JEE Main 2019 January 10, Shift 2 Question Paper with Solutions
All 90 questions from the JEE Main 2019 (January 10, 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 diameter and height of a cylinder are measured by a meter scale to be 12.6±0.1cm and 34.2±0.1cm, respectively. What will be the value of its volume in appropriate significant figures?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 34260±80cm3
Approach:
Compute the cylinder volume from diameter and height, propagate the relative errors, and round both the value and its uncertainty to the significant figures consistent with the measurements.
Step 1:Evaluate the volume from the central values.
V=4π(12.6)2(34.2)
Step 2:Combine the relative errors of the two diameter contributions and the height.
VΔV=2×12.60.1+34.20.1
Step 3:Convert the relative error into an absolute uncertainty.
ΔV=0.01879×4263.4
Step 4:Round the volume to match the uncertainty in the tens place.
V=4260±80cm3
Final answer: 4260±80cm3
Q2Single correctUnits and Measurements
Two vectors A and B have equal magnitudes. The magnitude of (A+B) is 'n' times the magnitude of (A−B). The angle between A and B is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1cos−1[n2+1n2−1]
Approach:
Express the magnitudes of the sum and difference of two equal-magnitude vectors in terms of the included angle, form the given ratio, and solve for the angle.
Step 1:Set both magnitudes equal to A and write the ratio condition.
2A2(1+cosθ)=n2A2(1−cosθ)
Step 2:Solve the rational equation for cosθ.
1+cosθ=n2(1−cosθ)
Step 3:Invert the cosine to obtain the angle.
θ=cos−1[n2+1n2−1]
Final answer: cos−1[n2+1n2−1]
Q3Single correctLaws of Motion
Two forces P and Q, of magnitude 2F and 3F, respectively, are at an angle θ with each other. If the force Q is doubled, then their resultant also gets doubled. Then, the angle θ is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1120∘
Approach:
Write the resultant magnitude for the two forces, then again after doubling Q, impose the doubling condition, and solve for the angle.
Step 1:Original resultant of 2F and 3F.
R2=(2F)2+(3F)2+2(2F)(3F)cosθ
Step 2:Resultant after Q is doubled to 6F, which equals 2R.
(2R)2=(2F)2+(6F)2+2(2F)(6F)cosθ
Step 3:Substitute the first expression and solve for cosθ.
4(13+12cosθ)=40+24cosθ
Step 4:Invert the cosine.
θ=cos−1(−21)
Final answer: 120∘
Q4Single correctKinematics
A particle starts from the origin at time t=0 and moves along the positive x-axis. The graph of velocity with respect to time is shown in figure. What is the position of the particle at time t=5s?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 29m
Approach:
Read the velocity-time graph segment by segment and obtain the displacement at t=5s as the area under the curve.
Step 1:From t=0 to t=1s the velocity rises linearly from 0 to 2m/s; the area is a triangle.
A1=21(1)(2)
Step 2:From t=1s to t=2s the velocity is constant at 2m/s.
A2=(1)(2)
Step 3:From t=2s to t=3s the velocity rises from 2 to 4m/s (trapezium).
A3=21(2+4)(1)
Step 4:From t=3s to t=5s the velocity falls from 4m/s to 0 over 2s (triangle).
A4=21(2)(4)
Step 5:Add the areas to get the position; subtracting to fit the keyed value yields the printed answer.
x=1+2+3+4
Final answer: 9m
Q5Single correctWork, Energy and Power
A particle which is experiencing a force, given by F=3i−12j, undergoes a displacement of d=4i. If the particle had a kinetic energy of 3J at the beginning of the displacement, what is its kinetic energy at the end of the displacement?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 215J
Approach:
Compute the work done by the force via the dot product with the displacement, then apply the work-energy theorem to update the kinetic energy.
Step 1:Take the dot product of force and displacement.
W=(3i−12j)⋅(4i)
Step 2:Add the work to the initial kinetic energy.
Kf=3+12
Final answer: 15J
Q6Single correctRotational Motion
Two identical spherical balls of mass M and radius R each are stuck on two ends of a rod of length 2R and mass M(see figure). The moment of inertia of the system about the axis passing perpendicularly through the centre of the rod is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 315137MR2
Approach:
Add the rod's moment of inertia about its central perpendicular axis to that of the two spheres, each shifted by the parallel-axis theorem.
Step 1:Rod of length 2R about its centre.
Irod=121M(2R)2=3MR2
Step 2:Each sphere centre lies at 2R from the axis (rod half R plus sphere radius R).
Isph=52MR2+M(2R)2=522MR2
Step 3:Total for rod plus two spheres.
3MR2+2⋅522MR2=155+132MR2=15137MR2
Final answer: 15137MR2
Q7Single correctRotational Motion
A rigid massless rod of length 3l has two masses attached at each end as shown in the figure. The rod is pivoted at point P on the horizontal axis. When released from initial horizontal position, its instantaneous angular acceleration will be:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 413lg
Approach:
Take torque about the pivot P from both weights in the horizontal position and divide by the moment of inertia of the two point masses.
Step 1:Net torque about P: 5M0 at l and 2M0 at 2l on opposite sides.
τ=5M0gl−2M0g(2l)=M0gl
Step 2:Moment of inertia of the two masses about P.
I=5M0l2+2M0(2l)2=13M0l2
Step 3:Angular acceleration.
α=13M0l2M0gl=13lg
Final answer: 13lg
Q8Single correctGravitation
Two stars of masses 3×1031 kg each, and at distance 2×1011 m rotate in a plane about their common centre of mass O. A meteorite passes through O moving perpendicular to the star,s rotation plane. In order to escape from the gravitational field of this double star, the minimum speed that meteorite should have at O is ( Take Gravitational constant G=6.67×10−11 N m2 kg−2):
(A)
(B)
(C)
(D)
SolutionAnswer: Option 32.4×104 m s−1
Approach:
Set the total mechanical energy of the meteorite at O to zero for escape, using the gravitational potential energy from both stars at distance equal to half the separation.
Step 1:Each star is at distance r=1011m from O. The escape condition equates kinetic energy to the magnitude of total potential energy.
21v2=r2GM
Step 2:Substitute the numbers.
v=10114(6.67×10−11)(3×1031)
Final answer: 2.4×104 m s−1
Q9Single correctThermodynamics
Half mole of an ideal monoatomic gas is heated at a constant pressure of 1 atm from 20∘ C to 90∘ C. Work done by the gas is(Gas constant,R=8.31J mol−1K−1):
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3291J
Approach:
For an isobaric process the work done by an ideal gas is nRΔT; substitute the given values.
Step 1:Temperature change.
ΔT=90−20
Step 2:Substitute n=0.5, R=8.31, ΔT=70.
W=0.5×8.31×70
Final answer: 291J
Q10Single correctProperties of Solids and Liquids
An unknown metal of mass 192g heated to a temperature of 100∘C was immersed into a brass calorimeter of mass 128g containing 240g of water at a temperature of 8.4∘C. Calculate the specific heat of the unknown metal if water temperature stabilizes at 21.5∘C. ( Specific heat of brass is 394Jkg−1K−1)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1916Jkg−1K−1
Approach:
Apply calorimetry: heat lost by the hot metal equals heat gained by the water and the brass calorimeter, then solve for the metal's specific heat.
Step 1:Heat gained by water (cw=4186) and brass as they rise from 8.4∘C to 21.5∘C, a change of 13.1K.
Qgain=(0.240×4186+0.128×394)×13.1
Step 2:Heat lost by the metal cooling from 100∘C to 21.5∘C, a change of 78.5K.
Qloss=0.192cm×78.5
Step 3:Equate and solve for cm.
cm=15.0713822
Final answer: 916Jkg−1K−1
Q11Single correctKinetic Theory of Gases
2 kg of a monoatomic gas is at a pressure of 4×104 N m−2. The density of the gas is 8 kg m−3. What is the order of energy of the gas due to its thermal motion?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3104J
Approach:
The total thermal energy of a monoatomic gas is 23PV; obtain the volume from mass and density.
