JEE ADVANCED (PAPER 2) SOLUTIONS 2017 PART I : PHYSICS. SECTION 1 (Maximum Marks : 21)

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1 JEE ADVANCED (PAPER ) SOLUTIONS 7 PART I : PHYSICS SECTION (Maximum Marks : ) 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 : +3 If only the bubble corresponding to the correct option is darkened. Zero Marks : If none of the bubbles is darkened. Negative Marks : In all other cases.. 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 is 3 5 times heavier than the Earth and is at a distance.5 4 times larger than the radius of the Earth. The escape velocity from Earth s gravitational field is v e =. km s. The minimum initial velocity (v s ) 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) (A) v s = km s (B) v s = 7 km s (C) v s = 6 km s (D) v s = 4 km s. Solution: (D) GmM mv e e =...(i) R Here v e =. km/sec Now GmM e GmM s s R d e e 5 GmM Gm 3 M e e s 4 e e mv R.5 R v v km/sec s e Hence (D) v s = 4 km/s is correct GmM 3. R e e 3. mv e. A photoelectric material having work-function is illuminated with light of wavelength hc. The fastest photoelectron has a de Broglie wavelength d. A change in wavelength of the incident light by results in a change d in d. Then the ratio d / is proportional to (A) / (B) 3 (C) (D) 3 d d / d / d /. Solution: (D) hc KE Hence d max and h hc m d h m KE IIT KALRASHUKLA CIVIL LINES, KAKADEO, SOUTH KANPUR []

2 JEE ADVANCED (PAPER ) SOLUTIONS 7 h d hc m m hc h d m hc.. 3 d d h d 3 d 3. 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 =. seconds and he measures the depth of the well to be L = meters. Take the acceleration due to gravity g = ms and the velocity of sound is 3 ms. Then the fractional error in the measurement, L/L, is closest to (A) % (B) 3% (C).% (D) 5% 3. Solution: (A) L L T g V sound L T. L L g V. =. = sound L. L 3 L 3 L =.875 L.875 L % L L 4. 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) I 6 3 4a (B) I 3 3 4a (C) I 3 3 4a (D) I 6 3 4a IIT KALRASHUKLA CIVIL LINES, KAKADEO, SOUTH KANPUR []

3 JEE ADVANCED (PAPER ) SOLUTIONS 7 4. Solution: (D) d = a sin3 o = a B C due to each segment There are segments Hence B I a I 3 I 3 4a 8a 5. Consider an expanding sphere of instantaneous radius R whose total mass remains constant. The expansion is such that the instantaneous density remains uniform throughout the volume. The rate of fractional change in density d dt is constant. The velocity v of any point on the surface of the expanding sphere is proportional to (A) R 5. Solution: (C) 4 M. R 3 M 4 R 3 M d 4 dr..3r. dt 3 dt Hence dr dt R R (B) R 3 (C) R (D) R /3 6. Consider regular polygons with number of sides n = 3, 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 (A) n hsin 6. Solution: (C) (B) n htan (C) h cos n (D) hsin n Hence h sec h n h sec n h cos n IIT KALRASHUKLA CIVIL LINES, KAKADEO, SOUTH KANPUR [3]

4 JEE ADVANCED (PAPER ) SOLUTIONS 7 7. Three vectors P, Q and R 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, Q and S is S b P bq (B) (A) S b P bq (C) S bp b Q (D) 7. Solution: (B) Q P R S P br P b Q P S P b bq S b P bq SECTION (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 option(s) 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 : + For darkening a bubble corresponding to each correct option, provided NO incorrect option is darkened. Zero Marks : 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 result in +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 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 (B) Total flux through the curved and the flat surface is Q (C) The component of the electric field normal to the flat surface is constant over the surface (D) The circumference of the flat surface is an equipotential IIT KALRASHUKLA CIVIL LINES, KAKADEO, SOUTH KANPUR [4] Q

