JEE Main 2019 April 09, Shift 1 Question Paper with Solutions
All 89 questions from the JEE Main 2019 (April 09, Shift 1) shift — Physics (30), Chemistry (29) 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.
In the density measurement of a cube, the mass and edge length are measured as (10.00±0.10) kg and (0.10±0.01) m, respectively. The error in the measurement of density is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 10.31
Approach:
The fractional error in density follows from the fractional errors in mass and edge length, since density equals mass divided by the cube of the edge length.
Step 1:Express the fractional error in density.
ρΔρ=mΔm+3LΔL
Step 2:Substitute the measured values.
ρΔρ=10.000.10+3×0.100.01
Step 3:Add the contributions.
ρΔρ=0.31
Final answer: 0.31
Q2Single correctLaws of Motion
A ball is thrown vertically up (taken as +x - axis) from the ground. The correct momentum-height (p−h) diagram is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3Graph (3)
Approach:
The momentum of the ball varies with height through energy conservation, and its sign changes between the upward and downward journeys.
Step 1:Relate momentum to height by conservation of mechanical energy.
2mp2=2mp02−mgh
Step 2:Identify the curve shape during ascent.
p=+p02−2m2gh
Step 3:Identify the curve shape during descent, where momentum is along the negative axis.
p=−p02−2m2gh
Final answer: Graph (3)
Q3Single correctKinematics
The stream of a river is flowing with a speed of 2 km h−1. A swimmer can swim at a speed of 4 km h−1. The direction of the swimmer with respect to the flow of the river, to cross the river straight, is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3120∘
Approach:
To cross straight, the upstream component of the swimming velocity must cancel the river flow, fixing the swimmer's heading.
Step 1:Set the upstream component of swim velocity equal to the stream speed, with ϕ measured from the cross-stream direction.
4sinϕ=2
Step 2:Solve for the angle from the cross-stream direction.
ϕ=30∘
Step 3:Convert to the angle measured from the direction of river flow.
θ=90∘+30∘=120∘
Final answer: 120∘
Q4Single correctWork, Energy and Power
A uniform cable of mass M and length L is placed on a horizontal surface such that its (n1)th part is hanging below the edge of the surface. To lift the hanging part of the cable upto the surface, the work done should be:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 12n2MgL
Approach:
The work equals the gain in gravitational potential energy of the hanging portion, lifted by the height of its centre of mass.
Step 1:Find the mass of the hanging part.
m=nM
Step 2:Locate its centre of mass below the edge, at half the hanging length.
hcm=21⋅nL=2nL
Step 3:Compute the work to raise the hanging part to the surface.
W=nMg2nL=2n2MgL
Final answer: 2n2MgL
Q5Single correctLaws of Motion
A body of mass 2 kg makes an elastic collision with a second body at rest and continues to move in the original direction but with one fourth of its original speed. What is the mass of the second body?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 21.2 kg
Approach:
For a one-dimensional elastic collision with a stationary target, the first body's final velocity is fixed by the mass ratio.
Step 1:Apply the elastic-collision result with the second body initially at rest, with the first body retaining a quarter of its speed.
4u=m1+m2m1−m2u
Step 2:Cross-multiply and solve for the second mass.
2+m2=4(2−m2)
Step 3:Obtain the mass.
m2=1.2kg
Final answer: 1.2 kg
Q6Single correctRotational Motion
A stationary horizontal disc is free to rotate about its axis. When a torque is applied on it, its kinetic energy as a function of θ, where θ is the angle by which it has rotated, is given as kθ2. If its moment of inertia is I then the angular acceleration of the disc is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1I2kθ
Approach:
Equate the given rotational kinetic energy to the standard expression to find angular speed, then differentiate to obtain angular acceleration.
Step 1:Set the kinetic energy expressions equal.
21Iω2=kθ2
Step 2:Differentiate with respect to angle.
2ωdθdω=I4kθ
Step 3:Identify the left side as angular acceleration; alternatively divide energy relation directly.
α=ωdθdω=I2kθ
Final answer: I2kθ
Q7Single correctRotational Motion
The following bodies are made to roll up (without slipping) the same inclined plane from a horizontal plane: (i) a ring of radius R, (ii) a solid cylinder of radius 2R and (iii) a solid sphere of radius 4R. If, in each case, the speed of the center of mass at the bottom of the incline is same, the ratio of the maximum heights they climb is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 220:15:14
Approach:
Energy conservation gives the climbing height in terms of the moment-of-inertia factor, which is independent of the radius for each shape.
Step 1:Express the height climbed in terms of the shape factor k=mR2I, which is the same speed for all.
h∝(1+k)
Step 2:Compute the factor for each body.
1+1=2,1+21=23,1+52=57
Step 3:Multiply through by 10 to clear fractions.
20:15:14
Final answer: 20:15:14
Q8Single correctGravitation
A solid sphere of mass M and radius a is surrounded by a uniform concentric spherical shell of thickness 2a and mass 2M. The gravitational field at distance 3a from the centre will be:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 23a2GM
Approach:
At a point outside both the sphere and the shell, the total enclosed mass acts as if concentrated at the centre.
Step 1:Identify the total mass enclosed within the radius 3a.
Menc=M+2M=3M
Step 2:Apply the inverse-square field law at distance 3a.
E=(3a)2G(3M)
Step 3:Simplify.
E=3a2GM
Final answer: 3a2GM
Q9Single correctProperties of Solids and Liquids
If 'M' is the mass of water that rises in a capillary tube of radius 'r', then mass of water which will rise in a capillary tube of radius '2r' is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 22 M
Approach:
The capillary rise varies inversely with radius, while the mass is the product of density, cross-sectional area and rise height.
Step 1:Express the mass in terms of radius.
m=ρπr2⋅ρgr2Tcosθ=g2πTcosθr
Step 2:Form the ratio for doubled radius.
Mm′=r2r=2
Step 3:State the mass for radius 2r.
m′=2M
Final answer: 2 M
Q10Single correctThermodynamics
Following figure shows two processes A and B for a gas. If ΔQA and ΔQB are the amount of heat absorbed by the system in two cases, and ΔUA and ΔUB are changes in internal energies, respectively, then:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3ΔQA>ΔQB;ΔUA=ΔUB
Approach:
Both processes share the same initial and final states, so the internal energy change is identical; the heat differs by the work done, which is the area under each path.
Step 1:Internal energy is a state function, so equal endpoints give equal changes.
ΔUA=ΔUB
Step 2:Path A lies above path B, enclosing greater area, so it does more work.
WA>WB
Step 3:Apply the first law; with equal ΔU and larger work, more heat is absorbed in A.
ΔQA>ΔQB
Final answer: ΔQA>ΔQB;ΔUA=ΔUB
Q11Single correctKinetic Theory of Gases
For given gas at 1 atm pressure, rms speed of the molecules is 200 m/s at 127∘C. At 2 atm pressure and at 227∘C, the rms speed of the molecules will be:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 41005 m/s
Approach:
Root-mean-square speed depends only on absolute temperature, not pressure, so the ratio of speeds follows the square root of the temperature ratio.
Step 1:Convert the temperatures to kelvin.
T1=400K,T2=500K
Step 2:Take the ratio of rms speeds, noting pressure is irrelevant.
v1v2=T1T2=400500=45
Step 3:Multiply by the given speed.
v2=200×25=1005m/s
Final answer: 1005 m/s
Q12Single correctKinetic Theory of Gases
An HCl molecule has rotational, translational and vibrational motions. If the rms velocity of HCl molecules in its gaseous phase is vˉ, m is its mass and kB is Boltzmann's constant, then its temperature will be:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 37kBmvˉ2
Approach:
The total average energy comes from all active degrees of freedom of the diatomic molecule, equated to the translational rms kinetic energy expression.
Step 1:Count the degrees of freedom: three translational, two rotational and two vibrational for HCl.
f=3+2+2=7
Step 2:Equate the equipartition energy to the kinetic energy.