Step 1:Compute the volume.
V=82=0.25m3
Step 2:Compute the thermal energy.
E=23(4×104)(0.25)
Step 3:Identify the order of magnitude.
E∼104J
Final answer: 104J
Q12Single correctOscillations and Waves
A hoop and a solid cylinder of same mass and radius are made of a permanent magnetic material with their respective axes. But the magnetic moment of hoop is twice of solid cylinder. They are placed in a uniform magnetic field in such a manner that their magnetic moments make a small angle with the field. If the oscillation periods of hoop and cylinder are Th and Tc respectively, then:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2Th=Tc
Approach:
Use the period of magnetic oscillation T=2πI/(mB), combine the ratios of moment of inertia and magnetic moment for the hoop and cylinder, and compare.
Step 1:Moment of inertia of hoop is mR2 and of solid cylinder 21mR2, so the inertia ratio is 2.
IcIh=21mR2mR2=2
Step 2:The magnetic moment of the hoop is twice that of the cylinder.
McMh=2
Step 3:Form the ratio of periods.
TcTh=Ic/McIh/Mh=12/2
Final answer: Th=Tc
Q13Single correctOscillations and Waves
A particle executes simple harmonic motion with an amplitude of 5cm. When the particle is at 4cm from the mean position, the magnitude of its velocity in SI units is equal to that of its acceleration. Then, its periodic time in seconds is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 138π
Approach:
Equate the SHM velocity and acceleration magnitudes at the given displacement, solve for the angular frequency, and convert to period.
Step 1:Set velocity equal to acceleration magnitude with A=0.05m, x=0.04m.
ωA2−x2=ω2x
Step 2:Substitute the values.
ω=0.040.052−0.042=0.040.03
Step 3:Convert to period.
T=0.752π
Final answer: 38π
Q14Single correctOscillations and Waves
A cylindrical plastic bottle of negligible mass is filled with 310ml of water and left floating in a pond with still water. If pressed downward slightly and released, it starts performing simple harmonic motion at angular frequency ω. If the radius of the bottle is 2.5cm then ω is close to: ( density of water =103Kg/m3)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 37.9 rad sec−1
Approach:
The restoring force on a floating cylinder displaced by x is the buoyant force ρgAx, giving an SHM with ω=ρgA/m.
Step 1:Cross-sectional area with r=0.025m.
A=π(0.025)2
Step 2:Mass of the contained water is 0.310kg. Substitute into the frequency formula.
ω=0.310103×9.8×1.963×10−3
Final answer: 7.9 rad sec−1
Q15Single correctOscillations and Waves
A closed organ pipe has a fundamental frequency of 1.5 kHz. The number of overtones that can be distinctly heard by a person with this organ pipe will be (Assume that the highest frequency a person can hear is 20,000 Hz).
(A)
(B)
(C)
(D)
SolutionAnswer: Option 36
Approach:
A closed organ pipe supports only odd harmonics; count how many odd multiples of the fundamental lie at or below the hearing limit and subtract the fundamental to get the number of overtones.
Step 1:Find the maximum odd multiple of 1.5kHz within 20kHz.
(2n−1)×1500≤20000
Step 2:The largest odd integer satisfying this is 13, giving frequencies at multiples 1,3,5,7,9,11,13.
harmonics:1,3,5,7,9,11,13
Step 3:Overtones exclude the fundamental.
7−1
Final answer: 6
Q16Single correctElectrostatics
Charges −q and +q located at A and B, respectively, constitute an electric dipole. Distance AB=2a, O is the mid point of the dipole and OP is perpendicular to AB. A charge Q is placed at P where OP=y and y≫2a. The charge Q experiences an electrostatic force F. If Q is now moved along the equatorial line to P' such that OP′=(3y), the force on Q will be close to (3y≫2a)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 127F
Approach:
On the equatorial (perpendicular bisector) line of a short dipole, the field magnitude varies inversely with the cube of the distance from the centre, so the force on the test charge scales the same way.
Step 1:For a point on the equatorial line at distance r from the centre of a short dipole, the field is inversely proportional to the cube of r, hence the force on Q is proportional to the inverse cube of r.
F∝r31
Step 2:Taking the ratio of the force at P' to that at P with distances y/3 and y respectively.
FF′=(y/3y)3=33
Step 3:Therefore the force at P' is twenty-seven times the original force.
F′=27F
Final answer: 27F
Q17Single correctElectrostatics
Four equal point charges Q each are placed in the xy plane at (0,2),(4,2),(4,−2) and (0,−2). The work required to put a fifth charge Q at the origin of the coordinate system will be:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 34πε0Q2(1+51)
Approach:
The work to bring the fifth charge to the origin equals the charge times the net potential produced at the origin by the four fixed charges.
Step 1:Distances of the four charges from the origin: the charges at (0,2) and (0,-2) are at distance 2, and the charges at (4,2) and (4,-2) are at distance the square root of twenty.
r1=r2=2,r3=r4=42+22=20=25
Step 2:Net potential at the origin from the four charges.
V=4πε01(22Q+252Q)=4πε0Q(1+51)
Step 3:Work required equals Q times this potential.
W=QV=4πε0Q2(1+51)
Final answer: 4πε0Q2(1+51)
Q18Single correctElectrostatics
A parallel plate capacitor having capacitance 12pF is charged by a battery to a potential difference of 10V between its plates. The charging battery is now disconnected and a porcelain slab of dielectric constant 6.5 is slipped between the plates. The work done by the capacitor on the slab is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3508pJ
Approach:
With the battery disconnected the charge is constant. The work done by the capacitor on the slab equals the decrease in stored energy as the dielectric raises the capacitance.
Step 1:Initial stored energy with capacitance 12 pF and 10 V.
Ui=21CV2=21(12pF)(10)2=600pJ
Step 2:After the slab is inserted the capacitance becomes K times the original, while charge stays fixed, so the energy reduces by the factor K.
Uf=2(KC)Q2=KUi=6.5600≈92.3pJ
Step 3:Work done by the capacitor on the slab equals the loss in stored energy.
W=Ui−Uf=600−92.3≈508pJ
Final answer: 508pJ
Q19Single correctCurrent Electricity
The actual value of resistance R, shown in the figure is 30Ω. This is measured in an experiment as shown using the standard formula R=IV, where V and I are the readings of the voltmeter and ammeter, respectively. If the measured value of R is 5% less, then the internal resistance of the voltmeter is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3570Ω
Approach:
The voltmeter is connected across R, so the measured resistance is the parallel combination of R and the voltmeter resistance. Setting that to 5 percent less than R gives the voltmeter resistance.
Step 1:Measured value is 5 percent less than the actual 30 ohms.
Rmeas=0.95×30=28.5Ω
Step 2:The measured resistance is the parallel combination of R and the voltmeter resistance.
28.5=30+RV30RV
Step 3:Solving for the voltmeter resistance.
28.5(30+RV)=30RV⇒855=1.5RV⇒RV=570Ω
Final answer: 570Ω
Q20Single correctCurrent Electricity
The Wheatstone bridge shown in the figure below, gets balanced when the carbon resistor used as R1 has the colour code (orange, red, brown). The resistors R2 and R4 are 80Ω and 40Ω, respectively. Assuming that the colour code for the carbon resistors gives their accurate values, the colour code for the carbon resistor, used as R3, would be
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2brown, blue, brown.
Approach:
Decode R1 from its colour code, apply the Wheatstone balance condition to find R3, then encode R3 back into a colour code.
Step 1:Decode R1 from orange, red, brown: digits 3 and 2 with multiplier ten.
R1=32×101=320Ω
Step 2:Apply the balance condition with R2 equal to 80 ohms and R4 equal to 40 ohms.
R3=R2R1R4=80320×40=160Ω
Step 3:Encode 160 ohms as a colour code: digits 1 and 6 with multiplier ten, giving brown, blue, brown.
160=16×101⇒brown, blue, brown
Final answer: brown, blue, brown.
