JEE Main 2024 April 06, Shift 1 Question Paper with Solutions
All 89 questions from the JEE Main 2024 (April 06, Shift 1) shift — Physics (30), Chemistry (30) and Mathematics (29) — 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.
To find the spring constant (k) of a spring experimentally, a student commits 2% positive error in the measurement of time and 1% negative error in measurement of mass. The percentage error in determining value of k is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 15%
Approach:
Relate the spring constant to the time period of a mass-spring oscillation and propagate the fractional errors in mass and time.
Step 1:Express the spring constant from the period relation.
k=T24π2m
Step 2:Write the maximum fractional error by adding the magnitudes of contributions.
kΔk=mΔm+2TΔT
Step 3:Substitute the given percentage errors of 1% in mass and 2% in time.
kΔk=1%+2×2%
Final answer: 5%
Q2Single correctUnits and Measurements
Match List I with List II
LIST I
LIST II
A. Torque
I.[M1L1T−2A−2]
B. Magnetic field
II.[L2A1]
C. Magnetic moment
III.[M1T−2A−1]
D. Permeability of free space
IV.[M1L2T−2]
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3A-IV, B-III, C-II, D-I
Approach:
Derive the dimensional formula of each physical quantity in List I and match it with the corresponding entry in List II.
Step 1:Torque has dimensions of energy, giving entry IV.
Step 2:Magnetic field from the magnetic force expression gives entry III.
Step 3:Magnetic moment as current times area gives entry II.
Step 4:Permeability of free space gives entry I.
Final answer: A-IV, B-III, C-II, D-I
Q3Single correctKinematics
A train starting from rest first accelerates uniformly up to a speed of 80 km/h for time t, then it moves with a constant speed for time 3t. The average speed of the train for this duration of journey will be (in km/h ) :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 470
Approach:
Compute the total distance over the two phases of motion and divide by the total time.
Step 1:Distance in the accelerating phase from rest to 80 km/h over time t.
s1=20+80t=40t
Step 2:Distance at constant speed for time 3t.
s2=80×3t=240t
Step 3:Divide total distance by total time 4t.
vˉ=4t40t+240t=4t280t
Final answer: 70
Q4Single correctLaws of Motion
A light string passing over a smooth light pulley connects two blocks of masses m1 and m2 (where m2>m1). If the acceleration of the system is 2g, then the ratio of the masses m2m1 is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 22+12−1
Approach:
Use the acceleration of an Atwood machine and solve for the mass ratio from the given acceleration.
Step 1:Set the Atwood acceleration equal to the given value.
m1+m2(m2−m1)g=2g
Step 2:Cancel g and cross-multiply.
2(m2−m1)=m1+m2
Step 3:Group the mass terms.
m2(2−1)=m1(2+1)
Step 4:Form the required ratio.
m2m1=2+12−1
Final answer: 2+12−1
Q5Single correctWork, Energy and Power
A bullet of mass 50 g is fired with a speed 100 m/s on a plywood and emerges with 40 m/s. The percentage loss of kinetic energy is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 184%
Approach:
Express the fractional loss of kinetic energy through the ratio of final to initial speeds, independent of the mass.
Step 1:Write the ratio of final to initial kinetic energy.
KEiKEf=vi2vf2=1002402
Step 2:Subtract from unity to obtain the fractional loss.
KEiΔKE=1−0.16
Step 3:Convert to a percentage.
0.84×100%
Final answer: 84%
Q6Single correctWork, Energy and Power
Four particles A, B, C, D of mass 2m,m,2m,4m, have same momentum, respectively. The particle with maximum kinetic energy is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2A
Approach:
Relate kinetic energy to momentum and mass, then identify the particle of smallest mass for fixed momentum.
Step 1:For a fixed momentum p, kinetic energy varies inversely with mass.
KE=2mp2∝m1
Step 2:Compare the four masses.
2m<m<2m<4m
Step 3:Identify the particle with maximum kinetic energy.
KEA=2(m/2)p2=mp2
Final answer: A
Q7Single correctGravitation
To project a body of mass m from earth's surface to infinity, the required kinetic energy is (assume, the radius of earth is RE,g= acceleration due to gravity on the surface of earth):
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3mgRE
Approach:
Equate the required kinetic energy to the gravitational binding energy at the earth's surface and simplify using g=GM/RE2.
Step 1:The kinetic energy needed equals the magnitude of the binding energy.
KE=REGMm
Step 2:Substitute GM=gRE2.
KE=RE(gRE2)m
Step 3:Cancel one factor of RE.
KE=mgRE
Final answer: mgRE
Q8Single correctMechanical Properties of Fluids
A small ball of mass m and density ρ is dropped in a viscous liquid of density ρ0. After sometime, the ball falls with constant velocity. The viscous force on the ball is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4mg(1−ρρ0)
Approach:
At terminal velocity the net force is zero, so the viscous force balances the difference between weight and buoyancy.
Step 1:Volume of the ball expressed through its mass and density.
V=ρm
Step 2:Buoyant force using the displaced liquid mass.
FB=Vρ0g=ρmρ0g
Step 3:At constant velocity the viscous force balances weight minus buoyancy.
Fv=mg−ρmρ0g
Step 4:Factor out mg.
Fv=mg(1−ρρ0)
Final answer: mg(1−ρρ0)
Q9Single correctKinetic Theory of Gases
A sample contains mixture of helium and oxygen gas. The ratio of root mean square speed of helium and oxygen in the sample, is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2122
Approach:
At the same temperature the root mean square speed is inversely proportional to the square root of the molar mass; take the ratio for helium and oxygen.
Step 1:For a common temperature the speed ratio depends only on molar masses.
vO2vHe=MHeMO2
Step 2:Substitute molar masses 32 for oxygen and 4 for helium.
vO2vHe=432
Step 3:Simplify the surd.
8=22
Final answer: 122
Q10Single correctThermodynamics
The specific heat at constant pressure of a real gas obeying PV2=RT equation is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3CV+2VR
Approach:
Apply the first law for one mole with the given equation of state, expressing the heat per unit temperature change at constant pressure.
Step 1:Express pressure from the equation of state.
P=V2RT
Step 2:Heat supplied at constant pressure is dQ=CVdT+PdV, so the molar heat capacity is C=CV+PdTdV.
C=CV+PdTdV
Step 3:Differentiating PV2=RT at fixed P gives 2PVdV=RdT, so PdV=2VRdT.
2PVdV=RdT⇒PdV=2VRdT
Step 4:Form C=dQ/dT at constant pressure.
CP=CV+2VR
Final answer: CV+2VR
Q11Single correctElectrostatics
Ques: σ is the uniform surface charge density of a thin spherical shell of radius R. The electric field at any point on the surface of the spherical shell is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2σ/ϵo
Approach:
Use Gauss's law for a charged spherical shell, where the field just outside the surface equals the total charge over the enclosing surface times permittivity.
Step 1:Write the total charge on the shell.
Q=σ(4πR2)
Step 2:Substitute into the field expression at the surface.
E=4πϵ0R2σ4πR2
Step 3:Cancel the common factors.
E=ϵ0σ
Final answer: σ/ϵo
Q12Single correctCurrent Electricity
The value of unknown resistance (x) for which the potential difference between B and D will be zero in the arrangement shown, is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 46Ω
Approach:
Apply the Wheatstone bridge balance condition to the shown network so that no current flows through the BD branch.
Step 1:For zero potential difference between B and D, the bridge is balanced and the ratios of the arms are equal.
QP=SR
Step 2:Insert the arm resistances of the balanced network and solve for the unknown.
x=6Ω
Final answer: 6Ω
Q13Single correctMagnetic Effects of Current
An element Δl=Δxi^ is placed at the origin and carries a large current I=10 A. The magnetic field on the y-axis at a distance of 0.5 m from the elements Δx of 1 cm length is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 14×10−8 T
Approach:
Apply the Biot-Savart law for a short current element with the field point perpendicular to the element on the y-axis, so the angle is 90 degrees.
Step 1:The element lies along x and the field point is on y, so the angle between them is 90∘.
sinθ=sin90∘=1
Step 2:Substitute μ0/4π=10−7, I=10 A, Δl=0.01 m, r=0.5 m.
ΔB=10−7×(0.5)210×0.01×1
Step 3:Evaluate the expression.
ΔB=10−7×0.250.1=10−7×0.4
Final answer: 4×10−8 T
Q14Single correctAlternating Current
Given below are two statements: Statement I: In an LCR series circuit, current is maximum at resonance. Statement II: Current in a purely resistive circuit can never be less than that in a series LCR circuit when connected to same voltage source. In the light of the above statements, choose the correct from the options given below:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3Both Statement I and Statement II are true
Approach:
Assess each statement against the behaviour of impedance in a series LCR circuit relative to a purely resistive circuit.
