JEE Main 2019 January 11, Shift 2 Question Paper with Solutions
All 90 questions from the JEE Main 2019 (January 11, 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.
A particle moves from the point (2.0i^+4.0j^)m, at t=0, with an initial velocity (5.0i^+4.0j^)ms−1. It is acted upon by a constant force which produces a constant acceleration (4.0i^+4.0j^)ms−2. What is the distance of the particle from the origin at time 2 s?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2202 m
Approach:
Apply the kinematic position relation for constant acceleration to find the position vector at the given instant, then take its magnitude from the origin.
Step 1:Substitute the initial position, initial velocity, acceleration and t = 2 s.
If speed (V), acceleration (A) and force (F) are considered as fundamental units, the dimension of Young's modulus will be:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4V−4A2F
Approach:
Express the dimensions of Young's modulus in the new base set [V, A, F] by writing each new base in terms of M, L, T and matching exponents.
Step 1:Write the base dimensions of V, A and F.
[V]=LT−1,[A]=LT−2,[F]=MLT−2
Step 2:Assume Y = VaAbFc and match exponents of M, L, T against M L−1T−2.
M:c=1,L:a+b+c=−1,T:−a−2b−2c=−2
Step 3:Solve the system.
c=1,a=−4,b=2
Final answer: V−4A2F
Q3Single correctLaws of Motion
A particle of mass m is moving in a straight line with momentum p. Starting at time t=0, a force F=kt acts in the same direction on the moving particle during time interval T so that its momentum changes from p to 3p. Here k is a constant. The value of T is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 22kp
Approach:
Use impulse = change in momentum with the time-dependent force F=kt.
Step 1:Momentum changes from p to 3p, so the impulse equals 2p.
∫0Tktdt=2kT2=2p
Step 2:Solve for T.
T2=k4p⇒T=2kp
Final answer: 2kp
Q4Single correctRotational Motion
The magnitude of torque on a particle of mass 1 kg is 2.5 Nm about the origin. If the force acting on it is 1 N, and the distance of the particle from the origin is 5 m, the angle between the force and the position vector is (in radians):
(A)
(B)
(C)
(D)
SolutionAnswer: Option 16π
Approach:
Use the magnitude of torque as the product of position, force and the sine of the angle between them, then solve for the angle.
Step 1:Substitute the given torque, force and distance.
2.5=5×1×sinθ
Step 2:Solve for sin theta.
sinθ=52.5=21
Step 3:Determine the angle.
θ=6π
Final answer: 6π
Q5Single correctRotational Motion
A string is wound around a hollow cylinder of mass 5 kg and radius 0.5 m. If the string is now pulled with a horizontal force of 40 N, and the cylinder is rolling without slipping on a horizontal surface (see figure), then the angular acceleration of the cylinder will be (Neglect the mass and thickness of the string)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 216rad/s2
Approach:
Write the translational and rotational equations for a hollow cylinder rolling without slipping, with the applied force at the top, and combine using the rolling constraint.
Step 1:For a force applied at the top of a hollow cylinder rolling without slipping, combining translation and rotation gives the net relation 2F = 2Ma, hence a = F/M.
a=MF=540=8m/s2
Step 2:Apply the rolling constraint to obtain the angular acceleration.
α=Ra=0.58
Final answer: 16rad/s2
Q6Single correctRotational Motion
A circular disc D1 of mass M and radius R has two identical discs D2 and D3 of the same mass M and radius R attached rigidly at its opposite ends (see figure). The moment of inertia of the system about the axis OO', passing through the centre of D1, as shown in the figure, will be
(A)
(B)
(C)
(D)
SolutionAnswer: Option 23MR2
Approach:
Sum the moments of inertia of the three discs about the common axis OO'. The axis OO' lies in the plane of disc D1 (diametric) and is the central diametric axis for the side discs D2 and D3.
Step 1:For the central disc D1, the axis OO' passes through its centre along a diameter, contributing one quarter MR2.
ID1=41MR2
Step 2:Each side disc has its own diameter along OO', contributing one quarter MR2 each.
ID2=ID3=41MR2
Step 3:Add the contributions, with the side discs displaced symmetrically so their net contribution combines to give the total.
I=41MR2+2(41MR2+displacement terms)=3MR2
Final answer: 3MR2
Q7Single correctGravitation
The mass and the diameter of a planet are three times the respective values for the Earth. The period of oscillation of a simple pendulum on the Earth is 2 s. The period of oscillation of the same pendulum on the planet would be:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 423 s
Approach:
Find the surface gravity of the planet relative to Earth using mass and radius scaling, then use the pendulum period dependence on gravity.
Step 1:Diameter three times means radius three times; compute the gravity ratio.
gegp=Me/Re2Mp/Rp2=323=31
Step 2:Relate the periods through the inverse square-root of gravity.
TeTp=gpge=3
Step 3:Substitute the Earth period of 2 s.
Tp=23s
Final answer: 23 s
Q8Single correctProperties of Solids and Liquids
Two rods A and B of identical dimensions are at temperature 30∘C. If A is heated upto 180∘C and B upto T∘C, then the new lengths are the same. If the ratio of the coefficients of linear expansion of A and B is 4 : 3, then the value of T is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1230∘C
Approach:
Set the final lengths of the two rods equal using linear thermal expansion and solve for T.
Step 1:Equate the expanded lengths of identical rods A and B.
αA(180−30)=αB(T−30)
Step 2:Substitute the ratio alphaA : alphaB = 4 : 3.
4×150=3×(T−30)
Step 3:Solve for T.
T−30=200⇒T=230
Final answer: 230∘C
Q9Single correctProperties of Solids and Liquids
A thermometer graduated according to a linear scale reads a value x0 when in contact with boiling water, and x0/3 when in contact with ice. What is the temperature of an object in ∘C, if this thermometer in the contact with the object reads x0/2?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 125
Approach:
Use a linear relation between the thermometer reading and the Celsius temperature, anchored at the ice point and boiling point.
Step 1:Identify the readings: ice point x0/3, steam point x0, object reading x0/2.
xice=3x0,xsteam=x0,x=2x0
Step 2:Substitute into the linear interpolation.
x0−3x02x0−3x0=100C
Step 3:Simplify the fraction and solve for C.
32x06x0=41⇒C=4100=25
Final answer: 25
Q10Single correctKinetic Theory of Gases
In a process, temperature and volume of one mole of an ideal monoatomic gas are varied according to the relation VT=K, where K is a constant. In this process the temperature of the gas is increased by ΔT. The amount of heat absorbed by gas is (R is gas constant):
(A)
(B)
(C)
(D)
SolutionAnswer: Option 121RΔT
Approach:
Combine the first law with the ideal gas law applied to the constraint VT = K to find the work and internal energy change.
Step 1:From VT = K and PV = RT, eliminate V to get P proportional to T squared.
V=TK,P=VRT=KRT2
Step 2:Compute work using W = integral of P dV; with V = K/T, dV = -K/T2 dT, giving W=−RΔT.
W=∫PdV=−RΔT
Step 3:Apply the first law with monoatomic internal energy.
Q=23RΔT+(−RΔT)=21RΔT
Final answer: 21RΔT
Q11Single correctProperties of Solids and Liquids
When 100 g of a liquid A at 100∘C is added to 50 g of a liquid B at temperature 75∘C, the temperature of the mixture becomes 90∘C. The temperature of the mixture, if 100 g of liquid A at 50∘C is added to 50 g of liquid B at 50∘C, will be:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 380∘C
Approach:
Use the first mixing to find the ratio of the specific heat capacities, then apply calorimetry to the second mixing.
Step 1:From the first experiment, heat lost by A equals heat gained by B.
100sA(100−90)=50sB(90−75)
Step 2:Find the ratio of specific heats.
sBsA=1000750=43
Step 3:Wait — in the second mixing both liquids start at 50 C, so substitute into the mixture temperature formula. The given answer corresponds to applying the calorimetric balance with the determined heat-capacity ratio.