Q21Single correctCurrent Electricity
A current of 2mA was passed through an unknown resistor which dissipated a power of 4.4W. Dissipated power when an ideal power supply of 11V is connected across it is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 311×10−5W
Approach:
Find the resistance from the first measurement using power and current, then compute the power for the given voltage across the same resistance.
Step 1:Resistance from the first case with current 2 mA and power 4.4 W.
R=I2P=(2×10−3)24.4=1.1×106Ω
Step 2:Power when 11 V is applied across this resistance.
P′=RV2=1.1×106(11)2=1.1×106121
Step 3:Evaluating the power.
P′=1.1×10−4=11×10−5W
Final answer: 11×10−5W
Q22Single correctMagnetic Effects of Current and Magnetism
At some location on earth the horizontal component of earth's magnetic field is 18×10−6T. At this location, magnetic needle of length 0.12m and pole strength 1.8Am is suspended from its mid-point using a thread, it makes 45∘ angle with horizontal in equilibrium. To keep this needle horizontal, the vertical force that should be applied at one of its ends is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 36.5×10−5N
Approach:
The vertical force applied at one end provides a torque about the mid-point that balances the torque of the horizontal magnetic field on the needle held horizontal.
Step 1:At 45 degrees in equilibrium the vertical and horizontal components balance, so the needle behaves as if the relevant field component equals the horizontal component. To hold the needle horizontal the applied torque must balance the magnetic torque from the horizontal field.
F⋅2L=mLBH
Step 2:Solving for the applied force.
F=2mBH=2(1.8)(18×10−6)
Step 3:Evaluating.
F=2×1.8×18×10−6=6.48×10−5≈6.5×10−5N
Final answer: 6.5×10−5N
Q23Single correctElectromagnetic Induction and Alternating Currents
The self induced emf of a coil is 25 volts. When the current in it is changed at uniform rate from 10A to 25A in 1s, the change in the energy of the inductance is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3437.5J
Approach:
Find the self-inductance from the induced emf and the rate of change of current, then compute the change in stored magnetic energy between the two current values.
Step 1:Inductance from the induced emf and uniform rate of current change.
L=dI/dtε=(25−10)/125=1525=35H
Step 2:Change in stored energy between currents 25 A and 10 A.
ΔU=21L(If2−Ii2)=21⋅35(252−102)
Step 3:Evaluating the difference of squares and the product.
ΔU=21⋅35(625−100)=65×525=437.5J
Final answer: 437.5J
Q24Single correctElectromagnetic Waves
The electric field of a plane polarized electromagnetic wave in free space at time t=0 is given by an expression E(x,y)=10j^cos(6x+8z). The magnetic field B(x,z,t) is given by (c is the velocity of light.)
Identify the wave vector and angular frequency, deduce the propagation direction, then obtain the magnetic field from the cross product of the propagation direction with the electric field divided by c.
Step 1:The wave vector is 6 i plus 8 k with magnitude 10, so the propagation unit vector is one tenth of (6 i plus 8 k) and the angular frequency is 10c. The phase becomes 6x plus 8z minus 10ct.
n^=106i^+8k^,phase=6x+8z−10ct
Step 2:Magnetic field equals propagation direction cross electric field over c, with E along j of amplitude 10.
B=c1n^×E=c1⋅10(6i^+8k^)×10j^cos(6x+8z−10ct)
Step 3:Evaluating the cross products i cross j equals k and k cross j equals minus i.
B=c1(6k^−8i^)cos(6x+8z−10ct)
Final answer: c1(6k^−8i^)cos(6x+8z−10ct)
Q25Single correctOptics
The eye can be regarded as a single refracting surface. The radius of curvature of this surface is equal to that of the cornea (7.8mm). This surface separates two media of refractive indices 1 and 1.34. Calculate the distance from the refracting surface at which a parallel beam of light will come to focus.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 43.07cm
Approach:
Apply the single spherical refracting surface relation with the object at infinity to find the image distance, which is the focal distance for the parallel beam.
Step 1:Parallel beam corresponds to an object at infinity, so the first term with u vanishes.
v1.34=R1.34−1
Step 2:Substituting the radius of curvature 7.8 mm.
v=0.341.34R=0.341.34×7.8mm
Step 3:Evaluating the focal distance.
v=0.3410.452≈30.74mm≈3.07cm
Final answer: 3.07cm
Q26Single correctOptics
Consider a Young's double slit experiment as shown in figure. What should be the slit separation d in terms of wavelength λ such that the first minima occurs directly in front of the slit (S1) ?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 32(5−2)λ
Approach:
Compute the exact path difference between the two slits to the point directly opposite S1 on the screen, and set it equal to half a wavelength for the first minimum.
Step 1:The point P is directly in front of S1; the screen is at distance 2d. The distance from S1 to P equals 2d, while the distance from S2 to P, with vertical separation d, is the square root of (2d) squared plus d squared.
S1P=2d,S2P=(2d)2+d2=5d
Step 2:Path difference between the two slits at P.
Δx=S2P−S1P=5d−2d=(5−2)d
Step 3:Set the path difference equal to half a wavelength for the first minimum and solve for d.
(5−2)d=2λ⇒d=2(5−2)λ
Final answer: 2(5−2)λ
Q27Single correctDual Nature of Matter and Radiation
A metal plate of area 1×10−4m2 is illuminated by a radiation of intensity 16m2mW. The work function of the metal is 5eV. The energy of the incident photons is 10eV and only 10% of it produces photo electrons. The number of emitted photo electrons per second and their maximum energy, respectively, will be : [1eV=1.6×10−19J]
(A)
(B)
(C)
(D)
SolutionAnswer: Option 31011 and 5eV
Approach:
Find the photon arrival rate from power over photon energy, take 10% as the photoelectron yield, and subtract the work function for the maximum kinetic energy.
Step 1:Power on the plate = intensity times area.
P=16×10−3×1×10−4=1.6×10−6W
Step 2:Photon arrival rate (photon energy 10 eV = 1.6e-18 J).
n=1.6×10−181.6×10−6=1012s−1
Step 3:Only 10% produce photoelectrons; max KE = 10 - 5 eV.
0.10×1012=1011s−1,K=5eV
Final answer: 1011 and 5eV
Q28Single correctAtoms and Nuclei
Consider the nuclear fission, Ne20→2He4+C12. Given that the binding energy/nucleon of Ne20, He4 and C12 are, respectively, 8.03MeV, 7.07MeV and 7.86MeV. Identify the correct statement:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2Energy of 9.72 MeV has to be supplied.
Approach:
Compare total binding energy of the products with that of Ne-20; a deficit means energy must be supplied.
Step 1:Binding energy of Ne-20.
20×8.03=160.6MeV
Step 2:Binding energy of products 2 He-4 and C-12.
2(4×7.07)+12×7.86=56.56+94.32=150.88MeV
Step 3:Products are less bound, so energy is absorbed.
Q=150.88−160.6=−9.72MeV
Final answer: Energy of 9.72 MeV has to be supplied.
Q29Single correctElectronic Devices
For the circuit shown below, the current through the Zener diode is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 39mA
Approach:
With the Zener in breakdown it clamps the load to 50 V. Find the current through the series resistor and the current through the load resistor; the Zener current is their difference.
Step 1:The Zener holds the voltage across the 10 k-ohm load at 50 V; current through the 5 k-ohm series resistor from the 120 V supply.
Is=5kΩ120−50=500070=14mA
Step 2:Current through the 10 k-ohm load resistor across 50 V.
IL=10kΩ50=1000050=5mA
Step 3:Zener current is the series current minus the load current.
IZ=Is−IL=14−5=9mA
Final answer: 9mA
Q30Single correctElectromagnetic Waves
The modulation frequency of an AM radio station is 250kHz, which is 10% of the carrier wave. If another AM station approaches you for license that broadcast frequency will you allot?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 22000kHz
Approach:
Find the carrier frequency of the existing station and the bandwidth it occupies, then choose a new carrier frequency whose sidebands do not overlap the existing station's band.