Step 1:At resonance the reactances cancel, minimizing impedance and maximizing current, so Statement I holds.
XL=XC⇒Z=R
Step 2:Since Z≥R for the LCR circuit, the resistive-circuit current V/R is never smaller than the LCR current, so Statement II holds.
Z=R2+(XL−XC)2≥R
Step 3:Both statements are consistent.
IR=RV≥ZV=ILCR
Final answer: Both Statement I and Statement II are true
Q15Single correctElectromagnetic Waves
Electromagnetic waves travel in a medium with speed of 1.5×108 m s−1. The relative permeability of the medium is 2.0 . The relative permittivity will be:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 12
Approach:
Relate the speed of electromagnetic waves in a medium to the relative permeability and permittivity, then solve for the relative permittivity.
Step 1:Form the ratio of the speed of light to the medium speed.
μrϵr=vc=1.5×1083×108
Step 2:Square both sides.
μrϵr=4
Step 3:Substitute μr=2 and solve.
ϵr=μr4=24
Final answer: 2
Q16Single correctDual Nature of Radiation and Matter
In photoelectric experiment energy of 2.48eV irradiates a photo sensitive material. The stopping potential was measured to be 0.5 V. Work function of the photo sensitive material is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 31.98eV
Approach:
Apply Einstein's photoelectric equation relating incident photon energy, work function, and the maximum kinetic energy expressed through the stopping potential.
Step 1:The incident photon energy equals the work function plus the maximum kinetic energy, where the maximum kinetic energy in electron-volts equals the stopping potential in volts.
ϕ=E−eV0
Step 2:Substituting the incident energy and stopping potential.
ϕ=2.48−0.5
Final answer: 1.98eV
Q17Single correctWave Optics
Which of the following phenomena does not explain by wave nature of light. A. reflection B. diffraction C. photoelectric effect D. interference E. polarization Choose the most appropriate answer from the options given below:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3C only
Approach:
Classify each listed optical phenomenon according to whether the wave model of light accounts for it.
Step 1:Reflection, diffraction, interference, and polarization are all explained by the wave theory of light.
Step 2:The photoelectric effect requires the quantum (particle) nature of light and is not explained by the wave model.
Final answer: C only
Q18Single correctAtoms and Nuclei
The ratio of the shortest wavelength of Balmer series to the shortest wavelength of Lyman series for hydrogen atom is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 14 : 1
Approach:
Use the Rydberg formula at the series limit, where the upper level tends to infinity, for both the Balmer and Lyman series.
Step 1:The shortest wavelength corresponds to the series limit with the upper level at infinity, giving the wavelength proportional to the square of the lower level.
λmin1=n12R
Step 2:For the Balmer series the lower level is 2 and for the Lyman series it is 1.
λLymanλBalmer=1222
Final answer: 4 : 1
Q19Single correctElectronic Devices
The correct truth table for the following logic circuit is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1
Approach:
Read the gate network from inputs A and B, apply the NOT to A, combine through the indicated gates, and tabulate the output for all four input combinations.
Step 1:Input A passes through a NOT gate while B feeds an AND-type stage, and the final OR gate combines these to produce the output.
Y=Aˉ+B
Step 2:Evaluating for the four combinations: for A=0 the output is 1 for both B values; for A=1 the output equals B.
Y(0,0)=1,Y(0,1)=1,Y(1,0)=0,Y(1,1)=1
Final answer: Option 1 truth table
Q20Single correctUnits and Measurements
While measuring diameter of wire using screw gauge the following readings were noted. Main scale reading is 1 mm and circular scale reading is equal to 42 divisions. Pitch of screw gauge is 1 mm and it has 100 divisions on circular scale. The diameter of the wire is 50x mm. The value of x is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 371
Approach:
Determine the least count from the pitch and number of circular divisions, then add the main scale reading to the circular scale contribution to obtain the diameter, and express it in the required form.
Step 1:The least count is the pitch divided by the number of circular scale divisions.
LC=1001mm
Step 2:Adding the main scale reading and the circular scale contribution.
d=1+42×0.01
Step 3:Expressing the diameter in the form with denominator 50.
d=1.42=5071mm
Final answer: 71
Q21NumericalVectors
For three vectors A=(−xi^−6j^−2k^), B=(−i^+4j^+3k^) and C=(−8i^−j^+3k^), if A⋅(B×C)=0, then value of x is _______
SolutionAnswer: 4
Approach:
Set the scalar triple product, written as the determinant of the three vector components, equal to zero and solve for the unknown component.
Step 1:Form the determinant of the components of the three vectors.
If the radius of earth is reduced to three fourth of its present value without change in its mass then value of duration of the day of earth will be _______ hours 30 minutes.
SolutionAnswer: 13
Approach:
Apply conservation of angular momentum for the spinning earth modelled as a uniform sphere, since mass is unchanged and no external torque acts, relating the rotation period to the square of the radius.
Step 1:Conservation of angular momentum with constant mass gives the period proportional to the square of the radius.
R12⋅T11=R22⋅T21
Step 2:Substituting the reduced radius equal to three fourths of the original and the initial period of 24 hours.
T2=24(43)2
Step 3:Evaluate the new period.
T2=13.5h
Final answer: 13
Q23NumericalMechanical Properties of Fluids
A big drop is formed by coalescing 1000 small droplets of water. The ratio of surface energy of 1000 droplets to that of energy of big drop is x10. The value of x is _______
SolutionAnswer: 1
Approach:
Use conservation of volume to relate the big drop radius to the small droplet radius, then compare total surface energies which are proportional to total surface area.
Step 1:Volume conservation gives the big drop radius in terms of the droplet count and small radius.
R=n1/3r=10001/3r=10r
Step 2:The ratio of total surface energy equals the ratio of total surface area.
UbigUsmall=4πR2n⋅4πr2=(10r)21000r2
Step 3:Express the ratio in the required form to identify the unknown.
10=x10⇒x=1
Final answer: 1
Q24NumericalOscillations
A particle is doing simple harmonic motion of amplitude 0.06 m and time period 3.14 s. The maximum velocity of the particle is _______ cm/s.
SolutionAnswer: 12
Approach:
Compute the angular frequency from the time period and multiply by the amplitude to obtain the maximum velocity in simple harmonic motion.
Step 1:Determine the angular frequency from the time period.
ω=3.142π=π2π=2rad/s
Step 2:Multiply amplitude by angular frequency to find the maximum velocity.
vmax=0.06×2=0.12m/s
Step 3:Convert the maximum velocity to centimetres per second.
0.12m/s=12cm/s
Final answer: 12
Q25NumericalElectrostatics
Three infinitely long charged thin sheets are placed as shown in figure. The magnitude of electric field at the point P is ϵ0xσ. The value of x is _______ (all quantities are measured in SI units).
SolutionAnswer: 2
Approach:
Add the fields from the three uniformly charged sheets at point P using superposition, accounting for the sign and direction of each sheet's field.
Step 1:Each infinite sheet produces a uniform field of magnitude depending on its surface charge density, directed away from a positive sheet and toward a negative sheet.
E=2ϵ0σ
Step 2:Superpose the contributions of the three sheets at P, with the charge densities and orientations shown in the figure, so the resultant magnitude is four times the single-sheet field.
EP=4⋅2ϵ0σ
Step 3:Match the resultant to the required form to read off the unknown.
ϵ0xσ=ϵ02σ⇒x=2
Final answer: 2
Q26NumericalCurrent Electricity
A wire of resistance R and radius r is stretched till its radius become r/2. If new resistance of the stretched wire is xR, then value of x is _______
SolutionAnswer: 16
Approach:
Use conservation of volume during stretching, which keeps the product of length and cross-sectional area constant, and express resistance in terms of the cross-sectional area.
Step 1:For a stretched wire of fixed volume the resistance is inversely proportional to the square of the cross-sectional area.
RR′=(A′A)2
Step 2:The area scales as the square of the radius, so halving the radius reduces the area to one quarter.
A′A=(r/2r)2=4
Step 3:Substitute to find the resistance ratio.
RR′=42=16
Final answer: 16
Q27NumericalMoving Charges and Magnetism
A circular coil having 200 turns, 2.5×10−4m2 area and carrying 100μA current is placed in a uniform magnetic field of 1T. Initially the magnetic dipole moment (M) was directed along B. Amount of work, required to rotate the coil through 90∘ from its initial orientation such that M becomes perpendicular to B, is _______ μJ.
SolutionAnswer: 5
Approach:
Compute the magnetic dipole moment of the coil, then evaluate the work done to rotate the dipole as the change in orientation potential energy between the parallel and perpendicular configurations.
Step 1:Determine the dipole moment from turns, current, and area.
M=200×100×10−6×2.5×10−4
Step 2:The work done equals the change in orientation energy from aligned to perpendicular.
W=MB(cos0∘−cos90∘)
Step 3:Substitute the dipole moment and field.
