JEE Main 2026 April 06, Shift 2 Question Paper with Solutions
All 74 questions from the JEE Main 2026 (April 06, Shift 2) shift — Physics (24), 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 percentage error in the calculated volume of a sphere, if there is 2% error in its diameter measurement, is _____ .
(A)
(B)
(C)
(D)
SolutionAnswer: Option 36
Approach:
Volume of a sphere is proportional to the cube of its diameter, so the percentage error in volume is three times the percentage error in diameter.
Step 1:Given: percentage error in diameter is 2%. Target: percentage error in the calculated volume of the sphere.
dΔd×100=2%
Step 2:Express volume in terms of diameter so the power of d is explicit.
V=6πd3⇒V∝d3
Step 3:Apply the error propagation rule for a quantity raised to the power three.
VΔV×100=3(dΔd×100)
Step 4:Substitute the diameter percentage error of 2%.
VΔV×100=3×2%=6%
Final answer: 6
Q27Single correctUnits and Measurements
Match List - I with List - II.
List - I
List - II
A. Boltzmann constant
I.[M−1L3T−2]
B. Stefan's constant
II.[ML2T−1]
C. Planck's constant
III.[ML2T−2K−1]
D. Gravitational constant
IV.[ML0T−3K−4]
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3A-III, B-IV, C-II, D-I
Approach:
Derive the dimensional formula of each constant from a characteristic physical relation and match it to the unique entry in List - II.
Step 1:Target: assign each List-I constant a dimensional formula from List-I. Boltzmann constant equals energy per unit temperature.
kB=[K][ML2T−2]=[ML2T−2K−1]
Step 2:Stefan's constant from the Stefan-Boltzmann law: power per unit area divided by temperature to the fourth power.
σ=[K4][ML2T−3]/[L2]=[ML0T−3K−4]
Step 3:Planck's constant equals energy divided by frequency.
h=[T−1][ML2T−2]=[ML2T−1]
Step 4:Gravitational constant from Newton's law of gravitation.
G=[M2][MLT−2][L2]=[M−1L3T−2]
Final answer: A-III, B-IV, C-II, D-I
Q28Single correctRotational Motion
A solid sphere (A) of mass 5m and a spherical shell (B) of mass m, both having same radius, are placed on a rough surface. When a force of same magnitude is applied tangentially at the highest points of A and B, they start rolling without slipping with an acceleration of aA and aB, respectively. The ratio of aA and aB is _____ .
(A)
(B)
(C)
(D)
SolutionAnswer: Option 15:21
Approach:
For a body rolling without slipping under a force applied at the top point, combine Newton's second law for translation with the torque equation about the centre to obtain the linear acceleration, then form the ratio for the two bodies.
Step 1:Given: equal force F applied at the top of each body; solid sphere A has mass 5m, spherical shell B has mass m, equal radii, rolling without slipping. Target: ratio aA:aB. Set up translation with friction f at the contact and torque about the centre; rolling gives α=a/R.
F+f=Ma,(F−f)R=IRa
Step 2:Eliminate f: the torque equation gives F−f=Ia/R2. Adding to the translation equation yields 2F=a(M+I/R2).
a=M(1+MR2I)2F
Step 3:Solid sphere A: I=52(5m)R2, so I/MR2=2/5.
aA=5m(1+52)2F=5m⋅572F=7m2F
Step 4:Spherical shell B: I=32mR2, so I/MR2=2/3.
aB=m(1+32)2F=m⋅352F=5m6F
Step 5:Form the ratio of the two accelerations.
aBaA=6F/5m2F/7m=72×65=4210=215
Final answer: 5:21
Q29Single correctWork, Energy and Power
A body of mass 1 kg moves along a straight line with a velocity v=2x2. The work done by the body during displacement from x=0 to 5 m is ___ J.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 31250
Approach:
By the work-energy theorem the net work equals the change in kinetic energy between the initial and final positions, using the position-dependent speed.
Step 1:Given: mass m=1 kg, speed v=2x2, displacement from x=0 to x=5 m. Target: work done.
m=1kg,v=2x2
Step 2:Evaluate the speed at the initial position x=0.
vi=2(0)2=0m/s
Step 3:Evaluate the speed at the final position x=5 m.
vf=2(5)2=50m/s
Step 4:Apply the work-energy theorem.
W=21(1)(50)2−21(1)(0)2=21(2500)
Final answer: 1250
Q30Single correctThermodynamics
A cylinder with adiabatic walls is closed at both ends and is divided into two compartments by a frictionless adiabatic piston. Ideal gas is filled in both (left and right) the compartments at same P, V, T. Heating is started from left side until pressure changes to 27P/8. If initial volume of each compartment was 9 litres then the final volume in right-hand side compartment is _____ litres. (for this ideal gas CP/CV=1.5)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 24
Approach:
The right compartment is bounded by adiabatic walls and an adiabatic piston, so as the rising common pressure compresses it the process is adiabatic; apply PVγ=constant between its initial and final states.
Step 1:Given: right gas initial state (P,9L), final pressure 27P/8 (equal on both sides at mechanical equilibrium), γ=3/2. Target: final right-side volume V2.
P1=P,V1=9L,Pf=827P,γ=23
Step 2:Apply the adiabatic relation between initial and final states of the right gas.
P(9)3/2=827PV23/2
Step 3:Evaluate (9)3/2=27 and simplify.
V23/2=278×27=8
Step 4:Solve for V2 by raising both sides to the power 2/3.
V2=82/3=(23)2/3=22=4
Final answer: 4
Q31Single correctElectromagnetic Waves
For an electromagnetic wave propagating through vacuum, k,E and ω represent propagation vector, electric field and angular frequency, respectively. The magnetic field associated with this wave is represented by :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2ωk×E
Approach:
For a plane electromagnetic wave the magnetic field is perpendicular to both k and E, the triad (E,B,k) is right-handed, and B=E/c; express this using k and ω.
Step 1:Given: propagation vector k, electric field E, angular frequency ω, vacuum propagation. Target: expression for B.
k,E,ωgiven
Step 2:Write the standard EM-wave relation: B points along k^×E with magnitude E/c.
B=ck^×E
Step 3:Replace the unit vector k^=k/k and the speed c=ω/k.
B=ω/k(k/k)×E=ωk×E
Step 4:Compare with the listed options.
B=ωk×E
Final answer: ωk×E
Q32Single correctKinematics
Two identical bodies A and B of equal masses have initial velocities v1=4i^m/s and v2=4j^m/s respectively. The body A has acceleration a1=6i^+6j^m/s2 while the acceleration of the other body B is zero. The centre of mass of the two bodies moves in ____ path.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3straight line
Approach:
Compute the velocity and acceleration of the centre of mass; if the centre-of-mass acceleration is parallel to its velocity, the motion has no transverse component and the path is a straight line.
Step 1:Given: equal masses; v1=4i^, v2=4j^, a1=6i^+6j^, a2=0. Target: shape of the centre-of-mass path.
v1=4i^,v2=4j^,a1=6i^+6j^,a2=0
Step 2:Compute the initial velocity of the centre of mass.
vcm=24i^+4j^=2i^+2j^
Step 3:Compute the acceleration of the centre of mass.
acm=2(6i^+6j^)+0=3i^+3j^
Step 4:Both vcm and acm lie along i^+j^, so they are parallel; with acceleration parallel to velocity the motion is rectilinear.
acm∥vcm(both along i^+j^)⇒straight line
Final answer: straight line
Q33Single correctProperties of Solids and Liquids
Figure represents the extension (Δl) of a wire of length 1 meter, suspended from the ceiling of the room at one end with a load W connected to the other end. If the cross-sectional area of the wire is 10−5m2 then the Young's modulus of the wire is ______ N/m2.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 31.0×1010
Approach:
Young's modulus equals stress over strain, Y=AΔlFL. The load-extension graph is linear, so the constant slope W/Δl is read from a clear grid point and combined with the given length and area.
Step 1:Known quantities: wire length L=1m, cross-sectional area A=10−5m2, applied force F=W, and extension Δl read from the graph.
L=1m,A=10−5m2,F=W
Step 2:Read the linear graph at the end grid point: at load W=60N the extension is Δl=6×10−4m. Compute the slope of the load-extension line.
ΔlW=6×10−460=105N/m
Step 3:Substitute the slope, length and area into the Young's modulus expression, since Y=ΔlW⋅AL.
Y=ΔlW⋅AL=105×10−51
Step 4:Evaluate the product to obtain Young's modulus.
Y=1.0×1010N/m2
Final answer: 1.0×1010N/m2
Q34Single correctProperties of Solids and Liquids
A cylindrical vessel of 40 cm radius is completely filled with water and its capacity is 528 dm3 (dm : decimeter) The vessel is placed on a solid block of exactly same height as vessel. If a small hole is made at 70 cm below the top of water level, then horizontal range of water falling on the ground in the beginning is ______ cm.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 21402
Approach:
Torricelli's theorem gives the efflux speed from the depth of the hole below the surface. The jet then undergoes horizontal projectile motion, falling the vertical height from the hole to the ground. The vessel height needed for that height is obtained from the cylinder volume.
Step 1:Set up the geometry. Volume V=528dm3=0.528m3, radius r=0.4m, depth of hole below the surface h=0.7m.
V=0.528m3,r=0.4m,h=0.7m
Step 2:Find the vessel height from the cylinder volume.
Hvessel=πr2V=(22/7)(0.4)20.528=1.05m=105cm
Step 3:Locate the hole above the ground. The hole is 70cm below the top, i.e. 105−70=35cm above the vessel base. The vessel rests on a block of equal height 105cm, so the hole is H=105+35=140cm=1.4m above the ground.
H=105+(105−70)=140cm=1.4m
Step 4:Combine efflux speed and fall time into the range R=2gh2H/g=2hH and substitute.
R=2hH=20.7×1.4=20.98=1.98m
Final answer: 1402cm
Q35Single correctKinetic Theory of Gases
If 2 mole of an ideal monoatomic gas at temperature T, is mixed with 6 mole of another ideal monoatomic gas at temperature 2T then the temperature of mixture is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 447T
Approach:
When two samples of the same monoatomic gas mix without heat loss, total internal energy is conserved. Since U=nCvT with the same Cv, the common Cv cancels and the mixture temperature is the mole-weighted mean of the two temperatures.
Step 1:Given data: n1=2 mol at T1=T and n2=6 mol at T2=2T, both monoatomic with the same Cv.
n1=2,T1=T,n2=6,T2=2T
Step 2:Apply conservation of internal energy; the common Cv cancels.
n1CvT1+n2CvT2=(n1+n2)CvTmix
Step 3:Substitute the values.
