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JEE Advanced 2016 Paper 2 Question Paper with Answers

54 questions · Physics, Chemistry & Mathematics

The complete JEE Advanced 2016 Paper 2 paper — every question with its correct answer, tagged to the chapter it tests. Free to read, no account needed.

Physics
18
Chemistry
18
Mathematics
18

Physics — JEE Advanced 2016 Paper 2

Q1·PhysicsSingle correct
The electrostatic energy of Z protons uniformly distributed throughout a spherical nucleus of radius R is given by E=35 Z(Z−1)e24πε0RE = \frac{3}{5}\,\frac{Z(Z-1)e^{2}}{4\pi\varepsilon_{0}R}E=53​4πε0​RZ(Z−1)e2​ The measured masses of the neutron, 11H^{1}_{1}\text{H}11​H, 715N^{15}_{7}\text{N}715​N and 815O^{15}_{8}\text{O}815​O are 1.008665 u, 1.007825 u, 15.000109 u and 15.003065u, respectively. Given that the radii of both the 715N^{15}_{7}\text{N}715​N and 815O^{15}_{8}\text{O}815​O nuclei are same, 1 u = 931.5 MeV/c2c^{2}c2 ( c is the speed of light) and e2/(4πε0)e^{2}/(4\pi\varepsilon_{0})e2/(4πε0​) = 1.44 MeV fm. Assuming that the difference between the binding energies of 715N^{15}_{7}\text{N}715​N and 815O^{15}_{8}\text{O}815​O is purely due to the electrostatic energy, the radius of either of the nuclei is (1 fm = 10−1510^{-15}10−15m)
  1. (A)2.85 fm
  2. (B)3.03 fm
  3. (C)3.42 fm
  4. (D)3.80 fm

Correct answer: (C)

Step-by-step solution →
Q2·PhysicsSingle correct
The ends Q and R of two thin wires, PQ and RS, are soldered (joined) together. Initially each of the wires has a length of 1 m at 10 0^{0}0C. Now the end P is maintained at 10 0^{0}0C, while the end S is heated and maintained at 400 0^{0}0C. The system is thermally insulated from its surroundings. If the thermal conductivity of wire PQ is twice that of the wire RS and the coefficient of linear thermal expansion of PQ is 1.2×10−5 K−11.2 \times 10^{-5}\ \text{K}^{-1}1.2×10−5 K−1, the change in length of the wire PQ is
  1. (A)0.78 mm
  2. (B)0.90 mm
  3. (C)1.56 mm
  4. (D)2.34 mm

Correct answer: (A)

Step-by-step solution →
Q3·PhysicsSingle correct
An accident in a nuclear laboratory resulted in deposition of a certain amount of radioactive material of half-life 18 days inside the laboratory. Tests revealed that the radiation was 64 times more than the permissible level required for safe operation of the laboratory. What is the minimum number of days after which the laboratory can be considered safe for use?
  1. (A)64
  2. (B)90
  3. (C)108
  4. (D)120

Correct answer: (C)

Step-by-step solution →
Q4·PhysicsSingle correct
There are two Vernier calipers both of which have 1 cm divided into 10 equal divisions on the main scale. The Vernier scale of one of the calipers (C1C_{1}C1​) has 10 equal divisions that correspond to 9 main scale divisions. The Vernier scale of the other caliper (C2C_{2}C2​) has 10 equal divisions that correspond to 11 main scale divisions. The readings of the two calipers are shown in the figure. The measured values (in cm) by calipers C1C_{1}C1​ and C2C_{2}C2​ respectively, are
  1. (A)2.87 and 2.86
  2. (B)2.87 and 2.87
  3. (C)2.87 and 2.83
  4. (D)2.85 and 2.82

Correct answer: (C)

Step-by-step solution →
Q5·PhysicsSingle correct
A gas is enclosed in a cylinder with a movable frictionless piston. Its initial thermodynamic state at pressure Pi=105P_{i} = 10^{5}Pi​=105 Pa and volume Vi=10−3 m3V_{i} = 10^{-3}\ \text{m}^{3}Vi​=10−3 m3 changes to a final state at Pf=(1/32)×105P_{f} = (1/32) \times 10^{5}Pf​=(1/32)×105 Pa and Vf=8×10−3 m3V_{f} = 8 \times 10^{-3}\ \text{m}^{3}Vf​=8×10−3 m3 in an adiabatic quasi-static process, such that P3V5P^{3}V^{5}P3V5 = constant. Consider another thermodynamic process that brings the system from the same initial state to the same final state in two steps: an isobaric expansion at PiP_{i}Pi​ followed by an isochoric (isovolumetric) process at volume VfV_{f}Vf​. The amount of heat supplied to the system in the two-step process is approximately
  1. (A)112 J
  2. (B)294 J
  3. (C)588 J
  4. (D)813 J

Correct answer: (C)

Step-by-step solution →
Q6·PhysicsSingle correct
A small object is placed 50 cm to the left of a thin convex lens of focal length 30 cm. A convex spherical mirror of radius of curvature 100 cm is placed to the right of the lens at a distance of 50 cm. The mirror is tilted such that the axis of the mirror is at an angle θ=300\theta = 30^{0}θ=300 to the axis of the lens, as shown in the figure. If the origin of the coordinate system is taken to be at the centre of the lens, the coordinates (in cm) of the point (x, y) at which the image is formed are
  1. (A)(25, 253)(25,\ 25\sqrt{3})(25, 253​)
  2. (B)(125/3, 25/3)(125/3,\ 25/\sqrt{3})(125/3, 25/3​)
  3. (C)(50−253, 25)(50 - 25\sqrt{3},\ 25)(50−253​, 25)
  4. (D)(0, 0)(0,\ 0)(0, 0)

Correct answer: (A)

Step-by-step solution →
Q7·PhysicsMultiple correct
While conducting the Young's double slit experiment, a student replaced the two slits with a large opaque plate in the x-y plane containing two small holes that act as two coherent point sources (S1S_{1}S1​, S2S_{2}S2​) emitting light of wavelength 600 nm. The student mistakenly placed the screen parallel to the x-z plane (for z > 0) at a distance D = 3 m from the mid-point of S1S2S_{1}S_{2}S1​S2​, as shown schematically in the figure. The distance between the sources d = 0.6003 mm. The origin O is at the intersection of the screen and the line joining S1S2S_{1}S_{2}S1​S2​. which of the following is(are) true of the intensity pattern on the screen?
  1. (A)Hyperbolic bright and dark bands with foci symmetrically placed about O in the x-direction
  2. (B)Semi circular bright and dark bands centred at point O
  3. (C)The region very close to the point O will be dark
  4. (D)Straight bright and dark bands parallel to the x-axis

Correct answer: (B), (C)