27kBT=21mvˉ2
Step 3:Solve for the temperature.
T=7kBmvˉ2
Final answer: 7kBmvˉ2
Q13Single correctOscillations and Waves
A simple pendulum oscillating in air has period T. The bob of the pendulum is completely immersed in a non-viscous liquid. The density of the liquid is 161th of the material of the bob. If the bob is inside liquid all the time, its period of oscillation in this liquid is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 34T151
Approach:
Buoyancy reduces the effective gravitational acceleration on the bob, lengthening the period by the inverse square root of the effective gravity factor.
Step 1:Compute the effective gravity using the density ratio.
geff=g(1−161)=1615g
Step 2:Relate the new period to the original through the gravity ratio.
TT′=geffg=1516
Step 3:Express the new period.
T′=4T151
Final answer: 4T151
Q14Single correctOscillations and Waves
A string is clamped at both the ends and it is vibrating in its 4th harmonic. The equation of the stationary wave is y=0.3sin(0.157x)cos(200πt). The length of the string is: (All quantities are in SI units.)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 480 m
Approach:
The wave number gives the wavelength, and for the fourth harmonic the string length equals two full wavelengths.
Step 1:Extract the wavelength from the wave number.
λ=0.1572π=40m
Step 2:Apply the harmonic condition for the fourth harmonic.
L=24λ=2λ
Step 3:Compute the length.
L=80m
Final answer: 80 m
Q15Single correctOscillations and Waves
The pressure wave, P=0.01sin[1000t−3x]N m−2, corresponds to the sound produced by a vibrating blade on a day when atmospheric temperature is 0∘C. On some other day when temperature is T, the speed of sound produced by the same blade and at the same frequency is found to be 336 m s−1. Approximate value of T is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 34∘ C
Approach:
The wave equation gives the speed of sound at zero degrees; the speed varies as the square root of absolute temperature, which fixes the new temperature.
Step 1:Determine the speed at zero degrees from the wave parameters.
v0=31000≈333m/s
Step 2:Relate the two speeds through absolute temperatures.
333336=273T
Step 3:Solve for the absolute temperature and convert to Celsius.
T≈273×1.018≈277K≈4∘C
Final answer: 4∘ C
Q16Single correctElectrostatics
A system of three charges are placed as shown in the figure: If D≫d, the potential energy of the system is best given by:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 14πϵ01[−dq2−D2qQd]
Approach:
The total potential energy is the sum of interaction energies for the three pairs: the two charges of the dipole (+q and −q separated by d) and their interactions with the distant charge Q. The dipole-charge interaction is expanded for D≫d.
Step 1:Interaction of the +q and −q pair separated by distance d.
U12=4πϵ01d(+q)(−q)=−4πϵ01dq2
Step 2:The pair +q,−q forms a dipole of moment p=qd; its interaction with charge Q at distance D along the axis varies as 1/D2.
Udipole−Q=−4πϵ01D2qQd
Step 3:Adding both contributions gives the total potential energy of the system.
U=4πϵ01[−dq2−D2qQd]
Final answer: 4πϵ01[−dq2−D2qQd]
Q17Single correctElectrostatics
A capacitor with capacitance 5 µF is charged to 5 µC. If the plates are pulled apart to reduce the capacitance to 2 µF, how much work is done?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 43.75×10−6J
Approach:
The charge is constant after disconnection. The energy stored is U=Q2/(2C), and the work done equals the change in stored energy as the capacitance changes from 5μF to 2μF.
Step 1:Compute initial stored energy with Q=5μC and Ci=5μF.
Ui=2×5×10−6(5×10−6)2=2.5×10−6J
Step 2:Compute final stored energy with Cf=2μF.
Uf=2×2×10−6(5×10−6)2=6.25×10−6J
Step 3:Work done equals the increase in stored energy.
W=Uf−Ui=(6.25−2.5)×10−6=3.75×10−6J
Final answer: 3.75×10−6J
Q18Single correctCurrent Electricity
A wire of resistance R is bent to form a square ABCD as shown in the figure. The effective resistance between E and C is: ( E is mid-point of arm CD )
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2647R
Approach:
Each side of the square has resistance R/4. Between E (midpoint of CD) and C, the wire forms two parallel paths: the short segment EC of length half a side, and the longer path E→D→A→B→C.
Step 1:Short path EC is half of one side, so its resistance is R/8.
R1=8R
Step 2:Long path E→D→A→B→C covers half a side plus three full sides, total length 87 of the wire.
Determine the charge on the capacitor in the following circuit:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4200μC
Approach:
In steady state no current flows through the capacitor branch. The current flows through the resistive loop, and the capacitor voltage equals the voltage across the 10Ω resistor it is connected across.
Step 1:No current passes through the 10μF branch in steady state, so the resistive loop carries current from the 72 V source through the 6Ω, 2Ω, 4Ω and 10Ω resistors. The 4Ω and 10Ω are in parallel across the central node.
R4∥10=4+104×10=1440Ω
Step 2:Solving the circuit gives a steady-state potential difference of 20 V across the 10Ω resistor that the capacitor spans.
VC=20V
Step 3:Charge on the capacitor.
Q=CVC=10×10−6×20=200×10−6C
Final answer: 200μC
Q20Single correctMoving Charges and Magnetism
A moving coil galvanometer has resistance 50 Ω and it indicates full deflection at 4 mA current. A voltmeter is made using this galvanometer and a 5 kΩ resistance. The maximum voltage, that can be measured using this voltmeter, will be close to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 320V
Approach:
A voltmeter is made by placing a high resistance in series with the galvanometer. The maximum voltage equals the full-scale current times the total series resistance.
Step 1:Total series resistance of galvanometer and added resistor.
G+R=50+5000=5050Ω
Step 2:Maximum voltage at full-scale current Ig=4mA.
V=4×10−3×5050=20.2V
Step 3:The value is closest to 20 V.
V≈20V
Final answer: 20V
Q21Single correctMoving Charges and Magnetism
A rigid square loop of side 'a' and carrying current I2 is lying on a horizontal surface near a long current I1 carrying wire in the same plane as shown in figure. The net force on the loop due to the wire will be:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2Repulsive and equal to 4πμ0I1I2
Approach:
Only the two sides of the loop parallel to the wire experience a net force, since forces on the perpendicular sides cancel. The near side (distance a) and far side (distance 2a) carry current in opposite senses relative to the wire, giving a net force.
Step 1:From the figure the near parallel side is at distance a and the far parallel side at distance 2a from the long wire.
r1=a,r2=2a
Step 2:Force on near side over length a.
F1=2πaμ0I1I2⋅a=2πμ0I1I2
Step 3:Force on far side over length a (opposite direction).
F2=2π(2a)μ0I1I2⋅a=4πμ0I1I2
Step 4:Net force is the difference; with the orientation shown the resultant is repulsive.
A rectangular coil (Dimension 5 cm × 2.5 cm ) with 100 turns, carrying a current of 3 A in the clock-wise direction, is kept centered at the origin and in the X−Z plane. A magnetic field of 1 T is applied along X− axis . If the coil is tilted through 45∘ about Z− axis , then the torque on the coil is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 40.27N m
Approach:
The torque on a current loop is τ=NIABsinθ, where θ is the angle between the magnetic moment and the field. Initially the coil lies in the X-Z plane so its normal is along Y, perpendicular to B (along X); tilting by 45∘ changes this angle.
Step 1:Area of the coil.
A=0.05×0.025=1.25×10−3m2
Step 2:Initially the normal is perpendicular to B (θ=90∘). After tilting 45∘ about Z, the angle between the moment and B becomes 45∘.
θ=45∘
Step 3:Compute the torque.
τ=100×3×1.25×10−3×1×21=20.375≈0.265N m
Final answer: 0.27N m
Q23Single correctElectromagnetic Induction
The total number of turns and cross-section area in a solenoid is fixed. However, its length L is varied by adjusting the separation between windings. The inductance of solenoid will be proportional to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3L1
Approach:
The self-inductance of a solenoid depends on the total number of turns N, cross-section area A, and length L. With N and A fixed, only the dependence on L remains.
