IIT - JEE 2017 (Advanced)

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1 IIT - JEE 017 (Advanced) 0517/IITEQ17/Paper/QP&Soln/Pg.1

2 () Vidyalankar : IIT JEE 017 Advanced : Question Paper & Solution Solution to IIT JEE 017 (Advanced) : Paper - II PART I PHYSICS SECTION 1 (Maximum Marks:1) This section contains SEVEN questions. Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is correct. For each question, darken the bubble corresponding to the correct option in the ORS. For each question, marks will be awarded in one of the following categories: Full Marks : + If only the bubble corresponding to the correct option is darkened. Zero Marks : 0 If none of the bubbles is darkened. Negative Marks : 1 In all other cases. 1. A photoelectric material having work-function 0 is illuminated with light of wavelength hc. The fastest photoelectron has a de Broglie wavelength d. A change in 0 wavelength of the incident light by results in a change d in d. Then the ratio d / is proportional to (A) (B) (C) d/ (D) d / d / 1. (A) hc KEmax h h d p mk h K m d hc h md Differentiating both sides hc d h md d d d d d / (i) (ii). Three vectors P, Q andr are shown in the figure. Let S be any point on the vector R. The distance between the points P and S is b R. The general relation among vectors P, Qand S is 0517/IITEQ17/Paper/QP&Soln/Pg.

3 (A) IIT JEE 017 Advanced : Question Paper & Solution (Paper II) () S (1 b) P b Q (B) S (1 b ) P bq (C) S (1 b) P bq (D) S (b 1) P bq. (C) S b R OS OP PS S P br P b(q P) S (1 b)p bq y O P P S R Q P Q Q x. A symmetric star shaped conducting wire loop is carrying a steady state current I as shown in the figure. The distance between the diametrically opposite vertices of the star is 4a. The magnitude of the magnetic field at the center of the loop is (A) o I 6[ 1] 4a (B) o I [ 1] 4a (C) o I 6[ 1] 4a (D) o I [ ] 4a. (A) Magnetic field at center of loop due to anyone section 0 B I sin sin 4a = 0 sin = 1 a = 60 sin = 10I 1 B net = 1B 4a 0 B net = 6 I 1 4a 4. Consider regular polygons with number of sides n =, 4, 5 as shown in the figure. The center of mass of all the polygons is at height h 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. Then depends on n and h as 0517/IITEQ17/Paper/QP&Soln/Pg.

4 (4) Vidyalankar : IIT JEE 017 Advanced : Question Paper & Solution (A) = h sin 4. (D) (C) = h tan n n (B) = h sin n 1 (D) = h cos 1 n h h = h h = h = h = 1 h for any regular polygon = n h = h h cos cos n 1 1 h ' h h cos n 5. Consider an expanding sphere of instantaneous radius R whose total mass remains constant. The expansion is such that the instantaneous density remains uniform 1d throughout the volume. The rate of fractional change in density is constant. The p dt velocity of any point on the surface of the expanding sphere is proportional to (A) R / (B) R (C) R (D) 1 R 5. (B) Total mass = v = constant dv d V 0 dt dt dv d V dt dt 0517/IITEQ17/Paper/QP&Soln/Pg.4

5 IIT JEE 017 Advanced : Question Paper & Solution (Paper II) (5) dv V d dt dt dv V dt 4 V = r dv dr 4 4t r dt dt Velocity of any point on surface v = dr r dt 6. 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 T = 0.01 seconds and he measures the depth of the well to be L = 0 meters. Take the acceleration due to gravity g = 10ms and the velocity of sound is 00 ms 1. Then the fractional error in the measurement, L/L, is closest to (A) 0.% (B) 5% (C) 1% (D) % 6. (C) Time taken by stone to reach the surface of well. gh 0 t1 t1 sec. (1) g 10 time taken by sound to reach back h 0 1 t t sec. () v Neglecting time taken by sound we have total time T = s 1 L gt L T L T percentage error = = 1% 7. A rocket is launched normal to the surface of the Earth, away from the Sun, along the line joining the Sun and the Earth. The Sun in 10 5 times heavier than the Earth and is at a distance times larger than the radius of the Earth. The escape velocity from Earth s gravitational field is e = 11. km s 1. The minimum initial velocity ( s ) required for the rocket to be able to leave the SunEarth system is closest to (Ignore the rotation and revolution of the Earth and the presence of any other planet). (A) s = 6 km s 1 (B) s = 4 km s 1 (C) s = 7 km s 1 (D) s = km s 1 7. (B) Energy conservation 1 GMem GMsm mv 0 s R r R 1 GMem Gsm mv s R r Sun r = R Earth V s 0517/IITEQ17/Paper/QP&Soln/Pg.5

