JEE Main 2016 April 09 Question Paper with Solutions
All 90 questions from the JEE Main 2016 (April 09) 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.
In the following 'I' refers to current and 'a' to acceleration of a point mass falling vertically in a viscous medium. Choose the option that corresponds to the dimensions of electrical conductivity :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3M−1L−3T3I2
Approach:
Express electrical conductivity through current density and electric field, then reduce to base dimensions.
Step 1:Resistance has dimensions from R = V/I, with V = work per charge.
[R]=ML2T−3I−2
Step 2:Resistivity equals resistance times area over length.
[ρ]=[R]LL2=ML3T−3I−2
Step 3:Conductivity is the reciprocal of resistivity.
[σ]=[ρ]−1=M−1L−3T3I2
Final answer: M−1L−3T3I2
Q2Single correctLaws of Motion
Which of the following option correctly describes the variation of the speed v and acceleration 'a' of a point mass falling vertically in a viscous medium that applies a force F=−kv, where 'k' is a constant, on the body ? (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2Graph in which v starts from zero and increases with time, saturating to a constant value, while a starts at a high value and decreases to zero.
Approach:
Write Newton's equation for fall with linear drag and analyse the time behaviour of speed and acceleration.
Step 1:At the instant of release the speed is zero, so the drag vanishes and the acceleration is maximum.
a(0)=g
Step 2:As speed grows the drag rises, reducing the net force and hence the acceleration.
a=g−mkv
Step 3:Speed rises from zero and asymptotically approaches the terminal value where acceleration becomes zero.
v→vt,a→0
Final answer: Graph in which v starts from zero and increases with time, saturating to a constant value, while a starts at a high value and decreases to zero.
Q3Single correctLaws of Motion
A rocket is fired vertically from the earth with an acceleration of 2g, where g is the gravitational acceleration. On an inclined plane inside the rocket, making an angle θ with the horizontal, a point object of mass m is kept. The minimum coefficient of friction μmin between the mass and the inclined surface such that the mass does not move is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1tanθ
Approach:
Work in the accelerating rocket frame, where an effective gravity acts, and balance forces along and perpendicular to the incline.
Step 1:In the rocket frame a pseudo-force adds to gravity, giving an effective downward field.
geff=3g
Step 2:Component of effective weight along the incline.
F∥=mgeffsinθ
Step 3:Maximum static friction from the normal reaction.
fmax=μmgeffcosθ
Step 4:For no motion the friction must balance the driving component; the common factor cancels.
μmgeffcosθ≥mgeffsinθ⇒μmin=tanθ
Final answer: tanθ
Q4Single correctWork, Energy and Power
A car of weight W is on an inclined road that rises by 100 m over a distance of 1 km and applies a constant frictional force 20W on the car. While moving uphill on the road at a speed of 10ms−1, the car needs power P. If it needs power 2P while moving downhill at speed v then value of v is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 215ms−1
Approach:
Find the driving force and power going uphill, then the driving force going downhill, and use the given power ratio to get the speed.
Step 1:Uphill the engine overcomes both the gravity component and friction.
Fup=Wsinθ+20W=0.1W+0.05W=0.15W
Step 2:Power required uphill at 10 m/s.
P=0.15W×10=1.5W
Step 3:Downhill gravity assists, so the magnitude of the engine force needed is the difference of gravity component and friction.
Fdown=∣0.1W−0.05W∣=0.05W
Step 4:Apply the given power downhill equal to half of P.
0.05Wv=21.5W=0.75W⇒v=15ms−1
Final answer: 15ms−1
Q5Single correctRotational Motion
A cubical block of side 30 cm is moving with velocity 2ms−1 on a smooth horizontal surface. The surface has a bump at a point O as shown in figure. The angular velocity (in rad/s) of the block immediately after it hits the bump, is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 15.0
Approach:
Conserve angular momentum about the bump O at the instant of impact, then use the cube's moment of inertia about an edge.
Step 1:Before impact the centre of mass moves horizontally at height a/2 above the bump.
Li=mv2a
Step 2:After impact the block rotates about the edge through O.
Lf=Iω=32ma2ω
Step 3:Equate the two and solve for the angular velocity.
ω=4a3v=4×0.33×2
Step 4:Evaluate numerically.
ω=1.26=5.0rad/s
Final answer: 5.0
Q6Single correctGravitation
Figure shows elliptical path abcd of a planet around the sun S such that the area of triangle csa is 41 the area of the ellipse. (See figure) With db as the semimajor axis, and ca as the semiminor axis. If t1 is the time taken for planet to go over path abc and t2 for path taken over cda then :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3t1=3t2
Approach:
Apply Kepler's second law, which makes the time spent proportional to the area swept by the radius from the sun.
Step 1:The major axis divides the ellipse into two equal halves, each of area A/2.
Ahalf=2A
Step 2:The triangle csa with the sun contributes one quarter of the ellipse area to the swept region.
Atriangle=4A
Step 3:Area swept along abc is half the ellipse plus the triangle; along cda it is half the ellipse minus the triangle.
Aabc=2A+4A=43A,Acda=2A−4A=4A
Step 4:Times are in the same ratio as the swept areas.
t2t1=A/43A/4=3⇒t1=3t2
Final answer: t1=3t2
Q7Single correctProperties of Solids and Liquids
Consider a water jar of radius R that has water filled up to height H and is kept on a stand of height h (see figure). Through a hole of radius r (r << R) at its bottom, the water leaks out and the stream of water coming down towards the ground has a shape like a funnel as shown in the figure. If the radius of the cross-section of water stream when it hits the ground is x. Then :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3x=r(H+hH)41
Approach:
Use Torricelli's theorem for the exit speed and the equation of continuity between the hole and the ground.
Step 1:Speed at the hole is set by the water column height H.
v1=2gH
Step 2:Falling a further height h to the ground, the speed increases.
v2=2g(H+h)
Step 3:Apply continuity of volume flow between hole and ground.
x2=r2v2v1=r2H+hH
Step 4:Take the square root.
x=r(H+hH)41
Final answer: x=r(H+hH)41
Q8Single correctThermodynamics
200 g water is heated from 40∘C to 60∘C. Ignoring the slight expansion of water, the change in its internal energy is (Given specific heat of water =4184J/kg/K) :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 316.7kJ
Approach:
With negligible expansion the work done is zero, so the change in internal energy equals the heat supplied.
Step 1:Convert the mass and find the temperature rise.
m=0.2kg,ΔT=20K
Step 2:Compute the heat supplied.
Q=0.2×4184×20=16736J
Step 3:Ignoring expansion, the internal energy change equals this heat.
ΔU=16.7kJ
Final answer: 16.7kJ
Q9Single correctThermodynamics
The ratio of work done by an ideal monoatomic gas to the heat supplied to it in an isobaric process is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 452
Approach:
Express the work and the heat for an isobaric process using the molar gas constant and the molar heat capacity at constant pressure.
Step 1:Form the ratio of work to heat.
QW=nCpΔTnRΔT=CpR
Step 2:Substitute the monoatomic value of Cp.
QW=25RR=52
Final answer: 52
Q10Single correctOscillations and Waves
Two particles are performing simple harmonic motion in a straight line about the same equilibrium point. The amplitude and time period for both particles are same and equal to A and T, respectively. At time t=0 one particle has displacement A while the other one has displacement 2−A and they are moving towards each other. If they cross each other at time t, then t is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 16T
Approach:
Write each displacement as a cosine with the correct phase, impose the meeting condition, and solve for the time.
Step 1:Particle 1 starts at the positive extreme.
x1=Acos(ωt)
Step 2:Particle 2 starts at -A/2 moving towards particle 1, fixing its phase at -2 pi/3.
x2=Acos(ωt−32π)
Step 3:Set the displacements equal so the cosines are equal at symmetric phases.
cos(ωt)=cos(ωt−32π)⇒2ωt=32π
Step 4:Solve for the time.
t=ωπ/3=6T
Final answer: 6T
Q11Single correctOscillations and Waves
Two engines pass each other moving in opposite directions with uniform speed of 30 m/s. One of them is blowing a whistle of frequency 540 Hz. Calculate the frequency heard by driver of second engine before they pass each other. Speed of sound is 330 m/sec is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3648Hz
Approach:
Apply the Doppler formula with both source and observer approaching each other before they meet.
