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

Transcription

1 DATE : /5/7 CODE 4 egd. Office : Aakash Tower, 8, Pusa oad, New Delhi-5 Ph.: Fax : Answers & Solutions Time : hrs. Max. Marks: 8 for JEE (Advanced)-7 PAPE - (Code - 4) INSTUCTIONS QUESTION PAPE FOMAT AND MAKING 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 PHYSICS This section contains SEVEN questions. SECTION - (Maximum Marks : ) Each question has FOU 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 OS. 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. 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 6 km s (B) v s km s (C) v s 7 km s (D) v s 4 km s Answer (D) Sol. v s m r E M M 4.5 S M 5 M loss in KE Gain in PE GM m GM m r mvs + GM G M vs GM vs. 4.4 km/s 4 km/s. A person measures the depth of a well by measuring the time interval between dropping a stone and receiving the sound of impact with the bottom of the well. The error in his measurement of time is δt. 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, δl/l, is closest to (A) 5% (B) % (C) % (D).%

3 Answer (B) Sol. t L g t L V T t + t L t t T L g L + V ΔT L L g L Δ + V Δ. + ΔL ΔL (5 + ) ΔL ΔL. 6 Δ L L 6 %. Consider an expanding sphere of instantaneous radius whose total mass remains constant. The expansion is such that the instantaneous density ρ remains uniform throughout the volume. The rate of fractional change in dρ density is constant. The velocity v of any point on the surface of the expanding sphere is proportional ρ dt to (A) (C) / (D) Answer (A) 4 Sol. M π ρ 4 d π ρ+ dt Dividing by ρ d dρ + dt ρ dt dρ dt (B) d K dt d dt

4 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) μ I 4πa (B) μ I 6 4πa μ I (C) 4πa Answer (B) (D) μ I 6 + 4πa 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. μi B net [ cos + cos ] 4πa μi 6 4πa 5. 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 Δ. Then Δ depends on n and h as h h h (A) π Δ hsin n (B) π Δ hsin n (C) Δ h π cos n (D) π Δ h tan n 4

5 Answer (C) π/n Sol. h θ l π π θ n π sinθ cos n l h sinθ Δ l h h sinθ h π cos n 6. Three vectors P, Q and are shown in the figure. Let S be any point on the vector. The distance between the points P and S is b. The general relation among vectors P, Q and S is Y P b P S Q S Q P Q O X (A) S ( b) P + bq (C) ( ) S b P + b Q (B) S ( b ) P + bq S b P + bq (D) ( ) Answer (A) Sol. S P + b ˆ P + b P + b P + b( Q P) ( bp ) + bq 5

6 7. A photoelectric material having work-function φ is illuminated with light of wavelength λ hc λ<. The fastest φ photoelectron has a de-broglie wavelength λ d. A change in wavelength of the incident light by Δλ result in a change Δλ d in λ d. Then the ratio Δλ d /Δλ is proportional to (A) λd / λ (B) λd / λ (C) λ / λ (D) λ / λ d Answer (B) d Sol. λ d h hc φm λ hc h m φ λ λ d hc h m f λ λ d h mhc dλ λ λd dλd dλ λ k λ d This section contains SEVEN questions. SECTION - (Maximum Marks : 8) Each question has FOU options (A), (B), (C) and (D). ONE O MOE 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 OS. 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. 6

7 8. A wheel of radius and mass M is placed at the bottom of a fixed step of height as shown in the figure. A constant force is continuously applied on the surface of the wheel so that it just climbs the step without slipping. Consider the torque τ 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 X (A) If the force is applied at point P tangentially then τ decreases continuously as the wheel climbs (B) If the force is applied tangentially at point S then τ but the wheel never climbs the step (C) If the force is applied normal to the circumference at point P then τ is zero (D) If the force is applied normal to the circumference at point X then τ is constant Answer (C, D) Sol. Correct options (C, D) [ Treating magnitude of force constant] For option (C): Applied force passes through point Q. So, its torque is zero. For option (D): Torque due to applied force at x remains constant. 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 θ. Which of the following statements about its motion is/are correct? A L θ B O (A) When the bar makes an angle θ with the vertical, the displacement of its midpoint from the initial position is proportional to ( cosθ) (B) The midpoint of the bar will fall vertically downward (C) Instantaneous torque about the point in contact with the floor is proportional to sinθ (D) The trajectory of the point A is a parabola 7