Tmix=2+62⋅T+6⋅2T=814T
Step 4:Simplify the fraction.
Tmix=814T=47T
Final answer: 47T
Q36Single correctOscillations and Waves
A spring stretches by 2 mm when it is loaded with a mass of 200 g . From equilibrium position the mass is further pulled down by 2 mm and released. The frequency associated with the system and maxmimum energy in the spring are ______ Hz and ______ J, respectively. (Take g = 10 m/s2)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1π550 and 8×10−3
Approach:
The static extension under the load gives the spring constant. The oscillation frequency follows from k and m. The maximum elastic energy stored in the spring occurs at the lowest point, where the total stretch is the static extension plus the pull-down amplitude.
Step 2:Compute the spring constant from the static equilibrium kx0=mg.
k=x0mg=2×10−30.2×10=1000N/m
Step 3:Compute the oscillation frequency.
f=2π10.21000=2π15000=π252=π550Hz
Step 4:Maximum energy stored in the spring occurs at the lowest point, where the total elongation is x0+A=4mm=4×10−3m.
E=21k(x0+A)2=21(1000)(4×10−3)2=8×10−3J
Final answer: π550 Hz and 8×10−3 J
Q37Single correctElectrostatics
The electric potential as a function of x, y is given by V=5(x2−y2) V. The electric field at a point (2,3) m is ______ V/m.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1(−20i^+30j^)
Approach:
The electric field is the negative gradient of the potential. Take the partial derivatives of V with respect to x and y, form E=−∇V, and evaluate at the given point.
Step 1:Set up the potential field and target point: V=5(x2−y2) V, evaluated at (x,y)=(2,3)m.
V=5(x2−y2),(x,y)=(2,3)
Step 2:Compute the partial derivatives of V.
∂x∂V=10x,∂y∂V=−10y
Step 3:Form the field as the negative gradient.
E=−10xi^+10yj^
Step 4:Evaluate at (2,3).
E=−10(2)i^+10(3)j^=(−20i^+30j^)V/m
Final answer: (−20i^+30j^) V/m
Q38Single correctMagnetic Effects of Current and Magnetism
A current of 30 A each flows in opposite directions in two conducting wires, placed parallel to each other at a distance of 8 cm. The magnetic field at the mid point between the two wires is ______ μT. (4πμ0=10−7N/A2)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2300
Approach:
Each long straight wire produces a field B=4πμ0r2I at the midpoint. Because the currents are antiparallel, the two field vectors at the midpoint point the same way and add, so the total is twice the single-wire field.
Step 1:Given data: separation 8cm, so midpoint distance r=4cm=0.04m; current I=30A; 4πμ0=10−7N/A2.
r=0.04m,I=30A
Step 2:Compute the field from one wire at the midpoint.
B1=4πμ0r2I=10−7×0.042×30=1.5×10−4T
Step 3:For antiparallel currents the two midpoint fields are codirectional, so they add.
B=2B1=2×1.5×10−4=3×10−4T
Step 4:Convert to microtesla.
B=3×10−4T=300μT
Final answer: 300μT
Q39Single correctElectromagnetic Induction and Alternating Currents
A square loop of side 2 cm is placed in a time varying magnetic field with magnitude as B=0.4sin(300t) Tesla. The normal to the plane of loop makes an angle of 60∘ with the field. The maximum induced emf produced in the loop is ______ mV.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 424
Approach:
Express the magnetic flux through the loop including the angle between the normal and the field, differentiate with respect to time to obtain the induced emf, and read off its amplitude.
Step 1:Set up the loop area for side 2 cm = 0.02 m.
A=(0.02)2=4×10−4m2
Step 2:Write the flux using B=0.4sin(300t) and the 60∘ angle between the normal and the field.
Φ=BAcos60∘=0.4sin(300t)⋅4×10−4⋅21
Step 3:Differentiate the flux to obtain the induced emf.
ε=−dtdΦ=−8×10−5⋅300cos(300t)
Step 4:Identify the amplitude of the cosine term as the maximum emf.
εmax=8×10−5⋅300=0.024V=24mV
Final answer: 24 mV
Q40Single correctElectrostatics
A sphere of capacitance 100 pF is charged to a potential of 100 V. Another identical uncharged metal sphere is brought in contact with the charged sphere, then the change in the total energy stored on these spheres, when they touch is α×10−7 J. The value of α is ______. (combined capacitance of spheres is 200 pF)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 225
Approach:
Find the initial energy on the charged sphere, then apply charge conservation on the combined 200 pF capacitance after contact, and take the magnitude of the energy change.
Step 1:Set up the initial charge on the first sphere.
Q=CV=100×10−12⋅100=10−8C
Step 2:Compute the initial stored energy.
Ui=21CQ2=21100×10−12(10−8)2=5×10−7J
Step 3:After contact the same charge Q resides on the combined 200 pF capacitance.
Uf=212CQ2=21200×10−12(10−8)2=2.5×10−7J
Step 4:Take the magnitude of the change in total energy and read off α.
ΔU=Ui−Uf=5×10−7−2.5×10−7=2.5×10−7J
Final answer: 5/2
Q41Single correctAtoms and Nuclei
The energy released if hydrogen atoms are combined to form 24He is ____ MeV. (Take binding energies per nucleon of 12H and 24He as 1.1 MeV and 7.2 MeV, respectively)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 224.4
Approach:
Two deuterium nuclei (2 nucleons each) fuse into helium-4. The energy released equals the total binding energy of the product minus the total binding energy of the reactants.
Step 1:Write the fusion reaction supplying the 4 nucleons of He-4 from two deuterons.
12H+12H→24He
Step 2:Compute the total binding energy of the two deuterons.
BEreactants=2×(2×1.1)=4.4MeV
Step 3:Compute the total binding energy of the He-4 product.
BEproduct=4×7.2=28.8MeV
Step 4:Subtract to obtain the energy released.
Q=28.8−4.4=24.4MeV
Final answer: 24.4 MeV
Q42Single correctOptics
Angle of minimum deviation is equal to the half of the angle of prism in an equilateral prism. The refractive index of the prism is ______.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 32
Approach:
Apply the prism formula relating refractive index to the angle of prism A and the angle of minimum deviation, with A=60∘ and Dm=A/2=30∘.
Step 1:State the angles: equilateral prism gives A=60∘ and Dm=A/2=30∘.
A=60∘,Dm=30∘
Step 2:Substitute the angles into the prism formula.
μ=sin(260)sin(260+30)=sin30∘sin45∘
Step 3:Evaluate the trigonometric values.
μ=2121=22=2
Step 4:Confirm the numerical value of the refractive index.
μ=2≈1.414
Final answer: 2
Q43Single correctElectronic Devices
Refer to the logic circuit given below. For two inputs (A=1, B=1) and (A=0, B=1), output (Y) will be.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 30, 0 respectively
Approach:
Trace each gate from the inputs: A passes through a NOT gate to give its complement, which together with B feeds a top OR gate and a bottom AND gate, whose outputs feed a final NOR gate. Form the Boolean expression for Y and evaluate it for both input pairs.
Step 1:Input A passes through the NOT gate to give its complement, which branches to both the top OR gate and the bottom AND gate; input B also feeds both gates.
A
Step 2:The top OR gate takes inputs A and B.
X1=A+B
Step 3:The bottom AND gate takes inputs A and B.
X2=A⋅B
Step 4:The final NOR gate combines X1 and X2. Since A⋅B is absorbed into A+B, the output reduces to Y=A+B=A⋅B.
Y=(A+B)+(A⋅B)=A+B=A⋅B
Step 5:Evaluate for A=1, B=1: A=0, so X1=0+1=1, X2=0, and Y=1=0.
Y=(0+1)+(0⋅1)=1=0
Step 6:Evaluate for A=0, B=1: A=1, so X1=1+1=1, X2=1, and Y=1=0.
Y=(1+1)+(1⋅1)=1=0
Final answer: 0, 0 respectively
Q44Single correctLaws of Motion
The velocity at which 6 kg mass (shown in figure) strikes the ground when it is released from a height of 6 m above the ground is ____ m/s. Assume pulley is massless and string is light and inextensible. (Take g=10m/s2)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 17.74
Approach:
Model the two blocks over a massless pulley as an Atwood machine. Determine the system acceleration, then apply kinematics to the 6 kg block falling 6 m to find its impact speed.
Step 1:Set up the system with m1=6 kg falling and m2=2 kg rising, and compute the acceleration.
a=6+2(6−2)⋅10=840=5m/s2
Step 2:The 6 kg block starts from rest and accelerates downward through h = 6 m.
v2=0+2⋅5⋅6=60
Step 3:Take the square root to obtain the impact speed.
v=60≈7.74m/s
Step 4:State the impact velocity of the 6 kg mass.
v≈7.74m/s
Final answer: 7.74 m/s
Q46NumericalLaws of Motion
A block takes t time to slide down a plane inclined at 45∘ to the horizontal. If the surface is made smooth (frictionless), the block takes time 2t to slide down the plane. The coefficient of friction between the block and the inclined plane is (100α). The value of α is _____.
SolutionAnswer: 75
Approach:
The block starts from rest in both cases and covers the same distance down the incline. The rough-surface acceleration is a1=g(sinθ−μcosθ) and the smooth-surface acceleration is a2=gsinθ. Equating the distances relates the accelerations through the given times, then solve for μ.
Step 1:Set up the situation: same distance s is travelled from rest in time t (rough) and time t/2 (smooth). Equate distances: 21a1t2=21a2(2t)2.
a1t2=a24t2
Step 2:Substitute the two accelerations with θ=45∘, where sinθ=cosθ=21.
gsinθ=4g(sinθ−μcosθ)
Step 3:Rearrange to isolate μ: 4μcosθ=3sinθ, giving μ=43tanθ.
μ=43tan45∘=43=0.75
Step 4:Since μ=100α, obtain α=100×0.75.
α=100×0.75=75
Final answer: 75
Q47NumericalDual Nature of Matter and Radiation
The de Broglie wavelength for an electron accelerated through the potential difference of V1 volt is λ1. When the potential difference is changed to V2 volt, the associated de Broglie wavelength is increased by 50%. If (V1/V2)=(9/α), then the value of α is _______.
SolutionAnswer: 4
Approach:
The de Broglie wavelength of an electron accelerated through potential V obeys λ=2meVh, so λ∝V1. Apply the 50% increase in wavelength to obtain the ratio of potentials.