Step 1:Express inductance using total turns N rather than turns per unit length.
Lind=μ0n2AL=μ0(LN)2AL=Lμ0N2A
Step 2:With N and A constant, inductance varies inversely with length.
Lind∝L1
Final answer: L1
Q24Single correctElectromagnetic Waves
The magnetic field of a plane electromagnetic wave is given by B=B0i^[cos(kz−ωt)]+B1j^cos(kz+ωt), where B0=3×10−5T and B1=2×10−6T. The RMS value of the force experienced by a stationary charge Q=10−4C at z=0 is closest to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 40.6N
Approach:
A stationary charge experiences force only from the electric field of the wave. The total magnetic amplitude gives the total electric amplitude via E0=cB0total, and the RMS force is QE0/2.
Step 1:Resultant magnetic amplitude from the two perpendicular components.
B=B02+B12=(3×10−5)2+(2×10−6)2≈3×10−5T
Step 2:Peak electric field at the charge.
E0=cB=3×108×3×10−5=9×103V/m
Step 3:Peak force on the charge.
F0=QE0=10−4×9×103=0.9N
Step 4:RMS value of the force.
Frms=20.9≈0.636N
Final answer: 0.6N
Q25Single correctRay Optics
A concave mirror for face viewing has a focal length of 0.4 m. The distance at which you hold the mirror from your face in order to see your image upright with a magnification of 5 is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 30.32m
Approach:
For an upright, magnified virtual image in a concave mirror, magnification m=−v/u=+5. Combining with the mirror equation gives the object distance.
Step 1:Upright magnified image gives m=+5, so v=−5u.
m=−uv=5⇒v=−5u
Step 2:Substitute into the mirror equation with f=−0.4m (concave).
−5u1+u1=−0.41
Step 3:Solve for the magnitude of object distance.
u=5×(−2.5)4=−0.32m
Final answer: 0.32m
Q26Single correctWave Optics
The figure shows a Young's double slit experimental setup. It is observed that when a thin transparent sheet of thickness t and refractive index μ is put in front of one of the slits, the central maximum gets shifted by a distance equal to n fringe width. If the wavelength of light used is λ, then t will be:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2(μ−1)nλ
Approach:
Introducing the sheet adds an extra optical path (μ−1)t, shifting the central maximum by aD(μ−1)t. Equating this shift to n fringe widths solves for t.
Step 1:Set the shift equal to n fringe widths.
aD(μ−1)t=nβ=naλD
Step 2:Cancel the common factor D/a.
(μ−1)t=nλ
Step 3:Solve for the thickness t.
t=(μ−1)nλ
Final answer: (μ−1)nλ
Q27Single correctDual Nature of Matter and Radiation
The electric field of light wave is given as E=10−3cos(5×10−72πx−2π×6×1014t)x^CN. This light falls on a metal plate of work function 2 eV . The stopping potential of the photo-electrons is: Given, E (in eV ) =λ(in A∘)12375
(A)
(B)
(C)
(D)
SolutionAnswer: Option 40.48V
Approach:
The wavelength is read from the spatial part of the wave. The photon energy is obtained from the given relation, and the stopping potential equals (photon energy minus work function) in volts.
Step 1:Identify the wavelength from the coefficient of x: k=5×10−72π, so λ=5×10−7m=5000A˚.
λ=5000A˚
Step 2:Compute the photon energy.
E=500012375=2.475eV
Step 3:Stopping potential from energy minus work function.
Vs=E−ϕ=2.475−2=0.475V
Final answer: 0.48V
Q28Single correctAtoms and Nuclei
Taking the wavelength of first Balmer line in hydrogen spectrum ( n = 3 to n = 2) as 660 nm , the wavelength of the 2nd Balmer line ( n = 4 to n = 2) will be:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2488.9nm
Approach:
The wave number for each Balmer transition follows the Rydberg formula. Taking the ratio of the two transitions eliminates the Rydberg constant and gives the unknown wavelength.
Step 1:First Balmer line (n=3→2).
λ11=R(41−91)=R365
Step 2:Second Balmer line (n=4→2).
λ21=R(41−161)=R163
Step 3:Take the ratio and solve for λ2.
λ2=λ13/165/36=660×365×316=488.9nm
Final answer: 488.9nm
Q29Single correctSemiconductor Electronics
An NPN transistor is used in common emitter configuration as an amplifier with 1 kΩ load resistance. Signal voltage of 10 mV is applied across the base-emitter. This produces a 3 mA change in the collector current and 15 µA change in the base current of the amplifier. The input resistance and voltage gain are:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 20.67kΩ,300
Approach:
Input resistance is the applied signal voltage divided by the base current change. Voltage gain is the current gain times the ratio of load to input resistance.
Step 1:Input resistance from signal voltage and base current change.
Rin=15×10−610×10−3=666.7Ω≈0.67kΩ
Step 2:Current gain from collector and base current changes.
β=ΔIBΔIC=15×10−63×10−3=200
Step 3:Voltage gain using load and input resistances.
AV=βRinRL=200×666.71000=300
Final answer: 0.67kΩ,300
Q30Single correctCommunication Systems
A signal Acosωt is transmitted using v0sinω0t as carrier wave. The correct amplitude modulated (AM) signal is:
An amplitude modulated wave is formed by varying the carrier amplitude with the message signal. Expanding the modulated carrier produces the carrier plus two side-band terms.
Step 1:Form the modulated carrier by adding the message to the carrier amplitude.
y=(v0+Acosωt)sinω0t
Step 2:Expand and apply the product-to-sum identity to the cross term.
y=v0sinω0t+Acosωtsinω0t
Step 3:Split the product term into two side bands.
y=v0sinω0t+2Asin(ω0−ω)t+2Asin(ω0+ω)t
Final answer: v0sinω0t+2Asin(ω0−ω)t+2Asin(ω0+ω)t
Chemistry29 questions
Q31Single correctSome Basic Concepts in Chemistry
For a reaction, N2(g)+3H2(g)⟶2NH3(g), identify di-hydrogen (H2) as a limiting reagent in the following reaction mixtures.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 356g of N2+10g of H2
Approach:
Compute moles of each reactant and compare with the 1:3 stoichiometric ratio; di-hydrogen is limiting when the available H2 is less than three times the available N2.
Step 1:Required ratio of moles of H2 to N2 for complete reaction is 3:1. H2 is limiting when moles of H2 are fewer than 3 times the moles of N2.
nN2nH2<3
Step 2:For 56 g N2 and 10 g H2: moles of N2 and moles of H2 are obtained from their molar masses 28 and 2.
nN2=2856=2,nH2=210=5
Step 3:Complete reaction of 2 mol N2 requires 6 mol H2, but only 5 mol H2 are present, so H2 runs out first.
5<3×2=6
Final answer: 56g of N2+10g of H2
Q32Single correctAtomic Structure
For any given series of spectral lines of atomic hydrogen, let Δvˉ=vˉmax−vˉmin be the difference in maximum and minimum wave number in cm−1. The ratio ΔvˉLyman/ΔvˉBalmer is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 49:4
Approach:
For each series the maximum wave number corresponds to the series limit and the minimum to the first line; the difference equals R times the inverse-square of the lower level minus the limiting term.
Step 1:Lyman series has n1=1. Maximum wave number occurs for n2 to infinity and minimum for n2=2.
ΔvˉLyman=R(1−∞1)−R(1−41)=4R
Step 2:Balmer series has n1=2. Maximum wave number occurs for n2 to infinity and minimum for n2=3.
ΔvˉBalmer=R(41−0)−R(41−91)=9R
Step 3:Taking the ratio of the two differences gives the required value.
ΔvˉBalmerΔvˉLyman=R/9R/4=49
Final answer: 9:4
Q33Single correctClassification of Elements and Periodicity in Properties
The element having greatest difference between its first and second ionization energies, is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2K
Approach:
Identify the element whose second electron is removed from a noble-gas core, producing the largest jump between first and second ionization energies.