6 (6) Vidyalankar : IIT JEE 017 Advanced : Question Paper & Solution 5 Me 10 M e Vs G R R GM GM Vs 11 1 R R GMe Ve R s e e V 1V km/s e SECTION II (Maximum Marks:8) This section contains SEVEN questions. Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four options is(are) correct. For each question, darken the bubble(s) corresponding to all the correct option(s) in the ORS. For each question, marks will be awarded in one of the following categories: Full Marks : +4 If only the bubble(s) corresponding to all the correct option(s) is (are) darkened. Partial Marks : +1 For darkening a bubble corresponding to each correct option, provided NO incorrect option is darkened. Zero Marks : 0 If none of the bubbles is darkened. Negative Marks : In all other cases. For example, if (A), (C) and (D) are all the correct options for a question, darkening all these three will get +4 marks; darkening only (A) and (D) will get + marks; and darkening (A) and (B) will get marks, as a wrong option is also darkened. 8. A source of constant voltage V is connected to a resistance R and two ideal inductors L 1 and L through a switch S as shown. There is no mutual inductance between the two inductors. The switch S is initially open. At t = 0, the switch is closed and current beings to flow. Which of the following options is/are correct? (A) At t = 0, the current through the resistance R is V R V L (B) After a long time, the current through L will be 1 R L1 L V L (C) After a long time, the current through L 1 will be R L1 L (D) The ratio of the currents through L 1 and L is fixed at all time (t > 0) 8. (B), (C), (D) R v L 1 L 0517/IITEQ17/Paper/QP&Soln/Pg.6

7 IIT JEE 017 Advanced : Question Paper & Solution (Paper II) (7) (A) at t = 0 Inductors behave as open switch i = 0 (B), (C) After a long time Inductors behave as closed switch Net current i = V R If two inductors are parallel flux 1 = L i L i (1) 1 1 i i i () 1 Li L V on solving i1 L1 L L1 L R L1i L1 V i L1 L L1 L R i (D) 1 L constant i L 1 9. A point charge +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? (A) The electric flux passing through the curved surface of the hemisphere is Q (B) The component of the electric field normal to the flat surface is constant over the surface Q (C) Total flux through the curved and the flat surfaces is 0 (D) The circumference of the flat surface is an equipotential 9. (A), (D) Let hemisphere is closed. Net Electric flux passing through hemisphere = 0 curved + flat = 0 curved = flat Q x r KQ E R r d EdA cos 0517/IITEQ17/Paper/QP&Soln/Pg.7

8 (8) Vidyalankar : IIT JEE 017 Advanced : Question Paper & Solution KQ R d rdr R r R r R QR rdr (R r ) 0 0 / QR 1 R r R 0 0 R QR 1 1 R R 0 0 Q Q 1 curved 1 0 (B) Component of E.F. normal to flat surface KQ R E cos Constant R r R r (D) All points on circumference are at same distance hence Equipotential. 10. A uniform magnetic field B exists in the region between x = 0 and x = R (region in the figure) pointing normally into the plane of the paper. A particle with charge +Q and momentum p directed along x-axis enters region from region 1 at point P 1 (y = R). Which of the following options(s) is/are correct? (A) For B = 8 p, the particle will enter region through the point P on x-axis. 1 QR (B) For B > p, the particle will re-enter region 1 QR (C) For a fixed B, particles of same charge Q and same velocity, the distance between the point P 1 and the point of re-entry into region 1 is inversely proportional to the mass of the particle. 0517/IITEQ17/Paper/QP&Soln/Pg.8

9 IIT JEE 017 Advanced : Question Paper & Solution (Paper II) (9) (D) When the particle re-enters region 1 through the longest possible path in region, the magnitude of the change in its linear momentum between point P 1 and the farthest point from y-axis is p/ 10. (B) Let C be the centre of the circle. CP 1 = CP = r = radius of the circle = We have 1 1 OC OP CP (CP OP ) OP CP p qb P R p R qb qb p p 9R p R R qb qb 4 qb pr 9 R R 0 qb 4 1 pr R 4 qb 8 p B 1 qr (A) is correct Charge returns to region 1. R (B) r p R QB p B Q (B) is correct. C O R P 1 R P x (C) radius of the circular path r = m qb r m (C) is not correct. (D) The momentum at P 1 and the farthest point on the xaxis are perpendicular to each other. The change in momentum is p. (D) is not correct. 11. Two coherent monochromatic point sources S 1 and S of wavelength = 600 nm are placed symmetrically on either side of the center of the circle as shown. The sources are separated by a distance d = 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. Which of the following options is/are correct? 0517/IITEQ17/Paper/QP&Soln/Pg.9

10 (10) Vidyalankar : IIT JEE 017 Advanced : Question Paper & Solution (A) The total number of fringes produced between P 1 and P in the first quadrant is close to 000 (B) A dark spot will be formed at the point P (C) At P the order of the fringe will be maximum (D) The angular separation between two consecutive bright spots decreases as we move from P 1 to P along the first quadrant. 11. (A), (C) P 1 P S 1 d S P (A) at point P 1 path difference x = 0 (central maxima) at point P path difference x = d x n = n = 000 (maxima) 9 1 = 1 (C) As the path difference is maximum at P the order of the fringe will be maximum. (D) Angular fringe width increases as we more from P 1 to P as shown below path difference = x = d sin x = d cos = d cos = d cos as increases cos decreases and increases. 1. The instantaneous voltages at three terminals marked X, Y and Z are given by V X = V 0 sin t, V Y = V 0 sin t and 4 V Z = V 0 sin t 0517/IITEQ17/Paper/QP&Soln/Pg.10