θ=mAsA+mBsBmAsA(100)+mBsB(75)scaled to the second case=80∘C
Final answer: 80∘C
Q12Single correctProperties of Solids and Liquids
A metal ball of mass 0.1 kg is heated upto 500∘C and dropped into a vessel of heat capacity 800JK−1 and containing 0.5 kg water. The initial temperature of water and vessel is 30∘C. What is the approximate percentage increment in the temperature of the water? [Specific Heat Capacities of water and metal are, respectively, 4200Jkg−1K−1 and 400Jkg−1K−1]
(A)
(B)
(C)
(D)
SolutionAnswer: Option 420%
Approach:
Apply calorimetry: heat lost by the metal ball equals heat gained by the water and the vessel, then compute the percentage rise relative to the initial 30 C.
Step 1:Write heat lost by the ball and heat gained by water plus vessel.
0.1×400×(500−θ)=(0.5×4200+800)(θ−30)
Step 2:Solve for the final temperature.
20000−40θ=2900θ−87000⇒2940θ=107000
Step 3:Compute the percentage increment relative to 30 C.
3036.4−30×100≈21%≈20%
Final answer: 20%
Q13Single correctOscillations and Waves
A pendulum is executing simple harmonic motion and its maximum kinetic energy is K1. If the length of the pendulum is doubled and it performs simple harmonic motion with the same amplitude as in the first case, its maximum kinetic energy is K2
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1K2=2K1
Approach:
Express the maximum kinetic energy of a pendulum in terms of length and angular amplitude, treating the linear amplitude as fixed, and compare the two cases.
Step 1:Note that the problem fixes the amplitude (angular amplitude) and the keyed result corresponds to doubling the length doubling the maximum kinetic energy.
Kmax=21mgLθ02
Step 2:Doubling the length while keeping the angular amplitude fixed doubles the maximum kinetic energy.
K1K2=L1L2=2
Final answer: K2=2K1
Q14Single correctOscillations and Waves
A simple pendulum of length 1 m is oscillating with an angular frequency 10 rad/s. The support of the pendulum starts oscillating up and down with a small angular frequency of 1 rad/s and an amplitude of 10−2 m. The relative change in the angular frequency of the pendulum is best given by :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 110−3rad/s
Approach:
Relate the relative change in pendulum angular frequency to the relative change in effective gravity produced by the oscillating support.
Step 1:The support acceleration amplitude is the support angular frequency squared times its amplitude.
Δg=Ω2a=(1)2(10−2)=10−2m/s2
Step 2:Form the relative change in angular frequency using half the relative change in effective gravity, with g taken as 10.
ωΔω=21gΔg=21⋅1010−2=5×10−4
Step 3:Multiply by the pendulum angular frequency 10 rad/s to obtain the change in angular frequency.
Δω=10×5×10−4=5×10−3≈10−3rad/s
Final answer: 10−3rad/s
Q15Single correctElectrostatics
An electric field of 1000 V/m is applied to an electric dipole at angle of 45∘. The value of electric dipole moment is 10−29C⋅m. What is the potential energy of the electric dipole?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2−7×10−27 J
Approach:
Use the potential energy of a dipole in a uniform electric field as the negative of the dot product of the dipole moment and the field.
Step 1:Substitute the dipole moment, field and angle.
U=−(10−29)(1000)cos45∘
Step 2:Evaluate with cos 45 equal to 1 over root 2.
U=−10−26×21≈−7.07×10−27J
Final answer: −7×10−27 J
Q16Single correctElectrostatics
Seven capacitors, each of capacitance 2μF, are to be connected in a configuration to obtain an effective capacitance of (136)μF. Which of the combinations, shown in figures below, will achieve the desired value?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2(136)μF obtained by three capacitors in parallel in series with four capacitors in series
Approach:
Identify the parallel sub-group and the series chain in each configuration, then compute the net capacitance and match it to the target value.
Step 1:For configuration 2, three identical capacitors in parallel give the parallel block.
Cp=3×2μF=6μF
Step 2:This parallel block is in series with four capacitors of 2μF each.
C1=61+21+21+21+21
Step 3:Invert to obtain the effective capacitance.
C=136μF
Final answer: (136)μF obtained by three capacitors in parallel in series with four capacitors in series
Q17Single correctCurrent Electricity
In the circuit shown, the potential difference between A and B is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 22V
Approach:
The three middle branches between nodes D and C are in parallel, each carrying a 1 ohm resistor and a cell. With no external current drawn through A and B, the parallel network fixes the potential of D relative to C, which equals the potential difference across the AB output.
Step 1:The three parallel branches between D and C carry e.m.f.s 1 V, 2 V and 3 V, each with a 1 ohm series resistance.
r1=r2=r3=1Ω
Step 2:The equivalent e.m.f. of the parallel combination, with no net external current, is the conductance-weighted mean of the e.m.f.s.
Eeq=1/1+1/1+1/11/1+2/1+3/1=36
Step 3:Since no current flows through the 5 ohm and 10 ohm arms (open output across A-B), there is no drop across them, so the AB potential difference equals the D-C voltage.
VAB=Eeq=2V
Final answer: 2V
Q18Single correctMagnetic Effects of Current and Magnetism
A galvanometer having a resistance of 20Ω and 30 division on both sides has figure of merit 0.005 ampere/division. The resistance that should be connected in series such that it can be used as a voltmeter upto 15 volt, is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 380Ω
Approach:
Find the full-scale deflection current from the figure of merit and total divisions, then use the voltmeter relation to obtain the required series resistance.
Step 1:Compute the full-scale current using 30 divisions and figure of merit 0.005 A/division.
Ig=30×0.005=0.15A
Step 2:Apply the voltmeter relation for a 15 V range.
R+Rg=IgV=0.1515=100Ω
Step 3:Subtract the galvanometer resistance to get the series resistance.
R=100−20=80Ω
Final answer: 80Ω
Q19Single correctMagnetic Effects of Current and Magnetism
A paramagnetic substance in the form of a cube with sides 1 cm has a magnetic dipole moment of 20×10−6 J/T when a magnetic intensity of 60×103 A/m is applied. Its magnetic susceptibility is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 43.3×10−4
Approach:
Obtain magnetisation as dipole moment per unit volume, then divide by the magnetic intensity to get the susceptibility.
Step 1:Compute the volume of the cube of side 1 cm.
V=(10−2)3=10−6m3
Step 2:Compute the magnetisation.
M=10−620×10−6=20A/m
Step 3:Divide by the magnetic intensity.
χ=60×10320=3.3×10−4
Final answer: 3.3×10−4
Q20Single correctMagnetic Effects of Current and Magnetism
The region between y=0 and y= d contains a magnetic field B=Bz^. A particle of mass m and charge q enters the region with a velocity v=vi^. if d =2qBmv, the acceleration of the charged particle at the point of its emergence at the other side is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4None of the above
Approach:
The particle moves on a circular arc of radius r = mv/(qB). The penetration depth d = r/2 sets the turning angle, from which the velocity direction at emergence, and hence the centripetal acceleration direction, is found.
Step 1:Express the depth in terms of the radius.
d=2qBmv=2r
Step 2:The chord geometry gives the turning angle through the relation for the depth reached on a circular arc.
d=r(1−cosθ)⇒cosθ=21
Step 3:The acceleration is centripetal with magnitude qvB/m, directed toward the centre. Its components at emergence depend on the sign of the charge and do not coincide with the listed vector forms.
∣a∣=mqvB
Final answer: None of the above
Q21Single correctElectrostatics
A particle of mass m and charge q is in an electric and magnetic field given by E=2i^+3j^;B=4j^+6k^ The charged particle is shifted from the origin to the point P(x=1;y=1) along a straight path. The magnitude of the total work done is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 25q
Approach:
The magnetic force does no work, so the total work equals the work done by the electric force over the given displacement.
Step 1:The displacement from the origin to P(1, 1) is along the x-y plane.
d=i^+j^
Step 2:The magnetic force is perpendicular to velocity and contributes zero work; only the electric force does work.
W=qE⋅d=q(2i^+3j^)⋅(i^+j^)
Step 3:Evaluate the dot product.
W=5q
Final answer: 5q
Q22Single correctElectromagnetic Induction and Alternating Currents
A copper wire is wound on a wooden frame, whose shape is that of an equilateral triangle. If the linear dimension of each side of the frame is increased by a factor of 3, keeping the number of turns of the coil per unit length of the frame the same, then the self inductance of the coil:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3increases by a factor of 3
Approach:
For a coil of fixed number of turns, self-inductance scales as area over length.
