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JEE Advanced 2017 Paper 2 Question Paper with Answers

53 questions · Physics, Chemistry & Mathematics

53 of the 54 questions from the JEE Advanced 2017 Paper 2 paper, each with its correct answer and tagged to the chapter it tests. Free to read, no account needed.

1 question is held back while we re-check the transcription or the answer key.

Physics
18
Chemistry
18
Mathematics
17

Physics — JEE Advanced 2017 Paper 2

Q1·PhysicsSingle correct
Consider an expanding sphere of instantaneous radius RRR whose total mass remains constant. The expansion is such that the instantaneous density ρ\rhoρ remains uniform throughout the volume. The rate of fractional change in density (1ρdρdt)\left(\dfrac{1}{\rho}\dfrac{d\rho}{dt}\right)(ρ1​dtdρ​) is constant. The velocity vvv of any point on the surface of the expanding sphere is proportional to
  1. (A)RRR
  2. (B)R3R^{3}R3
  3. (C)1R\dfrac{1}{R}R1​
  4. (D)R2/3R^{2/3}R2/3

Correct answer: (A)

Step-by-step solution →
Q2·PhysicsSingle correct
Consider regular polygons with number of sides n=3,4,5,…n = 3, 4, 5, \dotsn=3,4,5,… as shown in the figure. The center of mass of all the polygons is at height hhh from the ground. They roll on a horizontal surface about the leading vertex without slipping and sliding as depicted. The maximum increase in height of the locus of the center of mass for each polygon is Δ\DeltaΔ. Then Δ\DeltaΔ depends on nnn and hhh as
  1. (A)Δ=hsin⁡2 ⁣(πn)\Delta = h\sin^{2}\!\left(\dfrac{\pi}{n}\right)Δ=hsin2(nπ​)
  2. (B)Δ=h(1cos⁡(πn)−1)\Delta = h\left(\dfrac{1}{\cos\left(\frac{\pi}{n}\right)} - 1\right)Δ=h(cos(nπ​)1​−1)
  3. (C)Δ=hsin⁡ ⁣(2πn)\Delta = h\sin\!\left(\dfrac{2\pi}{n}\right)Δ=hsin(n2π​)
  4. (D)Δ=htan⁡2 ⁣(π2n)\Delta = h\tan^{2}\!\left(\dfrac{\pi}{2n}\right)Δ=htan2(2nπ​)

Correct answer: (B)

Step-by-step solution →
Q3·PhysicsSingle correct
A photoelectric material having work-function ϕ0\phi_{0}ϕ0​ is illuminated with light of wavelength λ\lambdaλ (λ<hcϕ0)\left(\lambda < \dfrac{hc}{\phi_{0}}\right)(λ<ϕ0​hc​). The fastest photoelectron has a de Broglie wavelength λd\lambda_{d}λd​. A change in wavelength of the incident light by Δλ\Delta\lambdaΔλ results in change Δλd\Delta\lambda_{d}Δλd​ in λd\lambda_{d}λd​. then the ratio Δλd/Δλ\Delta\lambda_{d}/\Delta\lambdaΔλd​/Δλ is proportional to
  1. (A)λd/λ\lambda_{d}/\lambdaλd​/λ
  2. (B)λd2/λ\lambda_{d}^{2}/\lambdaλd2​/λ
  3. (C)λd3/λ\lambda_{d}^{3}/\lambdaλd3​/λ
  4. (D)λd3/λ2\lambda_{d}^{3}/\lambda^{2}λd3​/λ2

Correct answer: (D)

Step-by-step solution →
Q4·PhysicsSingle correct
A symmetric star conducting wire loop is carrying a steady state current III as shown in figure. The distance between the diametrically opposite vertices of the star is 4a4a4a. The magnitude of the magnetic field at the center of the loop is
  1. (A)μ0I4πa 6[3−1]\dfrac{\mu_{0}I}{4\pi a}\,6[\sqrt{3}-1]4πaμ0​I​6[3​−1]
  2. (B)μ0I4πa 6[3+1]\dfrac{\mu_{0}I}{4\pi a}\,6[\sqrt{3}+1]4πaμ0​I​6[3​+1]
  3. (C)μ0I4πa 3[3−1]\dfrac{\mu_{0}I}{4\pi a}\,3[\sqrt{3}-1]4πaμ0​I​3[3​−1]
  4. (D)μ0I4πa 3[2−3]\dfrac{\mu_{0}I}{4\pi a}\,3[2-\sqrt{3}]4πaμ0​I​3[2−3​]

Correct answer: (A)

Step-by-step solution →
Q5·PhysicsSingle correct
Three vectors P⃗\vec{P}P, Q⃗\vec{Q}Q​ and R⃗\vec{R}R are shown in the figure. Let SSS be any point on the vector R⃗\vec{R}R. The distance between the point PPP ad SSS is b∣R⃗∣b|\vec{R}|b∣R∣. The general relation among vectors P⃗\vec{P}P, Q⃗\vec{Q}Q​ and S⃗\vec{S}S is
  1. (A)S⃗=(1−b)P⃗+bQ⃗\vec{S} = (1-b)\vec{P} + b\vec{Q}S=(1−b)P+bQ​
  2. (B)S⃗=(b−1)P⃗+bQ⃗\vec{S} = (b-1)\vec{P} + b\vec{Q}S=(b−1)P+bQ​
  3. (C)S⃗=(1−b2)P⃗+bQ⃗\vec{S} = (1-b^{2})\vec{P} + b\vec{Q}S=(1−b2)P+bQ​
  4. (D)S⃗=(1−b)P⃗+b2Q⃗\vec{S} = (1-b)\vec{P} + b^{2}\vec{Q}S=(1−b)P+b2Q​

Correct answer: (A)

Step-by-step solution →
Q6·PhysicsSingle correct
A rocket is launched normal to the surface of earth, away from the Sun, along the line joining the Sun and the Earth. The Sun is 3×1053\times10^{5}3×105 times heavier than the Earth and is at a distance 2.5×1042.5\times10^{4}2.5×104 times larger than the radius of the Earth. The escape velocity from Earth's gravitational field is ve=11.2 km s−1v_{e} = 11.2\ \text{km s}^{-1}ve​=11.2 km s−1. The minimum initial (vs)(v_{s})(vs​) required for the rocket to be able to leave the Sun-earth system is closest to (Ignore the rotation and revolution of the Earth and the presence of any other planet)
  1. (A)vs=22 km s−1v_{s} = 22\ \text{km s}^{-1}vs​=22 km s−1
  2. (B)vs=42 km s−1v_{s} = 42\ \text{km s}^{-1}vs​=42 km s−1
  3. (C)vs=62 km s−1v_{s} = 62\ \text{km s}^{-1}vs​=62 km s−1
  4. (D)vs=72 km s−1v_{s} = 72\ \text{km s}^{-1}vs​=72 km s−1