5 JEE ADVANCED (PAPER ) SOLUTIONS 7 8. Solution: (A, D) Q E. 4 R r Q R d = E.ds..r dr 4 R r R r QR R r dr d R r QR R r R QR R R Q 3/ Q Hence curved = Total flux will be zero The circumference of flat surface will be equipotential 9. 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 is motion is/are correct? (A) When the bar makes an angle with the vertical, the displacement of its midpoint from the initial position is proportional to ( cos) (B) Instantaneous torque about the point in contact with the floor is proportional to sin (C) The trajectory of the point A is a parabola (D) The midpoint of the bar will fall vertically downward 9. Solution: (A, D) L L L y cos cos Let coordinates of A(x, y) L x sin L L y cos Hence path will be circular. Instantaneous Torque due to mg and pseudo force will not be proportional to sin. IIT KALRASHUKLA CIVIL LINES, KAKADEO, SOUTH KANPUR [5]

6 JEE ADVANCED (PAPER ) SOLUTIONS 7. A uniform magnetic field B exists in the region between x = and x = 3R (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 at point P (y = R). (A) For B > p, the particle will re-enter region 3 QR (B) When the particle re-enters region through the longest possible path in region, the magnitude of the change in its linear momentum between point P and the farthest point from y-axis is p/ (C) For a fixed B, particles of same charge Q and same velocity v, the distance between the pint P and the point of re-entry into region is inversely proportional to the mass of the particle (D) For B = 8 p 3 QR, the particle will enter region 3 through the point P x-axis. Solution: (A, D) Particle will reach region if r < 3R p 3R QB P B 3QR Change in momentum will be equal to P distance between P and pint of sentry P d r m QB 3R r r R 9R r r R rr 4 3R rr 4 3R P r 8 qb 8P B 3QR IIT KALRASHUKLA CIVIL LINES, KAKADEO, SOUTH KANPUR [6]

7 JEE ADVANCED (PAPER ) SOLUTIONS 7. 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 pint Q. Which of the following option is/are correct (A) If the force is applied normal to the circumference at point P then is zero (B) If the force is applied normal to the circumference at point X then is constant (C) If the force is applied tangentially at point S then but the wheel never climbs the step (D) If the force is applied at point P tangentially then decreases continuously as the wheel climbs. Solution: (A, B) Considering force is constant in magnitude as per question.. Two coherent monochromatic point source S and S of wavelength = 6 m are placed symmetrically on either side of the center of the circle as shown. The sources are separated by a distance d =.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 (A) The total number of fringes produced between P and P in the first quadrant is close to 3 (B) A dark spot will be formed at the point P (C) The angular separation between two consecutive bright spots decreases as we more from P to P along the first quadrant. (D) At P the order of the fringe will be maximum. Solution: (A, D) x dcos, X at = 9 o min x max d.8mm n 3.8 n At P X max = n hence BF. x n dcos n d sin. if decrease increases. sin IIT KALRASHUKLA CIVIL LINES, KAKADEO, SOUTH KANPUR [7]

8 JEE ADVANCED (PAPER ) SOLUTIONS 7 3. A source of constant voltage V is connected to a resistance R and two ideal inductors L 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 =, the switch is closed and current begins to flow. Which of the following option is/are correct V L (A) After a long time, the current through L will be R L L (B) The ratio of the current through L and L is fixed at all times (t > ) V L (C) After a long time, the current through L will be R L L (D) At t =, the current through the resistance R is V/R 3. Solution: (A, B, C) I total = V R and it will be divided inversely proportional to L. V L V L Hence i., i. R L L R L L i i L L At t =, i = constant. 4. The instantaneous voltages at three terminals marked X, Y and Z are given by V V sin t X VY Vsin t 3 and 4 VZ Vsin t 3. 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 VXY V (B) independent of the choice of the two terminals (C) (D) V V rms XY 3 rms VXY V 4. Solution: (B, D) V XY = V sint V sin t 3. Vsin t Vsin t 3 IIT KALRASHUKLA CIVIL LINES, KAKADEO, SOUTH KANPUR [8]