Step 1:Set up the proportionality λ∝V1, which gives the ratio λ1λ2=V2V1.
λ1λ2=V2V1
Step 2:A 50% increase in wavelength means λ2=1.5λ1, so the wavelength ratio is 23.
λ1λ2=23
Step 3:Square both sides to obtain the potential ratio.
V2V1=(23)2=49
Step 4:Compare with V2V1=α9 to obtain the value of α.
α9=49⇒α=4
Final answer: 4
Q48NumericalMagnetic Effects of Current and Magnetism
A moving coil galvanometer when shunted with 2Ω resistance gives a full scale deflection for a current of 500 mA. When a resistance of 470Ω is connected in series it gives a full scale deflection for 10 V potential applied on it. The value of resistance of galvanometer coil is _____ Ω.
SolutionAnswer: 50
Approach:
Full-scale deflection corresponds to a fixed galvanometer current Ig. Express Ig from the series (voltmeter) case and from the shunted (ammeter) case, equate, and solve for the coil resistance G.
Step 1:Set up the series case: full-scale current Ig=G+RV=G+47010.
Ig=G+47010
Step 2:For the shunted case the line current is 500 mA=0.5 A and shunt S=2Ω. The galvanometer current satisfies IgG=(0.5−Ig)×2.
IgG=(0.5−Ig)×2
Step 3:Substitute Ig=G+47010 into Ig(G+2)=1.
G+47010(G+2)=1
Step 4:Solve the linear equation for G.
10G+20=G+470⇒9G=450⇒G=50
Final answer: 50
Q49NumericalCurrent Electricity
Two cells of emfs 1 V and 2 V and internal resistance 2Ω and 1Ω, respectively connected in parallel, gave a current of 1 A through an external resistance. If the polarity of one cell is reversed, then value of current through the external resistance will be 5α A. The value of α is _____.
SolutionAnswer: 3
Approach:
Replace the two parallel cells by a single equivalent emf and internal resistance. Determine the external resistance R from the given 1 A current, then recompute the current after reversing one cell's polarity.
Step 1:Set up the equivalent source with E1=1,r1=2,E2=2,r2=1: Eeq=2+11×1+2×2=35 V and req=32×1=32Ω.
Eeq=35 V,req=32Ω
Step 2:Apply the given current I=1 A to find R: 1=2/3+R5/3.
2/3+R5/3=1⇒32+R=35
Step 3:Reverse the polarity of the 1 V cell so E1=−1 V; the internal resistances are unchanged: Eeq′=3(−1)×1+2×2=33=1 V.
Eeq′=1 V,req=32Ω
Step 4:Compute the new current and compare with 5α.
I′=2/3+11=5/31=53 A⇒α=3
Final answer: 3
Q50NumericalOptics
A concave mirror of focal length 10 cm forms an image which is double the size of object when the object is placed at two different positions. The distance between the two positions of the object is _____ cm.
SolutionAnswer: 10
Approach:
A magnitude-2 magnification arises in two distinct cases for a concave mirror: a real inverted image (m=−2) and a virtual erect image (m=+2). Find the object distance in each case and take the difference.
Step 1:Set up Case 1 (real image): m=−2⇒v=2u. Apply the mirror formula with f=−10 cm: 2u1+u1=−101, so 2u3=−101.
Step 3:The two object positions lie at 15 cm and 5 cm from the mirror.
∣u1∣=15 cm,∣u2∣=5 cm
Step 4:Compute the separation between the two object positions.
Δu=15−5=10 cm
Final answer: 10
Chemistry25 questions
Q51Single correctSome Basic Concepts in Chemistry
Which of the following contain the same number of atoms? (Given : Molar mass in gmol−1 of H, He, O and S are 1, 4, 16 and 32 respectively) A. 2 g of O2 gas; B. 4 g of SO2 gas; C. 1400 mL of O2 at STP; D. 0.05L of He at STP; E. 0.0625 mol of H2 gas. Choose the correct answer from the options given below :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4A, C and E only
Approach:
For each sample find the number of moles, multiply by the number of atoms per molecule to get total atoms (in units of NA), and compare. STP molar volume is taken as 22400 mL/mol.
Step 1:A: 2 g O2=322=0.0625 mol; each molecule has 2 atoms, so atoms =0.0625×2=0.125NA.
NA×0.0625×2=0.125NA
Step 2:B: 4 g SO2=644=0.0625 mol with 3 atoms per molecule gives 0.1875NA. C: 1400 mL O2 at STP =224001400=0.0625 mol with 2 atoms gives 0.125NA.
B=0.1875NA,C=0.125NA
Step 3:D: 0.05 L =50 mL He at STP =2240050=0.00223 mol, monatomic, giving 0.00223NA. E: 0.0625 mol H2×2=0.125NA atoms.
D=0.00223NA,E=0.125NA
Step 4:A, C and E each contain 0.125NA atoms, so these three samples have the same number of atoms.
A=C=E=0.125NA
Final answer: A, C and E only
Q52Single correctAtomic Structure
The Bohr radius of a hydrogen like species is 70.53 pm. The species and the stationary state (n) are respectively (Given : Hydrogen atom Bohr radius is 52.9 pm)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4Li2+,2
Approach:
The radius of the n-th orbit of a hydrogen-like species is rn=Zn2a0 with a0=52.9 pm. Determine the required ratio Zn2 and identify the species and state that satisfy it.
Step 1:Set up the required ratio Zn2=a0rn=52.970.53≈1.333=34.
Zn2=52.970.53≈34
Step 2:For Li2+, Z=3. Then 3n2=34, giving n2=4.
3n2=34⇒n=2
Step 3:Check by direct substitution: r=322×52.9=34×52.9.
r=34×52.9=70.53 pm
Step 4:Rule out the alternatives: He+(Z=2),n=3 gives 29×52.9=238 pm, far from 70.53 pm. Compute the species and state.
He+,3⇒238 pm=70.53
Final answer: Li2+,2
Q53Single correctChemical Bonding and Molecular Structure
Given below are two statements : Statement I : The number of compounds among SO2,SO3,SF4,SF6 and H2S in which sulphur does not obey the Octet rule is 3. Statement II : Among [H2O,ClF3,SF4], [NH3,BrF5,SF4], [BrF5,ClF3,XeF4] and [XeF4,ClF3,H2O], the number of sets in which all the molecules have one lone pair of electrons on the central atom is 1. 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 false but Statement II is true
Approach:
Determine which sulphur species exceed the octet, then count central-atom lone pairs for each set to evaluate both statements.
Step 1:Statement I asks how many of SO2, SO3, SF4, SF6, H2S have sulphur not obeying the octet rule. Statement II asks how many of the four sets have every molecule possessing exactly one central-atom lone pair.
SO2,SO3,SF4,SF6,H2S
Step 2:Electron count around sulphur: SO2 has 10 electrons, SO3 has 12, SF4 has 10, SF6 has 12 (all expanded valence shell); H2S has 8 electrons (obeys octet). Thus 4 compounds violate the octet rule.
SO2(10),SO3(12),SF4(10),SF6(12),H2S(8)
Step 3:Central-atom lone pairs: H2O has 2, ClF3 has 2, SF4 has 1, NH3 has 1, BrF5 has 1, XeF4 has 2. Evaluate each set for all molecules having exactly one lone pair.
Step 4:Set [H2O, ClF3, SF4] gives (2,2,1) - fails. Set [NH3, BrF5, SF4] gives (1,1,1) - all one lone pair. Set [BrF5, ClF3, XeF4] gives (1,2,2) - fails. Set [XeF4, ClF3, H2O] gives (2,2,2) - fails. Exactly one set qualifies, so Statement II is true.
[NH3(1),BrF5(1),SF4(1)]
Step 5:Statement I is false and Statement II is true.
I false,II true
Final answer: Statement I is false but Statement II is true
Q54Single correctChemical Thermodynamics
Match List - I with List - II. Given V1 and V2 are initial and final volumes respectively. Choose the correct answer from the options given below :
List - I (Isothermal process)
List - II (Expression)
A. Reversible expansion
I.q=0
B. Free expansion
II.q=nRTlnV1V2
C. Irreversible Compression
III.w=−pext(V1−V2)
D. Cyclic reversible
IV.Tqrev=0
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3A-II, B-I, C-III, D-IV
Approach:
Apply the isothermal first-law result for each process type and match it to its characteristic expression.
Step 1:Each List-I process is an isothermal change; identify the heat or work expression that uniquely characterises it from List-II.
ΔT=0
Step 2:A Reversible expansion: for an isothermal reversible expansion the heat absorbed equals q=nRTln(V2/V1), which is expression II.
q=nRTlnV1V2
Step 3:B Free expansion: expansion into vacuum does zero work and for an ideal gas the temperature is unchanged, so q = 0, which is expression I.
w=0,q=0
Step 4:C Irreversible compression: work done against constant external pressure is w=−pext(V2−V1)=−pext(V1−V2) as written with final V2<V1, matching expression III. D Cyclic reversible process: the state returns to start so the reversible-heat integral gives qrev/T=0, matching expression IV.
w=−pext(V1−V2),Tqrev=0
Step 5:Collecting the pairings gives A-II, B-I, C-III, D-IV.
A-II,B-I,C-III,D-IV
Final answer: A-II, B-I, C-III, D-IV
Q55Single correctSolutions
Given below are two statements : Statement I: H2O molecules move from the chamber 1 to chamber 2. Statement II: The osmotic pressure of a solution prepared by dissolving 50 mg of potassium sulphate (molar mass = 174 g/mol) in 2 L of water (at 27∘C) is 0.0107 bar. (Given : R=0.083dm3 bar K−1mol−1 and assume complete dissociation of electrolyte). 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 false but Statement II is true
Approach:
Compare the molar glucose concentrations of the two chambers to fix the direction of solvent flow, then compute the osmotic pressure of the K2SO4 solution.
Step 1:Glucose molar mass is 180 g/mol. Chamber 1: moles = 18/180 = 0.1 mol in 0.100 L gives 1.0 M. Chamber 2: moles = 30/180 = 0.1667 mol in 0.250 L gives 0.667 M.
C1=0.1000.1=1.0M,C2=0.2500.1667=0.667M
Step 2:Across a semipermeable membrane water flows from the dilute side to the concentrated side, i.e. from chamber 2 to chamber 1. The claim that water moves from chamber 1 to chamber 2 is therefore wrong.
H2O:chamber 2→chamber 1
Step 3:K2SO4: moles = 0.050 g / 174 = 2.874e-4 mol; concentration C = 2.874e-4 / 2 = 1.437e-4 M. Complete dissociation gives van't Hoff factor i = 3.