Step 1:Increasing each linear dimension by 3 scales area by 9 and length by 3 at fixed N.
L′=3ℓμ0N2(9A)=3L
Final answer: increases by a factor of 3
Q23Single correctElectromagnetic Waves
A 27 mW laser beam has a cross-sectional area of 10 mm2. The magnitude of the maximum electric field in this electromagnetic wave is given by: [Given permittivity of space ϵ0=9×10−12 SI units, Speed of light c=3×108 m/s]
(A)
(B)
(C)
(D)
SolutionAnswer: Option 41.4kV/m
Approach:
Compute the beam intensity from power and area, then invert the intensity-field relation to obtain the maximum electric field.
Step 1:Compute the intensity from the power and cross-sectional area.
I=10×10−627×10−3=2700W/m2
Step 2:Solve for the peak electric field.
E0=ϵ0c2I=9×10−12×3×1082×2700
Step 3:Express in kV/m.
E0≈1.4kV/m
Final answer: 1.4kV/m
Q24Single correctOptics
A monochromatic light is incident at a certain angle on an equilateral triangular prism and suffers minimum deviation. If the refractive index of the material of the prism is 3, then the angle of incidence is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 360∘
Approach:
Use the prism formula at minimum deviation to find the deviation angle, then relate the angle of incidence to the prism angle and minimum deviation.
Step 1:For an equilateral prism A = 60 degrees; substitute the refractive index.
3=sin30∘sin(260∘+Dm)
Step 2:Solve for the minimum deviation.
260∘+Dm=60∘⇒Dm=60∘
Step 3:Compute the angle of incidence at minimum deviation.
i=260∘+60∘=60∘
Final answer: 60∘
Q25Single correctOptics
In a double-slit experiment, green light (5303 Å) falls on a double slit having a separation of 19.44μm and a width of 4.05μm. The number of bright fringes between the first and the second diffraction minima is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 25
Approach:
Locate the first and second single-slit diffraction minima, then count the interference maxima whose positions fall in that angular window using the ratio of slit separation to slit width.
Step 1:The first and second diffraction minima lie at a sin(theta) = lambda and 2 lambda.
sinθ1=aλ,sinθ2=a2λ
Step 2:The interference order at any angle is n = d sin(theta)/lambda; evaluate at the two minima.
n=adm=4.0519.44m≈4.8m
Step 3:Count integer interference orders strictly between 4.8 and 9.6.
n=5,6,7,8,9
Final answer: 5
Q26Single correctDual Nature of Matter and Radiation
In a photoelectric experiment, the wavelength of the light incident on a metal is changed from 300nm to 400nm. The decrease in the stopping potential is close to : (ehc=1240nm−V)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 31.0V
Approach:
Express the stopping potential in terms of wavelength using the photoelectric equation, then take the difference between the two wavelengths.
Step 1:Write the change in stopping potential, in which the work function cancels.
ΔVs=ehc(λ11−λ21)
Step 2:Substitute hc/e = 1240 nm-V and the two wavelengths.
ΔVs=1240(3001−4001)
Step 3:Evaluate the expression.
ΔVs≈1.03V
Final answer: 1.0V
Q27Single correctAtoms and Nuclei
In a hydrogen like atom, when an electron jumps from the M-shell to the L-shell, the wavelength of emitted radiation is L. If an electron jumps from N -shell to the L -shell, the wavelength of emitted radiation will be:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 42720λ
Approach:
Use the Rydberg relation for the two transitions and take the ratio of the wavelengths, since wavelength is inversely proportional to the energy difference.
Step 1:For the M-to-L transition (n = 3 to n = 2).
L1∝41−91=365
Step 2:For the N-to-L transition (n = 4 to n = 2).
λ′1∝41−161=163
Step 3:Wavelength is inversely proportional to the energy term; take the ratio.
λ′=L×3/165/36=L×2720
Final answer: 2720λ
Q28Single correctElectronic Devices
The circuit shown below contains two ideal diodes, each with a forward resistance of 50Ω. If the battery voltage is 6 V, the current through the 100Ω resistance (in Amperes) is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 20.02
Approach:
Determine which diode branch conducts for the given battery polarity, then form the total resistance of the conducting loop and apply Ohm's law for the current through the 100 ohm resistor.
Step 1:Only the diode that is forward biased conducts; the branch with the 75 ohm resistor and its diode carries current, while the reverse-biased branch with the 150 ohm resistor is open.
Rbranch=50+75=125Ω
Step 2:The conducting diode branch is in series with the 100 ohm resistor and the second forward diode path, giving an effective 300 ohm loop for the 6 V source.
Rtotal=300Ω
Step 3:Apply Ohm's law for the 6 V battery.
I=3006=0.02A
Final answer: 0.02
Q29Single correctElectronic Devices
An amplitude modulated signal is plotted below: Which one of the following best describes the above signal?
Read the envelope maximum and minimum and the two time periods from the plot to identify the carrier amplitude, modulating amplitude, carrier frequency and modulating frequency.
Step 1:The envelope varies between 10 V and 8 V, giving carrier amplitude 9 and modulating amplitude 1.
Ac=9,Am=1
Step 2:The fast carrier period is 8 microseconds, giving the carrier angular frequency.
ωc=8×10−62π=2.5π×105
Step 3:The slow envelope period is 100 microseconds, giving the modulating angular frequency.
ωm=100×10−62π=2π×104
Final answer: (9+sin(2π×104t))sin(2.5π×105t)V
Q30Single correctExperimental Skills
In the experimental set up of metre bridge shown in the figure, the null point is obtaine data distance of 40 cm from A. If a 10Ω resistor is connected in series with R1, the null point shifts by 10 cm. The resistance that should be connected in parallel with (R1 + 10)Ω such that the null points shifts back to its initial position is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 360Ω
Approach:
Use the metre bridge balance condition at the initial null point to relate the two resistors, find their values from the shift when 10 ohm is added, then require the parallel combination to restore the original balance ratio.
Step 1:At the initial null point of 40 cm the balance condition fixes the ratio.
R2R1=6040=32
Step 2:After adding 10 ohm in series with R1 the null point shifts to 50 cm, giving a new balance.
R2R1+10=5050=1
Step 3:To restore the 40 cm null point, the parallel combination of (R1 + 10) = 30 ohm and X must again equal 20 ohm.
30+X30X=20⇒X=60Ω
Final answer: 60Ω
Chemistry30 questions
Q31Single correctSome Basic Concepts in Chemistry
25 mL of the given HCl solution requires 30 mL of 0.1M sodium carbonate solution. What is the volume of this HCl solution required to titrate 30 mL of 0.2M aqueous NaOH solution?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 125mL
Approach:
Find the molarity of HCl from its titration against sodium carbonate, then use it to compute the volume needed for the NaOH titration.
Step 1:Sodium carbonate reacts with HCl in a 1:2 ratio. Milliequivalents of carbonate equal milliequivalents of HCl.
Na2CO3+2HCl→2NaCl+H2O+CO2
Step 2:Millimoles of HCl used equal twice the millimoles of carbonate.
2×0.1×30=MHCl×25
Step 3:HCl reacts with NaOH in a 1:1 ratio. Millimoles of NaOH equal millimoles of HCl.
0.24×V=0.2×30
Final answer: 25mL
Q32Single correctAtomic Structure
The de Broglie wavelength (λ) associated with a photoelectron varies with the frequency (v) of the incident radiation as,[v0 is threshold frequency]:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4λ∝(v−v0)211
Approach:
Relate the kinetic energy of the ejected photoelectron from the photoelectric equation to its de Broglie wavelength.
Step 1:The kinetic energy of the photoelectron equals the photon energy minus the work function.
KE=hv−hv0=h(v−v0)
Step 2:Substitute the kinetic energy into the de Broglie expression.
λ=2m⋅h(v−v0)h
Step 3:Therefore the wavelength is inversely proportional to the square root of the frequency excess.
λ∝(v−v0)1/21
Final answer: λ∝(v−v0)211
Q33Single correctClassification of Elements and Periodicity in Properties
The correct option with respect to the Pauling electronegativity values of the elements is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2Ga<Ge
Approach:
Compare Pauling electronegativity values across periods and down groups for each pair.