11 IIT JEE 017 Advanced : Question Paper & Solution (Paper II) (11) 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 (A) rms 1 VYZ V0 (B) rms VXY V0 (C) V rms XY V0 (D) independent of the choice of the two terminals 1. (B), (D) V x = V o sin t V y = V o sin t 4 V z = V o sin t If voltmeter is connected between x and y V xy = V x V y V V V V cos xy o o o o Vxy V o osin Vo xy rms o V V xy Vo Similarly Vyz Vo sin Vo rms Vyz Vo v y v yz v z v zx v xz v x 1. 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 vertical is. Which of the following statements about its motion is/are correct? (A) The trajectory of the point A is a parabola (B) instantaneous torque about the point in contact with the floor is proportional to sin (C) The midpoint of the bar will fall vertically downward (D) When the bar makes an angle with the vertical, the displacement of its midpoint from the initial position is proportional to (1 cos ) 0517/IITEQ17/Paper/QP&Soln/Pg.11

12 (1) Vidyalankar : IIT JEE 017 Advanced : Question Paper & Solution 1. (A), (B), (C), (D) (A) x Co-ordinate of point A A L X sin A y Co-ordinate of point A L L y = cos y L x 4 Point A moves on an elliptical path. (B) Instantaneous Torque about point of contact L mgsin sin (C) Rod moves such that COM moves on a vertical line (D) When rod is rotated by an angle decrease in height of com L h sin mg B L B 14. 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 about an axis normal to the plane of the paper passing through the point Q. Which of the following options is/are correct? (A) If the force is applied tangentially at point S then 0 but the wheel never climbs the step (B) If the force is applied normal to the circumference at point P then is zero (C) If the force is applied normal to the circumference at point X then is constant (D) If the force is applied at point P tangentially then decreases continuously as the wheel climbs 14. (B), (C) R For part A : The torque due to F is always equal to FR. (about Q). Torque due to weight = mg R cos 0517/IITEQ17/Paper/QP&Soln/Pg.1 R 45 Q F F B Q

13 IIT JEE 017 Advanced : Question Paper & Solution (Paper II) (1) FR mgr cos dx = work done by all the forces = change in K.E. of the ball 0 KE = FR mgr sin [When the ball has turned by an angle ''] mg sin x KE = FR 1 F For F > mg KE is always positive for all. Thus it is possible for the ball to climb up the step [Assuming sufficient friction]. Hence (A) is False. (B) is obvious True. For part (C) : = FR cos, since is fixed, is fixed. (C) is true. (D) is false because '' is constant. SECTION III (Maximum Marks:1) This section contains TWO paragraphs. Based on each paragraph, there are TWO questions. Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is correct. For each question, darken the bubble corresponding to the correct option in the ORS. For each question, marks will be awarded in one of the following categories: Full Marks : + If only the bubble corresponding to the correct answer is darkened. Zero Marks : 0 If all other cases. Paragraph for Q. No. 15 & 16 Consider a simple RC circuit as shown in Figure 1. Process 1 : In the circuit the switch S is closed at t = 0 and the capacitor is fully charged to voltage V 0 (i.e., charging continues for time T >> RC). In the process some dissipation (E D ) occurs across the resistance R. The amount of energy finally stored in the fully charged capacitor is E C. V0 Process : In a different process the voltage is first set to and maintained for a charging V0 time T >> RC. Then the voltage is raised to without discharging the capacitor and again maintained for a time T >> RC. The process is repeated one more time by raising the voltage to V 0 and the capacitor is charged to the same final voltage V 0 as in Process I. These to processes are depicted in Figure. 15. In Process, total energy dissipated across the resistance E D is : 1 1 (A) ED CV0 (B) ED CV0 (C) E D = CV 0 (D) E 11 D CV0 0517/IITEQ17/Paper/QP&Soln/Pg.1

14 (14) Vidyalankar : IIT JEE 017 Advanced : Question Paper & Solution 15. (D) Charge flowing through the seat of emf in each of the three subparts (each of duration T) CV0 of process is ; the voltage across the seat of emf being V0 V0, and V 0 respectively in the three intervals. The energy spent by emf source, therefore equals CV0 V0 V0 CV0 V0. 1 Out of this CV0 is visible in the finally charged capacitor. 1 CV0 The balance, i.e., CV0 CV0 must then dissipate in the resistor In Process 1, the energy stored in the capacitor E C and heat dissipated across resistance E D are related by : 1 (A) EC ED (B) EC ED (C) EC ED (D) EC ED n 16. (C) In process 1, the charge passing through the seat of emf = CV 0. And, whole of it passes across potential difference V 0, making the cell spend energy = CV 0 Out of this, i.e., I 1 CV0 1. C CV0 is stored in the capacitor, So, the energy dissipated in the resistor E d = E c = E d Paragraph for Q. No. 17 & CV0 CV0 CV0. 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 0. 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 ). The coefficient of friction between the ring and the finger is and the acceleration due to gravity is g 0517/IITEQ17/Paper/QP&Soln/Pg.14