9 JEE ADVANCED (PAPER ) SOLUTIONS 7 V 3 3 VXY V rms 4 VYZ V sin t V sin t Vsin t Vsin t 3 3 V YZrms V 3 SECTION 3 (Maximum Marks : ) 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 : +3 If only the bubble corresponding to the correct option is darkened. Zero Marks : In all other cases PARAGRAPH Consider a simple RC circuit as shown in figure. Process : In the circuit the switch S is closed at t = and the capacitor is fully charge to voltage V (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. v Process : In a different process the voltage is first set to and maintained for a charging time 3 v T>> RC. Then the voltage is raised to without discharging the capacitor and again 3 maintained for a time T >> RC. The process is repeated one more time by raising the voltage to V and the capacitor is charged to the same final voltage V as in Process. These two processes are depicted in Figure. 5. In Process, the energy stored in the capacitor E C and heat dissipated across resistance E D are related by: (A) E C = E D (B) E C = E D ln (C) E C = E D (D) E C = E D IIT KALRASHUKLA CIVIL LINES, KAKADEO, SOUTH KANPUR [9]

10 JEE ADVANCED (PAPER ) SOLUTIONS 7 5. Solution: (C) During charging, E C = E D = CV 6. In Process, total energy dissipated across the resistance E D is : (A) E 3CV (B) D 6. Solution: (C) From t = to t = T V D ED 3 CV E C CV 3 9 From t = T to t = T. V D E C CV 3 9 From t = T to t = E D CV 3 9 Hence ED ED E D E D CV 3 3 (C) E PARAGRAPH CV 3 D (D) E CV D One twirls a circular ring (of mass M and radius R) near the tip of one s finger as shown in Figure. 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. The rotating ring rolls without 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. 7. The total kinetic energy of the ring is (A) M R (B) 3 M R r (C) M R r (D) M R r 7. Solution: (BONUS) Centre of ring moves in a circular path of radius (R r) with same angular velocity o. This is motion of COM of ring and can be termed as Translatory Motion hence. KE COM = MVCM = M R r o As there is no slipping between finger and surface of ring hence angular speed of ring with respect to its COM is related with o as R = r o IIT KALRASHUKLA CIVIL LINES, KAKADEO, SOUTH KANPUR []

11 JEE ADVANCED (PAPER ) SOLUTIONS 7 hence = r R hence, KE ring, COM = o MR ' = M r KE total = KE ring, COM + KE COM = M R r r o o 8. The minimum value of below which the ring will drop down is (A) 3g R r 8. Solution: (D) M R r Mg o (B) g (R r) o Hence, (D) is correct answer. g R r (C) g Rr END OF PART I : PHYSICS (D) g Rr IIT KALRASHUKLA CIVIL LINES, KAKADEO, SOUTH KANPUR []

12 JEE ADVANCED (PAPER ) SOLUTIONS 7 PART II : CHEMISTRY SECTION (Maximum Marks : ) 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 : +3 If only the bubble corresponding to the correct option is darkened. Zero Marks : If none of the bubbles is darkened. Negative Marks : In all other cases. 9. Pure water freezes at 73 K and bar. The addition of 34.5 g of ethanol to 5 g of water changes the freezing point of the solution. Use the freezing point depression constant of water as K kg mol. The figures shown below represent plots of vapour pressure (V.P.) versus temperature (T). [molecular weight of ethanol is 46 g mol ] Among the following, the option representing change in the freezing point is (A) (B) (C) (D 9. Solution: (D) T f = K f m T f = T f = 3 So freezing point of solution = 7 K. The standard state Gibbs free energies of formation of C(Graphite) and C(diamond) at T = 98 K are f G [C(graphite)] = kj mol f G [C(diamond)] =.9 kj mol. The standard state means that the pressure should be bar, and substance should be pure at a given temperature. The conversion of graphite [C(graphite)] to diamond [C(diamond)] reduces IIT KALRASHUKLA CIVIL LINES, KAKADEO, SOUTH KANPUR []