C=20.050/174=1.437×10−4M,i=3
Step 4:Osmotic pressure pi = i C R T = 3 x 1.437e-4 x 0.083 x 300 K = 0.01073 bar, which rounds to 0.0107 bar.
π=3×1.437×10−4×0.083×300=0.0107bar
Step 5:Statement I is false and Statement II is true.
I false,II true
Final answer: Statement I is false but Statement II is true
Q56Single correctEquilibrium
Given is a concentrated solution of a weak electrolyte AxBy of concentration 'c' and dissociation constant 'K'. The degree of dissociation is given by :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2(cx+y−1xxyyK)x+y1
Approach:
Write the dissociation equilibrium of AxBy, express the equilibrium ion concentrations in terms of c and the degree of dissociation, then solve the equilibrium-constant expression for α.
Step 1:Set up the system: AxBy at concentration c dissociates with degree α into x cations and y anions; for a concentrated weak electrolyte α is small, so (1−α)≈1.
AxBy⇌xAy++yBx−
Step 2:Equilibrium concentrations: [A]=xcα, [B]=ycα, and undissociated species =c(1−α)≈c.
[A]=xcα,[B]=ycα,[AxBy]≈c
Step 3:Substitute into K=[A]x[B]y/[AxBy]=(xcα)x(ycα)y/c=xxyycx+y−1αx+y.
K=xxyycx+y−1αx+y
Step 4:Solve for α: αx+y=K/(cx+y−1xxyy), hence α=[K/(cx+y−1xxyy)]1/(x+y).
α=(cx+y−1xxyyK)x+y1
Step 5:Substituting x=y=1 reduces the result to α=K/c, the familiar Ostwald form, confirming the expression.
x=y=1⇒α=cK
Final answer: (cx+y−1xxyyK)x+y1
Q57Single correctRedox Reactions and Electrochemistry
For a general redox reaction Anode : Red1→O1n1++n1e− Cathode : Ox2+n2e−→Red2n2− Which of the following statement is incorrect?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4If the reaction is carried out reversibly, the electrical work done is equal to the ratio of charge and potential difference through which charge is moved.
Approach:
Test each statement against the balanced cell reaction, the Nernst equation, and the definition of electrical work to find the false one.
Step 1:Statement 1: multiplying the anode half-reaction by n2 and the cathode half-reaction by n1 balances the electrons (n1 n2 each side) and gives the stated overall reaction. This is correct.
n2Red1+n1Ox2⇌n2O1n1++n1Red2n2−
Step 2:Statement 2: because the electrons released at the anode equal those gained at the cathode after balancing, they cancel and do not appear in the overall equation. This is correct.
n1n2e− cancel
Step 3:Statement 3: rearranging the Nernst equation gives (E−E∘)/(RT/F)=−(1/n)lnQ, so a plot versus logQ is linear with slope proportional to 1/n. This is correct.
RT/FE−E∘=−n1lnQ,slope∝n1
Step 4:Statement 4: electrical work is the product of charge and potential difference, W = q x deltaV, not their ratio. The statement claims a ratio, which is dimensionally and physically wrong.
W=q×ΔV=ΔVq
Step 5:Only statement 4 is false, so it is the requested incorrect statement.
W=qΔV
Final answer: If the reaction is carried out reversibly, the electrical work done is equal to the ratio of charge and potential difference through which charge is moved.
Q58Single correctClassification of Elements and Periodicity in Properties
In a period, the first ionisation enthalpy of the element at extreme left and the negative electron gain enthalpy of the extreme right element, except noble gases, are respectively.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3lowest and highest
Approach:
Use the across-period trends of first ionisation enthalpy and electron gain enthalpy to characterise the leftmost and rightmost (non-noble-gas) elements.
Step 1:Identify the elements: the extreme left of a period is the group-1 alkali metal, and the extreme right excluding noble gases is the group-17 halogen.
left=alkali metal,right=halogen
Step 2:First ionisation enthalpy increases left to right, so the alkali metal at the extreme left has the lowest first ionisation enthalpy in the period.
IE1(alkali)=lowest in period
Step 3:Halogens have the most negative electron gain enthalpy, so the extreme-right element (excluding noble gases) has the highest negative electron gain enthalpy.
ΔegH(halogen)=most negative
Step 4:Combining the two, the species has the lowest first ionisation enthalpy and the highest negative electron gain enthalpy.
lowest and highest
Final answer: lowest and highest
Q59Single correctp-Block Elements
Given below are two statements : Statement I: F2O<H2O<Cl2O is the correct trend in terms of bond angle. Statement II: SiF4,SnF4 and PbF4 are ionic in nature. 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 true but Statement II is false
Approach:
Compare the central-oxygen bond angles of the three oxides and assess the bonding character of the group-14 tetrafluorides.
Step 1:Statement I concerns bond-angle ordering of F2O, H2O, Cl2O. Statement II concerns whether SiF4, SnF4 and PbF4 are all ionic.
F2O,H2O,Cl2O;SiF4,SnF4,PbF4
Step 2:Bond angles: F2O is about 103.2∘, H2O is 104.5∘, Cl2O is about 110.9∘. In F2O the highly electronegative F atoms draw bonding pairs away from O, contracting the angle; in Cl2O the large, less electronegative Cl atoms create bond-pair repulsion that widens it. The ordering F2O<H2O<Cl2O is therefore correct.
103.2∘<104.5∘<110.9∘
Step 3:For the tetrafluorides, SiF4 is a covalent molecular compound (high charge, small Si4+ gives strong covalent character per Fajans' rules). Although SnF4 and PbF4 have appreciable ionic character, SiF4 is not ionic, so the blanket statement that all three are ionic is false.
SiF4 covalent,SnF4/PbF4 ionic character
Step 4:Statement I is true and Statement II is false.
I true,II false
Final answer: Statement I is true but Statement II is false
Q60Single correctd- and f-Block Elements
The correct order of first (ΔiH1) and second (ΔiH2) ionisation enthalpy values of Cr and Mn are : A. ΔiH1:Cr>Mn B. ΔiH2:Cr>Mn C. ΔiH1:Mn>Cr D. ΔiH2:Mn>Cr Choose the correct answer from the options given below :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2B and C only
Approach:
Compare first and second ionisation enthalpies of Cr ([Ar]3d54s1) and Mn ([Ar]3d54s2) using their electronic configurations and the stability of half-filled 3d5.
Step 1:State the neutral configurations: Cr is [Ar]3d54s1 and Mn is [Ar]3d54s2. First ionisation enthalpy depends on how strongly the outermost electron is held.
Cr=[Ar]3d54s1,Mn=[Ar]3d54s2
Step 2:For first ionisation, Cr loses its single 4s1 electron forming the stable [Ar]3d5 ion, which releases that electron readily, giving Cr a lower ΔiH1. Mn must ionise from a filled 4s2 with higher effective nuclear charge, so ΔiH1(Mn)>ΔiH1(Cr). Statement C is correct and A is incorrect.
ΔiH1:Mn>Cr
Step 3:After first ionisation Cr+=[Ar]3d5 (stable half-filled) and Mn+=[Ar]3d54s1. The second electron from Cr+ comes from the stable 3d5 set and needs high energy, while Mn+ loses its loosely bound 4s1 electron easily, so ΔiH2(Cr)>ΔiH2(Mn). Statement B is correct and D is incorrect.
ΔiH2:Cr>Mn
Step 4:The valid statements are B and C, matching experimental values ΔiH1: Mn (717) > Cr (653) and ΔiH2: Cr (1592) > Mn (1509) kJ mol−1.
Correct statements: B and C
Final answer: B and C only
Q61Single correctCoordination Compounds
Which of the following sequences of hybridisation, geometry and magnetic nature are correct for the given coordination compounds ? A. [NiCl4]2−−sp3, tetrahedral, paramagnetic B. [Ni(NH3)6]2+−sp3d2, octahedral, paramagnetic C. [Ni(CO)4]−sp3, tetrahedral, paramagnetic D. [Ni(CN)4]2−−dsp2, square planar, diamagnetic Choose the correct answer from the options given below :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4A, B and D only
Approach:
Evaluate hybridisation, geometry and magnetic nature of each Ni complex using valence bond theory and the field strength of the ligands.
Step 1:Assign oxidation states and d-electron counts: in A, B and D nickel is Ni2+(3d8); in C nickel is Ni0(3d10).
Ni2+=3d8,Ni0=3d10
Step 2:A: [NiCl4]2− with weak-field Cl− uses sp3, tetrahedral, with 2 unpaired electrons, paramagnetic. B: [Ni(NH3)6]2+ uses sp3d2, octahedral, with 2 unpaired electrons, paramagnetic. Both labels are correct.
Step 3:C: [Ni(CO)4] has Ni0(3d10) fully paired, sp3, tetrahedral and diamagnetic; the label 'paramagnetic' is incorrect. D: [Ni(CN)4]2− with strong-field CN− pairs the 3d8 electrons, giving dsp2, square planar, diamagnetic, which is correct.
Q62Single correctPurification and Characterisation of Organic Compounds
Given below are two statements : Statement I: A mixture of C12H22O11 (sugar) and NaCl can be separated by dissolving sugar in alcohol, due to differential solubility. Statement II: Rose essence from rose petals is seperated by steam distillation due to its high volatility and insolubility in H2O. In the light of the above statements, choose the correct answer from the options given below :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1Both Statement I and Statement II are true
Approach:
Assess each statement against the principles of differential solubility and steam distillation.
Step 1:Set up the separation criteria: differential solubility separates two solids when one dissolves in a chosen solvent and the other does not; steam distillation separates a volatile, water-immiscible compound by co-distilling it with steam.
Ptotal=Pcompound+Pwater
Step 2:Statement I: Sugar is soluble in alcohol whereas NaCl is essentially insoluble in alcohol. Treating the mixture with alcohol dissolves only the sugar, allowing separation by filtration, so the statement is true.
Sugar (soluble in alcohol)=NaCl (insoluble in alcohol)
Step 3:Statement II: Rose essence (essential oil) is volatile in steam and immiscible with water, the exact requirement for steam distillation, so the statement is true.
Final answer: Both Statement I and Statement II are true
Q63Single correctSome Basic Principles of Organic Chemistry
Shown below is the structure of methyl acetate with three different α,β and γ carbon - oxygen bonds. The correct order of bond lengths of these bonds is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2α<β<γ
Approach:
Rank the three C-O bonds in methyl acetate by bond order, using the inverse relation between bond order and bond length and the partial double-bond character introduced by resonance.
