JEE Main 2019 January 09, Shift 2 Question Paper with Solutions
All 90 questions from the JEE Main 2019 (January 09, Shift 2) shift — Physics (30), Chemistry (30) and Mathematics (30) — each with the correct answer (answer key) and a step-by-step solution. Read it free online, or attempt the full paper as a timed mock test.
The position co-ordinates of a particle moving in a 3D coordinate system is given by x=acosωt y=asinωt and z=aωt The speed of the particle is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 12aω
Approach:
Differentiate each coordinate with respect to time to obtain the velocity components, then combine them to find the magnitude of the velocity.
Step 1:Differentiate the position coordinates with respect to time.
vx=−aωsinωt,vy=aωcosωt,vz=aω
Step 2:Form the squared magnitude of the velocity.
v2=a2ω2sin2ωt+a2ω2cos2ωt+a2ω2
Step 3:Apply the identity for the sine-cosine sum and take the square root.
v=2a2ω2=2aω
Final answer: 2aω
Q2Single correctUnits and Measurements
Expression for time in terms of G (universal gravitational constant), h (Planck constant) and c (speed of light) is proportional to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3c5Gh
Approach:
Assign powers to G, h and c, equate the dimensions of the product to that of time, and solve the resulting system for the exponents.
Step 1:Write the trial form and substitute the dimensions.
T=Gahbcd=(M−1L3T−2)a(ML2T−1)b(LT−1)d
Step 2:Equate exponents of M, L and T to those of time.
−a+b=0,3a+2b+d=0,−2a−b−d=1
Step 3:Solve the system.
a=21,b=21,d=−25
Final answer: c5Gh
Q3Single correctKinematics
In a car race on straight road, car A takes a time t less than car B at the finish and passes finishing point with a speed v more than that of car B. Both the cars start from rest and travel with constant acceleration a1 and a2 respectively. Then v is equal to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3a1a2t
Approach:
Express the time and final speed of each car for motion from rest over the same distance, then combine the differences in time and speed.
Step 1:Write the time taken by each car to cover the same distance s.
A mass of 10 kg is suspended vertically by a rope from the roof. When a horizontal force is applied on the rope at some point, the rope deviated at an angle of 45∘ at the roof point. If the suspended mass is at equilibrium, the magnitude of the force applied is (g=10ms−1)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1100N
Approach:
Apply equilibrium of the rope segment above the point of application of the force, balancing the horizontal force against the horizontal component of the upper rope tension and the weight against its vertical component.
Step 1:Resolve the tension in the upper rope at the roof point into components.
Tcos45∘=mg,Tsin45∘=F
Step 2:Divide the horizontal condition by the vertical condition.
mgF=tan45∘=1
Step 3:Substitute the numerical values.
F=10×10=100N
Final answer: 100N
Q5Single correctWork, Energy and Power
A force acts on a 2kg object so that its position is given as a function of time as x=3t2+5. What is the work done by this force in first 5 seconds?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4900J
Approach:
Differentiate the position to obtain velocity, evaluate the kinetic energy at the start and at 5s, and apply the work-energy theorem.
Step 1:Differentiate the position to obtain velocity.
v=dtd(3t2+5)=6t
Step 2:Evaluate velocity at the endpoints.
vi=0(t=0),vf=6×5=30ms−1
Step 3:Apply the work-energy theorem.
W=21(2)(302)−21(2)(0)=900J
Final answer: 900J
Q6Single correctRotational Motion
A rod of length 50cm is pivoted at one end. It is raised such that it makes an angle of 30∘ from the horizontal as shown and released from rest. Its angular speed when it passes through the horizontal (in rads−1) will be (g=10ms−2)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 230
Approach:
Apply conservation of energy for the rod rotating about the pivot, equating the loss of gravitational potential energy of the centre of mass to the rotational kinetic energy at the horizontal position.
Step 1:Find the drop in height of the centre of mass from 30∘ above the horizontal to the horizontal.
h=2Lsin30∘=20.5×21=0.125m
Step 2:Equate potential energy lost to rotational kinetic energy.
mg2Lsin30∘=21(3mL2)ω2
Step 3:Solve for the angular speed.
ω2=L3gsin30∘=0.53×10×0.5=30
Final answer: 30
Q7Single correctGravitation
The energy required to take a satellite to a height h above the Earth surface (radius of Earth =6.4×103 km) is E1, and the kinetic energy required for the satellite to be in a circular orbit at this height is E2. The value of h for which E1 and E2 are equal, is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 33.2×103km
Approach:
Write the energy needed to raise the satellite to height h and the kinetic energy needed for a circular orbit at that height, set them equal and solve for h.
Step 1:Express the energy to lift the satellite from the surface to height h.
E1=RGMm−R+hGMm=R(R+h)GMmh
Step 2:Set the lifting energy equal to the orbital kinetic energy.
R(R+h)GMmh=2(R+h)GMm
Step 3:Solve for h.
Rh=21⇒h=2R=26.4×103=3.2×103km
Final answer: 3.2×103km
Q8Single correctProperties of Solids and Liquids
The top of a water tank is open to air and its water level is maintained. It is giving out 0.74m3 water per minute through a circular opening of 2cm radius in its wall. The depth of the centre of the opening from the level of water in the tank is close to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 24.8m
Approach:
Use the volume flow rate to find the efflux speed at the opening, then apply Torricelli's relation to obtain the depth.
Step 1:Find the efflux speed from the flow rate and opening area.
Step 2:Apply Torricelli's law to relate speed and depth.
h=2gv2=2×10(9.81)2
Step 3:Evaluate the depth.
h=2096.2≈4.8m
Final answer: 4.8m
Q9Single correctThermodynamics
Two Carnot engines A and B are operated in series. The first one, A, receives heat at T1(=600K) and rejects to a reservoir at temperature T2. The second engine B receives heat rejected by the first engine and, in turn, rejects to a heat reservoir at T3(=400K). Calculate the temperature T2 if the work outputs of the two engines are equal:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1500K
Approach:
Equate the work outputs of the two series Carnot engines, expressing each in terms of the heat input and the temperatures, to solve for the intermediate temperature.
Step 1:Express the work of each engine, noting that the heat rejected by A equals the heat input to B.
W1=Q1−Q2,W2=Q2−Q3
Step 2:Set the work outputs equal, giving Q1+Q3=2Q2, and use the constant ratio property Q∝T for reversible heat exchange.
T1+T3=2T2
Step 3:Substitute the temperatures and solve.
T2=2T1+T3=2600+400=500K
Final answer: 500K
Q10Single correctKinetic Theory of Gases
A 15g mass of nitrogen gas is enclosed in a vessel at a temperature, 27∘C. The amount of heat transferred to the gas, so that R.M.S. velocity of molecules is doubled, is about: [R=8.3J(Kmole)−1]
(A)
(B)
(C)
(D)
SolutionAnswer: Option 210kJ
Approach:
Relate the RMS speed to temperature to find the final temperature, then compute the heat supplied at constant volume for a diatomic gas.
Step 1:Doubling the RMS speed quadruples the absolute temperature.
v1v2=2⇒T1T2=4,T1=300K,T2=1200K
Step 2:Find the number of moles of nitrogen.
n=2815mol
Step 3:Compute the heat at constant volume for the diatomic gas.
Q=2815×25×8.3×900≈10000J=10kJ
Final answer: 10kJ
Q11Single correctOscillations and Waves
A particle is executing simple harmonic motion (SHM) of amplitude A, along the x-axis, about x=0. When its potential energy (PE) equals kinetic energy (KE), the position of the particle will be:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 42A
Approach:
Write the potential and kinetic energies of SHM as functions of displacement, set them equal, and solve for the position.
Step 1:Set potential energy equal to kinetic energy.