8 Answer (A, B, C) l Sol. Torque about O, at any instant is mg. sinθ. Option (C) mg O θ l As no external force acts along x-axis, therefore centre of mass will fall vertically downward. Option (B) P (,) x y l cosθ θ O l l θ Displacement of centre of mass along y-axis l [ cosθ] Option (A) l x sin θ, y lcosθ x y + l l Trajectory is not parabola. The instantaneous voltages at three terminals marked X, Y and Z are given by V sinω x V t, π sin ω + VY V t and 4π VZ V sin ω t + 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) Independent of the choice of the two terminals (B) V rms XY V (C) rms VYZ V (D) V rms XY V 8

9 Answer (A, D) Sol. V XY V sin ωt V π sin ω t + V 6 V V V V rms VXY V YZ V π 4 π sin ω t + V sin ω t + V V 6 V V rms VYZ V A, D. A uniform magnetic field B exists in the region between x and x (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 ). Which of the following option(s) is/are correct? egion y egion egion B O +Q P P x ( y ) / (A) For p B >, the particle will re-enter region Q 8 p (B) For B Q, the particle will enter region through the point P on x-axis (C) 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 (D) 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/ 9

10 Answer (A, B) Sol. The particle will follow circular trajectory inside the magnetic field region. The magnetic field cannot change the magnitude of velocity and momentum. 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. Δ 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 mv B p QB p B Q 8 If B p ; r p Q QB 8 It passes through point P if r r cosθ rr - cosθ O θ r θ P sinθ r P 5 8

11 . Two coherent monochromatic point sources S and S of wavelength λ 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 Δθ. Which of the following options is/are correct? P Δθ S S P d (A) The angular separation between two consecutive bright spots decreases as we move from P to P along the first quadrant (B) At P the order of the fringe will be maximum (C) A dark spot will be formed at the point P (D) The total number of fringes produced between P and P in the first quadrant is close to Answer (B, D) Sol. d.8 m 8 4 m P P θ S d S P and λ 6 7 m Path difference at point P (as shown) Δx S P S P d sin θ, where θ angle is measured from vertical line as shown. For bright fringe d sin θ mλ...(i) Point P is the point of central maxima. At point P, path difference (Δx) d If P is the point of bright fringe, then d d mλ m λ On differentiating equation (i) d cos θ (Δθ) (Δm) λ constant for consecutive bright fringe π cos θ Δθ as θ varies from to

12 . A point charge +Q is placed just outside an imaginary hemispherical surface of radius as shown in the figure. Which of the following statements is/are correct? Q (A) The circumference of the flat surface is an equipotential (B) The component of the electric field normal to the flat surface is constant over the surface (C) Total flux through the curved and the flat surfaces is Q ε (D) The electric flux passing through the curved surface of the hemisphere is Answer (A, D) Sol. Net flux through curved surface and flat surface Q Q ε 45º φ Curved φ Plane Q ( cos ) θ ε Q ε The circumference points are equidistant from Q All points will be at the same potential. Option (A) and (D) are correct. 4. A source of constant voltage V is connected to a resistance 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? + V S L L (A) At t, the current through the resistance is V (B) The ratio of the currents through L and L is fixed at all times (t > ) (C) After a long time, the current through L will be V L L + L (D) After a long time, the current through L will be V L L + L

13 Answer (B, C, D) Sol. Final current through battery V Current through L Current through L V L L+ L V L L+ L At t current through source zero At any time i V V + e t LL L+ L Current through L L i L + L i Current through L i i L L il L + L i SECTION - (Maximum Marks : ) This section contains TWO Paragraphs. Based on each paragraph, there are TWO questions. Each question has FOU 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 OS. 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 PAAGAPH Consider a simple C circuit as shown in Figure. 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 >> C). In the process some dissipation (E D ) occurs across the resistance. 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 >> C. Then the voltage is raised to V without discharging the capacitor and again maintained for a time T >> C. 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.