Step 1:Identify the three bonds from the structure H3C−C(=O)−O−CH3: α is the carbonyl C=O, β is the carbonyl-carbon to ester-oxygen C−O, and γ is the ester-oxygen to methyl O−CH3 bond.
α=C=O,β=C−O(acyl),γ=O−CH3
Step 2:The α bond is a double bond (bond order ≈2), giving it the highest bond order and therefore the shortest length.
α:bond order≈2⇒shortest
Step 3:Resonance of the lone pair on the ester oxygen into the carbonyl gives the β bond partial double-bond character (bond order between 1 and 2), making it shorter than a pure single bond. The γ bond (O−CH3) has no double-bond character (bond order 1) and is the longest.
1<B.O.(β)<2,B.O.(γ)=1
Step 4:Increasing bond length follows the decreasing bond order, giving α<β<γ.
'x' is the product which is obtained by the hydrolysis of prop-1-yne in the presence of mercuric sulphate under dilute acidic medium at 333 K. 'y' is the product which is obtained by the reaction of ethane nitrile with methyl magnesium bromide in dry ether followed by hydrolysis. IUPAC name of product obtained from 'x' and 'y' in the presence of barium hydroxide followed by heating is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 24 - Methylpent-3-en-2-one
Approach:
Identify x and y, then carry out the aldol condensation between them in presence of barium hydroxide followed by dehydration on heating to obtain the α,β-unsaturated ketone, and name it.
Step 1:Identify x: Markovnikov hydration of prop-1-yne with HgSO4 in dilute acid at 333 K adds water across the triple bond and tautomerises to give acetone.
x=CH3COCH3(acetone)
Step 2:Identify y: ethane nitrile CH3CN adds CH3MgBr to form a ketimine magnesium salt, which on acidic hydrolysis gives acetone.
CH3CNCH3MgBrH3O+CH3COCH3
Step 3:x and y are both acetone. With Ba(OH)2, the α-carbon of one acetone adds to the carbonyl of the other (aldol), and heating eliminates water to give the α,β-unsaturated ketone mesityl oxide (CH3)2C=CHCOCH3.
2CH3COCH3Ba(OH)2,Δ(CH3)2C=CHCOCH3
Step 4:Numbering the five-carbon chain from the carbonyl end gives a ketone at C-2, a double bond at C-3 and a methyl branch at C-4, so the IUPAC name is 4-methylpent-3-en-2-one.
An optically active alkyl bromide C4H9Br, reacts with ethanolic KOH to form major compound [A] which reacts with bromine to give compound [B]. Compound [B] reacts with ethanolic KOH and sodamide to give compound [C]. One molecule of water adds to compound [C] on warming with mercuric sulphate and dilute sulphuric acid at 333 K to form compound [D]. The functional group in compound D will be confirmed by :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1Haloform test
Approach:
Track the sequence of transformations from the optically active 2-bromobutane to identify compound D and the test that confirms its functional group.
Step 1:Optically active C4H9Br is 2-bromobutane. With ethanolic KOH it undergoes dehydrohalogenation to give but-2-ene [A] (major, Saytzeff).
CH3CHBrCH2CH3alc.KOHCH3CH=CHCH3
Step 2:But-2-ene adds bromine to give 2,3-dibromobutane [B].
CH3CH=CHCH3+Br2→CH3CHBrCHBrCH3
Step 3:Double dehydrohalogenation with ethanolic KOH and sodamide gives but-2-yne [C].
CH3CHBrCHBrCH3alc.KOH/NaNH2CH3C≡CCH3
Step 4:Hydration of but-2-yne with HgSO4/dil.H2SO4 at 333 K gives butan-2-one [D], a methyl ketone, confirmed by the haloform (iodoform) test.
SolutionAnswer: Option 4Statement I is false but Statement II is true
Approach:
Apply the mechanism of ether cleavage by HI for an aryl-alkyl (benzyl phenyl) ether.
Step 1:In benzyl phenyl ether, the C-O bond on the side that forms the more stable carbocation (benzyl) breaks; the benzyl group leaves as benzyl iodide and phenol is retained because the C(aryl)-O bond is not cleaved by iodide.
C6H5−O−CH2C6H5+HI→C6H5OH+C6H5CH2I
Step 2:The bond cleaved is the alkyl side −O−CH2− bond, so Statement II is true.
−O−CH2−bondcleaved
Step 3:The claimed products (benzyl alcohol + iodobenzene) require cleavage of the aryl-O bond, which does not occur; therefore Statement I is false.
Products=benzylalcohol+iodobenzene
Final answer: Statement I is false but Statement II is true
Follow propanoic acid through amide formation, Hofmann bromamide degradation, diazotization, and coupling/decomposition.
Step 1:Propanoic acid with NH3/Δ gives propanamide.
CH3CH2COOHNH3/ΔCH3CH2CONH2
Step 2:Hofmann bromamide degradation with NaOH/Br2 removes one carbon, giving ethylamine.
CH3CH2CONH2NaOH/Br2CH3CH2NH2
Step 3:Reaction of the aliphatic primary amine with HNO2 at 0∘C gives an unstable diazonium salt that loses N2 to give ethanol.
CH3CH2NH2HNO2,0∘CCH3CH2OH+N2
Step 4:Benzene diazonium chloride with a reducing/active-hydrogen species undergoes loss of N2 (replacement of diazonium by H), giving benzene as the major product.
The number of compounds from the following which can undergo reaction with Br2/KOH (alcoholic) to give respective products and these respective products can also be obtained separately by Gabriel phthalimide reaction is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 33
Approach:
Identify which amides give Hofmann degradation to primary amines that could also be made by Gabriel synthesis (i.e. aliphatic primary amines without aryl-N or substituted amides).
Step 1:Only primary unsubstituted amides (−CONH2) undergo Hofmann bromamide degradation to primary amines. The N-substituted amides (C6H11CONHCH2CH3 and (CH3)3CCONHCH3) are excluded.
RCONH2Br2/KOHRNH2
Step 2:Of these, the Hofmann product must be an aliphatic primary amine to also be obtainable by Gabriel synthesis. Benzamide gives aniline (aryl amine) which cannot be made by Gabriel, so it is excluded.
C6H5CONH2→C6H5NH2(notGabriel)
Step 3:The remaining give aliphatic/benzylic primary amines obtainable by Gabriel: phenylacetamide gives benzylamine, acetamide gives methylamine, cyclohexanecarboxamide gives cyclohexylamine.
C6H5CH2CONH2,CH3CONH2,C6H11CONH2
Final answer: 3
Q69Single correctBiomolecules
Consider the following reactions. Total number of electrons in the π bonds and lone pair of electrons in the product (X) is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 418
Approach:
Identify the major product X by following each step of the reaction sequence, then count the total electrons present in π bonds plus the electrons in lone pairs of product X.
Step 1:Glucose on prolonged reaction with HI and heat is fully reduced; all the -OH groups and the -CHO group are removed to give the straight-chain alkane n-hexane.
CHO−(CHOH)4−CH2OHHI,ΔCH3−(CH2)4−CH3
Step 2:n-Hexane undergoes aromatization (cyclisation and dehydrogenation) over V2O5 at high temperature and pressure to give benzene.
C6H14V2O5/773KC6H6(benzene)
Step 3:Benzene undergoes Friedel-Crafts acylation with benzoyl chloride in presence of anhydrous AlCl3 to give diphenyl ketone (benzophenone), the product X.
C6H6+C6H5COClAlCl3C6H5−CO−C6H5
Step 4:Count the pi bonds in benzophenone: each of the two benzene rings contributes 3 pi bonds and the carbonyl group contributes 1 pi bond, giving 7 pi bonds in total. Electrons in pi bonds = 7 x 2 = 14.
(3+3+1)π bonds×2=14electrons
Step 5:Count the lone pairs: only the carbonyl oxygen carries lone pairs, namely 2 lone pairs. Electrons in lone pairs = 2 x 2 = 4.
2lone pairs×2=4electrons
Step 6:Add the pi-bond electrons and lone-pair electrons to obtain the required total.
14+4=18
Final answer: 18
Q70Single correctp-Block Elements
Treatment of a gas 'X' with a freshly prepared ferrous sulphate solution gives a compound 'Y' as a brown ring. The compounds X and Y are.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1NO and [Fe(NO)]SO4
Approach:
Recall the brown ring test for nitrates and the nature of the brown complex formed.
Step 1:In the brown ring test, nitrate is reduced by Fe2+ in presence of conc. H2SO4 to give nitric oxide (NO) gas, which is gas X.
NO3−+3Fe2++4H+→NO+3Fe3++2H2O
Step 2:NO combines with freshly prepared ferrous sulphate to form the brown complex (nitrosyl iron sulphate).
FeSO4+NO→[Fe(NO)]SO4
Step 3:Thus X is NO and Y is [Fe(NO)]SO4.
X=NO,Y=[Fe(NO)]SO4
Final answer: NO and [Fe(NO)]SO4
Q71NumericalCoordination Compounds
An excess of AgNO3 is added to 100 mL of a 0.05 M solution of tetraaquadichloridochromium (III) chloride. The number of moles of AgCl precipitated will be ______ ×10−3. (Nearest integer)
SolutionAnswer: 5
Approach:
Identify ionizable chloride ions outside the coordination sphere and compute moles of AgCl.
Step 1:Tetraaquadichloridochromium(III) chloride is [Cr(H2O)4Cl2]Cl. Two chlorides are coordinated (inside sphere) and one chloride is ionizable (outside sphere).
[Cr(H2O)4Cl2]Cl→[Cr(H2O)4Cl2]++Cl−
Step 2:Moles of complex = 0.05M×0.100L=0.005 mol.
n=0.05×0.100=5×10−3mol
Step 3:Only the 1 ionizable Cl− precipitates as AgCl, so moles AgCl = 1×0.005=5×10−3.
nAgCl=1×5×10−3=5×10−3mol
Final answer: 5
Q72NumericalHydrocarbons
An alkane (Y) requires 8 moles of oxygen for complete combustion and on chlorination with Cl2/hν, (Y) gives only one monochlorinated product (Z). The total number of primary carbon atoms in (Y) is ______.
SolutionAnswer: 4
Approach:
Use the combustion oxygen requirement to find the alkane formula, then use the single-monochloro-product condition to fix its structure and count primary carbons.
Step 1:Set O2 required equal to 8 moles: (3n+1)/2=8, so 3n+1=16, n=5. The alkane is C5H12.
23n+1=8⇒n=5
Step 2:Only one monochlorinated product is obtained when all hydrogens are equivalent. Among C5H12 isomers, neopentane (2,2-dimethylpropane) has all 12 H equivalent.