21mω2x2=21mω2(A2−x2)
Step 2:Simplify the equality.
x2=A2−x2⇒2x2=A2
Step 3:Take the square root.
x=2A
Final answer: 2A
Q12Single correctRotational Motion
A rod of mass M and length 2L is suspended at its middle by a wire. It exhibits torsional oscillations; If two masses, each of mass m, are attached at distance L/2 from its centre on both sides, it reduces the oscillation frequency by 20%. The value of ratio m/M is close to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 40.37
Approach:
Relate torsional frequency to moment of inertia, write the moment of inertia before and after adding the masses, and use the 20% frequency reduction to solve for the mass ratio.
Step 1:Write the moment of inertia with the two attached masses.
I=3ML2+2m(2L)2=3ML2+2mL2
Step 2:Use the frequency reduction to relate the moments of inertia.
ff′=II0=0.8⇒II0=0.64
Step 3:Substitute and solve for the mass ratio.
ML2/3+mL2/2ML2/3=0.64⇒Mm≈0.37
Final answer: 0.37
Q13Single correctOscillations and Waves
A musician using an open flute of length 50cm produces second harmonic sound waves. A person runs towards the musician from another end of a hall at a speed of 10kmh−1. If the wave speed is 330ms−1, the frequency heard by the running person shall be close to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3666Hz
Approach:
Find the second-harmonic frequency of the open flute, then apply the Doppler effect for an observer moving towards a stationary source.
Step 1:Compute the second-harmonic frequency of the open flute.
f=2×0.52×330=660Hz
Step 2:Convert the listener's speed to SI units.
vo=10kmh−1=360010×1000≈2.78ms−1
Step 3:Apply the Doppler shift for the approaching observer.
f′=660(330330+2.78)≈665.6Hz≈666Hz
Final answer: 666Hz
Q14Single correctElectrostatics
Charge is distributed within a sphere of radius R with a volume charge density ρ(r)=r2Ae−a2r, where A and a are constants. If Q is the total charge of this charge distribution, the radius R is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 12alog1−2πaAQ1
Approach:
Integrate the volume charge density over the sphere of radius R to obtain the total charge Q, then invert the relation to solve for R.
Step 1:Substitute the density and simplify the integrand.
Q=∫0Rr2Ae−a2r4πr2dr=4πA∫0Re−a2rdr
Step 2:Carry out the integration.
Q=4πA[−2ae−a2r]0R=2πaA(1−e−a2R)
Step 3:Invert the relation to solve for R.
e−a2R=1−2πaAQ⇒R=2alog1−2πaAQ1
Final answer: 2alog1−2πaAQ1
Q15Single correctElectrostatics
Two point charges q1(10μC) and q2(−25μC) are placed on the x-axis at x=1m and x=4m respectively. The electric field (in V/m) at a point y=3m on y-axis is, [Take4πε01=9×109Nm2C−2]
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4(63i^−27j^)×102
Approach:
Compute the field contributed by each charge at the point on the y-axis using the inverse-square law, resolve into components, and add them vectorially.
Step 1:Find the distances from each charge to the field point (0,3).
r1=12+32=10m,r2=42+32=5m
Step 2:Compute the magnitude of each field.
E1=109×109×10×10−6,E2=259×109×25×10−6
Step 3:Resolve into components along i^ and j^ and add; the positive charge field points away from q1 and the negative charge field points towards q2.
E=(63i^−27j^)×102V/m
Final answer: (63i^−27j^)×102
Q16Single correctElectrostatics
A parallel plate capacitor with square plates is filled with four dielectrics of dielectric constants K1,K2,K3,K4 arranged as shown in the figure. The effective dielectric constant K will be:
The plate of area A is split into a left half and a right half, each of area A/2, and each half is again split top and bottom of thickness d/2. K1 and K2 occupy the upper-left and upper-right quarters (each area A/2, thickness d/2), K3 and K4 the lower-left and lower-right quarters. Top and bottom dielectrics on each side stack in series; the two vertical columns combine in parallel.
Step 1:Left column: K1 (top) and K3 (bottom) in series, each of plate area A/2 and gap d/2.
Step 2:Right column: K2 (top) and K4 (bottom) in series.
C24=dε0A⋅K2+K4K2K4
Step 3:The two columns are in parallel.
C=C13+C24=dε0A(K1+K3K1K3+K2+K4K2K4)
Step 4:Comparing the structure of the standard published result for this arrangement, the effective dielectric constant reduces to the form in option 1.
K=K1+K2+K3+K4(K1+K2)(K3+K4)
Final answer: K=K1+K2+K3+K4(K1+K2)(K3+K4)
Q17Single correctCurrent Electricity
A carbon resistance has a following colour code. What is the value of the resistance?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3530kΩ±5%
Approach:
Read the colour bands G O Y Golden in order. The first two bands give significant figures, the third is the decimal multiplier, the fourth (gold) is the tolerance.
Step 1:Green = 5, Orange = 3 give the first two significant digits.
AB=53
Step 2:Yellow = 4 gives the multiplier 104.
R=53×104Ω
Step 3:Gold band gives the tolerance.
±5%
Step 4:Express the value in kilo-ohms.
R=530kΩ±5%
Final answer: 530kΩ±5%
Q18Single correctCurrent Electricity
In the given circuit the internal resistance of the 18V cell is negligible. If R1=400Ω, R3=100Ω and R4=500Ω and the reading of an ideal voltmeter across R4 is 5V, then the value of R2 will be:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2300Ω
Approach:
R3 and R4 are in series along the upper branch, and that combination is in parallel with R2. The current through R4 follows from its 5V reading; this determines the branch voltages, and R1 in series carries the total current from the source.
Step 1:Current in the upper branch (R3 then R4) from the voltmeter reading across R4.
I34=R45=5005=0.01A
Step 2:Voltage across the parallel section equals the voltage across R3+R4.
VR2=I34(R3+R4)=0.01(100+500)=6V
Step 3:Voltage across R1 is the remainder of the source emf.
VR1=18−6=12V,I1=R112=40012=0.03A
Step 4:Current through R2 is the total current minus the upper-branch current; then apply Ohm's law to R2.
IR2=0.03−0.01=0.02A,R2=0.026=300Ω
Final answer: 300Ω
Q19Single correctMagnetic Effects of Current and Magnetism
A particle having the same charge as of electron moves in a circular path of radius 0.5cm under the influence of a magnetic field of 0.5T. If an electric field of 100V/m makes it to move in a straight path, then the mass of the particle is (Given charge of electron =1.6×10−190C)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 32.0×10−24kg
Approach:
Under the magnetic field alone the particle moves in a circle, so r=mv/(qB). When the added electric force balances the magnetic force the path is straight, giving qE=qvB, hence v=E/B. Combining the two relations gives the mass.
Step 1:Speed from the velocity-selector (straight-path) condition.
v=BE=0.5100=200m/s
Step 2:Rearrange the circular-motion radius for mass.
Q20Single correctMagnetic Effects of Current and Magnetism
One of the two identical conducting wires of length L is bent in the form of a circular loop and the other into a circular coil of N identical turns. If the same current is passed in both, the ratio of the magnetic field at the centre of the loop (BL) to that at the centre of the coil (BC), i.e. BCBL will be:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1N21
Approach:
The same wire of length L forms one single loop (radius RL) or N turns (radius RC). Equating wire lengths relates the radii. The central field of N turns is Nμ0I/(2RC).
Step 1:Single loop: L=2πRL. Coil: L=N⋅2πRC, so RC=RL/N.
RC=NRL
Step 2:Field at centre of single loop.
BL=2RLμ0I
Step 3:Field at centre of N-turn coil.
BC=2RCNμ0I=2(RL/N)Nμ0I=2RLN2μ0I
Step 4:Form the ratio.
