FIITJEE SOLUTIONS TO JEE (ADVANCED) 2017 PAPER-2. Time : 3 Hours Maximum Marks : 183 COVER PAGE IS AS PER THE ACTUAL PAPER

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1 JEE(ADVANED)-07-Paper--PM- Note: For the benefit of the students, specially the aspiring ones, the question of JEE(advanced), 07 are also given in this booklet. Keeping the interest of students studying in class XI, the questions based on topics from class XI have been marked with *, which can be attempted as a test. For this test the time allocated in Physics, hemistry & Mathematics are minutes, minutes and 5 minutes respectively. FIITJEE SOLUTIONS TO JEE (ADVANED) 07 P-7-6- PAPER- ODE Time : 3 Hours Maximum Marks : 83 OVER PAGE IS AS PER THE ATUAL PAPER READ THE INSTRUTIONS AREFULLY GENERAL. This sealed booklet is your Question Paper. Do not break the seal till you are told to do so.. The paper ODE is printed on the right hand top corner of this sheet and the right hand top corner of the back cover of this booklet. 3. Use the Optical Response Sheet (ORS) provided separately for answering the questions. 4. The paper ODE is printed on the left part as well as the right part of the ORS. Ensure that both these codes are identical and same as that on the question paper booklet. If not, contact the invigilator for change of ORS. 5. Blank spaces are provided within this booklet for rough work. 6. Write your name, roll number and sign in the space provided on the back cover of this booklet. 7. After breaking the seal of the booklet at :00 pm, verify that the booklet contains 36 pages and that all the 54 questions along with the options are legible. If not, contact the invigilator for replacement of the booklet. 8. You are allowed to take away the Question Paper at the end of the examination. OPTIAL RESPONSE SHEET 9. The ORS (top sheet) will be provided with an attached andidate s Sheet (bottom sheet). The andidate s Sheet is a carbonless copy of the ORS. 0. Darken the appropriate bubbles on the ORS by applying sufficient pressure. This will leave an impression at the corresponding place on the andidate s Sheet.. The ORS will be collected by the invigilator at the end of the examination.. You will be allowed to take away the andidate s Sheet at the end of the examination. 3. Do not tamper with or mutilate the ORS. Do not use the ORS for rough work. 4. Write your name, roll number and code of the examination center, and sign with pen is the space provided for this purpose on the ORS. Do not write any of these details anywhere else on the ORS. Darken the appropriate bubble under each digit of your roll number. DARKENING THE BUBBLES ON THE ORS 5. Use a BLAK BALL POINT PEN to darken the bubbles on the ORS. 6. Darken the bubble OMPLETELY. 7. The correct way of darkening a bubble is as: 8. The ORS is machine-gradable. Ensure that the bubbles are darkened in the correct way. 9. Darken the bubbles ONLY IF you are sure of the answer. There is NO WAY to erase or un-darken a darkened bubble. Please see the last page of this booklet for rest of the instruction. DO NOT BREAK THE SEALS WITHOUT BEING INSTRUTED TO DO SO BY THE INVIGILATOR FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 65394

2 JEE(ADVANED)-07-Paper--PM- *Q. *Q. This section contains SEVEN questions. PART I: PHYSIS SETION (Maximum Marks: ) Each question has FOUR options [A], [B], [] and [D]. ONLY ONE of these four options is 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 : + 3 If only the bubble(s) corresponding to the correct option is darkened. Zero Marks : 0 If none of the bubbles is darkened. Negative Marks : In all other cases. onsider 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 d change in density is constant. The velocity v of any point on the surface of the expanding sphere is dt proportional to [A] R [B] R 3 [] R A 4 m = R =.4R dr 4 3 d R dt 3 dt d 3 dr dt R dt R d dr 3 dt dt dr R dt v R [D] R /3 onsider 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 I h [A] = h sin [] = h sin n n [B] = h cos n [D] = h tan n FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 65394

3 JEE(ADVANED)-07-Paper--PM-3 B For a regular polygon of n sides = h - h = h. cos cos n n h /n hc Q.3 A photoelectric material having work-function 0 is illuminated with light of wavelength. The 0 fastest photoelectron has a de Broglie wavelength d. A change in wavelength of the incident light by results in change d in d. then the ratio d / is proportional to [A] d / [B] 3 d / [] [D] 3 d / d / D P hc m h m. D 0 hc h hc d d m 0 3 D D 3 D D. Q.4 A symmetric star conducting wire loop is carrying a steady state current I as shown in 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 0 6[ 3 ] 4a I 0 3[ 3 ] 4a [B] [D] I 0 6[ 3 ] 4a I 0 3[ 3] 4a A 0I 4a B = sin 60 sin 30 I 4a FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 65394