Step 1:Both engines move towards each other, so observer speed adds and source speed subtracts.
v=330,vo=30,vs=30,f=540
Step 2:Substitute into the Doppler formula.
f′=540330−30330+30=540300360
Step 3:Evaluate.
f′=540×1.2=648Hz
Final answer: 648Hz
Q12Single correctElectrostatics
The potential (in volts) of a charge distribution is given by V(z)=30−5z2 for ∣z∣≤1m V(z)=35−10∣z∣ for ∣z∣≥1m V(z) does not depend on x and y. If this potential is generated by a constant charge per unit volume ρ0 (in units of ε0) which is spread over a certain region, then choose the correct statement.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1ρ0=10ε0 for ∣z∣≤1m and ρ0=0 elsewhere
Approach:
Apply Poisson's equation, which relates the second derivative of the potential to the charge density.
Step 1:Differentiate the inner potential twice.
dz2d2(30−5z2)=−10
Step 2:Relate to the charge density inside.
−10=−ε0ρ0⇒ρ0=10ε0
Step 3:Differentiate the outer potential twice.
dz2d2(35−10∣z∣)=0
Step 4:The charge density vanishes outside.
ρ0=0for ∣z∣≥1m
Final answer: ρ0=10ε0 for ∣z∣≤1m and ρ0=0 elsewhere
Q13Single correctElectrostatics
Three capacitors each of 4μF are to be connected in such a way that the effective capacitance is 6μF. This can be done by connecting them :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2two in series and one in parallel
Approach:
Test each configuration by computing its equivalent capacitance until 6 microfarad is reached.
Step 1:Combine two capacitors in series.
Cs=4+44×4=2μF
Step 2:Add the third capacitor in parallel with this series pair.
Ceq=2+4=6μF
Final answer: two in series and one in parallel
Q14Single correctCurrent Electricity
In the circuit shown, the resistance r is a variable resistance. If for r=fR, the heat generation in r is maximum then the value of f is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 221
Approach:
Write the power in r in terms of f using series R with the parallel combination of R and r, then maximize.
Step 1:Set R = 1 and r = f; find the parallel block and total resistance.
Rp=1+ff,Rtot=1+f1+2f
Step 2:Voltage across the parallel block from the current through it.
Vp=1+2ff
Step 3:Power dissipated in r.
Pr=fVp2=(1+2f)2f
Step 4:Differentiate and set to zero.
dfdPr=(1+2f)31−2f=0⇒f=21
Final answer: 21
Q15Single correctMagnetic Effects of Current and Magnetism
A magnetic dipole is acted upon by two magnetic fields which are inclined to each other at an angle of 75∘. One of the fields has a magnitude of 15 mT. The dipole attains stable equilibrium at an angle of 30∘ with this field. The magnitude of the other field (in mT) is close to :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 111
Approach:
At equilibrium the torques from the two fields balance; equate the two torque magnitudes and solve for the unknown field.
Step 1:The dipole makes 30 degrees with the first field, hence 45 degrees with the second.
75∘−30∘=45∘
Step 2:Balance the torques from the two fields.
mB1sin30∘=mB2sin45∘
Step 3:Solve for the second field magnitude.
B2=sin45∘B1sin30∘=0.70715×0.5
Step 4:Evaluate.
B2≈10.6mT≈11mT
Final answer: 11
Q16Single correctCurrent Electricity
A 50Ω resistance is connected to a battery of 5 V. A galvanometer of resistance 100Ω is to be used as an ammeter to measure current through the resistance, for this a resistance rs is connected to the galvanometer. Which of the following connections should be employed if the measured current is within 1% of the current without the ammeter in the circuit ?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1rs=0.5Ω in parallel with the galvanometer
Approach:
An ammeter must present a very small resistance so it does not alter the circuit current; a galvanometer is converted to an ammeter by placing a low-value shunt in parallel. The shunt is chosen so the extra series resistance kept in the loop changes the current by at most one percent.
Step 1:Compute the undisturbed current.
I0=505=0.1A
Step 2:For an ammeter the meter resistance must be added in series with the load, so it has to be made as small as possible by a parallel shunt across the 100Ω galvanometer.
Rp=rs+100rs×100
Step 3:With a parallel shunt of 0.5Ω the inserted resistance is about 0.4975Ω, changing the current to 5/50.4975≈0.099A, a deviation under one percent; a series resistance only raises the total resistance and a 1Ω parallel shunt adds roughly twice as much, so the 0.5Ω parallel choice satisfies the requirement.
Rp=100.50.5×100≈0.4975Ω
Final answer: rs=0.5Ω in parallel with the galvanometer
Q17Single correctAlternating Current
A series LR circuit is connected to a voltage source with V(t)=V0sinΩt. After very large time, current I(t) behaves as (t0≫RL) :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2I(t) is a sustained sinusoid of constant amplitude oscillating symmetrically about zero for t>t0
Approach:
The current in a driven series LR circuit is the sum of a transient term that decays with time constant L/R and a steady-state sinusoidal term at the driving frequency. After a time large compared with L/R the transient vanishes and only the constant-amplitude sinusoid survives.
Step 1:The full solution contains an exponential transient with time constant L/R plus a forced sinusoidal response.
I(t)=Ae−Rt/L+I0′sin(Ωt−ϕ)
Step 2:For t0≫L/R the exponential factor is negligibly small.
e−Rt0/L→0
Step 3:Only the steady-state sinusoid of fixed amplitude remains, oscillating about zero, matching the constant-amplitude wave.
I(t)≈I0′sin(Ωt−ϕ)
Final answer: I(t) is a sustained sinusoid of constant amplitude oscillating symmetrically about zero for t>t0
Q18Single correctElectromagnetic Waves
Microwave oven acts on the principle of :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2giving rotational energy to water molecules
Approach:
Microwave frequencies match the rotational resonance of polar water molecules; the oscillating field exerts a torque on the molecular dipole, exciting rotational motion that is then shared as heat.
Step 1:Water is a polar molecule with a permanent electric dipole moment.
p=0
Step 2:The microwave field oscillates at a frequency close to the rotational resonance of water, exerting an oscillating torque.
τ=pEsinθ
Step 3:The rotational energy spreads through molecular collisions, heating the food.
Erot→heat
Final answer: giving rotational energy to water molecules
Q19Single correctRay Optics and Optical Instruments
A convex lens, of focal length 30 cm, a concave lens of focal length 120 cm, and a plane mirror are arranged as shown. For an object kept at a distance of 60 cm from the convex lens, the final image, formed by the combination, is a real image, at a distance of :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 160 cm from the convex lens
Approach:
Trace the object through the convex lens, then the concave lens, reflect off the plane mirror and retrace back through both lenses. The intended retracing geometry makes the final real image coincide with the object plane at 60 cm from the convex lens.
Step 1:The convex lens images the object placed 60 cm away.
v1−−601=301
Step 2:This image lies at the concave-lens plane (the lens separation is 60 cm), so the rays converge to a point on the concave lens and pass essentially undeviated towards the plane mirror.
v=60cm=lens separation
Step 3:The plane mirror reflects the rays back through the concave and convex lenses; the symmetric return path reconstructs a real image at the object plane, 60 cm from the convex lens.
dfinal=60cm
Final answer: 60 cm from the convex lens
Q20Single correctWave Optics
In Young's double slit experiment, the distance between slits and the screen is 1.0 m and monochromatic light of 600 nm is being used. A person standing near the slits is looking at the fringe pattern. When the separation between the slits is varied, the interference pattern disappears for a particular distance d0 between the slits. If the angular resolution of the eye is 601∘, the value of d0 is close to :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 22 mm
Approach:
The fringe pattern stops being resolvable when the angular fringe spacing equals the angular resolution of the eye. Equate the angular fringe width to the given resolution to obtain the slit separation.
Step 1:Convert the eye's angular resolution to radians.
θ=601∘=601×180πrad=2.91×10−4rad
Step 2:Set the angular fringe spacing equal to this resolution.
d0λ=θ
Step 3:Solve for the slit separation.
d0=θλ=2.91×10−4600×10−9
Final answer: 2 mm
Q21Single correctDual Nature of Radiation and Matter
When photons of wavelength λ1 are incident on an isolated sphere, the corresponding stopping potential is found to be V. When photons of wavelength λ2 are used, the corresponding stopping potential was thrice that of the above value. If light of wavelength λ3 is used then find the stopping potential for this case :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3ehc[λ31+2λ21−2λ13]
Approach:
Apply Einstein's photoelectric equation for each wavelength to express the stopping potential in terms of the work function, eliminate the work function using the two known cases, then substitute for the third wavelength.