C(CH3)4=neopentane
Step 3:Neopentane has 4 terminal CH3 groups, i.e. 4 primary carbons (the central carbon is quaternary).
C(CH3)4→4primaryC
Final answer: 4
Q73NumericalRedox Reactions and Electrochemistry
500 mL of 0.2MMnO4− solution in basic medium when mixed with 500 mL of 1.5 M KI solution, oxidises iodide ions to liberate molecular iodine. This liberated iodine is then titrated with a standard xM thiosulphate solution in presence of starch till the end point. If 300 mL of thiosulphate was consumed, then the value of x is ______.
SolutionAnswer: 1
Approach:
In basic medium MnO4− is reduced to MnO2 (3-electron change); compute I2 liberated, then equate to thiosulphate by the iodine-thiosulphate reaction.
Step 1:Moles of MnO4− = 0.2×0.500=0.1 mol. In basic medium MnO4−+2H2O+3e−→MnO2+4OH−, gaining 3 electrons each, so total electrons = 0.1×3=0.3 mol.
ne=0.1×3=0.3mol
Step 2:Each I− loses 1 electron (2I−→I2+2e−), so moles of I2 = electrons/2 = 0.3/2=0.15 mol. (KI is in excess: 0.75 mol available.)
nI2=20.3=0.15mol
Step 3:I2 reacts with thiosulphate 1:2, so moles of S2O32− = 2×0.15=0.30 mol in 300 mL = 0.300 L.
nS2O3=2×0.15=0.30mol
Step 4:Concentration x = 0.30 mol / 0.300 L = 1 M.
x=0.3000.30=1M
Final answer: 1
Q74NumericalEquilibrium
In a closed flask at 600 K, one mole of X2Y4(g) attains equilibrium as given below : X2Y4(g)⇌2XY2(g). At equilibrium, 75%X2Y4(g) was dissociated and the total pressure is 1 atm. The magnitude of ΔrG⊖ (in kJmol−1) at this temperature is ______. (Nearest Integer) (Given: R=8.3Jmol−1K−1; ln10=2.3,log2=0.3,log3=0.48,log5=0.69,log7=0.84)
SolutionAnswer: 8
Approach:
Compute equilibrium mole fractions and partial pressures from 75% dissociation, find Kp, then use ΔG∘=−RTlnKp.
Step 1:Start with 1 mol X2Y4; 75% dissociates, so 0.75 mol reacts. At equilibrium: X2Y4 = 0.25 mol, XY2 = 2×0.75 = 1.5 mol. Total = 1.75 mol.
Decomposition of a hydrocarbon follows the equation k=(5.5×1011s−1)eT−28000K. The activation energy of reaction is ______ kJmol−1. (Nearest Integer) Given: R=8.3JK−1mol−1
SolutionAnswer: 232
Approach:
Compare the given Arrhenius expression with k = A e^(-Ea/RT) to extract Ea.
Step 1:Compare exponents: -Ea/(RT) = -28000/T, so Ea/R = 28000 K.
REa=28000K
Step 2:Solve for Ea: Ea=28000×R=28000×8.3=232400 J/mol.
Let f:R→R be defined as f(x)=3x2+x+32x2−3x+2. Then f is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4neither one-one nor onto
Approach:
Establish the domain via the denominator's sign, then test surjectivity by finding the bounded range using the discriminant and test injectivity from the structure of the rational function.
Step 1:The denominator 3x2+x+3 has discriminant 1−36=−35<0 with positive leading coefficient, so it is strictly positive and f is defined for every real x. Target: determine whether f is injective and/or surjective on R.
3x2+x+3>0∀x∈R
Step 2:Set y=f(x) and clear the denominator: y(3x2+x+3)=2x2−3x+2, giving (3y−2)x2+(y+3)x+(3y−2)=0.
(3y−2)x2+(y+3)x+(3y−2)=0
Step 3:For a real preimage x the discriminant in x must satisfy (y+3)2−4(3y−2)2≥0. This is a downward inequality that holds only on a bounded interval of y, so the range is a proper bounded subset of R and f is not onto.
(y+3)2−4(3y−2)2≥0
Step 4:The function has a horizontal asymptote y→32 as x→±∞ and attains interior extreme values, so it is non-monotonic; an interior output value is produced by two distinct inputs. Hence f is not injective.
f(x1)=f(x2)for some x1=x2
Step 5:Verification by consistency: a bounded range excludes surjectivity and a non-monotonic continuous rational function excludes injectivity, jointly fixing the classification.
Range(f)⊊R,f not injective
Final answer: neither one-one nor onto
Q2Single correctComplex Numbers and Quadratic Equations
Consider the quadratic equation (n2−2n+2)x2−3x+(n2−2n+2)2=0,n∈R. Let α be the minimum value of the product of its roots and β be the maximum value of the sum of its roots. Then the sum of the first six terms of the G.P., whose first term is α and the common ratio is βα, is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3243364
Approach:
Substitute k=n2−2n+2, express the product and sum of roots as functions of k, optimise them over k≥1, then sum the resulting geometric progression.
Step 1:Let k=n2−2n+2=(n−1)2+1, so k≥1 with minimum 1 at n=1. The equation is kx2−3x+k2=0. Target: α=min(product), β=max(sum).
k=(n−1)2+1≥1
Step 2:Product of roots =kk2=k and sum of roots =k3.
product=k,sum=k3
Step 3:The product k is minimised at k=1, so α=1. The sum k3 is maximised at the smallest k=1, so β=3. The common ratio is r=βα=31.
α=1,β=3,r=31
Step 4:Sum the first six terms: S6=1−1/31(1−(1/3)6)=2/31−1/729=2728/729⋅3=243364.
S6=1−1/31−(1/3)6=243364
Step 5:Verification by substitution: the six terms 1,31,91,271,811,2431 sum to 243243+81+27+9+3+1=243364.
243243+81+27+9+3+1=243364
Final answer: 243364
Q3Single correctComplex Numbers and Quadratic Equations
Let S={z∈C:z2+6iz−3=0}. Then z∈S∑z8 is equal to :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1162
Approach:
Solve the quadratic in z by the quadratic formula, convert each root to polar form, then apply De Moivre's theorem to the 8th power and add.
Step 1:Apply the quadratic formula with a=1,b=6i,c=−3. The discriminant is (6i)2−4(1)(−3)=−6+12=6.
z=2−6i±6
Step 2:Each root has modulus ∣z∣=(6/2)2+(6/2)2=3, hence ∣z∣8=34=81. Their arguments are −45∘ for z1 and −135∘ for z2.
∣z∣=3,∣z∣8=81
Step 3:By De Moivre, 8×(−45∘)=−360∘≡0∘ and 8×(−135∘)=−1080∘≡0∘. Thus z18=z28=81.
z18=z28=81
Step 4:Add the contributions: ∑z∈Sz8=81+81=162.
∑z∈Sz8=81+81=162
Step 5:Verification by substitution: z12=4(6−6i)2=46(1−i)2=46(−2i)=−3i, so z18=(−3i)4=81; the same holds for z2.
z12=−3i,z18=(−3i)4=81
Final answer: 162
Q4Single correctMatrices and Determinants
The sum of all possible values of θ∈[0,2π], for which the system of equations : xcos3θ−8y−12z=0, xcos2θ+3y+3z=0, x+y+3z=0 has a non-trivial solution, is equal to :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 44π
Approach:
A homogeneous system has a non-trivial solution exactly when its coefficient determinant vanishes; evaluate the determinant, solve the trigonometric equation, and add all roots in [0,2π].
Step 1:The coefficient matrix is cos3θcos2θ1−831−1233. Evaluating along the first column gives cos3θ(9−3)−cos2θ(−24+12)+(−24+36)=6cos3θ+12cos2θ+12.
det=6cos3θ+12cos2θ+12
Step 2:Using cos2θ=2cos2θ−1, the term 12cos2θ+12=24cos2θ. Setting det=0 gives 6cos3θ+24cos2θ=0.
6cos3θ+24cos2θ=0
Step 3:Substituting cos3θ=4cos3θ−3cosθ yields 24cos3θ+24cos2θ−18cosθ=0, i.e. 6cosθ(2cosθ−1)(2cosθ+3)=0. Since 2cosθ+3=0, either cosθ=0 or cosθ=21.
6cosθ(2cosθ−1)(2cosθ+3)=0
Step 4:In [0,2π]: cosθ=0 gives θ=2π,23π; cosθ=21 gives θ=3π,35π. Their sum is 2π+23π+3π+35π=2π+2π=4π.
2π+23π+3π+35π=4π
Step 5:Verification by substitution: at θ=3π, cosθ=21 and cos3θ=cosπ=−1, giving 6(−1)+24(41)=−6+6=0, confirming the root.
6(−1)+24(41)=0
Final answer: 4π
Q5Single correctMatrices and Determinants
Let A=139013001 and B=[bij],1≤i,j≤3. If B=A99−I, then the value of b32b31−b21 is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3149
Approach:
Write A=I+N with N strictly lower triangular and nilpotent of index 3, so the binomial expansion of A99 terminates after the N2 term.
Step 1:Set N=A−I=039003000. Then N2=009000000 and N3=0.
N2=009000000,N3=0
Step 2:Therefore A99=I+99N+(299)N2, with (299)=4851, so B=A99−I=99N+4851N2.
B=99N+4851N2
Step 3:Read off entries: b21=99⋅3=297, b32=99⋅3=297, and b31=99⋅9+4851⋅9=891+43659=44550.
b21=297,b32=297,b31=44550
Step 4:Evaluate the requested ratio: b32b31−b21=29744550−297=29744253=149.
29744550−297=149
Step 5:Verification by consistency: 44253=297×149 since 297×149=297×150−297=44550−297=44253.
297×149=44253
Final answer: 149
Q6Single correctSequence and Series
The sum 1+21(12+22)+31(12+22+32)+… upto 10 terms is equal to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 32315
Approach:
Express the general nth term using the sum-of-squares formula, then sum the resulting polynomial in n from 1 to 10 using standard summation identities.
Step 1:The nth term is Tn=n1∑k=1nk2=n1⋅6n(n+1)(2n+1)=6(n+1)(2n+1)=62n2+3n+1.
Tn=62n2+3n+1
Step 2:Sum over n=1 to 10: S=61(2∑110n2+3∑110n+∑1101) with ∑n2=385 and ∑n=55.
S=61(2(385)+3(55)+10)
Step 3:Combine the numerator: 770+165+10=945.
