3:Combine the two conclusions.
I false, II true
Final answer: Statement I is false but Statement II is true
Q11Single correctThermodynamics
The heat absorbed by a system in going through the given cyclic process is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 261.6 J
Approach:
For a complete cycle the internal energy change is zero, so the heat absorbed equals the work done, which equals the area enclosed by the circular loop on the P-V diagram.
Step 1:The loop is a circle with V-extent from 60 to 340 kPa and P-extent from 60 to 340 cc.
ΔV=340−60=280 kPa,ΔP=340−60=280 cc
Step 2:Compute the half-extents in SI units: 140 kPa =1.4×105 Pa, 140 cc =1.4×10−4m3.
2ΔP=1.4×105 Pa,2ΔV=1.4×10−4 m3
Step 3:Area enclosed equals the work done.
W=π(1.4×105)(1.4×10−4)=π×19.6
Step 4:Evaluate the heat absorbed.
Q=W=61.6 J
Final answer: 61.6 J
Q12Single correctKinetic Theory of Gases
If the collision frequency of hydrogen molecules in a closed chamber at 27∘C is Z, then the collision frequency of the same system at 127∘C is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 232Z
Approach:
At fixed volume and number density the collision frequency is proportional to the mean molecular speed, which scales as the square root of absolute temperature.
Step 1:Convert temperatures to kelvin.
T1=27+273=300 K,T2=127+273=400 K
Step 2:Use the square-root temperature dependence.
ZZ2=T1T2=300400=34
Step 3:Express the new collision frequency.
Z2=32Z
Final answer: 32Z
Q13Single correctCurrent Electricity
In the given figure R1=10Ω,R2=8Ω,R3=4Ω and R4=8Ω. Battery is ideal with emf 12 V. Equivalent resistant of the circuit and current supplied by battery are respectively :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 212Ω and 1 A
Approach:
The diagonal connecting wire places R2, R4 and R3 in parallel between the same two nodes; combine that parallel group, add R1 in series, then apply Ohm's law.
Step 1:Combine R2, R4 and R3 in parallel.
Rp1=81+81+41=81+1+2=84
Step 2:Add R1 in series with the parallel group.
Req=R1+Rp=10+2
Step 3:Apply Ohm's law for the battery current.
I=1212
Final answer: 12Ω and 1 A
Q14Single correctMagnetic Effects of Current and Magnetism
In a co-axial straight cable, the central conductor and the outer conductor carry equal currents in opposite directions. The magnetic field is zero :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1outside the cable
Approach:
Apply Ampere's circuital law to a circular loop in each region, using the net enclosed current.
Step 1:Between the conductors the loop encloses only the inner current, giving a non-zero field.
B⋅2πr=μ0I
Step 2:Outside the cable the loop encloses both currents, which are equal and opposite.
Ienc=I−I=0
Step 3:Therefore the field outside the cable is zero.
B⋅2πr=μ0(0)⇒B=0
Final answer: outside the cable
Q15Single correctElectromagnetic Induction and Alternating Currents
Two conducting circular loops A and B are placed in the same plane with their centers coinciding as shown in figure. The mutual inductance between them is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 42aμ0πb2
Approach:
Drive a current in the larger loop, treat its field as uniform over the small inner loop, compute the flux through the small loop, and identify the mutual inductance.
Step 1:The larger loop has radius a; its field near the common centre is nearly uniform over the small loop of radius b (b≪a).
B=2aμ0I
Step 2:Flux through the small inner loop of area πb2.
Φ=B⋅πb2=2aμ0I⋅πb2
Step 3:Divide by the current to obtain the mutual inductance.
M=IΦ=2aμ0πb2
Final answer: 2aμ0πb2
Q16Single correctAlternating Current
An alternating voltage of amplitude 40 V and frequency 4kHz is applied directly across the capacitor of 12μF. The maximum displacement current between the plates of the capacitor is nearly :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 212 A
Approach:
The displacement current between the plates of a capacitor equals the conduction current in the connecting wires, whose peak value is the applied voltage amplitude divided by the capacitive reactance.
Step 1:State the given quantities.
V0=40 V,f=4×103 Hz,C=12×10−6 F
Step 2:Express the peak displacement current as the peak conduction current.
I0=V0(2πfC)
Step 3:Substitute the values.
I0=40×2π×(4×103)×(12×10−6)
Step 4:Evaluate.
I0=40×0.3016=12.06 A
Final answer: 12 A
Q17Single correctWave Optics
Light emerges out of a convex lens when a source of light kept at its focus. The shape of wavefront of the light is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2plane
Approach:
A point source placed at the focus of a convex lens produces a parallel beam after refraction; the wavefront associated with a parallel (collimated) beam is planar.
Step 1:A source at the focus emits diverging spherical wavefronts toward the lens.
source at focus
Step 2:For object distance equal to the focal length, the refracted rays become parallel to the principal axis.
u=f⇒v=∞
Step 3:A wavefront is perpendicular to the rays; for parallel rays the surface of constant phase is a plane.
rays∥⇒plane wavefront
Final answer: plane
Q18Single correctDual Nature of Radiation and Matter
Given below are two statements : Statement I : Figure shows the variation of stopping potential with frequency (ν) for the two photosensitive materials M1 and M2. The slope gives value of eh, where h is Planck's constant, e is the charge of electron. Statement II : M2 will emit photoelectrons of greater kinetic energy for the incident radiation having same frequency. In the light of the above statements, choose the most appropriate answer from the options given below.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4Statement I is correct and Statement II is incorrect
Approach:
Apply Einstein's photoelectric equation expressed through stopping potential to interpret the slope of the stopping-potential versus frequency graph and the effect of work function on emitted kinetic energy.
Step 1:Rearrange the photoelectric equation for stopping potential.
V0=ehν−eϕ
Step 2:Identify the slope of the line, which is the same for all materials.
slope=eh
Step 3:From the graph, M2 has the larger threshold frequency, hence the larger work function.
ν0,M2>ν0,M1⇒ϕM2>ϕM1
Step 4:For the same incident frequency, the larger work function gives smaller kinetic energy.
KE=hν−ϕ⇒KEM2<KEM1
Final answer: Statement I is correct and Statement II is incorrect
Q19Single correctAtoms
An electron rotates in a circle around a nucleus having positive charge Ze. Correct relation between total energy (E) of electron to its potential energy (U) is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 22E = U
Approach:
Use the virial relation for an electron in a Coulomb orbit, where the kinetic energy is half the magnitude of the potential energy, to relate total energy and potential energy.
Step 1:Equate the Coulomb force to the centripetal force for the circular orbit.
r2kZe2=rmv2
Step 2:Write the kinetic and potential energies.
K=2rkZe2,U=−rkZe2
Step 3:Form the total energy.
E=K+U=−2U+U=2U
Step 4:Rearrange to match the options.
2E=U
Final answer: 2E = U
Q20Single correctSemiconductor Electronics
Following gates section is connected in a complete suitable circuit. For which of the following combination, bulb will glow (ON) :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3A = 1, B = 0, C = 0, D = 0
Approach:
Trace the inputs through the combinational logic network shown in the figure (NAND/NOR/AND stages feeding a final gate driving the bulb) and identify the input combination that produces a logic HIGH output.
Step 1:Evaluate the upper stage that combines inputs A and B.
P=A+B
Step 2:Evaluate the lower stage that combines inputs C and D.
Q=C+D
Step 3:Test the option that gives a HIGH final output, A = 1 with B = C = D = 0.