Step 1:Write the stopping potential for the first two wavelengths.
eV=λ1hc−ϕ,3eV=λ2hc−ϕ
Step 2:Eliminate V by combining the relations (3 times the first minus the second) to find the work function.
ϕ=2hc(λ13−λ21)
Step 3:Substitute into the stopping-potential expression for the third wavelength.
V3=e1[λ3hc−ϕ]=ehc[λ31+2λ21−2λ13]
Final answer: ehc[λ31+2λ21−2λ13]
Q22Single correctAtoms and Nuclei
A hydrogen atom makes a transition from n=2 to n=1 and emits a photon. This photon strikes a doubly ionized lithium atom (z=3) in excited state and completely removes the orbiting electron. The least quantum number for the excited state of the ion for the process is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 34
Approach:
Find the energy of the emitted hydrogen photon, then require it to be at least the binding energy of the lithium ion in level n so the electron is removed; solve for the smallest integer n.
Step 1:Compute the energy of the photon emitted in the n=2 to n=1 hydrogen transition.
E=13.6(1−41)=10.2eV
Step 2:Write the binding energy of the Li (Z=3) ion in level n.
∣En∣=13.6n29=n2122.4eV
Step 3:Require the photon energy to exceed the binding energy and solve for the least integer n.
10.2≥n2122.4⇒n2≥12⇒n≥3.46
Final answer: 4
Q23Single correctSemiconductor Electronics
The truth table given in fig. represents :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2OR - Gate
Approach:
Read the output column of the truth table for each input combination and match it to the standard logic-gate outputs.
Step 1:Tabulate the given outputs: inputs (0,0) give 0; (0,1) give 1; (1,0) give 1; (1,1) give 1.
Y=0,1,1,1
Step 2:The output is 0 only when both inputs are 0 and 1 otherwise.
Y=A+B
Step 3:This is the truth table of an OR gate.
Y=A+B
Final answer: OR - Gate
Q24Single correctCommunication Systems
An audio signal consists of two distinct sounds : one a human speech signal in the frequency band of 200 Hz to 2700 Hz, while the other is a high frequency music signal in the frequency band of 10200 Hz to 15200 Hz. The ratio of the AM signal bandwidth required to send both the signals together to the AM signal bandwidth required to send just the human speech is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 36
Approach:
The AM bandwidth equals twice the highest modulating frequency present. Compute the bandwidth needed for the combined signal and for speech alone, then take the ratio.
Step 1:For both signals together the highest frequency is 15200 Hz.
BWboth=2×15200=30400Hz
Step 2:For speech alone the highest frequency is 2700 Hz.
BWspeech=2×2700=5400Hz
Step 3:Take the ratio of the two bandwidths.
BWspeechBWboth=540030400≈5.6
Final answer: 6
Q25Single correctOscillations
A simple pendulum made of a bob of mass m and a metallic wire of negligible mass has time period 2 s at T=0∘C. If the temperature of the wire is increased and the corresponding change in its time period is plotted against its temperature, the resulting graph is a line of slope S. If the coefficient of linear expansion of metal is α then the value of S is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1α
Approach:
Express the pendulum period as a function of temperature using linear expansion of the wire, expand to first order, and identify the slope of period versus temperature.
Step 1:Write the period with the temperature-dependent length.
T=2πgL0(1+αΔθ)=T0(1+αΔθ)1/2
Step 2:Expand to first order in the small expansion term.
T≈T0(1+2αΔθ)
Step 3:The slope of period versus temperature is the derivative, with T0=2s.
S=dθdT=2T0α=22α=α
Final answer: α
Q26Single correctMechanical Properties of Solids
A uniformly tapering conical wire is made from a material of Young's modulus Y and has a normal, unextended length L. The radii, at the upper and lower ends of this conical wire, have values R and 3R, respectively. The upper end of the wire is fixed to a rigid support and a mass M is suspended from its lower end. The equilibrium extended length, of this wire, would equal :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2L(1+31πYR2Mg)
Approach:
Integrate the elemental extension along the cone, where the radius varies linearly from R to 3R, to obtain the total elongation, then add it to the natural length.
Step 1:Write the cross-sectional area as a function of position with radius varying from R to 3R.
A(x)=π(R+L2Rx)2
Step 2:Integrate the elemental extension over the full length with force Mg.
ΔL=∫0LπY(R+L2Rx)2Mgdx
Step 3:Add the elongation to the natural length to get the extended length.
Lext=L+3πYR2MgL=L(1+31πYR2Mg)
Final answer: L(1+31πYR2Mg)
Q27Single correctCurrent Electricity
To know the resistance G of a galvanometer by half deflection method, a battery of emf VE and resistance R is used to deflect the galvanometer by angle θ. If a shunt of resistance S is needed to get half deflection then G, R and S are related by the equation :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2S(R+G)=RG
Approach:
Compute the galvanometer current before the shunt and again after the shunt is connected in parallel, then impose that the second current is half of the first to find the relation between R, G and S.
Step 1:Before the shunt, the galvanometer carries the full current giving full deflection.
Ig=R+GVE
Step 2:With the shunt S across the galvanometer, the current through G for half deflection is half the initial value.
R+G+SGSVE⋅G+SS=21R+GVE
Step 3:Solving the condition yields the relation between the resistances.
S=R+GRG⇒S(R+G)=RG
Final answer: S(R+G)=RG
Q28Single correctRay Optics and Optical Instruments
To find the focal length of a convex mirror, a student records the following data : Object Pin : 22.2 cm; Convex Lens : 32.2 cm; Convex Mirror : 45.8 cm; Image Pin : 71.2 cm. The focal length of the convex lens is f1 and that of mirror is f2. Then taking index correction to be negligibly small, f1 and f2 are close to :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2f1=7.8cmf2=12.7cm
Approach:
Use the object pin and image pin positions with the lens to find the lens focal length, then use the fact that the rays retrace through the convex mirror only when they meet its centre of curvature to find the mirror radius and hence its focal length.
Step 1:Object distance for the lens is the separation of object pin and lens; the image distance is the separation of image pin and lens.
u=−(32.2−22.2)=−10cm,v=71.2−32.2=39cm
Step 2:Apply the lens equation to obtain the lens focal length.
f11=391+101⇒f1≈7.95cm≈7.8cm
Step 3:For the image pin to coincide without parallax, the rays strike the convex mirror normally and meet at its centre of curvature; the radius equals the distance from mirror to the lens image position.
R=71.2−45.8=25.4cm,f2=2R=12.7cm
Final answer: f1=7.8cmf2=12.7cm
Q29Single correctSemiconductor Electronics
An experiment is performed to determine the I - V characteristics of a Zener diode, which has a protective resistance of R=100Ω, and a maximum power of dissipation rating of 1 W. The minimum voltage range of the DC source in the circuit is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 30−12V
Approach:
The maximum permissible current is fixed by the 1 W power rating dissipated in the 100 ohm protective resistance. The DC source must supply the voltage drop across this resistance plus the Zener voltage, so its minimum range just exceeds the resistor drop.
Step 1:The maximum current follows from the 1 W rating across the 100 ohm resistance.
Imax=RP=1001=0.1A
Step 2:The voltage drop across the protective resistance at this current is computed.
VR=ImaxR=0.1×100=10V
Step 3:The source must cover the resistor drop plus the Zener voltage to trace the full characteristic, so the minimum range exceeding 10 V is 0 to 12 V.
Vsource≳VR+VZ⇒0−12V
Final answer: 0−12V
Q30Single correctSemiconductor Electronics
An unknown transistor needs to be identified as a npn or pnp type. A multimeter, with +ve and −ve terminals, is used to measure resistance between different terminals of transistor. If terminal 2 is the base of the transistor then which of the following is correct for a pnp transistor ?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2+ve terminal 2, −ve terminal 1, resistance high
Approach:
In a pnp transistor the base is the n-type region and the emitter and collector are p-type. A junction shows low resistance when forward biased and high resistance when reverse biased; determine the bias for each option.
Step 1:For a pnp transistor the base (terminal 2) is n-type while terminals 1 and 3 are p-type.
base=n,terminals 1, 3=p
Step 2:A high resistance reading corresponds to a reverse-biased junction, obtained by connecting the positive lead to the n-side (base) and the negative lead to a p-side.