Step 1:Potassium has the electronic configuration of argon plus one valence electron; removing the first electron leaves a stable noble-gas core.
K:[Ar]4s1
Step 2:The second electron must be pulled from this stable closed shell, requiring a very large energy and giving the greatest jump.
IE2≫IE1
Step 3:Ba, Ca and Sc each have at least two valence electrons, so their second ionization does not break a noble-gas core, giving smaller jumps.
Ca:[Ar]4s2
Final answer: K
Q34Single correctChemical Bonding and Molecular Structure
Among the following, the molecule expected to be stabilized by anion formation is: C2,O2,NO,F2
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2C2
Approach:
Compare the bond order of each molecule with that of its anion using molecular orbital filling; anion formation stabilizes a molecule when it increases the bond order.
Step 1:C2 has a bond order of 2 with the added electron entering a bonding pi orbital; the anion C2- gains a half-bond.
C2:B.O.=2→C2−:B.O.=2.5
Step 2:For O2, F2 and NO the additional electron enters an antibonding orbital, lowering the bond order on anion formation.
O2:2→1.5,F2:1→0.5
Step 3:Only C2 is stabilized by anion formation because its bond order rises.
Δ(B.O.)C2>0
Final answer: C2
Q35Single correctStates of Matter
Consider the van der Waal's constants, a and b, for the following gases.
Gas
Ar
Ne
Kr
Xe
a/(atm dm6mol−2)
1.3
0.2
5.1
4.1
b/(10−2dm3mol−1)
3.2
1.7
1.0
5.0
Which gas is expected to have the highest critical temperature?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1Kr
Approach:
Express the critical temperature in terms of the van der Waals constants and evaluate the a/b ratio for each gas, since critical temperature scales with a divided by b.
Step 1:Critical temperature is proportional to a/b, so the gas with the largest a/b ratio has the highest critical temperature.
Tc∝ba
Step 2:Evaluate a/b for Kr using its constants (b expressed in 0.01 dm3 mol-1 units).
baKr=1.0×10−25.1=510
Step 3:Xe gives 4.1/(5.0e-2)=82, Ar gives 1.3/(3.2e-2)=40.6, Ne gives 0.2/(1.7e-2)=11.8, all below Kr.
baXe=82,Ar=40.6,Ne=11.8
Final answer: Kr
Q37Single correctp-Block Elements
Magnesium powder burns in air to give
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1MgO and Mg3N2
Approach:
Determine the products formed when burning magnesium reacts with the principal components of air, oxygen and nitrogen.
Step 1:Magnesium reacts with atmospheric oxygen to form magnesium oxide.
2Mg+O2→2MgO
Step 2:At the high temperature of combustion magnesium also reacts with atmospheric nitrogen to form magnesium nitride.
3Mg+N2→Mg3N2
Step 3:Both products are obtained together when magnesium burns in air.
MgO+Mg3N2
Final answer: MgO and Mg3N2
Q38Single correctSome Basic Concepts in Chemistry
C60, an allotrope of carbon contains
(A)
(B)
(C)
(D)
SolutionAnswer: Option 420 hexagons and 12 pentagons.
Approach:
Use the known soccer-ball geometry of buckminsterfullerene, which combines hexagonal and pentagonal rings of carbon atoms.
Step 1:Buckminsterfullerene C60 has a closed cage shaped like a football.
C60
Step 2:The cage is built from twenty six-membered rings and twelve five-membered rings.
20 hexagons+12 pentagons
Step 3:This arrangement satisfies the count of sixty carbon atoms with each at a vertex.
60 vertices
Final answer: 20 hexagons and 12 pentagons.
Q39Single correctOrganic Chemistry - Some Basic Principles and Techniques
The correct IUPAC name of the following compound is:
Assign locants on the benzene ring so that the substituents get the lowest possible set of numbers, choosing the principal chain alphabetically when there is a tie.
Step 1:The ring carries a methyl group, a chloro group ortho to it, and a nitro group para to the methyl group.
CH3,Cl,NO2
Step 2:Numbering to give the lowest locant set places methyl at 1, chloro at 2 and nitro at 4.
{1,2,4}
Step 3:Listing substituents alphabetically gives the IUPAC name as 2-chloro-1-methyl-4-nitrobenzene.
2-chloro-1-methyl-4-nitrobenzene
Final answer: 2-chloro-1-methyl-4-nitrobenzene.
Q40Single correctHydrocarbons
The increasing order of reactivity of the following compounds towards aromatic electrophilic substitution reaction is: A: chlorobenzene; B: anisole (methoxybenzene); C: toluene; D: benzonitrile (cyanobenzene)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4D<A<C<B
Approach:
Rank the substituents by their activating or deactivating effect on the ring; stronger electron donation increases reactivity toward electrophiles while electron withdrawal decreases it.
Step 1:The cyano group is strongly electron-withdrawing, making benzonitrile (D) the least reactive.
−CNstrong deactivator
Step 2:Chlorine is weakly deactivating, methyl is weakly activating, so chlorobenzene (A) is below toluene (C).
−Cl<−CH3
Step 3:The methoxy group is a strong activator through resonance, making anisole (B) the most reactive.
−OCH3strong activator
Final answer: D<A<C<B
Q41Single correctHaloalkanes and Haloarenes
The major product of the following reaction is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1
Approach:
Alcoholic KOH causes dehydrohalogenation of the benzylic chloride to give a 4-chlorostyrene, which then undergoes free radical polymerization to a poly(4-chlorostyrene) chain.
Step 1:Alcoholic KOH eliminates HCl from the side-chain chloride to form a vinyl group attached to the para-chloro benzene ring.
ArCH2CH2ClKOH(alc.)ArCH=CH2
Step 2:Free radical polymerization of this styrene derivative links the vinyl units into a chain bearing pendant para-chlorophenyl groups.
nArCH=CH2→−(CH2CHAr)n−
Step 3:The repeating unit has the para-chlorophenyl group on alternate backbone carbons, matching option 1.
−(CH2−CH(C6H4Cl))n−
Final answer: −(CH2−CH(C6H4Cl))n−
Q42Single correctHydrocarbons
The major product of the following reaction is CH3C≡CH(i) DCl (1 equiv.)(ii) DI
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4CH3C(I)(Cl)CHD2
Approach:
Apply Markovnikov addition twice to the alkyne; each acid adds its proton (here deuterium) to the carbon bearing more hydrogens and its halide to the more substituted carbon.
Step 1:DCl adds across the triple bond with deuterium going to the terminal carbon and chlorine to the internal carbon by Markovnikov orientation.
CH3C≡CHDClCH3C(Cl)=CHD
Step 2:DI then adds to the remaining double bond, again placing iodine on the more substituted carbon and deuterium on the terminal carbon.
CH3C(Cl)=CHDDICH3C(I)(Cl)CHD2
Step 3:The final product carries both halogens on the central carbon and two deuteriums on the terminal carbon.
CH3C(I)(Cl)CHD2
Final answer: CH3C(I)(Cl)CHD2
Q43Single correctEnvironmental Chemistry
Excessive release of CO2 into the atmosphere results in
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1global warming.
Approach:
Identify the environmental consequence of carbon dioxide accumulation based on its role as a greenhouse gas.
Step 1:Carbon dioxide is a greenhouse gas that traps outgoing infrared radiation in the atmosphere.
CO2: greenhouse gas
Step 2:Increased atmospheric concentration raises the average surface temperature of the Earth.
[CO2]↑⇒T↑
Step 3:This temperature rise is known as global warming.
global warming
Final answer: global warming.
Q44Single correctSolutions
The osmotic pressure of a dilute solution of an ionic compound XY in water is four times that of a solution of 0.01M BaCl2 in water. Assuming complete dissociation of the given ionic compounds in water, the concentration of XY ( in mol L−1 ) in solution is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 46×10−2
Approach:
Express osmotic pressure as the product of the van't Hoff factor and concentration, then set the XY pressure equal to four times that of BaCl2 to solve for the XY concentration.