Step-by-step solution →
Q8·PhysicsMultiple correct
In an experiment to determine the acceleration due to gravity g, the formula used for the time period of a periodic motion is T=2π7(R−r)5gT = 2\pi\sqrt{\frac{7(R-r)}{5g}}T=2π5g7(R−r)​​. The values of R and r are measured to be (60 ± 1) mm and (10 ± 1) mm, respectively. In five successive measurements, the time period is found to be 0.52 s, 0.56 s, 0.57 s, 0.54 s and 0.59 s. The least count of the watch used for the measurement of time period is 0.01 s. Which of the following statement(s) is(are) true?
  1. (A)The error in the measurement of r is 10%
  2. (B)The error in the measurement of T is 3.57%
  3. (C)The error in the measurement of T is 2%
  4. (D)The error in the determined value of g is 11%

Correct answer: (A), (B), (D)

Step-by-step solution →
Q9·PhysicsMultiple correct
A rigid wire loop of square shape having side of length L and resistance R is moving along the x-axis with a constant velocity v0v_{0}v0​ in the plane of the paper. At t = 0, the right edge of the loop enters a region of length 3L where there is a uniform magnetic field B0B_{0}B0​ into the plane of the paper, as shown in the figure. For sufficiently large v0v_{0}v0​, the loop eventually crosses the region. Let x be the location of the right edge of the loop. Let v(x), I(x) and F(x) represent the velocity of the loop, current in the loop, and force on the loop, respectively, as a function of x. Counter-clockwise current is taken as positive. Which of the following schematic plot(s) is(are) correct? (Ignore gravity)
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (C), (D)

Step-by-step solution →
Q10·PhysicsMultiple correct
Light of wavelength λph\lambda_{ph}λph​ falls on a cathode plate inside a vacuum tube as shown in the figure. The work function of the cathode surface is ϕ\phiϕ and the anode is a wire mesh of conducting material kept at a distance d from the cathode. A potential difference V is maintained between the electrodes. If the minimum de Broglie wavelength of the electrons passing through the anode is λe\lambda_{e}λe​, which of the following statement(s) is(are) true?
  1. (A)For large potential difference (V >> ϕ/e\phi/eϕ/e), λe\lambda_{e}λe​ is approximately halved if V is made four times
  2. (B)λe\lambda_{e}λe​ increases at the same rate as λph\lambda_{ph}λph​ for λph<hc/ϕ\lambda_{ph} < hc/\phiλph​<hc/ϕ
  3. (C)λe\lambda_{e}λe​ is approximately halved, if d is doubled
  4. (D)λe\lambda_{e}λe​ decreases with increase in ϕ\phiϕ and λph\lambda_{ph}λph​

Correct answer: (A)

Step-by-step solution →
Q11·PhysicsMultiple correct
Two thin circular discs of mass m and 4m, having radii of a and 2a, respectively, are rigidly fixed by a massless, rigid rod of length ℓ=24 a\ell = \sqrt{24}\,aℓ=24​a through their centers. This assembly is laid on a firm and flat surface, and set rolling without slipping on the surface so that the angular speed about the axis of the rod is ω\omegaω. The angular momentum of the entire assembly about the point 'O' is L⃗\vec{L}L (see the figure). Which of the following statement(s) is(are) true?
  1. (A)The magnitude of angular momentum of the assembly about its center of mass is 17 ma2ω/217\,ma^{2}\omega/217ma2ω/2
  2. (B)The magnitude of the z-component of L⃗\vec{L}L is 55 ma2ω55\,ma^{2}\omega55ma2ω
  3. (C)The magnitude of angular momentum of center of mass of the assembly about the point O is 81 ma2ω81\,ma^{2}\omega81ma2ω
  4. (D)The center of mass of the assembly rotates about the z-axis with an angular speed of ω/5\omega/5ω/5

Correct answer: (D) or (A), (D)

Step-by-step solution →
Q12·PhysicsMultiple correct
Consider two identical galvanometers and two identical resistors with resistance R. If the internal resistance of the galvanometers RC<R/2R_{C} < R/2RC​<R/2, which of the following statement(s) about any one of the galvanometers is(are) true?
  1. (A)The maximum voltage range is obtained when all the components are connected in series
  2. (B)The maximum voltage range is obtained when the two resistors and one galvanometer are connected in series, and the second galvanometer is connected in parallel to the first galvanometer
  3. (C)The maximum current range is obtained when all the components are connected in parallel
  4. (D)The maximum current range is obtained when the two galvanometers are connected in series and the combination is connected in parallel with both the resistors

Correct answer: (B), (C)

Step-by-step solution →
Q13·PhysicsMultiple correct
In the circuit shown below, the key is pressed at time t = 0. Which of the following statement(s) is (are) true?
  1. (A)The voltmeter displays – 5 V as soon as the key is pressed, and displays +5 V after a long time
  2. (B)The voltmeter will display 0 V at time t=ln⁡2t = \ln 2t=ln2 seconds
  3. (C)The current in the ammeter becomes 1/e1/e1/e of the initial value after 1 second
  4. (D)The current in the ammeter becomes zero after a long time

Correct answer: (A), (B), (C), (D)

Step-by-step solution →
Q14·PhysicsMultiple correct
A block with mass M is connected by a massless spring with stiffness constant k to a rigid wall and moves without friction on a horizontal surface. The block oscillates with small amplitude A about an equilibrium position x0x_{0}x0​. Consider two cases: (i) when the block is at x0x_{0}x0​ ; and (ii) when the block is at x=x0+Ax = x_{0} + Ax=x0​+A. In both the cases, a particle with mass m (< M) is softly placed on the block after which they stick to each other. Which of the following statement(s) is (are) true about the motion after the mass m is placed on the mass M?
  1. (A)The amplitude of oscillation in the first case changes by a factor of Mm+M\sqrt{\frac{M}{m+M}}m+MM​​, whereas in the second case it remains unchanged
  2. (B)The final time period of oscillation in both the cases is same
  3. (C)The total energy decreases in both the cases
  4. (D)The instantaneous speed at x0x_{0}x0​ of the combined masses decreases in both the cases

Correct answer: (A), (B), (D)