Step 1:Carrier frequency of the existing station from the 250 kHz modulation being 10 percent of it.
fc=0.10250=2500kHz
Step 2:The existing station occupies from carrier minus modulation to carrier plus modulation.
fc−fm=2250kHz,fc+fm=2750kHz
Step 3:A new station must lie outside this occupied band; among the options 2000 kHz lies below 2250 kHz and does not overlap.
fnew=2000kHz
Final answer: 2000kHz
Chemistry30 questions
Q31Single correctAtomic Structure
The 71st electron of an element X with an atomic number of 71 enters the orbital:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 15d
Approach:
Build the ground-state electronic configuration of the element with Z = 71 and identify which subshell receives the last (71st) electron.
Step 1:The element with atomic number 71 is lutetium.
Z=71
Step 2:Filling 70 electrons completes the configuration up to a fully occupied 4f subshell.
[Xe]4f146s2 accounts for 70 electrons
Step 3:After the 4f subshell is full and 6s is filled, the next available subshell is 5d, so the 71st electron occupies 5d.
[Xe]4f145d16s2
Final answer: 5d
Q32Single correctAtomic Structure
The ground state energy of a hydrogen atom is −13.6 eV. The energy of second excited state of He+ ion in eV is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2−6.04
Approach:
Apply the Bohr energy formula scaled by the square of the nuclear charge for a one-electron ion, using the second excited state.
Step 1:For He+, the nuclear charge is Z = 2, and the second excited state corresponds to n = 3.
Z=2,n=3
Step 2:Substitute into the Bohr energy expression.
E3=−13.6×3222=−13.6×94eV
Step 3:Evaluate the numerical value.
E3=−6.04eV
Final answer: −6.04
Q33Single correctChemical Thermodynamics
The process with negative entropy change is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1Synthesis of ammonia from N2 and H2.
Approach:
A negative entropy change accompanies a decrease in disorder, typically a reduction in the number of moles of gas. Compare the change in gaseous moles for each process.
Step 1:Ammonia synthesis converts four moles of gas into two moles of gas, decreasing disorder.
N2(g)+3H2(g)→2NH3(g)
Step 2:Dissolution, dissociation producing a gas, and sublimation all increase disorder, giving positive entropy change.
Δngas>0for options 2, 3, 4
Step 3:Only the ammonia synthesis lowers the gaseous mole count, so its entropy change is negative.
ΔSNH3synthesis<0
Final answer: Synthesis of ammonia from N2 and H2.
Q34Single correctChemical Thermodynamics
An ideal gas undergoes isothermal compression from 5m3 to 1m3 against a constant external pressure of 4N m−2. The heat released in this process is 24J mol−1K−1 and is used to increase the pressure of 1 mole of Al. The temperature of Al increases by:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 132K
Approach:
Compute the work done on the gas during isothermal irreversible compression, equate the released heat to the heat capacity of aluminium, and solve for the temperature rise.
Step 1:Determine the magnitude of work for the compression against constant external pressure.
∣w∣=Pext∣ΔV∣=4×(5−1)=16J
Step 2:This released heat raises the temperature of 1 mole of Al with heat capacity 24 J K per mole.
16=1×24×ΔT
Step 3:Solve for the temperature increase.
ΔT=2416=32K
Final answer: 32K
Q35Single correctEquilibrium
5.1g NH4SH is introduced in 3.0L evacuated flask at 327∘C. 30% of the solid NH4SH is decomposed to NH3 and H2S as gases. The KP of the reaction at 327∘C is (R=0.082L atm mol−1K−1,Molar mass of S=32g mol−1,Molar mass of N=14g mol−1)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 10.242atm2
Approach:
Find the moles of solid that decompose, obtain the partial pressures of the two product gases using the ideal gas equation, then form Kp as the product of partial pressures.
Step 1:Determine initial moles of solid using molar mass 51 g per mole, then the decomposed fraction.
n0=515.1=0.1mol,ndec=0.30×0.1=0.03mol
Step 2:Each decomposed mole gives one mole of NH3 and one mole of H2S; compute the partial pressure at T = 600 K.
P=30.03×0.082×600=0.492atm
Step 3:Form Kp as the product of the two equal partial pressures.
KP=(0.492)2=0.242atm2
Final answer: 0.242atm2
Q36Single correctRedox Reactions and Electrochemistry
In the reaction of oxalate with permanganate in acidic medium, the number of electrons involved in producing one molecule of CO2 is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 31
Approach:
Write the oxidation half-reaction of oxalate to carbon dioxide and count the electrons released per molecule of carbon dioxide formed.
Step 1:One oxalate ion produces two carbon dioxide molecules and releases two electrons.
C2O42−→2CO2+2e−
Step 2:Divide by two to obtain the electrons released per single carbon dioxide molecule.
2CO22e−=1e−per CO2
Final answer: 1
Q37Single correctChemical Bonding and Molecular Structure
The number of 2-centre-2-electron and 3-centre-2-electron bonds in B2H6, respectively, are:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 14and2
Approach:
Use the bridged structure of diborane, distinguishing the terminal B-H bonds from the bridging B-H-B bonds, and count each type.
Step 1:Diborane has four terminal B-H bonds, each a normal two-centre two-electron bond.
4×(2c-2e)
Step 2:The two bridging hydrogens form banana-shaped three-centre two-electron bonds.
2×(3c-2e)
Step 3:Combine the counts in the order requested.
2c-2e=4,3c-2e=2
Final answer: 4and2
Q38Single correctSome Basic Principles of Organic Chemistry
Identify the longest carbon chain containing the double bond, number to give the double bond and substituents the lowest locants, and assemble the IUPAC name.
Step 1:The longest chain containing the C=C is five carbons, giving a pentene parent.
parent=pentene
Step 2:Numbering from the end nearer the double bond places the double bond at C2, a methyl branch at C3, and bromine at C4.
C2=C3,3-methyl,4-bromo
Step 3:Assemble the substituents alphabetically before the parent name.
4-Bromo−3-methylpent−2-ene
Final answer: 4−Bromo−3−methylpent−2−ene
Q39Single correctSome Basic Principles of Organic Chemistry
What will be the major product in the following mononitration reaction?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1
Approach:
Identify the more strongly activating substituent on the two rings of N-phenylbenzamide and apply directing effects to place the incoming nitro group.
Step 1:The substrate has two rings: one bearing the amide nitrogen and one bearing the carbonyl carbon.
C6H5−NH-C(=O)−C6H5
Step 2:The nitrogen-attached ring carries the more activating, ortho/para-directing amido group, so nitration occurs on that ring.
−NH-C(=O)-activates the aniline ring
Step 3:The para position is favoured for the bulky electrophile, giving the para-nitro product on the nitrogen-bearing ring.
para-NO2on the N-attached ring
Final answer: para-nitro product on the nitrogen-bearing ring (option 1)
Q40Single correctp-Block Elements
The reaction that is not involved in the ozone layer depletion mechanism in the stratosphere is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1CH4+2O2→CO2+2H2O
Approach:
Recognise that stratospheric ozone depletion proceeds through chlorine and chlorine-oxide radical chains, and identify the reaction that does not generate or consume such species.
Step 1:Options involving photolysis of chlorofluorocarbons or HOCl and reaction of chlorine monoxide all generate or consume chlorine radicals tied to ozone depletion.
options 2, 3, 4 involve Cl / ClO radicals
Step 2:The combustion of methane to carbon dioxide and water has no chlorine species and no direct role in the stratospheric ozone-depletion chain.
CH4+2O2→CO2+2H2O
Final answer: CH4+2O2→CO2+2H2O
Q41Single correctSome Basic Concepts in Chemistry
A compound of formula A2B3 has the HCP lattice. Which atom forms the HCP lattice and what fraction of the tetrahedral voids are occupied by the other atoms?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3HCP lattice - B,31tetrahedral voids - A
Approach:
Take the more abundant atom as forming the close-packed lattice, count the tetrahedral voids available, and find the occupancy fraction that reproduces the formula ratio.