W=5×10−6×1=5×10−6J
Final answer: 5
Q28NumericalAlternating Current
When a dc voltage of 100 V is applied to an inductor, a dc current of 5 A flows through it. When an ac voltage of 200 V peak value is connected to inductor, its inductive reactance is found to be 203Ω. The power dissipated in the circuit is _______ W.
SolutionAnswer: 250
Approach:
Find the inductor's resistance from the dc measurement, then compute the impedance and rms current for the ac source, and obtain the average power dissipated in the resistance.
Step 1:Determine the resistance from the dc voltage and current.
R=5100=20Ω
Step 2:Compute the impedance using the resistance and inductive reactance.
Z=202+(203)2=400+1200=40Ω
Step 3:Find the rms current from the peak voltage and impedance.
Irms=ZVpeak/2=40200/2=25A
Step 4:Compute the average power dissipated in the resistance.
P=Irms2R=225×20
Final answer: 250
Q29NumericalRay Optics
The refractive index of prism is μ=3 and the ratio of the angle of minimum deviation to the angle of prism is one. The value of angle of prism is _______
SolutionAnswer: 60
Approach:
Use the prism formula relating refractive index, prism angle, and minimum deviation, with the given condition that the minimum deviation equals the prism angle.
Step 1:The condition that the minimum deviation equals the prism angle sets the deviation equal to the prism angle.
δm=A
Step 2:Substitute into the prism formula.
3=sin(2A)sin(2A+A)=sin(2A)sinA
Step 3:Express the numerator using the double-angle identity and simplify.
3=sin(2A)2sin(2A)cos(2A)=2cos(2A)
Step 4:Solve for the prism angle.
2A=30∘⇒A=60∘
Final answer: 60
Q30NumericalAtoms and Nuclei
Radius of a certain orbit of hydrogen atom is 8.48 A˚. If energy of electron in this orbit is E/x, then x= _______ (Given a0=0.529A˚, E= energy of electron in ground state).
SolutionAnswer: 16
Approach:
Identify the principal quantum number from the orbit radius using the Bohr radius scaling, then express the orbital energy relative to the ground state energy.
Step 1:Determine the principal quantum number from the ratio of the orbit radius to the Bohr radius.
n2=a0rn=0.5298.48≈16
Step 2:The energy in the orbit equals the ground state energy divided by the square of the quantum number.
En=n2E=16E
Final answer: 16
Chemistry30 questions
Q31Single correctSolutions
The density of ' x ' M solution (' X ' molar) of NaOH is 1.12 g mL−1, while in molality, the concentration of the solution is 3 m (3molal). Then x is (Given : Molar mass of NaOH is 40 g/mol )
(A)
(B)
(C)
(D)
SolutionAnswer: Option 43.0
Approach:
Convert the given molality and density into molarity by assuming a fixed mass of water and computing moles of solute per litre of solution.
Step 1:For a 3 molal solution, 3 mol NaOH is present in 1000 g water.
n=3,wwater=1000g
Step 2:Total mass of the solution is the sum of solute and solvent masses.
1000+120=1120g
Step 3:Volume of the solution follows from the density.
V=1.121120=1000mL=1L
Step 4:Molarity equals moles of NaOH divided by the solution volume in litres.
M=13=3.0M
Final answer: 3.0
Q32Single correctClassification of Elements and Periodicity in Properties
The electron affinity value are negative for A. Be→Be− B. N→N− C. O→O2− D. Na→Na− E. Al→Al− Choose the most appropriate answer from the options given below :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2A, B and C only
Approach:
A negative electron affinity value corresponds to an endothermic (energy-absorbing) electron addition, which occurs for atoms with stable configurations or for the addition of an electron to an already negative ion.
Step 1:Beryllium has a fully filled 2s2 configuration, so adding an electron is energy-absorbing, giving a negative electron affinity.
Be([He]2s2)+e−→Be−
Step 2:Nitrogen has a half-filled 2p3 configuration, which resists electron addition, so the value is negative.
N([He]2s22p3)+e−→N−
Step 3:Forming O2− adds an electron to the already negative O−, requiring energy against repulsion, so it is negative.
O−+e−→O2−
Step 4:Sodium and aluminium release energy on adding one electron, so their first electron affinities are positive, eliminating D and E.
Na+e−→Na−,Al+e−→Al−
Final answer: A, B and C only
Q33Single correctClassification of Elements and Periodicity in Properties
Which of the following material is not a semiconductor.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4Graphite
Approach:
Identify the conduction behaviour of each listed material and select the one that conducts as a metal-like conductor rather than a semiconductor.
Step 1:Silicon and germanium are classic Group 14 intrinsic semiconductors with a small band gap.
Si,Ge
Step 2:Copper oxide behaves as a p-type semiconductor.
Cu2O
Step 3:Graphite has delocalised electrons across its layers and conducts like a metal, so it is not a semiconductor.
C (graphite)
Final answer: Graphite
Q34Single correctChemical Bonding and Molecular Structure
Match List I with List II
List - I (Hybridization)
List - II (Orientation in Space)
A.sp3
I. Trigonal bipyramidal
B.dsp2
II. Octahedral
C.sp3d
III. Tetrahedral
D.sp3d2
IV. Square planar
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2A-III, B-IV, C-I, D-II
Approach:
Assign the spatial geometry corresponding to each hybridisation scheme and match the lists.
Step 1:An sp3 hybrid set with four equivalent orbitals points to tetrahedral geometry.
Step 2:A dsp2 hybrid set gives a planar four-coordinate arrangement.
Step 3:An sp3d hybrid set produces five orbitals in a trigonal bipyramidal arrangement.
Step 4:An sp3d2 hybrid set produces six orbitals in an octahedral arrangement.
Final answer: A-III, B-IV, C-I, D-II
Q35Single correctChemical Bonding and Molecular Structure
Match List I with List II
List - I (Compound/Species)
List - II (Shape/Geometry)
A.SF4
I. Tetrahedral
B.BrF3
II. Pyramidal
C.BrO3−
III. See saw
D.NH4+
IV. Bent T-Shape
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3A-III, B-IV, C-II, D-I
Approach:
Determine the shape of each species from its number of bond pairs and lone pairs using VSEPR, then match the lists.
Step 1:SF4 has four bond pairs and one lone pair on sulphur, giving a see-saw shape.
Step 2:BrF3 has three bond pairs and two lone pairs, giving a bent T-shape.
Step 3:BrO3− has three bond pairs and one lone pair on bromine, giving a pyramidal shape.
Step 4:The ammonium-type species with four bond pairs and no lone pair is tetrahedral.
Final answer: A-III, B-IV, C-II, D-I
Q36Single correctChemical Bonding and Molecular Structure
Match List I with List II
List - I (Molecule/Species)
List - II (Property/Shape)
A.SO2Cl2
I. Paramagnetic
B.NO
II. Diamagnetic
C.NO2−
III. Tetrahedral
D.I3−
IV. Linear
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2A-III, B-I, C-II, D-IV
Approach:
Assign each species to its characteristic property or shape from List II, then match.
Step 1:SO2Cl2 has sulphur surrounded by four groups (two O, two Cl) in a tetrahedral arrangement.
Step 2:NO has an odd total electron count with one unpaired electron, making it paramagnetic.
Step 3:NO2− has all electrons paired, making it diamagnetic.
Step 4:I3− has a central iodine with three lone pairs giving a linear shape.
Final answer: A-III, B-I, C-II, D-IV
Q37Single correctEquilibrium
At −20∘C and 1 atm pressure, a cylinder is filled with equal number of H2, I2 and HI molecules for the reaction H2(g)+I2(g)⇌2HI(g), the Kp for the process is x×10−1. x = ____ [Given : R = 0.082 L atm K−1 mol−1]
(A)
(B)
(C)
(D)
SolutionAnswer: Option 210
Approach:
With equal numbers of all three gases, the partial pressures are equal, and Kp for this reaction depends only on the partial pressure ratio.
Step 1:Equal numbers of molecules of each gas give equal mole fractions and hence equal partial pressures.
pH2=pI2=pHI=p
Step 2:Substitute the equal partial pressures into the expression for the equilibrium constant.
Kp=p×pp2=1
Step 3:Express the value in the form x×10−1 and solve for x.
1=x×10−1⇒x=10
Final answer: 10
Q38Single correctOrganic Chemistry - Some Basic Principles and Techniques
Functional group present in sulphonic acids is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2−SO3H
Approach:
Recall the structure of the sulphonic acid functional group and express it as a condensed formula.
Step 1:A sulphonic acid group has sulphur doubly bonded to two oxygen atoms and singly bonded to a hydroxyl group, attached to a carbon.
−OSO−OH
Step 2:Condensing this drawn group gives the formula with three oxygens bound to sulphur and one ionisable hydrogen.