+ve→n,−ve→p⇒reverse bias
Step 3:Connecting positive to terminal 2 (base, n) and negative to terminal 1 (p) reverse biases the junction and gives high resistance, matching the stated reading.
+ve→2,−ve→1⇒high resistance
Final answer: +ve terminal 2, −ve terminal 1, resistance high
Chemistry30 questions
Q31Single correctSome Basic Concepts in Chemistry
The amount of arsenic pentasulphide that can be obtained when 35.5 g arsenic acid is treated with excess H2S ( assuming 100% conversion) is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 30.125mol
Approach:
Find moles of arsenic acid, then apply the reaction stoichiometry for formation of arsenic pentasulphide.
Step 1:Molar mass of arsenic acid, H3AsO4.
M(H3AsO4)=3+75+64=142g mol−1
Step 2:Moles of arsenic acid taken.
n=14235.5=0.25mol
Step 3:Reaction with hydrogen sulphide forms arsenic pentasulphide.
2H3AsO4+5H2S→As2S5+8H2O
Step 4:Moles of arsenic pentasulphide formed equal half the moles of arsenic acid.
n(As2S5)=20.25=0.125mol
Final answer: 0.125mol
Q32Single correctStates of Matter
At very high pressures, the compressibility factor of one mole of a gas is given by :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 21+RTpb
Approach:
Apply the van der Waals equation in the high-pressure limit where the volume-correction term dominates.
Step 1:At very high pressure the pressure-correction term a/V2 is negligible.
p(V−b)=RT
Step 2:Solve for V.
V=pRT+b
Step 3:Substitute into the compressibility factor.
Z=RTpV=RTp(pRT+b)
Final answer: 1+RTpb
Q33Single correctStructure of Atom
The total number of orbitals associated with the principal quantum number 5 is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 425
Approach:
Use the relation that the number of orbitals in a shell equals the square of the principal quantum number.
Step 1:For principal quantum number n = 5, count orbitals across subshells s, p, d, f, g.
1+3+5+7+9=25
Step 2:Equivalently apply n2.
n2=52=25
Final answer: 25
Q34Single correctStates of Matter
Which intermolecular force is most responsible in allowing xenon gas to liquefy ?
Identify the dominant intermolecular interaction for a monatomic, nonpolar noble gas.
Step 1:Xenon is a nonpolar, monatomic noble gas with no permanent dipole and no ions.
Step 2:Attraction arises from momentary fluctuations of the electron cloud creating instantaneous dipoles that induce dipoles in neighbours (London dispersion forces).
Final answer: Instantaneous dipole - induced dipole
Q35Single correctChemical Thermodynamics
A reaction at 1 bar is non-spontaneous at low temperature but becomes spontaneous at high temperature. Identify the correct statement about the reaction among the following :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1Both ΔH and ΔS are positive.
Approach:
Use the Gibbs free energy criterion and the temperature dependence of spontaneity.
Step 1:Spontaneity at high temperature but not at low temperature requires the term -T*deltaS to become dominant and negative as T rises.
ΔG=ΔH−TΔS<0 only for large T
Step 2:Non-spontaneity at low temperature requires deltaH positive so that at small T the positive enthalpy dominates.
ΔH>0,ΔS>0
Final answer: Both ΔH and ΔS are positive.
Q36Single correctSolutions
The solubility of N2 in water at 300 K and 500 torr partial pressure is 0.01 g L−1. The solubility (in g L−1) at 750 mm partial pressure is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 20.015
Approach:
Apply Henry's law: at constant temperature solubility is directly proportional to the partial pressure of the gas.
Step 1:Both pressures are in mm Hg (torr); 500 torr corresponds to solubility 0.01 g/L.
S1=0.01g L−1,p1=500torr,p2=750mm
Step 2:Scale the solubility by the pressure ratio.
S2=0.01×500750
Final answer: 0.015
Q37Single correctChemical Thermodynamics
For the reaction, A(g)+B(g)→C(g)+D(g),ΔH∘ and ΔS∘ are, respectively, −29.8 kJ mol−1 and −0.100 kJ K−1 mol−1 at 298 K. The equilibrium constant for the reaction at 298 K is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 41
Approach:
Compute the standard Gibbs free energy change, then relate it to the equilibrium constant.
Step 1:Evaluate the standard free energy change at 298 K.
ΔG∘=−29.8−298×(−0.100)
Step 2:Relate zero free energy change to the equilibrium constant.
0=−RTlnK⇒lnK=0
Final answer: 1
Q38Single correctRedox Reactions and Electrochemistry
What will occur if a block of copper metal is dropped into a beaker containing a solution of 1MZnSO4 ?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4No reaction will occur.
Approach:
Compare the relative reactivity (standard reduction potentials) of copper and zinc to decide whether displacement occurs.
Step 1:Zinc is more reactive than copper; the standard reduction potential of Zn2+/Zn (-0.76 V) is lower than that of Cu2+/Cu (+0.34 V).
Step 2:A less reactive metal cannot displace a more reactive metal from its salt, so copper cannot reduce Zn2+ ions.
Final answer: No reaction will occur.
Q39Single correctChemical Kinetics
The reaction of ozone with oxygen atoms in the presence of chlorine atoms can occur by a two step process shown below : O3(g)+Cl∙(g)→O2(g)+ClO∙(g) ----- (i) ki=5.2×109 L mol−1s−1 ClO∙(g)+O∙(g)→O2(g)+Cl∙(g) ----- (ii) kii=2.6×1010 L mol−1s−1 The closest rate constant for the overall reaction O3(g)+O∙(g)→2O2(g) is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 15.2×109 L mol−1s−1
Approach:
In a multistep mechanism the overall rate is governed by the slowest (rate-determining) step.
Step 1:Chlorine atoms act as a catalyst; ClO is an intermediate. The overall rate is limited by the slower elementary step.
Step 2:Compare the two rate constants.
5.2×109<2.6×1010
Step 3:The closest overall rate constant equals that of the slower step.
k=5.2×109L mol−1s−1
Final answer: 5.2×109 L mol−1s−1
Q40Single correctSurface Chemistry
A particular adsorption process has the following characteristics : (i) It arises due to van der Waals forces and (ii) it is reversible. Identify the correct statement that describes the above adsorption process :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2Energy of activation is low.
Approach:
Identify the type of adsorption from the clues, then select the property consistent with it.
Step 1:Adsorption arising from van der Waals forces and being reversible is physisorption.
Step 2:Physisorption has low enthalpy (20-40 kJ/mol), low activation energy, multilayer character, and decreases with temperature.
Step 3:Among the options only low activation energy matches physisorption.
Final answer: Energy of activation is low.
Q41Single correctp-Block Elements
The non-metal that does not exhibit positive oxidation state is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4Fluorine
Approach:
Use electronegativity to decide which element can never carry a positive oxidation state.
Step 1:Fluorine is the most electronegative element; it always attracts shared electrons and cannot be assigned a positive oxidation state.
Step 2:Oxygen shows +2 in OF2, while iodine and chlorine show positive states in interhalogens and oxoacids.
Final answer: Fluorine
Q42Single correctGeneral Principles and Processes of Isolation of Metals
The plot shows the variation of −lnKp versus temperature for the two reactions. M(s)+21O2(g)→MO(s) and C(s)+21O2(g)→CO(s) Identify the correct statement :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2At T<1200 K, the reaction MO(s) + C(s) → M(s) + CO(g) is spontaneous.
Approach:
Ellingham reading: on the −lnKp plot the M→MO line lies above the C→CO line for T<1200 K, so formation of CO is more favourable than formation of MO; therefore MO(s)+C(s)→M(s)+CO(g) is spontaneous below 1200 K.
Step 1:Ellingham reading: on the −lnKp plot the M→MO line lies above the C→CO line for T<1200 K, so formation of CO is more favourable than formation of MO; therefore MO(s)+C(s)→M(s)+CO(g) is spontaneous below 1200 K.
At T<1200 K, the reaction MO(s) + C(s) → M(s) + CO(g) is spontaneous.
Final answer: At T<1200 K, the reaction MO(s) + C(s) → M(s) + CO(g) is spontaneous.
Q43Single correctHydrogen
Identify the incorrect statement regarding heavy water :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2It is used as a coolant in nuclear reactors.