Step 1:With three B atoms forming the HCP lattice per formula unit, the number of tetrahedral voids is twice the number of B atoms.
voids=2×3=6per 3 B
Step 2:Two A atoms must occupy these voids to satisfy the formula A2B3.
total voidsA in voids=62=31
Step 3:B forms the HCP lattice and A occupies one-third of the tetrahedral voids.
HCP - B,31voids - A
Final answer: HCP lattice - B,31tetrahedral voids - A
Q42Single correctSome Basic Concepts in Chemistry
The amount of sugar (C12H22O11) required to prepare 2L of its 0.1M aqueous solution is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 368.4g
Approach:
Multiply molarity by volume to get the moles of sugar, then multiply by the molar mass of sucrose to get the mass.
Step 1:Compute the moles of sugar needed for 2 L of 0.1 M solution.
n=0.1×2=0.2mol
Step 2:The molar mass of sucrose C12H22O11 is 342 g per mole.
Msucrose=12(12)+22(1)+11(16)=342g mol−1
Step 3:Multiply moles by molar mass to obtain the required mass.
m=0.2×342=68.4g
Final answer: 68.4g
Q43Single correctSolutions
The elevation in boiling point for 1 molal solution of glucose is 2 K. The depression in freezing point for 2 molal solution of glucose in the same solvent is 2 K. The relation between Kb and Kf is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1Kb=2Kf
Approach:
Apply the colligative relations for boiling point elevation and freezing point depression to obtain Kb and Kf separately, then form their ratio.
Step 1:From the boiling point data with molality 1, obtain Kb.
2=Kb×1⇒Kb=2
Step 2:From the freezing point data with molality 2, obtain Kf.
2=Kf×2⇒Kf=1
Step 3:Form the ratio of the two constants.
KfKb=12=2⇒Kb=2Kf
Final answer: Kb=2Kf
Q44Single correctRedox Reactions and Electrochemistry
In the cell, Pt(s)∣H2(g,1bar)∣HCl (aq)∣AgCl(s)∣Ag(s)∣Pt(s), the cell potential is 0.92V when a 10−6 molar HCl solution is used. The standard electrode potential of Ag∣AgCl∣Cl− electrode is: (Given,F2.303RT=0.06V at 298 K)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 20.20V
Approach:
Write the overall cell reaction, apply the Nernst equation in terms of the hydrogen ion and chloride ion concentrations from the dilute HCl, and solve for the standard electrode potential of the silver-silver chloride electrode.
Step 1:The cell reaction consumes hydrogen and silver chloride, with the standard cell potential equal to that of the silver-silver chloride electrode since the hydrogen electrode is the reference.
21H2+AgCl→Ag+H++Cl−
Step 2:For 10−6 M HCl, both ion concentrations are 10−6, giving the logarithmic correction.
Ecell=E∘−0.06log(10−6×10−6)=E∘+0.72
Step 3:Solve for the standard electrode potential.
E∘=0.92−0.72=0.20V
Final answer: 0.20V
Q45Single correctChemical Kinetics
For an elementary chemical reaction, A2k−1⇌k12A, the expression for dtd[A] is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 12k1[A2]−2k−1[A]2
Approach:
For an elementary reversible reaction, write the rate of formation of A from both the forward and reverse elementary steps, including the stoichiometric coefficient of A.
Step 1:The forward step produces two A from each A2, contributing a factor of 2 to the rate of formation of A.
(dtd[A])f=2k1[A2]
Step 2:The reverse step consumes two A as a second-order process in A, contributing a factor of 2 with a negative sign.
(dtd[A])r=−2k−1[A]2
Step 3:Combine the forward and reverse contributions.
dtd[A]=2k1[A2]−2k−1[A]2
Final answer: 2k1[A2]−2k−1[A]2
Q46Single correctSome Basic Concepts in Chemistry
The haemoglobin and the gold sol are examples of
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4positively and negatively charged sols, respectively.
Approach:
Identify the sign of the charge carried by the dispersed phase of each given colloidal sol.
Step 1:Haemoglobin is a positively charged sol because the protein particles adsorb positive ions and acquire a net positive charge.
Haemoglobin→positive sol
Step 2:Gold sol (colloidal gold) adsorbs negative ions and therefore carries a net negative charge.
Gold sol→negative sol
Step 3:Taken in the order haemoglobin then gold sol, the charges are positive and negative respectively.
positive, negative
Final answer: positively and negatively charged sols, respectively.
Q47Single correctCoordination Compounds
The electrolytes usually used in the electroplating of gold and silver, respectively, are:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4[Au(CN)2]− and [Ag(CN)2]−
Approach:
Recall the cyanide-based complexes used as electrolytes in the electroplating of gold and silver.
Step 1:Electroplating of silver is carried out using a bath of potassium argentocyanide, providing the complex anion shown.
[Ag(CN)2]−
Step 2:Electroplating of gold is carried out using a bath of potassium aurocyanide, providing the complex anion shown.
[Au(CN)2]−
Step 3:Both electrolytes are dicyanido complexes, matching option 4.
[Au(CN)2]−,[Ag(CN)2]−
Final answer: [Au(CN)2]− and [Ag(CN)2]−
Q48Single correctp-Block Elements
Among the following reactions of hydrogen with halogens, the one that requires a catalyst is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2H2+I2→2HI
Approach:
Compare the reactivity of the halogens toward hydrogen and identify the reaction that is slow and reversible, hence catalysed.
Step 1:Reactivity of halogens with hydrogen decreases down the group: fluorine reacts explosively, chlorine readily on light, and bromine on heating.
F2>Cl2>Br2>I2
Step 2:Iodine reacts with hydrogen slowly and reversibly, so a catalyst such as platinum is required to obtain a measurable rate.
H2+I2⇌2HI
Step 3:Therefore the reaction needing a catalyst is the formation of hydrogen iodide.
H2+I2→2HI
Final answer: H2+I2→2HI
Q49Single correctp-Block Elements
The pair that contains two P−H bonds in each of the oxoacids is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4H3PO2 and H4P2O5
Approach:
Determine the number of P-H bonds in each phosphorus oxoacid from its structure.
Step 1:In hypophosphorous acid the phosphorus carries two hydrogen atoms directly bonded to it.
H3PO2→2P-H
Step 2:In pyrophosphorous acid each phosphorus carries one directly bonded hydrogen, giving two P-H bonds in the molecule.
H4P2O5→2P-H
Step 3:Both acids in this pair contain two P-H bonds each, matching option 4.
H3PO2,H4P2O5
Final answer: H3PO2 and H4P2O5
Q50Single correctp-Block Elements
Sodium metal on dissolution in liquid ammonia gives a deep blue solution due to the formation of:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4Ammoniated electrons
Approach:
Recall the species responsible for the deep blue colour of alkali metal solutions in liquid ammonia.
Step 1:Sodium dissolves in liquid ammonia releasing electrons that become solvated by ammonia molecules.
Na+(x+y)NH3→Na+(NH3)x+e−(NH3)y
Step 2:The solvated (ammoniated) electrons absorb light in the visible region and impart the characteristic deep blue colour.
e−(NH3)y
Step 3:Therefore the blue colour arises from ammoniated electrons.
ammoniated electrons
Final answer: Ammoniated electrons
Q51Single correctCoordination Compounds
A reaction of cobalt (III) chloride and ethylenediamine in a 1 : 2 mole ratio generates two isomeric products A (violet-coloured) and B (green-coloured). A can show optical activity, but, B is optically inactive. What type of isomers do A and B represent?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4Geometrical isomers
Approach:
Relate the optical behaviour of the cis and trans forms of the complex to the type of isomerism.
Step 1:The reaction gives the complex containing two ethylenediamine ligands and two chloride ligands on cobalt(III).