Step-by-step solution →
Q15·PhysicsSingle correct
A frame of reference that is accelerated with respect to an inertial frame of reference is called a non-inertial frame of reference. A coordinate system fixed on a circular disc rotating about a fixed axis with a constant angular velocity ω\omegaω is an example of a non-inertial frame of reference. The relationship between the force F⃗rot\vec{F}_{rot}Frot​ experienced by a particle of mass m moving on the rotating disc and the force F⃗in\vec{F}_{in}Fin​ experienced by the particle in an inertial frame of reference is F⃗rot=F⃗in+2m(v⃗rot×ω⃗)+m(ω⃗×r⃗)×ω⃗,\vec{F}_{rot} = \vec{F}_{in} + 2m\left(\vec{v}_{rot} \times \vec{\omega}\right) + m\left(\vec{\omega} \times \vec{r}\right) \times \vec{\omega},Frot​=Fin​+2m(vrot​×ω)+m(ω×r)×ω, where v⃗rot\vec{v}_{rot}vrot​ is the velocity of the particle in the rotating frame of reference and r⃗\vec{r}r is the position vector of the particle with respect to the centre of the disc. Now consider a smooth slot along a diameter of a disc of radius R rotating counter-clockwise with a constant angular speed ω\omegaω about its vertical axis through its center. We assign a coordinate system with the origin at the center of the disc, the x-axis along the slot, the y-axis perpendicular to the slot and the z-axis along the rotation axis (ω⃗=ωk^)\left(\vec{\omega} = \omega\hat{k}\right)(ω=ωk^). A small block of mass m is gently placed in the slot at r⃗=(R/2)i^\vec{r} = (R/2)\hat{i}r=(R/2)i^ at t = 0 and is constrained to move only along the slot. The distance r of the block at time t is
  1. (A)R4(e2ωt+e−2ωt)\frac{R}{4}\left(e^{2\omega t} + e^{-2\omega t}\right)4R​(e2ωt+e−2ωt)
  2. (B)R2cos⁡2ωt\frac{R}{2}\cos 2\omega t2R​cos2ωt
  3. (C)R2cos⁡ωt\frac{R}{2}\cos \omega t2R​cosωt
  4. (D)R4(eωt+e−ωt)\frac{R}{4}\left(e^{\omega t} + e^{-\omega t}\right)4R​(eωt+e−ωt)

Correct answer: (D)

Step-by-step solution →
Q16·PhysicsSingle correct
A frame of reference that is accelerated with respect to an inertial frame of reference is called a non-inertial frame of reference. A coordinate system fixed on a circular disc rotating about a fixed axis with a constant angular velocity ω\omegaω is an example of a non-inertial frame of reference. The relationship between the force F⃗rot\vec{F}_{rot}Frot​ experienced by a particle of mass m moving on the rotating disc and the force F⃗in\vec{F}_{in}Fin​ experienced by the particle in an inertial frame of reference is F⃗rot=F⃗in+2m(v⃗rot×ω⃗)+m(ω⃗×r⃗)×ω⃗,\vec{F}_{rot} = \vec{F}_{in} + 2m\left(\vec{v}_{rot} \times \vec{\omega}\right) + m\left(\vec{\omega} \times \vec{r}\right) \times \vec{\omega},Frot​=Fin​+2m(vrot​×ω)+m(ω×r)×ω, where v⃗rot\vec{v}_{rot}vrot​ is the velocity of the particle in the rotating frame of reference and r⃗\vec{r}r is the position vector of the particle with respect to the centre of the disc. Now consider a smooth slot along a diameter of a disc of radius R rotating counter-clockwise with a constant angular speed ω\omegaω about its vertical axis through its center. We assign a coordinate system with the origin at the center of the disc, the x-axis along the slot, the y-axis perpendicular to the slot and the z-axis along the rotation axis (ω⃗=ωk^)\left(\vec{\omega} = \omega\hat{k}\right)(ω=ωk^). A small block of mass m is gently placed in the slot at r⃗=(R/2)i^\vec{r} = (R/2)\hat{i}r=(R/2)i^ at t = 0 and is constrained to move only along the slot. The net reaction of the disc on the block is
  1. (A)−mω2Rcos⁡ωt j^−mgk^-m\omega^{2}R\cos \omega t\,\hat{j} - mg\hat{k}−mω2Rcosωtj^​−mgk^
  2. (B)mω2Rsin⁡ωt j^−mgk^m\omega^{2}R\sin \omega t\,\hat{j} - mg\hat{k}mω2Rsinωtj^​−mgk^
  3. (C)12mω2R(eωt−e−ωt)j^+mgk^\frac{1}{2}m\omega^{2}R\left(e^{\omega t} - e^{-\omega t}\right)\hat{j} + mg\hat{k}21​mω2R(eωt−e−ωt)j^​+mgk^
  4. (D)12mω2R(e2ωt−e−2ωt)j^+mgk^\frac{1}{2}m\omega^{2}R\left(e^{2\omega t} - e^{-2\omega t}\right)\hat{j} + mg\hat{k}21​mω2R(e2ωt−e−2ωt)j^​+mgk^

Correct answer: (C)

Step-by-step solution →
Q17·PhysicsSingle correct
Consider an evacuated cylindrical chamber of height h having rigid conducting plates at the ends and an insulating curved surface as shown in the figure. A number of spherical balls made of a light weight and soft material and coated with a conducting material are placed on the bottom plate. The balls have a radius r << h. Now a high voltage source (HV) is connected across the conducting plates such that the bottom plate is at +V0+V_{0}+V0​ and the top plate at −V0-V_{0}−V0​. Due to their conducting surface, the balls will get charged, will become equipotential with the plate and are repelled by it. The balls will eventually collide with the top plate, where the coefficient of restitution can be taken to be zero due to the soft nature of the material of the balls. The electric field in the chamber can be considered to be that of a parallel plate capacitor. Assume that there are no collisions between the balls and the interaction between them is negligible. (Ignore gravity) Which one of the following statements is correct?
  1. (A)The balls will bounce back to the bottom plate carrying the opposite charge they went up with
  2. (B)The balls will execute simple harmonic motion between the two plates
  3. (C)The balls will bounce back to the bottom plate carrying the same charge they went up with
  4. (D)The balls will stick to the top plate and remain there

Correct answer: (A)

Step-by-step solution →
Q18·PhysicsSingle correct
Consider an evacuated cylindrical chamber of height h having rigid conducting plates at the ends and an insulating curved surface as shown in the figure. A number of spherical balls made of a light weight and soft material and coated with a conducting material are placed on the bottom plate. The balls have a radius r << h. Now a high voltage source (HV) is connected across the conducting plates such that the bottom plate is at +V0+V_{0}+V0​ and the top plate at −V0-V_{0}−V0​. Due to their conducting surface, the balls will get charged, will become equipotential with the plate and are repelled by it. The balls will eventually collide with the top plate, where the coefficient of restitution can be taken to be zero due to the soft nature of the material of the balls. The electric field in the chamber can be considered to be that of a parallel plate capacitor. Assume that there are no collisions between the balls and the interaction between them is negligible. (Ignore gravity) The average current in the steady state registered by the ammeter in the circuit will be
  1. (A)proportional to V01/2V_{0}^{1/2}V01/2​
  2. (B)proportional to V02V_{0}^{2}V02​
  3. (C)proportional to the potential V0V_{0}V0​
  4. (D)zero

Correct answer: (B)