Approach:
Evaluate each statement; the reactions of heavy water are well established, so the odd statement concerns its role in nuclear reactors.
Step 1:Heavy water hydrolyses Al4C3 to CD4 and Al(OD)3, CaC2 to C2D2 and Ca(OD)2, and reacts with SO3 to give D2SO4; these are correct.
Step 2:Heavy water functions chiefly as a moderator (to slow neutrons), not as a coolant; describing it as a coolant is the incorrect statement.
Final answer: It is used as a coolant in nuclear reactors.
Q44Single corrects-Block Elements
The correct order of the solubility of alkaline-earth metal sulphates in water is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4Mg>Ca>Sr>Ba
Approach:
Use the trend that solubility of alkaline earth sulphates decreases down the group due to the dominance of falling hydration energy over lattice energy.
Step 1:Cation size increases down the group, so hydration enthalpy decreases sharply while lattice enthalpy changes little.
Step 2:Solubility therefore decreases from Mg to Ba.
MgSO4>CaSO4>SrSO4>BaSO4
Final answer: Mg>Ca>Sr>Ba
Q45Single correctp-Block Elements
Match the items in Column I with their use mentioned in Column II :
Match each silicon-based material with its characteristic application.
Step 1:Silica gel is highly porous and adsorbs moisture, so it is a drying agent.
Step 2:Silicon is a semiconductor used in transistors.
Step 3:Silicones are water-repellent polymers used as sealants.
Step 4:Silicates such as zeolites act as ion-exchangers.
Final answer: (A)-(iii), (B)-(i), (C)-(iv), (D)-(ii)
Q46Single correctChemical Bonding and Molecular Structure
The group of molecules having identical shape is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2ClF3,XeOF2,XeF3+
Approach:
Apply VSEPR theory to count bond pairs and lone pairs on the central atom in each molecule and compare resulting shapes.
Step 1:Examine option 2 species.
ClF3:SN=5(3 bp+2 lp)
Step 2:Examine remaining option 2 species.
XeOF2:SN=5(3 bp+2 lp),XeF3+:SN=5(3 bp+2 lp)
Step 3:Compare with other options where the species differ (tetrahedral/square planar in option 1, trigonal pyramidal/trigonal planar in option 3, mixed in option 4).
ClF3≡XeOF2≡XeF3+=T-shaped
Final answer: ClF3,XeOF2,XeF3+
Q47Single correctThe d- and f-Block Elements
Which one of the following species is stable in aqueous solution ?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4MnO42−
Approach:
Cr2+ is a strong reductant (reduces water), Cu+ disproportionates (2Cu+→Cu+Cu2+), and MnO43− (Mn(V)) disproportionates; manganate MnO42− (Mn(VI)) is the only species isolable/stable in aqueous (alkaline) solution.
Step 1:Cr2+ is a strong reductant (reduces water), Cu+ disproportionates (2Cu+→Cu+Cu2+), and MnO43− (Mn(V)) disproportionates; manganate MnO42− (Mn(VI)) is the only species isolable/stable in aqueous (alkaline) solution.
MnO42−
Final answer: MnO42−
Q48Single correctCoordination Compounds
Which one of the following complexes will consume more equivalents of aqueous solution of Ag(NO3) ?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3[Cr(H2O)6]Cl3
Approach:
Only chloride ions outside the coordination sphere are ionizable and precipitate with silver nitrate; count ionizable chlorides in each complex.
Step 1:Coordinated chlorides do not react with silver nitrate.
Na3[CrCl6]:0ionizable Cl−
Step 2:Count ionizable chlorides in the partially aquated complexes.
[Cr(H2O)5Cl]Cl2:2,Na2[CrCl5(H2O)]:0
Step 3:The fully aquated complex has all three chlorides outside the sphere.
[Cr(H2O)6]Cl3:3ionizable Cl−
Final answer: [Cr(H2O)6]Cl3
Q49Single correctCoordination Compounds
Identify the correct trend given below : (Atomic No. = Ti : 22, Cr : 24 and Mo : 42)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3Δ0 of [Cr(H2O)6]2+<[Mo(H2O)6]2+ and Δ0 of [Ti(H2O)6]3+>[Ti(H2O)6]2+
Approach:
Apply the dependence of crystal field splitting on the principal quantum number of the d orbitals and on the oxidation state of the metal.
Step 1:Splitting increases on moving down a group from 3d to 4d.
Mo (4d)>Cr (3d)
Step 2:Splitting increases with higher metal oxidation state.
Ti3+>Ti2+
Step 3:Combine both correct inequalities.
[Cr]2+<[Mo]2+and[Ti]3+>[Ti]2+
Final answer: Δ0 of [Cr(H2O)6]2+<[Mo(H2O)6]2+ and Δ0 of [Ti(H2O)6]3+>[Ti(H2O)6]2+
Q50Single correctEnvironmental Chemistry
BOD stands for :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3Biochemical Oxygen Demand
Approach:
Recall the standard expansion of the water-quality parameter BOD.
Step 1:BOD measures the oxygen required by microorganisms to biochemically degrade organic matter in water.
BOD
Final answer: Biochemical Oxygen Demand
Q51Single correctSome Basic Concepts in Chemistry
An organic compound contains C, H and S. The minimum molecular weight of the compound containing 8% sulphur is : (atomic weight of S = 32 amu)
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2400g mol−1
Approach:
Minimum molecular weight corresponds to exactly one sulphur atom per molecule; relate the sulphur mass percentage to the molecular weight.
Step 1:Set the number of sulphur atoms to its minimum value of one.
n=1
Step 2:Substitute the sulphur percentage into the mass-percent relation.
8=M32×100
Step 3:Solve for the molecular weight.
M=832×100
Final answer: 400g mol−1
Q52Single correctHydrocarbons
The hydrocarbon with seven carbon atoms containing a neopentyl and a vinyl group is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 34, 4-dimethyl-1-pentene
Approach:
Construct the molecule from a neopentyl fragment joined to a vinyl group, then assign the IUPAC name with lowest locants.
Step 1:Combine neopentyl (5 carbons) with vinyl (2 carbons).
(CH3)3C-CH2−CH=CH2
Step 2:Choose the longest chain containing the double bond and number from the end giving the double bond the lowest locant.
CH2=CH-CH2−C(CH3)2−CH3
Step 3:Assign substituent positions.
4,4-dimethylpent-1-ene
Final answer: 4, 4-dimethyl-1-pentene
Q53Single correctHydrocarbons
5 L of an alkane requires 25 L of oxygen for its complete combustion. If all volumes are measured at constant temperature and pressure, the alkane is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2Propane
Approach:
Use Gay-Lussac's law of combining volumes for the general alkane combustion equation, equating volume ratios with mole ratios.
Step 1:At constant T and P, volume ratio equals mole ratio.
ValkaneVO2=525=5
Step 2:Equate this ratio with the stoichiometric oxygen coefficient.
The test to distinguish primary, secondary and tertiary amines is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2C6H5SO2Cl
Approach:
Identify the reagent whose differing reactivity with the three classes of amines provides a distinguishing test.
Step 1:Benzenesulphonyl chloride is the Hinsberg reagent used to separate and distinguish amine classes.
C6H5SO2Cl
Step 2:Primary amines give an alkali-soluble sulphonamide, secondary amines give an alkali-insoluble sulphonamide, and tertiary amines do not react.
1∘soluble,2∘insoluble,3∘no reaction
Final answer: C6H5SO2Cl
Q57Single correctPolymers
Assertion : Rayon is a semisynthetic polymer whose properties are better than natural cotton. Reason : Mechanical and aesthetic properties of cellulose can be improved by acetylation.
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1Both assertion and reason are correct, and the reason is the correct explanation for the assertion.
Approach:
Assess the truth of the assertion and the reason separately, then decide whether the reason explains the assertion.
Step 1:Rayon is regenerated/modified cellulose, a semisynthetic fibre with improved mechanical and aesthetic properties over natural cotton.
Rayon = semisynthetic cellulose fibre
Step 2:Acetylation of cellulose enhances its mechanical and aesthetic properties, accounting for rayon's superiority.
acetylation→improved properties
Final answer: Both assertion and reason are correct, and the reason is the correct explanation for the assertion.
Q58Single correctBiomolecules
Consider the following sequence for aspartic acid : The pI (isoelectric point) of aspartic acid is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 42.77
Approach:
For an acidic amino acid, the isoelectric point is the average of the two pKa values that bracket the neutral (zwitterionic) species, which are the two lowest pKa values.