A=1,B=C=D=0
Step 4:The remaining listed combinations leave the final output LOW.
others⇒Y=0
Final answer: A = 1, B = 0, C = 0, D = 0
Q21NumericalMotion in a Straight Line
A body moves on a frictionless plane starting from rest. If Sn is distance moved between t=n−1 and t=n and Sn−1 is distance moved between t=n−2 and t=n−1, then the ratio SnSn−1 is (1−x2) for n=10. The value of x is _______.
SolutionAnswer: 19
Approach:
Use the distance covered in the nth second for motion from rest under uniform acceleration, form the ratio of successive intervals, and match it to the given expression.
Step 1:Write the distance covered in the nth and (n-1)th seconds.
Sn=2a(2n−1),Sn−1=2a(2n−3)
Step 2:Form the ratio.
SnSn−1=2n−12n−3
Step 3:Rewrite the ratio in the required form.
2n−12n−3=1−2n−12
Step 4:Substitute n=10.
x=2(10)−1=19
Final answer: 19
Q22NumericalLaws of Motion
Three blocks M1,M2,M3 having masses 4 kg, 6 kg and 10 kg respectively are hanging from a smooth pully using rope 1,2 and 3 as shown in figure. The tension in the rope 1, T1 when they are moving upward with acceleration of 2 ms−2 is _______N ( if g =10 m/s2 ).
SolutionAnswer: 240
Approach:
Treat the three hanging blocks as a single system supported by rope 1, then apply Newton's second law for vertical motion to find the tension that accelerates the whole system upward.
Step 1:Add the masses supported by rope 1.
M=M1+M2+M3=4+6+10=20 kg
Step 2:Apply Newton's second law along the vertical for upward acceleration.
T1−Mg=Ma
Step 3:Solve for the tension.
T1=M(g+a)=20(10+2)
Step 4:Evaluate.
T1=240 N
Final answer: 240
Q23NumericalMechanical Properties of Solids
The density and breaking stress of a wire are 6×104 kg/m3 and 1.2×108 N/m2 respectively. The wire is suspended from a rigid support on a planet where acceleration due to gravity is 31rd of the value on the surface of earth. The maximum length of the wire with breaking is _______ m (take, g =10 m/s2 ).
SolutionAnswer: 600
Approach:
The maximum length is reached when the stress from the wire's own weight equals the breaking stress; equate breaking stress to density times effective gravity times length.
Step 1:Determine the effective gravity on the planet.
geff=3g=310 m/s2
Step 2:Set breaking stress equal to self-weight stress at maximum length.
σmax=ρgeffLmax
Step 3:Solve for the length.
Lmax=ρgeffσmax=(6×104)(10/3)1.2×108
Step 4:Evaluate.
Lmax=2×1051.2×108=600 m
Final answer: 600
Q24NumericalElectrostatic Potential and Capacitance
Three capacitors of capacitances 25μF, 30μF and 45μF are connected in parallel to a supply of 100 V. Energy stored in the above combination is E. When these capacitors are connected in series to the same supply, the stored energy is x9E. The value of x is _____.
SolutionAnswer: 86
Approach:
Compute the equivalent capacitance for the parallel and series arrangements, form their ratio (which equals the energy ratio at the same voltage), and match it to the given fraction.
Step 1:Find the parallel equivalent capacitance.
Cp=25+30+45=100μF
Step 2:Find the series equivalent capacitance.
Cs1=251+301+451=90086
Step 3:At the same voltage, energy is proportional to capacitance.
UpUs=CpCs=100900/86=869
Step 4:Compare with the given expression.
x9=869⇒x=86
Final answer: 86
Q26NumericalExperimental Skills
In the experiment to determine the galvanometer resistance by half-deflection method, the plot of θ1 vs the resistance (R) of the resistance box is shown in the figure. The figure of merit of the galvanometer is _____ ×10−2 A / division. [The source has emf 2V]
SolutionAnswer: 5
Approach:
In the half-deflection (figure of merit) experiment the deflection relates to current as I = k(theta), giving 1/theta proportional to R through E = I(R + G). Read the slope of the 1/theta versus R line to obtain the figure of merit k.
Step 1:Express the inverse deflection as a linear function of R.
θ1=EkR+EkG
Step 2:Read the slope from the graph using the marked points.
Step 3:Relate slope to the figure of merit with E = 2 V.
k=E×slope
Step 4:Evaluate to the stated units.
k=5×10−2A/division
Final answer: 5
Q27NumericalMoving Charges and Magnetism
A 2 A current carrying straight metal wire of resistance 1Ω, resistivity 2×10−6Ωm, area of cross-section 10 mm2 and mass 500 g is suspended horizontally in mid air by applying a uniform magnetic field B. The magnitude of B is _____ ×10−1 T (given, g =10 m/s2 ).
SolutionAnswer: 5
Approach:
For the wire to be suspended, the upward magnetic force balances gravity. Find the wire length from its resistance, resistivity and cross-section, then equate magnetic force to weight to solve for B.
Step 1:Find the length from the resistance relation.
L=ρRA=2×10−6(1)(10×10−6)
Step 2:Apply force balance for suspension.
BIL=mg
Step 3:Solve for B.
B=ILmg=(2)(5)(0.5)(10)
Step 4:Evaluate.
B=0.5 T=5×10−1 T
Final answer: 5
Q28NumericalAlternating Current
An ac source is connected in given series LCR circuit. The rms potential difference across the capacitor of 20μF is _____V.
SolutionAnswer: 50
Approach:
Compute the reactances at the source angular frequency, find the circuit impedance and rms current, then multiply the rms current by the capacitive reactance to get the rms voltage across the capacitor.
Step 1:Identify source values: ω=100 rad/s and Vrms=50 V.
V=502sin(100t)⇒ω=100,Vrms=50 V
Step 2:Compute the reactances.
XL=100×1=100Ω,XC=100×20×10−61=500Ω
Step 3:Compute the impedance and rms current.
Z=3002+(100−500)2=90000+160000=500Ω,Irms=50050=0.1 A
Step 4:Compute the rms voltage across the capacitor.
VC=IrmsXC=0.1×500=50 V
Final answer: 50
Q29NumericalWave Optics
In Young's double slit experiment, carried out with light of wavelength 5000 Å, the distance between the slits is 0.3 mm and the screen is at 200 cm from the slits. The central maximum is at x=0 cm. The value of x for third maxima is _____mm.
SolutionAnswer: 10
Approach:
The position of the nth bright fringe in a double-slit pattern is n times the fringe width. Compute the fringe width from wavelength, screen distance and slit separation, then multiply by the order number.
Step 1:List the data in SI units.
λ=5×10−7 m,D=2 m,d=3×10−4 m,n=3
Step 2:Write the position of the third maximum.
x3=d3λD
Step 3:Substitute the values.
x3=3×10−43×(5×10−7)×2
Step 4:Evaluate and convert to millimetres.
x3=10−2 m=10 mm
Final answer: 10
Q30NumericalNuclei
If three helium nuclei combine to form a carbon nucleus then the energy released in this reaction is _____ ×10−2MeV. (Given 1u =931MeV/c2, atomic mass of helium =4.002603u )
SolutionAnswer: 727
Approach:
The energy released equals the mass defect (three helium masses minus the carbon-12 mass) converted to energy using the mass-energy equivalence with the given conversion factor. The carbon-12 mass is exactly 12 u by definition.
Step 1:Compute the total mass of three helium nuclei.
3mHe=3×4.002603=12.007809u
Step 2:Use the carbon-12 mass of exactly 12 u to find the mass defect.
Δm=12.007809−12=0.007809u
Step 3:Convert the mass defect to energy.