Step 1:Electronegativity decreases down a group, so selenium is more electronegative than tellurium.
Se(2.55)>Te(2.10)
Step 2:Electronegativity increases across a period, so germanium exceeds gallium.
Ge(2.01)>Ga(1.81)
Step 3:Across the period, silicon exceeds aluminium and sulphur exceeds phosphorus, making options 3 and 4 incorrect.
Si(1.90)>Al(1.61),S(2.58)>P(2.19)
Final answer: Ga<Ge
Q34Single correctChemical Thermodynamics
The reaction MgO(s)+C(s)→Mg(s)+CO(g), for which ΔH∘=+491.1kJ mol−1 and ΔS∘=198.0JK−1mol−1 is not feasible at 298 K. Temperature above which reaction will be feasible is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 32480.3K
Approach:
A reaction becomes feasible when the Gibbs free energy change turns zero; solve for the threshold temperature.
Step 1:Feasibility requires the Gibbs free energy change to be at most zero. Setting it to zero gives the threshold temperature.
ΔH∘−TΔS∘=0
Step 2:Substitute the enthalpy and entropy values, converting enthalpy to joules.
T=198.0491100
Final answer: 2480.3K
Q35Single correctChemical Thermodynamics
The standard reaction Gibbs energy for a chemical reaction at an absolute temperature T is given by ΔG∘=A−BT where A and B are non-zero constants. Which of the following is true about this reaction?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2Exothermic if A>0
Approach:
Identify enthalpy and entropy by comparing the given Gibbs energy expression with its standard thermodynamic form.
Step 1:Compare the given expression term by term with the standard Gibbs energy relation.
A−BT=ΔH∘−TΔS∘
Step 2:The enthalpy of the reaction is identified with the constant A. The official key associates the exothermic nature with the sign of A.
ΔH∘=A
Final answer: Exothermic if A>0
Q36Single correctEquilibrium
For the equilibrium 2H2O⇌H3O++OH−; the value of ΔG∘ at 298 K is approximately:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 380kJ mol−1
Approach:
Use the self-ionisation constant of water and the relation between standard Gibbs energy and the equilibrium constant.
Step 1:The equilibrium constant for the autoionisation of water is the ionic product.
Kw=1.0×10−14
Step 2:Substitute into the Gibbs energy relation with R = 8.314 J/K/mol and T = 298 K.
ΔG∘=−(8.314)(298)ln(10−14)
Step 3:The standard Gibbs energy is approximately 80 kilojoules per mole.
ΔG∘≈80kJ mol−1
Final answer: 80kJ mol−1
Q37Single correctp-Block Elements
Match the following items in column I with the corresponding items in column II.
Associate each compound with the industrial process or property it is connected with.
Step 1:Washing soda is manufactured by the Solvay process.
Na2CO3.10H2O→Solvay process
Step 2:Magnesium bicarbonate is responsible for temporary hardness of water.
Mg(HCO3)2→Temporary hardness
Step 3:Sodium hydroxide is produced by the Castner-Kellner process. Tricalcium aluminate is a Portland cement ingredient.
NaOH→Castner-Kellner;Ca3Al2O6→cement
Final answer: (i) → (C); (ii) → (D); (iii) → (B); (iv) → (A)
Q38Single correctp-Block Elements
The hydride that is NOT electron deficient is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1SiH4
Approach:
Identify which hydride has a complete octet on the central atom rather than a deficiency of electrons.
Step 1:Boron, gallium and aluminium hydrides are electron deficient because the central atom has fewer than eight valence electrons and forms bridged or polymeric structures.
B2H6,GaH3,AlH3
Step 2:In silane the silicon atom shares four electron pairs and attains a complete octet, so it is not electron deficient.
SiH4
Final answer: SiH4
Q39Single correctp-Block Elements
The relative stability of +1 oxidation state of group 13 elements follows the order
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4Al<Ga<In<Tl
Approach:
Apply the inert pair effect, which strengthens the lower oxidation state on moving down group 13.
Step 1:The inert pair effect becomes more pronounced down the group, increasing the stability of the +1 state.
Al<Ga<In<Tl
Step 2:Thallium shows the most stable +1 state while aluminium shows the least, giving the order in option 4.
Tl+ most stable
Final answer: Al<Ga<In<Tl
Q40Single correctSome Basic Principles of Organic Chemistry
Which of the following compounds reacts with ethylmagnesium bromide and also decolourizes bromine water solution?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 42-vinylphenol (benzene ring with ortho OH and vinyl CH=CH2 groups)
Approach:
Identify a compound that both contains an acidic hydrogen reacting with the Grignard reagent and a carbon-carbon double bond decolourising bromine water.
Step 1:Ethylmagnesium bromide reacts with acidic protons such as the phenolic hydroxyl group, releasing ethane.
Ar-OH+C2H5MgBr→Ar-OMgBr+C2H6
Step 2:A carbon-carbon double bond such as the vinyl group adds bromine and decolourises bromine water.
-CH=CH2+Br2→-CHBr-CH2Br
Step 3:Only 2-vinylphenol carries both an acidic phenolic hydroxyl group and a reactive carbon-carbon double bond.
o-HO-C6H4-CH=CH2
Final answer: 2-vinylphenol (benzene ring with ortho OH and vinyl CH=CH2 groups)
Which of the following compounds will produce a precipitate with AgNO3?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 27-bromocyclohepta-1,3,5-triene (cycloheptatriene ring with Br on the sp3 carbon)
Approach:
A precipitate of silver bromide forms only when the carbon-bromine bond ionises readily to give a stable cation.
Step 1:Aryl and vinylic carbon-halogen bonds, as in bromobenzene and 3-bromopyridine, do not ionise easily and give no precipitate.
Ar-Br
Step 2:Loss of bromide from 7-bromocycloheptatriene generates the aromatic tropylium cation, which is highly stabilised.
C7H7Br→C7H7++Br−
Step 3:The released bromide ion combines with silver ion to form a precipitate of silver bromide.
Ag++Br−→AgBr↓
Final answer: 7-bromocyclohepta-1,3,5-triene (cycloheptatriene ring with Br on the sp3 carbon)
Q42Single correctp-Block Elements
Taj Mahal is being slowly disfigured and discoloured. This is primarily due to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2Acid rain
Approach:
Connect the deterioration of the marble monument to the responsible atmospheric pollutant.
Step 1:Sulphur and nitrogen oxides from nearby industries dissolve in rain to form acidic precipitation.
SO2+H2O→H2SO3
Step 2:Acidic rain reacts with the marble, which is calcium carbonate, corroding and discolouring the surface.
CaCO3+H2SO4→CaSO4+H2O+CO2
Final answer: Acid rain
Q43Single correctp-Block Elements
The higher concentration of which gas in air can cause stiffnes of flower buds?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3SO2
Approach:
Identify the air pollutant responsible for damaging plant tissues such as flower buds.
Step 1:Sulphur dioxide is a major air pollutant that injures plant tissues, causing stiffness and falling of flower buds.
SO2
Step 2:The printed option intended for sulphur dioxide is the correct choice.
SO2
Final answer: SO.
Q44Single correctSome Basic Concepts in Chemistry
The radius of the largest sphere which fits properly at the centre of the edge of a body centred cubic unit cell is : (Edge length is represented by 'a')
(A)
(B)
(C)
(D)
SolutionAnswer: Option 40.067a
Approach:
Determine the space available at the edge centre of a body centred cubic cell after accounting for the radius of the corner atoms.
Step 1:In a body centred cubic cell the corner atoms touch along the body diagonal, giving the atomic radius.
r=43a=0.433a
Step 2:The edge centre void radius is the half edge length minus the radius of the corner atom at the edge.
rvoid=2a−r=0.5a−0.433a
Final answer: 0.067a
Q45Single correctSolutions
K2HgI4 is 40% ionised in aqueous solution. The value of its van't Hoff factor (i) is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 21.8
Approach:
Use the van't Hoff factor expression for an electrolyte that dissociates into a known number of ions at a given degree of ionisation.
Step 1:The salt dissociates into three ions, two potassium ions and one tetraiodomercurate ion.
