Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi Ph.: Fax : Answers & Solutions. Time : 3 hrs. Max.

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1 DATE : /5/7 CODE Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-5 Ph.: Fax : Answers & Solutions Time : hrs. Max. Marks: 8 for JEE (Advanced)-7 PAPER - (Code - ) INSTRUCTIONS QUESTION PAPER FORMAT AND MARKING SCHEME :. The question paper has three parts : Physics, Chemistry and Mathematics.. Each part has three sections as detailed in the following table : Section Question Type Single Correct Option One or more correct option(s) Comprehension Number of Questions Full Marks 7 + If only the bubble corresponding to the correct option is darkened 7 +4 If only the bubble(s) corresponding to all the correct option(s) is(are) darkened 4 + If only the bubble corresponding to the correct answer is darkened Category-wise Marks for Each Question Partial Marks Zero Marks Negative Marks If none of the In all other bubbles is cases darkened + For darkening a bubble corresponding to each correct option, provided NO incorrect option is darkened If none of the bubbles is darkened In all other cases In all other cases Maximum Marks of the Section 8

2 JEE (ADVANCED)-7 (PAPER-) CODE- PHYSICS This section contains SEVEN questions. SECTION - (Maximum Marks : ) 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 : If none of the bubbles is darkened Negative Marks : In all other cases. 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 d density is constant. The velocity v of any point on the surface of the expanding sphere is proportional dt to (A) R (B) R (C) R (D) R / Answer (A) 4 Sol. M R 4 dr R R dt Dividing by dr d R R dt dt d dt dr R R K dt dr R dt. 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 h h h

3 JEE (ADVANCED)-7 (PAPER-) CODE- (A) hsin n (C) hsin n Answer (B) /n (B) (D) h cos n htan n Sol. h n sincos n l h l sin = l h = h sin = h cos n hc. A photoelectric material having work-function is illuminated with light of wavelength. The fastest photoelectron has a de-broglie wavelength d. A change in wavelength of the incident light by result in a change d in d. Then the ratio d / is proportional to (A) d / (B) / (C) d / (D) d / Answer (D) d Sol. d h hc m hc h m hc h m f d h mhc d d d dd d d k

4 JEE (ADVANCED)-7 (PAPER-) CODE- 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 I 4a (A) I 6 4a (B) I 6 4a I (C) I 4a (D) 4a Answer (A) Sol. Considering one section out of symmetric star shaped conducting wire loop. From geometry : O (Center of loop) a a I Magnetic field at the center of the loop due to all identical sections is additive in nature. I B net = cos cos I = 6 4a 4a 5. 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 br. The general relation among vectors P, Q and S is Y P b R P S Q S R QP Q (A) S bp bq (C) S b P bq Answer (A) Sol. S = P b R Rˆ R = P b R = P br R O X (B) S bp bq (D) S b P b Q = P bq P = ( bp ) bq 4

5 JEE (ADVANCED)-7 (PAPER-) CODE- 6. 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 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 = 4 km s (C) v s = 6 km s (D) v s = 7 km s Answer (B) 4 m r =.5 R Sol. v s E S M = M M = 5 M loss in KE = Gain in PE GM m R GM m r mvs GM G M vs 4 R.5 R GM vs =. 4.4 km/s R 5 4 km/s 7. 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 ms. Then the fractional error in the measurement, L/L, is closest to (A).% (B) % (C) % (D) 5% Answer (B) Sol. t = L g t = L V T = t + t L t t T = L g L V T = L L g L V. = L 5. = L. = (5 ) L L = L = L 6 = %. 6 5

6 This section contains SEVEN questions. SECTION - (Maximum Marks : 8) JEE (ADVANCED)-7 (PAPER-) CODE- 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 : + 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 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 uniform magnetic field B exists in the region between x = 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 at point P (y = R). Which of the following option(s) is/are correct? Region y Region Region B O +Q P P x ( y= R) R/ (A) For p B, the particle will re-enter region QR (B) For 8 p B QR, the particle will enter region through the point P on x-axis (C) 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/ (D) For a fixed B, particles of same charge Q and same velocity v, the distance between the point P and the point of re-entry into region is inversely proportional to the mass of the particle Answer (A, B) 6