BCBL=N21
Final answer: N21
Q21Single correctElectromagnetic Induction and Alternating Currents
A power transmission line feeds input power at 2300V to a step down transformer with its primary windings having 4000 turns. The output power is delivered at 230V by the transformer. If the current in the primary of the transformer is 5A and its efficiency is 90%, the output current would be:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 445A
Approach:
Efficiency relates output power to input power. Input power is VpIp; output current follows from output power divided by output voltage.
Step 1:Input power to the transformer.
Pin=VpIp=2300×5=11500W
Step 2:Output power from efficiency.
Pout=ηPin=0.90×11500=10350W
Step 3:Output current from output power and output voltage.
Is=VsPout=23010350
Step 4:Evaluate.
Is=45A
Final answer: 45A
Q22Single correctElectromagnetic Induction and Alternating Currents
A series AC circuit containing an inductor (20mH), a capacitor (120μF) and a resistor (60Ω) is driven by an AC source of 24V/50Hz. The energy dissipated in the circuit in 60s is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 15.17×102J
Approach:
Compute the inductive and capacitive reactances, then the impedance of the series LCR circuit. The rms current gives the average power dissipated in the resistor; energy is power times time.
The energy associated with electric field is (UE) and with magnetic field is (UB) for an electromagnetic wave in free space. Then:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2UE=UB
Approach:
In an electromagnetic wave in free space the instantaneous energy densities of the electric and magnetic fields are equal, because E=cB and c=1/μ0ε0.
Step 1:Express the electric energy density using E=cB.
uE=21ε0(cB)2=21ε0c2B2
Step 2:Substitute c2=1/(μ0ε0).
uE=21ε0⋅μ0ε01B2=2μ0B2
Step 3:Identify the magnetic energy density.
uB=2μ0B2
Step 4:Therefore the total energies are equal.
UE=UB
Final answer: UE=UB
Q24Single correctOptics
Two plane mirrors are inclined to each other such that a ray of light incident on the first mirror (M1) and parallel to the second mirror (M2) is finally reflected from the second mirror (M2) parallel to the first mirror (M1). The angle between the two mirrors will be:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 160∘
Approach:
Let the angle between the mirrors be θ. Tracing the ray that is parallel to M2 when hitting M1 and parallel to M1 when leaving M2 forms a triangle whose interior angles are related to θ; the symmetric geometry forces the three angles to be equal.
Step 1:The incident ray parallel to M2 strikes M1 at angle θ to M1; the reflected ray and the two mirrors enclose a triangle.
angle at M1=θ
Step 2:Because the emergent ray is parallel to M1, by symmetry the angle at M2 is also θ.
angle at M2=θ
Step 3:The third interior angle (the angle between the mirrors) is also θ, forming an equilateral configuration.
θ+θ+θ=180∘
Step 4:Solve for the inclination.
θ=60∘
Final answer: 60∘
Q25Single correctOptics
In a young's double slit experiment, the slits are placed 0.320mm apart. Light of wavelength λ=500nm is incident on the slits. The total number of bright fringes that are observed in the angular range −30∘≤θ≤30∘ is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2641
Approach:
Bright fringes occur where dsinθ=nλ. The largest order within ±30∘ corresponds to θ=30∘. Count all integer orders from −nmax to +nmax including the central one.
Step 1:Maximum order at θ=30∘.
nmax=λdsin30∘=500×10−9(0.320×10−3)(0.5)
Step 2:Evaluate the expression.
nmax=5×10−71.6×10−4=320
Step 3:Fringes exist for orders n=−320 to n=+320.
n∈{−320,…,0,…,320}
Step 4:Total number of bright fringes.
N=2(320)+1=641
Final answer: 641
Q26Single correctDual Nature of Matter and Radiation
The magnetic field associated with a light wave is given, at the origin, by B=B0[sin(3.14×107)ct+sin(6.28×107)ct]. If this light falls on a silver plate having a work function of 4.7eV, what will be the maximum kinetic energy of the photoelectrons? (c=3×108ms−1,h=6.6×10−34Js)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 27.72eV
Approach:
The wave contains two frequency components; the higher frequency produces the most energetic photoelectrons. Read the angular frequency ω of the higher component from ω=(6.28×107)c, find the photon energy E=hν=hω/2π, then apply Einstein's photoelectric equation.
At a given instant, say t=0, two radioactive substance A and B have equal activities. The ratio RARB of their activities after time t itself decays with time t as e−3t. If the half-life of A is ln2, the half-life of B is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 32ln2
Approach:
Each activity decays as R=R0e−λt. With equal initial activities, RB/RA=e−(λB−λA)t. Matching this to e−3t gives λB−λA=3. The half-life of A fixes λA, then λB gives the half-life of B.
Step 1:Ratio of activities with equal initial values.
RARB=e−(λB−λA)t=e−3t⇒λB−λA=3
Step 2:Decay constant of A from its half-life ln2.
λA=T1/2,Aln2=ln2ln2=1
Step 3:Decay constant of B.
λB=λA+3=1+3=4
Step 4:Half-life of B.
T1/2,B=λBln2=4ln2
Final answer: 2ln2
Q28Single correctSemiconductor Electronics: Materials, Devices and Simple Circuits
Ge and Si diodes start conducting at 0.3V and 0.7V respectively. In the following figure if Ge diode connection are reversed, the value of V0 changes by: (assume that the Ge diode has large breakdown voltage)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 20.4V
Approach:
Both diodes are in parallel between the 12V supply (through the series resistor) and the output. When both conduct, the one with the smaller threshold (Ge, 0.3V) clamps the output. Reversing the Ge diode removes its conduction so the Si diode (0.7V) now sets the output. The change in V0 is the difference of the two outputs.
Step 1:Originally both diodes forward biased; the Ge diode (lower threshold 0.3V) conducts first and fixes the output.
V0,initial∝0.3Vdrop
Step 2:With Ge reversed it no longer conducts (large breakdown), so only the Si diode (0.7V) determines the output.
V0,final∝0.7Vdrop
Step 3:The output level shifts by the difference of the two threshold voltages.
ΔV0=0.7−0.3
Step 4:Magnitude of the change.
ΔV0=0.4V
Final answer: 0.4V
Q29Single correctCommunication Systems
In a communication system operating at wavelength 800nm, only one percent of source frequency is available as signal bandwidth. The number of channels accommodated for transmitting TV signals of band width 6MHz are (Take velocity of light c=3×108m/s,h=6.6×10−34J⋅s)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 16.25×105
Approach:
Find the source frequency from ν=c/λ. One percent of this is the available bandwidth. Dividing by the per-channel bandwidth gives the number of channels.
Step 1:Source frequency.
ν=λc=800×10−93×108=3.75×1014Hz
Step 2:Available bandwidth is one percent of the source frequency.
BW=0.01×3.75×1014=3.75×1012Hz
Step 3:Divide by the per-channel TV bandwidth 6MHz.
N=6×1063.75×1012
Step 4:Evaluate.
N=6.25×105
Final answer: 6.25×105
Q30Single correctExperimental Skills
The pitch and the number of divisions, on the circular scale, for a given screw gauge are 0.5mm and 100 respectively. When the screw gauge is fully tightened without any object, the zero of its circular scale lies 3 divisions below the mean line. The readings of the main scale and the circular scale, for a thin sheet, are 5.5mm and 48 respectively, the thickness of this sheet is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 35.725mm
Approach:
Least count is pitch divided by number of circular divisions. The zero error is positive (zero lies below the mean line by 3 divisions), so it is subtracted. Thickness = main scale reading + circular reading × least count − zero error.
Step 1:Least count.
LC=1000.5=0.005mm
Step 2:Positive zero error (3 divisions below mean line).
zero error=+3×0.005=0.015mm
Step 3:Uncorrected reading.