Step 1:BaCl2 dissociates into three ions, giving a van't Hoff factor of 3; its effective particle concentration is 3 times 0.01 M.
iBaCl2=3,πBaCl2∝3×0.01=0.03
Step 2:XY dissociates into two ions, giving a van't Hoff factor of 2, so its effective concentration is 2 times its molarity.
πXY∝2C
Step 3:Setting the XY pressure to four times that of BaCl2 and solving gives the XY concentration.
2C=4×0.03⇒C=0.06=6×10−2
Final answer: 6×10−2
Q45Single correctSolutions
Liquid M and liquid N form an ideal solution. The vapour pressures of pure liquids M and N are 450 and 700 mmHg, respectively, at the same temperature. Then correct statements is: ( xM = Mole fraction of 'M' in solution; xN = Mole fraction of 'N' in solution; yM = Mole fraction of 'M' in vapour phase; yN = Mole fraction of 'N' in vapour phase; )
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1xNxM>yNyM
Approach:
Relate the vapour-phase mole fractions to the liquid-phase mole fractions through Raoult's law and Dalton's law, then compare the two ratios using the relative volatilities.
Step 1:The ratio of vapour mole fractions equals the ratio of partial pressures, which depends on the pure vapour pressures.
yNyM=xNpN∘xMpM∘=xNxM⋅700450
Step 2:Since the pure vapour pressure of M is lower than that of N, the scaling factor is less than one.
700450<1
Step 3:Therefore the liquid ratio exceeds the vapour ratio for component M.
xNxM>yNyM
Final answer: xNxM>yNyM
Q46Single correctElectrochemistry
The standard Gibbs energy for the given cell reaction in kJ mol−1 at 298 K is: Zn(s)+Cu2+(aq)⟶Zn2+(aq)+Cu(s) E0=2 V at 298 K (Faraday's constant , F=96000 C mol−1)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4−384
Approach:
Apply the relation between standard Gibbs energy and standard cell potential, using the number of electrons transferred in the redox reaction.
Step 1:Identify the number of electrons transferred. Zinc loses two electrons and copper(II) gains two electrons.
n=2
Step 2:Substitute the values into the Gibbs energy relation.
ΔG0=−(2)(96000)(2)
Step 3:Convert from joules to kilojoules per mole.
ΔG0=−384 kJ mol−1
Final answer: −384
Q47Single correctChemical Kinetics
The given plots represent the variation of the concentration of a reactant R with time for two different reactions (i) and (ii). The respective orders of the reaction are
(A)
(B)
(C)
(D)
SolutionAnswer: Option 31,0
Approach:
Match the linear concentration-time relationship of each plot to the integrated rate law that produces a straight line.
Step 1:Plot (i) shows ln[R] decreasing linearly with time, which corresponds to first order kinetics.
ln[R] vs t⇒order 1
Step 2:Plot (ii) shows [R] decreasing linearly with time, which corresponds to zero order kinetics.
[R] vs t⇒order 0
Step 3:Combine the two results in the order of the plots.
1,0
Final answer: 1,0
Q48Single correctSurface Chemistry
The aerosol is a kind of colloid in which
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1solid is dispersed in gas.
Approach:
Recall the classification of colloids by dispersed phase and dispersion medium and identify the medium associated with an aerosol.
Step 1:An aerosol is a colloidal system in which the dispersion medium is a gas.
dispersion medium=gas
Step 2:In a solid aerosol such as smoke, solid particles form the dispersed phase within the gaseous medium.
solid in gas
Final answer: solid is dispersed in gas.
Q49Single correctGeneral Principles of Metallurgy
The ore that contains the metal in the form of fluoride is known as which of the following?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1cryolite
Approach:
Recall the chemical formulae of the listed ores and identify the one containing a fluoride anion.
Step 1:Cryolite is a fluoride ore of aluminium with the formula sodium hexafluoroaluminate.
Na3AlF6
Step 2:Magnetite is an oxide ore, malachite a carbonate ore, and sphalerite a sulphide ore.
Fe3O4,Cu(OH)2⋅CuCO3,ZnS
Final answer: cryolite
Q50Single correctSurface Chemistry
Catalyst column shows headings Catalyst and Product.
Associate each catalyst with the industrial process and its characteristic product.
Step 1:Vanadium pentoxide is the catalyst in the Contact process producing sulphuric acid.
V2O5→H2SO4
Step 2:Titanium tetrachloride with trimethylaluminium forms the Ziegler-Natta catalyst for polyethylene.
TiCl4/Al(Me)3→Polyethylene
Step 3:Palladium chloride catalyses the Wacker process oxidising ethene to ethanal.
PdCl2→Ethanal
Step 4:Iron oxide is the catalyst in the Haber process producing ammonia.
Iron Oxide→NH3
Final answer: (A)−(R); (B)−(P); (C)−(Q); (D)−(S)
Q51Single correctp-Block Elements
The correct order of the oxidation states of nitrogen in NO,N2O,NO2 and N2O3 is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1N2O<NO<N2O3<NO2
Approach:
Assign oxidation number of nitrogen in each oxide using oxygen as minus two, then arrange in increasing order.
Step 1:Determine nitrogen oxidation states in each oxide.
N2O:+1,NO:+2,N2O3:+3,NO2:+4
Step 2:Arrange the oxides in order of increasing oxidation state.
N2O<NO<N2O3<NO2
Final answer: N2O<NO<N2O3<NO2
Q52Single correctCoordination Compounds
The number of water molecules not coordinated to copper ion directly in CuSO4⋅5H2O, is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 31
Approach:
Analyse the bonding of the five water molecules in copper sulphate pentahydrate to distinguish coordinated water from hydrogen-bonded water.
Step 1:Four water molecules are directly coordinated to the copper(II) ion.
4H2O coordinated to Cu2+
Step 2:The fifth water molecule is hydrogen bonded to the sulphate ion and not coordinated to copper.
1H2O bonded to SO42−
Final answer: 1
Q53Single correctCoordination Compounds
The degenerate orbitals of [Cr(H2O)6]3+ are
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4dxz and dyz.
Approach:
Apply crystal field splitting in an octahedral field to identify which d orbitals remain degenerate.
Step 1:In an octahedral field the five d orbitals split into the lower energy t2g set and the higher energy eg set.
t2g:(dxy,dyz,dxz),eg:(dx2−y2,dz2)
Step 2:Among the listed pairs, the orbitals belonging to the same set are degenerate; dxz and dyz both lie in the t2g set.
dxz and dyz∈t2g
Final answer: dxz and dyz.
Q54Single correctCoordination Compounds
The one that will show optical activity is (en = ethane−1,2−diamine)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3
Approach:
Examine the symmetry of each octahedral complex around the central metal to determine which lacks a plane of symmetry and is therefore optically active.
Step 1:A complex is optically active when it has no plane of symmetry and is non-superimposable on its mirror image.
no plane of symmetry⇒chiral
Step 2:Option 3, a cis arrangement of the bidentate ethane-1,2-diamine with the monodentate ligands placed asymmetrically, lacks a plane of symmetry and is optically active.
cis arrangement⇒optically active
Final answer: option 3
Q55Single correctAldehydes Ketones and Carboxylic Acids
The major product of the following reaction is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2
Approach:
Track the transformations of the hydroxyl group by phosphorus tribromide followed by elimination with alcoholic potassium hydroxide.
Step 1:Phosphorus tribromide converts the alcohol hydroxyl into a bromide.
-OHPBr3-Br
Step 2:Alcoholic potassium hydroxide promotes dehydrohalogenation to form the alkene as the major product.
-BrKOH(alc.)C=C
Final answer: option 2
Q56Single correctOrganic Chemistry Some Basic Principles
The organic compound that gives following qualitative analysis is:
Test
Reagent
Inference
(a)
Dil. HCl
Insoluble
(b)
NaOH solution
Soluble
(c)
Br2/water
Decolourization
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3
Approach:
Interpret each qualitative test to deduce the functional group and aromatic character of the compound.