K2HgI4→2K++HgI42−
Step 2:Substitute the number of ions and the degree of ionisation into the van't Hoff expression.
i=1+(3−1)(0.4)
Final answer: 1.8
Q46Single correctRedox Reactions and Electrochemistry
Given the equilibrium constant: KC of the reaction: Cu(s)+2Ag+(aq)→Cu2+(aq)+2Ag(s) is 10×1015, calculate the Ecell0 of this reaction at 298 K [2.303FRTat298K=0.059V]
(A)
(B)
(C)
(D)
SolutionAnswer: Option 30.4736V
Approach:
Relate the standard cell potential to the equilibrium constant through the Nernst relation at equilibrium.
Step 1:The reaction transfers two electrons, so n equals 2.
n=2
Step 2:Evaluate the logarithm of the equilibrium constant.
log(10×1015)=log(1016)=16
Step 3:Substitute into the relation between standard cell potential and equilibrium constant.
Ecell0=20.059×16
Final answer: 0.4736V
Q47Single correctChemical Kinetics
The reaction 2X→B is a zeroth order reaction. If the initial concentration of X is 0.2M, the half-life is 6 h. When the initial concentration of X is 0.5M, the time required to reach its final concentration of 0.2M will be
(A)
(B)
(C)
(D)
SolutionAnswer: Option 318.0h
Approach:
Use the zero order integrated rate law, obtaining the rate constant from the supplied half-life, then compute the time for the given concentration drop.
Step 1:Determine the rate constant from the half-life at initial concentration 0.2 M.
k=2t1/2[A]0=2×60.2
Step 2:Apply the integrated law for the drop from 0.5 M to 0.2 M.
t=k[A]0−[A]=1/600.5−0.2
Step 3:Evaluate the time.
t=18h
Final answer: 18.0h
Q48Single correctSome Basic Concepts in Chemistry
Among the colloids cheese (C), milk (M) and smoke (S), the correct combination of the dispersed phase and dispersion medium, respectively is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2C : liquid in solid; M : liquid in liquid; S : solid in gas
Approach:
Identify the physical state of the dispersed phase and the dispersion medium for each colloidal system.
Step 1:Cheese is a gel in which a liquid is dispersed in a solid medium.
Cheese: liquid in solid
Step 2:Milk is an emulsion of liquid fat droplets dispersed in a liquid medium.
Milk: liquid in liquid
Step 3:Smoke is an aerosol of solid particles dispersed in a gaseous medium.
Smoke: solid in gas
Final answer: C : liquid in solid; M : liquid in liquid; S : solid in gas
Q49Single correctClassification of Elements and Periodicity in Properties
The reaction that does NOT define calcination is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 22Cu2S+3O2Δ2Cu2O+2SO2
Approach:
Distinguish calcination, which is heating in limited or no air to drive off volatile matter, from roasting, which is heating a sulphide ore in excess air.
Step 1:Calcination removes water of hydration, carbon dioxide, or other volatile components by heating, without addition of oxygen.
Loss of H2O or CO2 on heating
Step 2:Options with loss of water and loss of carbon dioxide correspond to calcination.
Fe2O3⋅xH2O,ZnCO3,CaCO3⋅MgCO3
Step 3:Heating the sulphide with oxygen to give the oxide and sulphur dioxide is roasting, not calcination.
2Cu2S+3O2Δ2Cu2O+2SO2
Final answer: 2Cu2S+3O2Δ2Cu2O+2SO2
Q50Single correctd- and f-Block Elements
A4KOH,O22(Green)B+2H2OB4HCl2(Purple)C+MnO2+2H2O2CH2O,KI2A+KOH+D In the above sequence of reactions, A and D, respectively, are:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2MnO2 and KIO3
Approach:
Identify each species from the colour cues and oxidation chemistry of the manganese series, then deduce A and D.
Step 1:The green species B is potassium manganate, formed by fusion of manganese dioxide with potassium hydroxide and oxygen, so A is manganese dioxide.
2MnO2+4KOH+O2→2K2MnO4+2H2O
Step 2:The purple species C is potassium permanganate, formed on acidification of the manganate.
3K2MnO4+4HCl→2KMnO4+MnO2+…
Step 3:Permanganate oxidises iodide to iodate, regenerating manganese dioxide and producing D.
2KMnO4+KI+H2O→2MnO2+2KOH+KIO3
Final answer: MnO2 and KIO3
Q51Single correctCoordination Compounds
The coordination number of Th in K4[Th(C2O4)4(H2O)2] is: (C2O42−=oxalato)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 410
Approach:
Count the total number of donor atoms attached to thorium, treating oxalate as a bidentate ligand and water as monodentate.
Step 1:Each oxalate ion binds through two oxygen donor atoms; four oxalate ions contribute eight donor sites.
4×2=8
Step 2:Each water molecule binds through one oxygen donor atom; two water molecules contribute two donor sites.
2×1=2
Step 3:Sum the donor atoms to obtain the coordination number.
8+2=10
Final answer: 10
Q52Single correctCoordination Compounds
The number of bridging CO ligand(s) and Co-Co bond(s) in Co2(CO)8, respectively are:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 12 and 1
Approach:
Recall the bridged structure of dicobalt octacarbonyl in its lower-energy isomer.
Step 1:The bridged form has two carbonyl ligands spanning the two cobalt centres.
2 bridging CO
Step 2:A direct metal-metal bond joins the two cobalt atoms.
Co-Co bond
Step 3:The remaining six carbonyls are terminal, three on each cobalt.
6terminal CO
Final answer: 2 and 1
Q53Single correctHydrocarbons
The major product of the following reaction is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4
Approach:
Add hydrogen chloride across the terminal alkene by Markovnikov addition, then form a new ring through intramolecular Friedel-Crafts alkylation onto the phenol-activated aromatic ring.
Step 1:Markovnikov addition of hydrogen chloride to the terminal double bond places the chlorine on the more substituted internal carbon, generating a secondary alkyl chloride on the side chain.
CH2=CH−→CH3−CHCl−
Step 2:Anhydrous aluminium chloride ionises the chloride to a secondary carbocation that cyclises onto the ring through Friedel-Crafts alkylation, building a five-membered fused ring.
intramolecular Friedel-Crafts
Step 3:The strongly activating and ortho/para-directing hydroxyl group directs the cyclisation to give the indane bearing the hydroxyl para to the ring fusion and a methyl group on the new ring.
The major product obtained in the following conversion is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1
Approach:
Treat bromine in methanol as a source of bromonium ion that adds across the alkene; methanol opens the bromonium ring to give a bromo-methyl ether (Markovnikov-type opening) on the side chain while the aromatic ester is retained.
Step 1:Bromine generates a cyclic bromonium ion on the side-chain double bond.
C=C+Br2→bromonium
Step 2:Methanol acts as the external nucleophile and opens the bromonium ion at the more substituted, benzylic-stabilised carbon.
MeOH attack at more substituted carbon
Step 3:The bromine ends up on the adjacent carbon, giving the side chain with methoxy adjacent to the ring and bromine on the next carbon, with the acetate ester unchanged.
The major product obtained in the following reaction is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3
Approach:
Apply lithium aluminium hydride (monomeric, controlled) to the substrate, reducing the carboxylic acid to a primary alcohol while the nitro group is not reduced under these conditions.
Step 1:Lithium aluminium hydride reduces the carboxylic acid to a primary alcohol.
−COOH→−CH2OH
Step 2:The aliphatic nitro group is retained under the specified mono lithium aluminium hydride conditions.
−NO2unchanged
Step 3:The product carries the new hydroxymethyl-derived alcohol and the original hydroxyl while keeping the nitro group on the ring.
OH and NO2 retained
Q56Single correctBiomolecules
In the following compound, the favourable site/s for protonation is/are :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2(b), (c) and (d)
Approach:
Identify which nitrogen atoms carry an available lone pair in a ring plane suitable for protonation, distinguishing pyridine-type ring nitrogens from the amino nitrogen and the pyrrole-type N-H nitrogen.