7 JEE (ADVANCED)-7 (PAPER-) CODE- Sol. The particle will follow circular trajectory inside the magnetic field region. The magnetic field cannot change the magnitude of velocity and momentum. R For longest possible path, the radius of circular motion can be. O P P At farthest point from y-axis, the momentum is directed upwards. p p The radius and hence separation between p and re-entry point is proportional to m, if Q, v, B are same. The particle will return to region only if it completes the half circle. R r mv B R p R QB p B QR 8 If B p ; r p R QR QB 8 r rr - cos O P R P It passes through point P if r r cos = R R sin r R 5 8 R R = R 7

8 9. The instantaneous voltages at three terminals marked X, Y and Z are given by V sin x V t, sin VY V t and 4 VZ V sin t JEE (ADVANCED)-7 (PAPER-) CODE- 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) V rms YZ V (C) V V (D) Independent of the choice of the two terminals rms XY Answer (A, D) Sol. V XY = V sin t V sin t V 6 V V V V rms VXY V YZ = V sin 4 t V sin t V V 6 V V V rms VYZ A, D.. 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? Q R 8

9 JEE (ADVANCED)-7 (PAPER-) CODE- (A) The electric flux passing through the curved surface of the hemisphere is Q Q (B) Total flux through the curved and the flat surfaces is (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 Answer (A, D) Sol. Net flux through curved surface and flat surface = Q R 45º Curved = Plane R Q cos Q The circumference points are equidistant from Q All points will be at the same potential. Option (A) and (D) are correct.. Two coherent monochromatic point sources S and S of wavelength = 6 nm are placed symmetrically on either side of the centre 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? P S S P d (A) A dark spot will be formed at the point P (B) At P the order of the fringe will be maximum (C) The total number of fringes produced between P and P in the first quadrant is close to (D) The angular separation between two consecutive bright spots decreases as we move from P to P along the first quadrant Answer (B, C) 9

10 Sol. d =.8 m = 8 4 m JEE (ADVANCED)-7 (PAPER-) CODE- P P S S P d and = 6 7 m Path difference at point P (as shown) x = S P S P = d sin, where angle is measured from vertical line as shown. For bright fringe d sin = m...(i) Point P is the point of central maxima. At point P, path difference (x) = d If P is the point of bright fringe, then d d m m On differentiating equation (i) d cos () = (m) = constant for consecutive bright fringe cos as varies from to. 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 options is/are correct? S + V R L L (A) After a long time, the current through L will be V L RL L (B) After a long time, the current through L will be V L RL L (C) The ratio of the currents through L and L is fixed at all times (t > ) (D) At t =, the current through the resistance R is V R Answer (A, B, C)

11 JEE (ADVANCED)-7 (PAPER-) CODE- Sol. Final current through battery = V R Current through L = V L R L L Current through L = V L R L L At t = current through source = zero At any time i = V V e R R tr LL L L Current through L = L i L L i Current through L = il L L i i i L L. 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 L B O (A) The midpoint of the bar will fall vertically downward (B) The trajectory of the point A is a parabola (C) Instantaneous torque about the point in contact with the floor is proportional to sin (D) When the bar makes an angle with the vertical, the displacement of its midpoint from the initial position is proportional to ( cos) Answer (A, C, D) l Sol. Torque about O, at any instant is mg. sin.

12 Option (C) JEE (ADVANCED)-7 (PAPER-) CODE- mg O l As no external force acts along x-axis, therefore centre of mass will fall vertically downward. Option (A) P (,) x y l cos O l l Displacement of centre of mass along y-axis l cos Option (D) l x sin, y lcos x y l l Trajectory is not parabola 4. 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? S P Q R X

13 JEE (ADVANCED)-7 (PAPER-) CODE- (A) If the force is applied at point P tangentially then decreases continuously as the wheel climbs (B) If the force is applied normal to the circumference at point X then is constant (C) If the force is applied normal to the circumference at point P then is zero (D) If the force is applied tangentially at point S then but the wheel never climbs the step Answer (B, C) Sol. Correct options (B, C) For option (C): Applied force passes through point Q. So, its torque is zero. For option (B): [ Treating magnitude of force constant] Torque due to applied force at x remains constant. SECTION - (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(s) 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 : In all other cases Consider a simple RC circuit as shown in Figure. PARAGRAPH Process : In the circuit the switch S is closed at t = and the capacitor is fully charged 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. Process : In a different process the voltage is first set to V and maintained for a charging time T >> RC. Then V 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 and the capacitor is charged to the same final voltage V as in Process. These two processes are depicted in Figure. S V v Process V + R C V / V / Process T>> RC Figure T Figure T t