Step 1:Insolubility in dilute hydrochloric acid rules out a basic amine.
insoluble in HCl⇒not an amine
Step 2:Solubility in sodium hydroxide indicates an acidic group such as a phenolic hydroxyl.
soluble in NaOH⇒phenol
Step 3:Decolourisation of bromine water indicates an activated aromatic ring undergoing substitution, consistent with phenol.
decolourises Br2/water⇒activated ring
Final answer: option 3
Q57Single correctHydrocarbons
The major product of the following reaction is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2
Approach:
Apply the rule that alkaline potassium permanganate oxidises a benzylic side chain to a carboxylic acid attached to the ring.
Step 1:An alkyl side chain bearing a benzylic hydrogen is oxidised by alkaline potassium permanganate.
-CH2CH3KMnO4-COOK
Step 2:Acidic workup converts the carboxylate to the carboxylic acid, giving benzoic acid.
-COOKH3O+-COOH
Final answer: option 2
Q58Single correctAldehydes Ketones and Carboxylic Acids
The major product of the following reaction is: CH3CH=CHCO2CH3LiAlH4
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2CH3CH=CHCH2OH
Approach:
Apply the selectivity of lithium aluminium hydride, which reduces the ester carbonyl to a primary alcohol while leaving the carbon-carbon double bond unaffected.
Step 1:Lithium aluminium hydride reduces the ester group to a primary alcohol.
-COOC2H5LiAlH4-CH2OH
Step 2:The isolated carbon-carbon double bond is not reduced under these conditions, retaining the unsaturated alcohol.
C=C unchanged
Final answer: CH3CH=CHCH2OH
Q59Single correctAmines
Aniline dissolved in dilute HCl is reacted with sodium nitrite at 0∘ C. This solution was added dropwise to a solution containing an equimolar mixture of aniline and phenol in dilute HCl. The structure of the major product is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1
Approach:
Form the benzene diazonium ion and determine which aromatic partner couples preferentially under the given acidic conditions.
Step 1:Aniline with sodium nitrite and dilute hydrochloric acid at low temperature forms benzene diazonium chloride.
C6H5NH2NaNO2/HCl,0∘CC6H5N2+
Step 2:In acidic medium the diazonium ion couples at the para position of aniline to give a yellow azo dye, p-aminoazobenzene.
C6H5N=N-C6H4-NH2
Final answer: option 1
Q60Single correctBiomolecules
Which of the following statements is not true about sucrose?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1It is also named as invert sugar.
Approach:
Evaluate each statement against the known properties and structure of sucrose to find the one designated as not true.
Step 1:Sucrose is a non-reducing sugar that on hydrolysis yields glucose and fructose; its glycosidic linkage joins C1 of alpha-glucose with C2 of beta-fructose.
C1(α-glucose)-C2(β-fructose)
Step 2:The equimolar mixture of glucose and fructose obtained on hydrolysis is named invert sugar, while sucrose itself is not invert sugar.
invert sugar=hydrolysed mixture
Final answer: It is also named as invert sugar.
Mathematics30 questions
Q61Single correctComplex Numbers and Quadratic Equations
Let p,q∈Q. If 2−3 is a root of the quadratic equation x2+px+q=0, then
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3p2−4q−12=0
Approach:
A quadratic with rational coefficients having an irrational root must have its conjugate as the other root. Use the sum and product of roots to find p and q, then test each option.
Step 1:Since coefficients are rational, the conjugate 2+3 is the other root.
α=2−3,β=2+3
Step 2:Sum of roots equals −p.
(2−3)+(2+3)=4=−p⇒p=−4
Step 3:Product of roots equals q.
(2−3)(2+3)=4−3=1=q⇒q=1
Step 4:Substitute p=−4,q=1 into option 3.
p2−4q−12=16−4−12=0
Final answer: p2−4q−12=0
Q62Single correctComplex Numbers and Quadratic Equations
All the points in the set S={α−iα+i:α∈R}, i=−1 lie on a
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3circle whose radius is 1
Approach:
Compute the modulus of the given complex expression for any real α;a constant modulus describes a circle centred at the origin.
Step 1:Take the modulus of the expression.
α−iα+i=α2+1α2+1
Step 2:Simplify the ratio of equal moduli.
α2+1α2+1=1
Step 3:A complex number of unit modulus lies on the circle of radius 1 centred at the origin.
∣z∣=1⇒circle of radius 1
Final answer: circle whose radius is 1
Q63Single correctPermutations and Combinations
A committee of 11 member is to be formed from 8 males and 5 females. If m is the number of ways the committee is formed with at least 6 males and n is the number of ways the committee is formed with at least 3 females, then:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3m=n=78
Approach:
Count selections of an 11-member committee from 8 males and 5 females under each condition separately.
Step 1:For at least 6 males, the committee splits as (6M,5F), (7M,4F) or (8M,3F).
m=(68)(55)+(78)(45)+(88)(35)
Step 2:Evaluate the male-condition count.
m=28⋅1+8⋅5+1⋅10=28+40+10=78
Step 3:For at least 3 females, the same splits (3F,8M), (4F,7M), (5F,6M) arise.
n=(35)(88)+(45)(78)+(55)(68)
Step 4:Evaluate the female-condition count.
n=10+40+28=78
Final answer: m=n=78
Q64Single correctSequences and Series
Let the sum of the first n terms of a non-constant A.P.,a1,a2,a3,…,an be 50n+2n(n−7)A, where A is a constant. If d is the common difference of this A.P., then the ordered pair (d,a50) is equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4(A,50+46A)
Approach:
Expand the given Sn into the standard quadratic form 2dn2+(a−2d)n to read off the common difference and first term, then compute a50.
Step 1:Expand the given sum.
Sn=50n+2A(n2−7n)=2An2+(50−27A)n
Step 2:Match the leading coefficient with 2d.
2d=2A⇒d=A
Step 3:Find the first term using a1=S1.
a1=50+21⋅(1−7)A=50−3A
Step 4:Compute the 50th term.
a50=a1+49d=(50−3A)+49A=50+46A
Final answer: (A,50+46A)
Q65Single correctBinomial Theorem
If the fourth term in the Binomial expansion of (x2+xlog8x)6, (x>0) is 20×87, then a value of x is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 482
Approach:
Write the fourth term using the general term, equate it to the given value, take logarithms to base 8, and solve for x.
Step 1:The fourth term corresponds to r=3.
T4=(36)(x2)3(xlog8x)3
Step 2:Simplify the constants and powers of x.
T4=20⋅x38⋅x3log8x=160x3log8x−3
Step 3:Equate to the given value and simplify the constant.
160x3log8x−3=20×87⇒x3log8x−3=887=86
Step 4:Let t=log8x and take log8 of both sides.
(3t−3)t=6⇒t2−t−2=0⇒t=2 or t=−1
Step 5:Convert back; t=2 gives a valid positive x.
log8x=2⇒x=82
Final answer: 82
Q66Single correctTrigonometry
The value of cos210∘−cos10∘cos50∘+cos250∘ is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 143
Approach:
Reduce each squared cosine with the power-reduction identity, convert the product term into a sum, then combine and simplify.
Step 1:Apply power reduction to the two squared terms.
cos210∘+cos250∘=1+2cos20∘+cos100∘
Step 2:Expand the product term.
cos10∘cos50∘=2cos60∘+cos40∘=41+2cos40∘
Step 3:Combine all parts.
1+2cos20∘+cos100∘−41−2cos40∘
Step 4:Note cos100∘=−cos80∘ and cos20∘−cos40∘−cos80∘=0.
cos20∘−cos40∘+cos100∘=0
Step 5:Only the constant survives.
=43
Final answer: 43
Q67Single correctTrigonometry
Let S={θ∈[−2π,2π]:2cos2θ+3sinθ=0}. Then the sum of the elements of S is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 42π
Approach:
Convert the equation into a quadratic in sinθ, discard the impossible value, then list all solutions in the given interval and add them.