Q=0.007809×931=7.270MeV
Step 4:Express in the stated units of 10−2 MeV.
Q=727×10−2MeV
Final answer: 727
Chemistry29 questions
Q31Single correctSome Basic Concepts of Chemistry
An organic compound has 42.1% carbon, 6.4% hydrogen and remainder is oxygen. If its molecular weight is 342, then its molecular formula is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3C12H22O11
Approach:
Determine percentage of oxygen, find moles of each element per 100 g, derive the empirical formula, then scale to the molecular weight 342.
Step 1:Oxygen percentage equals the remainder.
100−42.1−6.4=51.5%
Step 2:Moles per 100 g of compound.
C:1242.1=3.51,H:16.4=6.4,O:1651.5=3.22
Step 3:Divide by the smallest value 3.22.
C:1.09,H:1.99,O:1
Step 4:Empirical formula CH2O has mass 30; scale to 342.
k=30342≈11.4→11 for C12H22O11 (sucrose, M=342)
Final answer: C12H22O11
Q32Single correctSome Basic Concepts of Chemistry
The incorrect postulates of the Dalton's atomic theory are : (A) Atoms of different elements differ in mass. (B) Matter consists of divisible atoms. (C) Compounds are formed when atoms of different element combine in a fixed ratio. (D) All the atoms of given element have different properties including mass. (E) Chemical reactions involve reorganisation of atoms. Choose the correct answer from the options given below :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2(B), (D) only
Approach:
Evaluate each statement against Dalton's atomic theory and identify the ones that contradict it.
Step 1:Statement (A) is a correct postulate: atoms of different elements differ in mass and properties.
Step 2:Statement (B) is incorrect: Dalton stated atoms are indivisible, not divisible.
Step 3:Statement (C) is correct: compounds form when atoms combine in a fixed ratio. Statement (E) is correct: reactions involve rearrangement of atoms.
Step 4:Statement (D) is incorrect: atoms of a given element have identical mass and properties, not different ones.
Final answer: (B), (D) only
Q33Single correctp-Block Elements
Given below are two statements : Statement I : In group 13, the stability of +1 oxidation state increases down the group. Statement II : The atomic size of gallium is greater than that of aluminium. In the light of the above statements, choose the most appropriate answer from the options given below :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4Statement I is correct but Statement II is incorrect
Approach:
Assess each statement using the inert pair effect and the anomalous atomic radii caused by poor d-electron shielding in group 13.
Step 1:Statement I: due to the inert pair effect, the stability of the +1 oxidation state increases down group 13 (B, Al, Ga, In, Tl).
Step 2:Statement II: gallium follows the 3d transition series, so poor shielding by d-electrons gives gallium an atomic radius slightly smaller than that of aluminium.
rGa<rAl
Step 3:Statement I is correct while Statement II is incorrect.
Final answer: Statement I is correct but Statement II is incorrect
Q34Single correctClassification of Elements and Periodicity
The statement(s) that are correct about the species O2−,F−,Na+ and Mg2+. (A) All are isoelectronic (B) All have the same nuclear charge (C) O2− has the largest ionic radii (D) Mg2+ has the smallest ionic radii Choose the most appropriate answer from the options given below :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3(A), (C) and (D) only
Approach:
Count the electrons of each species, compare nuclear charges, and order the ionic radii of an isoelectronic series.
Step 1:Each species has 10 electrons, so all are isoelectronic. Statement (A) is correct.
O2−,F−,Na+,Mg2+:10e−
Step 2:Nuclear charges differ (8, 9, 11, 12), so Statement (B) is incorrect.
Z=8,9,11,12
Step 3:For an isoelectronic series, radius decreases as nuclear charge increases, so O2− is largest and Mg2+ is smallest.
O2−>F−>Na+>Mg2+
Final answer: (A), (C) and (D) only
Q35Single correctThermodynamics
Given below are two statements : One is labelled as Assertion (A) and the other is labelled as Reason (R) Assertion (A) : Enthalpy of neutralisation of strong monobasic acid with strong monoacidic base is always −57 kJ mol−1. Reason (R) : Enthalpy of neutralisation is the amount of heat liberated when one mole of H+ ions furnished by acid combine with one mole of OH− ions furnished by base to form one mole of water. In the light of the above statements, choose the correct answer from the options given below :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3Both (A) and (R) are true and (R) is the correct explanation of (A)
Approach:
Evaluate the truth of the Assertion and Reason about enthalpy of neutralisation, then judge whether the Reason explains the Assertion.
Step 1:Assertion: for strong acid and strong base, complete dissociation gives a constant heat release of about −57 kJ mol−1.
ΔHneut≈−57kJ mol−1
Step 2:Reason: the definition states one mole of H+ combining with one mole of OH− forms one mole of water with heat liberated.
Step 3:Because both strong acid and strong base fully ionise, the only common reaction is formation of water, which fixes the value, so (R) explains (A).
Final answer: Both (A) and (R) are true and (R) is the correct explanation of (A)
Q36Single correctEquilibrium
The following reaction occurs in the Blast furnance where iron ore is reduced to iron metal Fe2O3(s)+3CO(g)⇌Fe(l)+3CO2(g) Using the Le-chatelier's principle, predict which one of the following will not disturb the equilibrium.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3Addition of Fe2O3
Approach:
Apply Le-Chatelier's principle, recognising that pure solids and pure liquids do not appear in the equilibrium expression and therefore cannot shift the equilibrium.
Step 1:Only gaseous species CO and CO2 appear in the equilibrium expression; Fe2O3(s) and Fe(l) are condensed phases.
K=[CO]3[CO2]3
Step 2:Adding CO2, removing CO2, or removing CO changes a gaseous concentration and shifts the equilibrium.
Step 3:Adding solid Fe2O3 does not change its activity, so the equilibrium is undisturbed.
Final answer: Addition of Fe2O3
Q37Single correctp-Block Elements
The number of neutrons present in the more abundant isotope of boron is ' x '. Amorphous boron upon heating with air forms a product, in which the oxidation state of boron is ' y '. The value of x+y is ________
(A)
(B)
(C)
(D)
SolutionAnswer: Option 29
Approach:
Determine the neutron count of the most abundant boron isotope and the oxidation state of boron in the oxide formed on heating in air, then add the two values.
Step 1:The more abundant isotope of boron is 11B, with atomic number 5.
x=11−5=6
Step 2:Heating amorphous boron in air forms boron trioxide B2O3, where boron is in the +3 oxidation state.
4B+3O2→2B2O3
Step 3:Add the two values.
x+y=6+3=9
Final answer: 9
Q38Single correctChemical Bonding and Molecular Structure
Number of σ and π bonds present in ethylene molecule is respectively :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 45 and 1
Approach:
Count the sigma and pi bonds in the structure of ethylene C2H4.
Step 1:Ethylene has four C-H single bonds, each contributing one sigma bond.
4×(C-H σ)
Step 2:The carbon-carbon double bond contributes one sigma and one pi bond.
C=C:1σ+1π
Step 3:Total sigma bonds and pi bonds.
σ=4+1=5,π=1
Final answer: 5 and 1
Q39Single correctHydrocarbons
Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : Cis form of alkene is found to be more polar than the trans form. Reason (R): Dipole moment of trans isomer of 2-butene is zero. In the light of the above statements, choose the correct answer from the options given below :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2Both (A) and (R) are true and (R) is the correct explanation of (A)
Approach:
Compare the resultant dipole moments of cis and trans alkenes and judge whether the Reason justifies the Assertion.