Step-by-step solution →

Chemistry — JEE Advanced 2016 Paper 2

Q19·ChemistrySingle correct
The correct order of acidity for the following compounds is
  1. (A)I > II > III > IV
  2. (B)III > I > II > IV
  3. (C)III > IV > II > I
  4. (D)I > III > IV > II

Correct answer: (A)

Step-by-step solution →
Q20·ChemistrySingle correct
The geometries of the ammonia complexes of Ni2+Ni^{2+}Ni2+, Pt2+Pt^{2+}Pt2+ and Zn2+Zn^{2+}Zn2+, respectively, are
  1. (A)octahedral, square planar and tetrahedral
  2. (B)square planar, octahedral and tetrahedral
  3. (C)tetrahedral, square planar and octahedral
  4. (D)octahedral, tetrahedral and square planar

Correct answer: (A)

Step-by-step solution →
Q21·ChemistrySingle correct
For the following electrochemical cell at 298 K, Pt(s) ∣ H2(g,1 bar) ∣ H+(aq,1 M) ∣∣ M4+(aq),M2+(aq) ∣ Pt(s)Pt(s)\,|\,H_{2}(g, 1\ bar)\,|\,H^{+}(aq, 1\ M)\,||\,M^{4+}(aq), M^{2+}(aq)\,|\,Pt(s)Pt(s)∣H2​(g,1 bar)∣H+(aq,1 M)∣∣M4+(aq),M2+(aq)∣Pt(s) Ecell=0.092E_{cell} = 0.092Ecell​=0.092 V when [M2+(aq)][M4+(aq)]=10x\dfrac{[M^{2+}(aq)]}{[M^{4+}(aq)]} = 10^{x}[M4+(aq)][M2+(aq)]​=10x Given: EM4+/M2+0=0.151E^{0}_{M^{4+}/M^{2+}} = 0.151EM4+/M2+0​=0.151 V; 2.303RTF=0.0592.303\dfrac{RT}{F} = 0.0592.303FRT​=0.059 V The value of xxx is
  1. (A)−2-2−2
  2. (B)−1-1−1
  3. (C)1
  4. (D)2

Correct answer: (D)

Step-by-step solution →
Q22·ChemistrySingle correct
The major product of the following reaction sequence is
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (A)

Step-by-step solution →
Q23·ChemistrySingle correct
In the following reaction sequence in aqueous solution, the species X, Y and Z, respectively, are S2O32−→Ag+Xclear solution→Ag+Ywhite precipitate→with timeZblack precipitateS_{2}O_{3}^{2-} \xrightarrow{Ag^{+}} \underset{\text{clear solution}}{X} \xrightarrow{Ag^{+}} \underset{\text{white precipitate}}{Y} \xrightarrow{\text{with time}} \underset{\text{black precipitate}}{Z}S2​O32−​Ag+​clear solutionX​Ag+​white precipitateY​with time​black precipitateZ​
  1. (A)[Ag(S2O3)2]3−[Ag(S_{2}O_{3})_{2}]^{3-}[Ag(S2​O3​)2​]3−, Ag2S2O3Ag_{2}S_{2}O_{3}Ag2​S2​O3​, Ag2SAg_{2}SAg2​S
  2. (B)[Ag(S2O3)3]5−[Ag(S_{2}O_{3})_{3}]^{5-}[Ag(S2​O3​)3​]5−, Ag2SO3Ag_{2}SO_{3}Ag2​SO3​, Ag2SAg_{2}SAg2​S
  3. (C)[Ag(SO3)2]3−[Ag(SO_{3})_{2}]^{3-}[Ag(SO3​)2​]3−, Ag2S2O3Ag_{2}S_{2}O_{3}Ag2​S2​O3​, AgAgAg
  4. (D)[Ag(SO3)3]3−[Ag(SO_{3})_{3}]^{3-}[Ag(SO3​)3​]3−, Ag2SO4Ag_{2}SO_{4}Ag2​SO4​, AgAgAg

Correct answer: (A)

Step-by-step solution →
Q24·ChemistrySingle correct
The qualitative sketches I, II and III given below show the variation of surface tension with molar concentration of three different aqueous solutions of KCl, CH3OHCH_{3}OHCH3​OH and CH3(CH2)11OSO3−Na+CH_{3}(CH_{2})_{11}OSO_{3}^{-}Na^{+}CH3​(CH2​)11​OSO3−​Na+ at room temperature. The correct assignment of the sketches is
  1. (A)I : KCl; II : CH3OHCH_{3}OHCH3​OH; III : CH3(CH2)11OSO3−Na+CH_{3}(CH_{2})_{11}OSO_{3}^{-}Na^{+}CH3​(CH2​)11​OSO3−​Na+
  2. (B)I : CH3(CH2)11OSO3−Na+CH_{3}(CH_{2})_{11}OSO_{3}^{-}Na^{+}CH3​(CH2​)11​OSO3−​Na+; II : CH3OHCH_{3}OHCH3​OH; III : KCl
  3. (C)I : KCl; II : CH3(CH2)11OSO3−Na+CH_{3}(CH_{2})_{11}OSO_{3}^{-}Na^{+}CH3​(CH2​)11​OSO3−​Na+; III : CH3OHCH_{3}OHCH3​OH
  4. (D)I : CH3OHCH_{3}OHCH3​OH; II : KCl; III : CH3(CH2)11OSO3−Na+CH_{3}(CH_{2})_{11}OSO_{3}^{-}Na^{+}CH3​(CH2​)11​OSO3−​Na+

Correct answer: (D)

Step-by-step solution →
Q25·ChemistryMultiple correct
For ‘invert sugar’, the correct statement(s) is(are) (Given: specific rotations of (+)-sucrose, (+)-maltose, L-(–)-glucose and L-(+)-fructose in aqueous solution are +66°, +140°, –52° and +92°, respectively)
  1. (A)‘invert sugar’ is prepared by acid catalyzed hydrolysis of maltose
  2. (B)‘invert sugar’ is an equimolar mixture of D-(+)-glucose and D-(–)-fructose
  3. (C)specific rotation of ‘invert sugar’ is –20°
  4. (D)on reaction with Br2Br_{2}Br2​ water, ‘invert sugar’ forms saccharic acid as one of the products

Correct answer: (B), (C)

Step-by-step solution →
Q26·ChemistryMultiple correct
Among the following, reaction(s) which gives(give) tert-butyl benzene as the major product is(are)
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (B), (C), (D)

Step-by-step solution →
Q27·ChemistryMultiple correct
Extraction of copper from copper pyrite (CuFeS2CuFeS_{2}CuFeS2​) involves
  1. (A)crushing followed by concentration of the ore by froth flotation
  2. (B)removal of iron as slag
  3. (C)self-reduction step to produce ‘blister copper’ following evolution of SO2SO_{2}SO2​
  4. (D)refining of ‘blister copper’ by carbon reduction