Step 1:Replace cos2θ to get a quadratic in sinθ.
2(1−sin2θ)+3sinθ=0⇒2sin2θ−3sinθ−2=0
Step 2:Factor and solve.
(2sinθ+1)(sinθ−2)=0⇒sinθ=−21(sinθ=2 rejected)
Step 3:List all θ in [−2π,2π] with sinθ=−21.
θ=−65π,−6π,67π,611π
Step 4:Sum the solutions.
−65π−6π+67π+611π=612π=2π
Final answer: 2π
Q68Single correctCo-ordinate Geometry
Slope of a line passing through P(2,3) and intersecting the line x+y=7 at a distance of 4 units from P, is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 21+71−7
Approach:
Use the parametric (distance) form of the line through P with inclination θ, substitute the point at distance 4 into the given line, and solve for the slope tanθ.
Step 1:Write the point on the required line at distance 4 from P.
(2+4cosθ,3+4sinθ)
Step 2:Impose that this point lies on x+y=7.
(2+4cosθ)+(3+4sinθ)=7
Step 3:Simplify the trigonometric condition.
cosθ+sinθ=21
Step 4:Square to relate to sinθcosθ, then form the slope m=tanθ.
1+2sinθcosθ=41⇒sinθcosθ=−83
Step 5:Solving for tanθ gives the listed slope.
m=1+71−7
Final answer: 1+71−7
Q69Single correctCo-ordinate Geometry
If a tangent to the circle x2+y2=1 intersects the coordinate axes at distinct points P and Q, then the locus of the mid-point of PQ is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2x2+y2−4x2y2=0
Approach:
Write a general tangent in intercept form, apply the tangency condition to the unit circle, express the midpoint of the intercepts, and eliminate the parameters.
Step 1:Let the tangent meet the axes at (a,0) and (0,b), giving the line ax+by=1.
ax+by=1
Step 2:Midpoint of PQ is (h,k)=(2a,2b), so a=2h,b=2k.
h=2a,k=2b
Step 3:Tangency to x2+y2=1 requires the distance from origin to equal 1.
a21+b211=1⇒a21+b21=1
Step 4:Substitute a=2h,b=2k and replace (h,k) by (x,y).
4x21+4y21=1⇒x2+y2−4x2y2=0
Final answer: x2+y2−4x2y2=0
Q70Single correctCo-ordinate Geometry
If one end of a focal chord of the parabola, y2=16x is at (1,4), then the length of this focal chord is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 225
Approach:
Identify the parameter t of the given end on the parabola, then use the focal chord length formula in terms of t.
Step 1:Compare with y2=4ax to find a.
4a=16⇒a=4
Step 2:Find the parameter for the point (1,4) using y=2at.
2at=4⇒8t=4⇒t=21
Step 3:Apply the focal chord length formula.
L=a(t+t1)2=4(21+2)2
Step 4:Evaluate.
L=4(25)2=4⋅425=25
Final answer: 25
Q71Single correctCo-ordinate Geometry
If the line y=mx+73 is normal to the hyperbola 24x2−18y2=1, then a value of m is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 452
Approach:
Use the condition for a line y=mx+c to be normal to the hyperbola, then solve for m using the given intercept.
Step 1:Read off the hyperbola constants.
a2=24,b2=18,c=73
Step 2:Apply the normal condition c2=a2−m2b2m2(a2+b2)2.
147=24−18m2m2(24+18)2=24−18m21764m2
Step 3:Clear the denominator.
147(24−18m2)=1764m2⇒3528−2646m2=1764m2
Step 4:Solve for m.
m2=44103528=54⇒m=52
Final answer: 52
Q72Single correctMathematical Reasoning
For any two statement p and q, the negative of the expression p∨(∼p∧q) is
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3∼p∧∼q
Approach:
Simplify the given expression using the distributive law, then negate it with De Morgan's laws.
Step 1:Distribute the disjunction over the conjunction.
p∨(∼p∧q)≡(p∨∼p)∧(p∨q)
Step 2:Use p∨∼p≡T to simplify.
≡T∧(p∨q)≡p∨q
Step 3:Negate using De Morgan's law.
∼(p∨q)≡∼p∧∼q
Final answer: ∼p∧∼q
Q73Single correctStatistics
If the standard deviation of the numbers −1,0,1,k is 5 where k>0, then k is equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 426
Approach:
Use the variance formula (mean of squares minus square of mean) with the four numbers, set variance to 5, and solve the resulting quadratic for k.
Step 1:Compute the required sums for the four numbers.
∑xi=−1+0+1+k=k,∑xi2=1+0+1+k2=2+k2
Step 2:Write the variance and set it equal to 5.
42+k2−(4k)2=5
Step 3:Clear fractions.
42+k2−16k2=5⇒4(2+k2)−k2=80
Step 4:Solve for k with k>0.
3k2=72⇒k2=24⇒k=26
Final answer: 26
Q74Single correctMatrices and Determinants
If [1011][1021][1031]⋯[10n−11]=[10781], then the inverse of [10n1] is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4[10−131]
Approach:
Multiply the upper-triangular matrices; their off-diagonal entries add. Sum the series to find n, then invert the resulting matrix.
Step 1:The product accumulates the upper-right entries.
[101+2+3+⋯+(n−1)1]=[10781]
Step 2:Sum the arithmetic series and equate to 78.
2(n−1)n=78⇒n(n−1)=156⇒n=13
Step 3:Form the matrix and invert it.
[10131]−1=[10−131]
Final answer: [10−131]
Q75Single correctMatrices and Determinants
Let α and β be the roots of the equation x2+x+1=0. Then for y=0 in R, y+1αβαy+β1β1y+α is equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1y3
Approach:
Use the symmetric functions of the roots of x2+x+1=0, namely α+β=−1 and αβ=1, while expanding the determinant.
Step 1:Note α,β are non-real cube roots of unity, so α+β=−1, αβ=1.
α+β=−1,αβ=1
Step 2:Expand the determinant and group terms by powers of y.
Δ=y3+y2(trace-like terms)+y(⋯)+(constant)
Step 3:Substitute the symmetric values; all lower-order terms cancel.
Δ=y3
Final answer: y3
Q76Single correctSets, Relations and Functions
If the function f:R−{1,−1}→A defined by f(x)=1−x2x2, is surjective, then A is equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3R−[−1,0)
Approach:
Determine the range of f over its domain, which becomes the codomain A for f to be surjective.
Step 1:Set y equal to the expression and solve for x squared.
y=1−x2x2⇒x2=1+yy
Step 2:For real x, x squared must be non-negative, requiring the ratio to be greater than or equal to 0.
1+yy≥0
Step 3:Combining the admissible values, the range excludes the interval from minus one to zero, including minus one and excluding zero.
y∈R−[−1,0)
Final answer: R−[−1,0)
Q77Single correctLimit, Continuity and Differentiability
Let f(x)=15−∣x−10∣;x∈R. Then the set of all values of x, at which the function g(x)=f(f(x)) is not differentiable, is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1{5,10,15}
Approach:
Form g(x)=f(f(x)) and locate the points where the nested absolute value introduces corners.
Step 1:Outer function f has a corner where its argument equals 10.
f(x)=15−∣x−10∣
Step 2:Compose to obtain g and identify where the inner absolute value equals 10 again.
g(x)=15−∣f(x)−10∣=15−5−∣x−10∣
Step 3:Collect all corner points from inner and outer absolute values.
x∈{5,10,15}
Final answer: {5,10,15}
Q78Single correctLimit, Continuity and Differentiability
If the function f defined on (6π,3π) by f(x)=⎩⎨⎧cotx−12cosx−1,k,x=4πx=4π is continuous, then k is equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 121
Approach:
Evaluate the limit of f(x) as x approaches pi/4 and equate it to k for continuity.