13 JEE ADVANCED (PAPER ) SOLUTIONS 7 its volume by 6 m 3 mol. 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 : J = kg m s ; kg m s ; bar = 5 Pa] (A) 45 bar (B) 45 bar (C) 58 bar (D) 9 bar. Solution : (A) C graphite C diamond G =.9 KJ/mole P 6 = P = 9 Pa =.9 4 bar P = 45 bar P = 45 bar. For the following cell, Zn(s) ZnSO 4 (aq) CuSO 4 (aq) Cu(s) when the concentration of Zn + is times the concentration of Cu +, the expression for G (in J mol ) is [F is Faraday constant; R is gas constant; T is temperature; E (cell) =. V] (A).33 RT +. F (B).33 RT.F (C). F (D).F. Solution : (B) Zn + Cu + Zn + + Cu(s) E cell = E cell G = nf E cell G F.59 Zn log Cu =..59 G =.F +.59 F G =.33 RT. F log. The major product of the following reaction is (i) NaNO, HCl,C (ii) aq. NaOH (A) (B) (C) (D) IIT KALRASHUKLA CIVIL LINES, KAKADEO, SOUTH KANPUR [3]

14 JEE ADVANCED (PAPER ) SOLUTIONS 7. Solution: (B) 3. The order of basicity among the following compounds is (I) (II) (III) (IV) (A) II > I > IV > III (B) IV > I > II > III (C) IV > II > III > I (D) I > IV > III > I 3. Solution: (B) NH NH C NH is most basic as out of 3, nitrogen is always ready to donate lone pair and III is aromatic so least basic 4. Which of the following combination will produce H gas? (A) Au metal and NaCN(aq) in the presence of air (B) Fe metal and conc. HNO 3 (C) Zn metal and NaOH(aq) (D) Cu metal and conc. HNO 3 4. Solution: (C) Zn + NaOH Na ZnO + H 5. The order of the oxidation state of the phosphorus atom in H 3 PO, H 3 PO 4, H 3 PO 3, and H 4 P O 6 is (A) H 3 PO 4 > H 4 P O 6 > H 3 PO 3 > H 3 PO (B) H 3 PO 4 > H 3 PO > H 3 PO 3 > H 4 P O 6 (C) H 3 PO 3 > H 3 PO > H 3 PO 4 > H 4 P O 6 (D) H 3 PO > H 3 PO 3 > H 4 P O 6 > H 3 PO 4 5. Solution: (A) Oxidation no. of P in H 3 PO +, H 3 PO 4 + 5, H 3 PO & H 4 P O IIT KALRASHUKLA CIVIL LINES, KAKADEO, SOUTH KANPUR [4]

15 JEE ADVANCED (PAPER ) SOLUTIONS 7 SECTION (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 option(s) 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 : + For darkening a bubble corresponding to each correct option, provided NO incorrect option is darkened. Zero Marks : 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 result in +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. For a reaction taking place in a container in equilibrium 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 the entropy change of the system is negative. (C) With increase in temperature, the value of K for exothermic reaction decreases because favourable change in entropy of the surroundings decreases. (D) With increase in temperature, the value of K for endothermic reaction increases because unfavourable change in entropy of the surroundings decreases. 6. Solution: (D) G = H TS for exothermic reaction H < & on increasing temp. equilibrium constant decreases. for endothermic reaction H > & on increasing temp. equilibrium constant increases. 7. Among the following, the correct statement(s) is(are) (A) Al(CH 3 ) 3 has the three-centre two-electron bonds in its dimeric structure. (B) AlCl 3 has the three-centre two-electron bonds in its dimeric structure. (C) BH 3 has the three-centre two-electron bonds in its dimeric structure. (D) The Lewis acidity of BCl 3 is greater than that of AlCl 3 7. Solution: (A, C) AlCl 3 In dimer form has 3C 4e bond BCl 3 has less Lewis acidity due to back bonding. 8. The correct statement(s) about surface properties is(are) (A) Adsorption is accompanied by decrease in enthalpy and decrease in entropy of the system. (B) The critical temperatures of ethane and nitrogen are 563 K and 6 K, respectively. The adsorption of ethane will be more than that of nitrogen on same amount of activated charcoal at a given temperature. (C) Cloud is an emulsion type of colloid in which liquid is dispersed phase and gas is dispersion medium. (D) Brownian motion of colloidal particles does not depend on the size of the particles but depends on viscosity of the solution. IIT KALRASHUKLA CIVIL LINES, KAKADEO, SOUTH KANPUR [5]