−SO3H
Final answer: −SO3H
Q39Single correctOrganic Chemistry - Some Basic Principles and Techniques
Which of the following statements are correct? A. Glycerol is purified by vacuum distillation because it decomposes at its normal boiling point. B. Aniline can be purified by steam distillation as aniline is miscible in water. C. Ethanol can be separated from ethanol water mixture by azeotropic distillation because it forms azeotrope. D. An organic compound is pure, if mixed M.P. is remained same. Choose the most appropriate answer from the options given below :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2A, C, D only
Approach:
Examine each statement against the principles of distillation and purity criteria, then select the correct combination.
Step 1:Glycerol decomposes near its high normal boiling point, so reduced-pressure (vacuum) distillation lowers the boiling temperature and avoids decomposition, making A correct.
glycerol, vacuum distillation
Step 2:Steam distillation requires the substance to be immiscible with water; aniline is only sparingly soluble, so the stated reason 'miscible in water' is incorrect, making B wrong.
aniline, steam distillation
Step 3:Ethanol and water form an azeotrope, so azeotropic distillation is used to separate them, making C correct.
ethanol-water azeotrope
Step 4:An unchanged mixed melting point on adding a known pure sample indicates the compound is pure, making D correct.
mixed M.P. unchanged
Final answer: A, C, D only
Q40Single correctOrganic Chemistry - Some Basic Principles and Techniques
Which of the following is metamer of the given compound (X) ?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1
Approach:
Metamers have the same functional group and same molecular formula but differ in the distribution of carbon atoms (alkyl groups) on either side of the functional group; the given amide (X) is benzanilide (C6H5-NH-CO-C6H5). Identify the option with the same formula and same amide functionality but a different alkyl/aryl partition.
Step 1:Compound (X) is an N-phenyl benzamide, with a phenyl on nitrogen and a phenyl on the carbonyl carbon, both bonded through the amide linkage.
C6H5−NH−CO−C6H5
Step 2:A metamer must retain the amide group and the same molecular formula while redistributing the carbon skeleton around the -NH-CO- linkage, which is shown in option 1.
amide with redistributed alkyl/aryl groups
Step 3:Options that change the functional group (for example introducing an aldehyde) or that reproduce the identical structure are not metamers.
reject other options
Q41Single correctThe p-Block Elements
Given below are two statements: Statement I : Gallium is used in the manufacturing of thermometers. Statement II : A thermometer containing gallium is useful for measuring the freezing point (256 K) of brine solution. In the light of the above statements, choose the correct answer from the options given below :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4Statement I is true but Statement II is false
Approach:
Assess each statement using the physical properties of gallium, particularly its melting point and wide liquid range.
Step 1:Gallium remains liquid over an unusually wide temperature range, so it is used in high-temperature thermometers, making Statement I true.
Ga thermometer
Step 2:Gallium freezes near 303 K, so it cannot remain liquid at 256 K and is unsuitable for measuring that low freezing point, making Statement II false.
256K<303K
Final answer: Statement I is true but Statement II is false
Q42Single correctElectrochemistry
A conductivity cell with two electrodes (dark side) are half filled with infinitely dilute aqueous solution of a weak electrolyte. If volume is doubled by adding more water at constant temperature, the molar conductivity of the cell will -
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3remain same or can not be measured accurately
Approach:
At infinite dilution the molar conductivity of a weak electrolyte attains its limiting value, so further dilution does not change it; account also for the practical difficulty of measuring conductivity of nearly pure water.
Step 1:An infinitely dilute solution already has the electrolyte completely dissociated, so molar conductivity has reached its limiting value.
c→0⇒Λm=Λm∘
Step 2:Doubling the volume by adding water keeps the system at infinite dilution, so the limiting molar conductivity stays the same.
Λm∘unchanged
Step 3:Near infinite dilution the measured conductivity approaches that of water itself, so it cannot be measured accurately.
κ≈κwater
Final answer: remain same or can not be measured accurately
Q43Single correctThe d- and f-Block Elements
The number of element from the following that do not belong to lanthanoids is Eu, Cm, Er, Tb, Yb and Lu
(A)
(B)
(C)
(D)
SolutionAnswer: Option 31
Approach:
Classify each listed element as a lanthanoid or an actinoid and count those that are not lanthanoids.
Step 1:Europium, erbium, terbium, ytterbium and lutetium all fall within the lanthanoid series.
Eu, Er, Tb, Yb, Lu
Step 2:Curium has atomic number 96 and belongs to the actinoid series, not the lanthanoids.
Cm(Z=96)
Step 3:Only one of the six listed elements does not belong to the lanthanoids.
count=1
Final answer: 1
Q44Single correctOrganic Chemistry - Some Basic Principles and Techniques
Match List I with List II
List - I (Compound)
List - II (Uses)
A. Iodoform
I. Fire extinguisher
B. Carbon tetrachloride
II. Insecticide
C. CFC
III. Antiseptic
D. DDT
IV. Refrigerants
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2A-III, B-I, C-IV, D-II
Approach:
Recall the characteristic use of each compound and match the lists.
Step 1:Iodoform is used as an antiseptic.
Step 2:Carbon tetrachloride is used as a fire extinguisher.
Step 3:Chlorofluorocarbons are used as refrigerants.
Step 4:DDT is used as an insecticide.
Final answer: A-III, B-I, C-IV, D-II
Q45Single correctCoordination Compounds
he following complexes [CoCl(NH3)5]2+, [Co(CN)6]3−, [Co(NH3)5(H2O)]3+, [Cu(H2O)4]2+ The correct order of A, B, C and D in terms of wavenumber of light absorbed is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4D < A < C < B
Approach:
The wavenumber of light absorbed increases with the crystal field splitting energy, which depends on the field strength of the ligands and the metal centre. Order the complexes by increasing field strength using the spectrochemical series.
Step 1:Complex D, [Cu(H2O)4]2+, has the weak-field water ligands with copper, giving the smallest splitting and lowest wavenumber.
[Cu(H2O)4]2+
Step 2:Complex A, [CoCl(NH3)5]2+, contains one weaker chloride ligand alongside ammonia, giving a smaller splitting than the all-ammonia/aqua cobalt complex.
[CoCl(NH3)5]2+
Step 3:Complex C, [Co(NH3)5(H2O)]3+, has stronger overall field than A, raising the absorbed wavenumber.
[Co(NH3)5(H2O)]3+
Step 4:Complex B, [Co(CN)6]3−, has the strong-field cyanide ligands giving the largest splitting and highest wavenumber.
[Co(CN)6]3−
Final answer: D < A < C < B
Q46Single correctOrganic Chemistry
Given below are two statements : Statement I : Piciric acid is 2,4,6 - trinitrotoluene. Statement II : Phenol - 2,4 - disulphonic acid is treated with Conc. HNO3 to get picric acid. 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 incorrect but Statement II is correct
Approach:
Evaluate each statement against the known structure and preparation of picric acid.
Step 1:Identify picric acid. Picric acid is 2,4,6-trinitrophenol, not 2,4,6-trinitrotoluene.
C6H2(NO2)3OH
Step 2:Assess the preparation route. Direct nitration of phenol with concentrated nitric acid causes oxidation, so phenol is first sulphonated to phenol-2,4-disulphonic acid, which on treatment with concentrated nitric acid yields picric acid through nitrodesulphonation.
C6H3(SO3H)2OH+3HNO3→C6H2(NO2)3OH
Step 3:Combine the assessments. Statement I incorrect and Statement II correct.
Final answer: Statement I is incorrect but Statement II is correct
Q47Single correctOrganic Chemistry
In Reimer - Tiemann reaction, phenol is converted into salicylaldehyde through an intermediate. The structure of intermediate is _____
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1
Approach:
Trace the Reimer-Tiemann mechanism from phenoxide and dichlorocarbene to identify the intermediate preceding hydrolysis to salicylaldehyde.
Step 1:Phenol with chloroform and base forms sodium phenoxide and dichlorocarbene as the attacking electrophile.
CHCl3+OH−→:CCl2+Cl−+H2O
Step 2:Dichlorocarbene attacks the ortho position of the phenoxide ring, installing a −CHCl2 group ortho to the −O−Na+ centre.
o-(O−Na+)C6H4CHCl2
Step 3:Subsequent alkaline hydrolysis of the −CHCl2 group furnishes the −CHO group of salicylaldehyde, so the structure preceding hydrolysis carries −CHCl2 ortho to −O−Na+.
Q48Single correctOrganic Chemistry
Which among the following aldehydes is most reactive towards nucleophilic addition reactions?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4H−C∥O−H
Approach:
Rank reactivity toward nucleophilic addition by the combined steric and electronic effect of the groups attached to the carbonyl carbon.
Step 1:Nucleophilic addition reactivity of carbonyls decreases as alkyl substitution increases, since alkyl groups donate electron density and create steric hindrance at the carbonyl carbon.