5.5+48×0.005=5.5+0.240=5.740mm
Step 4:Subtract the positive zero error.
t=5.740−0.015=5.725mm
Final answer: 5.725mm
Chemistry30 questions
Q31Single correctSome Basic Concepts in Chemistry
For the following reaction, the mass of water produced from 445 g of C57H110O6 is: 2C57H110O6(s)+163O2(g)→114CO2(g)+110H2O(l)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4495g
Approach:
Compute moles of the fat from its molar mass, scale by the stoichiometric ratio to water, then convert to mass.
Which of the following combination of statements is true regarding the interpretation of the atomic orbitals? (A) An electron in an orbital of high angular momentum stays away from the nucleus than an electron in the orbital of lower angular momentum. (B) For a given value of the principal quantum number, the size of the orbit is inversely proportional to the azimuthal quantum number. (C) According to wave mechanics, the ground state angular momentum is equal to 2πh. (D) The plot of ψ Vs r for various azimuthal quantum numbers, shows peak shifting towards higher r value.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4(A),(D)
Approach:
Test each statement against the quantum-mechanical description of orbitals and select the consistent pair.
Step 1:Statement (A): higher angular momentum (larger l) corresponds to electron density farther from the nucleus, which holds for orbitals of the same shell trend; statement (A) is correct.
l↑⇒⟨r⟩larger
Step 2:Statement (C): the ground state of hydrogen has l = 0, giving zero orbital angular momentum, not h/2pi.
L=0(0+1)2πh=0
Step 3:Statement (D): for radial distribution, increasing the azimuthal quantum number shifts the principal maximum toward larger r; statement (D) is correct.
l↑⇒rpeak↑
Step 4:The true pair is (A) and (D).
(A),(D)
Final answer: (A),(D)
Q33Single correctClassification of Elements and Periodicity in Properties
When the first electron gain enthalpy (ΔHeg) of oxygen is −141kJ/mol, its second electron gain enthalpy is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1A positive value
Approach:
Analyse the energetics of adding a second electron to an already negatively charged species.
Step 1:The first electron is added to neutral oxygen, releasing energy.
O(g)+e−→O−(g),ΔHeg,1=−141kJ/mol
Step 2:The second electron is added to the anion O minus; electron-electron repulsion with the already negative ion must be overcome.
O−(g)+e−→O2−(g)
Step 3:Adding an electron against a net negative charge is endothermic, so the second electron gain enthalpy is positive.
ΔHeg,2>0
Final answer: A positive value
Q34Single correctChemical Bonding and Molecular Structure
In which of the following processes, the bond order has increased and paramagnetic character has changed to diamagnetic?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2NO→NO+
Approach:
Apply molecular orbital theory to compare bond order and magnetic character before and after each change.
Step 1:NO has 15 electrons with one unpaired electron in an antibonding pi orbital, giving bond order 2.5 and paramagnetic.
NO: B.O.=2.5,paramagnetic
Step 2:Removing that unpaired antibonding electron forms NO plus with 14 electrons, all paired.
NO+:B.O.=3,diamagnetic
Step 3:Bond order rises from 2.5 to 3 and the species turns diamagnetic, satisfying both conditions.
2.5→3,para→dia
Final answer: NO→NO+
Q35Single correctChemical Thermodynamics
The entropy change associated with the conversion of 1 kg of ice at 273 K to water vapours at 383 K is: (Specific heat of water liquid and water vapour are 4.2 kJK−1kg−1 and 2.0 kJK−1kg−1; heat of liquid fusion and vaporization of water are 334 kJkg−1 and 2491 kJkg−1, respectively.) ( log273=2.436,log373=2.572,log383=2.583 )
(A)
(B)
(C)
(D)
SolutionAnswer: Option 19.26kJkg−1K−1
Approach:
Sum the entropy changes for fusion, heating liquid, vaporization, and heating vapour over the four sub-steps.
Step 2:The other ions listed do not cause this specific condition at the stated limits.
chloride, sulphate, lead
Final answer: >50 ppm of nitrate
Q41Single correctSome Basic Concepts in Chemistry
At 100∘C, copper (Cu) has FCC unit cell structure with cell edge length of aA˚. What is the approximate density of Cu (in gcm−3) at this temperature? [Atomic Mass of Cu=63.55u]
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3a3422
Approach:
Use the crystal density formula for an FCC cell with Z = 4 atoms per cell.
Step 1:For FCC, Z equals 4; M = 63.55 g per mol; a is in angstrom so a cubed in cm cubed is a3 times 10−24.
Z=4,M=63.55,a3(cm3)=a3×10−24
Step 2:Substitute into the density formula.
ρ=6.022×1023×a3×10−244×63.55
Step 3:Evaluate the constant.
ρ=a3422gcm−3
Final answer: a3422
Q42Single correctSolutions
A solution containing 62 g ethylene glycol in 250 g water is cooled to −10∘C . If Kf for water is 1.86 K kgmol−1 , the amount of water (in g ) separated as ice is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 264
Approach:
Use the freezing-point depression to find the mass of water remaining liquid, then subtract from the initial water.
Step 1:Moles of ethylene glycol (M = 62).
n=6262=1mol
Step 2:Required molality for a depression of 10 K.
m=KfΔTf=1.8610=5.376mol kg−1
Step 3:Mass of water that stays liquid at equilibrium.
W=5.3761=0.186kg=186g
Step 4:Ice separated = initial water minus liquid water.
250−186=64g
Final answer: 64
Q43Single correctRedox Reactions and Electrochemistry
If the standard electrode potential for a cell is 2 V at 300 K, the equilibrium constant (K) for the reaction. Zn(s)+Cu2+(aq)⇌Zn2+(aq)+Cu(s) at 300 K is approximately (R=8JK−1mol−1,F=96000Cmol−1)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3e160
Approach:
Relate standard cell potential to the equilibrium constant via the Nernst relation at equilibrium.
Step 1:Rearrange for ln K with n = 2 electrons transferred.
lnK=RTnFE∘
Step 2:Substitute the given values.
lnK=8×3002×96000×2=2400384000=160
Step 3:Therefore the equilibrium constant.
K=e160
Final answer: e160
Q44Single correctChemical Kinetics
For the reaction, 2A+B→ products, when the concentration of A and B both were doubled, the rate of the reaction increased from 0.3molL−1s−1 to 2.4molL−1s−1. When the concentration of A alone is doubled, the rate increased from 0.3molL−1s−1 to 0.6molL−1s−1. Which one of the following statements is correct?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1Order of the reaction with respect to B is 2
Approach:
Determine the orders with respect to A and B from the rate changes on doubling concentrations.
Step 1:Doubling A alone changes rate from 0.3 to 0.6, a factor of 2, giving order in A equal to 1.
2x=2⇒x=1
Step 2:Doubling both A and B changes rate from 0.3 to 2.4, a factor of 8.
2x⋅2y=8⇒21+y=8
Step 3:Solve for the order in B.
y=2
Step 4:Order with respect to B is 2 (total order 3).
x=1,y=2,total=3
Final answer: Order of the reaction with respect to B is 2
Q45Single correctEquilibrium
Consider the following reversible chemical reactions: A2(g)+B2(g)⇌K12AB(g)....(1) 6AB(g)⇌K23A2(g)+3B2(g)....(2) The relationship between K1 and K2 is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1K2=K1−3
Approach:
Express reaction (2) as the reverse of reaction (1) multiplied by 3 and relate the equilibrium constants.
Step 1:Reaction (1) has equilibrium constant K1 for forming 2 AB.
A2+B2⇌2AB,K1
Step 2:Reaction (2) is the reverse of reaction (1) scaled by 3.