16 JEE ADVANCED (PAPER ) SOLUTIONS 7 8. Solution: (A, B) For Adsorption H < < & S < Easily liquefiable gases will be more adsorbed on charcoal. Cloud is aerosol (Liquid is Gas colloidal solution). Brownian motion is independent of nature of colloid but depends on size of particle and viscosity of the solution. 9. For the following compounds, the correct statement(s) with respect to nucleophilic substitution reactions is (are) (I) (II) CH3 (III) H3C C Br (IV) CH3 (A) I and III follow S N mechanism (B) I and II follow S N mechanism (C) Compound IV undergoes inversions of configuration. (D) The order of reactivity for I, III and IV is : IV > I > III 9. Solution: (A, B, C, D) can give S N as well as S N mechanism gives S N mechanism CH3 CH3 C Br gives S N mechanism CH3 (III) can give retention as well as invertion Order of reactivity IV > I > III 3. 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) Since P = 4.5, the reaction will not proceed unless an effective catalyst is used. (B) The activation energy of the reaction is unaffected by the value of the steric factor (C) Experimentally determined value of frequency factor is higher than that predicted by Arrhenius equation. (D) The value of frequency factor predicted by Arrhenius equation is higher than that determined experimentally. 3. Solution: (B, C) Aexperimental Static Factor P = A Theoretical IIT KALRASHUKLA CIVIL LINES, KAKADEO, SOUTH KANPUR [6]

17 JEE ADVANCED (PAPER ) SOLUTIONS 7 3. The option(s) with only amphoteric oxides is (are) (A) Cr O 3, BeO, SnO, SnO (B) ZnO, Al O 3, PbO, PbO (C) Cr O, CrO, SnO, PbO (D) NO, B O 3, PbO, SnO 3. Solution: (A, B) CrO is basic and NO is neutral oxide. 3. 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 3 /CHCl (i) O 3 /CHCl (i) P Q (ii) R S (ii) Zn/HO (C8H8O) (ii) Zn/H O (C H O) The option(s) with suitable combination of P and R, respectively, is (are) 8 8 (A) (B) (C) (D) 3. Solution: (A, B) Q is or S is IIT KALRASHUKLA CIVIL LINES, KAKADEO, SOUTH KANPUR [7]

18 JEE ADVANCED (PAPER ) SOLUTIONS 7 SECTION 3 (Maximum Marks : ) 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 : +3 If only the bubble corresponding to the correct option is darkened. Zero Marks : In all other cases PARAGRAPH - Upon heating KClO 3 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 3 gives Y and Z. 33. Y and Z are, respectively (A) N O 3 H 3 PO 4 (B) N O 5 and HPO 3 (C) N O 4 and HPO 3 (D) N O 4 and H 3 PO Solution: (B) HNO 3 + P 4 O HPO 3 + N O 5 N O 5 is anhydride of HNO W and X are, respectively (A) O and P 4 O (B) O and P 4 O 6 (C) O 3 and P 4 O 6 (D) O 3 and P 4 O 34. Solution: (A) KClO 3 KCl + 3O P 4 + O (excess) P 4 O PARAGRAPH - The reaction of compound P with CH 3 MgBr (excess) in (C H 5 ) followed by addition of H O gives Q. The compound Q on treatment with H SO 4 at C gives R. The reaction of R with CH 3 COCl in the presence of anhydrous AlCl 3 in CH Cl followed by treatment with H O produces compound S. [Et in compound P is ethyl group] Q R S 35. The reactions, Q to R and R to S, are (A) Dehydration and Friedel-Crafts acylation (B) Friedel-Crafts alkylation, dehydration and Friedel-Crafts acylation (C) Aromatic sulfonation and Friedel-Crafts acylation. (D) Friedel-Crafts alkylation and Friedel-Crafts acylation. IIT KALRASHUKLA CIVIL LINES, KAKADEO, SOUTH KANPUR [8]

19 JEE ADVANCED (PAPER ) SOLUTIONS The product S is (A) (B) (C) (D) 35. Solution: (A) 36. Solution: (B) END OF PART II : CHEMISTRY IIT KALRASHUKLA CIVIL LINES, KAKADEO, SOUTH KANPUR [9]