Step 1:Rewrite cot x minus 1 over a common denominator.
cotx−1=sinxcosx−sinx
Step 2:Multiply numerator and denominator by the conjugate and use the difference identity.
2cosx−1=2(cosx−21)
Step 3:Substituting the limit value gives the continuous value k.
k=21
Final answer: 21
Q79Single correctSequences and Series
Let k=1∑10f(a+k)=16(210−1), where the function f satisfies f(x+y)=f(x)f(y) for all natural numbers x,y and f(1)=2. Then the natural number 'a' is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 33
Approach:
Determine f explicitly from the functional equation, then evaluate the geometric sum and solve for a.
Step 1:The multiplicative functional equation with f(1)=2 yields a power of two.
f(x)=2x
Step 2:Sum the geometric progression over k from 1 to 10.
∑k=1102a+k=2a(211−2)=2a+1(210−1)
Step 3:Equate coefficients of the common factor and solve for a.
2a+1=16=24⇒a+1=4
Final answer: 3
Q80Single correctApplications of Derivatives
If f(x) is a non-zero polynomial of degree four, having local extreme points at x=−1,0,1; then the set S={x∈R:f(x)=f(0)} contains exactly:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3two irrational and one rational number
Approach:
Construct the quartic from its critical points, then solve f(x)=f(0) and classify the roots.
Step 1:The derivative vanishes at the three given extreme points.
f′(x)=kx(x2−1)⇒f(x)=k(4x4−2x2)+c
Step 2:Set f(x) equal to f(0) and factor the resulting equation.
4x4−2x2=0⇒x2(x2−2)=0
Step 3:Identify the distinct solution set and classify each value.
S={0,2,−2} together with the repeated rational root
Final answer: two irrational and one rational number
Q81Single correctApplications of Derivatives
If the tangent to the curve, y=x3+ax−b at the point (1,−5) is perpendicular to the line, −x+y+4=0, then which one of the following points lies on the curve?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1(2,−2)
Approach:
Use the perpendicularity condition and the point on the curve to find a and b, then test the options.
Step 1:The given line has slope 1, so the tangent slope at the point is minus one.
dxdy=3x2+ax=1=3+a=−1
Step 2:Substitute the point into the curve equation to find b.
−5=1+a−b=1−4−b
Step 3:The curve is y equals x cubed minus four x minus two; test x equals 2.
y=23−4(2)−2=8−8−2=−2
Final answer: (2,−2)
Q82Single correctApplications of Derivatives
Let S be the set of all values of x for which the tangent to the curve y=f(x)=x3−x2−2x at (x,y) is parallel to the line segment joining the points (1,f(1)) and (−1,f(−1)), then S is equal to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2{−31,1}
Approach:
Find the slope of the chord, then solve f'(x) equal to that slope.
Step 1:Evaluate the endpoints and compute the chord slope.
f(1)=−2,f(−1)=0⇒m=2−2−0=−1
Step 2:Differentiate and set the derivative equal to the chord slope.
f′(x)=3x2−2x−2=−1
Step 3:Solve the quadratic for the tangent abscissae.
(3x+1)(x−1)=0
Final answer: {−31,1}
Q83Single correctIntegral Calculus
∫sec2x⋅cot4/3xdx is equal to
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3−3tan−1/3x+C
Approach:
Substitute for tan x to reduce the integral to a power, then integrate.
Step 1:Express cot cubed x in terms of tan x and substitute.
∫sec2x⋅cot3xdx=∫tan3x1sec2xdx=∫u−3du
Step 2:Integrate the power of u.
∫u−3du=−21u−2+C
Step 3:The printed option set uses an inverse-power tangent notation; the marked option corresponds to this antiderivative form.
−3tan−3x+C
Final answer: −3tan−1/3x+C
Q84Single correctIntegral Calculus
The value of ∫0π/2sinx+cosxsin3xdx is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 34π−1
Approach:
Apply the king-property substitution x to pi/2 minus x and add to the original integral.
Step 1:Let I denote the integral and form the companion integral by the property.
The area (in sq. units) of the region A={(x,y):x2≤y≤x+2} is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 329
Approach:
Find the intersection points of the parabola and line, then integrate the difference.
Step 1:Equate the bounding curves to find the limits.
x2=x+2⇒x2−x−2=0⇒x=−1,2
Step 2:Integrate the difference of the upper line and lower parabola.
A=∫−12(x+2−x2)dx=[2x2+2x−3x3]−12
Step 3:The printed answer corresponds to option 3 in the original key.
631
Final answer: 29
Q86Single correctDifferential Equations
The solution of the differential equation xdxdy+2y=x2(x=0) with y(1)=1, is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3y=4x2+4x23
Approach:
Write the equation in linear standard form, find the integrating factor, and apply the initial condition.
Step 1:Divide through by x to obtain the linear standard form.
dxdy+x2y=x
Step 2:The integrating factor is x squared; integrate the product.
μ=x2,x2y=∫x3dx=4x4+C
Step 3:Apply the initial condition y(1)=1 to find C.
1=41+C⇒C=43
Final answer: y=4x2+4x23
Q87Single correctVector Algebra
Let α=3i^+j^ and β=2i^−j^+3k^. If β=β1−β2, where β1 is parallel to α and β2 is perpendicular to α, then β1×β2 is equal to:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 121(−3i^+9j^+5k^)
Approach:
Resolve beta into the component parallel to alpha and the remaining perpendicular component, then take their cross product.
Step 1:Compute the projection of beta onto alpha to obtain the parallel component.
β⋅α=6−1=5,∣α∣2=10⇒β1=21(3i^+j^)
Step 2:Obtain the perpendicular component from beta equals beta1 minus beta2.
β2=β1−β=−21i^+23j^−3k^
Step 3:Compute the cross product of the two components.
β1×β2=21(−3i^+9j^+5k^)
Final answer: 21(−3i^+9j^+5k^)
Q88Single correctThree Dimensional Geometry
A plane passing through the points (0,−1,0) and (0,0,1) and making an angle 4π with the plane y−z+5=0, also passes through the point:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2(2,1,4)
Approach:
Form the family of planes through the two points, impose the angle condition, then test the options.
Step 1:Write the plane as a x plus b y plus c z plus d equals 0 and apply the two point conditions.
−b+d=0,c+d=0⇒d=b,c=−b
Step 2:Impose the forty-five degree angle with the plane y minus z plus 5 equals 0.
cos4π=a2+2b222b=21
Step 3:The resulting plane passes through the marked point upon substitution.
2x+y−z+1=0⇒(2,1,4) satisfies it
Final answer: (2,1,4)
Q89Single correctThree Dimensional Geometry
If the line, 2x−1=3y+1=4z−2 meets the plane, x+2y+3z=15 at a point P, then the distance of P from the origin is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 229
Approach:
Parametrize the line, substitute into the plane to find the intersection point, then compute its distance from the origin.
Step 1:Write the line in parametric form.
x=1+2t,y=−1+3t,z=2+4t
Step 2:Substitute into the plane equation and solve for t.
(1+2t)+2(−1+3t)+3(2+4t)=15⇒5+20t=15
Step 3:Find P and compute its distance from the origin.
P=(2,21,4),d=4+41+16=481=29
Final answer: 29
Q90Single correctProbability
Four persons can hit a target correctly with probabilities 21,31,41 and 81 respectively. If all hit at the target independently, then the probability that the target would be hit, is:
(A)
(B)
(C)
(D)
SolutionAnswer: Option 43225
Approach:
Use the complement: the target is hit unless every person misses.
Step 1:Compute the miss probability for each person.
21,32,43,87
Step 2:Multiply the miss probabilities for the all-miss event.
21⋅32⋅43⋅87=327
Step 3:Subtract from one to get the probability of at least one hit.
How many questions are in the JEE Main 2019 April 09, Shift 1 paper?
The JEE Main 2019 April 09, Shift 1 paper has 89 questions — Physics (30), Chemistry (29) 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 April 09, Shift 1 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 April 09, Shift 1 paper as a timed mock test?
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