14 These two processes are depicted in Figure. S V v Process V + C V / V / Process T>> C Figure 5. In Process, total energy dissipated across the resistance E D is : (A) (C) ED CV (B) E CV (D) D Answer (B) Sol. E D W b ΔV ED ED T Figure CV T CV t CV V V + + V CV CV V + V + V CV CV [ V ] CV CV 6 CV 6. 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 ΔE c E C CV E CV D CV 4

15 PAAGAPH One twirls a circular ring (of mass M and radius ) 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 ω. 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 μ and the acceleration due to gravity is g. r Figure Figure 7. The minimum value of ω below which the ring will drop down is g (A) μ ( r) g (B) μ( r) g (C) μ ( r) g (D) μ( r) Answer (B) Sol. N f Mg N Mω ( r) f Mg f μn Mg μmω ( r) g ω μ ( r) 8. The total kinetic energy of the ring is (A) Mω ( r ) (B) M ( r ) ω M r (D) (C) ( ) ω M ω 5

16 Answer (D) Sol. k Mv + I ω c c Mω r + M ω ( ) V C ω ( r) r Mω M + ω r ω r << r k Mω + Mω Mω END OF PHYSICS 6

17 JEE (ADVANCED)-7 (PAPE-) CODE-4 CHEMISTY SECTION - (Maximum Marks : ) This section contains SEVEN questions. Each question has FOU 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 OS. 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. 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) 9 bar (C) 45 bar (B) 58 bar (D) 45 bar Answer (C) Sol. Gº V P 6 9 P 6 9 P Pa P F 45 bar P F 45bar. Which of the following combination will produce H gas? (A) Cu metal and conc. HNO (B) Zn metal and NaOH(aq) (C) Au metal and NaCN(aq) in the presence of air (D) Fe Metal and conc. HNO Answer (B) Sol. Zn + NaOH Na ZnO + H Iron become passive with conc. HNO. Copper liberate NO with HNO 7

18 JEE (ADVANCED)-7 (PAPE-) CODE-4. 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; is gas constant; T is temperature; E (cell). V] (A).T.F (B).F (C).T +.F (D).F Answer (A) Sol. Zn ZnSO 4 CuSO 4 Cu (aq) (aq) G G + T In Q G G +. T log Q [Zn ] Q [Cu ] G nf E Cell F.. F G. F +. T log G. T. F. The order of basicity among the following compounds is (A) IV > II > III > I (C) I > IV > III > II Answer (D) NH HC NH I N II NH NH HN N HN NH III IV (B) II > I > IV > III (D) IV > I > II > III NH Sol. HN NH IV esonance 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. 8

19 JEE (ADVANCED)-7 (PAPE-) CODE-4 N N This LPe is not involve in aromaticity. So more available H Also, 'N' is bonded to sp C on one side. II IV > I > II > III. 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 Answer (B) 7 7 T/K 7 7 T/K Sol. T f W ikf M W K 7 (K) T f (K) T f 7 K Also, with decrease in temperature, V.P. decreases. Graph (b) is correct. 4. 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 4 P O 6 > H PO 4 (C) H PO > H PO > H PO 4 > H 4 P O 6 (D) H PO 4 > H 4 P O 6 > H PO > H PO Answer (D) 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 9

20 5. The major product of the following reaction is JEE (ADVANCED)-7 (PAPE-) CODE-4 OH i) NaNO, HCl, C ii) aq.naoh NH + ONa OH (A) (B) NCl OH Cl N N OH (C) (D) Answer (C) N N OH OH O Sol. NH NaNO + HCl aq.naoh C + N NCl + N N OH O N N N N H This section contains SEVEN questions. SECTION - (Maximum Marks : 8) Each question has FOU options (A), (B), (C) and (D). ONE O MOE 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 OS. 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 (PAPE-) CODE-4 6. 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) 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) Adsorption is accompanied by decrease in enthalpy and decrease in entropy of the system (D) Brownian motion of colloidal particles does not depend on the size of the particles but depends on viscosity of the solution Answer (B, C) 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. Compounds P and 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) (ii) P i) O/CHCI ii) Zn/H O i) O/CHCI ii) Zn/H O (C HO) Q 8 8 (C HO) S 8 8 The option(s) with suitable combination of P and, respectively, is(are) (A) HC and HC CH (B) HC and CH CH HC (C) HC and CH CH (D) HC CH CH and CH CH CH Answer (B, C) 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