Step 1:In a cis alkene the bond dipoles point to the same side and add, giving a net dipole moment, so the cis form is more polar.
Step 2:In trans-2-butene the two methyl bond dipoles are oriented oppositely and cancel, giving a net dipole moment of zero.
μtrans-2-butene=0
Step 3:The zero dipole of the trans isomer accounts for the cis form being comparatively more polar, so (R) explains (A).
Final answer: Both (A) and (R) are true and (R) is the correct explanation of (A)
Q40Single correctSome Basic Principles of Organic Chemistry
For the Compounds : (A) H3C−CH2−O−CH2−CH2−CH3 (B) H3C−CH2−CH2−CH2−CH3 (C) (D) The increasing order of boiling point is : Choose the correct answer from the options given below :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2(B)<(A)<(C)<(D)
Approach:
Compare the intermolecular forces of the four compounds: alkane (B), ether (A), ketone (C) and alcohol (D), and order them by boiling point.
Step 1:Compound (B) is pentane, a non-polar alkane with only weak London forces, so it has the lowest boiling point.
Step 2:Compound (A) is an ether with a weak permanent dipole but no hydrogen bonding, placing it above the alkane.
Step 3:Compound (C) is a ketone (pentan-3-one) with a stronger dipole-dipole interaction from the C=O group.
Step 4:Compound (D) is an alcohol (pentan-2-ol) capable of intermolecular hydrogen bonding, giving the highest boiling point.
Final answer: (B)<(A)<(C)<(D)
Q41Single correctSome Basic Principles of Organic Chemistry
Given below are two statements: Statement I : Nitration of benzene involves the following step - Statement II : Use of Lewis base promotes the electrophilic substitution of benzene. In the light of the above statements, choose the most appropriate answer from the options given below :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1Statement I is correct but Statement II is incorrect
Approach:
Examine the generation of the nitronium ion in nitration and the role of Lewis acids versus Lewis bases in electrophilic aromatic substitution.
Step 1:Statement I shows protonated nitric acid losing water to generate the nitronium electrophile, which is the correct first step of nitration.
H2O+-NO2⇌H2O+NO2+
Step 2:Electrophilic substitution is promoted by Lewis acids (which generate or strengthen the electrophile), not by Lewis bases.
Step 3:Statement I is correct while Statement II is incorrect.
Final answer: Statement I is correct but Statement II is incorrect
Q42Single correctElectrochemistry
The reaction at cathode in the cells commonly used in clocks involves.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2reduction of Mn from +4 to +3
Approach:
Identify the dry (Leclanche) cell used in clocks and determine the manganese oxidation state change at its cathode.
Step 1:The cell commonly used in clocks is the dry cell, where the cathode reduces MnO2.
Step 2:Manganese in MnO2 is in the +4 state and is reduced to MnO(OH), where it is in the +3 state.
Mn+4+e−→Mn+3
Step 3:The cathode reaction reduces Mn from +4 to +3.
Final answer: reduction of Mn from +4 to +3
Q43Single correctElectrochemistry
Molar ionic conductivities of divalent cation and anion are 57 S cm2 mol−1 and 73 S cm2 mol−1 respectively. The molar conductivity of solution of an electrolyte with the above cation and anion will be :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3130 S cm2 mol−1
Approach:
Apply Kohlrausch's law of independent migration of ions for an electrolyte of divalent cation and divalent anion in a 1:1 ratio.
Step 1:An electrolyte of a divalent cation and divalent anion combines in a 1:1 ratio, giving one cation and one anion per formula unit.
ν+=1,ν−=1
Step 2:Sum the molar ionic conductivities of the cation and anion.
Λm∘=57+73
Step 3:The molar conductivity of the electrolyte solution.
Λm∘=130S cm2mol−1
Final answer: 130 S cm2 mol−1
Q44Single correctd- and f-Block Elements
The metal that shows highest and maximum number of oxidation state is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2Mn
Approach:
Compare the range of oxidation states available to the given first-row transition metals, which depends on the number of unpaired plus paired (n−1)d and ns electrons.
Step 1:Manganese has the configuration [Ar]3d54s2, giving seven electrons available for bonding.
Mn:[Ar]3d54s2
Step 2:Manganese therefore exhibits oxidation states from +2 up to +7, the highest among the listed metals.
+2to+7
Step 3:Fe, Co and Ti show smaller maximum oxidation states than +7, so Mn shows both the highest and the maximum number.
Final answer: Mn
Q45Single correctCoordination Compounds
Which one of the following complexes will exhibit the least paramagnetic behaviour? [Atomic number, Cr=24,Mn=25,Fe=26,Co=27 ]
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3[Co(H2O)6]2+
Approach:
Find the number of unpaired d-electrons in each M2+ ion with the weak-field aqua ligand (high spin), since paramagnetism increases with the number of unpaired electrons.
Step 1:Determine the d-electron count of each divalent ion.
Cr2+:d4,Fe2+:d6,Co2+:d7,Mn2+:d5
Step 2:With the weak-field H2O ligand all are high spin; count unpaired electrons.
Cr2+:4,Fe2+:4,Co2+:3,Mn2+:5
Step 3:The least number of unpaired electrons (3, for Co2+) gives the least paramagnetic behaviour.
μCo2+=3(5)=15≈3.87BM
Final answer: [Co(H2O)6]2+
Q46Single correctCoordination Compounds
The correct order of ligands arranged in increasing field strength.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2Br−<F−<H2O<NH3
Approach:
Rank the ligands by their position in the spectrochemical series, which lists ligands in order of increasing crystal field splitting (field strength).
Step 1:From the spectrochemical series the relevant fragment places the halide and oxygen/nitrogen donors in this order.
Br−<F−<H2O<NH3
Step 2:Check the other options against the series. In option 1 the halide order is reversed; in option 3 hydroxide is placed below water; in option 4 bromide is placed above chloride and hydroxide.
I−<Br−<Cl−<F−<OH−<H2O<NH3<CN−
Step 3:Only option 2 is fully consistent with the spectrochemical series.
Br−<F−<H2O<NH3
Final answer: Br−<F−<H2O<NH3
Q47Single correctHaloalkanes and Haloarenes
Given below are two statement: Statements I : Bromination of phenol in solvent with low polarity such as CHCl3 or CS2 requires Lewis acid catalyst. Statements II : The Lewis acid catalyst polarises the bromine to generate Br+. In the light of the above statements, choose the correct answer from the options given below :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3Statement I is false but Statement II is true
Approach:
Evaluate the truth of each statement about bromination of phenol in low-polarity solvents and the role of the Lewis acid.
Step 1:Phenol is strongly activated by the hydroxyl group, so monobromination in a low-polarity solvent proceeds without any Lewis acid catalyst.
−OH is a strong activating, o,p-directing group
Step 2:When a Lewis acid is used in electrophilic bromination, it polarises the bromine molecule to generate the electrophile.
Br2+Lewis acid→Br++[Lewis acid-Br]−
Step 3:Combining the assessments, Statement I is false and Statement II is true.
I false, II true
Final answer: Statement I is false but Statement II is true
Q48Single correctHaloalkanes and Haloarenes
Identify compound (Z) in the following reaction sequence.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1
Approach:
Trace the Dow process and the subsequent reactions: chlorobenzene with NaOH under high temperature and pressure gives sodium phenoxide, acidification gives phenol, and nitration gives the trinitro product.
Step 1:Chlorobenzene with NaOH at 623 K and 300 atm gives sodium phenoxide (X).