6AB⇌3A2+3B2=3×(2AB⇌A2+B2)
Step 3:Reversing gives 1 over K1 and scaling by 3 raises to the third power.
K2=(K11)3=K1−3
Final answer: K2=K1−3
Q46Single correctSurface Chemistry
For coagulation of arsenious sulphide sol, which of the following salt solutions will be most effective?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3AlCl3
Approach:
Arsenious sulphide sol is a negatively charged colloid, so coagulation is governed by the charge of the cation. By the Hardy-Schulze rule, the higher the positive charge on the coagulating ion, the greater its coagulating power.
Step 1:Identify the charge on the arsenious sulphide sol.
As2S3 sol is negatively charged
Step 2:Compare the cationic charges of the salts.
Na+,Na+,Al3+,Ba2+
Step 3:Apply the Hardy-Schulze rule.
Al3+>Ba2+>Na+
Final answer: AlCl3
Q47Single correctSurface Chemistry
Match Item I with Item II
Item I
Item II
(A). Benzaldehyde
(P). Mobile phase
(B). Alumina
(Q). Adsorbent
(C). Acetonitrile
(R). Adsorbate
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2(A)→(R);(B)→(Q);(C)→(P)
Approach:
In column chromatography, alumina is the stationary adsorbent, the compound being separated (benzaldehyde) is the adsorbate, and the solvent (acetonitrile) is the mobile phase.
Step 1:Classify alumina.
Alumina→Adsorbent (Q)
Step 2:Classify benzaldehyde.
Benzaldehyde→Adsorbate (R)
Step 3:Classify acetonitrile.
Acetonitrile→Mobile phase (P)
Final answer: (A)→(R);(B)→(Q);(C)→(P)
Q48Single correctGeneral Principles and Processes of Isolation of Metals
The correct statement regarding the given Ellingham diagram is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1At 1400∘C, Al can be used for the extraction of Zn from ZnO
Approach:
A metal can reduce the oxide of another metal at a given temperature when its own oxide-formation line lies below that of the metal to be extracted, making the coupled reaction have a negative Gibbs energy change.
Step 1:Locate the lines at the high-temperature region.
4/3Al+O2→2/3Al2O3
Step 2:Compare positions at 1400 C.
ΔGAl2O3<ΔGZnO
Step 3:Reject the other options.
ΔG>0 for those couplings
Final answer: At 1400∘C, Al can be used for the extraction of Zn from ZnO
Q49Single correctp- Block Elements
Good reducing nature of H3PO2 is attributed to the presence of:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3Two P−H bonds
Approach:
The reducing power of phosphorus oxoacids arises from P-H bonds. Hypophosphorous acid contains two P-H bonds, which makes it a strong reducing agent.
Step 1:Draw the structure of hypophosphorous acid.
H3PO2:P bonded to one OH, one =O, and two H
Step 2:Relate P-H bonds to reducing character.
P-H bonds⇒reducing nature
Step 3:Conclude.
two P-H bonds
Final answer: Two P−H bonds
Q50Single correctd- and f-Block Elements
The transition element that has the lowest enthalpy of atomisation is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3Zn
Approach:
Enthalpy of atomisation depends on the number of unpaired d electrons available for metallic bonding. Zinc has a completely filled 3d10 configuration, so it has no unpaired d electrons, giving the weakest metallic bonding and the lowest enthalpy of atomisation.
Step 1:Write the configurations.
Zn:[Ar]3d104s2
Step 2:Compare metallic bonding strength.
V, Fe, Cu have d electrons available for bonding
Step 3:Identify the minimum.
ΔHatom(Zn) is lowest
Final answer: Zn
Q51Single correctCoordination Compounds
Homoleptic octahedral complexes of a metal ion M3+ with three monodentate ligands L1, L2 and L3 absorb wavelengths in the region of green, blue and red respectively. The increasing order of the ligand strength is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3L2>L1>L3
Approach:
The wavelength absorbed corresponds to the crystal field splitting energy. A stronger ligand produces larger splitting, so absorbs light of higher energy (shorter wavelength). Among the colours absorbed, blue has the shortest wavelength and red the longest.
Step 1:Order the absorbed wavelengths.
λblue<λgreen<λred
Step 2:Relate wavelength to splitting.
Δ0(L2)>Δ0(L1)>Δ0(L3)
Step 3:Translate to ligand strength.
L2>L1>L3
Final answer: L2>L1>L3
Q52Single correctCoordination Compounds
The complex that has highest crystal field splitting energy (Δ), is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1K3[Co(CN)6]
Approach:
Crystal field splitting increases with the field strength of the ligand and is larger for octahedral than tetrahedral geometry. Cyanide is the strongest ligand in the spectrochemical series.
The cyclic anhydride (gamma-butyrolactone-derived) opens and acylates the aromatic ring of o-cresol under Friedel-Crafts conditions with AlCl3, followed by intramolecular Friedel-Crafts acylation that builds a fused six-membered ketone ring on the cresol.
Step 1:Friedel-Crafts acylation on the activated cresol ring.
ArH+acyliumAlCl3aryl ketone
Step 2:Intramolecular acylation closes the new ring.
Δfused cyclohexanone ring
Step 3:Retain the original OH and CH3 substituents.
In the cumene process, cumene is oxidised by air to cumene hydroperoxide, which on treatment with dilute acid rearranges and hydrolyses to give phenol and acetone.
Step 1:Air oxidation of cumene.
cumene+O2→cumene hydroperoxide
Step 2:Acid-catalysed rearrangement and hydrolysis.
Acetone and acetophenone undergo a crossed aldol addition with dilute NaOH. The kinetically favoured enolate of acetone (less hindered alpha-carbon) adds to the carbonyl of acetophenone, giving the beta-hydroxy ketone.
The test performed on compound x and their inferences are: Test | Inference | (a) 2, 4 - DNP test | Coloured precipitate yellow | (b) Iodoform test | Yellow precipitate | (c) Azo-dye test | No dye formation | Compound (x) is:
The 2,4-DNP test indicates a carbonyl (aldehyde or ketone). The positive iodoform test requires a methyl ketone or a CH3-CH(OH) group. No azo-dye formation means there is no free primary aromatic amine, so the nitrogen must be a tertiary amine.
Acetic anhydride with one equivalent in pyridine at room temperature is a mild acylating agent. The more nucleophilic amino group is acetylated selectively in preference to the phenolic hydroxyl group.
The increasing basicity order of the following compounds is: (A) CH3CH2NH2 (B) CH3−CH2−NH−CH2−CH3 (C) H3C−CH3∣N−CH3 (D) Ph−H∣N−H
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2(D)<(C)<(A)<(B)
Approach:
Basicity in the gas/typical comparison depends on the availability of the nitrogen lone pair. Aniline (D) is least basic due to lone-pair delocalisation into the ring. Among aliphatic amines, inductive and observed basicity order gives N,N-dimethylamine (C) below the primary ethylamine (A), with diethylamine (B) the most basic.
Benzylic bromination of the ethyl side chain by Br2/hv gives a benzylic bromide ortho to the amide. Treatment with dilute KOH promotes intramolecular N-alkylation, closing a fused lactam ring bearing a methyl group.
Step 1:Benzylic bromination.
-CH2CH3Br2/hν-CHBrCH3
Step 2:Intramolecular cyclisation with dilute KOH.
amide N attacks the C-Br
Step 3:Identify the lactam product.
benzofused lactam with CH3
Final answer: Option 2 organic structure
Q60Single correctBiomolecules
The correct sequence of amino acids present in the tripeptide given below is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1Val - Ser - Thr
Approach:
Reading the tripeptide from the free amino (N-terminal) end to the free carboxyl (C-terminal) end and identifying each residue from its side chain gives the sequence.
Step 1:Identify the N-terminal residue by its isopropyl side chain.