22 CH O/CHCl Zn/H O O C H O + CH C H JEE (ADVANCED)-7 (PAPE-) CODE-4 CH CH Undergoes Cannizzaro's reaction CH CH O /CH Cl Zn/H O C + CH CH O CH Undergoes Haloform reaction CH CH 8. For the following compounds, the correct statement(s) with respect to nucleophilic substitution reaction is(are) O CH CH Br Br H C C Br Br CH I II III IV (A) I and II follow S N mechanism (B) Compound IV undergoes inversion of configuration (C) The order of reactivity for I, III and IV is : IV > I > III (D) I and III follow S N mechanism Answer (A, B, D) Sol. (A) I & II will follow S N when medium is polar aprotic and nucleophile is strong in high concentration. Option (B) is correct as Br Nu S N Nu Inversion in case of S N. (C) is incorrect for both S N and S N conditions. When medium is highly polar and protic I & III will follow S N. Hence, (D) is correct. 9. 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 Answer (A, C) 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]

23 JEE (ADVANCED)-7 (PAPE-) CODE-4. The option(s) with only amphoteric oxides is(are) (A) Cr O, BeO, SnO, SnO (B) ZnO, AI O, PbO, PbO (C) NO, B O, PbO, SnO (D) Cr O, CrO, SnO, PbO Answer (A, B) 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. Among the following, the correct statement(s) is(are) (A) AlCl 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) The Lewis acidity of BCl is greater than that of AlCl (D) Al(CH ) has the three-centre two-electron bonds in its dimeric structure Answer (A, B, C) 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.. 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 endothermic reaction increases because unfavourable change in entropy of the surroundings decreases (B) With increase in temperature, the value of K for exothermic reaction decreases because favourable change in entropy of the surroundings decreases (C) With increase in temperature, the value of K for exothermic reaction decreases because the entropy change of the system is positive (D) With increase in temperature, the value of K for endothermic reaction increases because the entropy change of the system is negative Answer (A, B, C) 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.

24 This section contains TWO Paragraphs. Based on each paragraph, there are TWO questions. SECTION - (Maximum Marks : ) JEE (ADVANCED)-7 (PAPE-) CODE-4 Each question has FOU 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 OS. 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 PAAGAPH 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.. Y and Z are, respectively (A) N O 5 and HPO (B) N O and H PO 4 (C) N O 4 and H PO (D) N O 4 and HPO Answer (A) 4. W and X are, respectively (A) O and P 4 O 6 (B) O and P 4 O (C) O and P 4 O 6 (D) O and P 4 O Answer (A) Solutions of Q.No () & (4) [MnO ] Sol. KClO KCl O (W) 5O P4 P4O (X) P4O 4HNO NO54HPO (X) (Z) PAAGAPH 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. The reaction of 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 S P 5. The reactions, Q to and to S, are (A) Friedel-Crafts alkylation, dehydration and Friedel-Crafts acylation (B) Dehydration and Friedel-Crafts acylation (C) Friedel-Crafts alkylation and Friedel-Crafts acylation (D) Aromatic sulfonation and Friedel-Crafts acylation 4

25 JEE (ADVANCED)-7 (PAPE-) CODE-4 Answer (C) 6. The product S is (A) (H C) C COCH CH (B) (H C) C HCOC HC CH (H C) C HC CH (H C) C HO S O CH (C) (D) COCH COCH Answer (C) Solutions of Q. 5 and 6 O C(CH ) C OEt C(CH ) CH MgBr (excess)/(c H) O 5 + 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) () O CH C Cl/AlCl (Acylation) CH C(CH ) CH COCH (S) END OF CHEMISTY 5

26 JEE (ADVANCED)-7 (PAPE-) CODE-4 This section contains SEVEN questions. MATHEMATICS SECTION - (Maximum Marks : ) Each question has FOU 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 OS. 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. How many matrices M with entries from {,, } are there, for which the sum of the diagonal entries of M T M is 5? (A) 5 (B) 98 (C) 6 (D) 6 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! 8. 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 5 (A) (B) (C) (D) Answer (B) Sol. x + y + z n(s) + C C 66 6

27 JEE (ADVANCED)-7 (PAPE-) CODE-4 Let z n, where n,,,, 4, 5 x + y + n x + y n Total such solution 5 ( n) 6 n P(E) 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 (D) Sol. equired equation of plane is x y z 6 4(x ) (y ) + ( 5)(y ) 4x y 5y 8 x 9 x dy 4 9 x dx, x and y 7, then 4. If y y(x) satisfies the differential equation y(56) (A) (B) 6 (C) 9 (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 at x 56, y x 4. If f : is a twice differentiable function such that f (x) > for all x, and (A) f () (C) f () (B) (D) f () f () f, f (), then 7