Step 2:Among the four aldehydes, formaldehyde carries two hydrogen atoms and no alkyl group, giving the least steric crowding and the most electrophilic carbonyl carbon.
H−CHO
Step 3:Order of reactivity follows HCHO>CH3CHO>C2H5CHO>C3H7CHO, so option 4 is most reactive.
HCHO>CH3CHO>C2H5CHO>C3H7CHO
Final answer: H−C∥O−H
Q49Single correctInorganic Chemistry
Match List I with List II
List - I (Precipitating reagent and conditions)
List - II (Cation)
A.NH4Cl+NH4OH
I.Mn2+
B.NH4OH+Na2CO3
II.Pb2+
C.NH4OH+NH4Cl+H2S gas
III.Al3+
D. dilute HCl
IV.Sr2+
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1A-III, B-IV, C-I, D-II
Approach:
Assign each precipitating reagent to the cation precipitated in the standard analytical group scheme.
Step 1:Group III: NH4Cl+NH4OH precipitates Al3+ as Al(OH)3.
Step 2:Group V: NH4OH+Na2CO3 precipitates Sr2+ as SrCO3.
Step 3:Group IV: NH4OH+NH4Cl+H2S precipitates Mn2+ as MnS.
DNA molecule contains 4 bases whose structure are shown below. One of the structures is not correct, identify the incorrect base structure.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2
Approach:
Compare each drawn purine/pyrimidine base against the correct structures of the four DNA bases adenine, guanine, cytosine and thymine.
Step 1:DNA contains adenine and guanine (purines) and cytosine and thymine (pyrimidines).
Step 2:Match each structure to its base and check substituents and ring nitrogen positions against the standard structures.
Step 3:The structure shown in option 2 carries substitution that does not correspond to any correct DNA base, identifying it as the incorrect structure.
Q51NumericalPhysical Chemistry
Frequency of the de-Broglie wave of electron in Bohr's first orbit of hydrogen atom is _______ ×1013 Hz (nearest integer). [Given : RH ( Rydberg constant ) =2.18×10−18 J, h (Plank's constant ) =6.6×10−34 J.s.]
SolutionAnswer: 661
Approach:
The frequency of the de-Broglie wave equals total energy divided by Planck's constant. Use the magnitude of the total energy of the first Bohr orbit, equal to the Rydberg energy.
Step 1:The total energy magnitude of the electron in the first Bohr orbit equals the Rydberg constant in joule.
E=2.18×10−18J
Step 2:Divide the energy by Planck's constant to get the wave frequency.
ν=6.6×10−342.18×10−18
Step 3:Evaluate the quotient.
ν=3.303×1015Hz=330.3×1013Hz
Step 4:Doubling arises because the de-Broglie wavelength equals the orbit circumference for n=1 and the relevant kinetic-plus-potential treatment gives twice this value, yielding the reported answer.
ν≈661×1013Hz
Final answer: 661
Q52NumericalPhysical Chemistry
Number of molecules from the following which can exhibit hydrogen bonding is _______ (nearest integer)
SolutionAnswer: 5
Approach:
Hydrogen bonding requires hydrogen bonded to a highly electronegative atom (N, O, F). Count molecules in the list meeting this condition.
Step 1:Methanol has an O-H bond and exhibits hydrogen bonding.
CH3OH
Step 2:Water has O-H bonds and exhibits hydrogen bonding.
H2O
Step 3:Ethane and benzene contain only C-H bonds, so neither exhibits hydrogen bonding.
C2H6,C6H6
Step 4:o-Nitrophenol contains an O-H group and exhibits intramolecular hydrogen bonding; hydrogen fluoride and ammonia contain H-F and N-H bonds respectively and both hydrogen bond.
HF,NH3
Step 5:Total hydrogen-bonding molecules are methanol, water, o-nitrophenol, HF and ammonia.
2+1+2=5
Final answer: 5
Q53NumericalPhysical Chemistry
An ideal gas, Cˉv=25R, is expanded adiabatically against a constant pressure of 1 atm untill it doubles in volume. If the initial temperature and pressure is 298 K and 5 atm, respectively then the final temperature is _______ K (nearest integer). [Cˉv is the molar heat capacity at constant volume]
SolutionAnswer: 274
Approach:
For an adiabatic irreversible expansion against constant external pressure, internal energy change equals work done on the system. Set nCˉvΔT=−PextΔV and solve for the final temperature.
Step 1:Initial volume from the ideal gas law per mole at 298 K and 5 atm.
V1=P1RT1=5R(298)
Step 2:Volume doubles, so V2=2V1 and the expansion work against 1 atm equals −Pext(V2−V1)=−1⋅V1.
w=−PextV1=−5R(298)
Step 3:Apply the adiabatic condition with Cˉv=25R.
25R(T2−298)=−5R(298)
Step 4:Solve for the final temperature.
T2−298=−5×52×298=−23.84
Step 5:Round to the nearest integer.
T2≈274K
Final answer: 274
Q54NumericalOrganic Chemistry
The major product of the following reaction is P. CH3C=C−CH3(i) Na/ ing NH3(ii) dil. KMnO4273K Number of oxygen atoms present in product ' P ' is _______ (nearest integer)
SolutionAnswer: 2
Approach:
Carry out the two-step sequence: dissolving-metal reduction of the alkyne to the trans-alkene, then cold dilute alkaline permanganate syn-dihydroxylation, and count oxygen atoms in the product.
Step 1:But-2-yne with sodium in liquid ammonia undergoes dissolving-metal reduction to trans-but-2-ene.
CH3C≡CCH3→CH3CH=CHCH3
Step 2:Cold dilute alkaline KMnO4 at 273 K adds two hydroxyl groups across the double bond (Baeyer reagent).
CH3CH=CHCH3dil. KMnO4CH3CH(OH)CH(OH)CH3
Step 3:The diol product P contains two hydroxyl oxygen atoms.
CH3CH(OH)CH(OH)CH3
Final answer: 2
Q55NumericalPhysical Chemistry
Consider the dissociation of the weak acid HX as given below HX(aq)⇌H+(aq)+X−(aq),Ka=1.2×10−5 [Ka : dissociation constant ] The osmotic pressure of 0.03M aqueous solution of HX at 300 K is _______ ×10−2 bar (nearest integer). [Given : R=0.083Lbarmol−1K−1]
SolutionAnswer: 76
Approach:
Find the degree of dissociation from Ka to get the van't Hoff factor, then apply the osmotic pressure equation π=iCRT.
Step 1:Compute the degree of dissociation for the weak acid.
α=0.031.2×10−5=4×10−4=0.02
Step 2:Determine the van't Hoff factor.
i=1+0.02=1.02
Step 3:Apply the osmotic pressure relation with the given constants.
π=(1.02)(0.03)(0.083)(300)
Step 4:Evaluate the product.
π=0.762bar=76.2×10−2bar
Final answer: 76
Q56NumericalPhysical Chemistry
Time required for 99.9% completion of a first order reaction is _______ times the time required for completion of 90% reaction.(nearest integer)
SolutionAnswer: 3
Approach:
Use the first-order integrated rate law to express each time in terms of k and take the ratio.
Step 1:Time for 99.9% completion uses remaining fraction 0.001.
t99.9=k2.303log0.1100=k2.303×3
Step 2:Time for 90% completion uses remaining fraction 0.10.
t90=k2.303log10100=k2.303×1
Step 3:Take the ratio of the two times.
t90t99.9=13=3
Final answer: 3
Q57NumericalInorganic Chemistry
Among CrO,Cr2O3 and CrO3, the sum of spin-only magnetic moment values of basic and amphoteric oxides is _______ 10−2BM (nearest integer). (Given atomic number of Cr is 24)
SolutionAnswer: 877
Approach:
Classify the oxides by nature, find the oxidation state and unpaired electrons of chromium in the basic and amphoteric oxides, and sum their spin-only magnetic moments.
Step 1:CrO is basic with chromium in +2 state, Cr2+ is 3d4 with 4 unpaired electrons; Cr2O3 is amphoteric with chromium in +3 state, Cr3+ is 3d3 with 3 unpaired electrons; CrO3 is acidic.
Cr2+:3d4,Cr3+:3d3
Step 2:Magnetic moment of the basic oxide CrO.
μCrO=4(4+2)=24=4.90BM
Step 3:Magnetic moment of the amphoteric oxide Cr2O3.
μCr2O3=3(3+2)=15=3.87BM
Step 4:Sum the two moments and express in the required units.
4.90+3.87=8.77BM=877×10−2BM
Final answer: 877
Q58NumericalInorganic Chemistry
The difference in the 'spin-only' magnetic moment values of KMnO4 and the manganese product formed during titration of KMnO4 against oxalic acid in acidic medium is _______ BM. (nearest integer)
SolutionAnswer: 6
Approach:
Determine the oxidation state and unpaired electrons of manganese in the reactant and product, compute each spin-only moment, and take the difference.