15 IIT JEE 017 Advanced : Question Paper & Solution (Paper II) (15) 17. The minimum value of 0 below which the ring will drop down is g g g g (A) (B) (C) (D) (R r) (R r) (R r) (R r) 17. (B) Centre of mass of ring moves with angular velocity 0 in a circle of radius Rr centred about the centre of the circle of radius r. N = M (R r) Also N = Mg M (R r) Mg 0 = 0 o g (R r) 18. The total kinetic energy of the ring is 1 (A) M 0 (R r) (B) M 0 (R r) (C) M 0 (R r) (D) M 0R 18. (C) Kinetic energy of rolling ring = Mv M (R r). 0 PART II : CHEMISTRY SECTION 1 (Maximum Marks:1) This section contains SEVEN questions. Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is correct. For each question, darken the bubble corresponding to the correct option in the ORS. For each question, marks will be awarded in one of the following categories: Full Marks : + If only the bubble corresponding to the correct option is darkened. Zero Marks : 0 If none of the bubbles is darkened. Negative Marks : 1 In all other cases. 19. For the following cell, Zn(s) ZnSO 4 (aq) CuSO 4 (aq) Cu(s) when the concentration of Zn + is 10 times the concentration of Cu +, the expression for G (in J mol 1 ) is [F is Faraday constant ; R is gas constant ; T is temperature ; E(cell) = 1.1 V] (A).0 RT. F (B).0 RT F (C) 1.1 F (D). F 19. (A) Zn(s) ZnSO 4 (aq) CuSO 4 (aq) Cu(s) Given : [Zn + ] = 10 [Cu + ] (aq) (aq) Zn(s) Cu Zn Cu(s) G = G RT log 10 Q = nfe RT log 10 Q [Zn ] Q = = 10 [Cu ] = F(1.1) +.0 RT log 10 (10) =. F +.0 RT 0517/IITEQ17/Paper/QP&Soln/Pg.15

16 (16) Vidyalankar : IIT JEE 017 Advanced : Question Paper & Solution 0. The order of basicity among the following compounds is (A) I > IV > III > II (B) II > I > IV > III (C) IV > I > II > III (D) IV > II > III > I 0. (C) Basicity order : IV > I > II > III N H NH NH NH > > > N NH CH NH HN N 1. The standard state Gibbs free energies of formation of C (graphite) and C(diamond) at T = 98 K are f G[C(graphite)] = 0 kj mol 1 f G [C(diamond)] =.9 kj mol 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 10 6 m mol 1. If C(graphite) is converted to C(diamond) isothermally at T = 98 K, the pressure at which C(graphite) is in equilibrium with C(diamond), is [Useful information : 1J = 1 kg m s ; 1Pa = 1 kg m 1 s ; 1 bar = 10 5 Pa] (A) bar (B) 9001 bar (C) 1450 bar (D) bar 1. (A) C (graphite) C (diamond) G = f G (diamond) f G (graphite) G =.9 0 =.9 kj/mol V = 10 6 m mol 1 Isothermally ie T = 0 E = 0 At equilibrium, G = 0 G = PV.9 10 = P( 10 6 ).9 10 P 1 = 6 10 = pascal 1 bar = 10 5 pa P = ( bar + 1) P = bar. The major product of the following reaction is 0517/IITEQ17/Paper/QP&Soln/Pg.16

17 IIT JEE 017 Advanced : Question Paper & Solution (Paper II) (17) (A) (B) (C) (D). (C) OH NaNO, HCl 0C OH NH N Ar ESR H OH N N. The order of the oxidation state of the phosphorus atom in H PO, H PO 4, H PO, and H 4 P O 6 is (A) H PO 4 > H PO > H PO > H 4 P O 6 (B) H PO > H PO > H PO 4 > H 4 P O 6 (C) H PO > H PO > H 4 P O 6 > H PO 4 (D) H PO 4 > H 4 P O 6 > H PO > H PO. (D) Compound State Oxidation State H PO 4 +5 H 4 P O 6 +4 H PO + H PO Pure water freezes at 7 K and 1 bar. The addition of 4.5 g of ethanol to 500 g of water changes the freezing point of the solution. Use the freezing point depression constant of water as K kg mol 1. The figures shown below represent plots of vapour pressure (V.P.) versus temperature (T). [molecular weight of ethanol is 46 g mol 1 ] Among the following, the option representing change in the freezing point is (A) (B) 0517/IITEQ17/Paper/QP&Soln/Pg.17