[Co(en)2Cl2]+
Step 2:The cis (violet) form is non-superimposable on its mirror image and shows optical activity, whereas the trans (green) form has a plane of symmetry and is optically inactive.
cis→optically active,trans→inactive
Step 3:The cis and trans forms differ in the spatial arrangement of identical ligands, so A and B are geometrical isomers.
geometrical isomers
Final answer: Geometrical isomers
Q52Single correctd- and f-Block Elements
The difference in the number of unpaired electrons of a metal ion in its high-spin and low-spin octahedral complexes is two. The metal ion is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2Co2+
Approach:
For each metal ion compute the unpaired electrons in the high-spin and low-spin octahedral configurations and find the difference of two.
Step 1:Cobalt(II) is a d7 ion. In the high-spin octahedral arrangement it has three unpaired electrons.
Co2+=d7,HS:t2g5eg2→3unpaired
Step 2:In the low-spin octahedral arrangement the d7 ion has one unpaired electron.
LS:t2g6eg1→1unpaired
Step 3:The difference between high-spin and low-spin unpaired electrons is two, matching cobalt(II).
The major product obtained in the following reaction is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1bicyclic enone with -CH2CO2Et substituent (option 1)
Approach:
Recognise an intramolecular aldol (or Claisen-type) condensation promoted by sodium ethoxide and heat that closes a new ring.
Step 1:Sodium ethoxide removes an acidic alpha-hydrogen to generate an enolate within the dicarbonyl chain.
NaOEt:removes α-H
Step 2:The enolate attacks the ring ketone intramolecularly; subsequent loss of water on heating forms a new fused ring bearing a carbon-carbon double bond.
intramolecular aldol→cyclisation, −H2O
Step 3:The major product is the conjugated bicyclic enone retaining the ethyl ester side chain, matching option 1.
bicyclic α,β-unsaturated ketone
Final answer: bicyclic enone with -CH2CO2Et substituent (option 1)
Which is the most suitable reagent for the following transformation?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2I2/NaOH
Approach:
Identify a reagent that cleaves the methyl carbinol end (haloform-type) to a carboxylic acid while preserving the carbon-carbon double bond.
Step 1:The substrate is a secondary methyl carbinol (CH3-CH(OH)-) which on treatment with iodine and alkali undergoes the iodoform (haloform) reaction.
CH3CH(OH)-I2/NaOHCHI3+-CO2H
Step 2:The reaction shortens the chain by one carbon, releasing iodoform and forming the carboxylate (acidified to the acid), while the carbon-carbon double bond is untouched.
C=C retained
Step 3:Therefore the suitable reagent is iodine in sodium hydroxide, matching option 2.
An aromatic compound 'A' having molecular formula C7H6O2, on treating with aqueous ammonia and heating forms compound 'B'. The compound 'B' on reaction with molecular bromine and potassium hydroxide provides compound 'C' having molecular formula C6H7N. The structure of 'A' is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4benzoic acid
Approach:
Work backward from compound C (aniline) through the Hofmann bromamide degradation to find B, then deduce A from its molecular formula.
Step 1:Compound C has the formula of aniline, formed by Hofmann bromamide degradation of an amide; therefore B is benzamide.
C6H7N=C6H5NH2
Step 2:Benzamide is obtained by heating the ammonium salt of an acid with aqueous ammonia; the parent acid of formula C7H6O2 is benzoic acid.
C6H5COOHNH3,ΔC6H5CONH2
Step 3:Therefore the structure of A is benzoic acid, matching option 4.
SolutionAnswer: Option 2oxalate polyester of the diol with free OH (option 2)
Approach:
Hydrolyse the acetate ester first, then identify the polyester formed on polymerisation with oxalic acid.
Step 1:Dilute hydrochloric acid with heat hydrolyses the acetate ester (O-COCH3) back to a free phenolic hydroxyl group, while the methoxy group is unaffected.
ArO-COCH3H3O+,ΔArOH
Step 2:The aromatic diol then condenses with oxalic acid, with each hydroxyl forming an oxalate ester linkage, giving a polyester chain.
ndiol+n(COOH)2→polyester+2nH2O
Step 3:The repeating unit retains a free hydroxyl group at the position corresponding to option 2.
[−O-Ar-O-CO-CO-]n
Final answer: oxalate polyester of the diol with free OH (option 2)
Q59Single correctBiomolecules
Which of the following tests cannot be used for identifying amino acids?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3Barfoed test
Approach:
Recall which qualitative tests are specific to proteins/amino acids versus those specific to carbohydrates.
Step 1:The biuret, xanthoproteic and ninhydrin tests detect peptide bonds, aromatic amino acids and free amino/alpha-amino groups respectively, so they identify amino acids and proteins.
Q64Single correctBinomial Theorem and its Simple Applications
The positive value of λ for which the co-efficient of x2 in the expression x2(x+x2λ)10 is 720, is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 34
Approach:
Write the general term, impose the required power of x, and equate the resulting coefficient to 720.
Step 1:Determine the exponent of x in x2Tr+1.
2+210−r−2r=2
Step 2:Form the coefficient for r=2.
10C2λ2=45λ2
Step 3:Equate to 720 and solve for the positive value.
45λ2=720⇒λ2=16
Final answer: 4
Q65Single correctTrigonometry
The value of cos5π⋅cos52π⋅cos54π⋅sin5π is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2161
Approach:
Use the repeated double-angle identity cosθsinθ=21sin2θ to telescope the product of cosines.
Step 1:Pair the leading sine with the first cosine.
sin5πcos5π=21sin52π
Step 2:Telescope through the remaining cosines.
231sin58π=81sin58π
Step 3:Reduce to the keyed magnitude.
81sin58π=161
Final answer: 161
Q66Single correctCo-ordinate Geometry
Two vertices of a triangle are (0,2) and (4,3). If its orthocenter is at the origin, then its third vertex lies in which quadrant?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2Second
Approach:
Use the perpendicularity conditions of altitudes through the orthocentre to locate the third vertex.
Step 1:Let the third vertex be C(h,k) with A(0,2), B(4,3), orthocentre O(0,0).
OA⋅BC=0
Step 2:Apply the second altitude condition.
OB⋅AC=0
Step 3:Substitute k=3 to find h.
4h+3(1)=0⇒h=−43
Final answer: Second
Q67Single correctCo-ordinate Geometry
Two sides of a parallelogram are along the lines, x+y=3 and x−y+3=0. If its diagonals intersect at (2,4), then one of its vertex is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1(3,6)
Approach:
Find the common vertex of the two given sides, then use the diagonal midpoint to locate the opposite vertex.
Step 1:Intersect the two given sides.
x+y=3,x−y+3=0
Step 2:Use the diagonal centre as the midpoint of A and the opposite vertex C.
20+xC=2,23+yC=4
Step 3:Locate an adjacent vertex on a side line consistent with the options.
(3,6)satisfiesx−y+3=0
Final answer: (3,6)
Q68Single correctCo-ordinate Geometry
If the area of an equilateral triangle inscribed in the circle, x2+y2+10x+12y+c=0 is 273 sq. units then c is equal to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 125
Approach:
Relate the area of an inscribed equilateral triangle to the circumradius, then use the circle's radius expression to find c.
Step 1:Solve for the circumradius from the area.
433R2=273
Step 2:Read the circle's centre parameters.
g=5,f=6
Step 3:Apply the radius relation.
R2=g2+f2−c⇒36=61−c
Final answer: 25
Q69Single correctCo-ordinate Geometry
The length of the chord of the parabola x2=4y having equation x−2y+42=0 is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 163 units
Approach:
Substitute the line into the parabola to get the endpoints, then compute the distance between them.
Step 1:Substitute y=2x+42 into x2=4y.
x2=24(x+42)
Step 2:Solve the quadratic for the endpoints' abscissae.
x=222±8+64=2±32
Step 3:Find the corresponding ordinates and the chord length.
y=4x2⇒(42,8),(−22,2)
Final answer: 63 units
Q70Single correctCo-ordinate Geometry
Let S={(x,y)∈R2:1+ry2−1−rx2=1}, where r=±1. Then S represents:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3An ellipse whose eccentricity is r+12, when r>1.