20 JEE ADVANCED (PAPER ) SOLUTIONS 7 PART III : MATHEMATICS SECTION (Maximum Marks : ) 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 : +3 If only the bubble corresponding to the correct option is darkened. Zero Marks : If none of the bubbles is darkened. Negative Marks : In all other cases. 37. The equation of the plane passing through the point (,, ) and perpendicular to the planes x + y z = 5 and 3x 6y z = 7, is (A) 4x + y + 5z = 3 (B) 4x y + 5z = 7 (C) 4x + y 5z = (D) 4x + y + 5z = Solution: (A) x y z 3 6 4x y 5z Let O 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) Centroid (B) Circumcentre (C) Orthocentre (D) Incenter 38. Solution: (C) p.q r.s r.p q.s p s. q r PS QR Similarly QS PR and RS PQ S is orthocentre of PQR 39. Let S = {,, 3,..,9}. For k =,,, 5, let N k be the number of subsets of S, each containing five elements out of which exactly k are odd. Then N + N + N 3 + N 4 + N 5 = (A) 5 (B) 5 (C) (D) Solution: (D) N + N + N 3 + N 4 + N 5 = [Total number of subsets which contain exactly 5 elements] [N o ] = 9 C 5 = 6 4. Three randomly chosen non-negative integers x, y and z are found to satisfy the equation x + y + z =. Then the probability that z is even, is (A) (B) 6 (C) 36 (D) Solution: (B) x + y + z =, x, y, z Total number of solutions = C C 66 No. of solutions when z is even = Required probability = z z z4 z6 z8 z IIT KALRASHUKLA CIVIL LINES, KAKADEO, SOUTH KANPUR []

21 JEE ADVANCED (PAPER ) SOLUTIONS 7 4. How many 3 3 matrices M with entries from {,, } are there, for which the sum of the diagonal entries of M T M is 5? (A) 35 (B) 6 (C) 6 (D) Solution: (D) a p x Let M = b q y c r z t r (M T M) = a + b + c + p + q + r + x + y + z = 5 Total no. of ways = 9 P + 9 C 5 = = 98, where 9 P = No. of matrices when comes once, comes once and comes 7 times. 9 C 5 = No. of matrices when comes 5 times and comes 4 times. 4. If y = y(x) satisfies the differential equation 8 x 9 x dy 4 9 x dx, x And y() = 7, then y(56) = (A) 6 (B) 9 (C) 3 (D) 8 4. Solution: (C) dy dx 8 x 9 x 4 9 x y 56 dx dy 7 8 x 9 x 4 9 x 56 y x 3 7 y = If f : is a twice differentiable function such that f (x) > for all x, and f,f (), then (A) f () > (B) f () (C) < f () 43. Solution: (A) f (x) > and f, f() = Applying LMVT on, we get f () f f '(c) for some c, Also f (x) is a strictly increasing function (f (x) > ) f () > (D) < f () IIT KALRASHUKLA CIVIL LINES, KAKADEO, SOUTH KANPUR []

22 JEE ADVANCED (PAPER ) SOLUTIONS 7 SECTION (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 option(s) 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 : + For darkening a bubble corresponding to each correct option, provided NO incorrect option is darkened. Zero Marks : 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 result in +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. 44. Let and be nonzero real numbers such that (cos cos ) + cos cos. Then which of the following is/are true? (A) 3 tan tan (B) tan 3 tan (C) 3 tan tan (D) tan 3 tan 44. Solution: (A, C) (cos cos) + cos cos = cos cos + cos 3 cos = [use cos = cos ] = 3 sec sec 3 tan = tan [use sec = + tan ] sin(x) 45. If g(x) sin (t) dt, then sin x (A) g ' (B) g ' (C) g ' (D) g ' 45. Solution: (BONUS) g(x) = cos x sin (sin x) cos x sin (sin x) at x =, g(x) = ( x) cos x x cos x g ' at x =, g(x) = ( + x) cos x x cos x g ' 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) attains its maximum at x = (B) f (x) = at exactly three points in (, ) (C) f (x) = at more than three points in (, ) (D) f (x) attains its minimum at x = IIT KALRASHUKLA CIVIL LINES, KAKADEO, SOUTH KANPUR []