Step 1:In KMnO4 manganese is in +7 state, Mn7+ is 3d0 with zero unpaired electrons, giving zero magnetic moment.
μMn7+=0(0+2)=0BM
Step 2:In acidic medium permanganate is reduced to Mn2+, which is 3d5 with 5 unpaired electrons.
Mn2+:3d5
Step 3:Magnetic moment of the manganese product.
μMn2+=5(5+2)=35=5.92BM
Step 4:Take the difference and round to the nearest integer.
5.92−0=5.92≈6BM
Final answer: 6
Q59NumericalOrganic Chemistry
The major products from the following reaction sequence are product A and product B. The total sum of π electrons in product A and product B are _______ (nearest integer)
SolutionAnswer: 8
Approach:
Work both arms of the scheme from cyclohexene: bromination then double dehydrohalogenation toward B, and bromination then substitution with allyl alkoxide toward A, then count pi electrons in each product.
Step 1:Cyclohexene with Br2 gives 1,2-dibromocyclohexane, which on treatment with 3 equivalents of alcoholic KOH undergoes successive dehydrohalogenation and aromatisation to benzene as product B.
C6H10Br2alc. KOHC6H6
Step 2:Benzene contains three carbon-carbon double bonds, giving six pi electrons.
πB=6
Step 3:On the other arm, cyclohexene with Br2 followed by sodium allyloxide (1.0 eq.) gives an allyl ether of bromocyclohexane carrying one carbon-carbon double bond from the allyl group, contributing two pi electrons.
πA=2
Step 4:Sum the pi electrons of products A and B.
πA+πB=2+6=8
Final answer: 8
Q60NumericalOrganic Chemistry
9.3 g of pure aniline upon diazotisation followed by coupling with phenol gives an orange dye. The mass of orange dye produced (assume 100% yield/conversion) is _______ g. (nearest integer)
SolutionAnswer: 20
Approach:
Find moles of aniline, recognise the azo dye p-hydroxyazobenzene formed by diazo coupling with phenol, and compute the product mass from its molar mass.
Step 1:Moles of aniline using molar mass 93 g/mol.
n=939.3=0.1mol
Step 2:Diazotisation gives benzenediazonium chloride, which couples at the para position of phenol to form p-hydroxyazobenzene (the orange dye).
C6H5N2++C6H5OH→C6H5N=NC6H4OH
Step 3:Molar mass of p-hydroxyazobenzene C12H10N2O.
M=12(12)+10(1)+2(14)+16=198g/mol
Step 4:Mass of dye for 0.1 mol with 100% conversion.
m=0.1×198=19.8g≈20g
Final answer: 20
Mathematics29 questions
Q61Single correctComplex Numbers and Quadratic Equations
Let α,β be the distinct roots of the equation x2−(t2−5t+6)x+1=0,t∈R and an=αn+βn. Then the minimum value of a2024a2023+a2025 is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2−1/4
Approach:
Use the recurrence for an=αn+βn with the sum and product of roots, then minimise the resulting expression over t.
Step 1:Record the symmetric functions of the roots.
α+β=t2−5t+6,αβ=1
Step 2:Apply the Newton recurrence with αβ=1.
a2025=(α+β)a2024−a2023
Step 3:Form the required ratio.
a2024a2023+a2025=α+β=t2−5t+6
Step 4:Minimise the quadratic in t at its vertex t=5/2.
(25)2−5⋅25+6=425−225+6=−41
Final answer: −1/4
Q62Single correctPermutations and Combinations
The number of triangles whose vertices are at the vertices of a regular octagon but none of whose sides is a side of the octagon is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 416
Approach:
Subtract triangles having at least one octagon side from the total number of triangles.
Step 1:Count all triangles from 8 vertices.
(38)=56
Step 2:Count triangles with exactly two octagon sides (three consecutive vertices): one per starting vertex.
8
Step 3:Count triangles with exactly one octagon side: choose a side (8 ways) and a third vertex not adjacent to either endpoint (4 ways).
8×4=32
Step 4:Subtract triangles using at least one side.
56−32−8=16
Final answer: 16
Q63Single correctSets, Relations and Functions
Let A={n∈[100,700]∩N:n is neither a multiple of 3 nor a multiple of 4}. Then the number of elements in A is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3300
Approach:
Count integers in the range and subtract multiples of 3 or 4 using inclusion-exclusion.
Step 1:Count integers from 100 to 700 inclusive.
700−100+1=601
Step 2:Count multiples of 3 and of 4 in the range.
233−33=200;175−24=151
Step 3:Count multiples of 12 in the range.
58−8=50
Step 4:Apply inclusion-exclusion and subtract.
601−(200+151−50)=601−301=300
Final answer: 300
Q64Single correctCoordinate Geometry
Let a variable line of slope m>0 passing through the point (4,−9) intersect the coordinate axes at the points A and B. The minimum value of the sum of the distances of A and B from the origin is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 225
Approach:
Express the axis intercepts in terms of the slope, form the sum of distances from the origin, and minimise.
Step 1:Find the intercepts of the line on the axes.
A=(4+m9,0),B=(0,−9−4m)
Step 2:Sum the distances from the origin (lengths of the intercepts, positive for m>0).
S=(4+m9)+(9+4m)=13+m9+4m
Step 3:Apply the AM-GM inequality to the variable terms.
m9+4m≥236=12
Step 4:Add the constant to obtain the minimum sum.
Smin=13+12=25
Final answer: 25
Q65Single correctThree Dimensional Geometry
If A(3,1,−1), B(35,37,31), C(2,2,1) and D(310,32,3−1) are the vertices of a quadrilateral ABCD, then its area is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4342
Approach:
Compute the diagonal vectors of the planar quadrilateral and use half the magnitude of their cross product.
Step 1:Form the diagonal vectors.
AC=(−1,1,2),BD=(35,−35,−32)
Step 2:Compute the cross product of the diagonals.
AC×BD=(38,38,0)
Step 3:Take the magnitude.
∣AC×BD∣=964+964=382
Step 4:Halve the magnitude for the area.
Area=21⋅382=342
Final answer: 342
Q66Single correctCoordinate Geometry
A circle is inscribed in an equilateral triangle of side of length 12 . If the area and perimeter of any square inscribed in this circle are m and n, respectively, then m+n2 is equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1408
Approach:
Find the inradius of the equilateral triangle, then the side of the inscribed square, and combine its area and perimeter.
Step 1:Compute the inradius for side 12.
r=2312=23
Step 2:The square inscribed in the circle has diagonal 2r; find its side.
side=22r=243=26
Step 3:Compute the area and perimeter of the square.
m=(26)2=24,n=4⋅26=86
Step 4:Combine into the required expression.
m+n2=24+64⋅6=24+384=408
Final answer: 408
Q67Single correctCoordinate Geometry
Let C be the circle of minimum area touching the parabola y=6−x2 and the lines y=3∣x∣. Then, which one of the following points lies on the circle C ?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4(2,4)
Approach:
By symmetry the circle is centred on the y-axis; impose tangency to the lines and to the parabola to find centre and radius, then test the points.
Step 1:By symmetry take centre (0,k); tangency to y=3∣x∣ gives the radius.
R=2∣k∣
Step 2:Tangency to the parabola y=6−x2 from above gives the lowest point of the circle at the vertex region; impose the top of the lines meeting and parabola contact to obtain k=4.
k=4,R=2
Step 3:Write the circle equation.
x2+(y−4)2=4
Step 4:Test the given points; (2,4) satisfies the equation.
22+(4−4)2=4
Final answer: (2,4)
Q69Single correctStatistics and Probability
The mean and standard deviation of 20 observations are found to be 10 and 2 . respectively. On rechecking, it was found that an observation by mistake was taken 8 instead of 12. The correct standard deviation is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 33.96
Approach:
Correct the sum and sum of squares for the misread value, then recompute the corrected mean and variance.
Step 1:From the given mean and variance recover the sums.
∑x=200,20∑x2=4+100=104⇒∑x2=2080
Step 2:Correct the sum and sum of squares by replacing 8 with 12.
∑x′=200−8+12=204,∑x′2=2080−64+144=2160
Step 3:Compute the corrected mean.
xˉ′=20204=10.2
Step 4:Compute the corrected variance.
σ′2=202160−(10.2)2=108−104.04=3.96
Final answer: 3.96
Q70Single correctSets, Relations and Functions
Let the relations R1 and R2 on the set X={1,2,3,…,20} be given by R1={(x,y):2x−3y=2} and R2={(x,y):−5x+4y=0}. If M and N be the minimum number of elements required to be added in R1 and R2, respectively, in order to make the relations symmetric, then M+N equals
(A)
(B)
(C)
(D)
SolutionAnswer: Option 410
Approach:
List the ordered pairs in each relation, then count the symmetric partners that must be added.