14 JEE (ADVANCED)-7 (PAPER-) CODE- 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 In (C) E C = E D (D) E C = E D Answer (A) Sol. Final charge on capacitor = CV W b = CV E c = CV E D = W b E c = CV CV = CV E C E D 6. In Process, total energy dissipated across the resistance E D is : (A) ED CV (B) ED CV (C) ED CV (D) ED CV Answer (C) Sol. E D = W b V = CV V V V CV = CV V V V CV CV = V CV = CV CV = 6 4

15 JEE (ADVANCED)-7 (PAPER-) CODE- PARAGRAPH 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 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. R r R Figure Figure 7. The total kinetic energy of the ring is (A) M R (B) M R r (C) M R r (D) M R r Answer (A) Sol. k Mv I c c M R r MR ( ) V C = ( R r) r M R M R R R r r << R r R k M R M R M R 8. The minimum value of below which the ring will drop down is g (A) R r g (C) R r g (B) R r g (D) R r Answer (A) 5

16 JEE (ADVANCED)-7 (PAPER-) CODE- Sol. N f Mg N = M (R r) f = Mg f N Mg M (R r) R g r END OF PHYSICS 6

17 JEE (ADVANCED)-7 (PAPER-) CODE- CHEMISTRY This section contains SEVEN questions. SECTION - (Maximum Marks : ) 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 : If none of the bubbles is darkened Negative Marks : In all other cases 9. Pure water freezes at 7 K and bar. The addition of 4.5 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) V.P./bar Ice Water Water + Ethanol (B) V.P./bar Ice Water Water + Ethanol 7 7 T/K 7 7 T/K (C) V.P./bar Ice Water Water + Ethanol (D) V.P./bar Ice Water Water + Ethanol 7 7 T/K 7 7 T/K Answer (C) Sol. T f = W ikf M W 4.5 = 46 5 = K 7 (K) T f = (K) T f = 7 K Also, with decrease in temperature, V.P. decreases. Graph (C) is correct. 7

18 . For the following cell, JEE (ADVANCED)-7 (PAPER-) CODE- 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).F (B).RT.F (C).RT +.F (D).F Answer (B) Sol. Zn ZnSO 4 CuSO 4 Cu (aq) (aq) G =G + RT In Q G = G +. RT log Q [Zn ] Q [Cu ] G = nf E Cell = F. =. F G =. F +. RT log G. RT. F. 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 its volume by 6 m 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 ; Pa = kg m s ; bar = 5 Pa] (A) 45 bar (B) 58 bar (C) 45 bar (D) 9 bar Answer (A) Sol. Gº = V P 6 9 P 6 9 P Pa P F = 45 bar P F = 45bar 8

19 JEE (ADVANCED)-7 (PAPER-) CODE-. Which of the following combination will produce H gas? (A) Fe Metal and conc. HNO (B) Cu metal and conc. HNO (C) Zn metal and NaOH(aq) (D) Au metal and NaCN(aq) in the presence of air Answer (C) Sol. Zn + NaOH Na ZnO + H Iron become passive with conc. HNO. Copper liberate NO with HNO. 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 > H PO > H PO 4 > H 4 P O 6 (B) H PO 4 > H PO > H PO > H 4 P O 6 (C) H PO 4 > H 4 P O 6 > H PO > H PO (D) H PO > H PO > H 4 P O 6 > H PO 4 Answer (C) Sol. Oxidation state H PO 4 P = + 5 H 4 P O 6 P = + 4 H PO P = + H PO P = + H PO 4 > H 4 P O 6 > H PO > H PO 4. The major product of the following reaction is OH i) NaNO, HCl, C ii) aq.naoh NH OH + ONa (A) (B) Cl OH NCl N = N OH (C) (D) Answer (C) N = N OH OH O Sol. NaNO + HCl aq.naoh C NH + N NCl + N N OH O N = N 9 N = N H

20 5. The order of basicity among the following compounds is (A) II > I > IV > III (C) IV > I > II > III Answer (C) NH HC NH I N II NH JEE (ADVANCED)-7 (PAPER-) CODE- NH HN N HN NH III IV (B) IV > II > III > I (D) I > IV > III > II NH Sol. HN NH IV Resonance with two NH groups increases electron density on 'N' of NH CH NH NH Lesser increase of electron density on = NH due to only one resonance with one NH N sp N H III This LPe is not available as it is involve in aromatic Sextet. 'N' is bonded to sp C on both sides. N N H II This LPe is not involve in aromaticity. So more available Also, 'N' is bonded to sp C on one side. IV > I > II > III This section contains SEVEN questions. SECTION - (Maximum Marks : 8) 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 : + 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 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.