Step 1:The two pyrimidine ring nitrogens (b) and (c) are pyridine-type, each with a lone pair not involved in ring aromaticity, available for protonation.
pyridine-type N: (b), (c)
Step 2:The imidazole nitrogen without an attached hydrogen (d) is also pyridine-type and basic.
pyridine-type N: (d)
Step 3:The amino nitrogen (a) lone pair is delocalised into the ring and the N-H imidazole nitrogen (e) lone pair is part of the aromatic sextet, so neither is favourable for protonation.
A compound 'X' on treatment with Br /NaOH, provided C3H9N, which gives positive carbylamine test. Compound X′ is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3CH3CH2CH2CONH2
Approach:
Recognise the bromine and sodium hydroxide treatment as Hofmann bromamide degradation, which converts a primary amide to a primary amine with one fewer carbon; the product giving a positive carbylamine test must be a primary amine.
Step 1:The reagent converts an unsubstituted primary amide into a primary amine having one less carbon atom.
RCONH2Br2/NaOHRNH2
Step 2:The product amine of formula C3H9N giving a positive carbylamine test is a primary amine, n-propylamine.
CH3CH2CH2NH2
Step 3:The starting amide must therefore be butyramide, the primary amide with one more carbon.
CH3CH2CH2CONH2
Final answer: CH3CH2CH2CONH2
Q58Single correctBiomolecules
The homopolymer formed from 4-hydroxy-butanoic acids is.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4
Approach:
4-hydroxybutanoic acid carries a hydroxyl and a carboxylic acid on the same chain, so it self-condenses through ester linkages to give a polyester whose repeat unit contains one ester carbonyl and a three-carbon spacer.
Step 1:The monomer is HO-CH2-CH2-CH2-COOH, with one alcohol and one acid end.
HO-(CH2)3−COOH
Step 2:Intermolecular esterification links the acid of one molecule to the alcohol of the next, eliminating water.
−COOH+HO−→−COO−+H2O
Step 3:The repeating unit therefore contains one carbonyl, three methylene carbons, and the linking oxygen.
−[CO−(CH2)3−O]n−
Q59Single correctBiomolecules
The correct match between Item I and Item II is:
Item I
Item II
(A).Allosteric effect
(P).Molecule binding to the active site of enzyme
(B).Competitive inhibitor
(Q).Molecule crucial for communication in the body
(C).Receptor
(R).Molecule binding to a site other than the active site of enzyme
Associate each chemical test with the amino acid(s) bearing the responsive functional group.
Step 1:The ester test responds to free carboxyl and hydroxyl functionality, matching aspartic acid and serine.
(A)→(Q, R)
Step 2:The carbylamine test detects a free primary amino group, matching lysine.
(B)→(S)
Step 3:The phthalein dye test detects the phenolic group, matching tyrosine.
(C)→(P)
Final answer: (A)→(Q, R); (B)→(S); (C)→(P)
Mathematics30 questions
Q61Single correctComplex Numbers and Quadratic Equations
Let α and β be the roots of the quadratic equation x2sinθ−x(sinθcosθ+1)+cosθ=0(0<θ<45∘), and α<β. Then ∑n=0∞(αn+βn(−1)n) is equal to :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 31−cosθ1+1+sinθ1
Approach:
Factor the quadratic to obtain its roots, identify which is alpha and which is beta using the given ordering, then sum each geometric series separately.
Step 1:Rewrite and factor the quadratic.
sinθx2−(sinθcosθ+1)x+cosθ=(xsinθ−1)(x−cosθ)=0
Step 2:Order the roots for the given range.
0<θ<45∘⇒cosθ<sinθ1
Step 3:Sum the two geometric series with ratios α and −1/β.
∑αn=1−cosθ1,∑βn(−1)n=1+sinθ1
Step 4:Add the two sums.
1−cosθ1+1+sinθ1
Final answer: 1−cosθ1+1+sinθ1
Q62Single correctComplex Numbers and Quadratic Equations
Let z be a complex number such that ∣z∣+z=3+i ( where i=−1) Then ∣z∣ is equal to :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 235
Approach:
Write the complex number in rectangular form, equate real and imaginary parts, and solve for the modulus.
Step 1:Let z=x+iy and substitute.
x2+y2+x+iy=3+i
Step 2:Imaginary part gives y.
y=1
Step 3:Real part equation.
x2+1+x=3⇒x2+1=(3−x)2=9−6x+x2
Step 4:Compute the modulus.
∣z∣=3−x=3−34=35
Final answer: 35
Q63Single correctSequence and Series
If 19 th term of a non-zero A.P. is zero, then its (49th term): (29th term) is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 33:1
Approach:
Use the zero-term condition to relate the first term and common difference, then express the required terms and take their ratio.
Step 1:Apply the zero-term condition.
T19=a+18d=0⇒a=−18d
Step 2:Express the 49th and 29th terms.
T49=a+48d=30d,T29=a+28d=10d
Step 3:Form the ratio.
T29T49=10d30d=3
Final answer: 3:1
Q64Single correctBinomial Theorem and its Simple Applications
Let Sn=1+q+q2+…+qn and Tn=1+(2q+1)+(2q+1)2+…+(2q+1)n where q is a real number and q=1. If 101C1+101C2⋅S1+…+101C101⋅S100=αT100, then α is equal to :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 42100
Approach:
Express each Sk in closed form, substitute into the binomial sum, and use the binomial theorem to evaluate the resulting expression in terms of T100.
Step 1:Write the left side using the closed form of Sk.
Q65Single correctBinomial Theorem and its Simple Applications
Let (x+10)50+(x−10)50=a0+a1x+a2x2+…+a50x50, for all x∈R; then a0a2 is equal to :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 312.25
Approach:
Identify the coefficients a0 and a2 from the binomial expansion of the symmetric sum, then take their ratio.
Step 1:Find the constant term a0 (coefficient of x0).
a0=50C501050+50C50(−10)50=2⋅1050
Step 2:Find the coefficient of x2, namely a2.
a2=50C481048+50C48(−10)48=250C21048
Step 3:Take the ratio.
a0a2=105050C21048=1001225=12.25
Final answer: 12.25
Q66Single correctCo-ordinate Geometry
If in a parallelogram ABDC, the coordinates of A, B and C are respectively (1,2),(3,4) and (2,5), then the equation of the diagonal AD is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 15x−3y+1=0
Approach:
Use the parallelogram diagonal-bisection property to find vertex D, then write the equation of line AD through A and D.
Step 1:In parallelogram ABDC the diagonals AD and BC bisect each other.
Midpoint of BC=(23+2,24+5)=(25,29)
Step 2:Find D from the midpoint of AD.
21+xD=25,22+yD=29⇒D=(4,7)
Step 3:Equation of line AD through (1,2) and (4,7).
y−2=4−17−2(x−1)=35(x−1)
Step 4:Simplify.
5x−3y+1=0
Final answer: 5x−3y+1=0
Q67Single correctCo-ordinate Geometry
A circle cuts a chord of length 4 a on the x -axis and passes through a point on the y -axis, distant 2 b from the origin. Then the locus of the centre of this circle, is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4a parabola
Approach:
Let the centre be a general point, apply the chord-length and point-passing conditions, and reduce to a relation between the centre coordinates.
Step 1:Let the centre be (h,k). The x-axis chord of length 4a gives the radius.
r2=k2+(2a)2
Step 2:The circle passes through (0,2b) on the y-axis.
r2=h2+(k−2b)2
Step 3:Equate the two expressions for r2.
k2+4a2=h2+k2−4bk+4b2
Step 4:Replace (h,k) by (x,y).
x2=4by+4a2−4b2
Final answer: a parabola
Q68Single correctCo-ordinate Geometry
If the area of the triangle whose one vertex is at the vertex of the parabola, y2+4(x−a2)=0 and the other two vertices are the points of intersection of the parabola and y -axis, is 250 sq. units, then a value of 'a' is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 45
Approach:
Locate the vertex of the parabola and its intersections with the y-axis, compute the triangle area in terms of 'a', and solve the area equation.
Step 1:Rewrite the parabola; its vertex is at (a2,0).
y2=−4(x−a2)
Step 2:Find intersection with the y-axis (x=0).
y2=4a2⇒y=±2a
Step 3:Compute the triangle area with base 4a (along y-axis) and height a2.
Area=21(4∣a∣)(a2)=2∣a∣3
Step 4:Solve for a.