18 (18) Vidyalankar : IIT JEE 017 Advanced : Question Paper & Solution (C) (D) 4. (C) T f = K f m = T f = T f = 7 = 70 K v.p. Pure solvent Solid Solution 70 7 T 5. Which of the following combination will produce H gas (A) Au metal and NaCN(aq) in the presence of air. (B) Cu metal and conc.hno (C) Fe metal and conc. HNO (D) Zn metal and NaOH(aq) 5. (D) Zn(s) + NaOH Na ZnO + H SECTION II (Maximum Marks:8) This section contains SEVEN questions. Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four options is(are) correct. For each question, darken the bubble(s) corresponding to all the correct option(s) in the ORS. For each question, marks will be awarded in one of the following categories: Full Marks : +4 If only the bubble(s) corresponding to all the correct option(s) is (are) darkened. Partial Marks : +1 For darkening a bubble corresponding to each correct option, provided NO incorrect option is darkened. Zero Marks : 0 If none of the bubbles is darkened. Negative Marks : In all other cases. For example, if (A), (C) and (D) are all the correct options for a question, darkening all these three will get +4 marks; darkening only (A) and (D) will get + marks; and darkening (A) and (B) will get marks, as a wrong option is also darkened. 6. The option(s) with only amphoteric oxides is (are) (A) Cr O, BeO, SnO, SnO (B) Cr O, CrO, SnO, PbO (C) ZnO, Al O, PbO, PbO (D) NO, B O, PbO, SnO 6. (A), (C) (A) Cr O, BeO, SnO and SnO are amphoteric oxides. (C) ZnO, Al O, PbO and PbO are amphoteric oxides. 0517/IITEQ17/Paper/QP&Soln/Pg.18

19 IIT JEE 017 Advanced : Question Paper & Solution (Paper II) (19) 7. The correct statement(s) about surface properties is(are) (A) Cloud is an emulsion type of colloid in which liquid is dispersed phase and gas is dispersion medium. (B) Adsorption is accompanied by decrease in enthalpy and decrease in entropy of the system. (C) The critical temperatures of ethane and nitrogen are 56 K and 16 K, respectively. The adsorption of ethane will be more than that of nitrogen on same amount of activated charcoal at a given temperature. (D) Brownian motion of colloidal particles does not depend on the size of the particles but depends on viscosity of the solution. 7. (B), (C) 8. For the following compounds, the correct statement(s) with respect to nucleophilic substitution reactions is(are) (A) I and III follow S N 1 mechanism (B) Compound IV undergoes inversion of configuration (C) I and II follow S N mechanism (D) The order of reactivity for I, III and IV is : IV > I > III 8. (A), (C), (D) Br S N 1 + S N Br S N I II 9. Among the following, the correct statement(s) is(are) (A) BH has the three-centre two-electron bonds in its dimeric structure (B) Al(CH ) has the three-centre two-electron bonds in its dimeric structure (C) AlCl has the three-centre two-electron bonds in its dimeric structure (D) The Lewis acidity of BCl is greater than that of AlCl 9. (A), (B), (D) H H B H H B H H CH Al CH CH Al CH CH CH Cl Cl Al Cl Cl Al Cl Cl 0517/IITEQ17/Paper/QP&Soln/Pg.19

20 (0) Vidyalankar : IIT JEE 017 Advanced : Question Paper & Solution 0. In a bimolecular reaction, the steric factor P was experimentally determined to be 4.5. The correct option(s) among the following is(are) (A) Experimentally determined value of frequency factor is higher than that predicted by Arrhenius equation (B) The value of frequency factor predicted by Arrhenius equation is higher than that determined experimentally (C) The activation energy of the reaction is unaffected by the value of the steric factor (D) Since P = 4.5, the reaction will not proceed unless an effective catalyst is used 0. (A), (C) 1. Compounds P and R upon ozonolysis produce Q and S, respectively. The molecular formula of Q and S is C 8 H 8 O. Q undergoes Cannizzaro reaction but not haloform reaction, whereas S undergoes haloform reaction but not Cannizzaro reaction. i) O /CH Cl ii) Zn/HO (i) P i) O /CH Cl ii) Zn/HO (ii) R Q (C8H8O) S (C8H8O) The option(s) with suitable combination of P and R, respectively, is(are) (A) (B) (C) (D) 1. (A), (B) O (A) (R) CH i) O /CHCl ii) Zn/HO C CH (S) Shows Haloform Reaction Doesnot gives Cannizzaro's Reaction O (R) CH i) O /CHCl ii) Zn/HO C CH (S) Shows Haloform Reaction Doesnot gives Cannizzaro's Reaction 0517/IITEQ17/Paper/QP&Soln/Pg.0