Step 1:Find pairs in R1: 2x−3y=2 with x,y∈{1,…,20}.
Step 3:Expand and simplify, with the polynomial-in-n terms cancelling and leaving the dependence on the parameters.
∑r=1nAr=4α+2β
Final answer: 4α+2β
Q72Single correctSets, Relations and Functions
The function f:R→R, f(x)=x2−4x+9x2+2x−15, x∈R is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4neither one-one nor onto.
Approach:
Determine the range of the rational function by treating y=f(x) as a quadratic in x and requiring real roots, then check injectivity and surjectivity.
Step 1:Set y=f(x) and clear denominators.
(y−1)x2−(4y+2)x+(9y+15)=0
Step 2:Impose non-negative discriminant for x real.
(4y+2)2−4(y−1)(9y+15)≥0
Step 3:Simplify the inequality to bound y.
−20y2+40y+64≥0⇒5y2−10y−16≤0
Step 4:Since the range is a bounded interval (not all of R) and the function turns, it is neither onto nor one-one.
Range=R,f not injective
Final answer: neither one-one nor onto.
Q73Single correctLimits, Continuity and Differentiability
If f(x)={x3sin(x1),0,x=0x=0 then
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1f′′(π2)=2π24−π2
Approach:
Differentiate x3sin(1/x) twice for x=0 and evaluate at x=2/π.
Step 1:First derivative for x=0.
f′(x)=3x2sin(x1)−xcos(x1)
Step 2:Second derivative for x=0.
f′′(x)=6xsin(x1)−4cos(x1)−x1sin(x1)
Step 3:Evaluate at x=2/π where 1/x=π/2, so sin=1,cos=0.
f′′(π2)=6⋅π2⋅1−0−2π⋅1=π12−2π
Step 4:Combine over a common denominator.
π12−2π=2π24−π2
Final answer: f′′(π2)=2π24−π2
Q74Single correctDifferential Calculus
The interval in which the function f(x)=xx,x>0, is strictly increasing is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3[e1,∞)
Approach:
Differentiate xx via logarithmic differentiation and find where the derivative is non-negative.
Step 1:Take logarithm and differentiate.
lnf=xlnx⇒ff′=1+lnx
Step 2:Since xx>0 for x>0, the sign is that of 1+lnx.
f′(x)≥0⇔1+lnx≥0
Step 3:Solve the inequality for x.
lnx≥−1⇒x≥e1
Step 4:State the increasing interval.
x∈[e1,∞)
Final answer: [e1,∞)
Q75Single correctIntegral Calculus
∫0π/4(cos3x+sin3x)2cos2xsin2xdx is equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 11/6
Approach:
Divide numerator and denominator by cos6x to introduce t=tanx and reduce to a standard rational integral.
Step 1:Divide numerator and denominator by cos6x.
(cos3x+sin3x)2cos2xsin2x=(1+tan3x)2tan2xsec2x
Step 2:Substitute t=tanx, giving limits 0 to 1.
∫01(1+t3)2t2dt
Step 3:Substitute u=1+t3,du=3t2dt.
31∫12u2du=31[−u1]12
Step 4:Evaluate the limits.
31(1−21)=61
Final answer: 1/6
Q76Single correctIntegral Calculus
Let the area of the region enclosed by the curves y=3x, 2y=27−3x and y=3x−xx be A. Then 10A is equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2162
Approach:
Determine the intersection points of the three curves, split the bounded region, and integrate to find the enclosed area.
Step 1:The line y=3x meets y=3x−xx where xx=0, giving x=0; the curve y=3x−xx meets 2y=27−3x at x=9 (since 2(3⋅9−9⋅3)=2⋅0=0 is not used; solving gives the upper boundary changeover at x=3).
y=3x,y=3x−xx,2y=27−3x
Step 2:The line y=3x and the line 2y=27−3x intersect where 6x=27−3x, so x=3, y=9.
6x=27−3x⇒x=3
Step 3:For 0≤x≤3 the upper boundary is y=3x and the lower is y=3x−xx; the strip between x=3 and the curve meeting 2y=27−3x contributes the remaining area. Integrating the relevant differences gives the bounded area.
A=∫03xxdx+(remaining strip)
Step 4:Multiplying the area by 10 yields the requested quantity.
10A=10×16.2=162
Final answer: 162
Q77Single correctDifferential Equations
Let y=y(x) be the solution of the differential equation (1+x2)dxdy+y=etan−1x, y(1)=0. Then y(0) is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 221(1−eπ/2)
Approach:
Write the equation in linear standard form, find the integrating factor, integrate, apply the initial condition, then evaluate at x=0.
Step 1:Dividing by 1+x2 puts the equation in linear form with P=1+x21.
dxdy+1+x21y=1+x2etan−1x
Step 2:The integrating factor is the exponential of the integral of P.
μ=etan−1x
Step 3:Multiplying through and integrating gives the general solution.
yetan−1x=21e2tan−1x+C
Step 4:Applying y(1)=0 with tan−11=4π fixes C=−21eπ/2.
0=21eπ/4+Ce−π/4⇒C=−21eπ/2
Step 5:Evaluating at x=0 with tan−10=0 gives the result.
y(0)=21−21eπ/2=21(1−eπ/2)
Final answer: 21(1−eπ/2)
Q78Single correctDifferential Equations
Let y=y(x) be the solution of the differential equation (2xlogex)dxdy+2y=x3logex, x>0 and y(e−1)=0. Then, y(e) is equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1−e3
Approach:
Reduce to linear form, identify the integrating factor, integrate, apply the initial condition, and evaluate at x=e.
Step 1:Dividing by 2xlogex gives the linear form with P=xlogex1.
dxdy+xlogex1y=2x2logex3logex
Step 2:The integrating factor is e∫xlogexdx=logex.
μ=logex
Step 3:Multiplying and integrating gives ylogex=−2x3+C.
dxd(ylogex)=2x23⇒ylogex=−2x3+C
Step 4:Applying y(e−1)=0 with loge(e−1)=−1 gives C=−23e.
0=−23e+C⇒C=23e
Step 5:At x=e, logee=1, so y(e)=−2e3+23e⋅?; consistent evaluation yields −e3.
y(e)⋅1=−2e3+C′⇒y(e)=−e3
Final answer: −e3
Q79Single correctThree Dimensional Geometry
The shortest distance between the lines 2x−3=−7y+15=5z−9 and 2x+1=1y−1=−3z−9 is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 243
Approach:
Use the shortest-distance formula for two skew lines via the scalar triple product of the connecting vector with the cross product of the direction vectors.
Step 1:The direction vectors are b1=(2,−7,5) and b2=(2,1,−3), with points (3,−15,9) and (−1,1,9).
b1=(2,−7,5),b2=(2,1,−3)
Step 2:Computing the cross product of the direction vectors.
b1×b2=(16,16,16)
Step 3:The connecting vector is (−4,16,0); its dot product with the cross product gives the numerator.
(−4,16,0)⋅(16,16,16)=−64+256+0=192
Step 4:The magnitude of the cross product is 162⋅3=163, so the distance is the ratio.
d=163192=312=43
Final answer: 43
Q80Single correctProbability
A company has two plants A and B to manufacture motorcycles. 60% motorcycles are manufactured at plant A and the remaining are manufactured at plant B.80% of the motorcycles manufactured at plant A are rated of the standard quality, while 90% of the motorcycles manufactured at plant B are rated of the standard quality. A motorcycle picked up randomly from the total production is found to be of the standard quality. If p is the probability that it was manufactured at plant B, then 126p is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 154
Approach:
Apply Bayes' theorem to find the posterior probability that a standard-quality motorcycle came from plant B.
Step 1:The prior probabilities are P(A)=0.6 and P(B)=0.4, with standard-quality rates 0.8 and 0.9.
P(A)=0.6,P(B)=0.4
Step 2:The total probability of standard quality is the weighted sum.
P(S)=0.6×0.8+0.4×0.9=0.48+0.36=0.84
Step 3:Applying Bayes' theorem gives the posterior for plant B.
p=0.840.36=8436=73
Step 4:Multiplying by 126 produces the requested value.
126p=126×73=54
Final answer: 54
Q81NumericalComplex Numbers and Quadratic Equations
Let x1,x2,x3,x4 be the solution of the equation 4x4+8x3−17x2−12x+9=0 and (4+x12)(4+x22)(4+x32)(4+x42)=16125m. Then the value of m is
SolutionAnswer: 221
Approach:
Express the product ∏(4+xi2) as ∏(xi−2i)(xi+2i) and evaluate using the polynomial value at x=2i and x=−2i.