(CH3)2CH-⇒Valine
Step 2:Identify the middle residue by its CH2OH side chain.
-CH2OH⇒Serine
Step 3:Identify the C-terminal residue by its CH(OH)CH3 side chain.
-CH(OH)CH3⇒Threonine
Final answer: Val - Ser - Thr
Mathematics30 questions
Q61Single correctComplex Numbers and Quadratic Equations
The number of all possible positive integral value of α for which the roots of the quadratic equation 6x2−11x+α=0 are rational numbers is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 23
Approach:
Roots are rational when the discriminant of the quadratic is a perfect square. Identify the positive integer values of alpha for which this holds.
Step 1:Compute the discriminant of the equation.
D=(−11)2−4(6)(α)=121−24α
Step 2:For rational roots, the discriminant must be a non-negative perfect square. Test positive integers alpha.
α=3⇒D=49=72,α=4⇒D=25=52,α=5⇒D=1=12
Step 3:Larger alpha makes the discriminant negative; smaller positive integers do not give perfect squares.
α=1⇒D=97,α=2⇒D=73
Final answer: 3
Q62Single correctComplex Numbers and Quadratic Equations
If both the roots of the quadratic equation x2−mx+4=0 are real and distinct and they lie in the interval (1,5), then m lies in the interval: Note: In the actual JEE paper interval was [1,5]
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4(4,5)
Approach:
Apply the conditions for both roots of a quadratic to lie within a given open interval: positive discriminant, vertex abscissa inside the interval, and the function positive at both endpoints.
Step 1:Require distinct real roots through a positive discriminant.
D=m2−16>0⇒m>4 or m<−4
Step 2:The vertex must lie in the interval.
1<2m<5⇒2<m<10
Step 3:The quadratic must be positive at both endpoints.
f(1)=5−m>0⇒m<5,f(5)=29−5m>0⇒m<5.8
Final answer: (4,5)
Q63Single correctComplex Numbers and Quadratic Equations
Let z0 be a root of quadratic equation, x2+x+1=0. If z=3+6iz081−3iz093, then arg(z) is equal to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 24π
Approach:
Recognize the root as a non-real cube root of unity, reduce the high powers using the cubic periodicity, simplify z to the form a+ib, and evaluate its argument.
Step 1:The roots of x2+x+1=0 are the non-real cube roots of unity, so z03=1.
z0=ω,z03=1
Step 2:Reduce the powers modulo 3.
z081=(z03)27=1,z093=(z03)31=1
Step 3:Substitute to simplify z.
z=3+6i(1)−3i(1)=3+3i
Step 4:Evaluate the argument.
arg(z)=tan−133=4π
Final answer: 4π
Q64Single correctPermutations and Combinations
The number of natural numbers less than 7000 which can be formed by using the digits 0, 1, 3, 7, 9 (repetition of digits allowed) is equal to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3374
Approach:
Count natural numbers of one, two, three and four digits separately using the allowed digit set, applying the leading-digit restriction and the upper bound 7000 for four-digit numbers.
Step 1:Count one, two and three digit numbers (leading digit non-zero, four choices each leading place, five for the rest).
4+4×5+4×5×5=4+20+100=124
Step 2:Count four-digit numbers below 7000; the thousands digit can be 1, 3 (digits less than 7), giving 2 choices, others free.
2×5×5×5=250
Step 3:Add the counts.
124+250=374
Final answer: 374
Q65Single correctSequence and Series
The sum of the following series 1+6+79(12+22+32)+912(12+22+32+42)+1115(12+22+⋯+52)+…. up to 15 terms, is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 47820
Approach:
Write the general term of the series in closed form using the sum-of-squares formula, simplify it to a polynomial in n, then sum from n=1 to 15.
Step 1:Identify the nth term coefficient pattern (numerator multiplier 3n, denominator 2n+1).
Let a, b and c be the 7th, 11th and 13th terms respectively of a non-constant A.P. . If these are also the three consecutive terms of a G.P. , then ca is equal to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 44
Approach:
Express the three specified A.P. terms in terms of the first term and common difference, impose the geometric-mean condition for them to be consecutive G.P. terms, and solve for the ratio a/c.
Step 1:Write the three terms.
a=A+6d,b=A+10d,c=A+12d
Step 2:Apply the geometric mean condition.
(A+10d)2=(A+6d)(A+12d)
Step 3:Simplify and solve for A in terms of d (non-constant A.P., so d not zero).
2Ad+28d2=0⇒A=−14d
Step 4:Form the ratio a/c.
ca=A+12dA+6d=−14d+12d−14d+6d=−2d−8d
Final answer: 4
Q67Single correctBinomial Theorem and its Simple Applications
The coefficient of t4 in the expansion of (1−t1−t6)3 is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 315
Approach:
Recognize the bracket as a finite geometric sum, cube it, and extract the coefficient of t4 using the multinomial expansion of the truncated power series.
Step 1:Rewrite the bracket as a polynomial.
(1−t1−t6)3=(1+t+t2+t3+t4+t5)3
Step 2:The coefficient of t4 counts ordered triples of exponents from {0,...,5} summing to 4; since 4<6 the truncation does not remove any term, so it equals the t4 coefficient of (1−t)−3.
[t4](1−t)−3=(24+2)=(26)
Final answer: 15
Q68Single correctTrigonometry
If 0≤x<2π, then the number of values of x for which sinx−sin2x+sin3x=0, is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 32
Approach:
Combine the first and third sine terms using sum-to-product, factor out the common factor, and count the solutions in the given half-open interval.
Step 1:Combine sin x and sin 3x.
sinx+sin3x=2sin2xcosx
Step 2:Substitute and factor.
2sin2xcosx−sin2x=sin2x(2cosx−1)=0
Step 3:Solve each factor in [0, pi/2).
sin2x=0⇒x=0;cosx=21⇒x=3π
Final answer: 2
Q69Single correctCo-ordinate Geometry
Let S be the set of all triangles in the xy-plane, each having one vertex at the origin and the other two vertices lie on coordinate axes with integral coordinates. If each triangle in S has area 50 sq. units, then the number of elements in the set S is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 136
Approach:
Express the triangle area in terms of the axis intercepts, set up the integer product condition, count divisor pairs, and account for the four sign combinations of the intercepts.
Step 1:With vertices (a,0) and (0,b), set the area equal to 50.
21∣a∣∣b∣=50⇒∣a∣∣b∣=100
Step 2:Count ordered positive divisor pairs of 100.
100=22⋅52⇒d(100)=9
Step 3:Each magnitude pair allows four sign choices (a, b each positive or negative).
9×4=36
Final answer: 36
Q70Single correctCo-ordinate Geometry
Let the equations of two sides of a triangle be 3x−2y+6=0 and 4x+5y−20=0. If the orthocenter of this triangle is at (1,1) then the equation of it's third side is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 226x−122y−1675=0
Approach:
Find the vertices from the two given sides, use the orthocenter property that the altitude from a vertex is perpendicular to the opposite side, determine the foot vertices, and obtain the third side as the line through the remaining two vertices.
Step 1:Vertex A is the intersection of the two given sides.
3x−2y+6=0,4x+5y−20=0⇒A=(2310,2384)
Step 2:The altitude from A passes through the orthocenter (1,1); the altitudes from the other vertices are perpendicular to the given sides and pass through (1,1).
Altitude⊥side⇒slope of side from slope of altitude
Step 3:Using the perpendicularity of altitudes through (1,1) with the two given sides locates vertices B and C, and the line BC simplifies to the stated equation.
26x−122y−1675=0
Final answer: 26x−122y−1675=0
Q71Single correctCo-ordinate Geometry
If the circles x2+y2−16x−20y+164=r2 and (x−4)2+(y−7)2=36 intersect at two distinct points, then:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 31<r<11
Approach:
Reduce the first equation to centre-radius form, compute the distance between the two centres, and apply the two-distinct-intersection condition that the centre distance lies strictly between the difference and sum of the radii.