C6H5Cl+NaOH→C6H5ONa(X)
Step 2:Treatment of sodium phenoxide with HCl liberates phenol (Y).
C6H5ONa+HCl→C6H5OH(Y)
Step 3:Nitration of phenol with concentrated nitric acid gives 2,4,6-trinitrophenol (picric acid) as Z.
C6H5OHConc. HNO32,4,6-trinitrophenol(Z)
Final answer: 2,4,6-trinitrophenol (picric acid)
Q49Single correctAldehydes, Ketones and Carboxylic Acids
Identify ' A ' in the following reaction:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4
Approach:
The reagents hydrazine followed by base (Wolff-Kishner reduction) convert the carbonyl group of the ketone to a methylene group.
Step 1:The starting carbonyl is butan-2-one (a methyl ethyl ketone framework drawn as CH3-CO-CH2-CH3).
CH3COCH2CH3
Step 2:Hydrazine forms the hydrazone, and ethylene glycol/KOH on heating decomposes it, replacing the C=O with CH2.
C=O→C=N−NH2→CH2
Step 3:The carbon skeleton is retained, so butan-2-one is reduced to butane.
CH3CH2CH2CH3
Final answer: Butane (option 4)
Q50Single correctBiomolecules
Which of the following gives a positive test with ninhydrin?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2Egg albumin
Approach:
The ninhydrin test detects free amino groups of amino acids and proteins, giving a violet colour.
Step 1:Ninhydrin reacts with the free amino groups present in amino acids and proteins.
−NH2+ninhydrin
Step 2:Starch and cellulose are polysaccharides and polyvinyl chloride is a synthetic polymer; none contain amino groups.
no -NH2
Step 3:Egg albumin is a protein and contains free amino groups, so it responds positively.
protein → positive
Final answer: Egg albumin
Q51NumericalAmines
9.3 g of pure aniline is treated with bromine water at room temperature to give a white precipitate of the product ' P '. The mass of product ' P ' obtaind is 26.4 g. The percentage yield is _________ %.
SolutionAnswer: 80
Approach:
Aniline reacts with bromine water to give 2,4,6-tribromoaniline. Compute moles of aniline, the theoretical mass of product, and compare with the actual mass.
Step 1:Moles of aniline from its molar mass 93 g/mol.
n=939.3=0.1mol
Step 2:Each mole of aniline gives one mole of 2,4,6-tribromoaniline, whose molar mass is 330 g/mol.
M(C6H2Br3NH2)=330g/mol
Step 3:Substitute the actual and theoretical masses into the yield expression.
%yield=3326.4×100
Final answer: 80
Q52NumericalStructure of Atom
The value of Rydberg constant (RH) is 2.18×10−18 J. The velocity of electron having mass 9.1×10−31 kg in Bohr's first orbit of hydrogen atom = _________ ×105 ms−1 (nearest integer).
SolutionAnswer: 22
Approach:
Equate the Rydberg constant (in joules) to the ionisation energy of hydrogen's ground state, which equals the kinetic energy of the electron, and solve for the speed.
Step 1:The magnitude of the total energy of the first orbit equals the kinetic energy.
21mv2=RH=2.18×10−18J
Step 2:Solve for the speed of the electron.
v=m2RH=9.1×10−312×2.18×10−18
Step 3:Evaluate the square root and express in the required units.
v≈2.19×106=21.9×105ms−1
Final answer: 22
Q53NumericalChemical Bonding and Molecular Structure
In the lewis dot structure for NO2−, total number of valence electrons around nitrogen is _________
SolutionAnswer: 8
Approach:
Draw the Lewis structure of the nitrite ion and count all valence electrons (bonding plus lone pair) surrounding the central nitrogen atom.
Step 1:In nitrite the nitrogen forms one double bond and one single bond with the two oxygen atoms and carries one lone pair.
O=N−O−
Step 2:Count bonding electrons: a double bond contributes 4 electrons and a single bond contributes 2 electrons.
4+2=6
Step 3:Add the 2 electrons of the lone pair on nitrogen.
6+2=8
Final answer: 8
Q54NumericalThermodynamics
The heat of combustion of solid benzoic acid at constant volume is −321.30 kJ at 27∘C. The heat of combustion at constant pressure is (−321.30−xR)kJ, the value of x is _________.
SolutionAnswer: 150
Approach:
Relate the enthalpy of combustion to the internal energy change using the change in moles of gas for benzoic acid combustion, then identify the coefficient of R.
Step 1:Determine the change in moles of gas: 7 moles of gaseous products minus 7.5 moles of gaseous reactants.
Δng=7−215=−21
Step 2:Apply the enthalpy relation at T = 300 K.
ΔH=ΔU+ΔngRT=−321.30+(−0.5)(R)(300)
Step 3:Comparing with (−321.30−xR) kJ gives the coefficient of R.
x=0.5×300=150
Final answer: 150
Q55NumericalSolutions
An artificial cell is made by encapsulating 0.2M glucose solution within a semipermeable membrane. The osmotic pressure developed when the artificial cell is placed within a 0.05M solution of NaCl at 300 K is _________ ×10−1 bar. (nearest integer). [Given : R = 0.083 Lbarmol−1K−1] Assume complete dissociation of NaCl
SolutionAnswer: 25
Approach:
Compute the effective osmolarity inside and outside the cell, take the difference, and apply the osmotic pressure equation to the net concentration.
Step 1:Glucose does not dissociate so its osmolarity is 0.2 M; NaCl dissociates completely (i = 2) giving 0.1 M.
Cin=0.2M,Cout=2×0.05=0.1M
Step 2:The net concentration driving osmosis across the membrane is the difference.
ΔC=0.2−0.1=0.1M
Step 3:Apply the osmotic pressure equation at 300 K.
π=0.1×0.083×300=2.49bar=24.9×10−1bar
Final answer: 25
Q56NumericalChemical Kinetics
During Kinetic study of reaction 2A+B→C+D, the following results were obtained : | A [M] | B [M] | initial rate of formation of D I | 0.1 | 0.1 | 6.0×10−3 II | 0.3 | 0.2 | 7.20×10−2 III | 0.3 | 0.4 | 2.88×10−1 IV | 0.4 | 0.1 | 2.40×10−2 Based on above data, overall order of the reaction is _________
SolutionAnswer: 3
Approach:
Determine the order with respect to each reactant by comparing experiments where one concentration is held constant, then sum the orders.
Step 1:Compare experiments I and IV where B is constant: A increases from 0.1 to 0.4 (factor 4) and the rate increases from 6.0×10−3 to 2.40×10−2 (factor 4).
4m=4⇒m=1
Step 2:Compare experiments II and III where A is constant: B increases from 0.2 to 0.4 (factor 2) and the rate increases from 7.20×10−2 to 2.88×10−1 (factor 4).
2n=4⇒n=2
Step 3:Add the individual orders to obtain the overall order.
m+n=1+2=3
Final answer: 3
Q57NumericalThe d- and f-Block Elements
The spin-only magnetic moment value of the ion among Ti2+, V2+, Co3+ and Cr2+, that acts as strong oxidising agent in aqueous solution is _________ BM (Near integer). (Given atomic numbers : Ti : 22, V : 23, Cr : 24, Co : 27)
SolutionAnswer: 5
Approach:
Identify which of the listed ions is a strong oxidising agent in aqueous solution, determine its number of unpaired electrons, then apply the spin-only formula.
Step 1:Among the ions, Co3+ is the strong oxidising agent in aqueous solution as it tends to be reduced to the more stable Co2+.
