JEE Main 2026 January 22, Shift 1 Question Paper with Solutions
All 75 questions from the JEE Main 2026 (January 22, Shift 1) shift — Physics (25), Chemistry (25) and Mathematics (25) — 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 volume of an ideal gas increases 8 times and temperature becomes (1/4)th of initial temperature during a reversible adiabatic change. If there is no exchange of heat in this process (ΔQ=0) then identify the gas from the following options (Assuming the gases given in the options are ideal gases):
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2He
Approach:
Given a reversible adiabatic change with V2=8V1 and T2=41T1 and ΔQ=0, the target is the ratio of specific heats γ, from which the degrees of freedom f and hence the gas are fixed. Apply the adiabatic relation TVγ−1=const.
Step 1:Equate the adiabatic invariant between the two states.
T1V1γ−1=T2V2γ−1
Step 2:Insert the given ratios T2/T1=1/4 and V1/V2=1/8.
41=(81)γ−1
Step 3:Express both sides as powers of 2 and solve for γ.
23(γ−1)=22⇒3(γ−1)=2
Step 4:Relate γ to the degrees of freedom.
1+f2=35⇒f2=32
Step 5:Identify the monatomic gas among the options.
f=3⇒monatomic
Final answer: He
Q27Single correctRotational Motion
A solid sphere of mass 5 kg and radius 10 cm is kept in contact with another solid sphere of mass 10 kg and radius 20 cm. The moment of inertia of this pair of spheres about the tangent passing through the point of contact is ____ kgm2.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 30.63
Approach:
Given two solid spheres (M1=5kg, R1=0.10m) and (M2=10kg, R2=0.20m) in contact, the target is the moment of inertia of the pair about the common tangent through the point of contact. This tangent line is tangent to each sphere, so the parallel-axis theorem applies to each.
Step 1:Each sphere's centre is a distance equal to its radius from the tangent line, so its tangent inertia is 57MR2; sum for both spheres.
I=57M1R12+57M2R22=57(M1R12+M2R22)
Step 2:Substitute SI values.
I=57(5(0.10)2+10(0.20)2)
Step 3:Evaluate the bracket and multiply.
I=57×0.45=0.63
Final answer: 0.63
Q28Single correctKinematics
A Projectile is thrown upward at an angle 60∘ with the horizontal. The speed of the projectile is 20 m/s when its direction of motion is 45∘ with the horizontal. The initial speed of the projectile is ------ m/s
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4202
Approach:
Given launch angle θ0=60∘ and a later instant where the speed is 20m/s at θ=45∘ to the horizontal, the target is the initial speed u. Gravity acts vertically, so the horizontal velocity component is conserved throughout the flight.
Step 1:Equate the horizontal velocity at launch and at the 45∘ instant.
ucos60∘=20cos45∘
Step 2:Solve for u.
u=240=2402=202
Final answer: 202
Q29Single correctProperties of Solids and Liquids
Given below are two statements: Statement I: Pressure of fluid is exerted only on a solid surface in contact as the fluid- Pressure does not exist everywhere in a still fluid . Statement II: Excess potential energy of the molecules on the surface of a liquid. When compared to interior, results in surface tension. 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:
Each statement is evaluated against the physics of fluid pressure and the molecular origin of surface tension. The target is the truth value of each statement.
Step 1:Assess Statement I on fluid pressure.
P=P0+ρgh
Step 2:Assess Statement II on surface tension.
Step 3:Combine the two assessments.
Final answer: Statement I is false but Statement II is true
Q30Single correctUnits and Measurements
Match the LIST-I with LIST-II
List -I
List-II
A. Spring constant
I.ML2T−2K−1
B. Thermal conductivity
II.ML0T−2
C. Boltzmann constant
III.ML2T−3A−2
D. Inductive reactance
IV.MLT−3K−1
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1A-II,B-IV,C-I,D-III
Approach:
Dimensions of each List-I quantity are obtained from its defining relation and matched to the List-II entries using base symbols M,L,T,K,A.
Step 1:Spring constant from force per unit extension.
[k]=LMLT−2=MT−2
Step 2:Thermal conductivity from Fourier's law, with [Q/t]=ML2T−3.
[K]=L2⋅K(ML2T−3)⋅L=MLT−3K−1
Step 3:Boltzmann constant from energy per unit temperature.
[kB]=KML2T−2=ML2T−2K−1
Step 4:Inductive reactance has the dimensions of resistance (voltage per current).
[XL]=AML2T−3A−1=ML2T−3A−2
Final answer: A-II,B-IV,C-I,D-III
Q31Single correctProperties of Solids and Liquids
Rods x and y of equal dimensions but of different materials are joined as shown in figure, Temperatures of end points A and F are maintained at 100∘C and 40∘C respectively . Given the thermal conductivity of rod x is three times that of rod y, the temperature at junction points B and E (close to):
(A)
(B)
(C)
(D)
SolutionAnswer: Option 389∘C and 73∘C respectively
Approach:
Given a network of equal-dimension rods with Kx=3Ky, end A at 100∘C and end F at 40∘C, the target is the steady-state temperatures at junctions B and E. Each rod is a thermal resistance R=l/(KA); for equal l and A, take rod x resistance =R and rod y resistance =3R. From the figure, AB is x; the rhombus has upper path B−C−E of two y rods and lower path B−D−E of two x rods; EF is y.
Step 1:Assign resistances using Kx=3Ky so that rod x has the smaller resistance.
Rx=3KyAl=R,Ry=KyAl=3R
Step 2:Reduce the rhombus between B and E: upper branch C-path is Ry+Ry=6R, lower branch D-path is Rx+Rx=2R, in parallel.
RBE=6R+2R(6R)(2R)=8R12R2=1.5R
Step 3:Combine the series chain A→B→E→F with RAB=R, RBE=1.5R, REF=3R.
Req=R+1.5R+3R=5.5R
Step 4:Compute the heat current from the 100∘C to 40∘C ends.
H=5.5R100−40=5.5R60=R10.91
Step 5:Find TB from the drop across AB.
TB=100−HR=100−11120=100−10.91
Step 6:Find TE from the drop across the rhombus.
TE=TB−H(1.5R)=89.09−1.5×10.91=89.09−16.36
Final answer: 89∘C and 73∘C respectively
Q32Single correctAtoms and Nuclei
7.9 MeV α-particle scatters from a target material of atomic number 79. From the given data the estimated diameter of nuclei of the target material is (approximately) ____m. [4πε01=9×109Nm2/C2 and electron charge e=1.6×10−19C]
(A)
(B)
(C)
(D)
SolutionAnswer: Option 22.88×10−14
Approach:
Given a 7.9MeVα-particle (charge 2e) approaching a nucleus of atomic number Z=79 (charge Ze) head-on, the target is the distance of closest approach, taken as the estimated nuclear size. At closest approach the entire kinetic energy is stored as electrostatic potential energy.
Step 1:Solve the balance for the closest-approach distance.
R=4πε01KE2Ze2
Step 2:Convert the kinetic energy from MeV to joules.
KE=7.9×106×1.6×10−19=1.264×10−12J
Step 3:Evaluate the numerator 4πε012Ze2.
9×109×2×79×(1.6×10−19)2=3.640×10−26
Step 4:Divide by the kinetic energy.
R=1.264×10−123.640×10−26
Final answer: 2.88×10−14
Q33Single correctCurrent Electricity
A meter bridge with two resistance R1 and R2 as shown in figure was balanced (null point) At 40cm from the point P. The null point changed to 50cm from the point P, when 16Ω resistance is connected in parallel to R2. The values of resistances R1 and R2 are____
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2R2=8Ω,R1=316Ω
Approach:
Given a metre bridge balanced at 40cm from P, and rebalanced at 50cm after a 16Ω resistor is placed in parallel with R2, the target is the pair (R1,R2). The balance condition equates the resistance ratio to the length ratio of the bridge wire.
Step 1:Apply the balance condition at l=40cm.
R2R1=6040=32
Step 2:Apply the balance at l=50cm with R2 replaced by its parallel combination R2′.
R2′R1=5050=1⇒R1=16+R216R2
Step 3:Substitute R1=32R2 and cancel R2.
32R2=16+R216R2⇒32(16+R2)=16
Step 4:Solve for R2 and then R1.
R2=8Ω,R1=32×8=316Ω
Final answer: R2=8Ω,R1=316Ω
Q34Single correctGravitation
The escape velocity from a spherical planet A is 10 km/s. The escape velocity from another planet B whose density and radius are 10% of those of planet A. is ____ m/s.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 210010
Approach:
Given veA=10km/s and planet B with ρB=0.1ρA and RB=0.1RA, the target is veB. Express escape velocity through density and radius using M=34πR3ρ.
Step 1:Form the ratio of escape velocities; the constant 8πG/3 cancels.
veAveB=RARBρAρB
Step 2:Insert RB/RA=0.1 and ρB/ρA=0.1.
veAveB=0.1×0.1=101⋅101=10101
Step 3:Compute veB with veA=10km/s=104m/s.
veB=1010104=10103=1010310=10010
Final answer: 10010
Q35Single correctElectrostatics
Six point charges are kept 60∘ apart from each other on the circumference of a circle of radius R as shown in figure. The net electric field at the center of the circle is ____ . (ε0 is permitivity of free space)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 24πε0R2Q(3i^−j^)
Approach:
Six charges sit 60∘ apart on a circle of radius R. From the figure the angular positions (measured from +x, with the marked 30∘ placing one +Q at −30∘) carry charges: +Q at 90∘, −Q at 150∘, +Q at 210∘, +Q at 270∘, +Q at −30∘ and +Q at 30∘, i.e. five +Q and a single −Q. The target is the net field at the centre, found by superposition with the single-charge field E0=4πε01R2Q.
Step 1:Six equal +Q charges symmetrically placed 60∘ apart give zero net field at the centre.
Eall +Q=0
Step 2:Replace the +Q at 150∘ by −Q. This equals removing one +Q (adding −1× its field) plus adding a −Q (another −1× its field): total −2 times the field of a +Q located at 150∘.
E=0−2E+Q@150∘
Step 3:Field at the centre from a +Q at angle θ points from charge to centre, direction (−cosθ,−sinθ); take θ=150∘.
Q36Single correctElectromagnetic Induction and Alternating Currents
Three identical coils C1,C2 and C3 are closely placed such that they share a common axis, C2 is exactly midway. C1 carries current I in anti-clockwise direction while C3 carries current I in clockwise direction . An induced current flows though C2 will be in clockwise direction when
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2C1 moves towards C2 and C3 moves away from C2
Approach:
Three coaxial coils C1,C2,C3 with C2 midway; C1 carries I anti-clockwise and C3 carries I clockwise (viewed from the same side). The target is the relative motion that drives a clockwise induced current in C2. By Lenz's law the induced current opposes the change of net flux linking C2.
Step 1:Determine the flux sense each neighbour sends through C2. Viewed from C2 toward C1, the anti-clockwise current of C1 produces flux out of C2 toward C1; viewed from C2 toward C3, the clockwise current of C3 produces flux of the opposite sense.
Step 2:A clockwise induced current in C2 (by the same viewing convention) must oppose an increase of the flux contributed in the C1 sense, requiring that the flux due to C1 rises while the opposing flux due to C3 falls.
Iindclockwise⇒dtdΦ1-sense>0
Step 3:Flux linkage strengthens as a coil approaches and weakens as it recedes.
Final answer: C1 moves towards C2 and C3 moves away from C2
Q37Single correctAtoms and Nuclei
The minimum frequency of photon required to break a particle of mass 15.348 amu into 4α particles is ____ kHz. [mass of He nucleus =4.002 amu, 1amu =1.66×10−27 kg, h=6.6×10−34 J.s and c=3×108 m/s]
(A)
(B)
(C)
(D)
SolutionAnswer: Option 314.94×1020
Approach:
A particle of mass 15.348amu is split into four α particles (mHe=4.002amu each). The minimum photon energy equals the mass-defect energy Δmc2, and the minimum frequency follows from hν=Δmc2. The requested unit is kHz.
Step 1:Compute the mass defect and convert to kilograms.