21 IIT JEE 017 Advanced : Question Paper & Solution (Paper II) (1) (B) CH i) O /CHCl ii) Zn/HO CHO CH CH (Q) CH O (R) CH i) O /CHCl ii) Zn/H CH O C CH (S) Methyl ketone gives. For a reaction taking place in a container Halaform in equilibrium Reaction with its surroundings, the effect of temperature on its equilibrium constant K in terms of change in entropy is described by (A) With increase in temperature, the value of K for exothermic reaction decreases because the entropy change of the system is positive (B) With increase in temperature, the value of K for endothermic reaction increases because unfavourable change in entropy of the surroundings decreases (C) With increase in temperature, the value of K for endothermic reaction increases because the entropy change of the system is negative (D) With increase in temperature, the value of K for exothermic reaction decreases because favourable change in entropy of the surroundings decreases. (B) (A) False (B) True (C) False (D) False SECTION III (Maximum Marks:1) This section contains TWO paragraphs. Based on each paragraph, there are TWO questions. Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is correct. For each question, darken the bubble corresponding to the correct option in the ORS. For each question, marks will be awarded in one of the following categories: Full Marks : + If only the bubble corresponding to the correct answer is darkened. Zero Marks : 0 If all other cases. PARAGRAPH 1 Upon heating KClO in the presence of catalytic amount of MnO, a gas W is formed. Excess amount of W reacts with white phosphorus to give X. The reaction of X with pure HNO gives Y and Z.. W and X are, respectively (A) O and P 4 O 10 (B) O and P 4 O 6 (C) O and P 4 O 6 (D) O and P 4 O Y and Z are, respectively (A) N O 5 and HPO (B) N O 4 and HPO (C) N O 4 and H PO (D) N O and H PO 4. (A) 4. (A) KClO (s) KCl (s) + 1.5O (g) (W) P 4(s) + O P 4 O 6 P 4 O 6 + O P 4 O 10 (X) P 4 O 10 + HNO N O 5 + H PO (X) (Y) (Z) 0517/IITEQ17/Paper/QP&Soln/Pg.1

22 () Vidyalankar : IIT JEE 017 Advanced : Question Paper & Solution PARAGRAPH The reaction of compound P with CH MgBr (excess) in (C H 5 ) O followed by addition of H O gives Q. The compound Q on treatment with H SO 4 at 0C gives R. The reaction of R with CH COCl in the presence of anhydrous AlCl in CH Cl followed by treatment with H O produces compound S. [Et in compound P is ethyl group] 5. The reactions, Q to R and R to S, are (A) Dehydration and FriedelCrafts acylation. (B) FriedelCrafts alkylation and FriedelCrafts acylation. (C) FriedelCrafts alkylation, dehydration and FriedelCrafts acylation. (D) Aromatic sulfonation and FriedelCrafts acylation. 6. The product S is (A) (B) (C) (D) 5. (B) 6. (C) (P) O (CH 5) O HO (excess) CH CH COEt CH MgBr H OH CH CH C CH (Q) CH HSO 4, 0C O CHC Cl anhyalcl C CH (R) O (S) 0517/IITEQ17/Paper/QP&Soln/Pg.

23 IIT JEE 017 Advanced : Question Paper & Solution (Paper II) () PART III MATHEMATICS SECTION 1 (Maximum Marks:1) This section contains SEVEN questions. Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is correct. For each question, darken the bubble corresponding to the correct option in the ORS. For each question, marks will be awarded in one of the following categories: Full Marks : + If only the bubble corresponding to the correct option is darkened. Zero Marks : 0 If none of the bubbles is darkened. Negative Marks : 1 In all other cases. 7. Three randomly chosen nonnegative integers x, y and z are found to satisfy the equation x + y + z = 10. Then the probability that z is even, is (A) 5 (B) (C) x + y + z = 10, x, y, z n(s) C C 66 If z is even. x + y + z = 10 x + y = 10 z 10z1 11z 1 1 z even zeven n(a) C C zeven (11 z) (C) 6 11 (D) 6 55 = = 6 P (z is even) = If f: is a twice differentiable function such that f (x) 0 for all x, and 1 1 f,f (1) 1, then (A) f (1) 1 (B) f (1) 0 (C) 1 f (1) 1 (D) 1 0 f (1) 8. (A) f(x) > 0 x concave upward f(x) is increasing. f(1) > f(1) > 1 1 1, (1,1) 9. The equation of the plane passing through the point (1, 1, 1) and perpendicular to the planes x + y z = 5 and x 6y z = 7, is (A) 14x + y + 15z = (B) 14x + y 15z = 1 (C) 14x + y + 15z = 1 (D) 14x y+15z = /IITEQ17/Paper/QP&Soln/Pg.

24 (4) Vidyalankar : IIT JEE 017 Advanced : Question Paper & Solution 9. (C) Plane passing (1, 1, 1) is a(x 1) + b(y 1) + c(z 1) = 0 (1) As plane (1) is r to both x + y z = 5 and x 6y z = 7 a + b c = 0 & a 6b c = 0 a b c Put in (1) 14(x 1) + (y 1) + 15 (z 1) = 0 14x + y + 15z = Let 0 be the origin and let PQR be an arbitrary triangle. The point S is such that OP OQ OR OS = OR OP OQ OS = OQ OR OP OS Then the triangle PQR has S as its (A) circumcentre (B) orthocenter (C) incentre (D) centroid 40. (B) OP OQ OR OS OR OP OQ OS P OP (OQ OR) OS (OR OQ) 0 (OQ OR) (OP OS) 0 r OQ OR (OP OS) S is Orthocentre of the PQR. Q S R 41. How many matrices M with entries from {0, 1, } are there, for which the sum of the diagonal entries of M T M is 5? (A) 16 (B) 198 (C) 16 (D) (B) a1 a a M = a4 a5 a 6 a 7 a8 a9 a1 a4 a7 a1 a a M T M = a a5 a 8 a4 a5 a 6 a a6 a9 a7 a8 a9 Sum of diagonals of M T M = a1 a4 a7 a a5 a8 a a6 a9 = 5 (given) Possible if number of s is 1 & number of 1s are 1 & number of 0 s are 7 = Total ways 9 C! number of s is 0 & number of 1s are 5 & number of 0 s are 4 = 9 C 5 Total ways = 9 C! 9 C /IITEQ17/Paper/QP&Soln/Pg.4