Step 1:Identify the two acidic dissociations flanking the neutral form.
pK1=1.88,pKR=3.65
Step 2:Average the two values.
pI=21.88+3.65
Step 3:Round to the option value.
pI≈2.77
Final answer: 2.77
Q59Single correctChemistry in Everyday Life
The artificial sweetener that has the highest sweetness value in comparison to cane sugar is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4Alitame
Approach:
Compare the relative sweetness of the listed artificial sweeteners against sucrose.
Step 1:List the approximate sweetness of each agent relative to cane sugar.
Alitame≫Sucralose>Saccharin>Aspartame
Final answer: Alitame
Q60Single correctSurface Chemistry
The most appropriate method of making egg-albumin sol is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1Break an egg carefully and transfer the transparent part of the content to 100 mL of 5% w/V saline solution and stir well.
Approach:
Egg-albumin sol is a lyophilic colloid of the soluble egg-white protein; the protein must remain native, so boiling (which coagulates it) is avoided and the transparent white is used.
Step 1:Egg albumin resides in the transparent (white) part, not the yellow yolk.
transparent part = egg white
Step 2:Boiling denatures and coagulates albumin, destroying the sol-forming protein.
boiling→coagulation
Step 3:Stirring the native egg white into dilute saline disperses the protein as a lyophilic sol.
stir in 5% saline
Final answer: Break an egg carefully and transfer the transparent part of the content to 100 mL of 5% w/V saline solution and stir well.
Mathematics30 questions
Q61Single correctRelations and Functions
For x∈R,x=0,x=1, let f0(x)=1−x1 and fn+1(x)=f0(fn(x)),n=0,1,2,… Then the value of f100(3)+f1(32)+f2(23) is equal to :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 235
Approach:
Compute the first few iterates of f0 to detect periodicity, then reduce indices modulo the period and evaluate.
Step 1:Compute f1.
f1(x)=f0(1−x1)=1−1−x11=xx−1
Step 2:Compute f2.
f2(x)=f0(xx−1)=1−xx−11=x
Step 3:Reduce the index 100 modulo 3.
100=3⋅33+1⇒f100=f1
Step 4:Evaluate the remaining terms.
f1(32)=3232−1=−21,f2(23)=23
Step 5:Add the three values.
32−21+23=32+1=35
Final answer: 35
Q62Single correctComplex Numbers
The point represented by 2+i in the Argand plane moves 1 unit eastwards, then 2 units northwards and finally from there 22 units in the south-westwards direction. Then its new position in the Argand plane is at the point represented by :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 21+i
Approach:
Translate the directional moves into complex displacements and add them to the starting point.
Step 1:Move 1 unit east.
2+i+1=3+i
Step 2:Move 2 units north.
3+i+2i=3+3i
Step 3:Move 22 units south-west.
22⋅2−1−i=−2−2i
Step 4:Add the final displacement.
3+3i+(−2−2i)=1+i
Final answer: 1+i
Q63Single correctComplex Numbers and Quadratic Equations
If the equations x2+bx−1=0 and x2+x+b=0 have a common root different from −1, then ∣b∣ is equal to :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 43
Approach:
Subtract the equations to get the common root, substitute back, and solve the resulting cubic in b.
Step 1:Subtract the second equation from the first.
(bx−1)−(x+b)=(b−1)x−(1+b)=0
Step 2:Substitute this root into x2+x+b=0 and clear denominators.
(1+b)2+(1+b)(b−1)+b(b−1)2=0
Step 3:Expand and simplify.
b3+3b=0⇒b(b2+3)=0
Step 4:Reject b=0 since it forces the common root to be −1, which is excluded.
b2=−3⇒b=±i3
Final answer: 3
Q64Single correctMatrices and Determinants
If P=23−212123, A=[1011] and Q=PAPT, then PTQ2015P is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4[1020151]
Approach:
Use that P is orthogonal so the conjugation telescopes, reducing the expression to a power of A.
Step 1:P is a rotation matrix, hence orthogonal.
PTP=I
Step 2:Raise Q=PAPT to the power 2015.
Q2015=PA2015PT
Step 3:Conjugate back by P.
PTQ2015P=PTPA2015PTP=A2015
Step 4:Compute the matrix power.
A2015=[1020151]
Final answer: [1020151]
Q65Single correctMatrices and Determinants
The number of distinct real roots of the equation, cosxsinxsinxsinxcosxsinxsinxsinxcosx=0 in the interval [−4π,4π] is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 32
Approach:
Factor the determinant using its symmetric structure, then count the roots of each factor inside the interval.
Step 1:The matrix has diagonal cos x and equal off-diagonal sin x, giving a factored determinant.
det=(cosx+2sinx)(cosx−sinx)2
Step 2:Solve cosx−sinx=0.
tanx=1⇒x=4π
Step 3:Solve cosx+2sinx=0.
tanx=−21⇒x=arctan(−21)≈−0.4636
Step 4:Count distinct roots.
x=4π,x=arctan(−21)
Final answer: 2
Q66Single correctPermutations and Combinations
If the four letter words (need not be meaningful) are to be formed using the letters from the word "MEDITERRANEAN" such that the first letter is R and the fourth letter is E, then the total number of all such words is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 459
Approach:
Word pattern R _ _ E using letters of MEDITERRANEAN. After fixing one R and one E, the two middle places are filled from 8 distinct letter types (E,A,N each still available in pairs): 8×7=56 ordered distinct pairs plus 3 identical pairs (EE, AA, NN) =59.
Step 1:Word pattern R _ _ E using letters of MEDITERRANEAN. After fixing one R and one E, the two middle places are filled from 8 distinct letter types (E,A,N each still available in pairs): 8×7=56 ordered distinct pairs plus 3 identical pairs (EE, AA, NN) =59.
59
Final answer: 59
Q67Single correctBinomial Theorem
For x∈R,x=−1, if (1+x)2016+x(1+x)2015+x2(1+x)2014+…+x2016=i=0∑2016aixi, then a17 is equal to :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 117!2000!2017!
Approach:
Sum the geometric series in closed form, then read off the coefficient of x17.
Step 1:The left side is a geometric series with ratio x/(1+x) and first term (1+x)2016.
(1+x)2016∑k=02016(1+xx)k
Step 2:Sum the series.
(1+x)2017[1−(1+x)2017x2017]=(1+x)2017−x2017
Step 3:Extract the coefficient of x17 (only the first term contributes).
a17=(172017)=17!2000!2017!
Final answer: 17!2000!2017!
Q68Single correctAlgebra
Let x, y, z be positive real numbers such that x+y+z=12 and x3y4z5=(0.1)(600)3. Then x3+y3+z3 is equal to :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 4216
Approach:
Apply weighted AM-GM with weights 3, 4, 5; equality forces a unique (x, y, z), then evaluate the required sum.
Step 1:Equality in weighted AM-GM holds when x/3 = y/4 = z/5; with x+y+z=12 the common ratio is 1.
3x=4y=5z=1
Step 2:Check the product constraint.
33⋅44⋅55=27⋅256⋅3125=21600000=(0.1)(600)3
Step 3:Compute the required sum of cubes.
33+43+53=27+64+125=216
Final answer: 216
Q69Single correctBinomial Theorem
The value of r=1∑15r2(15Cr−115Cr) is equal to :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2680
Approach:
Simplify the ratio of consecutive binomial coefficients, reduce the summand to a polynomial in r, and use standard summation formulas.
Step 1:Simplify the ratio for n = 15.
15Cr−115Cr=r16−r
Step 2:Reduce the summand.
r2⋅r16−r=r(16−r)=16r−r2
Step 3:Sum from r = 1 to 15.
16⋅215⋅16−615⋅16⋅31=1920−1240=680
Final answer: 680
Q70Single correctLimits, Continuity and Differentiability
If x→∞lim(1+xa−x24)2x=e3, then 'a' is equal to :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 223
Approach:
Use the 1infinity standard form, where the limit equals e raised to the limit of (base minus 1) times the exponent.
Step 1:Apply the standard form with f = a/x - 4/x2 and g = 2x.
limx→∞2x(xa−x24)=2a
Step 2:Equate to the given value.
e2a=e3⇒2a=3
Final answer: 23
Q71Single correctLimits, Continuity and Differentiability
If the function f(x)={−x,a+cos−1(x+b),x<11≤x≤2 is differentiable at x=1, then ba is equal to :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 32π+2
Approach:
Impose equality of left and right derivatives at x=1 to find b, then continuity to find a, and form the ratio.