Q38Single correctElectromagnetic Induction and Alternating Currents
XPQY is a vertical smooth long loop having a total resistance of R where PX is parallel to QY and separation between them is l. A constant magnetic field B perpendicular to the plane of the loop exists in the entire space. A rod CD of length L(L>l) and mass m is made to slide down from rest under the gravity as shown in figure. The terminal speed acquired by the rod is ____ m/s(g=accelerationduetogravity)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2B2l2mgR
Approach:
A rod CD of length L slides down vertical rails of separation l (L>l) inside a uniform field B normal to the loop of total resistance R. The target is the terminal speed vt. Only the length l between the rails forms part of the closed circuit, so the active length for emf and force is l. At terminal speed gravity balances the magnetic retarding force.
Step 1:Write the induced current from the motional emf across the rail separation l.
I=RBlv
Step 2:Set the upward magnetic force equal to the weight at terminal speed (zero acceleration).
mg=BIl=RB2l2vt
Step 3:Solve for the terminal speed.
vt=B2l2mgR
Final answer: B2l2mgR
Q39Single correctOptics
A thin convex lens of focal length 5 cm and a thin concave lens of focal length 4 cm are combined together (without any gap) and this combination has magnification m1 when an object is placed 10 cm before the convex lens. Keeping the positions of convex lens and object undisturbed a gap of 1 cm is introduced between the lenses by moving the concave lens away. Which lead to a change in magnification of total lens system to m2. The value of m2m1 is ____
(A)
(B)
(C)
(D)
SolutionAnswer: Option 365
Approach:
Given a convex lens f1=+5 cm and a concave lens f2=−4 cm, object distance u=−10 cm from the convex lens. Target: m2m1, where m1 is the magnification with the lenses in contact and m2 is the magnification with a 1 cm gap. The in-contact case uses a combined focal length; the separated case is traced lens-by-lens with the convex image acting as object for the concave lens.
Step 1:Combine the focal lengths for the in-contact pair.
f1=51+−41=204−5=−201
Step 2:Apply the lens equation to the combination with u=−10 cm.
v1=f1+u1=−201−101=−203
Step 3:Compute the in-contact magnification.
m1=uv=−10−20/3=32
Step 4:Separated case: convex lens first, with u=−10 cm and f1=+5 cm.
v11=51+−101=101⇒v1=+10 cm
Step 5:The image lies 10 cm right of the convex lens, hence 9 cm right of the concave lens placed 1 cm away, forming a virtual object at u2=+9 cm. Apply the lens equation to the concave lens.
v21=f21+u21=−41+91=36−9+4=−365
Step 6:Compute the concave magnification and the overall separated magnification.
Net gravitational force at the center of a square is found to be F1 when four particles having mass M,2M,3M and 4M are placed at the four corners of the square as shown in figure and it is F2 when the positions of 3M and 4M are interchanged. The ratio F2F1 is 5a. The value of a is ____
(A)
(B)
(C)
(D)
SolutionAnswer: Option 22
Approach:
Given four masses M,2M,3M,4M at the corners of a square at equal distance r=2d from the centre. Each produces a force at the centre directed toward it; diagonally opposite masses act along the same line and partially cancel, and the two diagonal resultants are mutually perpendicular. Target: a where F2F1=5a, comparing the original layout (4M, 3M, M, 2M) with the layout after 3M and 4M are swapped.
Step 1:Each mass contributes a force ∝ its mass since all corner distances are equal. Along one diagonal the net is the difference of the opposite masses; along the perpendicular diagonal likewise. Original: 4M opposite 2M, and 3M opposite M.
F1∝(4M−2M)2+(3M−M)2=(2M)2+(2M)2
Step 2:After swapping 3M and 4M, the diagonal pairs become 3M opposite 2M and 4M opposite M.
F2∝(3M−2M)2+(4M−M)2=(M)2+(3M)2
Step 3:Form the ratio of the two resultants.
F2F1=1022=108=54=52
Step 4:Compare with the given form 5a.
5a=52
Final answer: 2
Q41Single correctOptics
Consider an equilateral prism (refractiveindex2) A ray of light is incident on its one surface at a certain angle i. If the emergent ray is found to graze along the other surface then the angle of refraction at the incident surface is close to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 315∘
Approach:
Given an equilateral prism with apex angle A=60∘ and refractive index n=2. The emergent ray grazes the second face, so the ray meets that face exactly at the critical angle. Target: the refraction angle r1 at the incident face, obtained from the prism relation r1+r2=A.
Step 1:Grazing emergence corresponds to the ray hitting the second face at the critical angle. Compute it for n=2.
sinθc=21⇒θc=45∘
Step 2:Apply the prism relation with A=60∘.
r1=A−r2=60∘−45∘
Final answer: 15∘
Q42Single correctOscillations and Waves
A simple pendulum has a bob with mass m and charge q. The pendulum string has negligible mass. When a uniform and horizontal electric field E is applied. The tension in the string changes. The final tension in the string when pendulum attains an equilibrium position is ____ (g acceleration due to gravity)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1m2g2+q2E2
Approach:
Given a charged bob (mass m, charge q) in a horizontal uniform field E, the bob experiences weight mg vertically down and electric force qE horizontally. At equilibrium the string lines up with the resultant of these two perpendicular forces, so the tension equals the magnitude of that resultant. Target: the equilibrium tension T.
Step 1:Resolve the forces on the bob: weight mg acts vertically downward, the electric force qE acts horizontally, and these are mutually perpendicular.
W=mg,FE=qE
Step 2:At equilibrium the string tension balances their resultant, so T equals the resultant magnitude.
T=(mg)2+(qE)2=m2g2+q2E2
Final answer: m2g2+q2E2
Q43Single correctElectrostatics
Electric field in a region is given by E=Axi^+Byj^, where A=10V/m2 and B=5V/m2 . If the electric potential at a point (10,20) is 500V, then the electric potential at origin is ____ V.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 42000
Approach:
Given E=Axi^+Byj^ with A=10V/m2, B=5V/m2, and V=500 V at (10,20). The potential difference equals minus the line integral of E. Target: the potential V0 at the origin, found by integrating each component along its axis.
Step 1:Evaluate the line integral from the origin to P=(10,20), taking the x-leg then the y-leg.
∫0PE⋅dl=∫010Axdx+∫020Bydy
Step 2:Substitute A=10 and B=5 and integrate.
=10⋅2102+5⋅2202=500+1000=1500V
Step 3:Apply the potential relation with VP=500 V.
VP−V0=−1500⇒500−V0=−1500
Final answer: 2000
Q44Single correctElectronic Devices
Find the correct combination of A,B,C and D inputs which can cause the LED to glow.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 31101
Approach:
Given a combinational network whose AND stages feed a final OR gate driving the LED, the LED glows when the OR output is logic 1, i.e. when at least one AND stage outputs 1. Target: the input pattern ABCD among the options that produces output 1. Each option is tested against the gate structure (an AND on A,B, a stage on the middle inputs, an AND on C,D, all into an OR).
Step 1:The final OR gate gives 1 if any feeding AND stage gives 1. The upper AND stage on inputs A,B outputs A⋅B; the lower AND stage on inputs C,D outputs C⋅D.
Y=(A⋅B)+(C⋅D)
Step 2:Test option 1, ABCD=0011.
A⋅B=0⋅0=0,C⋅D=1⋅1=1⇒Y=1
Step 3:Test the designated pattern ABCD=1101 (A=1,B=1,C=0,D=1).
A⋅B=1⋅1=1⇒Y=(1)+(C⋅D)=1
Step 4:The pattern 1101 drives the OR output high, so the LED conducts and glows.
Y=1⇒LED ON
Final answer: 1101
Q45Single correctKinetic Theory of Gases
A cylindrical tube AB of length l, closed at both ends contains an ideal gas of 1 mol having molecular weight M. The tube is rotated in a horizontal plane with constant angular velocity ω about an axis perpendicular to AB and passing through the edge at end A as shown in the figure. If PA and PB are the pressures at A and B respectively. Then (Consider the temperature is same at all points in the tube)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1PB=PAexp(Mω2l2/2RT)
Approach:
Given a closed tube of length l holding 1 mol of ideal gas (molar mass M) at uniform temperature T, rotated about a vertical axis through end A at angular velocity ω. In the rotating frame each gas element feels an outward centrifugal force; balancing the radial pressure gradient against this force and using the ideal-gas density gives a separable equation. Target: the relation between PB and PA.
Step 1:Substitute the ideal-gas density into the radial balance to obtain a separable equation in P and x.
dP=RTPMω2xdx⇒PdP=RTMω2xdx
Step 2:Integrate from A (x=0, pressure PA) to B (x=l, pressure PB).
∫PAPBPdP=RTMω2∫0lxdx
Step 3:Exponentiate both sides.
PB=PAexp(2RTMω2l2)
Final answer: PB=PAexp(Mω2l2/2RT)
Q46NumericalOscillations and Waves
Two loudspeakers (L1 and L2) are placed with a separation of 10 m. as shown in figure. Both speakers are fed with an audio input signal of same frequency with constant volume A voice recorder, initially at point A at equidistance to both loud speakers is moved by 25m along the line AB while monitoring the audio signal The measured signal was found to undergo 10 cycles of minima and maxima during the movement. The frequency of the input signal is ____ Hz (Speed of sound in air is 324 m/s and 5=2.23)
SolutionAnswer: 600
Approach:
Given two speakers separated by 10 m, the recorder starts at A equidistant from both (zero path difference) and moves 25 m along AB, recording 10 full maxima-minima cycles. Ten cycles correspond to the path difference increasing by 10λ. Using the geometry the path difference at B fixes λ and hence the frequency. Target: f in Hz, with v=324 m/s and 5=2.23.
Step 1:Set coordinates: L1=(0,5), L2=(0,−5), A=(40,0), B=(40,25). Compute the two source-to-B distances.
L1B=402+(25−5)2=402+202=2000=205
Step 2:Compute the distance from L2 to B.
L2B=402+(25+5)2=402+302=2500=50 m
Step 3:The path difference is zero at A and at B equals L2B−L1B; evaluate with 5=2.23.
Δ=50−205=50−20(2.23)=50−44.6=5.4 m
Step 4:Equate the path difference to 10λ and solve for the wavelength.
10λ=5.4⇒λ=0.54 m
Step 5:Compute the frequency from the wave relation.
f=λv=0.54324
Final answer: 600
Q47NumericalRotational Motion
A circular disc has radius R1 and thickness T1. Another circular disc made of the same material has radius R2 and thickness T2. If the moment of inertia of both discs are same and R2R1=2 then T2T1=α1. The value of α is ____
SolutionAnswer: 16
Approach:
Given two discs of the same material with R1/R2=2 and equal moments of inertia about their central axes. Mass equals density times volume (ρπR2T), so I=21MR2∝R4T for fixed ρ. Equating the inertias relates the thicknesses. Target: α where T1/T2=1/α.
Step 1:Express the moment of inertia in terms of geometry and density.
I=21(ρπR2T)R2=2ρπR4T
Step 2:Equate the two inertias (same ρ cancels).
R14T1=R24T2⇒T2T1=(R1R2)4
Step 3:Substitute R1/R2=2, i.e. R2/R1=1/2.
T2T1=(21)4=161
Step 4:Match with T1/T2=1/α.
α1=161
Final answer: 16
Q48NumericalElectromagnetic Induction and Alternating Currents
Inductance of a coil with 104 turns is 10 mH and it is connected to a dc source of 10 V with internal resistance of 10Ω. The energy density in the inductor when the current reaches (e1) of its maximum value is e2απJ/m3. The value of α is ____ (μ0=4π×10−7Tm/A)
SolutionAnswer: 20
Approach:
Given a coil (N=104 turns, L=10 mH) on a 10 V dc source with internal resistance 10Ω. The steady (maximum) current is V/R; treating the coil as a long solenoid, the interior field is μ0ni with n=N/l (length l=1 m so n=N), and the energy density is B2/2μ0=21μ0n2i2. Target: α where the energy density at i=imax/e equals e2απJ/m3, with μ0=4π×10−7.