Correct answer: (B)

Step-by-step solution →
Q7·PhysicsSingle correct
A person measures the depth of a well by measuring the time interval between dropping a stone and receiving the sound of impact with the bottom of the well. The error in his measurement of time is δT=0.01\delta T = 0.01δT=0.01 seconds and he measures the depth of the well to be L=20L = 20L=20 meters. Take the acceleration due to gravity g=10 ms−2g = 10\ \text{ms}^{-2}g=10 ms−2 and the velocity of sound is 300 ms−1300\ \text{ms}^{-1}300 ms−1. Then the fractional error in the measurement, δL/L\delta L/LδL/L, is closest to
  1. (A)0.2 %0.2\ \%0.2 %
  2. (B)1 %1\ \%1 %
  3. (C)3 %3\ \%3 %
  4. (D)5 %5\ \%5 %

Correct answer: (B)

Step-by-step solution →
Q8·PhysicsMultiple correct
A uniform magnetic field BBB exists in the region between x=0x = 0x=0 and x=3R2x = \dfrac{3R}{2}x=23R​ (region 2 in the figure) pointing normally into the plane of the paper. A particle with charge +Q+Q+Q and momentum ppp directed along x-axis enters region 2 from region 1 at point P1P_{1}P1​ (y=−R)(y = -R)(y=−R). Which of the following option(s) is/are correct?
  1. (A)For B>23pQRB > \dfrac{2}{3}\dfrac{p}{QR}B>32​QRp​, the particle will re-enter region 1
  2. (B)For B=813pQRB = \dfrac{8}{13}\dfrac{p}{QR}B=138​QRp​, the particle will enter region 3 through the point P2P_{2}P2​ on x-axis
  3. (C)When the particle re-enters region 1 through the longest possible path in region 2, the magnitude of the change in its linear momentum between point P1P_{1}P1​ and the farthest point from y-axis is p/2p/\sqrt{2}p/2​
  4. (D)For a fixed BBB, particles of same charge QQQ and same velocity vvv, the distance between the point P1P_{1}P1​ and the point of re-entry into region 1 is inversely proportional to the mass of the particle

Correct answer: (A), (B)

Step-by-step solution →
Q9·PhysicsMultiple correct
The instantaneous voltages at three terminals marked X, Y and Z are given by VX=V0sin⁡ωtV_{X} = V_{0}\sin\omega tVX​=V0​sinωt VY=V0sin⁡ ⁣(ωt+2π3)V_{Y} = V_{0}\sin\!\left(\omega t + \dfrac{2\pi}{3}\right)VY​=V0​sin(ωt+32π​) and VZ=V0sin⁡ ⁣(ωt+4π3)V_{Z} = V_{0}\sin\!\left(\omega t + \dfrac{4\pi}{3}\right)VZ​=V0​sin(ωt+34π​). An ideal voltmeter is configured to read rms value of the potential difference between its terminals. It is connected between points X and Y and then between Y and Z. The reading(s) of the voltmeter will be
  1. (A)VXYrms=V032V_{XY}^{rms} = V_{0}\sqrt{\dfrac{3}{2}}VXYrms​=V0​23​​
  2. (B)VYZrms=V012V_{YZ}^{rms} = V_{0}\sqrt{\dfrac{1}{2}}VYZrms​=V0​21​​
  3. (C)VXYrms=V0V_{XY}^{rms} = V_{0}VXYrms​=V0​
  4. (D)independent of the choice of the two terminals

Correct answer: (A), (D)

Step-by-step solution →
Q10·PhysicsMultiple correct
A point charge +Q+Q+Q is placed just outside an imaginary hemispherical surface of radius R as shown in the figure. Which of the following statements is/are correct?
  1. (A)The electric flux passing through the *curved* surface of the hemisphere is −Q2ε0(1−12)-\dfrac{Q}{2\varepsilon_{0}}\left(1 - \dfrac{1}{\sqrt{2}}\right)−2ε0​Q​(1−2​1​)
  2. (B)Total flux through the curved and the flat surfaces is Qε0\dfrac{Q}{\varepsilon_{0}}ε0​Q​
  3. (C)The component of the electric field normal to the flat surface is constant over the surface
  4. (D)The circumference of the flat surface is an equipotential

Correct answer: (A), (D)

Step-by-step solution →
Q11·PhysicsMultiple correct
Two coherent monochromatic point sources S1S_{1}S1​ and S2S_{2}S2​ of wavelength λ=600\lambda = 600λ=600 nm are placed symmetrically on either side of the centre of the circle as shown. The sources are separated by a distance d=1.8d = 1.8d=1.8 mm. This arrangement produces interference fringes visible as alternate bright and dark spots on the circumference of the circle. The angular separation between two consecutive bright spots is Δθ\Delta\thetaΔθ. Which of the following options is/are correct?
  1. (A)A dark spot will be formed at the point P2P_{2}P2​
  2. (B)At P2P_{2}P2​ the order of the fringe will be maximum
  3. (C)The total number of fringes produced between P1P_{1}P1​ and P2P_{2}P2​ in the first quadrant is close to 3000
  4. (D)The angular separation between two consecutive bright spots decreases as we move from P1P_{1}P1​ to P2P_{2}P2​ along the first quadrant

Correct answer: (B), (C)

Step-by-step solution →
Q12·PhysicsMultiple correct
A source of constant voltage V is connected to a resistance R and two ideal inductors L1L_{1}L1​ and L2L_{2}L2​ through a switch S as shown. There is no mutual inductance between the two inductors. The switch S is initially open. At t=0t = 0t=0, the switch is closed and current begins to flow. Which of the following options is/are correct?
  1. (A)After a long time, the current through L1L_{1}L1​ will be VRL2L1+L2\dfrac{V}{R}\dfrac{L_{2}}{L_{1}+L_{2}}RV​L1​+L2​L2​​
  2. (B)After a long time, the current through L2L_{2}L2​ will be VRL1L1+L2\dfrac{V}{R}\dfrac{L_{1}}{L_{1}+L_{2}}RV​L1​+L2​L1​​
  3. (C)The ratio of the currents through L1L_{1}L1​ and L2L_{2}L2​ is fixed at all times (t>0)(t > 0)(t>0)
  4. (D)At t=0t = 0t=0, the current through the resistance R is VR\dfrac{V}{R}RV​