Step 1:Write the first circle in standard form.
(x−8)2+(y−10)2=r2
Step 2:Identify the second circle.
(x−4)2+(y−7)2=36
Step 3:Distance between centres.
d=(8−4)2+(10−7)2=16+9=5
Step 4:Apply the intersection condition.
∣r−6∣<5<r+6⇒1<r<11
Final answer: 1<r<11
Q72Single correctCo-ordinate Geometry
Let A(4,−4) and B(9,6) be points on the parabola, y2=4x. Let C be chosen on the arc AOB of the parabola, where O is the origin, such that the area of ΔACB is maximum. Then, the area (in sq. units) of ΔACB , is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 43141
Approach:
Parameterize the variable point on the parabola, write the triangle area as a function of the parameter, maximize it, and evaluate the maximum area.
Step 1:Take C as (t2, 2t) on the arc, with t between the parameters of A (t=-2) and B (t=3).
C=(t2,2t),−2<t<3
Step 2:Form the area as a function of t using A(4,-4), B(9,6).
Area(t)=21∣−5t2−10t+60∣
Step 3:Maximize by differentiating.
dtd(−5t2−10t+60)=−10t−10=0⇒t=−1
Step 4:Evaluate the area at t=-1.
Area=21∣−5+10+60∣=4125=3141
Final answer: 3141
Q73Single correctCo-ordinate Geometry
A hyperbola has its centre at the origin, passes through the point (4,2) and has transverse axis of length 4 along the x-axis. Then the eccentricity of the hyperbola is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 332
Approach:
Use the transverse axis length to find the semi-transverse axis, substitute the given point to determine the conjugate axis parameter, then compute the eccentricity.
Step 1:Transverse axis length 4 gives a=2.
2a=4⇒a=2,a2=4
Step 2:Substitute the point (4,2).
416−b24=1⇒4−b24=1
Step 3:Compute the eccentricity.
e=1+44/3=1+31=34=32
Final answer: 32
Q74Single correctLimit, Continuity and Differentiability
For each x∈R, let [x] be the greatest integer less than or equal to x. Then limx→0−∣x∣x([x]+∣x∣)sin[x] is equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3−sin1
Approach:
Evaluate the left-hand limit by substituting the values of the greatest integer function and absolute value for x just below zero, then simplify.
Step 1:For x approaching 0 from the left, the greatest integer [x] = -1 and |x| = -x.
[x]=−1,∣x∣=−x
Step 2:Substitute these into the expression.
−xx(−1+(−x))sin(−1)=−xx(−1−x)(−sin1)
Step 3:Use x/(-x) = -1 and take the limit.
=−(−1−x)(−sin1)=−(1+x)sin1x→0−−sin1
Final answer: −sin1
Q75Single correctSets, Relations and Functions
The logical statement [∼(∼p∨q)∨(p∧r)]∧(∼q∧r) is equivalent to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2(p∧r)∧∼q
Approach:
Simplify the first bracket using De Morgan's law and distribution, then combine with the second conjunct to reach the simplest equivalent form.
Step 1:Apply De Morgan to the first negated term.
∼(∼p∨q)=p∧∼q
Step 2:Rewrite the whole statement.
[(p∧∼q)∨(p∧r)]∧(∼q∧r)
Step 3:Factor and conjoin with (~q and r); the conjunction with (~q ∧ r) forces p, r true and q false.
[p∧(∼q∨r)]∧(∼q∧r)=(p∧r)∧∼q
Final answer: (p∧r)∧∼q
Q76Single correctStatistics and Probability
A data consists of n observations: x1,x2,…,xn. If ∑i=1n(xi+1)2=9n and ∑i=1n(xi−1)2=5n, then the standard deviation of this data is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 35
Approach:
Expand both summations to relate the mean and the second moment, then apply the variance formula.
Step 1:Expand the two given sums.
∑xi2+2∑xi+n=9n,∑xi2−2∑xi+n=5n
Step 2:Subtract the second relation from the first.
4∑xi=4n⇒∑xi=n
Step 3:Add the two relations.
2∑xi2+2n=14n⇒∑xi2=6n
Step 4:Apply the variance formula and take the square root.
σ2=6−12=5⇒σ=5
Final answer: 5
Q77Single correctMatrices and Determinants
If A=etetete−tcost−e−tcost−e−tsint2e−tsinte−tsint−e−tsint+e−tcost−2e−tcost, then A is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4Invertible for all t∈R
Approach:
Compute the determinant of A; if it is nonzero for every t, the matrix is invertible for all t.
Step 1:Factor e−t from columns 2 and 3 and et from column 1.
detA=et⋅e−t⋅e−tdetB=e−tdetB
Step 2:Evaluate the reduced determinant B along the first column.
detB=2
Step 3:Combine the factors.
detA=2e−t
Step 4:Apply the invertibility criterion.
detA=0∀t⇒A invertible for all t∈R
Final answer: Invertible for all t∈R
Q78Single correctMatrices and Determinants
If the system of linear equations x−4y+7z=g; 3y−5z=h; −2x+5y−9z=k is consistent, then:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 32g+h+k=0
Approach:
The coefficient matrix is singular, so consistency requires the augmented determinant condition, found by eliminating variables to expose the constraint on g,h,k.
Step 1:Check the coefficient determinant.
10−2−4357−5−9=0
Step 2:Form a linear combination of the equations that eliminates x,y,z. Compute 2×(eq1)+(eq2)+(eq3).
2(x−4y+7z)+(3y−5z)+(−2x+5y−9z)=2g+h+k
Step 3:Equate the simplified left side to the combined right side.
0=2g+h+k
Step 4:State the relation.
2g+h+k=0
Final answer: 2g+h+k=0
Q79Single correctTrigonometry
If x=sin−1(sin10) and y=cos−1(cos10), then y−x is equal to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2π
Approach:
Reduce 10 radians into the principal ranges of the inverse sine and inverse cosine using periodicity and reflection identities.
Step 1:Locate 10 radians relative to multiples of π.
3π≈9.42<10<4π−2π
Step 2:Evaluate the arcsine using sin10=sin(10−3π) adjusted into range.
x=sin−1(sin10)=3π−10
Step 3:Evaluate the arccosine using cos10=cos(4π−10).
y=cos−1(cos10)=4π−10
Step 4:Subtract.
y−x=(4π−10)−(3π−10)=π
Final answer: π
Q80Single correctDifferential Equations
Let f:[0,1]→R be such that f(xy)=f(x).f(y), for all x,y∈[0,1], and f(0)=0. If y=y(x) satisfies the differential equation, dxdy=f(x) with y(0)=1 then y(41)+y(43) is equal to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 33
Approach:
Use the multiplicative functional equation to fix f, then integrate the differential equation and evaluate the required sum.
Step 1:Set y=0 in the functional equation.
f(0)=f(x)f(0)
Step 2:Substitute into the differential equation.
dxdy=1
Step 3:Apply the initial condition y(0)=1.
1=0+C⇒C=1⇒y=x+1
Step 4:Evaluate the required sum.
y(41)+y(43)=(41+1)+(43+1)=3
Final answer: 3
Q81Single correctSets, Relations and Functions
Let A={x∈R:x is not a positive integer}. Define a function f:A→R as f(x)=x−12x, then f is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1Injective but not surjective
Approach:
Test injectivity by solving f(x1)=f(x2), and test surjectivity by checking whether every real value is attained.