Correct answer: (A), (B), (C)

Step-by-step solution →
Q28·ChemistryMultiple correct
The CORRECT statement(s) for cubic close packed (ccp) three dimensional structure is(are)
  1. (A)The number of the nearest neighbours of an atom present in the topmost layer is 12
  2. (B)The efficiency of atom packing is 74%
  3. (C)The number of octahedral and tetrahedral voids per atom are 1 and 2, respectively
  4. (D)The unit cell edge length is 222\sqrt{2}22​ times the radius of the atom

Correct answer: (B), (C), (D)

Step-by-step solution →
Q29·ChemistryMultiple correct
Reagent(s) which can be used to bring about the following transformation is(are)
  1. (A)LiAlH4LiAlH_{4}LiAlH4​ in (C2H5)2O(C_{2}H_{5})_{2}O(C2​H5​)2​O
  2. (B)BH3BH_{3}BH3​ in THF
  3. (C)NaBH4NaBH_{4}NaBH4​ in C2H5OHC_{2}H_{5}OHC2​H5​OH
  4. (D)Raney Ni/H2H_{2}H2​ in THF

Correct answer: (C), (D)

Step-by-step solution →
Q30·ChemistryMultiple correct
Mixture (s) showing positive deviation from Raoult’s law at 35°C is (are)
  1. (A)carbon tetrachloride + methanol
  2. (B)carbon disulphide + acetone
  3. (C)benzene + toluene
  4. (D)phenol + aniline

Correct answer: (A), (B)

Step-by-step solution →
Q31·ChemistryMultiple correct
The nitrogen containing compound produced in the reaction of HNO3HNO_{3}HNO3​ with P4O10P_{4}O_{10}P4​O10​
  1. (A)can also be prepared by reaction of P4P_{4}P4​ and HNO3HNO_{3}HNO3​
  2. (B)is diamagnetic
  3. (C)contains one N – N bond
  4. (D)reacts with Na metal producing a brown gas

Correct answer: (B), (D)

Step-by-step solution →
Q32·ChemistryMultiple correct
According to Molecular Orbital Theory
  1. (A)C22−C_{2}^{2-}C22−​ is expected to be diamagnetic
  2. (B)O22+O_{2}^{2+}O22+​ expected to have a longer bond length than O2O_{2}O2​
  3. (C)N2+N_{2}^{+}N2+​ and N2−N_{2}^{-}N2−​ have the same bond order
  4. (D)He2+He_{2}^{+}He2+​ has the same energy as two isolated He atoms

Correct answer: (A), (C)

Step-by-step solution →
Q33·ChemistrySingle correct
Thermal decomposition of gaseous X2X_{2}X2​ to gaseous X at 298 K takes place according to the following equation: X2(g)⇌2X(g)X_{2}(g) \rightleftharpoons 2X(g)X2​(g)⇌2X(g) The standard reaction Gibbs energy, ΔrG0\Delta_{r}G^{0}Δr​G0, of this reaction is positive. At the start of the reaction, there is one mole of X2X_{2}X2​ and no X. As the reaction proceeds, the number of moles of X formed is given by β. Thus, βequilibrium\beta_{equilibrium}βequilibrium​ is the number of moles of X formed at equilibrium. The reaction is carried out at a constant total pressure of 2 bar. Consider the gases to behave ideally. (Given: R = 0.083 L bar K−1K^{-1}K−1 mol−1mol^{-1}mol−1) The equilibrium constant KpK_{p}Kp​ for this reaction at 298 K, in terms of βequilibrium\beta_{equilibrium}βequilibrium​, is
  1. (A)8βequilibrium22−βequilibrium\dfrac{8\beta_{equilibrium}^{2}}{2-\beta_{equilibrium}}2−βequilibrium​8βequilibrium2​​
  2. (B)8βequilibrium24−βequilibrium2\dfrac{8\beta_{equilibrium}^{2}}{4-\beta_{equilibrium}^{2}}4−βequilibrium2​8βequilibrium2​​
  3. (C)4βequilibrium22−βequilibrium\dfrac{4\beta_{equilibrium}^{2}}{2-\beta_{equilibrium}}2−βequilibrium​4βequilibrium2​​
  4. (D)4βequilibrium24−βequilibrium2\dfrac{4\beta_{equilibrium}^{2}}{4-\beta_{equilibrium}^{2}}4−βequilibrium2​4βequilibrium2​​

Correct answer: (B)

Step-by-step solution →
Q34·ChemistrySingle correct
Thermal decomposition of gaseous X2X_{2}X2​ to gaseous X at 298 K takes place according to the following equation: X2(g)⇌2X(g)X_{2}(g) \rightleftharpoons 2X(g)X2​(g)⇌2X(g) The standard reaction Gibbs energy, ΔrG0\Delta_{r}G^{0}Δr​G0, of this reaction is positive. At the start of the reaction, there is one mole of X2X_{2}X2​ and no X. As the reaction proceeds, the number of moles of X formed is given by β. Thus, βequilibrium\beta_{equilibrium}βequilibrium​ is the number of moles of X formed at equilibrium. The reaction is carried out at a constant total pressure of 2 bar. Consider the gases to behave ideally. (Given: R = 0.083 L bar K−1K^{-1}K−1 mol−1mol^{-1}mol−1) The INCORRECT statement among the following, for this reaction is
  1. (A)Decrease in the total pressure will result in formation of more moles of gaseous X
  2. (B)At the start of the reaction, dissociation of gaseous X2X_{2}X2​ takes place spontaneously
  3. (C)βequilibrium=0.7\beta_{equilibrium} = 0.7βequilibrium​=0.7
  4. (D)Kc<1K_{c} < 1Kc​<1

Correct answer: (C)

Step-by-step solution →
Q35·ChemistrySingle correct
Treatment of compound O with KMnO4/H+KMnO_{4}/H^{+}KMnO4​/H+ gave P, which on heating with ammonia gave Q. The compound Q on treatment with Br2Br_{2}Br2​/NaOH produced R. On strong heating, Q gave S, which on further treatment with ethyl 2-bromopropanoate in the presence of KOH followed by acidification, gave a compound T. The compound R is
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (A)

Step-by-step solution →
Q36·ChemistrySingle correct
Treatment of compound O with KMnO4/H+KMnO_{4}/H^{+}KMnO4​/H+ gave P, which on heating with ammonia gave Q. The compound Q on treatment with Br2Br_{2}Br2​/NaOH produced R. On strong heating, Q gave S, which on further treatment with ethyl 2-bromopropanoate in the presence of KOH followed by acidification, gave a compound T. The compound T is
  1. (A)glycine
  2. (B)alanine
  3. (C)valine
  4. (D)serine

Correct answer: (B)