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

Step-by-step solution →
Q13·PhysicsMultiple correct
A rigid uniform bar AB of length L is slipping from its vertical position on a frictionless floor (as shown in the figure). At some instant of time, the angle made by the bar with the vortical is θ\thetaθ. Which of the following statements about its motion is/are correct?
  1. (A)The midpoint of the bar will fall vertically downward
  2. (B)The trajectory of the point A is a parabola
  3. (C)Instantaneous torque about the point in contact with the floor is proportional to sin⁡θ\sin\thetasinθ
  4. (D)When the bar makes an angle θ\thetaθ with the vertical, the displacement of its midpoint from the initial position is proportional to (1−cos⁡θ)(1-\cos\theta)(1−cosθ)

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

Step-by-step solution →
Q14·PhysicsMultiple correct
A wheel of radius R and mass M is placed at the bottom of a fixed step of height R as shown in the figure. A constant force is continuously applied on the surface of the wheel so that it just climbs the step without slipping. Consider the torque τ\tauτ about an axis normal to the plane of the paper passing through the point Q. Which of the following options is/are correct?
  1. (A)If the force is applied at point P tangentially then τ\tauτ decreases continuously as the wheel climbs
  2. (B)If the force is applied normal to the circumference at point X then τ\tauτ is constant
  3. (C)If the force is applied normal to the circumference at point P then τ\tauτ is zero
  4. (D)If the force is applied tangentially at point S then τ≠0\tau \neq 0τ=0 but the wheel never climbs the step

Correct answer: (A)

Step-by-step solution →
Q15·PhysicsSingle correct
PARAGRAPH 1 Consider a simple RC circuit as shown in Figure 1. Process 1: In the circuit the switch S is closed at t=0t = 0t=0 and the capacitor is fully charged to voltage V0V_{0}V0​ (i.e. charging continues for time T≫RCT \gg RCT≫RC). In the process some dissipation (ED)(E_{D})(ED​) occurs across the resistance R. The amount of energy finally stored in the fully charged capacitor is ECE_{C}EC​. Process 2: In a different process the voltage is first set to V03\dfrac{V_{0}}{3}3V0​​ and maintained for a charging time T≫RCT \gg RCT≫RC. Then the voltage is raised to 2V03\dfrac{2V_{0}}{3}32V0​​ without discharging the capacitor and again maintained for time T≫RCT \gg RCT≫RC. The process is repeated one more time by raising the voltage to V0V_{0}V0​ and the capacitor is charged to the same final voltage V0V_{0}V0​ as in Process 1. These two processes are depicted in Figure 2. In Process 1, the energy stored in the capacitor ECE_{C}EC​ and heat dissipated across resistance EDE_{D}ED​ are related by:
  1. (A)EC=EDE_{C} = E_{D}EC​=ED​
  2. (B)EC=EDln⁡2E_{C} = E_{D}\ln 2EC​=ED​ln2
  3. (C)EC=12EDE_{C} = \dfrac{1}{2}E_{D}EC​=21​ED​
  4. (D)EC=2EDE_{C} = 2E_{D}EC​=2ED​

Correct answer: (A)

Step-by-step solution →
Q16·PhysicsSingle correct
PARAGRAPH 1 Consider a simple RC circuit as shown in Figure 1. Process 1: In the circuit the switch S is closed at t=0t = 0t=0 and the capacitor is fully charged to voltage V0V_{0}V0​ (i.e. charging continues for time T≫RCT \gg RCT≫RC). In the process some dissipation (ED)(E_{D})(ED​) occurs across the resistance R. The amount of energy finally stored in the fully charged capacitor is ECE_{C}EC​. Process 2: In a different process the voltage is first set to V03\dfrac{V_{0}}{3}3V0​​ and maintained for a charging time T≫RCT \gg RCT≫RC. Then the voltage is raised to 2V03\dfrac{2V_{0}}{3}32V0​​ without discharging the capacitor and again maintained for time T≫RCT \gg RCT≫RC. The process is repeated one more time by raising the voltage to V0V_{0}V0​ and the capacitor is charged to the same final voltage V0V_{0}V0​ as in Process 1. These two processes are depicted in Figure 2. In Process 2, total energy dissipated across the resistance EDE_{D}ED​ is:
  1. (A)ED=12CV02E_{D} = \dfrac{1}{2}CV_{0}^{2}ED​=21​CV02​
  2. (B)ED=3(12CV02)E_{D} = 3\left(\dfrac{1}{2}CV_{0}^{2}\right)ED​=3(21​CV02​)
  3. (C)ED=13(12CV02)E_{D} = \dfrac{1}{3}\left(\dfrac{1}{2}CV_{0}^{2}\right)ED​=31​(21​CV02​)
  4. (D)ED=3CV02E_{D} = 3CV_{0}^{2}ED​=3CV02​

Correct answer: (C)

Step-by-step solution →
Q17·PhysicsSingle correct
PARAGRAPH 2 One twirls a circular ring (of mass M and radius R) near the tip of one's finger as shown in Figure 1. In the process the finger never loses contact with the inner rim of the ring. The finger traces out the surface of a cone, shown by the dotted line. The radius of the path traced out by the point where the ring and the finger is in contact is r. The finger rotates with an angular velocity ω0\omega_{0}ω0​. The rotating ring rolls without slipping on the outside of a smaller circle described by the point where the ring and the finger is in contact (Figure 2). The coefficient of friction between the ring and the finger is μ\muμ and the acceleration due to gravity is g. The total kinetic energy of the ring is
  1. (A)Mω02R2M\omega_{0}^{2}R^{2}Mω02​R2
  2. (B)12Mω02(R−r)2\dfrac{1}{2}M\omega_{0}^{2}(R-r)^{2}21​Mω02​(R−r)2
  3. (C)Mω02(R−r)2M\omega_{0}^{2}(R-r)^{2}Mω02​(R−r)2
  4. (D)32Mω02(R−r)2\dfrac{3}{2}M\omega_{0}^{2}(R-r)^{2}23​Mω02​(R−r)2

Correct answer: (C)