23 JEE ADVANCED (PAPER ) SOLUTIONS Solution: (A, C) cos x cos x sin x f(x) = cos x cos x sin x cos x cos 4x sin x sin x cos x f(x) attains its maximum when cos x = i.e., x = f (x) = sin x ( + 4 cos x) = at exactly 7 points in (, ) x( x ) 47. Let f(x) = cos for x. Then x x (A) lim f (x) (B) lim x (C) lim f (x) = (D) lim x x x f (x) does not exist f (x) does not exist 47. Solution: (AD) x( x ) f(x) = cos x x x( x) lim f (x) lim cos lim x cos x x x x x x x( x ) lim f (x) lim cos lim x cos does not exist x x x x x x 48. If the line x = divides the area of region R = {(x, y) : x 3 y x, x } into two equal parts, then (A) (B) (C) (D) Solution: (A, B) 3 3 (x x )dx (x x )dx = A root of above equation lies in, [apply IVT on x4 4x + in, ] 49. If f : is a differentiable function such that f (x) > f(x) for all x, and f() =, then (A) f(x) > e x in (, ) (B) f(x) is increasing in (, ) (C) f (x) < e x in (, ) (D) f(x) is decreasing in (, ) 49. Solution: (A, B) f (x) > f(x) e x f (x) e x f(x) > x e. f (x) ' g(x) = e x f(x) is an increasing function x > g(x) > g() e x f(x) > f(x) > e x Now f(x) > e x x (, ) f(x) > x (, ) f (x) > f(x) f (x) > x (, ) f is increasing on (, ) IIT KALRASHUKLA CIVIL LINES, KAKADEO, SOUTH KANPUR [3]

24 JEE ADVANCED (PAPER ) SOLUTIONS 7 5. If I 98 k k k k dx, then x(x) (A) I > log e 99 (B) I < log e 99 (C) I < Solution: (B, D) x + > k + [As k < x < k +] x k x(x ) x(k ) Also, k k dx dx x(x ) x(k ) k k 98 k 98 k (k )dx dx x(x ) x k k k k 98 k 98 (k )dx k ln x(x ) k k k k I < ln(99) k k or x(x) (k) k k k k dx dx x(x) 98 k I k 49 I 5 k [As k 98] (D) I > 49 5 SECTION 3 (Maximum Marks : ) 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 : +3 If only the bubble corresponding to the correct option is darkened. Zero Marks : In all other cases PARAGRAPH - Let O be the origin, and OX, OYand OZbe three unit vectors in the directions of the sides QR, RP, PQ, respectively, of a triangle PQR. 5. OX OY = (A) sin R (B) sin (P + R) (C) sin (P + Q) (D) sin (Q + R) IIT KALRASHUKLA CIVIL LINES, KAKADEO, SOUTH KANPUR [4]

25 JEE ADVANCED (PAPER ) SOLUTIONS 7 5. Solution: (C) OX OY OX. OY sin XOY = sin R = sin (P + Q) 5. If the triangle PQR varies, then the minimum value of cos(p + Q) + cos (Q + R) + cos (R + P) is (A) 5 5 (B) (C) 3 3 (D) Solution: (D) cos (P + Q) + cos (Q + R) + cos (R + P) = (cos P + cos Q + cos R) 3, Let x = (cos P + cos Q + cos R) P Q P Q R x = cos cos sin R P Q R sin cos sin ( x) P Q D 4cos 8( x) P Q x cos 3 3 x PARAGRAPH - Let p, q be integers and let, be the roots of the equation, x x =, where. For n =,,,.., let a n = p n + q n. Fact : If a and b are rational numbers and a + b 5 =, then a = = b. 53. a = (A) a + a (B) a + a (C) a + a (D) a a 53. Solution: (B) x x, and and p q a a a 54. If a 4 = 8, then p + q = (A) (B) 7 (C) 4 (D) IIT KALRASHUKLA CIVIL LINES, KAKADEO, SOUTH KANPUR [5]

26 JEE ADVANCED (PAPER ) SOLUTIONS Solution: (A) 4 3 a 4 = 8 p(3 + ) + q (3 + ) = p q p q p q 8 p q = and p + q = 8 p = 4, q = 4 p + q = 4 END OF PART III : MATHEMATICS IIT KALRASHUKLA CIVIL LINES, KAKADEO, SOUTH KANPUR [6]