Co3++e−→Co2+
Step 2:Cobalt is [Ar]3d74s2, so Co3+ is 3d6; in the weak aqueous field this is high spin with 4 unpaired electrons.
Co3+:3d6⇒n=4
Step 3:Apply the spin-only formula.
μ=4(4+2)=24=4.9BM
Final answer: 5
Q58NumericalAmines
The number of halobenzenes from the following that can be prepared by Sandmeyer's reaction is _________
SolutionAnswer: 2
Approach:
Recall which aryl halides are accessible by the Sandmeyer reaction, which converts an arenediazonium salt to an aryl halide using cuprous halides.
Step 1:The Sandmeyer reaction uses cuprous chloride (CuCl) and cuprous bromide (CuBr) to give chlorobenzene and bromobenzene.
ArN2++CuCl/CuBr→ArCl/ArBr
Step 2:Fluorobenzene is prepared by the Balz-Schiemann reaction and iodobenzene by reaction with potassium iodide; astatine compounds are not prepared this way.
ArF←Balz-Schiemann,ArI←KI
Step 3:Count the halobenzenes (chlorobenzene and bromobenzene) obtainable by the Sandmeyer reaction.
2
Final answer: 2
Q59NumericalAlcohols, Phenols and Ethers
Consider the given chemical reaction sequence : Total sum of oxygen atoms in Product A and Product B are _________
SolutionAnswer: 14
Approach:
Determine Product A from sulphonation of phenol and Product B from nitration of that product, then count the oxygen atoms in each.
Step 1:Phenol with concentrated sulphuric acid undergoes disulphonation to give 4-hydroxybenzene-1,3-disulphonic acid (Product A), with the hydroxyl oxygen plus two sulphonic acid groups.
A:−OH+2(−SO3H)
Step 2:Nitration of Product A with concentrated nitric acid replaces the sulphonic acid groups, yielding 2,4,6-trinitrophenol (Product B) with the hydroxyl oxygen and three nitro groups.
B:−OH+3(−NO2)
Step 3:Sum the oxygen atoms of Products A and B.
7+7=14
Final answer: 14
Mathematics30 questions
Q61Single correctComplex Numbers and Quadratic Equations
Consider the following two statements : Statement I : For any two non-zero complex numbers z1,z2, (∣z1∣+∣z2∣)∣z1∣z1+∣z2∣z2≤2(∣z1∣+∣z2∣), and Statement II : If x, y, z are three distinct complex numbers and a, b, c are three positive real numbers such that ∣y−z∣a=∣z−x∣b=∣x−y∣c, then y−za2+z−xb2+x−yc2=1. Between the above two statements,
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1Statement I is correct but Statement II is incorrect.
Approach:
Statement I is examined with the triangle inequality applied to unit-modulus vectors; Statement II is tested by checking whether the stated identity equals 1.
Step 1:Each term has unit modulus, so the sum of the two unit vectors has modulus at most 2.
∣z1∣z1+∣z2∣z2≤1+1=2
Step 2:Multiplying both sides by the positive quantity preserves the inequality, confirming Statement I.
(∣z1∣+∣z2∣)∣z1∣z1+∣z2∣z2≤2(∣z1∣+∣z2∣)
Step 3:In Statement II the common ratio condition makes the three fractions complex quantities summing to a complex value that is not generally equal to the real number 1.
y−za2+z−xb2+x−yc2=1
Step 4:Combining the two evaluations identifies the matching option.
I correct, II incorrect
Final answer: Statement I is correct but Statement II is incorrect.
Q62Single correctSequences and Series
If 1+21+2+31+…+99+1001=m and 1⋅21+2⋅31+…+99⋅1001=n, then the point (m, n) lies on the line
(A)
(B)
(C)
(D)
SolutionAnswer: Option 211x−100y=0
Approach:
Each series telescopes; the resulting values of m and n are substituted into the candidate lines.
Step 1:Rationalising each term of the first series gives a telescoping difference of square roots.
m=∑k=199(k+1−k)=100−1=9
Step 2:Splitting each term of the second series by partial fractions telescopes the sum.
n=∑k=199(k1−k+11)=1−1001=10099
Step 3:Substituting the point into the second candidate line.
11(9)−100(10099)=99−99=0
Step 4:The point satisfies the line 11x−100y=0.
(m,n)=(9,10099)
Final answer: 11x−100y=0
Q63Single correctTrigonometry
Suppose θ∈[0,4π] is a solution of 4cosθ−3sinθ=1. Then cosθ is equal to :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3(36−2)4
Approach:
The equation is written in terms of cosθ using sinθ=1−cos2θ on the given interval, then solved as a quadratic.
Step 1:Isolating the sine term and squaring.
3sinθ=4cosθ−1⇒9sin2θ=(4cosθ−1)2
Step 2:Replacing sin2θ and expanding gives a quadratic in c=cosθ.
9(1−c2)=16c2−8c+1⇒25c2−8c−8=0
Step 3:Solving the quadratic and keeping the positive root valid on [0,π/4].
c=508+64+800=508+126=254+66
Step 4:Rewriting the root in the offered form by rationalising against 36−2.
36−24=54−44(36+2)=50126+8=254+66
Final answer: (36−2)4
Q64Single correctCoordinate Geometry
Let two straight lines drawn from the origin O intersect the line 3x+4y=12 at the points P and Q such that △OPQ is an isosceles triangle and ∠POQ=90∘. If l=OP2+PQ2+QO2, then the greatest integer less than or equal to l is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 246
Approach:
The perpendicular distance from the origin to the line is the altitude of the right isosceles triangle, which fixes the equal legs and the hypotenuse.
Step 1:Distance from O to the line gives the altitude from the right angle to the hypotenuse PQ.
d=9+16∣−12∣=512
Step 2:For a right isosceles triangle the altitude to the hypotenuse equals half the hypotenuse.
Let a circle C of radius 1 and closer to the origin be such that the lines passing through the point (3,2) and parallel to the coordinate axes touch it. Then the shortest distance of the circle C from the point (5,5) is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 34
Approach:
The two lines through (3,2) parallel to the axes are tangents, so the centre lies one radius inside each; the centre closer to the origin is selected, then the distance to the external point is reduced by the radius.
Step 1:The tangent lines x=3 and y=2 place the centre at distance 1 from each; the centre nearer the origin is at (2,1).
C=(3−1,2−1)=(2,1)
Step 2:Distance from the centre to the point (5,5).
∣CP∣=(5−2)2+(5−1)2=9+16=5
Step 3:Subtracting the radius gives the shortest distance from the circle.
dmin=5−1=4
Step 4:The shortest distance equals 4.
dmin=4
Final answer: 4
Q67Single correctCoordinate Geometry
If the line 2x+3y−k=0,k>0, intersect the x-axis and y-axis at the points A and B, respectively. If the equation of the circle having the line segment AB as a diameter is x2+y2−3x−2y=0 and the length of the latus rectum of the ellipse x2+9y2=k2 is nm, where m and n are coprime, then 2m+n is equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 111
Approach:
The intercepts A and B are found from the diameter circle, fixing k; the ellipse is then put in standard form and its latus rectum computed.
Step 1:The diameter circle passes through the intercepts; its x and y intercepts are A(3,0) and B(0,2), giving the line through them.
2(3)+3(0)=k⇒k=6
Step 2:Substituting k=6 into the ellipse and dividing through.
x2+9y2=36⇒36x2+4y2=1
Step 3:Computing the latus rectum.
L=a2b2=62⋅4=68=34
Step 4:With m=4 and n=3 coprime, evaluating 2m+n.