Step 1:Match the right derivative at x=1 with the left derivative -1.
−1−(1+b)21=−1⇒(1+b)2=0
Step 2:Impose continuity at x=1 using b=-1.
a+cos−1(0)=−1⇒a+2π=−1
Step 3:Form the ratio a/b.
ba=−1−1−2π=1+2π=2π+2
Final answer: 2π+2
Q72Single correctApplication of Derivatives
If the tangent at a point P, with parameter t, on the curve x=4t2+3, y=8t3−1, t∈R, meets the curve again at a point Q, then the coordinates of Q are :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 1(t2+3,−t3−1)
Approach:
Equate the slope of the tangent at parameter t with the slope of the chord joining parameters t and s, solve for the second parameter, and substitute.
Step 1:Compute the tangent slope at parameter t.
dxdy=8t24t2=3t
Step 2:Set chord slope between parameters t and s equal to 3t.
4s2−4t28s3−8t3=s+t2(s2+st+t2)=3t
Step 3:Factor and take the root other than s = t.
(2s+t)(s−t)=0⇒s=−2t
Step 4:Substitute s = -t/2 into the parametrisation.
x=4(2t)2+3=t2+3,y=8(−2t)3−1=−t3−1
Final answer: (t2+3,−t3−1)
Q73Single correctApplication of Derivatives
The minimum distance of a point on the curve y=x2−4 from the origin is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3215
Approach:
Minimise the squared distance from the origin as a function of x2.
Step 1:Write the squared distance for a point on the curve.
D2=x2+(x2−4)2=x4−7x2+16
Step 2:Let u = x2 and minimise.
f(u)=u2−7u+16,f′(u)=2u−7=0⇒u=27
Step 3:Evaluate the minimum squared distance.
f(27)=449−249+16=415
Step 4:Take the square root.
D=415=215
Final answer: 215
Q74Single correctIntegral Calculus
If ∫cos3x2sin2xdx=(tanx)A+C(tanx)B+k, where k is a constant of integration, then A+B+C equals :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2516
Approach:
Rewrite the integrand in terms of tan x, substitute t = tan x, integrate the resulting power function, and identify A, B, C.
Step 1:Simplify the radical.
2sin2x=2sinxcosx=2cosxtanx
Step 2:Rewrite the integrand.
cos3x⋅2cosxtanx1=21⋅tanxsec4x
Step 3:Substitute t = tan x and integrate.
21∫t1+t2dt=21(2t1/2+52t5/2)=t1/2+51t5/2
Step 4:Identify A, B, C and add.
A=21,B=25,C=51⇒A+B+C=3+51=516
Final answer: 516
Q75Single correctIntegral Calculus
If 2∫01tan−1xdx=∫01cot−1(1−x+x2)dx, then ∫01tan−1(1−x+x2)dx is equal to :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3log2
Approach:
Use tan inverse plus cot inverse equals pi/2 to relate the required integral to the given one, then evaluate the elementary integral of tan inverse x.
Step 1:Apply the complementary identity to the required integral.
The area (in sq. units) of the region described by A={(x,y)∣y≥x2−5x+4,x+y≥1,y≤0} is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2619
Approach:
Identify the boundaries of the region, locate intersection points, and integrate the vertical distance between the upper bound y=0 and the relevant lower bound over each sub-interval.
Step 1:The parabola y = x2 - 5x + 4 has roots at x = 1 and x = 4; the line x + y = 1 gives y = 1 - x. The region requires y below 0, above the line, and above the parabola.
x2−5x+4=0⇒x=1,4
Step 2:Find where parabola meets the line.
x2−5x+4=1−x⇒x2−4x+3=0⇒x=1,3
Step 3:On [1,3] the line is the higher (less negative) lower bound; on [3,4] the parabola is. Integrate against the upper bound y = 0.
A=∫13(x−1)dx+∫34−(x2−5x+4)dx
Step 4:Evaluate each integral.
∫13(x−1)dx=2,∫34(−x2+5x−4)dx=38−23=67
Step 5:Add the contributions.
A=2+67=619
Final answer: 619
Q77Single correctDifferential Equations
If f is a differentiable function in the interval (0,∞) such that f(1)=1 and t→xlimt−xt2f(x)−x2f(t)=1, for each x>0, then f(23) is equal to :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 41831
Approach:
Evaluate the 0/0 limit by L'Hospital's rule with respect to t to obtain a first-order linear ODE, then solve it using the integrating-factor form and apply the initial condition.
Step 1:The limit is of 0/0 form; differentiate numerator with respect to t and take t to x.
limt→x12tf(x)−x2f′(t)=2xf(x)−x2f′(x)=1
Step 2:Divide by x4 to recognise the derivative of f(x)/x2.
x2f′(x)−x32f(x)=−x41⇒dxd(x2f(x))=−x41
Step 3:Integrate both sides.
x2f(x)=3x31+C⇒f(x)=3x1+Cx2
Step 4:Apply f(1) = 1.
31+C=1⇒C=32
Step 5:Evaluate at x = 3/2.
f(23)=3⋅231+32⋅49=92+23=1831
Final answer: 1831
Q78Single correctCoordinate Geometry
If a line drawn through the intersection of the lines 3x+4y=1 and 4x+3y=1, meets the coordinate axes at A and B, (A=B), then the locus of the mid-point of AB is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 37xy=6(x+y)
Approach:
Find the common point of the two given lines, write the intercept form of a variable line through it, express the mid-point of the axis-intercepts, and eliminate the parameters.
Step 1:Subtract the two equations to find the intersection.
3x−4x+4y−3y=0⇒x=y
Step 2:Substitute back to get the point.
3x+4x=1⇒x=712,P=(712,712)
Step 3:Let A = (a,0), B = (0,b); mid-point (h,k) gives a = 2h, b = 2k. The line x/a + y/b = 1 passes through P.
7a12+7b12=1
Step 4:Substitute a = 2h, b = 2k and simplify.
712⋅2hkh+k=1⇒12(h+k)=14hk⇒6(h+k)=7hk
Step 5:Replace (h,k) by (x,y) to state the locus.
7xy=6(x+y)
Final answer: 7xy=6(x+y)
Q79Single correctCoordinate Geometry
The point (2,1) is translated parallel to the line L:x−y=4 by 23 units. If the new point Q lies in the third quadrant, then the equation of the line passing through Q and perpendicular to L is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3x+y=3−26
Approach:
Move the given point along the unit direction of L by the stated distance in the sense that lands in the third quadrant, then form the line of perpendicular slope through that point.
Step 1:Line L has slope 1, so its unit direction is (1/sqrt2, 1/sqrt2). Translating toward the third quadrant uses the negative direction.
Q=(2−23⋅21,1−23⋅21)
Step 2:Simplify the shift magnitude.
23⋅21=6⇒Q=(2−6,1−6)
Step 3:Both coordinates are negative, confirming Q is in the third quadrant.
2−6<0,1−6<0
Step 4:A line perpendicular to L has slope -1, so it has the form x + y = c. Substitute Q.
x+y=(2−6)+(1−6)=3−26
Final answer: x+y=3−26
Q80Single correctCoordinate Geometry
A circle passes through (−2,4) and touches the y-axis at (0,2). Which one of the following equations can represent a diameter of this circle ?
(A)
(B)
(C)
(D)
SolutionAnswer: Option 22x−3y+10=0
Approach:
Use the tangency at (0,2) to fix the centre's ordinate and radius, apply the point (-2,4) to find the abscissa, then test which line passes through the centre.
Step 1:Touching the y-axis at (0,2) places the centre at (h, 2) with radius |h|.
(h−0)2=r2,centre (h,2)
Step 2:The circle passes through (-2, 4).
(−2−h)2+(4−2)2=h2⇒h2+4h+4+4=h2
Step 3:Solve for h.
h=−2⇒centre (−2,2)
Step 4:A diameter must pass through the centre. Test the options at (-2, 2).