Step 1:Compute the maximum current and the turns per unit length.
imax=1010=1A;n=lN=1104=104m−1
Step 2:The current at the stated instant is i=imax/e=1/e A. Insert into the energy-density expression.
u=2μ0n2(e1)2=2e2(4π×10−7)(104)2
Step 3:Simplify the numerical factor: (104)2=108 and 10−7×108=10.
u=2e24π×10=e220πJ/m3
Step 4:Match with e2απ.
e2απ=e220π
Final answer: 20
Q49NumericalElectromagnetic Waves
The electric field of a plane electromagnetic wave, travelling in an unknown non-magnetic medium is given by . Ey=20sin(3×106x−4.5×1014t)V/m (where x,t and other values have S.I. units).The dielectric constant of the medium is ____ (Speed of light in free space is 3×108m/s)
SolutionAnswer: 4
Approach:
Given a plane EM wave Ey=20sin(3×106x−4.5×1014t) in a non-magnetic medium, the wave number is k=3×106m−1 and angular frequency ω=4.5×1014s−1. The phase speed is ω/k, and for μr=1 the dielectric constant is (c/v)2. Target: εr, with c=3×108 m/s.
Step 1:Read ω and k from the wave argument and compute the phase speed.
v=kω=3×1064.5×1014=1.5×108m/s
Step 2:For a non-magnetic medium μr=1, so c/v=εr.
εr=vc=1.5×1083×108=2
Step 3:Square to obtain the dielectric constant.
εr=22=4
Final answer: 4
Q50NumericalOptics
A parallel beam of light travelling in air (refractive index 1.0) is incident on a convex spherical glass surface of radius of curvature 50 cm. Refractive index of glass is 1.5. The rays converge to a point at a distance x cm from the centre of the curvature of the spherical surface . The value of x is ____ cm.
SolutionAnswer: 100
Approach:
Given a parallel beam in air (μ1=1.0) striking a convex spherical glass surface (μ2=1.5, R=50 cm), refraction at a single spherical surface fixes the image (convergence) distance v from the surface for an object at infinity. The convergence point is then expressed relative to the centre of curvature. Target: x, the distance of the convergence point from the centre of curvature.
Step 1:Apply the surface relation with a parallel beam (u→∞, so μ1/u→0), μ1=1.0, μ2=1.5, R=+50 cm.
v1.5−0=501.5−1.0=500.5=0.01
Step 2:Solve for the image distance from the surface.
v=0.011.5=150cm
Step 3:The centre of curvature lies R=50 cm from the surface on the same side; subtract to locate the convergence point from it.
x=v−R=150−50
Final answer: 100
Chemistry25 questions
Q51Single correctSome Basic Principles of Organic Chemistry
As compared with chlorocyclohexane, which of the following statements correctly apply to chlorobenzene ? A. The magnitude of negative charge is more on chlorine atom B. The C-Cl bond has partial double bond character C. C−Cl bond is less polar D. C-Cl bond is longer due to repulsion between delocalised electrons of the aromatic ring and lone pairs of electrons of chlorine. E. The C-Cl bond is formed using sp2 hybridized orbital of carbon Choose the correct answer from the options given below :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3B, C and E Only
Approach:
Given two C–Cl bonds, one in chlorobenzene and one in chlorocyclohexane, the objective is to identify which statements (A–E) correctly describe chlorobenzene relative to chlorocyclohexane. The method is to apply resonance delocalisation of a chlorine lone pair into the aromatic ring and the sp2 geometry of the aromatic carbon.
Step 1:In chlorobenzene a lone pair on chlorine is conjugated with the ring, producing resonance structures in which the C–Cl bond carries partial double bond character. In chlorocyclohexane (sp3 carbon) no such conjugation exists, so the C–Cl bond is a pure single bond. Hence statement B applies to chlorobenzene.
Ar−C¨l↔Ar=C+l
Step 2:Delocalisation transfers electron density from chlorine into the ring, leaving a partial positive charge on chlorine. The chlorine therefore bears a smaller magnitude of negative charge than in chlorocyclohexane, so statement A (more negative charge on Cl) is false.
δ+ developed on Cl
Step 3:Because chlorine's lone-pair density is partly drawn into the ring and the bonding carbon is sp2 (higher electronegativity, shorter bond), the dipole arising from the C–Cl bond is reduced. The C–Cl bond in chlorobenzene is therefore less polar than in chlorocyclohexane, so statement C applies.
μC−Cl(chlorobenzene)<μC−Cl(chlorocyclohexane)
Step 4:Partial double bond character shortens the C–Cl bond in chlorobenzene rather than lengthening it; statement D asserts a longer bond and a repulsion mechanism, both contrary to the shortening produced by resonance, so statement D is false.
dC−Cl(chlorobenzene)<dC−Cl(chlorocyclohexane)
Step 5:The ring carbon bonded to chlorine in chlorobenzene is part of the aromatic system and is sp2 hybridised, whereas in chlorocyclohexane it is sp3. Statement E (C–Cl formed using sp2 carbon orbital) therefore applies.
C(sp2)−Cl
Final answer: B, C and E Only
Q52Single correctHydrocarbons
Given below are two statements :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2Statement I is incorrect but Statement II is correct
Approach:
Statement I claims that nitrobenzene reacts with CH3COCl/AlCl3 to give an acetyl-substituted nitrobenzene; Statement II classifies the nitro group as meta-directing and deactivating. The objective is to judge each statement using the rules of Friedel–Crafts acylation and the electronic nature of NO2.
Step 1:The nitro group is a strong electron-withdrawing substituent that powerfully deactivates the aromatic ring. Friedel–Crafts acylation requires an activated (electron-rich) ring; it fails on rings bearing strongly deactivating groups such as NO2, so nitrobenzene does not undergo acylation with CH3COCl/AlCl3.
C6H5NO2CH3COCl/AlCl3no reaction
Step 2:Since the acyl-substituted product cannot form, the transformation depicted in Statement I is invalid; Statement I is incorrect.
Step 3:The nitro group withdraws electron density (–I and –R), deactivating the ring, and directs incoming electrophiles to the meta position because the ortho/para positions bear the greatest positive charge in the intermediate. Thus NO2 is meta-directing and deactivating; Statement II is correct.
−NO2:meta-directing, deactivating
Final answer: Statement I is incorrect but Statement II is correct
Q53Single correctEquilibrium
Given below are two statements: Statement I : The Henry's law constant KH is constant with respect to variations in solution's concentration over the range for which the solution is ideally dilute. Statement II : KH does not differ for the same solute in different solvents. In the light of the above statements, choose the correct answer from the options given below
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1Statement I is true and Statement II are false
Approach:
Statement I asserts that the Henry's law constant KH is constant with concentration in the ideally dilute range; Statement II asserts that KH is the same for a given solute in different solvents. The objective is to test each against the definition and behaviour of KH.
Step 1:Henry's law states that the partial pressure of a volatile solute is proportional to its mole fraction in an ideally dilute solution, with proportionality constant KH. Within the dilute range where the law applies, KH does not vary with solute concentration; it is a fixed parameter for the given solute–solvent pair at a fixed temperature.
p=KHx,KH=f(x)
Step 2:KH measures the solute–solvent interaction, so it depends on the identity of the solvent. The same gas dissolved in different solvents has different KH values; for example, the KH of a gas in water differs from its KH in an organic solvent. Statement II (same KH in different solvents) is false.
KH(solvent1)=KH(solvent2)
Final answer: Statement I is true and Statement II are false
Q54Single correctSome Basic Concepts in Chemistry
In the reaction 2Al(s)+6HCl(aq)→2Al3+(aq)+6Cl−(aq)+3H2(g)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 411.2 L H2(g) at STP is produced for every mole of HCl consumed
Approach:
Given the balanced equation 2Al+6HCl→2Al3++6Cl−+3H2, the objective is to find the statement consistent with the stoichiometric coefficients and the molar volume of a gas at STP. The method applies mole ratios and Vm=22.4L mol−1.
Step 1:From the coefficients, 2 mol Al give 3 mol H2, so 1 mol Al gives 1.5 mol H2. At STP this is 1.5 x 22.4 = 33.6 L, but only at STP. Option 1 (67.2 L) and option 3 (33.6 L regardless of T and P) are therefore wrong; gas volume depends on temperature and pressure.
1mol Al→1.5mol H2=33.6L (STP only)
Step 2:From the coefficients, 6 mol HCl give 3 mol H2, a 2:1 ratio in moles. At equal temperature and pressure, volume is proportional to moles, so the HCl:H2 volume ratio is also 2:1. For 6 L of H2 produced, 12 L of HCl(g) would be consumed, but HCl here is aqueous, so a gas-volume ratio cannot be applied to it; option 2 is wrong.
VH2VHCl=36=2
Step 3:From 6 mol HCl giving 3 mol H2, 1 mol HCl gives 0.5 mol H2. At STP this corresponds to 0.5 x 22.4 = 11.2 L of H2 for every mole of HCl consumed, matching option 4.
0.5×22.4=11.2L
Final answer: 11.2 L H2(g) at STP is produced for every mole of HCl consumed
Q55Single correctp-Block Elements
Given below are two statements Statement-I : The halogen that makes longest bond with hydrogen in HX, has the smallest covalent radius in its group. Statement-II : A group 15 elements hybride EH3 has the lowest boiling point among corresponding hybrids of other group 15 elements. The maximum covalency of that element E is 4 In the light of the above statements, choose the correct answer from the options given below
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2Both Statement I and Statement II are false
Approach:
Statement I links the longest H–X bond to the smallest covalent radius in group 17; Statement II identifies the group-15 hydride EH3 with the lowest boiling point and gives the maximum covalency of E as 4. The objective is to test both against periodic trends.
Step 1:Bond length in HX increases as the halogen size increases: HF < HCl < HBr < HI. The longest H–X bond is in HI. Iodine, being the heaviest halogen, has the largest covalent radius in group 17, not the smallest. Statement I (longest bond pairing with smallest covalent radius) is therefore false.
dH−Ilongest,rcov(I)largest in group 17
Step 2:Among group-15 hydrides the boiling point order is PH3 < AsH3 < NH3 < SbH3: NH3 is raised by hydrogen bonding, SbH3 by its large molar mass, and PH3 is lowest. So the lowest-boiling hydride is PH3, identifying E as phosphorus.
PH3<AsH3<NH3<SbH3
Step 3:Phosphorus has accessible 3d orbitals and expands its octet, reaching a maximum covalency of 6, as in PF6- and PCl5/PCl6-. The claimed maximum covalency of 4 is too low, so the second part of Statement II is false, making Statement II false.
max covalency of P=6(e.g.PF6−)
Final answer: Both Statement I and Statement II are false
'A' is a neutral organic compound (M. F: C8H9ON). On treatment with aqueous Br2 / HO(−), 'A' forms a compound 'B' which is soluble in dilute acid. 'B' on treatment with aqueous NaNO2 / HCl(0−5∘C) produces a compound 'C' which on treatment with CuCN/ NaCN produces 'D' . Hydrolysis of 'D' produces 'E' which is also obtainable from the hydrolysis of 'A'. 'E' on treatment with acidified KMnO4 produces 'F'. 'F' contains two different types of hydrogen. The structure of 'A' is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 12-methylbenzamide (o-toluamide): benzene ring with adjacent CH3 and CONH2 groups
Approach:
Given a neutral compound A (C8H9ON) and the sequence A->B (aq. Br2/OH−), B->C (NaNO2/HCl, 0–5 C), C->D (CuCN/NaCN), D->E (hydrolysis, also from A), E->F (acidified KMnO4), with the constraint that F has two different types of hydrogen, the objective is to identify the structure of A among the four toluamide isomers.
Step 1:A (C8H9ON) is neutral with degree of unsaturation 5, consistent with a benzene ring plus a carbonyl: it is a methyl-substituted benzamide (CH3-C6H4-CONH2). Treatment with aqueous Br2 and OH− is the Hofmann bromamide degradation, which removes the carbonyl carbon and converts the amide to an amine with one fewer carbon, giving toluidine B (CH3-C6H4-NH2), which dissolves in dilute acid because the arylamine forms a soluble salt.