Step-by-step solution →
Q18·PhysicsSingle correct
PARAGRAPH 2 One twirls a circular ring (of mass M and radius R) near the tip of one's finger as shown in Figure 1. In the process the finger never loses contact with the inner rim of the ring. The finger traces out the surface of a cone, shown by the dotted line. The radius of the path traced out by the point where the ring and the finger is in contact is r. The finger rotates with an angular velocity ω0\omega_{0}ω0​. The rotating ring rolls without slipping on the outside of a smaller circle described by the point where the ring and the finger is in contact (Figure 2). The coefficient of friction between the ring and the finger is μ\muμ and the acceleration due to gravity is g. The minimum value of ω0\omega_{0}ω0​ below which the ring will drop down is
  1. (A)gμ(R−r)\sqrt{\dfrac{g}{\mu(R-r)}}μ(R−r)g​​
  2. (B)2gμ(R−r)\sqrt{\dfrac{2g}{\mu(R-r)}}μ(R−r)2g​​
  3. (C)3g2μ(R−r)\sqrt{\dfrac{3g}{2\mu(R-r)}}2μ(R−r)3g​​
  4. (D)g2μ(R−r)\sqrt{\dfrac{g}{2\mu(R-r)}}2μ(R−r)g​​

Correct answer: (A)

Step-by-step solution →

Chemistry — JEE Advanced 2017 Paper 2

Q19·ChemistrySingle correct
Pure water freezes at 273 K and 1 bar. The addition of 34.5 g of ethanol to 500 g of water changes the freezing point of the solution. Use the freezing point depression constant of water as 2 K kg mol−1^{-1}−1. The figures shown below represent plots of vapour pressure (V.P.) versus temperature (T). [molecular weight of ethanol is 46 g mol−1^{-1}−1] Among the following, the option representing change the freezing point is
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (C)

Step-by-step solution →
Q20·ChemistrySingle correct
For the following cell, Zn(s) ∣ ZnSO4(aq) ∥ CuSO4(aq) ∣ Cu(s)\mathrm{Zn}(s)\,|\,\mathrm{ZnSO_{4}}(aq)\,\|\,\mathrm{CuSO_{4}}(aq)\,|\,\mathrm{Cu}(s)Zn(s)∣ZnSO4​(aq)∥CuSO4​(aq)∣Cu(s) When the concentration of Zn2+\mathrm{Zn^{2+}}Zn2+ is 10 times the concentration of Cu2+\mathrm{Cu^{2+}}Cu2+, the expression for ΔG\Delta GΔG (in J mol−1^{-1}−1) is [F is Faraday constant; R is gas constant; T is temperature; E0(cell)E^{0}(cell)E0(cell) = 1.1 V]
  1. (A)1.1 F1.1\,F1.1F
  2. (B)2.303 RT−2.2 F2.303\,RT - 2.2\,F2.303RT−2.2F
  3. (C)2.303 RT+1.1 F2.303\,RT + 1.1\,F2.303RT+1.1F
  4. (D)−2.2 F-2.2\,F−2.2F

Correct answer: (B)

Step-by-step solution →
Q21·ChemistrySingle correct
The standard state Gibbs free energies of formation of C(graphite) and C(diamond) at T = 298 K are ΔfG0[C(graphite)]=0 kJ mol−1\Delta_{f}G^{0}\left[\mathrm{C(graphite)}\right] = 0\ \mathrm{kJ\,mol^{-1}}Δf​G0[C(graphite)]=0 kJmol−1 ΔfG0[C(diamond)]=2.9 kJ mol−1\Delta_{f}G^{0}\left[\mathrm{C(diamond)}\right] = 2.9\ \mathrm{kJ\,mol^{-1}}Δf​G0[C(diamond)]=2.9 kJmol−1 The standard state means that the pressure should be 1 bar, and substance should be pure at a given temperature. The conversion of graphite [C(graphite)] to diamond [C(diamond)] reduces its volume by 2×10−6 m3 mol−12\times10^{-6}\ \mathrm{m^{3}\,mol^{-1}}2×10−6 m3mol−1. If C(graphite) is converted to C(diamond) isothermally at T = 298 K, the pressure at which C(graphite) is in equilibrium with C(diamond), is [Useful information: 1 J = 1 kg m2^{2}2s−2^{-2}−2; 1 Pa = 1 kg m−1^{-1}−1s−2^{-2}−2; 1 bar = 10510^{5}105 Pa]
  1. (A)14501 bar
  2. (B)58001 bar
  3. (C)1450 bar
  4. (D)29001 bar

Correct answer: (A)

Step-by-step solution →
Q22·ChemistrySingle correct
Which of the following combination will produce H2\mathrm{H_{2}}H2​ gas?
  1. (A)Fe metal and conc. HNO3\mathrm{HNO_{3}}HNO3​
  2. (B)Cu metal and conc. HNO3\mathrm{HNO_{3}}HNO3​
  3. (C)Zn metal and NaOH(aq)
  4. (D)Au metal and NaCN (aq) in the presence of air

Correct answer: (C)

Step-by-step solution →
Q23·ChemistrySingle correct
The order of the oxidation state of the phosphorus atom in H3PO2\mathrm{H_{3}PO_{2}}H3​PO2​, H3PO4\mathrm{H_{3}PO_{4}}H3​PO4​, H3PO3\mathrm{H_{3}PO_{3}}H3​PO3​, and H4P2O6\mathrm{H_{4}P_{2}O_{6}}H4​P2​O6​ is
  1. (A)H3PO3>H3PO2>H3PO4>H4P2O6\mathrm{H_{3}PO_{3}} > \mathrm{H_{3}PO_{2}} > \mathrm{H_{3}PO_{4}} > \mathrm{H_{4}P_{2}O_{6}}H3​PO3​>H3​PO2​>H3​PO4​>H4​P2​O6​
  2. (B)H3PO4>H3PO2>H3PO3>H4P2O6\mathrm{H_{3}PO_{4}} > \mathrm{H_{3}PO_{2}} > \mathrm{H_{3}PO_{3}} > \mathrm{H_{4}P_{2}O_{6}}H3​PO4​>H3​PO2​>H3​PO3​>H4​P2​O6​
  3. (C)H3PO4>H4P2O6>H3PO3>H3PO2\mathrm{H_{3}PO_{4}} > \mathrm{H_{4}P_{2}O_{6}} > \mathrm{H_{3}PO_{3}} > \mathrm{H_{3}PO_{2}}H3​PO4​>H4​P2​O6​>H3​PO3​>H3​PO2​
  4. (D)H3PO2>H3PO3>H4P2O6>H3PO4\mathrm{H_{3}PO_{2}} > \mathrm{H_{3}PO_{3}} > \mathrm{H_{4}P_{2}O_{6}} > \mathrm{H_{3}PO_{4}}H3​PO2​>H3​PO3​>H4​P2​O6​>H3​PO4​