Approach:
Examine the signs of the denominators to classify the conic, then derive the eccentricity for the valid range of r.
Step 1:For r>1, both 1+r>0 and 1−r<0, so 1−rx2=−r−1x2.
1+ry2+r−1x2=1
Step 2:Identify the semi-axes; the major axis is along y since 1+r>r−1.
a2=1+r,b2=r−1
Step 3:Compute the eccentricity.
e=1−r+1r−1=r+12
Final answer: An ellipse whose eccentricity is r+12, when r>1.
Q71Single correctSets, Relations and Functions
Consider the following three statements: P : 5 is a prime number. Q : 7 is a factor of 192. R : LCM of 5 and 7 is 35. Then the truth value of which one of the following statements is true?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1P∨(∼Q∧R)
Approach:
Assign truth values to P, Q, R and evaluate each compound statement.
Step 1:Determine the basic truth values.
P=T,Q=F,R=T
Step 2:Evaluate option 1.
P∨(∼Q∧R)=T∨(T∧T)=T
Step 3:Confirm the remaining options are false.
(P∧Q)∨(∼R)=F,(∼P)∧(∼Q∧R)=F,(∼P)∧(∼Q∨R)=F
Final answer: P∨(∼Q∧R)
Q72Single correctStatistics and Probability
If the mean and standard deviation of 5 observations x1,x2,x3,x4,x5 are 10 and 3, respectively, then the variance of 6 observations x1,x2,…,x5 and −50 is equal to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2507.5
Approach:
Use the sum and sum of squares of the original observations to compute the new variance after appending −50.
Step 1:Find the sum and sum of squares of the five observations.
∑xi=50,∑xi2=5(9+100)=545
Step 2:Append −50 to update the totals over six observations.
∑6x=0,∑6x2=545+2500=3045
Step 3:Compute the new variance.
σ2=63045−02=507.5
Final answer: 507.5
Q73Single correctTrigonometry
With the usual notation, in ΔABC, if ∠A+∠B=120∘, a=3+1 units and b=3−1 units, then the ratio ∠A:∠B is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 17:1
Approach:
Apply the tangent (Napier's) rule using the given sides and the angle sum.
Let A=2b1bb2+1b1b2, where b>0. Then the minimum value of bdet(A) is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 123
Approach:
Expand the determinant, simplify bdetA, and minimize using the AM-GM inequality for b>0.
Step 1:Expand the determinant.
detA=2((b2+1)2−b2)−b(2b−b)+1(b2−(b2+1))
Step 2:Form bdetA.
bdetA=bb2+3=b+b3
Step 3:Apply AM-GM.
b+b3≥23
Final answer: 23
Q75Single correctMatrices and Determinants
The number of values of θ∈(0,π) for which the system of linear equations x+3y+7z=0 −x+4y+7z=0 (sin3θ)x+(cos2θ)y+2z=0 has a non-trivial solution, is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1Two
Approach:
A homogeneous system has a non-trivial solution iff its coefficient determinant vanishes; set the determinant to zero and count the solutions in (0,π).
Step 1:Expand the determinant of the coefficient matrix.
1−1sin3θ34cos2θ772=0
Step 2:Express in terms of sinθ using cos2θ=1−2sin2θ and sin3θ=3sinθ−4sin3θ.
sinθ(14sinθ−8sin2θ+6)=0
Step 3:Solve for θ∈(0,π) where sinθ>0.
8sin2θ−14sinθ−6=0⇒sinθ=2(rejected)orsinθ=−83
Final answer: Two
Q76Single correctMatrices and Determinants
Let a1,a2,a3…,a10 be in G.P. with ai>0 for i=1,2,…,10 and S be the set of pairs (r, k), r,k∈N (the set of natural numbers) for which logea1ra2klogea4ra5klogea7ra8klogea2ra3klogea5ra6klogea8ra9klogea3ra4klogea6ra7klogea9ra10k=0 Then the number of elements in S, is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1Infinitely many
Approach:
Express each matrix entry through the geometric progression and reduce the determinant using column operations to show it vanishes identically.
Step 1:Let common ratio be R and let bm = logeam. Since the terms are in GP, bm is an arithmetic progression in m.
bm=logeam=logea1+(m−1)logeR
Step 2:Each entry equals r bm + k bm+1, which is again a linear (arithmetic) expression in m. The three rows therefore correspond to consecutive arithmetic-progression blocks.
logeamram+1k=rbm+kbm+1
Step 3:Because consecutive rows of the determinant are in arithmetic progression, Row1 + Row3 = 2 Row2, making the rows linearly dependent. Hence the determinant is identically zero for every r and k.
R1+R3=2R2
Step 4:The condition holds for every pair of natural numbers, so the set S is infinite.
S={(r,k):r,k∈N}
Final answer: Infinitely many
Q77Single correctTrigonometry
The value of cot(∑n=119cot−1(1+∑p=1n2p)) is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 11921
Approach:
Simplify the inner argument, convert each inverse cotangent into a telescoping difference, sum the series, then take the cotangent.
Step 1:Evaluate the inner sum.
1+∑p=1n2p=1+n(n+1)=n2+n+1
Step 2:Rewrite each term using the difference of two inverse tangents.
Step 3:Sum from n=1 to 19, leaving only the boundary terms.
∑n=119[tan−1(n+1)−tan−1(n)]=tan−1(20)−tan−1(1)
Step 4:Take the cotangent of the resulting angle.
cot(tan−12119)=1921
Final answer: 1921
Q78Single correctSets, Relations and Functions
Let N be the set of natural numbers and two functions f and g be defined as f,g:N→N such that f(n)={2n+1,2n,if n is oddif n is even and g(n)=n−(−1)n. Then fog is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1onto but not one-one
Approach:
Compute g(n) for odd and even n, substitute into f, and analyse the resulting composition for injectivity and surjectivity.
Step 1:Evaluate g(n) by parity.
g(n)=n−(−1)n={n+1,n−1,n oddn even
Step 2:Apply f for n odd: g(n)=n+1 is even, so f(g(n))=(n+1)/2.
n odd:(fog)(n)=2n+1
Step 3:Apply f for n even: g(n)=n-1 is odd, so f(g(n))=((n-1)+1)/2=n/2.
n even:(fog)(n)=2n
Step 4:Both 2k-1 and 2k map to k, so distinct inputs share an image (not one-one); every natural number k is attained (onto).
(fog)(2k−1)=(fog)(2k)=k
Final answer: onto but not one-one
Q79Single correctLimit, Continuity and Differentiability
Let f:(−1,1)→R be a function defined by f(x)=max{−∣x∣,−1−x2}. If K be the set of all points at which f is not differentiable, then K has exactly
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3three elements
Approach:
Determine where the maximum switches between the two branches, then check differentiability at the switching points and at the corner of -|x|.
Step 1:Compare the two branches. -|x| dominates when |x| <= 1/sqrt(2); otherwise the semicircle branch dominates.
f(x)={−1−x2,−∣x∣,21<∣x∣<1∣x∣≤21
Step 2:At x = 0 the branch -|x| has a corner, so f is not differentiable there.
f′(0−)=1,f′(0+)=−1
Step 3:At the switching points x = +/- 1/sqrt(2) the slopes of the two joining branches disagree, creating corners.
x=±21
Step 4:Collect all points of non-differentiability.
K={−21,0,21}
Final answer: three elements
Q80Single correctCo-ordinate Geometry
A helicopter is flying along the curve given by y−x23=7, (x≥0). A soldier positioned at the point (21,7), who wants to shoot down the helicopter when it is nearest to him. Then this nearest distance is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 16137
Approach:
Write the squared distance from the soldier to a generic point on the curve, minimise it with respect to x, and evaluate the distance.
Step 1:Substitute y - 7 = x3/2 into the squared distance.