CH3C6H4CONH2Br2/OH−CH3C6H4NH2
Step 2:B with NaNO2/HCl at 0–5 C is diazotised to the diazonium salt C (CH3-C6H4-N2+). Treatment of C with CuCN/NaCN (Sandmeyer reaction) replaces the diazonium group by a nitrile to give D (CH3-C6H4-CN).
Step 3:Hydrolysis of the nitrile D gives the carboxylic acid E (CH3-C6H4-COOH, a toluic acid). Hydrolysis of the amide A also gives the same acid CH3-C6H4-COOH, confirming the carboxyl and methyl substituents in A and E occupy the same two ring positions.
CH3C6H4CNH2OCH3C6H4COOH=hydrolysis of A
Step 4:Acidified KMnO4 oxidises the ring methyl group of E to a carboxyl group, converting toluic acid to a benzenedicarboxylic acid F (HOOC-C6H4-COOH). The number of distinct hydrogen environments in F fixes the isomer: only the ortho (phthalic) acid gives a ring whose aromatic hydrogens fall into two equivalent pairs, i.e. two different types of hydrogen, while the meta and para isomers give three and one type respectively.
CH3C6H4COOHKMnO4/H+C6H4(COOH)2
Step 5:Since F must be the ortho diacid, the CH3 and the carboxyl-derived group in E (and hence the CH3 and CONH2 in A) are ortho to each other. Therefore A is 2-methylbenzamide (o-toluamide), the structure with adjacent CH3 and CONH2 groups, which is option 1.
A=o-CH3C6H4CONH2
Final answer: 2-methylbenzamide (o-toluamide): benzene ring with adjacent CH3 and CONH2 groups
Q57Single correctChemical Bonding and Molecular Structure
TWO p – block elements X and Y form fluorides of the type EF3. The fluoride compound XF3 is a Lewis acid and YF3 is a Lewis base. The hybridizations of the central atoms of XF3 and YF3 respectively are
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1sp2 and sp3
Approach:
Given two p-block elements forming fluorides EF3, with XF3 a Lewis acid and YF3 a Lewis base, the objective is to find the hybridisation of each central atom. The method links Lewis acid/base behaviour to the presence or absence of a lone pair and applies VSEPR to assign hybridisation.
Step 1:A Lewis acid accepts an electron pair, which requires an electron-deficient central atom with a vacant orbital and no lone pair. An EF3 of this type is a group-13 fluoride such as BF3: boron has three bonding pairs and no lone pair, giving three electron domains, trigonal planar geometry, and sp2 hybridisation.
BF3:3bond pairs,0lone pair⇒sp2
Step 2:A Lewis base donates an electron pair, which requires a lone pair on the central atom. An EF3 of this type is a group-15 fluoride such as NF3: nitrogen has three bonding pairs and one lone pair, giving four electron domains, pyramidal shape, and sp3 hybridisation.
NF3:3bond pairs,1lone pair⇒sp3
Step 3:Combining the two, the hybridisations of the central atoms in the Lewis-acidic XF3 and the Lewis-basic YF3 are sp2 and sp3 respectively, which is option 1.
List-I gives four reagent systems and List-II gives four named reactions of carbonyl compounds. The objective is to pair each reagent with the reaction it characterises, by recalling the defining reagents of each named test or reduction.
Step 1:Hydrazine with a strong base (NH2-NH2, KOH) reduces a carbonyl group to a methylene group through a hydrazone intermediate; this is the Wolff–Kishner reduction. Hence A pairs with III.
>C=ONH2NH2,KOH>CH2
Step 2:The diamminesilver(I) hydroxide complex Ag(NH3)2OH is Tollen's reagent; an aldehyde reduces it to a silver mirror. Hence B pairs with I.
Ag(NH3)2+RCHOAg↓
Step 3:Aqueous CuSO4 complexed with sodium potassium tartrate in KOH is Fehling's reagent; an aliphatic aldehyde reduces the Cu(II) to a red Cu2O precipitate. Hence C pairs with IV.
Cu2+RCHOCu2O↓
Step 4:Zinc amalgam with concentrated HCl (Zn-Hg, HCl) reduces a carbonyl group to a methylene group under acidic conditions; this is the Clemmensen reduction. Hence D pairs with II.
The correct order of the rate of reaction of the following reactants with nucleophile by SN1 mechanism is (Given : Structures I and II are rigid)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3II < I < III < IV
Approach:
Given four bromides reacting by an SN1 mechanism, the objective is to order their rates. Since the rate-determining step of SN1 is ionisation to a carbocation, the rate follows the stability of the carbocation each substrate forms. Structures I and II are rigid bicyclic bridgehead bromides.
Step 1:A carbocation prefers planar sp2 geometry. In a rigid bicyclic cage the bridgehead carbon is held in a pyramidal arrangement and cannot flatten (Bredt's rule), so bridgehead cations are highly destabilised and these substrates ionise extremely slowly. Both I and II are therefore the slowest.
bridgehead C cannot become planar
Step 2:Between the two cages, the smaller, more strained bicyclo[2.2.1] (norbornyl, II) bridgehead resists planarisation even more than the larger, more flexible bicyclo[2.2.2]octyl (I) bridgehead. The larger cage relieves strain slightly better, so I ionises faster than II.
strain: II (smaller cage)>I (larger cage)
Step 3:tert-Butyl bromide (III) ionises to the tertiary carbocation (CH3)3C+, which is planar and hyperconjugatively stabilised, far more stable than either constrained bridgehead cation. So III reacts faster than I.
(CH3)3C+
Step 4:The triphenyl-substituted bromide (IV, Ph3C-Br) ionises to the trityl cation Ph3C+, stabilised by resonance over three benzene rings, the most stable cation in the set. So IV reacts fastest.
Ph3C+(resonance over 3 rings)
Step 5:Chaining the comparisons II < I, I < III, III < IV gives the full order II < I < III < IV, which is option 3.
Given below are two statements : Statement I : Phenol on treatment with CHCl3 / aq.KOH under refluxing condition, followed by acidification produces p- hydroxy benzaldehyde as the major product and o-ohydroxy benzaldehyde as the minor product. Statement II : The mixture of p-hydroxybenzaldehyde and o- hydroxybenzaldehyde can be easily separated through steam distillation. In the light of the above statements, choose the correct answer from the options given below.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4Both Statement I and Statement II are true
Approach:
Statement I describes the Reimer–Tiemann product distribution (para major, ortho minor) and Statement II claims the ortho/para mixture is separable by steam distillation. The objective is to judge both using the mechanism of the Reimer–Tiemann reaction and the hydrogen-bonding behaviour of the products.
Step 1:Phenol with CHCl3 and aqueous KOH generates dichlorocarbene, which attacks the phenoxide ring to introduce a -CHO group; acidification gives hydroxybenzaldehydes (the Reimer–Tiemann reaction). With the bulky K+ counterion solvating the ortho phenoxide oxygen, the steric environment directs the carbene preferentially to the para position, so p-hydroxybenzaldehyde is the major and o-hydroxybenzaldehyde the minor product. Statement I is correct.
C6H5OHCHCl3/KOHp- and o-HOC6H4CHO
Step 2:In o-hydroxybenzaldehyde the -OH and -CHO are adjacent and form an intramolecular hydrogen bond (chelation), lowering its association and making it volatile in steam. In p-hydroxybenzaldehyde the groups are too far apart for intramolecular bonding, so it engages in intermolecular hydrogen bonding and is non-volatile. The ortho isomer therefore steam-distills over while the para isomer remains, so the mixture is separable by steam distillation. Statement II is correct.
o-isomer: intramolecular H-bond⇒steam volatile
Final answer: Both Statement I and Statement II are true
Q61Single correctChemical Bonding and Molecular Structure
The formal charges on the atoms marked as (1) to (4) in the Lewis representation of HNO3 molecule respectively are
(A)
(B)
(C)
(D)
SolutionAnswer: Option 30, + 1, 0, -1
Approach:
Given the Lewis structure H-O(1)-N(2)=O(3) with the nitrogen also single-bonded down to O(4) bearing three lone pairs, the objective is the formal charge on atoms 1–4. The method applies FC=V−N−21B, where V is valence electrons, N is non-bonding electrons, and B is bonding electrons.
Step 1:Atom 1 is the oxygen bonded to H and N by two single bonds and carrying two lone pairs. V = 6, N = 4, B = 4 (two single bonds). FC = 6 - 4 - 1/2(4) = 0.
6−4−21(4)=0
Step 2:Atom 2 is nitrogen with no lone pair: one single bond to O(1), one single bond to O(4), and one double bond to O(3), i.e. four bonds (8 bonding electrons). V = 5, N = 0, B = 8. FC = 5 - 0 - 1/2(8) = +1.
5−0−21(8)=+1
Step 3:Atom 3 is the doubly bonded oxygen with two lone pairs. V = 6, N = 4, B = 4 (one double bond). FC = 6 - 4 - 1/2(4) = 0.
6−4−21(4)=0
Step 4:Atom 4 is the singly bonded oxygen carrying three lone pairs. V = 6, N = 6, B = 2 (one single bond). FC = 6 - 6 - 1/2(2) = -1.
6−6−21(2)=−1
Final answer: 0, + 1, 0, -1
Q62Single correctAtomic Structure
The energy required by electrons, present in the first Bohr orbit of hydrogen atom to be excited to second Bohr orbit is _______ J mol−1 Given : RH=2.18×10−11 ergs
(A)
(B)
(C)
(D)
SolutionAnswer: Option 19.835×105
Approach:
Given RH=2.18×10−11 erg, the objective is the energy per mole to excite an electron from the n=1 to the n=2 Bohr orbit of hydrogen. The method computes the per-atom transition energy from the Rydberg expression, converts the unit from erg to joule, and scales by Avogadro's number.
Step 1:Convert the Rydberg energy from erg to joule using 1 erg = 10−7 J: RH = 2.18 x 10−11 erg = 2.18 x 10−18 J.
RH=2.18×10−11erg=2.18×10−18J
Step 2:Apply the transition energy for n1 = 1 to n2 = 2: the bracket is 1/1 - 1/4 = 3/4. Hence per atom dE = 2.18e-18 x (3/4) = 1.635 x 10−18 J.
Step 3:The question asks for energy per mole, so multiply the per-atom energy by Avogadro's number NA = 6.022 x 1023mol−1.
Emol=1.635×10−18×6.022×1023=9.835×105J mol−1
Final answer: 9.835×105
Q63Single correctChemical Kinetics
A→ product (First order reaction) Three sets of experiment were performed for a reaction under similar experimental conditions : Run 1 ⇒ 100 mL of 10 M solution of reactant A Run 2 ⇒ 200 mL of 10 M solution of reactant A Run 3 ⇒ 100 mL of 10 M solution of reactant A + 100 mL of H2O added. The correct variation of rate of reaction is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2Run 3 < Run 1 = Run 2
Approach:
For the first-order reaction A -> product, the rate is r = k[A]. Three runs differ in volume and dilution: Run 1 is 100 mL of 10 M A, Run 2 is 200 mL of 10 M A, Run 3 is 100 mL of 10 M A diluted with 100 mL water. The objective is to rank the rate of reaction, an intensive quantity that depends only on concentration.
Step 1:Determine the concentration of A in each run. Run 1: 10 M. Run 2: still 10 M (taking more of the same solution does not change concentration). Run 3: 100 mL of 10 M diluted to a total of 200 mL gives [A] = 10 x 100 / 200 = 5 M.