S=6945
Step 4:Simplify: 6945=2315.
S=6945=2315
Step 5:Verification by substitution: the first three terms T1=1, T2=6(3)(5)=25, T3=6(4)(7)=314 match 1, 21(5), 31(14) from the series, confirming the term formula.
T1=1,T2=25,T3=314
Final answer: 2315
Q7Single correctPermutations and Combinations
A building has ground floor and 10 more floors. Nine persons enter in a lift at the ground floor. The lift goes up to the 10th floor. The number of ways, in which any 4 persons exit at a floor and the remaining 5 persons exit at a different floor, if the lift does not stop at the first and the second floors, is equal to :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 37056
Approach:
Count the floors at which the lift can stop, choose which persons form the group of 4, then assign distinct floors to the two groups using the multiplication principle.
Step 1:The lift cannot stop at the 1st or 2nd floor, so the usable floors are 3, 4, 5, 6, 7, 8, 9, 10, giving 8 stopping floors.
Available floors=8
Step 2:Choose which 4 of the 9 persons exit together: (49)=126; the remaining 5 form the other group automatically.
(49)=126
Step 3:Assign a floor to the group of 4 and a different floor to the group of 5; these are ordered distinct choices: 8×7=56.
8×7=56
Step 4:By the multiplication principle the total count is 126×56=7056.
126×56=7056
Step 5:Verification by consistency: the two groups are distinguishable (sizes 4 and 5), so floor assignment is ordered, matching 8×7; the product 126×56=7056 is integral and within range.
126×56=7056
Final answer: 7056
Q8Single correctStatistics and Probability
Let the mean and the variance of seven observations 2,4,α,8,β,12,14,α<β, be 8 and 16 respectively. Then the quadratic equation whose roots are 3α+2 and 2β+1 is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2x2−41x+420=0
Approach:
Use the mean condition to relate α and β, the variance condition to fix their values, then construct the quadratic from the transformed roots.
Step 1:Given seven observations with mean 8 and variance 16; unknowns α and β with α<β. Target: quadratic with roots 3α+2 and 2β+1.
xˉ=8,σ2=16,α<β
Step 2:Apply the mean condition.
72+4+α+8+β+12+14=8⇒40+α+β=56
Step 3:Apply the variance condition to find the sum of squares.
7∑xi2−64=16⇒∑xi2=560
Step 4:Subtract the fixed squares 4+16+64+144+196=424 to isolate the unknowns.
α2+β2=560−424=136
Step 5:Use the identity to find the product, then solve for the values.
(α+β)2=256⇒2αβ=256−136=120⇒αβ=60
Step 6:Form the transformed roots and build the quadratic.
3α+2=20,2β+1=21⇒x2−41x+420=0
Final answer: x2−41x+420=0
Q9Single correctStatistics and Probability
A bag contains 6 blue and 6 green balls. Pairs of balls are drawn without replacement until the bag is empty. The probability that each drawn pair consists of one blue and one green ball is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 323116
Approach:
Count the total number of ways to partition the 12 balls into 6 unordered pairs and the number of those partitions in which every pair has one blue and one green ball.
Step 1:Treat the 12 balls as distinct; the process partitions them into 6 unordered pairs. Target: probability each pair is one blue and one green.
6 blue,6 green,6 pairs
Step 2:Total number of ways to split 12 distinct balls into 6 unordered pairs.
(2!)66!12!=10395
Step 3:Favourable count: pairing each blue ball with a distinct green ball is a bijection of the 6 blue onto the 6 green balls.
6!=720
Step 4:Form the probability and simplify.
P=10395720=23116
Final answer: 23116
Q10Single correctCo-ordinate Geometry
Let C be a circle having centre in the first quadrant and touching the x-axis at a distance of 3 units from the origin. If the circle C has an intercept of length 63 on y-axis, then the length of the chord of the circle C on the line x−y=3 is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 362
Approach:
Determine the circle from its tangency to the x-axis and the given y-axis intercept length, then compute the chord on the given line using the perpendicular distance from the centre.
Step 1:The circle touches the x-axis at distance 3 from the origin with centre in the first quadrant, so the centre is (3,k) and the radius equals k.
centre (3,k),r=k
Step 2:Apply the y-axis intercept condition with the centre x-coordinate equal to 3.
2k2−9=63⇒k2−9=27⇒k2=36
Step 3:Compute the perpendicular distance from the centre (3,6) to the line x−y−3=0.
d=1+1∣3−6−3∣=26=32
Step 4:Apply the chord-length formula.
2r2−d2=236−18=218=62
Final answer: 62
Q11Single correctCo-ordinate Geometry
The eccentricity of an ellipse E with centre at the origin O is 23 and its directrices are x=±346. Let H:a2x2−b2y2=1 be a hyperbola whose eccentricity is equal to the length of semi-major axis of E, and whose length of latus rectum is equal to the length of minor axis of E. Then the distance between the foci of H is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 478
Approach:
Determine the ellipse axes from its eccentricity and directrix, transfer the semi-major axis and minor axis to the hyperbola's eccentricity and latus rectum, then compute the distance between the hyperbola's foci.
Step 1:Ellipse E has eccentricity e=3/2 and directrix A/e=46/3. Target: distance between the foci of hyperbola H.
e=23,eA=346
Step 2:Solve for the semi-major axis A of the ellipse.
A=23⋅346=6418=22
Step 3:Compute the minor semi-axis and full minor axis of the ellipse.
B2=8(1−43)=2⇒2B=22
Step 4:Transfer the values to the hyperbola: its eccentricity equals A and its latus rectum equals the minor axis.
eH=22,a2b2=22⇒ab2=2
Step 5:Use the eccentricity relation to find the ratio of axes, then solve for a.
eH2=1+a2b2=8⇒b2=7a2;7a2=2a⇒a=72
Step 6:Compute the distance between the foci of H.
2aeH=2⋅72⋅22=78
Final answer: 78
Q12Single correctCo-ordinate Geometry
Let x=9 be a directrix of an ellipse E, whose centre is at the origin and eccentricity is 31. Let P(α,0), α>0, be a focus of E and AB be a chord passing through P. Then the locus of the mid point of AB is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 19y2=8x(1−x)
Approach:
Find the ellipse from its directrix and eccentricity, locate the positive focus, then derive the locus of midpoints of focal chords using the midpoint-chord relation T=S1.
Step 1:Ellipse E centred at the origin with eccentricity 1/3 and directrix x=9. Target: locus of the midpoint of a chord through the positive focus.
e=31,ea=9
Step 2:Solve for the ellipse parameters.
a=3,a2=9,b2=a2(1−e2)=9⋅98=8
Step 3:Locate the focus with positive abscissa.
ae=3⋅31=1⇒P(1,0)
Step 4:For a midpoint (h,k) the chord is T=S1; impose that it passes through P(1,0).
9hx+8ky=9h2+8k2;at (1,0):9h=9h2+8k2
Step 5:Simplify and replace (h,k) by (x,y).
9x−x2=8y2⇒9y2=8x(1−x)
Final answer: 9y2=8x(1−x)
Q13Single correctTrigonometry
If sin(tan−1(x2))=cot(sin−11−x2), x∈(0,1), then the value of x is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 121
Approach:
Convert each inverse-trigonometric expression into an algebraic expression in x and solve the resulting equation on the interval (0,1).
Step 1:Given the equation in x on (0,1). Target: the value of x. Set t=x2 for the left side and s=1−x2 for the right side.
t=x2,s=1−x2
Step 2:Evaluate the left-hand side.
sin(tan−1(x2))=1+2x2x2
Step 3:Evaluate the right-hand side using 1−s2=x for x>0.
cot(sin−11−x2)=1−x2x
Step 4:Equate the two sides and cancel the positive factor x.
1+2x22=1−x21⇒2(1−x2)=1+2x2
Step 5:Solve for x.
4x2=1⇒x=21
Final answer: x=21
Q14Single correctThree Dimensional Geometry
The shortest distance between the lines 1x−4=2y−3=−3z−2 and 2x+2=4y−6=−5z−5 is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 335
Approach:
Apply the shortest-distance formula for skew lines using the two direction vectors and the vector joining points on the lines.
Step 1:Read off points and direction vectors of the two lines, and form the joining vector.
Step 3:Compute the magnitude of the cross product.
∣d1×d2∣=4+1+0=5
Step 4:Compute the scalar triple product in the numerator.
(−6,3,3)⋅(2,−1,0)=−12−3+0=−15
Step 5:Apply the shortest-distance formula.
D=515=35
Final answer: 35
Q15Single correctVector Algebra
Let a=2i^+3j^+3k^ and b=6i^+3j^+3k^. Then the square of the area of the triangle with adjacent sides determined by the vectors (2a+3b) and (a−b) is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 31800
Approach:
Use bilinearity to reduce the cross product of the side vectors to a multiple of a cross b, compute that cross product, then form the triangle area and square it.
Step 1:Given a=(2,3,3) and b=(6,3,3); side vectors are 2a+3b and a−b. Target: square of the triangle area.
a=(2,3,3),b=(6,3,3)
Step 2:Evaluate the cross product of the side vectors using bilinearity; the self terms vanish.
(2a+3b)×(a−b)=−2(a×b)+3(b×a)=−5(a×b)
Step 3:Compute the cross product a cross b.
a×b=(3⋅3−3⋅3,3⋅6−2⋅3,2⋅3−3⋅6)=(0,12,−12)
Step 4:Compute the magnitude and hence the area of the triangle.
∣a×b∣=0+144+144=122;Area=21⋅5⋅122=302
Step 5:Square the area.
(302)2=900⋅2=1800
Final answer: 1800
Q16Single correctLimit, Continuity and Differentiability
Let limx→2(x−2)2(tan(x−2))(rx2+(p−2)x−2p)=5 for some r,p∈R. If the set of all possible values of q, such that the roots of the equation rx2−px+q=0 lie in (0,2), be the interval (α,β), then 4(α+β) equals:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 317
Approach:
Use the tangent approximation to reduce the limit, force the numerator to carry a factor (x-2) so the limit is finite, determine r and p, then impose the conditions for both roots of the quadratic to lie in (0,2).
Step 1:Since tan(x−2)∼(x−2) as x→2, the expression behaves like x−2rx2+(p−2)x−2p. For a finite limit the numerator must vanish at x=2.
r(4)+(p−2)(2)−2p=4r−4=0⇒r=1
Step 2:With r=1 the numerator factors as x2+(p−2)x−2p=(x−2)(x+p), so the limit equals limx→2(x+p)=2+p.