2m+n=2(4)+3=11
Final answer: 11
Q68Single correctMatrices and Determinants
Let A and B be two square matrices of order 3 such that ∣A∣=3 and ∣B∣=2. Then ATA(adj(2A))−1(adj(4B))(adj(AB))−1AAT is equal to :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 464
Approach:
The determinant of the product is evaluated using the scalar-multiple, adjoint, and inverse determinant rules for order-3 matrices.
Step 1:Evaluating the determinant of each factor for order n=3.
Step 2:Writing the inverse adjoint determinants as reciprocals.
∣(adj(2A))−1∣=(24)21,∣(adj(AB))−1∣=(6)21
Step 3:Multiplying all determinant contributions.
9⋅9⋅5761⋅(128)2⋅361=576⋅3681⋅16384
Step 4:Simplifying the product.
207361327104=64
Final answer: 64
Q69Single correctMatrices and Determinants
If the system of equations 11x+y+λz=−5, 2x+3y+5z=3, 8x−19y−39z=μ has infinitely many solutions, then λ4−μ is equal to :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 347
Approach:
For infinitely many solutions the coefficient determinant vanishes, fixing λ; consistency of the augmented system then fixes μ.
Step 1:Expressing the third row as a combination of the first two to satisfy dependence in the coefficients.
8=11a+2b,−19=a+3b,−39=λa+5b
Step 2:Using the third coefficient relation to find λ.
−39=2λ+5(−7)⇒2λ=−4⇒λ=−2
Step 3:Applying the same combination to the constants gives μ.
μ=2(−5)+(−7)(3)=−10−21=−31
Step 4:Evaluating the required expression.
λ4−μ=(−2)4−(−31)=16+31=47
Final answer: 47
Q70Single correctPermutations and Combinations
Let A={1,3,7,9,11} and B={2,4,5,7,8,10,12}. Then the total number of one-one maps f:A→B, such that f(1)+f(3)=14, is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2240
Approach:
Ordered pairs of distinct elements of B summing to 14 are counted for f(1),f(3), then the remaining three elements of A are mapped injectively into the leftover elements of B.
Step 1:Listing distinct ordered pairs from B with sum 14.
(2,12),(12,2),(4,10),(10,4)
Step 2:Each choice fixes two images, leaving five elements of B for the remaining three elements of A.
∣B∣−2=5 available for f(7),f(9),f(11)
Step 3:Injectively assigning the remaining three elements.
5P3=5⋅4⋅3=60
Step 4:Applying the multiplication principle.
4×60=240
Final answer: 240
Q71Single correctDifferential Calculus
Let f(x)=x5+2x3+3x+1,x∈R, and g(x) be a function such that g(f(x))=x for all x∈R. Then g′(7)g(7) is equal to :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 114
Approach:
g is the inverse of f; the value at 7 and the inverse-derivative rule give g(7) and g′(7).
Step 1:Finding the argument that maps to 7.
f(1)=1+2+3+1=7⇒g(7)=1
Step 2:Differentiating f.
f′(x)=5x4+6x2+3
Step 3:Applying the inverse-derivative rule at 7.
g′(7)=f′(1)1=141
Step 4:Forming the required ratio.
g′(7)g(7)=1/141=14
Final answer: 14
Q72Single correctDifferential Calculus
If the function f(x)=x3sin3x+αsinx−βcos3x,x∈R, is continuous at x=0, then f(0) is equal to :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4-4
Approach:
The numerator is expanded as a Maclaurin series; the constant, x, and x2 coefficients must vanish for a finite limit, after which the x3 coefficient gives f(0).
Step 1:Expanding each term to order x3.
sin3x=3x−627x3,αsinx=αx−6αx3,βcos3x=β−29βx2
Step 2:Setting the constant and x2 terms of the numerator to zero, and the x term to zero for a finite cube-order limit.
−β=0,29β=0,3+α=0
Step 3:Collecting the x3 coefficient of the numerator.
−627−6α=−627+63=−624=−4
Step 4:Dividing by x3 and taking the limit defines f(0).
f(0)=limx→0x3−4x3=−4
Final answer: -4
Q73Single correctDifferential Calculus
Let a rectangle ABCD of sides 2 and 4 be inscribed in another rectangle PQRS such that the vertices of the rectangle ABCD lie on the sides of the rectangle PQRS. Let a and b be the sides of the rectangle PQRS when its area is maximum. Then (a+b)2 is equal to :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 172
Approach:
The sides of PQRS are written in terms of the inclination θ of ABCD, the area is maximised over θ, and the resulting sides are summed.
Step 1:The outer sides are projections of the inner rectangle's sides at angle θ.
a=4cosθ+2sinθ,b=4sinθ+2cosθ
Step 2:Writing the area and simplifying.
ab=(4cosθ+2sinθ)(4sinθ+2cosθ)=8+10sin2θ
Step 3:Area is maximal when sin2θ=1, i.e. θ=4π.
a=b=26=32
Step 4:Evaluating (a+b)2.
(a+b)2=(62)2=72
Final answer: 72
Q74Single correctDifferential Calculus
For the function f(x)=sinx+3x−π2(x2+x), where x∈[0,2π], consider the following two statements : (I) f is increasing in (0,2π). (II) f' is decreasing in (0,2π). Between the above two statements,
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4both (I) and (II) are true
Approach:
The first derivative is tested for positivity on the interval, and the second derivative is tested for negativity to assess monotonicity of f′.
Step 1:Differentiating f.
f′(x)=cosx+3−π2(2x+1)
Step 2:Differentiating again gives a strictly negative second derivative on the interval.
f′′(x)=−sinx−π4<0 on (0,2π)
Step 3:Since f′ decreases, its minimum on the closed interval is at the right end, where it stays positive.
f′(2π)=0+3−π2(π+1)=1−π2>0
Step 4:Positive derivative throughout makes f increasing, so both statements hold.
f′(x)>0⇒f increasing
Final answer: both (I) and (II) are true
Q75Single correctIntegral Calculus
The value of ∫−ππ1+cos2y2y(1+siny)dy is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4π2
Approach:
The integrand is split by parity over the symmetric interval; the odd part integrates to zero and the even part is handled with a king-property substitution.
Step 1:Separating the integrand into two pieces.
I=∫−ππ1+cos2y2ydy+∫−ππ1+cos2y2ysinydy
Step 2:The first integrand is odd and vanishes over the symmetric interval.
∫−ππ1+cos2y2ydy=0
Step 3:The second integrand is even, so it doubles the integral over [0,π]; applying y→π−y reduces it.
I=4∫0π1+cos2yysinydy=2π∫0π1+cos2ysinydy
Step 4:Evaluating with t=cosy gives an arctangent.
2π[−tan−1(cosy)]0π=2π(4π+4π)=π2
Final answer: π2
Q76Single correctIntegral Calculus
The integral ∫0π/43sinx+5cosx136sinxdx is equal to :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 13π−50loge2+20loge5
Approach:
Express the numerator as a linear combination of the denominator and its derivative, then integrate term by term.
Step 1:Match coefficients of sine and cosine.
3A−5B=136,5A+3B=0
Step 2:Split the integrand using the decomposition.
3sinx+5cosx136sinx=12−203sinx+5cosx3cosx−5sinx
Step 3:Integrate over the interval.
∫0π/412dx−20[ln(3sinx+5cosx)]0π/4
Step 4:Simplify the logarithm.
3π−20ln(42)+20ln5=3π−50ln2+20ln5
Final answer: 3π−50loge2+20loge5
Q77Single correctDifferential Equations
If y=y(x) is the solution of the differential equation dxdy+2y=sin(2x),y(0)=43, then y(8π) is equal to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3e−π/4
Approach:
Solve the first-order linear ODE using an integrating factor, apply the initial condition, then evaluate at the required point.