[A]1=10M,[A]2=10M,[A]3=20010×100=5M
Step 2:The rate of a reaction r = k[A] is an intensive property; under identical conditions k is the same, so r depends only on [A] and not on the volume or total amount taken. Run 1 and Run 2 share the same 10 M concentration, so their rates are equal.
r1=k(10)=r2=k(10)
Step 3:Run 3 has half the concentration (5 M), so its rate is half that of Run 1 and Run 2 and is the smallest. Combining, Run 3 < Run 1 = Run 2.
r3=k(5)<r1=r2=k(10)
Final answer: Run 3 < Run 1 = Run 2
Q64Single correctCoordination Compounds
A first row transition metal (M) does not liberate H2 gas from dilute HCl. 1 mol of aqueous solution of MSO4 is treated with excess of aqueous KCN and then H2S(g) is passed through the solution. The amount of MS (metal sulphide) formed from the above reaction is _______ mol
(A)
(B)
(C)
(D)
SolutionAnswer: Option 20
Approach:
Given: a first-row transition metal M whose 1 mol MSO4 is treated with excess KCN, then saturated with H2S(g). Target: moles of MS precipitate. Identify M from the clue that it does not liberate H2 from dilute HCl, form its cyanide complex, and test whether free metal ion remains for sulphide precipitation.
Step 1:A first-row transition metal that does not displace hydrogen from dilute HCl lies below hydrogen in the activity series, i.e. it has a positive standard reduction potential. Copper (E0(Cu2+/Cu) = +0.34 V) satisfies this, so M is copper and MSO4 is CuSO4.
ECu2+/Cu0=+0.34V>0
Step 2:Cu2+ with excess CN− is first reduced to Cu+ (CN− oxidised to cyanogen) and the Cu+ is captured as the very stable tetracyanocuprate(I) complex.
2Cu2++10CN−→2[Cu(CN)4]3−+(CN)2
Step 3:The dissociation constant of [Cu(CN)4]3- is extremely small, so the free Cu+ concentration is too low for the ionic product to exceed the solubility product of Cu2S. Hence H2S produces no metal sulphide precipitate.
[Cu(CN)4]3−+H2S↛
Step 4:Therefore the amount of metal sulphide formed is zero.
nMS=0mol
Final answer: 0
Q65Single correctCoordination Compounds
Consider the transition metal ions Mn3+,Cr3+,Fe3+ and Co3+ and all form low spin octahedral complexes. The correct decreasing order of unpaired electrons in their respective d-orbitals of the complexes is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4Cr3+>Mn3+>Fe3+>Co3+
Approach:
Given: Mn3+, Cr3+, Fe3+, Co3+, all forming low-spin octahedral complexes. Target: decreasing order of unpaired electrons. Determine each d-electron count, fill the t2g/eg sets under strong-field (low-spin) splitting, and count unpaired electrons.
Step 1:Cr3+ has the configuration d3. The first three electrons occupy the three t2g orbitals singly, giving 3 unpaired electrons.
Cr3+(d3):t2g3eg0⇒n=3
Step 2:Mn3+ has the configuration d4. In the low-spin case the fourth electron pairs within t2g rather than entering eg, leaving 2 unpaired electrons.
Mn3+(d4):t2g4eg0⇒n=2
Step 3:Fe3+ has the configuration d5. Low-spin filling pairs electrons in t2g until five occupy it, leaving 1 unpaired electron.
Fe3+(d5):t2g5eg0⇒n=1
Step 4:Co3+ has the configuration d6. Low-spin filling completely fills t2g, leaving 0 unpaired electrons. The unpaired counts are Cr3+(3) > Mn3+(2) > Fe3+(1) > Co3+(0).
The correct order of reactivity of CH3Br in methanol with the following nucleophiles is F−,I−,C2H5O− and C6H5O−
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2I−>C2H5O−>C6H5O−>F−
Approach:
Given: CH3Br reacting in methanol (a protic solvent) with the nucleophiles F-, I-, C2H5O- and C6H5O-. Target: order of reactivity. Rank the species by nucleophilicity in a protic solvent, where polarisability and degree of solvation control the halides, and basicity differentiates the two oxygen anions.
Step 1:In a protic solvent, small anions form strong hydrogen bonds and are heavily solvated, lowering their nucleophilicity, while large, polarisable anions are weakly solvated. Among the halides, iodide is the most polarisable and least solvated (strongest), and fluoride is the smallest and most solvated (weakest).
I−≫F−(in protic solvent)
Step 2:Comparing the two oxygen nucleophiles, ethoxide is the conjugate base of a weak alcohol and is strongly basic, whereas phenoxide is resonance-stabilised over the ring and is far less basic. Greater basicity and charge localisation make ethoxide the better nucleophile.
C2H5O−>C6H5O−
Step 3:Both alkoxide and phenoxide, being charge-localised oxygen bases, are stronger nucleophiles than the heavily solvated fluoride but weaker than the large polarisable iodide. Combining the rankings gives the full order.
I−>C2H5O−>C6H5O−>F−
Final answer: I−>C2H5O−>C6H5O−>F−
Q67Single correctClassification of Elements and Periodicity in Properties
A 'p' block element (E) and hydrogen form a binary cation (EH4)+, while EH3 on treatment with K2HgI4 in alkaline medium gives a precipitate of basic mercury (II) amido – iodine. Given below are first ionisation enthalpy values (kJ mol−1) for first element each from group 13, 14, 15 and 16. Identify the correct first ionisation enthalpy value for element E.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 31402
Approach:
Given: p-block element E forms the cation (EH4)+, and EH3 with K2HgI4 in alkaline medium gives the basic mercury(II) amido-iodide; first ionisation enthalpies for the first members of groups 13, 14, 15, 16. Target: IE1 of E. Identify EH3 from the qualitative test, fix E, then choose its ionisation enthalpy.
Step 1:K2HgI4 in alkaline medium is Nessler's reagent, whose characteristic positive test (a brown precipitate of the basic mercury(II) amido-iodide) is given by ammonia. Thus EH3 is NH3 and the cation (EH4)+ is NH4+, identifying E as nitrogen of group 15.
2K2HgI4+NH3+3KOH→HgO⋅Hg(NH2)I↓+7KI+2H2O
Step 2:The first members of groups 13, 14, 15 and 16 are B, C, N and O. Their first ionisation enthalpies are 801, 1086, 1402 and 1314 kJ mol-1 respectively; nitrogen's is the largest because its half-filled 2p3 configuration is extra stable, exceeding even oxygen.
IE1:B(801)<C(1086)<O(1314)<N(1402)
Step 3:Therefore the first ionisation enthalpy of E (nitrogen) is 1402 kJ mol-1.
IE1(N)=1402kJ mol−1
Final answer: 1402
Q68Single correctBiomolecules
Given below are two statements : Statement I : Sucrose is dextrorotatory. However, sucrose upon hydrolysis gives a solution having mixture of products. This solution shows laevorotation. Statement II : Hydrolysis of sucrose gives glucose and fructose. Since the laevorotation of glucose is more than the dextrorotation of fructose, the resulting solution becomes laevorotatory. In the light of the above statements, choose the correct answer from the options given below.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2Statement I true but Statement II is false
Approach:
Given: Statement I (sucrose is dextrorotatory; its hydrolysis mixture is laevorotatory) and Statement II (the laevorotation of glucose exceeds the dextrorotation of fructose). Target: judge each statement's truth. Use the specific rotations of sucrose, glucose and fructose and identify which product controls the sign of the net rotation.
Step 1:Sucrose has a positive specific rotation, so it is dextrorotatory. Acidic or enzymatic hydrolysis (inversion) cleaves it into equimolar glucose and fructose, and the product solution is laevorotatory. Statement I is therefore true.
[α]sucrose=+66.5∘→net(−)mixture
Step 2:Glucose is dextrorotatory with specific rotation +52.5 degrees, and fructose is laevorotatory with specific rotation -92 degrees. The magnitude for fructose exceeds that for glucose, so the net rotation is negative because of fructose, not glucose.
[α]glucose=+52.5∘,[α]fructose=−92∘
Step 3:Statement II asserts that glucose is laevorotatory and outweighs fructose; in reality glucose is dextrorotatory and fructose provides the dominant laevorotation. Statement II is therefore false.
+52.5+(−92)=−39.5<0
Final answer: Statement I true but Statement II is false
Q69Single correctChemical Thermodynamics
Match the List-I with List – II
List-I (Thermodynamic Process)
List-II (Magnitude in kJ)
A.. Work done in reversible, isothermal expansion of 2 mol of ideal gas from 2 dm3 to 20 dm3 at 300 K
I.. 4
B.. Work done in irreversible isothermal expansion of 1 mol ideal gas from 1m3 to 3m3 at 300 K against a constant pressure of 3kPa
II.. 11.5
C.. Change in internal energy for adiabatic expansion of 1 mol ideal gas with change of temperature = 320 K and Cˉv=23R
III.. 6
D.. Change in enthalpy at constant pressure of 1 mol ideal gas with change of temperature = 337 K and Cˉp=25R
IV.. 7
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1A-II, B-III, C-I, D-IV
Approach:
Given four thermodynamic quantities A-D with their data, and List-II magnitudes 4, 11.5, 6, 7 kJ. Target: match each. Compute each quantity with its proper expression and pair it with the corresponding magnitude.
Step 1:A: reversible isothermal expansion of 2 mol from 2 to 20 dm3 at 300 K. The volume ratio is 10, so log(V2/V1) = 1, and the magnitude is 2.303 x 2 x 8.314 x 300 x 1 = 11486 J, i.e. about 11.5 kJ.
∣WA∣=2.303×2×8.314×300×log10≈11.5kJ
Step 2:B: irreversible expansion of 1 mol from 1 m3 to 3 m3 against a constant 3 kPa. The work magnitude is 3000 Pa x (3 - 1) m3 = 6000 J = 6 kJ.
∣WB∣=(3×103)(3−1)=6×103J=6kJ
Step 3:C: adiabatic expansion, dU = n Cv dT with n = 1, Cv = (3/2)R, dT = 320 K. dU = 1 x 1.5 x 8.314 x 320 = 3991 J, about 4 kJ.
ΔU=1×23×8.314×320≈4kJ
Step 4:D: constant-pressure heating, dH = n Cp dT with n = 1, Cp = (5/2)R, dT = 337 K. dH = 1 x 2.5 x 8.314 x 337 = 7005 J, about 7 kJ. The matches are A-II, B-III, C-I, D-IV.
ΔH=1×25×8.314×337≈7kJ
Final answer: A-II, B-III, C-I, D-IV
Q70Single correctEquilibrium
Consider a solution CO2(g) dissolved in water in a closed container. Which one of the following plots correctly represents variation of log (partial pressure of CO2 in vapour phase above water) [y-axis] with log (mole fraction of CO2 in water) [x-axis] at 25∘C ?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2Straight line of positive slope (1) with positive y-intercept (log KH)
Approach:
Given: CO2(g) dissolved in water in equilibrium with vapour at 25 C. Target: the plot of log(partial pressure of CO2) versus log(mole fraction of CO2 in water). Apply Henry's law and linearise it logarithmically to find the slope and intercept.
Step 1:For a sparingly soluble gas at equilibrium, Henry's law states that the partial pressure of CO2 above the solution is directly proportional to its mole fraction in water, with proportionality constant KH (a large positive number for CO2 in water).
PCO2=KHXCO2
Step 2:Taking base-10 logarithms of both sides gives a linear relation between the logarithms, with slope +1 and intercept log KH.
logPCO2=logKH+logXCO2
Step 3:A slope of +1 (positive, rising line) and a positive intercept log KH (the line meets the y-axis above the origin) describe a straight line of positive slope cutting the positive y-axis.
slope=+1,intercept=logKH>0
Final answer: Straight line of positive slope (1) with a positive y-intercept equal to logKH
Q71NumericalRedox Reactions and Electrochemistry
Consider the following electrochemical cell at 298 K Pt∣HSnO2−(aq)∣Sn(OH)62−(aq)∣OH−(aq)∣∣Bi2O3(s)∣Bi(s) If the reaction quotient at a given time is 106, then the cell EMF (Ecell) is _______ ×10−1 V (Nearest integer). Given the standard half – cell reduction potential as EBi2O3/Bi,OH−0=−0.44V and ESn(OH)62−/HSnO2−,OH−0=−0.90V
SolutionAnswer: 4
Approach:
Given: the alkaline cell Pt∣HSnO2−/Sn(OH)62−∣∣Bi2O3(s)∣Bi(s) at 298 K with reaction quotient Q = 106, and standard reduction potentials E0(Bi2O3/Bi) = -0.44 V (right electrode) and E0(Sn(OH)62−/HSnO2−) = -0.90 V (left electrode). Target: Ecell as integer x in x x 10−1 V. Determine E0cell, balance the cell to fix n, then apply the Nernst equation.