4 JEE(ADVANED)-07-Paper--PM-4 *Q.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 point P ad S is b R. The general relation among vectors P, Q and S is Y P P S b R R= Q - P S Q Q [A] S ( b)p bq [] S ( b )P bq O X [B] S (b )P bq [D] S ( b)p b Q *Q.6 A S P br and Q S ( b)r where R Q P From these two equations S ( b)p bq A rocket is launched normal to the surface of earth, away from the Sun, along the line joining the Sun and the Earth. The Sun is times heavier than the Earth and is at a distance times larger than the radius of the Earth. The escape velocity from Earth s gravitational field is v e =. km s -. The minimum initial (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 - [] v s = 6 km s - [D] v s = 7 km s - B 5 GMm G(30 M) m mves 0 4 R.50 R V es = 3 v e = 40.3 km/s Q.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 = 0.0 seconds and he measures the depth of the well to be L = 0 meters. Take the acceleration due to gravity g = 0 ms - and the velocity of sound is 300 ms -. Then the fractional error in the measurement, L/L, is closest to [A] 0. % [B] % [] 3 % [D] 5 % B t = L g L dt L dl g L FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 65394

5 JEE(ADVANED)-07-Paper--PM-5 dl = dt gl dl dt L L gl 5 = 6% % This section contains SEVEN questions. SETION (Maximum Marks: 8) Each question has FOUR options [A], [B], [] 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 : 0 If none of the bubbles is darkened. Negative Marks : In all other cases. For example, if [A], [] 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. 3R Q.8 A uniform magnetic field B exists in the region between x = 0 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 = R). Which of the following option(s) is/are correct? y Region Region Region 3 O +Q P (y = R) 3R/ B P x FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 65394

6 JEE(ADVANED)-07-Paper--PM-6 [A] For p B, the particle will re-enter region 3 QR 8 p [B] For B, the particle will enter region 3 through the point P on x-axis 3 QR [] 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 A, B p R qb (R R) R R 3 R R 8 R R R R Q.9 The instantaneous voltages at three terminals marked X, Y and Z are given by V V sin t x 0 Vy V0 sin t 3 and 4 Vz V0 sin 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 3 VXY V0 [B] rms VYZ V0 rms [] VXY V0 [D] independent of the choice of the two terminals A, D VXYO VYO VZXO 3V0 3 V V V rms rms XX YZ 0 V 0 V 0 V 0 FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 65394

7 JEE(ADVANED)-07-Paper--PM-7 Q.0 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 Q [A] The electric flux passing through the curved surface of the hemisphere is 0 Q [B] Total flux through the curved and the flat surfaces is 0 [] 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 A, D Solid angle subtended by flat surface at position of charge = ( cos ) = ( cos 45) = Flux entering through curved surface = flux leaving through flat surface Q Flux entering through curved surface = 4 Q 0 Also all points of circumference are at equal distance from the charge 0 Q. Two coherent monochromatic point sources S and S of wavelength = 600 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 +Q 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 [] The total number of fringes produced between P and P in the first quadrant is close to 3000 [D] The angular separation between two consecutive bright spots decreases as we move from P to P along the first quadrant FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 65394

8 JEE(ADVANED)-07-Paper--PM-8 B, Path difference at point P x = d cos At P, = 90 x = 0 maxima At P, = 0 x = d = n 3.80 n 3000 maxima For maxima at P, d cos = n or d[sin ] d = (dn) d Angular fringe width, dn d sin as d (i) P S S P d Q. 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 = 0, the switch is closed and current begins to flow. Which of the following options is/are correct? R S + V L L V L [A] After a long time, the current through L will be R L L V L [B] After a long time, the current through L will be R L L [] The ratio of the currents through L and L is fixed at all times (t > 0) [D] At t = 0, the current through the resistance R is V R A, B, Let current through R, L and L is i, i and i respectively L i = L i (i) V i i i R (ii) After long time L V L V Solving, i, i L L R L L R FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 65394

9 JEE(ADVANED)-07-Paper--PM-9 *Q.3 A rigid uniform bar AB of length L is slipping from its vertical position on a frictionless floor (as shown in the figure). At some instant of time, the angle made by the bar with the vortical is. 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 [] 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 ) A, D There is no horizontal force on rod during its motion.m. will fall vertically downwards mg N ma...(i) L N sin I...(ii) L a sin...(iii) L mg sin Solving (i), (ii) and (iii) ml I sin 4 ml Torque about point of contact 3 L L Displacement mid point cos FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 65394

10 JEE(ADVANED)-07-Paper--PM-0 *Q.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. onsider 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 X R [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 [] 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 0 but the wheel never climbs the step A, Let angualr displacement of wheel about Q is When force is applied at X, torque of force will increase while torque of mg will decrease Net torque will increase When force is applied at P normally, torque of force = 0 wheel will not rotate When force is applied at S, = FR SETION 3 (Maximum Marks : ) This section contains TWO paragraphs Based on each table, there are TWO questions Each question has FOUR options [A], [B], [], 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 catogories: Full Marks : +3 If only the bubble corresponding to the correct option is darkened Zero Marks : 0 In all other cases PARAGRAPH onsider a simple R circuit as shown in Figure. Process : In the circuit the switch S is closed at t = 0 and the capacitor is fully charged to voltage V 0 (i.e. charging continues for time T >> R). 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. Process : In a different process the voltage is first set to V 0 3 and maintained for a charging time T>> R. Then the voltage is raised to V 0 3 without discharging the capacitor and again maintained for time T >> R. The FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 65394