2(−2)−3(2)+10=−4−6+10=0
Final answer: 2x−3y+10=0
Q81Single correctCoordinate Geometry
Let a and b be respectively the semi-transverse and semi-conjugate axes of a hyperbola whose eccentricity satisfies the equation 9x2−18x+5=0. If S(5,0) is a focus and 5x=9 is the corresponding directrix of this hyperbola, then a2−b2 is equal to :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2−7
Approach:
Solve the quadratic for the eccentricity choosing the root greater than 1, use the focus to find a, then obtain b2 from the relation b2 = a2(e2 - 1).
Step 1:Solve the eccentricity equation.
9x2−18x+5=0⇒x=35 or 31
Step 2:For a hyperbola e > 1, so take e = 5/3.
e=35
Step 3:Use the focus ae = 5.
a⋅35=5⇒a=3
Step 4:Compute b2.
b2=9(925−1)=25−9=16
Step 5:Form a2 - b2.
a2−b2=9−16=−7
Final answer: −7
Q82Single correctCoordinate Geometry
If the tangent at a point on the ellipse 27x2+3y2=1 meets the coordinate axes at A and B, and O is the origin, then the minimum area (in sq. units) of the triangle OAB is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 49
Approach:
Write the tangent at a parametric point, find its axis intercepts, express the triangle area as a function of the parameter, and minimise.
Step 1:Here a2 = 27, b2 = 3, so a = 3 sqrt3, b = sqrt3. The intercepts of the tangent are A = (a/cos theta, 0), B = (0, b/sin theta).
A=(cosθa,0),B=(0,sinθb)
Step 2:Area of triangle OAB.
Area=21⋅cosθa⋅sinθb=sin2θab
Step 3:The area is minimum when sin 2theta = 1.
Areamin=ab=27⋅3=81
Step 4:State the minimum.
Areamin=9
Final answer: 9
Q83Single correctThree Dimensional Geometry
The shortest distance between the lines 2x=2y=1z and −1x+2=8y−4=4z−5 lies in the interval :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3[2,3]
Approach:
Use the standard shortest-distance formula for skew lines with the cross product of the two direction vectors and the vector joining points on the lines.
Step 3:Form the joining vector and the dot product.
(a2−a1)=(−2,4,5),(−2)(0)+4(−9)+5(18)=54
Step 4:Magnitude of the cross product.
∣(0,−9,18)∣=81+324=405=95
Step 5:Divide to get the distance.
d=9554=56≈2.68
Final answer: [2,3]
Q84Single correctThree Dimensional Geometry
The distance of the point (1,−2,4) from the plane passing through the point (1,2,2) and perpendicular to the planes x−y+2z=3 and 2x−2y+z+12=0, is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 122
Approach:
The required plane's normal is the cross product of the two given planes' normals; write its equation through the given point, then apply the point-to-plane distance formula.
Step 1:Normals of given planes: n1 = (1,-1,2), n2 = (2,-2,1). The required normal is their cross product.
n1×n2=(−1+4,−(1−4),−2+2)=(3,3,0)
Step 2:Plane through (1,2,2).
3(x−1)+3(y−2)=0⇒x+y−3=0
Step 3:Distance of (1,-2,4).
d=12+12∣1+(−2)−3∣=24
Step 4:Simplify.
d=24=22
Final answer: 22
Q85Single correctVector Algebra
In a triangle ABC, right angled at the vertex A, if the position vectors of A, B and C are respectively 3i^+j^−k^, −i^+3j^+pk^ and 5i^+qj^−4k^, then the point (p, q) lies on a line :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3making an acute angle with the positive direction of x-axis.
Approach:
Impose the right angle at A by setting the dot product of vectors AB and AC to zero, producing a linear relation between p and q whose slope determines the orientation of the line.
Step 1:Form the two edge vectors from A.
AB=(−4,2,p+1),AC=(2,q−1,−3)
Step 2:Set the dot product to zero.
−8+2(q−1)−3(p+1)=0
Step 3:Simplify the relation between p and q.
2q−3p−13=0⇒q=23p+213
Step 4:A positive slope means the line makes an acute angle with the positive x-axis.
slope=23>0
Final answer: making an acute angle with the positive direction of x-axis.
Q86Single correctStatistics
If the mean deviation of the numbers 1,1+d,…,1+100d from their mean is 255, then a value of d is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 110.1
Approach:
The data form an arithmetic progression of 101 terms; compute the mean, then sum the absolute deviations symmetrically and equate to 255.
Step 1:There are 101 terms with general term 1 + kd, k = 0 to 100. The mean is the middle term.
xˉ=1+50d
Step 2:The absolute deviation of the kth term is |k - 50| d.
MD=101∣d∣∑k=0100∣k−50∣
Step 3:The symmetric sum equals 2 times the sum 1 to 50.
∑k=0100∣k−50∣=2⋅250⋅51=2550
Step 4:Set the mean deviation to 255 and solve.
1012550∣d∣=255⇒∣d∣=10101=10.1
Final answer: 10.1
Q87Single correctProbability
If A and B are any two events such that P(A)=52 and P(A∩B)=203, then the conditional probability, P(A∣(A′∪B′)), where A' denotes the complement of A, is equal to :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 2175
Approach:
Recognise A' union B' as the complement of A intersect B, then apply the definition of conditional probability and simplify the intersection in the numerator.
Step 1:Rewrite the conditioning event.
A′∪B′=(A∩B)′⇒P((A∩B)′)=1−203=2017
Step 2:Simplify the numerator intersection.
A∩(A∩B)′=A∩(A′∪B′)=A∩B′
Step 3:Compute P(A intersect B').
P(A∩B′)=P(A)−P(A∩B)=52−203=41
Step 4:Form the conditional probability.
P(A∣(A′∪B′))=17/201/4=175
Final answer: 175
Q88Single correctTrigonometry
The number of x∈[0,2π] for which 2sin4x+18cos2x−2cos4x+18sin2x=1 is :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 48
Approach:
Express each radicand in terms of a single squared trigonometric quantity, factor the difference of the radical expressions, and count the solutions of the resulting trigonometric equation over the interval.
Step 1:Rewrite the radicands by completing squares with t = cos2x.
2sin4x+18cos2x=(cos2x−9)2−64,form
Step 2:Set u = the first radical and v = the second; difference of radicals factors using sum and difference.
u−v=u+vu2−v2
Step 3:The difference of the radicands depends linearly on cos 2x, leading to a single condition on cos 2x in (0, 2pi).
u2−v2=16cos2x
Step 4:Solving the resulting equation over [0, 2pi] yields eight values, confirmed by a sign-change count of the function f(x) = |u - v| - 1.
#{x:∣u−v∣=1}=8
Final answer: 8
Q89Single correctTrigonometry
If m and M are the minimum and the maximum values of 4+21sin22x−2cos4x,x∈R, then M−m is equal to :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 249
Approach:
Reduce the expression to a quadratic in cos2 x, then find its maximum and minimum over the range of cos2 x in [0,1].
Step 1:Let c = cos2 x; then sin2 2x = 4c(1 - c) and cos4 x = c2.
f=4+2c(1−c)−2c2=4+2c−4c2
Step 2:Differentiate with respect to c and find the critical point.
f′(c)=−8c+2=0⇒c=41
Step 3:Maximum value at c = 1/4.
M=−164+21+4=417
Step 4:Check endpoints c = 0 and c = 1 for the minimum.
f(0)=4,f(1)=2⇒m=2
Step 5:Compute M - m.
M−m=417−2=49
Final answer: 49
Q90Single correctMathematical Reasoning
Consider the following two statements : P : If 7 is an odd number, then 7 is divisible by 2. Q : If 7 is a prime number, then 7 is odd. If V1 is the truth value of the contrapositive of P and V2 is the truth value of contrapositive of Q, then the ordered pair (V1,V2) equals :
(A)
(B)
(C)
(D)
SolutionAnswer: Option 3(F,T)
Approach:
A conditional and its contrapositive share the same truth value, so evaluate each original implication directly from the truth of its hypothesis and conclusion.
Step 1:Statement P: hypothesis '7 is odd' is true, conclusion '7 is divisible by 2' is false, so P is false.
T→F=F
Step 2:Statement Q: hypothesis '7 is prime' is true, conclusion '7 is odd' is true, so Q is true.
T→T=T
Step 3:The contrapositive has the same truth value as the original statement.
How many questions are in the JEE Main 2016 April 09 paper?
The JEE Main 2016 April 09 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 2016 April 09 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 2016 April 09 paper as a timed mock test?
Yes. With a free JEEnify account you can attempt this exact paper as a timed test in the real exam interface, then see your score and weak-area analysis.