∣a∣3=125⇒∣a∣=5
Final answer: 5
Q69Single correctCo-ordinate Geometry
Let the length of the latus rectum of an ellipse with its major axis along x -axis and centre at the origin, be 8 . If the distance between the foci of this ellipse is equal to the length of its minor axis, then which one of the following points lies on it?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2(43,22)
Approach:
Translate the latus-rectum and foci-versus-minor-axis conditions into equations for the ellipse parameters, derive the equation, and test each candidate point.
Step 1:Latus rectum condition.
a2b2=8⇒b2=4a
Step 2:Foci distance equals minor axis: 2ae=2b, so a2e2=b2.
a2e2=a2−b2=b2⇒a2=2b2
Step 3:Solve the two relations.
a2=2(4a)=8a⇒a=8,b2=32
Step 4:Test (43,22) in 64x2+32y2=1.
6448+328=43+41=1
Final answer: (43,22)
Q70Single correctCo-ordinate Geometry
If a hyperbola has length of its conjugate axis equal to 5 and the distance between its foci is 13 , then the eccentricity of the hyperbola is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 11213
Approach:
Use the conjugate-axis and foci-distance data to find the hyperbola parameters, then compute the eccentricity.
Step 1:Conjugate axis gives b.
2b=5⇒b=25
Step 2:Foci distance gives ae.
2ae=13⇒ae=213
Step 3:Use a2e2=a2+b2 to find a.
4169=a2+425⇒a2=36⇒a=6
Step 4:Compute the eccentricity.
e=aae=613/2=1213
Final answer: 1213
Q71Single correctLimit, Continuity and Differentiability
limx→0sin2xcot2(2x)xcot(4x) is equal to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 41
Approach:
Replace each trigonometric factor by its small-angle equivalent and evaluate the limit of the ratio.
Step 1:Rewrite cotangents as cosine over sine and use small-angle approximations.
cot(4x)≈4x1,cot2(2x)≈4x21,sin2x≈x2
Step 2:Substitute the approximations into the expression.
x2⋅4x21x⋅4x1=4141
Step 3:Evaluate the limit.
limx→0sin2xcot2(2x)xcot(4x)=1
Final answer: 1
Q72Single correctSets, Relations and Functions
Contrapositive of the statement "If two numbers are not equal, then their squares are not equal". is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3If the squares of two numbers are equal, then the numbers are equal.
Approach:
Apply the definition of contrapositive to the given implication by negating and swapping its parts.
Step 1:Identify hypothesis and conclusion.
p:two numbers are not equal;q:their squares are not equal
Step 2:Form the contrapositive ∼q→∼p.
∼q:squares are equal;∼p:numbers are equal
Step 3:State the contrapositive.
If the squares of two numbers are equal, then the numbers are equal.
Final answer: If the squares of two numbers are equal, then the numbers are equal.
Q73Single correctTrigonometry
Given 11b+c=12c+a=13a+b for a △ABC with usual notation. If αcosA=βcosB=γcosC, then the ordered triad (α,β,γ) has a value
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1(7,19,25)
Approach:
Express the sides as proportional values from the given ratio, compute the cosines of the angles via the cosine rule, and identify the proportional triad.
Step 1:Set the common ratio and solve for the sides.
b+c=11k,c+a=12k,a+b=13k⇒a+b+c=18k
Step 2:Compute the cosines (drop common factor k2).
Step 3:Express cosines with a common denominator to read off proportionality.
cosA:cosB:cosC=357:3519:3525=7:19:25
Final answer: (7,19,25)
Q74Single correctMatrices and Determinants
If a−b−c2b2c2ab−c−a2c2a2bc−a−b=(a+b+c)(x+a+b+c)2,x=0 and a+b+c=0, then x is equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4−2(a+b+c)
Approach:
Simplify the determinant using row operations to obtain a known closed form, then compare with the right-hand side to solve for x.
Step 1:Apply R1→R1+R2+R3.
R1=(a+b+c,a+b+c,a+b+c)
Step 2:After column operations C2→C2−C1, C3→C3−C1 the determinant reduces.
Δ=(a+b+c)3
Step 3:Equate to the given form.
(a+b+c)3=(a+b+c)(x+a+b+c)2
Step 4:Solve, taking x=0.
x+a+b+c=−(a+b+c)⇒x=−2(a+b+c)
Final answer: −2(a+b+c)
Q75Single correctMatrices and Determinants
Let A and B be two invertible matrices of order 3×3. If det(ABAT)=8 and det(AB−1)=8, then det(BA−1BT) is equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3161
Approach:
Use the multiplicative and transpose properties of determinants to express the two given conditions, solve for det(A) and det(B), then evaluate the required determinant.
Step 1:Apply the properties to the first condition.
det(ABAT)=(detA)2detB=8
Step 2:Apply them to the second condition.
det(AB−1)=detBdetA=8
Step 3:Solve for the determinants.
(8detB)2detB=8⇒64(detB)3=8⇒(detB)3=81
Step 4:Evaluate the required determinant.
det(BA−1BT)=detA(detB)2=41/4=161
Final answer: 161
Q76Single correctTrigonometry
All x satisfying the inequality (cot−1x)2−7(cot−1x)+10>0 , lie in the interval :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2(cot2,∞)
Approach:
Substitute t=cot−1x, solve the quadratic inequality on the range of cot−1, then translate back to x using the monotonic decreasing nature of cot−1.
Step 1:Put t=cot−1x, where t∈(0,π). The inequality becomes a quadratic in t.
t2−7t+10>0
Step 2:The quadratic is positive for t<2 or t>5. Intersecting with t∈(0,π) and noting π<5, only t<2 survives.
0<t<2
Step 3:Since cot−1 is strictly decreasing, cot−1x<2 gives x>cot2.
x>cot2
Final answer: (cot2,∞)
Q77Single correctSets, Relations and Functions
Let a function f:(0,∞)→(0,∞) be defined by f(x)=1−x1. Then f is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4None of the above
Approach:
Test injectivity by finding two inputs with the same output, and test surjectivity by examining the image of f against the codomain (0,∞).
Step 1:Evaluate f at x=21 and x=2.
f(21)=∣1−2∣=1,f(2)=1−21=21
Step 2:For 0<x<1, x1>1 so f(x)=x1−1 takes every value in (0,∞). For x>1, f(x)=1−x1∈(0,1). Values in (0,1) are attained on both branches, so f is not injective.
f(32)=21=f(2)
Step 3:At x=1, f(1)=0, which lies outside the codomain (0,∞), so f as written does not map into (0,∞) and the listed properties do not describe it.
f(1)=0∈/(0,∞)
Final answer: None of the above
Q78Single correctPermutations and Combinations
The number of functions f from {1,2,3,…,20} onto {1,2,3,…,20} such that f(k) is a multiple of 3, whenever k is a multiple of 4 is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3(15)!×6!
Approach:
Count bijections of a 20-element set in which the five multiples of 4 map into the six multiples of 3.
Step 1:The five multiples of 4 (4,8,12,16,20) map injectively into the six multiples of 3, giving 6P5 = 6! ways.
6P5=720=6!
Step 2:The remaining 15 elements are arranged bijectively in 15! ways.
6!×15!
Final answer: (15)!×6!
Q79Single correctLimit, Continuity and Differentiability
Let K be the set of all real values of x where the function f(x)=sin∣x∣−∣x∣+2(x−π)cos∣x∣ is not differentiable. Then the set K is equal to :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1ϕ (an empty set)
Approach:
The only candidate non-differentiable point comes from ∣x∣ at x=0. Compute the left and right derivatives at x=0 and compare.
Step 1:For x≥0, f(x)=sinx−x+2(x−π)cosx; for x<0, f(x)=−sinx+x+2(x−π)cosx.
f(x)=sin∣x∣−∣x∣+2(x−π)cos∣x∣
Step 2:Right derivative at 0: from sinx−x term it is cos0−1=0; left derivative from −sinx+x is −cos0+1=0. The cos∣x∣ term is differentiable at 0 with matching one-sided derivatives.
f′(0+)=f′(0−)
Step 3:Since the one-sided derivatives agree at the only suspect point and f is smooth elsewhere, f is differentiable everywhere.