Correct answer: (C)

Step-by-step solution →
Q24·ChemistrySingle correct
The major product of the following reaction is
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (C)

Step-by-step solution →
Q25·ChemistrySingle correct
The order of basicity among the following compound is
  1. (A)II > I > IV > III
  2. (B)IV > II > III > I
  3. (C)IV > I > II > III
  4. (D)I > IV > III > II

Correct answer: (C)

Step-by-step solution →
Q26·ChemistryMultiple correct
The correct statement(s) about surface properties is(are)
  1. (A)Adsorption is accompanied by decrease in enthalpy and decrease in entropy of the system
  2. (B)The critical temperatures of ethane and nitrogen are 563 K and 126 K, respectively. The adsorption of ethane will be more than that of nitrogen on same amount of activated charcoal at a given temperature
  3. (C)Cloud is an emulsion type of colloid in which liquid is dispersed phase and gas is dispersion medium
  4. (D)Brownian motion of colloidal particles does not depend on the size of the particles but depends on viscosity of the solution

Correct answer: (A), (B)

Step-by-step solution →
Q27·ChemistryMultiple correct
For a reaction taking place in a container in equilibrium with its surroundings, the effect of temperature on its equilibrium constant KKK in terms of change in entropy is described by
  1. (A)With increase in temperature, the value of KKK for exothermic reaction decreases because entropy change of the system is positive
  2. (B)With increase in temperature, the value of KKK for endothermic reaction increases because unfavourable change in entropy of the surroundings decreases
  3. (C)With increase in temperature, the value of KKK for endothermic reaction increases because the entropy change of the system is negative
  4. (D)With increase in temperature, the value of KKK for exothermic reaction decreases because favourable change in entropy of the surrounding decreases

Correct answer: (B), (D)

Step-by-step solution →
Q28·ChemistryMultiple correct
In a bimolecular reaction, the steric factor P was experimentally determined to be 4.5. The correct option(s) among the following is(are)
  1. (A)The activation energy of the reaction is unaffected by the value of the steric factor
  2. (B)Experimentally determined value of frequency factor is higher than that predicted by Arrhenius equation
  3. (C)Since P = 4.5, the reaction will not proceed unless an effective catalyst is used
  4. (D)The value of frequency factor predicted by Arrhenius equation is higher than that determined experimentally

Correct answer: (A), (B)

Step-by-step solution →
Q29·ChemistryMultiple correct
For the following compounds, the correct statement(s) with respect to nucleophilic substitution reactions is(are)
  1. (A)I and III follow SN1\mathrm{S_{N}}1SN​1 mechanism
  2. (B)I and II follow SN2\mathrm{S_{N}}2SN​2 mechanism
  3. (C)Compound IV undergoes inversion of configuration
  4. (D)The order of reactivity for I, III and IV is: IV > I > III

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

Step-by-step solution →
Q30·ChemistryMultiple correct
Among the following, the correct statement(s) is(are)
  1. (A)Al(CH3)3\mathrm{Al(CH_{3})_{3}}Al(CH3​)3​ has the three-centre two-electron bonds in its dimeric structure
  2. (B)BH3\mathrm{BH_{3}}BH3​ has the three-centre two-electron bonds in its dimeric structure
  3. (C)AlCl3\mathrm{AlCl_{3}}AlCl3​ has the three-centre two-electron bonds in its dimeric structure
  4. (D)The Lewis acidity of BCl3\mathrm{BCl_{3}}BCl3​ is greater than that of AlCl3\mathrm{AlCl_{3}}AlCl3​

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

Step-by-step solution →
Q31·ChemistryMultiple correct
The option(s) with only amphoteric oxides is(are)
  1. (A)Cr2O3\mathrm{Cr_{2}O_{3}}Cr2​O3​, BeO, SnO, SnO2\mathrm{SnO_{2}}SnO2​
  2. (B)Cr2O3\mathrm{Cr_{2}O_{3}}Cr2​O3​, CrO, SnO, PbO
  3. (C)NO, B2O3\mathrm{B_{2}O_{3}}B2​O3​, PbO, SnO2\mathrm{SnO_{2}}SnO2​
  4. (D)ZnO, Al2O3\mathrm{Al_{2}O_{3}}Al2​O3​, PbO, PbO2\mathrm{PbO_{2}}PbO2​

Correct answer: (A), (D)

Step-by-step solution →
Q32·ChemistryMultiple correct
Compounds P and R upon ozonolysis produce Q and S, respectively. The molecular formula of Q and S is C8H8O\mathrm{C_{8}H_{8}O}C8​H8​O. Q undergoes Cannizzaro reaction but not haloform reaction, whereas S undergoes haloform reaction but not Cannizzaro reaction. The option(s) with suitable combination of P and R, respectively, is(are)
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (A), (B)

Step-by-step solution →
Q33·ChemistrySingle correct
PARAGRAPH 1 Upon heating KClO3\mathrm{KClO_{3}}KClO3​ in the presence of catalytic amount of MnO2\mathrm{MnO_{2}}MnO2​, a gas W is formed. Excess amount of W reacts with white phosphorus to give X. The reaction of X with HNO3\mathrm{HNO_{3}}HNO3​ gives Y and Z. W and X are, respectively
  1. (A)O3\mathrm{O_{3}}O3​ and P4O6\mathrm{P_{4}O_{6}}P4​O6​
  2. (B)O2\mathrm{O_{2}}O2​ and P4O6\mathrm{P_{4}O_{6}}P4​O6​
  3. (C)O2\mathrm{O_{2}}O2​ and P4O10\mathrm{P_{4}O_{10}}P4​O10​
  4. (D)O3\mathrm{O_{3}}O3​ and P4O10\mathrm{P_{4}O_{10}}P4​O10​

Correct answer: (C)