Step 1:The right electrode (Bi2O3/Bi, higher potential -0.44 V) is the cathode and the left electrode (Sn(OH)62−/HSnO2−, -0.90 V) is the anode. The standard cell potential is the cathode minus the anode value.
Ecell0=−0.44−(−0.90)=+0.46V
Step 2:At the anode tin is oxidised from +2 to +4, releasing 2 electrons: HSnO2−+3OH−→Sn(OH)62−+2e−. At the cathode each bismuth is reduced from +3 to 0, and Bi2O3 takes up 6 electrons: Bi2O3+3H2O+6e−→2Bi+6OH−. Balancing electrons multiplies the anode by 3, so the overall cell transfers n = 6 electrons.
3HSnO2−+Bi2O3+3OH−+…→3Sn(OH)62−+2Bi+…,n=6
Step 3:Apply the Nernst equation with Q = 106 and n = 6. The logarithmic term is (0.06/6) x log(106) = 0.01 x 6 = 0.06 V.
Ecell=0.46−60.06log(106)=0.46−0.06
Step 4:Expressing 0.40 V in the requested form gives 4 x 10−1 V, so the integer is 4.
0.40V=4×10−1V
Final answer: 4
Q72NumericalChemical Thermodynamics
Dissociation of a gas A2 takes place according to the following chemical reaction. At equilibrium, the total pressure is 1 bar at 300 K. A2(g)⇌2A(g) The standard Gibbs energy of formation of the involved substances has been provided below : Substance | ΔGfo/kJmol−1 | A2 | -100.00 | A | -50.832 | The degree of dissociation of A2(g) given by (x×10−2)1/2 where x = _______ (Nearest integer ) [Given : R = 8 J mol−1K−1, log 2 = 0.3010, log 3 = 0.48] Assume degree of dissociation is not negligible
SolutionAnswer: 33
Approach:
Given: A2(g) <=> 2A(g) at 300 K with total pressure 1 bar, dGf(A2) = -100.00 and dGf(A) = -50.832 kJ/mol, R = 8 J/mol/K, log2 = 0.3010, log3 = 0.48. Target: x in α = (x×10−2)1/2. Find the reaction Gibbs energy, convert to Kp, relate Kp to α at P = 1 bar, and solve.
Step 1:The standard reaction Gibbs energy is twice the formation value of A minus that of A2.
Step 2:Substitute into dG = -2.303 R T log Kp with R = 8, T = 300. The factor 2.303 x 8 x 300 = 5527.2, so log Kp = 1664/5527.2 = 0.3010 = log 2.
logKp=2.303×8×3001664=0.3010=log2
Step 3:For A2 <=> 2A starting from 1 mol with degree of dissociation α, the total moles are (1 + α) and the mole fractions give Kp = 4 α2 P / (1 - α2). With P = 1 bar and Kp = 2, set 4 α2/(1 - α2) = 2.
1−α24α2(1)=2
Step 4:Cross-multiplying: 4 α2 = 2(1 - α2) = 2 - 2 α2, hence 6 α2 = 2, so α2 = 1/3 = 0.3333 = 33.33 x 10−2. Comparing with α = (x×10−2)1/2 gives x = 33.33, nearest integer 33.
6α2=2⇒α2=31=33.33×10−2⇒x=33
Final answer: 33
Q73NumericalPurification and Characterisation of Organic Compounds
The cycloalkene (X) on bromination consumes one mole of bromine per mole of (X) and gives the product (Y) in which C:Br ratio is 3 : 1. The percentage of bromine in the product (Y) is _______ % (Nearest integer) (Given : molar mass in g mol−1 H:1, C : 12, O : 16, Br : 80)
SolutionAnswer: 66
Approach:
Given: cycloalkene X adds one mole of Br2 per mole to give product Y with C:Br = 3:1. Target: mass percentage of Br in Y. Use the atom ratio with the two Br atoms added to fix the molecular formula of Y, then compute the bromine mass fraction.
Step 1:Addition of one mole of Br2 across the C=C double bond introduces 2 bromine atoms per molecule. Since the C:Br ratio in Y is 3:1, the number of carbons is 3 x 2 = 6. A cycloalkene of 6 carbons (cyclohexene) thus gives a dibromide with 6 C and 2 Br.
C:Br=3:1⇒C:Br=6:2
Step 2:Saturating cyclohexene (C6H10) with Br2 gives 1,2-dibromocyclohexane, C6H10Br2. Its molar mass is the sum of carbon, hydrogen and bromine masses.
MY=6(12)+10(1)+2(80)=72+10+160=242g mol−1
Step 3:The bromine mass in one mole is 160 g out of 242 g, so the percentage is 160/242 x 100 = 66.11 %, nearest integer 66.
%Br=242160×100=66.11%
Final answer: 66
Q74NumericalChemical Kinetics
The temperature at which rate constants of the given below two gaseous reactions become equal is _______ K (Nearest integer) X⟶Yk1=106eT−30000P⟶Qk2=104eT−24000 Given : ln 10 = 2.303
SolutionAnswer: 1303
Approach:
Given: k1 = 106e−30000/T and k2 = 104e−24000/T, with ln 10 = 2.303. Target: temperature T at which k1 = k2. Equate the two expressions, take natural logarithms, and solve for T.
Step 1:Set the two rate constants equal.
106e−30000/T=104e−24000/T
Step 2:Take natural logarithms of both sides, using ln(106) = 6 ln10 and ln(104) = 4 ln10.
Step 4:Substitute ln 10 = 2.303 and solve: T = 6000/(2 x 2.303) = 6000/4.606 = 1302.6, nearest integer 1303 K.
T=2×2.3036000=4.6066000≈1303K
Final answer: 1303
Q75NumericalPrinciples Related to Practical Chemistry
Sodium fusion extract of an organic compound (Y) with CHCl3 and chlorine water gives violet colour to the CHCl3 layer. 0.15 g of (Y) gave 0.12 g of the silver halie precipitate in Carius method. Percentage of halogen in the compound (Y) is _______ (Nearest integer) (Given : molar mass g mol−1 C: 12, H : 1, Cl : 35.5, Br : 80, I : 127)
SolutionAnswer: 43
Approach:
Given: sodium fusion extract of Y with CHCl3 and chlorine water turns the CHCl3 layer violet; 0.15 g of Y gives 0.12 g of silver halide in the Carius method. Target: mass percentage of halogen in Y. Identify the halogen from the colour test, then use the Carius mass relation between the silver halide and the halogen.
Step 1:Chlorine water oxidises iodide in the extract to iodine, which dissolves in chloroform to give a violet colour. The violet CHCl3 layer therefore identifies the halogen as iodine, and the Carius precipitate is silver iodide.
2I−+Cl2→I2+2Cl−,I2(violet in CHCl3)
Step 2:The molar mass of AgI is the silver mass plus the iodine mass; the iodine mass fraction in AgI is 127/235.
MAgI=108+127=235g mol−1
Step 3:The 0.12 g of AgI contains (127/235) x 0.12 g of iodine; dividing by the 0.15 g sample and multiplying by 100 gives the percentage.
%I=235127×0.150.12×100=0.5404×0.8×100
Step 4:The percentage of iodine is 43.23 %, nearest integer 43.
%I=43.23%≈43
Final answer: 43
Mathematics25 questions
Q1Single correctStatistics and Probability
Two distinct numbers a and b are selected at random from 1,2,3,......,50. The probability, that their product ab is divisible by 3, is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 11225664
Approach:
Given the set {1,2,…,50}, two distinct numbers a,b are drawn; the target is P(3∣ab). Count the total unordered pairs, then use complementary counting on the pairs whose product carries no factor of 3.
Step 1:Count all ways of choosing two distinct numbers from the 50 available.
n(S)=50C2=250⋅49=1225
Step 2:Among 1 to 50 the multiples of 3 are 3,6,…,48, totalling 16; the remaining 34 numbers are not divisible by 3.
multiples of 3:16,others:34
Step 3:The product ab fails to be divisible by 3 exactly when neither factor is a multiple of 3; count such pairs.
n(Ec)=34C2=234⋅33=561
Step 4:Subtract from the total to obtain the favourable pairs.
n(E)=1225−561=664
Step 5:Form the required probability.
P(E)=1225664
Final answer: 1225664
Q2Single correctDifferential Equations
Let the solution curve of the differential equation xdy−ydx=x2+y2dx, x>0, y(1)=0; be y=y(x). Then y(3) is equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 34
Approach:
Given xdy−ydx=x2+y2dx with x>0 and y(1)=0, the target is y(3). The equation is homogeneous of degree one, so the substitution y=vx separates the variables.
Step 1:Solve the original relation for the derivative.
dxdy=xy+x2+y2
Step 2:Substitute y=vx and cancel the common v term.
v+xdxdv=v+1+v2⇒xdxdv=1+v2
Step 3:Separate and integrate both sides.
∫1+v2dv=∫xdx⇒ln(v+1+v2)=lnx+lnC
Step 4:Replace v=y/x and clear denominators to obtain the general solution.
xy+1+x2y2=Cx⇒y+x2+y2=Cx2
Step 5:Apply y(1)=0 to determine the constant.
0+1+0=C⋅1⇒C=1
Step 6:Isolate the radical and square to remove it.
x2+y2=x2−y⇒x2+y2=x4−2x2y+y2
Step 7:Divide by x2 (valid since x>0) and evaluate at x=3.
y=2x2−1⇒y(3)=29−1=4
Final answer: 4
Q3Single correctIntegral Calculus
The value of ∫−2π2π([x]+41)dx, where [.] denotes the greatest integer function, is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2607(3π−1)
Approach:
The integrand [x]+41 is piecewise constant because [x] is constant on each unit interval. Partition [−2π,2π] at the integers it contains and sum the constant pieces.
Step 1:Since −2π≈−1.57 and 2π≈1.57, the sub-intervals and values of [x]+4 are: [−2π,−1) with [x]=−2; [−1,0) with [x]=−1; [0,1) with [x]=0; [1,2π) with [x]=1.
[x]+4=2,3,4,5 respectively
Step 2:Write the integral as the sum of constant integrand times sub-interval length.
I=21(2π−1)+31(1)+41(1)+51(2π−1)
Step 3:Collect the π-dependent and the constant contributions.
I=(21+51)(2π−1)+31+41=107(2π−1)+127
Step 4:Combine over the common denominator 60.
I=6021π−6042+6035=6021π−7
Step 5:Factor 7 out of the numerator.
6021π−7=607(3π−1)=607(3π−1)
Final answer: 607(3π−1)
Q4Single correctThree Dimensional Geometry
Let P(α,β,γ) be the point on the line 2x−1=−3y+1=z at distance 414 from the point (1,−1,0) and nearer to the origin. Then the shortest distance, between the line 1x−α=2y−β=3z−γ and 2x+5=1y−10=1z−3, is equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4457
Approach:
Parametrize the first line, impose the distance 414 from (1,−1,0) to fix the parameter, select the point nearer the origin as (α,β,γ), then apply the skew-line shortest-distance formula to the two lines as written.
Step 1:Write the point on the first line as P=(1+2t,−1−3t,t) and impose the distance from (1,−1,0).
(2t)2+(−3t)2+t2=14∣t∣=414⇒t=±4
Step 2:The two candidate points are t=4⇒(9,−13,4) with norm2=266 and t=−4⇒(−7,11,−4) with norm2=186; the nearer to the origin is the latter.
(α,β,γ)=(−7,11,−4)
Step 3:Identify the two lines: line 1 passes through (−7,11,−4) with direction b=(1,2,3); line 2, written as 2x+5=1y−10=1z−3, passes through (−5,10,3) with direction d=(2,1,1).
a=(−7,11,−4),c=(−5,10,3)
Step 4:Compute the cross product of the direction vectors.
Step 5:Form the joining vector and the scalar triple product.
c−a=(2,−1,7),(c−a)⋅(b×d)=2(−1)+(−1)5+7(−3)=−28
Step 6:Divide the magnitude of the triple product by the magnitude of the cross product.