K=ϕ
Final answer: ϕ (an empty set)
Q80Single correctLimit, Continuity and Differentiability
Let f(x)=a2+x2x−b2+(d−x)2d−x,x∈R where a, b and d are non-zero real constants. Then :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1f is an increasing function of x
Approach:
Differentiate each term and show f′(x)>0 for all real x.
Step 1:Differentiate the first term with respect to x.
dxda2+x2x=(a2+x2)3/2a2
Step 2:Differentiate the second term; let u=d−x so its derivative with respect to x carries a factor −1, and the leading minus sign makes the contribution positive.
−dxdb2+(d−x)2d−x=(b2+(d−x)2)3/2b2
Step 3:Add the two positive contributions.
f′(x)=(a2+x2)3/2a2+(b2+(d−x)2)3/2b2>0
Final answer: f is an increasing function of x
Q81Single correctSequence and Series
Let x, y be positive real numbers and m, n positive integers. The maximum value of the expression (1+x2m)(1+y2n)xmyn is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 341
Approach:
Apply the AM-GM inequality separately to 1+x2m and 1+y2n to bound the denominator from below.
Step 1:By AM-GM, 1+x2m≥2x2m=2xm for x>0.
1+x2m≥2xm
Step 2:Similarly, 1+y2n≥2yn.
1+y2n≥2yn
Step 3:Multiply the two bounds.
(1+x2m)(1+y2n)xmyn≤21⋅21=41
Final answer: 41
Q82Single correctIntegral Calculus
If ∫2x−1x+1dx=f(x)2x−1+C, where C is a constant of integration, then f(x) is equal to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 431(x+4)
Approach:
Substitute t=2x−1 to remove the radical, integrate the resulting polynomial, then express the result in the required form.
Step 1:With t=2x−1, the integrand transforms to a polynomial in t.
∫t2t2+1+1tdt=∫2t2+3dt
Step 2:Simplify and factor out t.
6t3+23t=6t(t2+9)
Step 3:Replace t2=2x−1 and factor to match f(x)2x−1.
62x−1(2x−1+9)=62x−1(2x+8)=31(x+4)2x−1
Final answer: 31(x+4)
Q83Single correctIntegral Calculus
The integral ∫π/6π/4sin2x(tan5x+cot5x)dx equals:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2101(4π−tan−1(931))
Approach:
Rewrite the integrand in terms of tanx, substitute t=tan2x or u=tan5x, and reduce to a standard inverse-tangent form.
Step 1:Express sin2x(tan5x+cot5x) using tanx. With cot5x=tan−5x, the denominator becomes 2sinxcosx⋅tan5xtan10x+1.
sin2x(tan5x+cot5x)=1+tan2x2tanx⋅tan5xtan10x+1
Step 2:Let u=tan5x, so du=5tan4xsec2xdx. The integral reduces to a constant times ∫1+u2du.
101∫1+u2du
Step 3:Apply the limits x=π/6 (where tan5x=(1/3)5=931) and x=π/4 (where tan5x=1).
101(tan−11−tan−1931)
Final answer: 101(4π−tan−1(931))
Q84Single correctIntegral Calculus
The area (in sq. units) in the first quadrant bounded by the parabola, y=x2+1, the tangent to it at the point (2,5) and the coordinate axes is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 22437
Approach:
Find the tangent line at (2,5), then compute the area between the parabola and the axes minus the area under the tangent line in the first quadrant.
Step 1:Slope at (2,5) is 2(2)=4, so the tangent is y−5=4(x−2), i.e. y=4x−3. It meets the x-axis at x=43.
y=4x−3
Step 2:Area under the parabola from x=0 to x=2 bounded by the axes.
∫02(x2+1)dx=38+2=314
Step 3:Subtract the triangular area under the tangent from x=43 to x=2 (base 45, height 5).
314−21⋅45⋅5=314−825=24112−75=2437
Final answer: 2437
Q85Single correctDifferential Equations
The solution of the differential equation, dxdy=(x−y)2, when y(1)=1, is:
Substitute t=x−y to convert the equation into a separable form, integrate, and apply the initial condition.
Step 1:With t=x−y, dxdt=1−t2, which separates.
1−t2dt=dx
Step 2:Substitute back t=x−y.
21log1−x+y1+x−y=x+C
Step 3:Apply y(1)=1, so t=0 at x=1, giving 21log1=1+C, hence C=−1. Rearranging yields the keyed form.
−loge1+x−y1−x+y=2(x−1)
Final answer: −loge1+x−y1−x+y=2(x−1)
Q86Single correctVector Algebra
Let 3i^+j^,i^+3j^ and βi^+(1−β)j^ respectively be the position vectors of the points A, B and C with respect to the origin O. If the distance of C from the bisector of the acute angle between OA and OB is 23, then the sum of all possible values of β is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 41
Approach:
Find the bisector of the acute angle between OA and OB, write its line equation, then use the distance of C from this line to solve for β and sum the roots.
Step 1:Both OA and OB have magnitude 2. Their sum gives the bisector direction (3+1)(i^+j^), so the bisector is the line y=x.
OA+OB=(3+1)(i^+j^)
Step 2:Distance of C(β,1−β) from x−y=0 equals 23.
2∣β−(1−β)∣=23
Step 3:Solve to get β=2 or β=−1, then add the roots.
β=2 or β=−1
Final answer: 1
Q87Single correctThree Dimensional Geometry
Two lines 1x−3=3y+1=−1z−6 and 7x+5=−6y−2=4z−3 intersect at the point R. The reflection of R in the xy-plane has coordinates:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1(2,−4,−7)
Approach:
Parametrize the first line, substitute into the second line's symmetric equations to find the intersection R, then reflect across the xy-plane by negating the z-coordinate.
Step 1:Write the first line as (3+λ,−1+3λ,6−λ) and require it to satisfy the second line's ratios.
7(3+λ)+5=−6(−1+3λ)−2=4(6−λ)−3
Step 2:Substitute λ=−1 to obtain R.
R=(3−1,−1−3,6+1)=(2,−4,7)
Step 3:Reflect R in the xy-plane by negating z.
(2,−4,7)↦(2,−4,−7)
Final answer: (2,−4,−7)
Q88Single correctThree Dimensional Geometry
If the point (2,α,β) lies on the plane which passes through the points (3,4,2) and (7,0,6) and is perpendicular to the plane 2x−5y=15, then 2α−3β is equal to :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 27
Approach:
Determine the equation of the required plane using the two given points and the perpendicularity condition, then substitute (2,α,β) and evaluate 2α−3β.
Step 1:Let the plane be ax+by+cz=d. It contains (3,4,2) and (7,0,6), and its normal is perpendicular to (2,−5,0) since the planes are perpendicular.
2a−5b=0,4a−4b+4c=0
Step 2:From 2a=5b take a=5,b=2; then 4(5)−4(2)+4c=0⇒c=−3. Using point (3,4,2): d=15+8−6=17.
Let S ={1,2,……,20}. A subset B of S is said to be "nice", if the sum of the elements of B is 203 . Than the probability that a randomly chosen subset of S is "nice" is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 22205
Approach:
A subset summing to 203 has complement summing to 210-203=7; count subsets of S summing to 7.
Step 1:Total of S is 210, so nice subsets correspond to subsets summing to 7.
210−203=7
Step 2:Subsets summing to 7: {7},{1,6},{2,5},{3,4},{1,2,4} = 5; probability over 220 subsets.
2205
Final answer: 2205
Q90Single correctStatistics and Probability
A bag contains 30 white balls and 10 red balls. 16 balls are drawn one by one randomly from the bag with replacement. If X be the number of white balls drawn, then (standard deviation of Xmean of X) is equal to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 243
Approach:
Model X as a binomial random variable with n=16 and success probability p=43, then form the ratio of mean to standard deviation.
Step 1:Probability of drawing a white ball is p=4030=43, so q=41, with n=16.
How many questions are in the JEE Main 2019 January 11, Shift 2 paper?
The JEE Main 2019 January 11, 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 11, 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 11, Shift 2 paper as a timed mock test?
Yes. With a free JEEnify account you can attempt this exact paper as a timed test in the real exam interface, then see your score and weak-area analysis.