2+p=5⇒p=3
Step 3:The quadratic is x2−3x+q=0. Both roots lie in (0,2) when f(0)=q>0, f(2)=q−2>0, the vertex x=23∈(0,2), and D=9−4q≥0.
q>2andq≤49
Step 4:Thus (α,β)=(2,49), and 4(α+β)=4(2+49).
4(2+49)=8+9=17
Final answer: 17
Q17Single correctMatrices and Determinants
Let A=120311−1α−1 be a singular matrix. Let f(x)=∫0x(t2+2t+3)dt,x∈[1,α]. If M and m are respectively the maximum and the minimum values of f in [1,α], then 3(M−m) is equal to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 268
Approach:
Determine α from the singularity condition det(A)=0, then evaluate f explicitly and use its monotonicity on [1,α] to locate the extrema.
Step 1:Evaluate the determinant along the first row and set it to zero.
detA=1(−1−α)−3(−2−0)+(−1)(2−0)=−1−α+6−2=3−α
Step 2:On [1,3], f(x)=3x3+x2+3x has f′(x)=x2+2x+3=(x+1)2+2>0, so f is strictly increasing; the minimum is at x=1 and the maximum at x=3.
f′(x)=x2+2x+3>0∀x
Step 3:Evaluate f at the endpoints.
m=f(1)=31+1+3=313,M=f(3)=9+9+9=27
Step 4:Compute 3(M−m).
3(27−313)=81−13=68
Final answer: 68
Q18Single correctIntegral Calculus
Let f:R→R be such that f(xy)=f(x)f(y), for all x,y∈R and f(0)=0. Let g:[1,∞)→R be a differentiable function such that x2g(x)=∫1x(t2f(t)−tg(t))dt. Then g(2) is equal to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 33215
Approach:
Use the multiplicative functional equation to deduce f(x)=1, differentiate the integral relation to obtain a first-order linear ODE in g, and solve it with the initial value forced at x=1.
Step 1:Setting y=0 gives f(0)=f(x)f(0); dividing by f(0)=0 yields f(x)=1 for all x.
f(0)=f(x)f(0)⇒f(x)=1
Step 2:With f=1 the relation is x2g(x)=∫1x(t2−tg(t))dt. Differentiate both sides.
2xg+x2g′=x2−xg⇒x2g′+3xg=x2
Step 3:Multiply by the integrating factor x3 to get (x3g)′=x3. Integrating, x3g=4x4+C. At x=1 the right side of the original relation is 0, so g(1)=0, giving C=−41.
x3g=4x4−41⇒g(x)=4x−4x31
Step 4:Evaluate at x=2.
g(2)=42−321=3216−1=3215
Final answer: 3215
Q19Single correctIntegral Calculus
The area of the region {(x,y):x2−8x≤y≤−x} is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 16343
Approach:
Find the intersection points of the parabola y=x2−8x and the line y=−x, then integrate the difference of the upper and lower boundaries over that interval.
Step 1:Set the boundaries equal to find limits of integration.
x2−8x=−x⇒x2−7x=0⇒x=0,7
Step 2:On [0,7] the line y=−x lies above the parabola y=x2−8x, so the integrand is their difference.
A=∫07[(−x)−(x2−8x)]dx=∫07(7x−x2)dx
Step 3:Integrate term by term.
A=[27x2−3x3]07=2343−3343
Step 4:Combine the fractions.
A=343(21−31)=6343
Final answer: 6343
Q20Single correctIntegral Calculus
The value of the integral ∫−11(x2+2∣x∣+1x3+∣x∣+1)dx is equal to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 22\loge 2
Approach:
Rewrite the denominator as a perfect square, split the numerator into odd and even parts, discard the odd part by symmetry, and evaluate the remaining even integral.
Step 1:The denominator is x2+2∣x∣+1=(∣x∣+1)2, an even function. The numerator splits into the odd term x3 and the even term ∣x∣+1.
∫−11(∣x∣+1)2x3dx=0
Step 2:Apply even symmetry to the remaining part; for x≥0, ∣x∣=x.
I=2∫01(x+1)2x+1dx=2∫01x+1dx
Step 3:Integrate.
I=2[ln(x+1)]01=2(ln2−ln1)
Step 4:Simplify.
I=2loge2
Final answer: 2loge2
Q21NumericalSets, Relations and Functions
Let R={(x,y)∈N×N:loge(x+y)≤2}. Then the minimum number of elements, required to be added in R to make it a transitive relation, is ______.
SolutionAnswer: 15
Approach:
Translate the inequality into a bound on x+y to list R, then compute its transitive closure and subtract the existing pairs to count the additions needed.
Step 1:The condition loge(x+y)≤2 gives x+y≤e2≈7.389, so x+y≤7 with x,y≥1. Counting ordered pairs gives ∣R∣=6+5+4+3+2+1=21.
x+y≤7,x,y≥1⇒∣R∣=21
Step 2:The valid coordinate values are 1,2,…,6. Whenever (a,b) and (b,c) both lie in R, transitivity forces (a,c), which need not satisfy a+c≤7.
(a,b),(b,c)∈R⇒(a,c)required
Step 3:The transitive closure consists of all ordered pairs (a,c) with a,c∈{1,…,6} reachable through a common middle value; this yields 36 pairs in total.
∣R+∣=36
Step 4:The minimum number of pairs to add equals the closure size minus the existing size.
∣R+∣−∣R∣=36−21=15
Final answer: 15
Q22NumericalBinomial Theorem and its Simple Applications
If (1−x3)10=∑r=010arxr(1−x)30−2r, then a109a9 is equal to ______.
SolutionAnswer: 30
Approach:
Factor 1−x3 and re-express it as a sum of x(1−x) and (1−x)3, then apply the binomial theorem and match the exponent of (1−x) to identify the coefficients ar.
Step 1:Since 1+x+x2=(1−x)2+3x, it follows that 1−x3=(1−x)[(1−x)2+3x]=(1−x)3+3x(1−x).
1−x3=3x(1−x)+(1−x)3
Step 2:Raise to the 10th power with p=3x(1−x) and q=(1−x)3.
(1−x3)10=∑r=010(r10)(3x)r(1−x)r(1−x)3(10−r)
Step 3:The total power of (1−x) is r+30−3r=30−2r, matching the given form, so ar=(r10)3r.
ar=(r10)3r
Step 4:Compute the required ratio using a9=(910)39=10⋅39 and a10=310.
a109a9=9⋅31010⋅39=9⋅310=30
Final answer: 30
Q23NumericalCo-ordinate Geometry
Let the line x−y=4 intersect the circle C:(x−4)2+(y+3)2=9 at the points Q and R. If P(α,β) is a point on C such that PQ=PR, then (6α+8β)2 is equal to ______.
SolutionAnswer: 18
Approach:
A point P on the circle equidistant from chord endpoints Q and R lies on the perpendicular bisector of QR, which passes through the centre perpendicular to the chord; intersecting this diameter with the circle gives the candidate points.
Step 1:The circle has centre (4,−3) and radius 3. The chord lies on x−y=4 (slope 1); PQ=PR forces P onto the diameter through the centre perpendicular to the chord (slope −1).
y+3=−1(x−4)⇒y=−x+1
Step 2:Substitute y=−x+1 into the circle to locate P.
(x−4)2+(−x+4)2=9⇒2(x−4)2=9
Step 3:Solve for the coordinates of the two candidate points.
x=4±23,y=1−x=−3∓23
Step 4:Evaluate 6α+8β=6(4±23)+8(−3∓23)=24−24±218−24=∓26, then square.
(6α+8β)2=(26)2=236=18
Final answer: 18
Q24NumericalThree Dimensional Geometry
Let the image of the point P(0,−5,0) in the line 2x−1=1y=−2z+1 be the point R and the image of the point Q(0,2−1,0) in the line −1x−1=4y+9=1z+1 be the point S. Then the square of the area of the parallelogram PQRS is ______.
SolutionAnswer: 162
Approach:
Reflect each point across its line by finding the foot of the perpendicular and using R = 2F - P, then compute the parallelogram area from the cross product of adjacent side vectors.
Step 1:For P(0,-5,0) and line through A1=(1,0,−1) with direction (2,1,−2): t=9(P−A1)⋅d=9(−1)(2)+(−5)(1)+(1)(−2)=−1, so F1=(−1,−1,1) and R=2F1−P=(−2,3,2).
R=2F1−P=(−2,3,2)
Step 2:For Q(0,-1/2,0) and line through A2=(1,−9,−1) with direction (−1,4,1): t=18(Q−A2)⋅d=18(−1)(−1)+(8.5)(4)+(1)(1)=2, so F2=(−1,−1,1) and S=2F2−Q=(−2,−23,2).
S=2F2−Q=(−2,−23,2)
Step 3:Adjacent sides from P are PQ=(0,29,0) and PS=(−2,27,2); their cross product gives the area vector.
PQ×PS=(9,0,9)
Step 4:The area is the magnitude of this vector; square it.
(Area)2=92+02+92=162
Final answer: 162
Q25NumericalLimit, Continuity and Differentiability
Let f(x)={x3+8;x2−4;x<0,x≥0, and g(x)={(x−8)1/3;(x+4)1/2;x<0,x≥0. Then the number of points, where the function g∘f is discontinuous, is ______.
SolutionAnswer: 3
Approach:
Identify the points where f changes definition (x=0) and where f(x) crosses 0 (the branch point of g), then test the one-sided limits of g(f(x)) at each such candidate.
Step 1:The branch of g switches when its argument f(x) crosses 0. For x<0, f=x3+8=0 at x=−2; for x≥0, f=x2−4=0 at x=2; and f itself changes definition at x=0.
f=0 at x=−2,x=2;branch change at x=0
Step 2:At x=−2: as x→−2−, f→0− so g(f)→(0−8)1/3=−2; at x=−2, f=0 so g(f)=(0+4)1/2=2. Left limit −2=2= value.
limx→−2−g(f(x))=−2=2=g(f(−2))
Step 3:At x=0: as x→0−, f→8 so g(f)→(8+4)1/2=23; as x→0+, f→−4 so g(f)→(−4−8)1/3=−121/3. The two one-sided limits differ.
23=−121/3
Step 4:At x=2: as x→2−, f→0− so g(f)→(0−8)1/3=−2; at x=2, f=0 so g(f)=(0+4)1/2=2. Left limit differs from the value. The three discontinuities are at x=−2,0,2.
How many questions are in the JEE Main 2026 April 06, Shift 2 paper?
The JEE Main 2026 April 06, Shift 2 paper has 74 questions — Physics (24), 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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