Step-by-step solution →
Q34·ChemistrySingle correct
PARAGRAPH 1 Upon heating KClO3\mathrm{KClO_{3}}KClO3​ in the presence of catalytic amount of MnO2\mathrm{MnO_{2}}MnO2​, a gas W is formed. Excess amount of W reacts with white phosphorus to give X. The reaction of X with HNO3\mathrm{HNO_{3}}HNO3​ gives Y and Z. Y and Z are, respectively
  1. (A)N2O3\mathrm{N_{2}O_{3}}N2​O3​ and H3PO4\mathrm{H_{3}PO_{4}}H3​PO4​
  2. (B)N2O5\mathrm{N_{2}O_{5}}N2​O5​ and HPO3\mathrm{HPO_{3}}HPO3​
  3. (C)N2O4\mathrm{N_{2}O_{4}}N2​O4​ and HPO3\mathrm{HPO_{3}}HPO3​
  4. (D)N2O4\mathrm{N_{2}O_{4}}N2​O4​ and H3PO3\mathrm{H_{3}PO_{3}}H3​PO3​

Correct answer: (B)

Step-by-step solution →
Q35·ChemistrySingle correct
PARAGRAPH 2 The reaction of compound P with CH3MgBr\mathrm{CH_{3}MgBr}CH3​MgBr(excess) in (C2H5)2O\mathrm{(C_{2}H_{5})_{2}O}(C2​H5​)2​O followed by addition of H2O\mathrm{H_{2}O}H2​O gives Q. The compound Q on treatment with H2SO4\mathrm{H_{2}SO_{4}}H2​SO4​ at 0∘^{\circ}∘C gives R. The reaction of R with CH3COCl\mathrm{CH_{3}COCl}CH3​COCl in the presence of anhydrous AlCl3\mathrm{AlCl_{3}}AlCl3​ in CH2Cl2\mathrm{CH_{2}Cl_{2}}CH2​Cl2​ followed by treatment with H2O\mathrm{H_{2}O}H2​O produces compound S [Et in compound P is ethyl group] The product S is
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (A)

Step-by-step solution →
Q36·ChemistrySingle correct
PARAGRAPH 2 The reaction of compound P with CH3MgBr\mathrm{CH_{3}MgBr}CH3​MgBr(excess) in (C2H5)2O\mathrm{(C_{2}H_{5})_{2}O}(C2​H5​)2​O followed by addition of H2O\mathrm{H_{2}O}H2​O gives Q. The compound Q on treatment with H2SO4\mathrm{H_{2}SO_{4}}H2​SO4​ at 0∘^{\circ}∘C gives R. The reaction of R with CH3COCl\mathrm{CH_{3}COCl}CH3​COCl in the presence of anhydrous AlCl3\mathrm{AlCl_{3}}AlCl3​ in CH2Cl2\mathrm{CH_{2}Cl_{2}}CH2​Cl2​ followed by treatment with H2O\mathrm{H_{2}O}H2​O produces compound S [Et in compound P is ethyl group] The reactions, Q to R and R to S, are
  1. (A)Dehydration and Friedel-Crafts acylation
  2. (B)Aromatic sulfonation and Friedel-Crafts acylation
  3. (C)Friedel-Crafts alkylation, dehydration and Friedel-Crafts acylation
  4. (D)Friedel-Crafts alkylation and Friedel-Crafts acylation

Correct answer: (C)

Step-by-step solution →

Mathematics — JEE Advanced 2017 Paper 2

Q37·MathematicsSingle correct
The equation of the plane passing through the point (1,1,1)(1, 1, 1)(1,1,1) and perpendicular to the planes 2x+y−2z=52x + y - 2z = 52x+y−2z=5 and 3x−6y−2z=73x - 6y - 2z = 73x−6y−2z=7, is
  1. (A)14x+2y−15z=114x + 2y - 15z = 114x+2y−15z=1
  2. (B)14x−2y+15z=2714x - 2y + 15z = 2714x−2y+15z=27
  3. (C)14x+2y+15z=3114x + 2y + 15z = 3114x+2y+15z=31
  4. (D)−14x+2y+15z=3-14x + 2y + 15z = 3−14x+2y+15z=3

Correct answer: (C)

Step-by-step solution →
Q38·MathematicsSingle correct
Let OOO be the origin and let PQRPQRPQR be an arbitrary triangle. The point SSS is such that OP→⋅OQ→+OR→⋅OS→=OR→⋅OP→+OQ→⋅OS→=OQ→⋅OR→+OP→⋅OS→\overrightarrow{OP}\cdot\overrightarrow{OQ} + \overrightarrow{OR}\cdot\overrightarrow{OS} = \overrightarrow{OR}\cdot\overrightarrow{OP} + \overrightarrow{OQ}\cdot\overrightarrow{OS} = \overrightarrow{OQ}\cdot\overrightarrow{OR} + \overrightarrow{OP}\cdot\overrightarrow{OS}OP⋅OQ​+OR⋅OS=OR⋅OP+OQ​⋅OS=OQ​⋅OR+OP⋅OS Then the triangle PQRPQRPQR has SSS as its
  1. (A)centroid
  2. (B)circumcentre
  3. (C)incentre
  4. (D)orthocenter

Correct answer: (D)

Step-by-step solution →
Q39·MathematicsSingle correct
If y=y(x)y = y(x)y=y(x) satisfies the differential equation 8x(9+x)dy=(4+9+x)−1dx8\sqrt{x}\left(\sqrt{9+\sqrt{x}}\right)dy = \left(\sqrt{4+\sqrt{9+\sqrt{x}}}\right)^{-1}dx8x​(9+x​​)dy=(4+9+x​​​)−1dx, x>0\quad x > 0x>0 and y(0)=7y(0) = \sqrt{7}y(0)=7​, then y(256)=y(256) =y(256)=
  1. (A)333
  2. (B)999
  3. (C)161616
  4. (D)808080

Correct answer: (A)

Step-by-step solution →
Q40·MathematicsSingle correct
If f:R→Rf : \mathbb{R} \to \mathbb{R}f:R→R is a twice differentiable function such that f′′(x)>0f''(x) > 0f′′(x)>0 for all x∈Rx \in \mathbb{R}x∈R, and f(12)=12f\left(\dfrac{1}{2}\right) = \dfrac{1}{2}f(21​)=21​, f(1)=1f(1) = 1f(1)=1, then
  1. (A)f′(1)≤0f'(1) \le 0f′(1)≤0
  2. (B)0<f′(1)≤120 < f'(1) \le \dfrac{1}{2}0<f′(1)≤21​
  3. (C)12<f′(1)≤1\dfrac{1}{2} < f'(1) \le 121​<f′(1)≤1
  4. (D)f′(1)>1f'(1) > 1f′(1)>1

Correct answer: (D)