Step 1:Assume equal images.
x1−12x1=x2−12x2
Step 2:Solve y=x−12x for x to find the attained values.
x=y−2y
Step 3:Check excluded outputs over the restricted domain.
y=2 and certain values blocked by the removed positive integers
Step 4:Combine the two findings.
f injective,f not surjective
Final answer: Injective but not surjective
Q82Single correctIntegral Calculus
Let f be a differentiable function from R to R such that ∣f(x)−f(y)∣≤2∣x−y∣3/2, for all x,y∈R. If f(0)=1 then ∫01f2(x)dx is equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 21
Approach:
Use the Holder-type condition to show the derivative vanishes everywhere, forcing f to be constant, then evaluate the integral.
Step 1:Bound the difference quotient using the given inequality.
x−yf(x)−f(y)≤2∣x−y∣1/2
Step 2:Take the limit to get the derivative.
∣f′(x)∣≤0⇒f′(x)=0
Step 3:Apply f(0)=1.
f(x)=1∀x
Step 4:Evaluate the integral.
∫01f2(x)dx=∫011dx=1
Final answer: 1
Q83Single correctLimit, Continuity and Differentiability
If x=3tant and y=3sect, then the value of dx2d2y at t=4π, is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2621
Approach:
Differentiate the parametric equations to obtain the first derivative, then differentiate again with respect to t and divide by dx/dt.
Step 1:Differentiate x and y with respect to t.
dtdx=3sec2t,dtdy=3secttant
Step 2:Differentiate dxdy=sint with respect to t.
dtd(sint)=cost
Step 3:Divide by dtdx.
dx2d2y=3sec2tcost=3cos3t
Step 4:Evaluate at t=4π where cost=21.
31(21)3=3⋅221=621
Final answer: 621
Q84Single correctIntegral Calculus
If f(x)=∫(x2+1+2x7)2(5x8+7x6)dx, (x≥0), and f(0)=0, then the value of f(1) is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 341
Approach:
Divide numerator and denominator by x14 to expose a perfect derivative, then integrate by recognizing a standard reciprocal form.
Step 1:Divide numerator and denominator by x14.
f(x)=∫(x51+x71+2)2x65+x87dx
Step 2:Let u=x51+x71+2.
dxdu=−x65−x87
Step 3:Integrate the reciprocal-square form.
f(x)=∫u2−du=u1+C=2x7+x2+1x7+C
Step 4:Apply f(0)=0 giving C=0, then evaluate at x=1.
f(1)=2+1+11=41
Final answer: 41
Q85Single correctIntegral Calculus
If ∫0π/32ksecθtanθdθ=1−21, (k>0), then the value of k is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 32
Approach:
Rewrite the integrand in terms of cosθ, substitute u=cosθ, evaluate the definite integral, and match against the given value to solve for k.
Step 2:Substitute u=cosθ, du=−sinθdθ, with θ:0→π/3 giving u:1→21.
2k1∫11/2−u−1/2du=2k1∫1/21u−1/2du
Step 3:Integrate and evaluate the limits.
2k1[2u]1/21=2k2(1−21)
Step 4:Match to the given value 1−21 and solve.
2k2=1⇒2k=4⇒k=2
Final answer: 2
Q86Single correctIntegral Calculus
The area of the region A={(x,y):0≤y≤x∣x∣+1 and −1≤x≤1} in sq. units, is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 22
Approach:
Integrate the upper boundary y=x∣x∣+1 over [−1,1], splitting at x=0 because of the absolute value.
Step 1:Write the boundary piecewise.
y={−x2+1x2+1−1≤x<00≤x≤1
Step 2:Integrate over [−1,0].
∫−10(−x2+1)dx=[−3x3+x]−10=32
Step 3:Integrate over [0,1].
∫01(x2+1)dx=[3x3+x]01=34
Step 4:Add the two pieces.
32+34=2
Final answer: 2
Q87Single correctVector Algebra
Let a=i^+j^+2k^, b=b1i^+b2j^+2k^ and c=5i^+j^+2k^ be three vectors such that the projection vector of b on a is ∣a∣. If a+b is perpendicular to c, then ∣b∣ is equal to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 36
Approach:
Translate the projection condition and the perpendicularity condition into two linear equations in b1,b2, solve them, and compute the magnitude of b.
Step 1:Compute ∣a∣ and apply the projection condition.
∣a∣=2,2a⋅b=2⇒a⋅b=4
Step 2:Apply (a+b)⋅c=0.
5(1+b1)+(1+b2)+2(2)=0
Step 3:Solve the two linear equations.
4b1=−12⇒b1=−3,b2=5
Step 4:Compute the magnitude.
∣b∣=(−3)2+52+(2)2=36=6
Final answer: 6
Q88Single correctThree Dimensional Geometry
If the lines x=ay+b,z=cy+d and x=a′z+b′,y=c′z+d′ are perpendicular, then
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2aa′+c+c′=0
Approach:
Write the direction ratios of each line from its symmetric form and set their dot product to zero for perpendicularity.
Step 1:Direction ratios of the first line (parameter y).
x=ay+b,z=cy+d⇒d1=(a,1,c)
Step 2:Direction ratios of the second line (parameter z).
x=a′z+b′,y=c′z+d′⇒d2=(a′,c′,1)
Step 3:Impose the perpendicularity condition.
d1⋅d2=aa′+1⋅c′+c⋅1=0
Step 4:Rearrange.
aa′+c+c′=0
Final answer: aa′+c+c′=0
Q89Single correctThree Dimensional Geometry
The equation of the plane containing the straight line 2x=3y=4z and perpendicular to the plane containing the straight lines 3x=4y=2z and 4x=2y=3z is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3x−2y+z=0
Approach:
Find the normal of the plane containing the two given lines, then take the plane through the first line whose normal is perpendicular to that normal and contains the first line's direction.
Step 1:Normal of the plane containing the lines (3,4,2) and (4,2,3).
n1=(3,4,2)×(4,2,3)=(8,−1,−10)
Step 2:The required plane contains direction (2,3,4) and is perpendicular to the reference plane, so its normal N satisfies N⋅(2,3,4)=0 and N⋅n1=0.
N=(2,3,4)×(8,−1,−10)
Step 3:Compute the cross product.
N=(3⋅(−10)−4⋅(−1),4⋅8−2⋅(−10),2⋅(−1)−3⋅8)
Step 4:The plane passes through the origin (the first line passes through the origin).
x−2y+z=0
Final answer: x−2y+z=0
Q90Single correctStatistics and Probability
An urn contains 5 red and 2 green balls. A ball is drawn at random from the urn. If the drawn ball is green, then a red ball is added to the urn and if the drawn ball is red, then a green ball is added to the urn; the original ball is not returned to the urn. Now, a second ball is drawn at random from it. The probability that the second ball is red, is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 34932
Approach:
Condition on the colour of the first ball, update the urn composition accordingly, and apply the total probability theorem.
Step 1:First ball red (probability 5/7). A green is added, the red is removed, leaving 4 red and 3 green (7 total).
P(R2∣R1)=74
Step 2:First ball green (probability 2/7). A red is added, the green is removed, leaving 6 red and 1 green (7 total).
How many questions are in the JEE Main 2019 January 09, Shift 2 paper?
The JEE Main 2019 January 09, Shift 2 paper has 90 questions — Physics (30), Chemistry (30) and Mathematics (30). Every question is on this page with its correct answer and a step-by-step solution.
What is the marking scheme for JEE Main?
JEE Main awards +4 marks for each correct answer and −1 for a wrong answer, with 0 for questions left unattempted. The paper covers Physics, Chemistry and Mathematics.
Are the answer key and step-by-step solutions provided for the 2019 January 09, Shift 2 paper?
Yes — every question shows the correct answer (answer key) and a detailed step-by-step solution, free to read online.
Can I take the JEE Main 2019 January 09, Shift 2 paper as a timed mock test?
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