21 JEE (ADVANCED)-7 (PAPER-) CODE- 6. 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 56 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 Answer (A, B) Sol. Adsorption is an exothermic process and is accompanied by decrease is entropy, H, S sys sys More is critical temperature (T c ), more are intermolecular forces of attraction. More is extent of adsorption. 7. 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 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 Answer (A, B, D) Sol. Whether reaction is endothermic or exothermic in forward direction increase in temperature cause intake of heat from surrounding to system in endothermic direction due to which entropy change in system is positive and S of surrounding is negative. 8. 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) The activation energy of the reaction is unaffected by the value of the steric factor (B) Experimentally determined value of frequency factor is higher than that predicted by Arrhenius equation (C) Since P = 4.5, the reaction will not proceed unless an effective catalyst is used (D) The value of frequency factor predicted by Arrhenius equation is higher than that determined experimentally Answer (A, B) Sol. Aexperimental Steric factor Acalculated Steric factor = 4.5 It means A experimental > A calculated [This seems that reaction occurs more quickly than particles collide, thus concept of steric factor was introduced] 9. For the following compounds, the correct statement(s) with respect to nucleophilic substitution reaction is(are) CH CH Br Br H C C Br Br I II CH III IV

22 JEE (ADVANCED)-7 (PAPER-) CODE- (A) I and III follow S N mechanism (B) I and II follow S N mechanism (C) Compound IV undergoes inversion of configuration (D) The order of reactivity for I, III and IV is : IV > I > III Answer (A, B, C) Sol. When medium is highly polar and protic I & III will follow S N. Hence, (A) is correct. (B) I & II will follow S N when medium is polar aprotic and nucleophile is strong in high concentration. Option (C) is correct as Br Nu S N Nu Inversion in case of S N. (D) is incorrect for both S N and S N conditions.. Among the following, the correct statement(s) is(are) (A) Al(CH ) has the three-centre two-electron bonds in its dimeric structure (B) BH 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 Answer (A, B, D) Sol. H H B e H e e H e B H H It has two c-e bonds. Cl AlCl Al Cl Cl Cl Al Cl Cl No c-e bond CH CH Al Al CH CH CH CH Has two c-e bonds. Also BCl is stronger lewis acid than AlCl.. The option(s) with only amphoteric oxides is(are) (A) Cr O, BeO, SnO, SnO (B) Cr O, CrO, SnO, PbO (C) NO, B O, PbO, SnO (D) ZnO, AI O, PbO, PbO Answer (A, D) Sol. ZnO, Al O, PbO, PbO, Cr O, BeO, SnO and SnO are amphoteric oxides. NO is neutral oxide CrO is basic oxide B O is acidic oxide

23 JEE (ADVANCED)-7 (PAPER-) CODE-. 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) P i) O /CHCI Q ii) Zn/H O (C8H8O) (ii) R i) O /CHCI S ii) Zn/H O (C8H8O) The option(s) with suitable combination of P and R, respectively, is(are) (A) HC and CH CH HC (B) HC and CH CH (C) HC CH CH and CH CH CH (D) HC and Answer (A, B) HC CH Sol. CH (i) O /CH Cl (ii) Zn/H O CH O O C H + H C H Undergoes Cannizzaro's reaction CH C CH O /CH Cl Zn/H O Undergoes Haloform reaction O O C + H C H CH CH O/CHCl Zn/H O O C H O + CH C H CH CH Undergoes Cannizzaro's reaction CH CH O /CH Cl Zn/H O C + CH CH O CH Undergoes Haloform reaction CH CH O

24 This section contains TWO Paragraphs. Based on each paragraph, there are TWO questions. SECTION - (Maximum Marks : ) JEE (ADVANCED)-7 (PAPER-) CODE- Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is correct. For each question, darken the bubble(s) 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 : In all other cases PARAGRAPH Upon heating KCIO 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 6 (B) O and P 4 O 6 (C) O and P 4 O (D) O and P 4 O Answer (C) 4. Y and Z are, respectively (A) N O and H PO 4 (B) N O 5 and HPO (C) N O 4 and HPO (D) N O 4 and H PO Answer (B) Solutions of Q.No () & (4) [MnO ] Sol. KClO KCl O (W) 5O P4 P4O (X) P4O 4HNO NO54HPO (X) (Z) 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 ºC 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] (H C) C CO Et Q R S P 4