Step-by-step solution →
Q41·MathematicsSingle correct
How many 3×33 \times 33×3 matrices MMM with entries from {0,1,2}\{0, 1, 2\}{0,1,2} are there, for which the sum of the diagonal entries of MTMM^{T}MMTM is 555 ?
  1. (A)126126126
  2. (B)198198198
  3. (C)162162162
  4. (D)135135135

Correct answer: (B)

Step-by-step solution →
Q42·MathematicsSingle correct
Let S={1,2,3,…,9}S = \{1, 2, 3, \dots, 9\}S={1,2,3,…,9}. For k=1,2,…,5k = 1, 2, \dots, 5k=1,2,…,5, let NkN_{k}Nk​ be the number of subsets of SSS, each containing five elements out of which exactly kkk are odd. Then N1+N2+N3+N4+N5=N_{1} + N_{2} + N_{3} + N_{4} + N_{5} =N1​+N2​+N3​+N4​+N5​=
  1. (A)210210210
  2. (B)252252252
  3. (C)125125125
  4. (D)126126126

Correct answer: (D)

Step-by-step solution →
Q43·MathematicsSingle correct
Three randomly chosen nonnegative integers xxx, yyy and zzz are found to satisfy the equation x+y+z=10x + y + z = 10x+y+z=10. Then the probability that zzz is even, is
  1. (A)3655\dfrac{36}{55}5536​
  2. (B)611\dfrac{6}{11}116​
  3. (C)12\dfrac{1}{2}21​
  4. (D)511\dfrac{5}{11}115​

Correct answer: (B)

Step-by-step solution →
Q44·MathematicsMultiple correct
Let α\alphaα and β\betaβ be nonzero real numbers such that 2(cos⁡β−cos⁡α)+cos⁡αcos⁡β=12(\cos\beta - \cos\alpha) + \cos\alpha\cos\beta = 12(cosβ−cosα)+cosαcosβ=1. Then which of the following is/are true ?
  1. (A)tan⁡(α2)+3tan⁡(β2)=0\tan\left(\dfrac{\alpha}{2}\right) + \sqrt{3}\tan\left(\dfrac{\beta}{2}\right) = 0tan(2α​)+3​tan(2β​)=0
  2. (B)3tan⁡(α2)+tan⁡(β2)=0\sqrt{3}\tan\left(\dfrac{\alpha}{2}\right) + \tan\left(\dfrac{\beta}{2}\right) = 03​tan(2α​)+tan(2β​)=0
  3. (C)tan⁡(α2)−3tan⁡(β2)=0\tan\left(\dfrac{\alpha}{2}\right) - \sqrt{3}\tan\left(\dfrac{\beta}{2}\right) = 0tan(2α​)−3​tan(2β​)=0
  4. (D)3tan⁡(α2)−tan⁡(β2)=0\sqrt{3}\tan\left(\dfrac{\alpha}{2}\right) - \tan\left(\dfrac{\beta}{2}\right) = 03​tan(2α​)−tan(2β​)=0

Correct answer: (A), (C)

Step-by-step solution →
Q45·MathematicsMultiple correct
If f:R→Rf : \mathbb{R} \to \mathbb{R}f:R→R is a differentiable function such that f′(x)>2f(x)f'(x) > 2f(x)f′(x)>2f(x) for all x∈Rx \in \mathbb{R}x∈R, and f(0)=1f(0) = 1f(0)=1, then
  1. (A)f(x)f(x)f(x) is increasing in (0,∞)(0, \infty)(0,∞)
  2. (B)f(x)f(x)f(x) is decreasing in (0,∞)(0, \infty)(0,∞)
  3. (C)f(x)>e2xf(x) > e^{2x}f(x)>e2x in (0,∞)(0, \infty)(0,∞)
  4. (D)f′(x)<e2xf'(x) < e^{2x}f′(x)<e2x in (0,∞)(0, \infty)(0,∞)

Correct answer: (A), (C)

Step-by-step solution →
Q46·MathematicsMultiple correct
Let f(x)=1−x(1+∣1−x∣)∣1−x∣cos⁡(11−x)f(x) = \dfrac{1 - x\left(1 + |1-x|\right)}{|1-x|}\cos\left(\dfrac{1}{1-x}\right)f(x)=∣1−x∣1−x(1+∣1−x∣)​cos(1−x1​) for x≠1x \ne 1x=1. Then
  1. (A)lim⁡x→1−f(x)=0\lim\limits_{x \to 1^{-}} f(x) = 0x→1−lim​f(x)=0
  2. (B)lim⁡x→1−f(x)\lim\limits_{x \to 1^{-}} f(x)x→1−lim​f(x) does not exist
  3. (C)lim⁡x→1+f(x)=0\lim\limits_{x \to 1^{+}} f(x) = 0x→1+lim​f(x)=0
  4. (D)lim⁡x→1+f(x)\lim\limits_{x \to 1^{+}} f(x)x→1+lim​f(x) does not exist

Correct answer: (A), (D)

Step-by-step solution →
Q47·MathematicsMultiple correct
If f(x)=∣cos⁡(2x)cos⁡(2x)sin⁡(2x)−cos⁡xcos⁡x−sin⁡xsin⁡xsin⁡xcos⁡x∣f(x) = \begin{vmatrix} \cos(2x) & \cos(2x) & \sin(2x) \\ -\cos x & \cos x & -\sin x \\ \sin x & \sin x & \cos x \end{vmatrix}f(x)=​cos(2x)−cosxsinx​cos(2x)cosxsinx​sin(2x)−sinxcosx​​, then
  1. (A)f′(x)=0f'(x) = 0f′(x)=0 at exactly three points in (−π,π)(-\pi, \pi)(−π,π)
  2. (B)f′(x)=0f'(x) = 0f′(x)=0 at more than three points in (−π,π)(-\pi, \pi)(−π,π)
  3. (C)f(x)f(x)f(x) attains its maximum at x=0x = 0x=0
  4. (D)f(x)f(x)f(x) attains its minimum at x=0x = 0x=0

Correct answer: (B), (C)

Step-by-step solution →
Q48·MathematicsMultiple correct
If the line x=αx = \alphax=α divides the area of region R={(x,y)∈R2:x3≤y≤x, 0≤x≤1}R = \{(x, y) \in \mathbb{R}^{2} : x^{3} \le y \le x,\ 0 \le x \le 1\}R={(x,y)∈R2:x3≤y≤x, 0≤x≤1} into two equal parts, then
  1. (A)0<α≤120 < \alpha \le \dfrac{1}{2}0<α≤21​
  2. (B)12<α<1\dfrac{1}{2} < \alpha < 121​<α<1
  3. (C)2α4−4α2+1=02\alpha^{4} - 4\alpha^{2} + 1 = 02α4−4α2+1=0
  4. (D)α4+4α2−1=0\alpha^{4} + 4\alpha^{2} - 1 = 0α4+4α2−1=0