Step 1:Multiply through by the integrating factor.
dxd(e2xy)=e2xsin(2x)
Step 2:Integrate the right side.
∫e2xsin(2x)dx=4e2x(sin2x−cos2x)
Step 3:Apply y(0)=43.
43=41(0−1)+C⇒C=1
Step 4:Evaluate at x=8π.
y=41(sin4π−cos4π)+e−π/4=e−π/4
Final answer: e−π/4
Q78Single correctThree Dimensional Geometry
If the line 32−x=4λ+13y−2=4−z makes a right angle with the line 3μx+3=61−2y=75−z, then 4λ+9μ is equal to :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 46
Approach:
Write each line in standard symmetric form to read off direction ratios, then set the dot product to zero for perpendicularity.
Step 1:Rewrite the first line so each coordinate has unit coefficient.
−3x−2=34λ+1y−32=−1z−4
Step 2:Rewrite the second line similarly.
3μx+3=−3y−21=−7z−5
Step 3:Apply the perpendicularity condition.
(−3)(3μ)+(4λ+1)(−3)+(−3)(−7)=0
Step 4:Simplify.
12λ+9μ=18⇒4λ+9μ=6
Final answer: 6
Q79Single correctThree Dimensional Geometry
Let d be the distance of the point of intersection of the lines 3x+6=2y=1z+1 and 4x−7=3y−9=2z−4 from the point (7,8,9). Then d2+6 is equal to :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 475
Approach:
Parameterize both lines, equate to find the common point, then compute the squared distance to the given point.
Step 1:Parameterize the lines.
L1:(3t−6,2t,t−1),L2:(4s+7,3s+9,2s+4)
Step 2:Equate coordinates to find the intersection.
3t−6=4s+7,2t=3s+9,t−1=2s+4
Step 3:Substitute to obtain the point of intersection.
(3,6,2)
Step 4:Compute the squared distance from (7,8,9) and add 6.
d2=42+22+72=69,d2+6=75
Final answer: 75
Q80Single correctProbability
The coefficients a, b, c in the quadratic equation ax2+bx+c=0 are chosen from the set {1,2,3,4,5,6,7,8}. The probability of this equation having repeated roots is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2641
Approach:
Count ordered triples (a,b,c) from the set that satisfy the repeated-root condition b2=4ac, then divide by the total.
Step 1:Count the total number of ordered selections.
8×8×8=512
Step 2:Require b2=4ac, so b must be even and 4ac a perfect square.
Step 4:Count favourable triples and form the probability.
1+3+4=8⇒P=5128=641
Final answer: 641
Q81NumericalPermutations and Combinations
The number of ways of getting a sum 16 on throwing a dice four times is ______
SolutionAnswer: 125
Approach:
Count ordered solutions of x1+x2+x3+x4=16 with each xi∈{1,…,6} using the coefficient of x16 in the generating function.
Step 1:Shift variables by setting yi=xi−1, requiring y1+y2+y3+y4=12 with 0≤yi≤5.
∑yi=12,0≤yi≤5
Step 2:Apply inclusion-exclusion on the upper bound yi≤5.
(315)−(14)(39)+(24)(33)
Step 3:Combine the terms.
455−336+6=125
Final answer: 125
Q82NumericalSequences and Series
Let a1,a2,a3,… be in an arithmetic progression of positive terms. Let Ak=a12−a22+a32−a42+…+a2k−12−a2k2. If A3=−153, A5=−435 and a12+a22+a32=66, then a17−A7 is equal to ______
SolutionAnswer: 910
Approach:
Use the difference-of-squares pairing in Ak, express it through the first term and common difference, solve the system, then evaluate.
Step 3:Subtract to get 4d2=36 and use the positivity and a12+a22+a32=66 to fix signs.
d=3,a=1
Step 4:Compute a17 and A7.
a17=49,A7=−7⋅3[2+13⋅3]=−861
Final answer: 910
Q83NumericalBinomial Theorem
If the constant term in the expansion of (1+2x−3x3)(23x2−3x1)9 is p, then 108p is equal to ______
SolutionAnswer: 54
Approach:
Find the general term of the binomial power, identify which multiplying factor combines with it to give degree zero, and sum the contributions.
Step 1:The power of x in Tr+1 is 2(9−r)−r=18−3r.
x18−3r
Step 2:Combine with 1 (degree 0): need 18−3r=0⇒r=6.
(69)(23)3(−31)6=84⋅827⋅7291
Step 3:Combine with 2x (degree 1): need 18−3r=−1, no integer r; combine with −3x3: need 18−3r=−3⇒r=7.
−3(79)(23)2(−31)7=−3⋅36⋅49⋅(−21871)
Step 4:Add the contributions and multiply by 108.
p=187+91=21,108p=54
Final answer: 54
Q84NumericalConic Sections
Suppose AB is a focal chord of the parabola y2=12x of length l and slope m<3. If the distance of the chord AB from the origin is d, then ld2 is equal to ______
SolutionAnswer: 108
Approach:
Express the focal chord length and the perpendicular distance from the origin in terms of the slope, then multiply.
Step 1:For y2=12x, 4a=12, so a=3 and the focus is (3,0).
a=3
Step 2:With slope m=tanθ, the focal chord length is l=sin2θ12=m212(1+m2).
l=m212(1+m2)
Step 3:The chord line through (3,0) has distance d=m2+13∣m∣ from the origin.
d2=m2+19m2
Step 4:Multiply.
ld2=m212(1+m2)⋅m2+19m2=108
Final answer: 108
Q85NumericalDifferential Equations
Let f be a differentiable function in the interval (0,∞) such that f(1)=1 and limt→xt−xt2f(x)−x2f(t)=1 for each x>0. Then 2f(2)+3f(3) is equal to ______
SolutionAnswer: 24
Approach:
Evaluate the 00 limit by differentiation to obtain a first-order linear ODE, solve it with the initial condition, then evaluate.
Step 1:The numerator vanishes at t=x; differentiating with respect to t and taking t→x gives the limit value.
2xf(x)−x2f′(x)=1
Step 2:Rewrite in standard linear form.
f′(x)−x2f(x)=−x21
Step 3:Solve using the integrating factor.
dxd(x−2f)=−x−4⇒x−2f=3x31+C
Step 4:Apply f(1)=1 to find C=32, then evaluate.
f(x)=3x1+32x2,2f(2)+3f(3)=2⋅617+3⋅319=24
Final answer: 24
Q86NumericalProbability
From a lot of 10 items, which include 3 defective items, a sample of 5 items is drawn at random. Let the random variable X denote the number of defective items in the sample. If the variance of X is σ2, then 96σ2 is equal to ______
SolutionAnswer: 56
Approach:
Recognize X as hypergeometric (sampling without replacement) and apply the variance formula.
If S={a∈R:∣2a−1∣=3[a]+2{a}}, where [t] denotes the greatest integer less than or equal to t and {t} represents the fractional part of t, then 72∑a∈Sa is equal to ______
SolutionAnswer: 18
Approach:
Write a=n+f with n=[a] integer and f={a}∈[0,1), then solve the modulus equation case by case.
How many questions are in the JEE Main 2024 April 05, Shift 1 paper?
The JEE Main 2024 April 05, Shift 1 paper has 88 questions — Physics (29), Chemistry (29) and Mathematics (30). Every question is on this page with its correct answer and a step-by-step solution.
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