25 JEE (ADVANCED)-7 (PAPER-) CODE- 5. The product S is (H C) C HC CH COCH (A) (B) (H C) C CH (C) (H C) C HCOC COCH HC CH (D) (H C) C HO S O CH Answer (A) 6. The reactions, Q to R and R to S, are (A) Dehydration and Friedel-Crafts acylation (B) Aromatic sulfonation and Friedel-Crafts acylation (C) Friedel-Crafts alkylation, dehydration and Friedel-Crafts acylation (D) Friedel-Crafts alkylation and Friedel-Crafts acylation Answer (C) Solutions of Q. 5 and 6 O C(CH ) C OEt C(CH ) CH MgBr (excess)/(c H ) O 5 COCH + OMgBr CH C CH (P) H O H HO O H C(CH ) C(CH ) HSO 4 / C(CH ) CH C CH C H O (dehydration) (Q) CH C(CH ) C(CH ) C(CH ) CH (alkylation) (R) O CH C Cl/AlCl (Acylation) CH C(CH ) CH COCH (S) END OF CHEMISTRY 5

26 JEE (ADVANCED)-7 (PAPER-) CODE- This section contains SEVEN questions. MATHEMATICS SECTION - (Maximum Marks : ) 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 : If none of the bubbles is darkened Negative Marks : In all other cases 7. The equation of the plane passing through the point (,, ) and perpendicular to the planes x + y z = 5 and x 6y z = 7, is (A) 4x + y 5z = (B) 4x y + 5z = 7 (C) 4x + y + 5z = (D) 4x + y + 5z = Answer (C) Sol. Required equation of plane is x y z 6 4(x ) (y ) + ( 5)(y ) = 4x y 5y 8. 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) Incentre (D) Orthocenter Answer (D) Sol. OP OQ OR OS OR OP OQ OS OQ OR OP OS OP OQ OR OS OR OP OQ OS OP ( OQ OR) OS ( OR OQ) 6 Q P S R

27 JEE (ADVANCED)-7 (PAPER-) CODE- RQ ( OP OS) RQ SP RQ SP and similarly from OR OP OQ OS OQ OR OP OS SR PQ S is the orthocentre. 8 x 9 x dy 4 9 x dx, x and y 7, then 9. If y = y(x) satisfies the differential equation y(56) = (A) (B) 9 (C) 6 (D) 8 Answer (A) Sol. As, dx dy 8 x 9 x 4 9 x Integrating, y 4 9 x c at x =, y 7 c = so, y 4 9 x at x = 56, y = 4. If f : R R is a twice differentiable function such that f (x) > for all x R, and (A) f () (C) Answer (D) (B) f () f () (D) f () f, f (), then Sol. f( x), f and f() = f ( x) is always increasing B f() f f () / A / f () 7

28 Slope of tangent at B > Slope of chord AB. JEE (ADVANCED)-7 (PAPER-) CODE- 4. How many matrices M with entries from {,, } are there, for which the sum of the diagonal entries of M T M is 5? (A) 6 (B) 98 (C) 6 (D) 5 Answer (B) Sol. Let a b c a a a T M a b c, M b b b a b c c c c Sum of diagonal entries = ai bi ci i 5 Possible cases are (one, one and seven zeros) or (five 's and four 's) 9! 9! ! 5!4! 4. Let S = {,,,..., 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 + N 4 + N 5 = (A) (B) 5 (C) 5 (D) 6 Answer (D) Sol. Required number of subsets = 5 C 4 C C 4 C + 5 C 4 C + 5 C 4 4 C + 5 C 5 4 C = = 6 Alternate method Coefficient of x 5 in ( + x) 5 ( + x) 4 = 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 6 (A) 55 (B) (C) 6 8

29 JEE (ADVANCED)-7 (PAPER-) CODE- (D) 5 Answer (B) Sol. x + y + z = n(s) = + C = C = 66 Let z = n, where n =,,,, 4, 5 x + y + n = x + y = n Total such solution P(E) = ( n) 6 n This section contains SEVEN questions. SECTION - (Maximum Marks : 8) 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 : + 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 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. 44. If sin( x ) ( ) sin ( tdt ), sin x gx then (C) g (A) g (B) g (D) g Answer (No options is correct) Sol. g ( x ) sin sinx cos x sin sin x cos x g, g None of the given options is correct. 9