Correct answer: (B), (C)

Step-by-step solution →
Q49·MathematicsMultiple correct
If I=∑k=198∫kk+1k+1x(x+1) dxI = \displaystyle\sum_{k=1}^{98} \int_{k}^{k+1} \dfrac{k+1}{x(x+1)}\,dxI=k=1∑98​∫kk+1​x(x+1)k+1​dx, then
  1. (A)I>log⁡e99I > \log_{e} 99I>loge​99
  2. (B)I<log⁡e99I < \log_{e} 99I<loge​99
  3. (C)I<4950I < \dfrac{49}{50}I<5049​
  4. (D)I>4950I > \dfrac{49}{50}I>5049​

Correct answer: (B), (D)

Step-by-step solution →
Q50·MathematicsSingle correct
PARAGRAPH 1 Let OOO be the origin, and OX→\overrightarrow{OX}OX, OY→\overrightarrow{OY}OY, OZ→\overrightarrow{OZ}OZ be three unit vectors in the directions of the sides QR→\overrightarrow{QR}QR​, RP→\overrightarrow{RP}RP, PQ→\overrightarrow{PQ}PQ​, respectively, of a triangle PQRPQRPQR. ∣OX→×OY→∣=\left|\overrightarrow{OX} \times \overrightarrow{OY}\right| =​OX×OY​=
  1. (A)sin⁡(P+Q)\sin(P + Q)sin(P+Q)
  2. (B)sin⁡2R\sin 2Rsin2R
  3. (C)sin⁡(P+R)\sin(P + R)sin(P+R)
  4. (D)sin⁡(Q+R)\sin(Q + R)sin(Q+R)

Correct answer: (A)

Step-by-step solution →
Q51·MathematicsSingle correct
PARAGRAPH 1 Let OOO be the origin, and OX→\overrightarrow{OX}OX, OY→\overrightarrow{OY}OY, OZ→\overrightarrow{OZ}OZ be three unit vectors in the directions of the sides QR→\overrightarrow{QR}QR​, RP→\overrightarrow{RP}RP, PQ→\overrightarrow{PQ}PQ​, respectively, of a triangle PQRPQRPQR. If the triangle PQRPQRPQR varies, then the minimum value of cos⁡(P+Q)+cos⁡(Q+R)+cos⁡(R+P)\cos(P+Q) + \cos(Q+R) + \cos(R+P)cos(P+Q)+cos(Q+R)+cos(R+P) is
  1. (A)−53-\dfrac{5}{3}−35​
  2. (B)−32-\dfrac{3}{2}−23​
  3. (C)32\dfrac{3}{2}23​
  4. (D)53\dfrac{5}{3}35​

Correct answer: (B)

Step-by-step solution →
Q52·MathematicsSingle correct
PARAGRAPH 2 Let ppp, qqq be integers and let α\alphaα, β\betaβ be the roots of the equation x2−x−1=0x^{2} - x - 1 = 0x2−x−1=0, where α≠β\alpha \ne \betaα=β. For n=0,1,2,…n = 0, 1, 2, \dotsn=0,1,2,…, let an=pαn+qβna_{n} = p\alpha^{n} + q\beta^{n}an​=pαn+qβn. FACT: If aaa and bbb are rational numbers and a+b5=0a + b\sqrt{5} = 0a+b5​=0, then a=0=ba = 0 = ba=0=b. a12=a_{12} =a12​=
  1. (A)a11−a10a_{11} - a_{10}a11​−a10​
  2. (B)a11+a10a_{11} + a_{10}a11​+a10​
  3. (C)2a11+a102a_{11} + a_{10}2a11​+a10​
  4. (D)a11+2a10a_{11} + 2a_{10}a11​+2a10​

Correct answer: (B)

Step-by-step solution →
Q53·MathematicsSingle correct
PARAGRAPH 2 Let ppp, qqq be integers and let α\alphaα, β\betaβ be the roots of the equation x2−x−1=0x^{2} - x - 1 = 0x2−x−1=0, where α≠β\alpha \ne \betaα=β. For n=0,1,2,…n = 0, 1, 2, \dotsn=0,1,2,…, let an=pαn+qβna_{n} = p\alpha^{n} + q\beta^{n}an​=pαn+qβn. FACT: If aaa and bbb are rational numbers and a+b5=0a + b\sqrt{5} = 0a+b5​=0, then a=0=ba = 0 = ba=0=b. If a4=28a_{4} = 28a4​=28, then p+2q=p + 2q =p+2q=
  1. (A)212121
  2. (B)141414
  3. (C)777
  4. (D)121212

Correct answer: (D)

Step-by-step solution →

Chapters tested in this paper

  • Three Dimensional Geometry 176/186
  • Matrices and Determinants 180/186
  • Sequence and Series 164/186
  • p-Block Elements 164/186
  • Definite Integration 168/186
  • Rotational Motion 172/186
  • Redox Reactions and Electrochemistry 177/186
  • Kinematics 156/186
  • Vector Algebra 173/186
  • Differential Equations 167/186
  • Probability 176/186
  • Permutations and Combinations 162/186
  • Magnetic Field of Current 147/186
  • Chemical Bonding and Molecular Structure 151/186
  • Application of Derivatives 139/186
  • Limits and Continuity 149/186
  • Aldehydes and Ketones 135/186
  • Equilibrium 163/186
  • Solutions 158/186
  • Chemical Thermodynamics 165/186
  • Units and Measurements 149/186
  • Hydrocarbons 126/186
  • Trigonometric Functions 144/186
  • Chemical Kinetics 169/186
  • Gravitation 152/186
  • Electric Field and Coulomb's Law 133/186
  • Dual Nature of Matter and Radiation 155/186
  • Amines 133/186
  • Quadratic Equations 148/186
  • Electromagnetic Induction 120/186
  • Wave Optics 130/186
  • Area Under Curves 139/186
  • Alternating Currents 108/186
  • Organic Compounds Containing Halogens 109/186
  • Capacitors and Dielectrics 115/186
  • Classification of Elements and Periodicity in Properties 113/186
  • Surface Chemistry 98/186
  • Hydrogen 81/186
  • Diazonium Salts and Reactions 53/186
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