Step-by-step solution →

Mathematics — JEE Advanced 2016 Paper 2

Q37·MathematicsSingle correct
Let P=[1004101641]P = \begin{bmatrix} 1 & 0 & 0 \\ 4 & 1 & 0 \\ 16 & 4 & 1 \end{bmatrix}P=​1416​014​001​​ and I be the identity matrix of order 3. If Q=[qij]Q = \left[ q_{ij} \right]Q=[qij​] is a matrix such that P50−Q=IP^{50} - Q = IP50−Q=I, then q31+q32q21\frac{q_{31} + q_{32}}{q_{21}}q21​q31​+q32​​ equals
  1. (A)52
  2. (B)103
  3. (C)201
  4. (D)205

Correct answer: (B)

Step-by-step solution →
Q38·MathematicsSingle correct
Area of the region {(x,y)∈R2:y≥∣x+3∣,5y≤x+9≤15}\left\{ (x, y) \in \mathbb{R}^{2} : y \geq \sqrt{|x + 3|}, 5y \leq x + 9 \leq 15 \right\}{(x,y)∈R2:y≥∣x+3∣​,5y≤x+9≤15} is equal to
  1. (A)16\frac{1}{6}61​
  2. (B)43\frac{4}{3}34​
  3. (C)32\frac{3}{2}23​
  4. (D)53\frac{5}{3}35​

Correct answer: (C)

Step-by-step solution →
Q39·MathematicsSingle correct
The value of ∑k=1131sin⁡(π4+(k−1)π6)sin⁡(π4+kπ6)\sum_{k=1}^{13} \frac{1}{\sin\left( \frac{\pi}{4} + \frac{(k-1)\pi}{6} \right) \sin\left( \frac{\pi}{4} + \frac{k\pi}{6} \right)}∑k=113​sin(4π​+6(k−1)π​)sin(4π​+6kπ​)1​ is equal to
  1. (A)3−33 - \sqrt{3}3−3​
  2. (B)2(3−3)2\left( 3 - \sqrt{3} \right)2(3−3​)
  3. (C)2(3−1)2\left( \sqrt{3} - 1 \right)2(3​−1)
  4. (D)2(2+3)2\left( 2 + \sqrt{3} \right)2(2+3​)

Correct answer: (C)

Step-by-step solution →
Q40·MathematicsSingle correct
Let bi>1b_{i} > 1bi​>1 for i=1,2,…,101i = 1, 2, \ldots, 101i=1,2,…,101. Suppose log⁡eb1,log⁡eb2,…,log⁡eb101\log_{e} b_{1}, \log_{e} b_{2}, \ldots, \log_{e} b_{101}loge​b1​,loge​b2​,…,loge​b101​ are in Arithmetic Progression (A. P.) with the common difference log⁡e2\log_{e} 2loge​2. Suppose a1,a2,…,a101a_{1}, a_{2}, \ldots, a_{101}a1​,a2​,…,a101​ are in A.P. such that a1=b1a_{1} = b_{1}a1​=b1​ and a51=b51a_{51} = b_{51}a51​=b51​. If t=b1+b2+…+b51t = b_{1} + b_{2} + \ldots + b_{51}t=b1​+b2​+…+b51​ and s=a1+a2+…+a51s = a_{1} + a_{2} + \ldots + a_{51}s=a1​+a2​+…+a51​, then
  1. (A)s>ts > ts>t and a101>b101a_{101} > b_{101}a101​>b101​
  2. (B)s>ts > ts>t and a101<b101a_{101} < b_{101}a101​<b101​
  3. (C)s<ts < ts<t and a101>b101a_{101} > b_{101}a101​>b101​
  4. (D)s<ts < ts<t and a101<b101a_{101} < b_{101}a101​<b101​

Correct answer: (B)

Step-by-step solution →
Q41·MathematicsSingle correct
The value of ∫−π2π2x2cos⁡x1+ex dx\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{x^{2} \cos x}{1 + e^{x}} \, dx∫−2π​2π​​1+exx2cosx​dx is equal to
  1. (A)π24−2\frac{\pi^{2}}{4} - 24π2​−2
  2. (B)π24+2\frac{\pi^{2}}{4} + 24π2​+2
  3. (C)π2−eπ2\pi^{2} - e^{\frac{\pi}{2}}π2−e2π​
  4. (D)π2+eπ2\pi^{2} + e^{\frac{\pi}{2}}π2+e2π​

Correct answer: (A)

Step-by-step solution →
Q42·MathematicsSingle correct
Let P be the image of the point (3, 1, 7) with respect to the plane x−y+z=3x - y + z = 3x−y+z=3. Then the equation of the plane passing through P and containing the straight line x1=y2=z1\frac{x}{1} = \frac{y}{2} = \frac{z}{1}1x​=2y​=1z​ is
  1. (A)x+y−3z=0x + y - 3z = 0x+y−3z=0
  2. (B)3x+z=03x + z = 03x+z=0
  3. (C)x−4y+7z=0x - 4y + 7z = 0x−4y+7z=0
  4. (D)2x−y=02x - y = 02x−y=0

Correct answer: (C)

Step-by-step solution →
Q43·MathematicsMultiple correct
Let a,b∈Ra, b \in \mathbb{R}a,b∈R and f:R→Rf : \mathbb{R} \to \mathbb{R}f:R→R be defined by f(x)=acos⁡(∣x3−x∣)+b∣x∣sin⁡(∣x3+x∣)f(x) = a \cos(|x^{3} - x|) + b|x| \sin(|x^{3} + x|)f(x)=acos(∣x3−x∣)+b∣x∣sin(∣x3+x∣). Then f is
  1. (A)differentiable at x=0x = 0x=0 if a=0a = 0a=0 and b=1b = 1b=1
  2. (B)differentiable at x=1x = 1x=1 if a=1a = 1a=1 and b=0b = 0b=0
  3. (C)NOT differentiable at x=0x = 0x=0 if a=1a = 1a=1 and b=0b = 0b=0
  4. (D)NOT differentiable at x=1x = 1x=1 if a=1a = 1a=1 and b=1b = 1b=1

Correct answer: (A), (B)

Step-by-step solution →
Q44·MathematicsMultiple correct
Let f(x)=lim⁡n→∞(nn(x+n)(x+n2)…(x+nn)n!(x2+n2)(x2+n24)…(x2+n2n2))xnf(x) = \lim_{n \to \infty} \left( \frac{n^{n} (x + n) \left( x + \frac{n}{2} \right) \ldots \left( x + \frac{n}{n} \right)}{n! \left( x^{2} + n^{2} \right) \left( x^{2} + \frac{n^{2}}{4} \right) \ldots \left( x^{2} + \frac{n^{2}}{n^{2}} \right)} \right)^{\frac{x}{n}}f(x)=limn→∞​(n!(x2+n2)(x2+4n2​)…(x2+n2n2​)nn(x+n)(x+2n​)…(x+nn​)​)nx​, for all x>0x > 0x>0. Then
  1. (A)f(12)≥f(1)f\left( \frac{1}{2} \right) \geq f(1)f(21​)≥f(1)
  2. (B)f(13)≤f(23)f\left( \frac{1}{3} \right) \leq f\left( \frac{2}{3} \right)f(31​)≤f(32​)
  3. (C)f′(2)≤0f'(2) \leq 0f′(2)≤0
  4. (D)f′(3)f(3)≥f′(2)f(2)\frac{f'(3)}{f(3)} \geq \frac{f'(2)}{f(2)}f(3)f′(3)​≥f(2)f′(2)​