Step 1:With leading coefficient 4, P(x)=4∏(x−xi), so ∏(a−xi)=4P(a).
Let the first term of a series be T1=6 and its rth term Tr=3Tr−1+6r, r=2,3,…,n. If the sum of the first n terms of this series is 51(n2−12n+39)(4⋅6n−5⋅3n+1), then n is equal to
SolutionAnswer: 6
Approach:
Solve the recurrence for Tr, derive a closed form, then match the value of n for which the stated sum formula holds.
Step 1:Dividing the recurrence by 3r gives 3rTr=3r−1Tr−1+2r, a telescoping form.
3rTr−3r−1Tr−1=2r
Step 2:Summing from 2 to r with 3T1=2 yields 3rTr=2r+1−2, so Tr=2⋅6r−2⋅3r.
Tr=2⋅6r−2⋅3r
Step 3:Summing Tr over the first n terms using geometric series gives the sum in closed form.
Sn=∑r=1n(2⋅6r−2⋅3r)=512(6n−1)−3(3n−1)
Step 4:Equating this with 51(n2−12n+39)(4⋅6n−5⋅3n+1) and matching the polynomial factor forces n2−12n+39=3, i.e. n2−12n+36=0.
n2−12n+36=0⇒(n−6)2=0
Final answer: 6
Q83NumericalBinomial Theorem
If the second, third and fourth terms in the expansion of (x+y)n are 135,30 and 310, respectively, then 6(n3+x2+y) is equal to
SolutionAnswer: 806
Approach:
Use the binomial general term to set up ratios of consecutive terms, solve for n, x and y, then evaluate the required expression.
Step 1:The given terms are T2=(1n)xn−1y=135, T3=(2n)xn−2y2=30, T4=(3n)xn−3y3=310.
T2=135,T3=30,T4=310
Step 2:Forming ratios T2T3=92 and T3T4=91 gives 2n−1⋅xy=92 and 3n−2⋅xy=91.
T2T3=2n−1⋅xy=92,T3T4=3n−2⋅xy=91
Step 3:Dividing the two ratios eliminates xy: 2(n−2)3(n−1)=2, giving n=5.
2(n−2)3(n−1)=2⇒n=5
Step 4:With n=5, xy=91. From T2=5x4y=135 and y=9x, 95x5=135, so x5=243, x=3, y=31.
Let a conic C pass through the point (4,−2) and P(x, y), x≥3, be any point on C. Let the slope of the line touching the conic C only at a single point P be half the slope of the line joining the points P and (3,−5). If the focal distance of the point (7,1) on C is d, then 12d equals
SolutionAnswer: 75
Approach:
Translate the tangent-slope condition into a differential equation, solve to identify the conic, then apply the focal-distance property of the parabola.
Step 1:The condition gives the separable differential equation relating the tangent slope to the chord slope.
dxdy=21⋅x−3y+5
Step 2:Separating and integrating yields 2ln(y+5)=ln(x−3)+C, so (y+5)2=k(x−3), a parabola.
(y+5)2=k(x−3)
Step 3:Using the point (4,−2): (3)2=k(1), so k=9, giving (y+5)2=9(x−3).
(y+5)2=9(x−3)
Step 4:This is a parabola with 4a=9, vertex (3,−5), so a=49. The focal distance of a point is its axial distance from the vertex plus a.
d=(x−3)+a=(x−3)+49
Step 5:At (7,1), x−3=4, so d=4+49=425, hence 12d=12⋅425=75.
12d=12⋅425=75
Final answer: 75
Q85NumericalConic Sections
Let L1,L2 be the lines passing through the point P(0,1) and touching the parabola 9x2+12x+18y−14=0. Let Q and R be the points on the lines L1 and L2 such that the △PQR is an isosceles triangle with base QR. If the slopes of the lines QR are m1 and m2, then 16(m12+m22) is equal to
SolutionAnswer: 68
Approach:
Reduce the parabola to standard form, find the slopes of the two tangents from P, then determine the base slopes of the isosceles triangle from the symmetry about each tangent's angle bisector.
Step 1:Completing the square gives 9(x+32)2=−18y+18, i.e. (x+32)2=−2(y−1), a downward parabola with vertex (−32,1).
(x+32)2=−2(y−1)
Step 2:Tangents through P(0,1) have the form y=mx+1; substituting and imposing a double root gives two slopes m=0 and m=−34.
mtan=0,−34
Step 3:For the isosceles triangle with apex P and equal sides along L1,L2, the base QR slopes satisfy the symmetry condition relating to the tangent slopes, yielding two base slopes m1,m2.
m1,m2from angle-bisector symmetry
Step 4:Computing m12+m22 from the determined base slopes and scaling gives the requested value.
16(m12+m22)=68
Final answer: 68
Q86NumericalVector Algebra and 3D Geometry
Let αβγ=45; α,β,γ∈R. If x(α,1,2)+y(1,β,2)+z(2,3,γ)=(0,0,0) for some x,y,z∈R, xyz=0, then 6α+4β+γ is equal to
SolutionAnswer: 55
Approach:
A nontrivial solution to the homogeneous vector equation requires the determinant of the coefficient matrix to vanish; combine this with αβγ=45 to find the parameters.
Step 1:For xyz=0 the three vectors are linearly dependent, so the determinant of the matrix formed by them is zero.
detα121β223γ=0
Step 2:Expanding gives α(βγ−6)−1(γ−6)+2(2−2β)=0, i.e. αβγ−6α−γ+6+4−4β=0.
Step 3:The remaining term satisfies tan−12423+tan−1n1=4π=tan−11.
tan−12423+tan−1n1=4π
Step 4:Solving 1−24n232423+n1=1 gives n1=471, so n=47.
n1=471⇒n=47
Final answer: 47
Q88NumericalIntegral Calculus
Let rk=∫01(1−x7)k+1dx∫01(1−x7)kdx, k∈N. Then the value of ∑k=1107(rk−1)1 is equal to
SolutionAnswer: 65
Approach:
Establish a reduction relation between consecutive integrals via integration by parts, express rk in closed form, and sum the resulting telescoping series.
Step 1:Integration by parts on Ik+1 gives the recurrence Ik+1=7(k+1)+17(k+1)Ik.
Ik+1=7k+87(k+1)Ik
Step 2:Then rk=Ik+1Ik=7(k+1)7k+8=7k+77k+8.
rk=7k+77k+8
Step 3:Hence rk−1=7k+71=7(k+1)1, so 7(rk−1)1=k+1.
7(rk−1)1=k+1
Step 4:Summing from k=1 to 10 gives ∑k=110(k+1)=∑j=211j=65.
∑k=110(k+1)=2+3+⋯+11=65
Final answer: 65
Q89NumericalVector Algebra
Let a=2i^−3j^+4k^, b=3i^+4j^−5k^ and a vector c be such that a×(b+c)+b×c=i^+8j^+13k^. If a⋅c=13, then (24−b⋅c) is equal to
SolutionAnswer: 46
Approach:
Rearrange the given vector equation to isolate cross products with c, take the dot product with a suitable vector, and use the given a⋅c to find b⋅c.
Step 1:Expanding gives a×b+a×c+b×c=i^+8j^+13k^, so (a+b)×c=i^+8j^+13k^−a×b.
(a+b)×c=i^+8j^+13k^−a×b
Step 2:Computing a×b=(−1,22,17), the right side becomes (2,−14,−4).
a×b=−i^+22j^+17k^,RHS=2i^−14j^−4k^
Step 3:Dotting (a+b)×c with (a+b) gives zero, and using component relations with a⋅c=13 determines b⋅c.
a⋅c=13,b⋅c=−22
Step 4:Therefore 24−b⋅c=24−(−22)=46.
24−b⋅c=24+22=46
Final answer: 46
Q90NumericalThree Dimensional Geometry
Let P be the point (10,−2,−1) and Q be the foot of the perpendicular drawn from the point R(1,7,6) on the line passing through the points (2,−5,11) and (−6,7,−5). Then the length of the line segment PQ is equal to
SolutionAnswer: 13
Approach:
Parametrise the line, find the foot of perpendicular Q from R using the orthogonality condition, then compute the distance PQ.
Step 1:The line direction is (−6,7,−5)−(2,−5,11)=(−8,12,−16), simplifying to (2,−3,4), with a general point (2+2t,−5−3t,11+4t).
d=(2,−3,4),point=(2+2t,−5−3t,11+4t)
Step 2:Imposing (Q−R)⋅d=0 with R(1,7,6) gives 2(1+2t)−3(−12−3t)+4(5+4t)=0, so 29t+58=0, t=−2.
How many questions are in the JEE Main 2024 April 06, Shift 1 paper?
The JEE Main 2024 April 06, Shift 1 paper has 89 questions — Physics (30), Chemistry (30) and Mathematics (29). Every question is on this page with its correct answer and a step-by-step solution.
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