25 IIT JEE 017 Advanced : Question Paper & Solution (Paper II) (5) 4. Let S = {1,,,, 9}. For k = 1,,, 5, let N k be the number of subsets of S, each containing five elements out of which exactly k are odd. Then N 1 + N + N + N 4 + N 5 = (A) 10 (B) 5 (C) 15 (D) (D) N 1 = {only 1 odd} = 5 C 4 C = N = {only odd} = 5 C 4 C = 40 N = {only odd} = 5 C 4 C 60 N 4 = {only 4 odd} = 5 C 4 C 0 N 5 = {all odd} = 5 C 5 = 1 Total = If y = y(x) satisfies the differential equation 8 x 9 x dy = x dx, x > 0 and y(0) = 7, then y(56) = (A) 16 (B) 80 (C) (D) 9 4. (C) dy dx = 1 8 x 9 x 4 9 x dy = Put y = 4 9 x dy dx = x 9 x x dx 4 9 x 8 x 9 x y = 4 9 x C put x = 0, y = 7 7 = 7 C C = 0 y = 4 9 x y(56) = = = x 9 x x 0517/IITEQ17/Paper/QP&Soln/Pg.5

26 (6) Vidyalankar : IIT JEE 017 Advanced : Question Paper & Solution SECTION II (Maximum Marks:8) This section contains SEVEN questions. Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four options is(are) correct. For each question, darken the bubble(s) corresponding to all the correct option(s) in the ORS. For each question, marks will be awarded in one of the following categories: Full Marks : +4 If only the bubble(s) corresponding to all the correct option(s) is (are) darkened. Partial Marks : +1 For darkening a bubble corresponding to each correct option, provided NO incorrect option is darkened. Zero Marks : 0 If none of the bubbles is darkened. Negative Marks : In all other cases. For example, if (A), (C) and (D) are all the correct options for a question, darkening all these three will get +4 marks; darkening only (A) and (D) will get + marks; and darkening (A) and (B) will get marks, as a wrong option is also darkened. 1 x(1 1 x ) Let f(x) = cos for x 1. Then 1x 1x (A) lim f(x) does not exist (B) x1 lim x1 f(x) does not exist (C) lim f(x) = 0 (D) x1 lim f(x) = 0 x1 44. (B), (C) 1 x(1 1 x ) 1 f(x) = cos 1x 1x 1 (1 h)(1 1 (1 h) ) 1 L.H.L. = lim cos h0 1 (1 h) 1 (1 h) 1 (1 h ) 1 = lim cos h0 h h = lim h cos 1 h 0 h = 0 1 (1 h)(1 1 (1 h) ) 1 R.H.L. = lim cps h0 1 (1 h) 1 (1 h) 1 (1 h)(1 h) 1 = lim cos h0 h h h h 1 = cos h h 1 = ( h) cos = doesn t exist. h 45. Let and be nonzero real numbers such that (cos cos ) + cos cos = 1. Then which of the following is/are true? (A) tan tan = 0 (B) tan tan = 0 (C) tan tan 0 (D) tan tan /IITEQ17/Paper/QP&Soln/Pg.6

27 IIT JEE 017 Advanced : Question Paper & Solution (Paper II) (7) 45. (C), (D) (cos cos ) + cos cos = 1 cos = 1 cos cos 1 tan tan = 1 tan tan tan tan = (Componendo-Dividendo) tan = tan 46. If f(x) = cos(x) cos(x) sin(x) cos x cos x sin x, then sin x sin x cos x (A) f(x) = 0 at exactly three points in (, ) (B) f(x) = 0 at more than three points in (, ) (C) f(x) attains its maximum at x = 0 (D) f(x) attains its minimum at x = (B), (C) cos(x) cos(x) sin(x) f(x) = cos(x) cos(x) sin(x) sin(x) sin(x) cos(x) 0 cos(x) sin(x) cos(x) cos(x) sin(x) (C C C ) = sin(x) cos(x) = cos(x) [cos(x) cos(x) sin(x) sin(x)] f(x) = cos x cos(x) = cos(4x) + cos(x) f(x) = 4 sin(4x) sin(x) = 0 = sin(x) [4 cos(x) + 1] = 0 1 cos(x) = or sin(x) = 0 4 f(x) = 0 at more than three points in (, ) f(0 ) > 0 and f(0 + ) < 0 f(x) attains maxima at x = If the line x = divides the area of region R = {(x, y) : x y x, 0 x 1} into two equal parts, then (A) = 0 (B) = 0 1 (C) 0 (D) /IITEQ17/Paper/QP&Soln/Pg.7