Correct answer: (B), (C)

Step-by-step solution →
Q45·MathematicsMultiple correct
Let f:R→(0,∞)f : \mathbb{R} \to (0, \infty)f:R→(0,∞) and g:R→Rg : \mathbb{R} \to \mathbb{R}g:R→R be twice differentiable functions such that f′′f''f′′ and g′′g''g′′ are continuous functions on R\mathbb{R}R. Suppose f′(2)=g(2)=0f'(2) = g(2) = 0f′(2)=g(2)=0, f′′(2)≠0f''(2) \neq 0f′′(2)=0 and g′(2)≠0g'(2) \neq 0g′(2)=0. If lim⁡x→2f(x)g(x)f′(x)g′(x)=1\lim_{x \to 2} \frac{f(x) g(x)}{f'(x) g'(x)} = 1limx→2​f′(x)g′(x)f(x)g(x)​=1, then
  1. (A)f has a local minimum at x=2x = 2x=2
  2. (B)f has a local maximum at x=2x = 2x=2
  3. (C)f′′(2)>f(2)f''(2) > f(2)f′′(2)>f(2)
  4. (D)f(x)−f′′(x)=0f(x) - f''(x) = 0f(x)−f′′(x)=0 for at least one x∈Rx \in \mathbb{R}x∈R

Correct answer: (A), (D)

Step-by-step solution →
Q46·MathematicsMultiple correct
Let u^=u1i^+u2j^+u3k^\hat{u} = u_{1}\hat{i} + u_{2}\hat{j} + u_{3}\hat{k}u^=u1​i^+u2​j^​+u3​k^ be a unit vector in R3\mathbb{R}^{3}R3 and w^=16(i^+j^+2k^)\hat{w} = \frac{1}{\sqrt{6}} \left( \hat{i} + \hat{j} + 2\hat{k} \right)w^=6​1​(i^+j^​+2k^). Given that there exists a vector v⃗\vec{v}v in R3\mathbb{R}^{3}R3 such that ∣u^×v⃗∣=1|\hat{u} \times \vec{v}| = 1∣u^×v∣=1 and w^⋅(u^×v⃗)=1\hat{w} \cdot (\hat{u} \times \vec{v}) = 1w^⋅(u^×v)=1. Which of the following statement(s) is(are) correct?
  1. (A)There is exactly one choice for such v⃗\vec{v}v
  2. (B)There are infinitely many choices for such v⃗\vec{v}v
  3. (C)If u^\hat{u}u^ lies in the xy-plane then ∣u1∣=∣u2∣|u_{1}| = |u_{2}|∣u1​∣=∣u2​∣
  4. (D)If u^\hat{u}u^ lies in the xz-plane then 2∣u1∣=∣u3∣2|u_{1}| = |u_{3}|2∣u1​∣=∣u3​∣

Correct answer: (B), (C)

Step-by-step solution →
Q47·MathematicsMultiple correct
Let P be the point on the parabola y2=4xy^{2} = 4xy2=4x which is at the shortest distance from the center S of the circle x2+y2−4x−16y+64=0x^{2} + y^{2} - 4x - 16y + 64 = 0x2+y2−4x−16y+64=0. Let Q be the point on the circle dividing the line segment SP internally. Then
  1. (A)SP=25SP = 2\sqrt{5}SP=25​
  2. (B)SQ:QP=(5+1):2SQ : QP = \left( \sqrt{5} + 1 \right) : 2SQ:QP=(5​+1):2
  3. (C)the x-intercept of the normal to the parabola at P is 6
  4. (D)the slope of the tangent to the circle at Q is 12\frac{1}{2}21​

Correct answer: (A), (C), (D)

Step-by-step solution →
Q48·MathematicsMultiple correct
Let a,b∈Ra, b \in \mathbb{R}a,b∈R and a2+b2≠0a^{2} + b^{2} \neq 0a2+b2=0. Suppose S={z∈C:z=1a+ibt,t∈R,t≠0}S = \left\{ z \in \mathbb{C} : z = \frac{1}{a + ibt}, t \in \mathbb{R}, t \neq 0 \right\}S={z∈C:z=a+ibt1​,t∈R,t=0}, where i=−1i = \sqrt{-1}i=−1​. If z=x+iyz = x + iyz=x+iy and z∈Sz \in Sz∈S, then (x, y) lies on
  1. (A)the circle with radius 12a\frac{1}{2a}2a1​ and centre (12a,0)\left( \frac{1}{2a}, 0 \right)(2a1​,0) for a>0,b≠0a > 0, b \neq 0a>0,b=0
  2. (B)the circle with radius −12a-\frac{1}{2a}−2a1​ and centre (−12a,0)\left( -\frac{1}{2a}, 0 \right)(−2a1​,0) for a<0,b≠0a < 0, b \neq 0a<0,b=0
  3. (C)the x-axis for a≠0,b=0a \neq 0, b = 0a=0,b=0
  4. (D)the y-axis for a=0,b≠0a = 0, b \neq 0a=0,b=0

Correct answer: (A), (C), (D)

Step-by-step solution →
Q49·MathematicsMultiple correct
Let a,λ,μ∈Ra, \lambda, \mu \in \mathbb{R}a,λ,μ∈R. Consider the system of linear equations ax+2y=λax + 2y = \lambdaax+2y=λ 3x−2y=μ3x - 2y = \mu3x−2y=μ Which of the following statement(s) is(are) correct?
  1. (A)If a=−3a = -3a=−3, then the system has infinitely many solutions for all values of λ\lambdaλ and μ\muμ
  2. (B)If a≠−3a \neq -3a=−3, then the system has a unique solution for all values of λ\lambdaλ and μ\muμ
  3. (C)If λ+μ=0\lambda + \mu = 0λ+μ=0, then the system has infinitely many solutions for a=−3a = -3a=−3
  4. (D)If λ+μ≠0\lambda + \mu \neq 0λ+μ=0, then the system has no solution for a=−3a = -3a=−3

Correct answer: (B), (C), (D)