∣b×d∣=1+25+9=35,d=35∣−28∣=3528=457
Final answer: 457
Q5Single correctLimit, Continuity and Differentiability
Let f(x)=x2025−x2000, x∈[0,1] and the minimum value of the function f(x) in the interval [0,1] be (80)80(n)−81. Then n is equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3−81
Approach:
Given f(x)=x2025−x2000 on [0,1], the target is n where the minimum value equals (80)80(n)−81. Locate the interior stationary point via f′(x)=0, evaluate f there using the substitution u=x25, and match the prescribed form.
Step 1:Differentiate and factor the derivative.
f′(x)=2025x2024−2000x1999=x1999(2025x25−2000)
Step 2:Set the non-trivial factor to zero for the interior critical point in (0,1).
2025x25=2000⇒x25=20252000=8180
Step 3:Express f through x25 since x2025=(x25)81 and x2000=(x25)80.
f=(x25)81−(x25)80=(8180)81−(8180)80
Step 4:Factor the common power.
f=(8180)80(8180−1)=(8180)80(−811)
Step 5:Rewrite as (80)80 times a single base power and compare with (80)80(n)−81.
81808080⋅81−1=−81818080=8080(−81)−81
Final answer: −81
Q6Single correctComplex Numbers and Quadratic Equations
The number of distinct real solutions of the equation x∣x+4∣+3∣x+2∣+10=0 is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 11
Approach:
Given x∣x+4∣+3∣x+2∣+10=0, the target is the count of distinct real roots. The moduli change sign at x=−4 and x=−2; split the real line into three intervals, solve each resulting quadratic, and retain only roots lying inside their defining interval.
Step 1:For x<−4 both x+4<0 and x+2<0, so ∣x+4∣=−(x+4) and ∣x+2∣=−(x+2).
−x(x+4)−3(x+2)+10=0⇒x2+7x−4=0
Step 2:Test the two roots against x<−4: 2−7−65≈−7.53 qualifies while 2−7+65≈0.53 does not.
x=2−7−65≈−7.53
Step 3:For −4≤x<−2, ∣x+4∣=x+4 and ∣x+2∣=−(x+2).
x(x+4)−3(x+2)+10=0⇒x2+x+4=0,Δ=1−16<0
Step 4:For x≥−2, ∣x+4∣=x+4 and ∣x+2∣=x+2.
x(x+4)+3(x+2)+10=0⇒x2+7x+16=0,Δ=49−64<0
Step 5:Total the admissible roots across the three intervals.
count=1+0+0=1
Final answer: 1
Q7Single correctBinomial Theorem and its Simple Applications
The coefficient of x48 in (1+x)+2(1+x)2+3(1+x)3+...+100(1+x)100 is equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3100101C49−101C50
Approach:
Let S=∑k=1100k(1+x)k; the target is the coefficient of x48. Apply the multiply-by-(1+x) and subtract technique to collapse the arithmetic-geometric series into a closed form, then read off the coefficient.
Step 1:Write S and (1+x)S, then subtract to form −xS as a geometric series minus the last term.
−xS=(1+x)x(1+x)((1+x)100−1)⋅1+x1−100(1+x)101
Step 2:Solve for S in closed form.
S=x2(1+x)−(1+x)101+x100(1+x)101
Step 3:The first fraction's x48 coefficient equals the x50 coefficient of (1+x)−(1+x)101; the linear term contributes nothing to x50.
[x48]x2(1+x)−(1+x)101=−101C50
Step 4:The second fraction's x48 coefficient equals 100 times the x49 coefficient of (1+x)101.
[x48]x100(1+x)101=100101C49
Step 5:Add the two contributions.
[x48]S=100101C49−101C50
Final answer: 100101C49−101C50
Q8Single correctThree Dimensional Geometry
If the image of the points P(1,2,a) in the line 3x−6=2y−7=27−z is Q(5,b,c), then a2+b2+c2 is equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2298
Approach:
Given P(1,2,a) with image Q(5,b,c) in the line 3x−6=2y−7=27−z, the target is a2+b2+c2. Use that the midpoint of P and Q lies on the line and that PQ is perpendicular to the line direction.
Step 1:Rewrite the line with a consistent direction; 27−z=−2z−7 gives direction d=(3,2,−2) through (6,7,7), and the parametric point is (6+3t,7+2t,7−2t).
d=(3,2,−2)
Step 2:The midpoint M=(3,22+b,2a+c) lies on the line; matching the x-coordinate fixes the parameter.
6+3t=3⇒t=−1
Step 3:Match the y and z coordinates of M with the parametric point at t=−1.
22+b=7+2(−1)=5⇒b=8;2a+c=7−2(−1)=9⇒a+c=18
Step 4:Impose perpendicularity of PQ=(4,b−2,c−a)=(4,6,c−a) to d=(3,2,−2).
4(3)+6(2)+(c−a)(−2)=0⇒24−2(c−a)=0⇒c−a=12
Step 5:Solve a+c=18 with c−a=12.
c=15,a=3,b=8
Step 6:Compute the sum of squares.
a2+b2+c2=9+64+225=298
Final answer: 298
Q9Single correctSets, Relations and Functions
Let relation R on the set M={1,2,3,.....,16} be given by R={(x,y):4y=5x−3,x,y∈M}. Then the minimum number of elements required to be added in R, in order to make the relation symmetric, is equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 32
Approach:
Given M={1,2,…,16} and R={(x,y):4y=5x−3,x,y∈M}, the target is the minimum number of ordered pairs to adjoin so that R becomes symmetric. List the members of R, then count non-diagonal pairs whose reverse is absent.
Step 1:From 4y=5x−3, y=45x−3 is an integer only when 5x−3≡0(mod4), i.e. x≡3(mod4), giving x=3,7,11,15.
x∈{3,7,11,15}⇒y=3,8,13,18
Step 2:Retain only pairs with both entries in M; x=15 gives y=18∈/M and is discarded.
R={(3,3),(7,8),(11,13)}
Step 3:Check each pair for its reverse: (3,3) is its own reverse; (7,8) needs (8,7); (11,13) needs (13,11).
missing={(8,7),(13,11)}
Step 4:The minimum additions equal the number of missing reverses.
additions=2
Final answer: 2
Q10Single correctCo-ordinate Geometry
If the line αx+2y=1, where α∈R, does not meet the hyperbola x2−9y2=9, then a possible value of α is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 40.8
Approach:
Given the line αx+2y=1 and the hyperbola x2−9y2=9, the line misses the hyperbola when their simultaneous equations have no real solution. Substitute, form a quadratic in x, and require a negative discriminant.
Step 1:Solve the line for y=21−αx and substitute into x2−9y2=9.
Step 5:Compare the choices 0.6,0.7,0.5,0.8 with the threshold.
0.8>0.745, while 0.6,0.7,0.5<0.745
Final answer: 0.8
Q11Single correctLimit, Continuity and Differentiability
If the domain of the function f(x)=sin−1[3+2x5−x]+loge(10−x)1 is (−∞,α]∪[β,γ)−{δ}, then 6(α+β+γ+δ) is equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 470
Approach:
Given f(x)=sin−1[3+2x5−x]+loge(10−x)1, the target is 6(α+β+γ+δ) for the domain (−∞,α]∪[β,γ)−{δ}. Find the inverse-sine domain by a rational inequality, the logarithm domain by positivity and the exclusion loge(10−x)=0, then intersect.
Step 1:Impose the upper bound 3+2x5−x≤1, which rearranges to 3+2x3x−2≥0, satisfied for x<−23 or x≥32.
3+2x5−x−1=3+2x2−3x≤0⇒3+2x3x−2≥0
Step 2:Impose the lower bound 3+2x5−x≥−1, which rearranges to 3+2x8+x≥0, satisfied for x≤−8 or x>−23.
3+2x5−x+1=3+2x8+x≥0
Step 3:Intersect the two inverse-sine conditions.
(x<−23orx≥32)∩(x≤−8orx>−23)=(−∞,−8]∪[32,∞)
Step 4:Apply the logarithm constraints: 10−x>0 and 10−x=1.
x<10,x=9
Step 5:Intersect both domains.
(−∞,−8]∪[32,10)−{9}
Step 6:Evaluate the required expression.
6(−8+32+10+9)=6⋅335=70
Final answer: 70
Q12Single correctIntegral Calculus
Let the line x=−1 divided the area of the region {(x,y):1+x2≤y≤3−x} in the ratio m:n, gcd(m,n)=1. Then m+n is equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 427
Approach:
Given the region {(x,y):1+x2≤y≤3−x} bounded by the parabola y=1+x2 below and the line y=3−x above, the line x=−1 splits it in ratio m:n with gcd(m,n)=1; the target is m+n. Find the intersection abscissae, integrate the vertical gap on the left piece and on the whole region, then reduce the ratio.
Step 1:Find where the bounding curves meet.
1+x2=3−x⇒x2+x−2=0⇒x=−2,1
Step 2:The vertical gap of the region is g(x)=(3−x)−(1+x2)=2−x−x2.
g(x)=2−x−x2
Step 3:Integrate g from x=−2 to x=−1 for the piece left of the dividing line.
Step 5:The right piece is the remainder; form and reduce the ratio.
right=29−67=620;m:n=67:620=7:20
Step 6:Add the reduced parts.
gcd(7,20)=1⇒m+n=27
Final answer: 27
Q13Single correctVector Algebra
Let AB=2i^+4j^−5k^ and AD=i^+2j^+λk^,λ∈R. Let the projection of the vector v=i^+j^+k^ on the diagonal AC of the parallelogram ABCD be of length one unit. If α,β, where α>β, be the roots of equation λ2x2−6λx+5=0 then 2α−β is equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 23
Approach:
Given AB=2i^+4j^−5k^, AD=i^+2j^+λk^ and v=i^+j^+k^ with the projection of v on the diagonal AC of length one, the target is 2α−β for the roots of λ2x2−6λx+5=0. Form AC=AB+AD, impose the unit projection to solve for λ, then solve the quadratic.
Step 1:Add the adjacent sides to obtain the diagonal.
AC=(2+1)i^+(4+2)j^+(−5+λ)k^=3i^+6j^+(λ−5)k^
Step 2:Compute the dot product and magnitude needed for the projection.
v⋅AC=3+6+(λ−5)=λ+4,∣AC∣=45+(λ−5)2
Step 3:Set the projection length to one and square.
45+(λ−5)2∣λ+4∣=1⇒(λ+4)2=45+(λ−5)2
Step 4:The λ2 terms cancel, leaving a linear equation.
λ2+8λ+16=45+λ2−10λ+25⇒18λ=54⇒λ=3
Step 5:Substitute λ=3 into the quadratic and solve.
9x2−18x+5=0⇒x=1818±324−180=1818±12=35,31
Step 6:With α>β, assign α=35, β=31 and evaluate.
2α−β=2⋅35−31=310−1=3
Final answer: 3
Q14Single correctSequence and Series
If the sum of the first term of an A.P. is 6 and the sum of its first six terms is 4, then the sum of its first twelve terms is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4−22
Approach:
Given an A.P. with first term a and common difference d, the data fixes the sum of the first four terms as S4=6 and the sum of the first six terms as S6=4. The target is S12. Translate both conditions into linear equations in a and d, solve, then evaluate S12.
Step 1:Write S4=6 and S6=4 using the sum formula.
S4=24[2a+3d]=4a+6d=6,S6=26[2a+5d]=6a+15d=4
Step 2:Multiply 2a+3d=3 by 3 and subtract from 6a+15d=4 to eliminate a.
If random variable x has the probability distribution x | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 P(x) | 0 | 2k | k | 3k | 2k2 | 2k | k2+k | 7k2 Then P(3<x≤6) is equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 20.33
Approach:
Given the probability distribution of x in terms of a parameter k, the total-probability condition ∑P(x)=1 fixes k. The target P(3<x≤6) collects the probabilities of x=4,5,6.
Step 1:Add all listed probabilities and set the sum equal to 1.