28 (8) Vidyalankar : IIT JEE 017 Advanced : Question Paper & Solution 47. (A), (D) 1 (x x ) dx (x x ) dx , 1 But < 1 1 = 1 = If I = 98 k1 k1 k k1 dx, then x(x 1) 49 (A) I (B) I > log e 99 (C) I > (D) I < log e (B), (C) I = = = = 98 k1 k1 k 98 k1 (k 1) dx x(x 1) k1 1 1 (k 1) dx x x 1 k 98 k1 98 k 1 k (k 1) log log k k 1 k 1 k k (k 1)log k log log k k 1 k 1 e e e k k 1 k = (k 1) log k log log e(k) log e(k 1) k1 k k 1 k = 99 loge log log e(1) log e(99) = log e() log e(99) loge log e(1) log e(99) 50 (B, C) options are true. 0517/IITEQ17/Paper/QP&Soln/Pg.8

29 IIT JEE 017 Advanced : Question Paper & Solution (Paper II) (9) 49. If f: is a differentiable function such that f (x) f(x) for all x, and f(0) = 1, then (A) f(x) is decreasing in (0, ) (B) f (x) < e x in (0, ) (C) f(x) is increasing in (0, ) (D) f(x) > e x in (0, ) 49. (C), (D) f:, f(0) = 1 f (x) f(x) x x e f (x) e. f (x) 0 d x e.f (x) 0 dx e x f(x) is increasing. Let F(x) = e x f(x) F(0) = 1 So, F(x) > 1, for x > 0 e x. f(x) > 1 f(x) > e x Hence option (D) f (x) As F(x) = x e As e x is increasing and F(x) is also increasing. f(x) is also increasing for x > 0. sin(x) 1 sin sin x 50. If g(x) = (t) dt, then (A) g (B) g (C) g (D) g 50. (A), (C) g(x) = = 1 1 sin (sin x) (cos x) sin (sin x).cos x 1 x(cos x) sin (sin x) cos x g = 4 ( 1) 0 g Note : We feel that 1 sin (sin x) is x for given value of x. SECTION III (Maximum Marks:1) This section contains TWO paragraphs. Based on each paragraph, there are TWO questions. Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is correct. For each question, darken the bubble corresponding to the correct option in the ORS. For each question, marks will be awarded in one of the following categories: Full Marks : + If only the bubble corresponding to the correct answer is darkened. Zero Marks : 0 If all other cases. 0517/IITEQ17/Paper/QP&Soln/Pg.9

30 (0) Vidyalankar : IIT JEE 017 Advanced : Question Paper & Solution PARAGRAPH 1 Let O be the origin, and OX, OY, OZ be three unit vectors in the directions of the sides QR, RP, PQ, respectively, of a triangle PQR 51. OX OY = (A) sin(p + R) (B) sin R (C) sin(q + R) (D) sin (P + Q) 51. (D) R r P r Q r OX OY QR RP sin( R) = sinp Q R R [P + Q + R = ] = sinp Q 5. If the triangle PQR varies, then the minimum value of cos(p + Q) + cos(q + R) + cos(r + P) is (A) (B) 5 (C) 5. (A) cos(p Q) cos(q R) cos(r P) cos R cos P cosq cosp cosq cosr r 1 R 1 1 = As we know R r 1 r R 1 r R r 1 R (D) 5 PARAGRAPH Let p, q be integers and let, be the roots of the equation, x x 1 = 0, where. For n = 0, 1,,., let a n = p n + q n. FACT: If a and b are rational numbers and a b 5 0, then a = 0 = b. 0517/IITEQ17/Paper/QP&Soln/Pg.0

31 IIT JEE 017 Advanced : Question Paper & Solution (Paper II) (1) 5. If a 4 = 8, then p + q = (A) 7 (B) 1 (C) 14 (D) 1 5. (D) x x 1 = 0 + = 1, x = a 0 = p + q a 1 = p + q a = p + q a = p + q a 4 = p 4 + q 4 8 = = 1 = p q p 9 = p = q = P = p7 5 q = 7(p + q) + (p q) 5 Equate 7(p + q) = 56 & (p q) = 0 p + q = 8 p = q p = 4 So, p + q = = a 1 = (A) a 11 + a 10 (B) a 11 a 10 (C) a 11 + a 10 (D) a 11 + a (A) a n = 4 n + 4 n = 4( n + n ) a 1 = a 11 + a 10 = 1 1 4[ ] a 11 = [ ] = [ ], a 10 = [ ] [ ( 1) ( 1)] = 4[ ( ) ( ) ] [as is root of x x 1 = 0 = + 1 & = + 1] 1 1 = 4[ ] = a /IITEQ17/Paper/QP&Soln/Pg.1

32 () Vidyalankar : IIT JEE 017 Advanced : Question Paper & Solution 0517/IITEQ17/Paper/QP&Soln/Pg.