28 Answer (C) JEE (ADVANCED)-7 (PAPE-) CODE-4 Sol. f( x), f and f() f ( x) is always increasing B f() f f () / A / f () Slope of tangent at B > Slope of chord AB. 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) 6 (B) 5 (C) (D) 5 Answer (A) Sol. equired number of subsets 5 C 4 C C 4 C + 5 C 4 C + 5 C 4 4 C + 5 C 5 4 C Alternate method Coefficient of x 5 in ( + x) 5 ( + x) 4 9 C Let O be the origin and let PQ be an arbitrary triangle. The point S is such that OP OQ O OS O OP OQ OS OQ O OP OS Then the triangle PQ has S as its (A) Circumcentre (C) Centroid (B) Incentre (D) Orthocenter Answer (D) Sol. OP OQ O OS O OP OQ OS OQ O OP OS OP OQ O OS O OP OQ OS OP ( OQ O) OS ( O OQ) Q ( OP OS) Q SP 8 Q P S

29 JEE (ADVANCED)-7 (PAPE-) CODE-4 Q SP and similarly from O OP OQ OS OQ O OP OS S PQ S is the orthocentre. This section contains SEVEN questions. SECTION - (Maximum Marks : 8) Each question has FOU options (A), (B), (C) and (D). ONE O MOE 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 OS. 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 I 98 k k k k dx, xx ( ) then (A) 49 I 5 (B) 49 I 5 (C) I < log e 99 (D) I > log e 99 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) 9

30 45. If the line x divides the area of region (A) 4 4 JEE (ADVANCED)-7 (PAPE-) CODE-4 x, y : x y x, x into two equal parts, then (B) (C) 4 4 (D) Answer (C, D) Sol. ( x x ) dx ( x x ) dx y y x y x 4 4 x x x x x O (, ) (, ) x 4 4 x Let f ( ) 4 f (), f () f If, sin( x ) ( ) sin ( tdt ), sin x gx then (A) g (B) g (C) g (D) g

31 JEE (ADVANCED)-7 (PAPE-) CODE-4 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. 47. 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) attains its maximum at x (B) f(x) attains its minimum at x (C) f(x) at exactly three points in (, ) (D) f (x) at more than three points in (, ) Answer (A, D) 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 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

32 Answer (A, B) Sol. As (cos cos) coscos cos cos cos Using componendo and dividendo JEE (ADVANCED)-7 (PAPE-) CODE-4 cos cos cos cos tan tan So, tan tan Or tan tan 49. Let x( x) f( x) cos for x. x x Then (A) lim f( x) does not exist x (B) lim f( x) x (C) lim f( x) x (D) lim f( x) does not exist x Answer (B, 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

33 JEE (ADVANCED)-7 (PAPE-) CODE-4 5. 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) is decreasing in (, ) (D) f(x) > e x in (, ) Answer (B, D) 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 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(, ) This section contains TWO Paragraphs. Based on each paragraph, there are TWO questions. SECTION - (Maximum Marks : ) Each question has FOU 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 OS. 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 PAAGAPH- Let O be the origin, and OX, OY, OZ be three unit vectors in the directions of the sides Q, P, PQ, respectively, of a triangle PQ. 5. If the triangle PQ varies, then the minimum value of cos( P Q) cos( Q) cos( P) is (A) (B) 5 (C) 5 (D)

34 Answer (D) JEE (ADVANCED)-7 (PAPE-) CODE-4 Sol. cos(p + Q) + cos(q + ) + cos( + P) (cosp + cosq + cos) Maximum value of cosp + cosq + cos Hence minimum of (cosp + cosq + cos) 5. OX OY (A) sin (P + ) (B) sin (Q + ) (C) sin (P + Q) (D) sin Answer (C) Sol. OX OY Q P pq pq sin pq sin( P Q) PAAGAPH- 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. If a 4 8, then p + q (A) (B) 4 (C) 7 (D) Answer (A) Sol. a p 5 q ( pq) 4 5( pq) 8 6 p q 4 4

35 JEE (ADVANCED)-7 (PAPE-) CODE a a (A) a (C) a (B) a a a (D) a a Answer (C) Sol. +...(i) and +...(ii) Multiplying (i) by p and (ii) by q and adding, a a + a END OF MATHEMATICS 5