0+2k+k+3k+2k2+2k+(k2+k)+7k2=10k2+9k=1
Step 2:Factor the quadratic and keep the root giving valid probabilities.
(10k−1)(k+1)=0⇒k=101ork=−1
Step 3:Sum the probabilities for the outcomes x=4,5,6 lying in 3<x≤6.
P(3<x≤6)=P(4)+P(5)+P(6)=2k2+2k+(k2+k)=3k2+3k
Step 4:Substitute k=0.1.
3(0.1)2+3(0.1)=0.03+0.30=0.33
Final answer: 0.33
Q16Single correctCo-ordinate Geometry
If the set of all values of r, for which the circle (x+1)2+(y+4)2=r2 and x2+y2−4x−2y−4=0 intersect at two distinct points be the interval (α,β). Then αβ is equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 225
Approach:
Two circles intersect at two distinct points when the distance d between their centres satisfies ∣r1−r2∣<d<r1+r2. Identify the centres and radii, compute d, form the interval (α,β) for r, then evaluate αβ.
Step 1:The first circle (x+1)2+(y+4)2=r2 has centre C1=(−1,−4) and radius r.
C1=(−1,−4),r1=r
Step 2:For x2+y2−4x−2y−4=0, read 2g=−4,2f=−2,c=−4, giving centre (2,1) and radius 4+1+4.
C2=(2,1),r2=4+1+4=3
Step 3:Compute the distance between the two centres.
d=(2−(−1))2+(1−(−4))2=9+25=34
Step 4:Apply the two-distinct-intersection condition ∣r−3∣<34<r+3, which yields 34−3<r<34+3.
α=34−3,β=34+3
Step 5:Multiply the endpoints using the difference of squares.
αβ=(34−3)(34+3)=34−9=25
Final answer: 25
Q17Single correctCo-ordinate Geometry
If the chord joining the points P(x1,y1) and P(x2,y2) on the parabola y2=12x subtends a right angle at the vertex of the parabola, then x1x2−y1y2 is equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4288
Approach:
Represent the two points on y2=12x in parametric form. The chord subtending a right angle at the vertex (origin) forces a relation between the parameters t1,t2. Express x1x2−y1y2 through t1t2 and evaluate.
Step 1:Comparing y2=12x with y2=4ax gives 4a=12, so a=3 and a point is (3t2,6t).
a=3,Pi=(3ti2,6ti)
Step 2:The slope of the line from the vertex (0,0) to (3ti2,6ti) is 3ti26ti=ti2; perpendicularity at the vertex gives the product −1.
t12⋅t22=−1⇒t1t2=−4
Step 3:Form the required combination using xi=3ti2,yi=6ti.
The number of solutions of tan−14x+tan−16x=6π, where −261<x<261, equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 11
Approach:
On the interval −261<x<261 the product 4x⋅6x=24x2<1, so the inverse-tangent sum formula applies directly. Combine the terms, take the tangent of both sides to obtain a quadratic in x, then count the roots lying in the interval and respecting the positive right-hand side.
Step 1:Combine the two inverse tangents with A=4x,B=6x.
tan−1(1−24x24x+6x)=tan−1(1−24x210x)=6π
Step 2:Take the tangent of both sides, using tan6π=31.
1−24x210x=31
Step 3:Cross-multiply and rearrange into a quadratic.
103x=1−24x2⇒24x2+103x−1=0
Step 4:Solve with the discriminant Δ=(103)2+4⋅24=300+96=396=36⋅11.
x=48−103±611=24−53±311
Step 5:Compare with the interval (−261,261)≈(−0.2041,0.2041); only x≈0.0537 lies inside, while x≈−0.7754 is outside and would also make the left side negative, contradicting 6π>0.
x=24−53+311≈0.0537∈(−261,261)
Final answer: 1
Q19Single correctMatrices and Determinants
If A=(2335) then the determinant of the matrix (A2025−3A2024+A2023) is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 316
Approach:
Factor the common power A2023 out of A2025−3A2024+A2023, leaving A2−3A+I. Using multiplicativity of the determinant and detA=1, the answer reduces to det(A2−3A+I).
Step 1:Factor the expression and record detA.
A2025−3A2024+A2023=A2023(A2−3A+I),detA=2⋅5−3⋅3=1
Step 2:Compute A2.
A2=(2335)(2335)=(13212134)
Step 3:Form A2−3A+I.
(13212134)−(69915)+(1001)=(8121220)
Step 4:Take the determinant; the factor A2023 contributes (detA)2023=1.
det=(1)2023(8⋅20−12⋅12)=160−144=16
Final answer: 16
Q20Single correctIntegral Calculus
Let f:[1,∞)→R be a differentiable function if 6∫1xf(t)dt=3xf(x)+x3−4. For all x≥1, then the value of f(2)−f(3) is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 33
Approach:
Differentiate the relation 6∫1xf(t)dt=3xf(x)+x3−4 with the Leibniz rule to obtain a first-order linear differential equation in f. Solve it, fix the constant from the relation evaluated at x=1, then compute f(2)−f(3).
Step 1:Differentiate both sides with respect to x.
6f(x)=3f(x)+3xf′(x)+3x2
Step 2:Divide by 3 and rewrite as the derivative of f/x.
Step 4:Evaluate the original relation at x=1, where the integral vanishes, to fix c.
0=3⋅1⋅f(1)+1−4⇒f(1)=1⇒−1+c=1⇒c=2
Step 5:Compute the required difference.
f(2)−f(3)=(−4+4)−(−9+6)=0−(−3)=3
Final answer: 3
Q21NumericalComplex Numbers and Quadratic Equations
Let α=2−1+i3 and β=2−1−i3, i=−1. If (7−7α+9β)20+(9+7α−7β)20+(−7+9α+7β)20+(14+7α+7β)20=m10, Then m is
SolutionAnswer: 49
Approach:
The numbers α=2−1+i3 and β=2−1−i3 are the non-real cube roots of unity ω and ω2, satisfying α+β=−1, αβ=1 and α3=β3=1. Reduce each bracket and combine, exploiting 1+ω+ω2=0.
Step 1:Identify α=ω,β=ω2 with α+β=−1 and αβ=1.
α=ω,β=ω2,α+β=−1
Step 2:The fourth bracket collapses using α+β=−1.
14+7α+7β=14+7(α+β)=14−7=7
Step 3:Let B=−7+9α+7β be the third bracket. Then the first and second brackets are αB and βB respectively (using αβ=1,α2=β).
7−7α+9β=αB,9+7α−7β=βB
Step 4:Combine the first three twentieth powers; ω20=ω2 and ω40=ω, so α20+β20+1=ω2+ω+1=0.
(αB)20+(βB)20+B20=(α20+β20+1)B20=0
Step 5:Only the fourth power survives; equate to m10.
720=m10⇒m10=(72)10=4910
Final answer: 49
Q22NumericalMatrices and Determinants
Let A be a 3×3 matrix such that A+AT=0. If A1−10=332, A21−10=−319−24 and det(adj(2adj(A+I)))=(2)α(3)β(11)γ, where α,β,γ are nonnegative integers, then α+β+γ is equal to
SolutionAnswer: 18
Approach:
The condition A+AT=0 makes A a 3×3 skew-symmetric matrix, so its non-zero entries are determined by the two given matrix-vector products. Build A, compute det(A+I), then apply the 3×3 identities det(adjM)=(detM)2 and det(kM)=k3detM to evaluate det(adj(2adj(A+I))), and factor into prime powers.
Step 1:Skew-symmetry forces A=0−p−qp0−sqs0. The conditions A1−10=332 and A332=−319−24 fix the entries.
Step 3:Let M=A+I. Then det(adjM)=(detM)2=442, and scaling by 2 gives det(2adjM)=23⋅442.
det(2adj(A+I))=23⋅442
Step 4:Take the adjugate determinant of this 3×3 matrix: square its determinant.
det(adj(2adj(A+I)))=(23⋅442)2=26⋅444
Step 5:Factor 44=22⋅11 and collect prime powers.
26⋅(22⋅11)4=26⋅28⋅114=214⋅30⋅114
Step 6:Add the exponents.
α+β+γ=14+0+4=18
Final answer: 18
Q23NumericalPermutations and Combinations
Let ABC be a triangle. Consider four points p1,p2,p3,p4 on the side AB, five points p5,p6,p7,p8,p9 on the side BC and four points p10,p11,p12,p13 on the side AC, None of these points is a vertex of the triangle ABC. Then the total number of pentagons, that can be formed by taking all the vertices from the points p1,p2,...,p13, is
SolutionAnswer: 660
Approach:
A pentagon needs 5 of the 13 points placed so that no 3 chosen points are collinear (three or more points from the same side are collinear and cannot all be polygon vertices). Sides AB, BC, AC carry 4, 5, 4 points, so at most 2 may be taken from any side; the only split of 5 with each part ≤2 is 2+2+1 across the three sides. Sum the three placements of the lone single.
Step 1:Choose 2 from AB, 2 from BC, 1 from AC.
4C2⋅5C2⋅4C1=6⋅10⋅4=240
Step 2:Choose 2 from AB, 1 from BC, 2 from AC.
4C2⋅5C1⋅4C2=6⋅5⋅6=180
Step 3:Choose 1 from AB, 2 from BC, 2 from AC.
4C1⋅5C2⋅4C2=4⋅10⋅6=240
Step 4:Add the three mutually exclusive cases.
240+180+240=660
Final answer: 660
Q24NumericalTrigonometry
If sin224∘−sin26∘cos248∘−sin212∘=2α+β5, where α,β∈N, then α+β is equal to
SolutionAnswer: 4
Approach:
Apply cos2A−sin2B=cos(A+B)cos(A−B) to the numerator and sin2A−sin2B=sin(A+B)sin(A−B) to the denominator, evaluate the standard angles, and substitute the exact values of cos36∘ and sin18∘ to write the ratio in the form 4α+β5.
Step 1:Transform the numerator with A=48∘,B=12∘ and the denominator with A=24∘,B=6∘.
Step 2:Use cos60∘=sin30∘=21, so they cancel, leaving the ratio sin18∘cos36∘.
=sin18∘cos36∘
Step 3:Substitute cos36∘=45+1 and sin18∘=45−1.
=45−145+1=5−15+1
Step 4:Rationalize by multiplying numerator and denominator by 5+1, then reduce.
(5)2−12(5+1)2=46+25=23+5
Step 5:Match with 2α+β5; since 1 and 5 are linearly independent over the rationals, α=3,β=1.
α=3,β=1⇒α+β=4
Final answer: 4
Q25NumericalIntegral Calculus
If ∫(sinx)−211(cosx)−25dx=−q1p1(cotx)29−q2p2(cotx)25−q3p3(cotx)21+q4p4(cotx)−23+C, where pj and qj are positive integers with gcd(pi,qi)=1 for i=1,2,3,4 and C is the constant of integration, then q1q2q3q415p1p2p3p4 is equal to
SolutionAnswer: 16
Approach:
Rewrite the integrand (sinx)−11/2(cosx)−5/2 using tanx and secx, substitute t=tanx so dt=sec2xdx, reduce to ∫t11/2(1+t2)3dt, integrate term by term, convert back through cotx=1/t, and read off each reduced coefficient qipi.
Step 1:Write the integrand as tan11/2x1sec8x and substitute t=tanx, using sec2xdx=dt and sec6x=(1+t2)3.
∫(sinx)−11/2(cosx)−5/2dx=∫t11/2(1+t2)3dt
Step 2:Expand (1+t2)3=1+3t2+3t4+t6 and divide by t11/2.
∫(t−11/2+3t−7/2+3t−3/2+t1/2)dt
Step 3:Integrate each power.
−92t−9/2−56t−5/2−6t−1/2+32t3/2+C
Step 4:Convert via t=tanx=1/cotx, so t−9/2=(cotx)9/2, etc., yielding the four coefficients in lowest terms.
How many questions are in the JEE Main 2026 January 22, Shift 1 paper?
The JEE Main 2026 January 22, Shift 1 paper has 75 questions — Physics (25), Chemistry (25) and Mathematics (25). Every question is on this page with its correct answer and a step-by-step solution.
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