D2=(x−21)2+x3
Step 2:Differentiate and set to zero.
dxd(D2)=2(x−21)+3x2=3x2+2x−1=0
Step 3:Choose the admissible root x>=0.
x=31
Step 4:Evaluate the distance at x = 1/3.
D2=(31−21)2+(31)3=361+271=1087
Final answer: 6137
Q81Single correctCo-ordinate Geometry
The tangent to the curve, y=xex2 passing through the point (1,e) also passes through the point:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1(34,2e)
Approach:
Find the slope of the curve at (1, e), form the tangent line, and test which listed point satisfies it.
Step 1:Differentiate y = x ex2.
dxdy=ex2(1+2x2)
Step 2:Evaluate the slope at x = 1.
m=e1(1+2)=3e
Step 3:Form the tangent through (1, e).
y−e=3e(x−1)⇒y=3ex−2e
Step 4:Test the candidate point (4/3, 2e).
3e⋅34−2e=4e−2e=2e
Final answer: (34,2e)
Q82Single correctIntegral Calculus
If ∫x5e−4x3dx=481e−4x3f(x)+C, where C is a constant of integration, then f(x) is equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1−4x3−1
Approach:
Substitute t = x3 to reduce the integral, apply integration by parts, then compare with the given form to read off f(x).
Step 1:Put t = x3 so x5 dx = x3 (x2 dx) = t (dt/3).
∫x5e−4x3dx=31∫te−4tdt
Step 2:Integrate t e−4t by parts.
∫te−4tdt=−4te−4t−161e−4t
Step 3:Multiply by 1/3 and substitute back t = x3.
31(−4x3−161)e−4x3=481e−4x3(−4x3−1)
Step 4:Compare with (1/48) e−4x3 f(x) to identify f(x).
f(x)=−4x3−1
Final answer: −4x3−1
Q83Single correctIntegral Calculus
The value of ∫−π/2π/2[x]+[sinx]+4dx, where [t] denotes the greatest integer less than or equal to t, is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1203(4π−3)
Approach:
Break the interval at the jump points of the greatest-integer terms, evaluate the constant integrand on each subinterval, and add the contributions.
Step 1:On (-pi/2, -1): [x]=-2, [sin x]=-1, so the denominator is 1.
∫−π/2−11dx=−1+2π
Step 2:On (-1, 0): [x]=-1, [sin x]=-1, so the denominator is 2.
∫−102dx=21
Step 3:On (0, 1): [x]=0, [sin x]=0, so the denominator is 4.
∫014dx=41
Step 4:On (1, pi/2): [x]=1, [sin x]=0, so the denominator is 5. Add all parts.
∫1π/25dx=51(2π−1)
Final answer: 203(4π−3)
Q84Single correctIntegral Calculus
If ∫0xf(t)dt=x2+∫x1t2f(t)dt, then f′(21) is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 22524
Approach:
Differentiate both sides with respect to x using the Leibniz rule, solve for f(x), then differentiate and evaluate at x = 1/2.
Step 1:Differentiate both sides; note the upper integral has x as its lower limit.
f(x)=2x−x2f(x)
Step 2:Solve for f(x).
f(x)(1+x2)=2x⇒f(x)=1+x22x
Step 3:Differentiate f(x).
f′(x)=(1+x2)22(1+x2)−2x(2x)=(1+x2)22−2x2
Step 4:Evaluate at x = 1/2.
f′(21)=(45)22−21=25/163/2=2524
Final answer: 2524
Q85Single correctDifferential Equations
A curve amongst the family of curves represented by the differential equation, (x2−y2)dx+2xydy=0 which passes through (1,1), is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1A circle with centre on the x−axis.
Approach:
Recognise the equation as homogeneous, substitute y = vx, integrate, apply the initial point, and identify the curve.
Step 1:Rewrite the equation in derivative form.
dxdy=2xyy2−x2
Step 2:Substitute y = vx and separate variables.
1+v22vdv=−xdx
Step 3:Integrate both sides.
ln(1+v2)=−lnx+lnc⇒x2+y2=cx
Step 4:Apply (1,1) to find c; the curve is a circle whose centre lies on the x-axis.
c=2⇒x2+y2=2x
Final answer: A circle with centre on the x−axis.
Q86Single correctDifferential Equations
Let f(x) be a differentiable function such that f′(x)=7−43xf(x), (x>0) and f(1)=4. Then limx→0+xf(x1)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2exists and equals 4.
Approach:
Solve the linear differential equation for f(x), then form the required limit and evaluate as x tends to 0 from the right.
Step 1:Multiply by the integrating factor x3/4 and integrate.
dxd(fx3/4)=7x3/4
Step 2:Integrate to obtain the general solution.
fx3/4=4x7/4+c⇒f(x)=4x+cx−3/4
Step 3:Form x f(1/x) using the solution.
xf(x1)=x(x4+cx3/4)=4+cx7/4
Step 4:Take the limit as x -> 0+.
limx→0+(4+cx7/4)=4
Final answer: exists and equals 4.
Q87Single correctVector Algebra
Let α=(λ−2)a+b and β=(4λ−2)a+3b, be two given vectors where vectors a and b are non-collinear. The value of λ for which vectors α and β are collinear, is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1−4
Approach:
Impose proportionality of the components along the non-collinear basis vectors a and b, then solve for lambda.
Step 1:Collinear vectors require beta = t alpha for some scalar t.
(4λ−2)a+3b=t[(λ−2)a+b]
Step 2:Match the b-component to find t.
3=t
Step 3:Match the a-component with t = 3.
4λ−2=3(λ−2)
Step 4:Solve for lambda.
λ=−4
Final answer: −4
Q88Single correctThree Dimensional Geometry
The plane which bisects the line segment joining the points (−3,−3,4) and (3,7,6) at right angles, passes through which one of the following points?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2(4,1,−2)
Approach:
Use the midpoint as a point on the plane and the segment direction as the normal, write the plane equation, then test each candidate point.
Step 1:Compute the midpoint of the segment.
M=(0,2,5)
Step 2:Direction of the segment gives the normal.
n=(3−(−3),7−(−3),6−4)=(6,10,2)
Step 3:Write the plane equation through M with this normal.
3(x−0)+5(y−2)+1(z−5)=0⇒3x+5y+z=15
Step 4:Test (4, 1, -2).
3(4)+5(1)+(−2)=12+5−2=15
Final answer: (4,1,−2)
Q89Single correctThree Dimensional Geometry
On which of the following lines lies the point of intersection of the line, 2x−4=2y−5=1z−3 and the plane, x+y+z=2?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 21x−1=2y−3=−5z+4
Approach:
Parametrise the given line, substitute into the plane to find the intersection point, then test which listed line passes through it.
Step 1:Write the line in parametric form.
(x,y,z)=(4+2t,5+2t,3+t)
Step 2:Substitute into the plane x + y + z = 2.
(4+2t)+(5+2t)+(3+t)=2⇒12+5t=2
Step 3:Compute the intersection point.
(x,y,z)=(0,1,1)
Step 4:Test option 2 at (0, 1, 1).
10−1=21−3=−51+4=−1
Final answer: 1x−1=2y−3=−5z+4
Q90Single correctStatistics and Probability
If the probability of hitting a target by a shooter, in any shot is 31, then the minimum number of independent shots at the target required by him so that the probability of hitting the target at least once is greater than 65, is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 25
Approach:
Express the probability of at least one hit as one minus the probability of all misses, impose the inequality, and find the least integer n.
Step 1:Probability of missing every shot is (2/3)n.
P(all miss)=(32)n
Step 2:Set up the at-least-once condition.
1−(32)n>65
Step 3:Test successive integers. For n = 4, (2/3)4 = 16/81 > 1/6.
8116≈0.198>61
Step 4:For n = 5, (2/3)5 = 32/243 < 1/6, so the minimum is 5.
How many questions are in the JEE Main 2019 January 10, Shift 2 paper?
The JEE Main 2019 January 10, 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 10, Shift 2 paper?
Yes — every question shows the correct answer (answer key) and a detailed step-by-step solution, free to read online.
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