Step-by-step solution →
Q50·MathematicsMultiple correct
Let f:[−12,2]→Rf : \left[ -\frac{1}{2}, 2 \right] \to \mathbb{R}f:[−21​,2]→R and g:[−12,2]→Rg : \left[ -\frac{1}{2}, 2 \right] \to \mathbb{R}g:[−21​,2]→R be functions defined by f(x)=[x2−3]f(x) = [x^{2} - 3]f(x)=[x2−3] and g(x)=∣x∣f(x)+∣4x−7∣f(x)g(x) = |x| f(x) + |4x - 7| f(x)g(x)=∣x∣f(x)+∣4x−7∣f(x), where [y][y][y] denotes the greatest integer less than or equal to y for y∈Ry \in \mathbb{R}y∈R. Then
  1. (A)f is discontinuous exactly at three points in [−12,2]\left[ -\frac{1}{2}, 2 \right][−21​,2]
  2. (B)f is discontinuous exactly at four points in [−12,2]\left[ -\frac{1}{2}, 2 \right][−21​,2]
  3. (C)g is NOT differentiable exactly at four points in (−12,2)\left( -\frac{1}{2}, 2 \right)(−21​,2)
  4. (D)g is NOT differentiable exactly at five points in (−12,2)\left( -\frac{1}{2}, 2 \right)(−21​,2)

Correct answer: (B), (C)

Step-by-step solution →
Q51·MathematicsSingle correct
Football teams T1T_{1}T1​ and T2T_{2}T2​ have to play two games against each other. It is assumed that the outcomes of the two games are independent. The probabilities of T1T_{1}T1​ winning, drawing and losing a game against T2T_{2}T2​ are 12\frac{1}{2}21​, 16\frac{1}{6}61​ and 13\frac{1}{3}31​, respectively. Each team gets 3 points for a win, 1 point for a draw and 0 point for a loss in a game. Let X and Y denote the total points scored by teams T1T_{1}T1​ and T2T_{2}T2​, respectively, after two games. P(X>Y)P(X > Y)P(X>Y) is
  1. (A)14\frac{1}{4}41​
  2. (B)512\frac{5}{12}125​
  3. (C)12\frac{1}{2}21​
  4. (D)712\frac{7}{12}127​

Correct answer: (B)

Step-by-step solution →
Q52·MathematicsSingle correct
Football teams T1T_{1}T1​ and T2T_{2}T2​ have to play two games against each other. It is assumed that the outcomes of the two games are independent. The probabilities of T1T_{1}T1​ winning, drawing and losing a game against T2T_{2}T2​ are 12\frac{1}{2}21​, 16\frac{1}{6}61​ and 13\frac{1}{3}31​, respectively. Each team gets 3 points for a win, 1 point for a draw and 0 point for a loss in a game. Let X and Y denote the total points scored by teams T1T_{1}T1​ and T2T_{2}T2​, respectively, after two games. P(X=Y)P(X = Y)P(X=Y) is
  1. (A)1136\frac{11}{36}3611​
  2. (B)13\frac{1}{3}31​
  3. (C)1336\frac{13}{36}3613​
  4. (D)12\frac{1}{2}21​

Correct answer: (C)

Step-by-step solution →
Q53·MathematicsSingle correct
Let F1(x1,0)F_{1}(x_{1}, 0)F1​(x1​,0) and F2(x2,0)F_{2}(x_{2}, 0)F2​(x2​,0), for x1<0x_{1} < 0x1​<0 and x2>0x_{2} > 0x2​>0, be the foci of the ellipse x29+y28=1\frac{x^{2}}{9} + \frac{y^{2}}{8} = 19x2​+8y2​=1. Suppose a parabola having vertex at the origin and focus at F2F_{2}F2​ intersects the ellipse at point M in the first quadrant and at point N in the fourth quadrant. The orthocentre of the triangle F1MNF_{1}MNF1​MN is
  1. (A)(−910,0)\left( -\frac{9}{10}, 0 \right)(−109​,0)
  2. (B)(23,0)\left( \frac{2}{3}, 0 \right)(32​,0)
  3. (C)(910,0)\left( \frac{9}{10}, 0 \right)(109​,0)
  4. (D)(23,6)\left( \frac{2}{3}, \sqrt{6} \right)(32​,6​)

Correct answer: (A)

Step-by-step solution →
Q54·MathematicsSingle correct
Let F1(x1,0)F_{1}(x_{1}, 0)F1​(x1​,0) and F2(x2,0)F_{2}(x_{2}, 0)F2​(x2​,0), for x1<0x_{1} < 0x1​<0 and x2>0x_{2} > 0x2​>0, be the foci of the ellipse x29+y28=1\frac{x^{2}}{9} + \frac{y^{2}}{8} = 19x2​+8y2​=1. Suppose a parabola having vertex at the origin and focus at F2F_{2}F2​ intersects the ellipse at point M in the first quadrant and at point N in the fourth quadrant. If the tangents to the ellipse at M and N meet at R and the normal to the parabola at M meets the x-axis at Q, then the ratio of area of the triangle MQR to area of the quadrilateral MF1NF2MF_{1}NF_{2}MF1​NF2​ is
  1. (A)3 : 4
  2. (B)4 : 5
  3. (C)5 : 8
  4. (D)2 : 3

Correct answer: (C)

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Chapters tested in this paper

  • Properties of Solids and Liquids 172/186
  • Three Dimensional Geometry 176/186
  • Matrices and Determinants 180/186
  • Coordination Compounds 176/186
  • Current Electricity 160/186
  • Sequence and Series 164/186
  • p-Block Elements 164/186
  • Definite Integration 168/186
  • Rotational Motion 172/186
  • Redox Reactions and Electrochemistry 177/186
  • Geometrical Optics 172/186
  • Vector Algebra 173/186
  • Probability 176/186
  • Chemical Bonding and Molecular Structure 151/186
  • Application of Derivatives 139/186
  • Thermodynamics 154/186
  • Aldehydes and Ketones 135/186
  • Equilibrium 163/186
  • Solutions 158/186
  • Laws of Motion 130/186
  • Units and Measurements 149/186
  • Hydrocarbons 126/186
  • Complex Numbers 165/186
  • Trigonometric Functions 144/186
  • Biomolecules 162/186
  • Electric Field and Coulomb's Law 133/186
  • Dual Nature of Matter and Radiation 155/186
  • Amines 133/186
  • Electromagnetic Induction 120/186
  • Wave Optics 130/186
  • Area Under Curves 139/186
  • Oscillations 117/186
  • Nuclei 116/186
  • Capacitors and Dielectrics 115/186
  • Parabola 101/186
  • Ellipse 103/186
  • Isolation of Metals 106/186
  • Differentiability 91/186
  • Surface Chemistry 98/186
  • Experimental Skills 68/186
  • Carboxylic Acids and Derivatives 54/186
  • Solid State 63/186
  • Principles of Qualitative Analysis 58/186
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