30 JEE (ADVANCED)-7 (PAPER-) CODE- 45. Let and be non-zero real numbers such that (cos cos) + cos cos =. Then which of the following is/are true? (A) tan tan (B) tan tan (C) tan tan (D) tan tan Answer (A, C) Sol. As (cos cos) = coscos cos cos cos Using componendo and dividendo cos cos cos cos tan tan So, tan tan Or tan tan 46. If f : R R is a differentiable function such that f(x) > f(x) for all x R, and f() =, then (A) f(x) is increasing in (, ) (B) f(x) is decreasing in (, ) (C) f(x) > e x in (, ) (D) f(x) < e x in (, ) Answer (A, C) Sol. f( x) f( x) x x e f( x) e f( x) d e x f ( x ) e dx e x f( x) e f( x) for all x (, ) x x f( x) is increasing function. x f( x) f( x) e

31 JEE (ADVANCED)-7 (PAPER-) CODE- f(x) is increasing Also as, f( x) f() f( x) x f( x) f( x) x x i.e., f( x) e x(, ) x e x(, ) 47. Let x( x) f( x) cos for x. x x Then (A) lim f( x) x (B) lim f( x) does not exist x (C) lim f( x) x (D) lim f( x) does not exist x Answer (A, D) Sol. x( x ) f( x) cos x x x( x ) lim cos ( x ) x x x x lim cos ( x ) x x x lim ( )cos x = a number lying between and Hence, limit does not exist. x( ( x)) lim cos ( x) x x x( x) lim cos ( x) x x lim( x)cos x x 48. If cos( x) cos( x) sin( x) f( x) cos x cos x sin x, then sin x sin x cos x (A) f(x) = at exactly three points in (, ) (B) f (x) = at more than three points in (, ) (C) f(x) attains its maximum at x = (D) f(x) attains its minimum at x =

32 Answer (B, C) JEE (ADVANCED)-7 (PAPER-) CODE- Sol. C CC cosx sinx f( x) cosx cosx sinx sinx cosx f( x) cos x(cosxcos x sin xsin x) f( x) cosxcosx; (f() = maximum at x = ) f( x) cos4x cosx f( x) sin x(4cos x ) sinx or x,, cosx 4 x,, and cosx gives 4 solutions in (, ) 4 Total number of solutions = If the line x = divides the area of region R x, y R : x y x, x into two equal parts, then (A) (B) (C) 4 4 (D) 4 4 Answer (B, C) Sol. ( x x ) dx ( x x ) dx y y = x y = x 4 4 x x x x x O (, ) (, ) x 4 4 x = 4 4

33 JEE (ADVANCED)-7 (PAPER-) CODE- 4 Let f ( ) 4 f (), f () f 8, 5. If I 98 k k k k dx, xx ( ) then (A) I > log e 99 (B) I < log e 99 (C) (D) 49 I 5 49 I 5 Answer (B, D) Sol. I = 98 k k dx ( k ) xx ( ) k 98 k 98 x k k ( k ) log ( k ) log log x k k k k k 98 k k ( k )log klog log( k ) logk k k k 99 99log log log99 log 99 99log log log e(99) SECTION - (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(s) 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 : In all other cases

34 PARAGRAPH- JEE (ADVANCED)-7 (PAPER-) CODE- 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. 5. OX OY (A) sin (P + Q) (B) sin R (C) sin (P + R) (D) sin (Q + R) Answer (A) Sol. OX OY QR RP pq pq sinr pq sin( P Q) 5. If the triangle PQR varies, then the minimum value of cos( P Q) cos( QR) cos( R P) is (A) (B) (C) 5 (D) 5 Answer (B) Sol. cos(p + Q) + cos(q + R) + cos(r + P) = (cosp + cosq + cosr) Maximum value of cosp + cosq + cosr = Hence minimum of (cosp + cosq + cosr) = 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 ab 5, then a = = b. 5. a = a (A) a (C) a (B) a a a (D) a a 4

35 JEE (ADVANCED)-7 (PAPER-) CODE- Answer (B) Sol. = = +...(i) and = +...(ii) Multiplying (i) by p and (ii) by q and adding, a = a + a 54. If a 4 = 8, then p + q = (A) (B) 4 (C) 7 (D) Answer (D) Sol. a 4 = p 5 q ( pq) 4 5( pq) 8 6 p = q = 4 END OF MATHEMATICS 5