11 JEE(ADVANED)-07-Paper--PM- process is repeated one more time by raising the voltage to V 0 and the capacitor is charged to the same final voltage V 0 as in Process. These two processes are depicted in Figure. S V V Process 0 V + R V 0 /3 V 0 /3 Process T T>>R T t Figure Figure Q.5. In Process, the energy stored in the capacitor E and heat dissipated across resistance E D are released by: [A] E ED [B] E ED ln [] E ED [D] E ED A W V V b 0 0 U V E 0 Heat dissipated Wb E or E E V D 0 Q.6 In Process, total energy dissipated across the resistance E D is: [A] ED V0 [B] ED 3 V0 [] ED V0 [D] ED 3V0 3 If capacitor is charged from H Wbattery U Vi to f i f f i V f ; heat dissipated is V V V V V V f V i Total heat dissipated V0 V0 V0 V 0 V V0 6 FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 65394

12 JEE(ADVANED)-07-Paper--PM- 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 0. The rotating ring rolls without slipping on the outside of a smaller circle described by the point where the ring and the finger is in contact (Figure ). The coefficient of friction between the ring and the finger is and the acceleration due to gravity is g. R R Figure Figure *Q.7 The total kinetic energy of the ring is [A] M 0R M 0 R r 3 M 0 R r [B] 0 [] M R r [D] *Q.8 The minimum value of 0 below which the ring will drop down is [A] [] A g R r 3g R r [B] [D] g R r g R r FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 65394

13 JEE(ADVANED)-07-Paper--PM-3 PART II: HEMISTRY Q.9. Q.0. SETION [Maximum Marks: ] This section contains SEVEN questions Each question has FOUR options [A], [B], [] 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 : 0 If none of the bubbles is darkened Negative Marks : In all other cases Pure water freezes at 73 K and bar. The addition of 34.5 g of ethanol to 500 g of water changes the freezing point of the solution. Use the freezing point depression constant of water as 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 the freezing point is (A) (B) () V.P./bar V.P./bar T K m f f Ice T/K Ice T/K Water Water + Ethanol Water Water + Ethanol (D) =.5 = 3 Freezing point of ethanol + water mixture = 73-3 = 70 For the following cell, V.P./bar V.P./bar Ice Zn s ZnSO aq uso aq u s 4 4 Ice 7 73 T/K 7 73 T/K Water Water + Ethanol Water Water + Ethanol When the concentration of Zn + is 0 times the concentration of u +, the expression for G (in J mol ) is [F is Faraday constant; R is gas constant; T is temperature; E 0 (cell) =. V] [A]. F [B].303 RT. F [].303 RT +. F [D]. F B Zn e Zn u e u Zn u Zn u FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 65394

14 JEE(ADVANED)-07-Paper--PM-4.303RT Zn E E log nf u.303rt Ecell. log0 F.303RT. F G nfe cell.303rt F. F =.303RT. F o cell cell *Q.. The standard state Gibbs free energies of formation of (graphite) and (diamond) at T = 98 K are G 0 graphite 0 kjmol f G 0 diamond.9 kjmol f The standard state means that the pressure should be bar, and substance should be pure at a given temperature. The conversion of graphite [(graphite)] to diamond [(diamond)] reduces its volume by 0 6 m 3 mol. If (graphite) is converted to (diamond) isothermally at T = 98 K, the pressure at which (graphite) is in equilibrium with (diamond), is [Useful information: J = kg m s ; Pa = kg m s ; bar = 0 5 Pa] [A] 450 bar [] 450 bar [B] 5800 bar [D] 900 bar Q.. A G PdV o o G f diamond G f graphite PdV.90 J mol P 0 m mol P Pa 9 5 P bar P = bar P = 4500 bar Which of the following combination will produce H gas? [A] Fe metal and conc. HNO 3 [B] u metal and conc. HNO 3 [] Zn metal and NaOH(aq) [D] Au metal and NaN (aq) in the presence of air Zn NaOH Na ZnO H *Q.3. 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 3 > H 3 PO > H 3 PO 4 > H 4 P O 6 [B] H 3 PO 4 > H 3 PO > H 3 PO 3 > H 4 P O 6 [] H 3 PO 4 > H 4 P O 6 > H 3 PO 3 > H 3 PO [D] H 3 PO > H 3 PO 3 > H 4 P O 6 > H 3 PO 4 Species Oxidation state of P H 3 PO 4 +5 H 4 P O 6 +4 H 3 PO 3 +3 H 3 PO + FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 65394

15 JEE(ADVANED)-07-Paper--PM-5 Q.4. The major product of the following reaction is OH i) NaNO 0,Hl, 0 ii) aq.naoh NH OH ONa + [A] [B] l N l OH N N OH [] [D] N N OH OH OH i) NaNO Hl ii) 0 o Aq. NaOH NH N l N N *Q.5. The order of basicity among the following compound is NH HN N N NH H 3 NH III I II NH NH IV NH [A] II > I > IV > III [] IV > I > II > III [B] IV > II > III > I [D] I > IV > III > II NH NH H N NH > onjugate acid has three equivalent canonical forms H 3 NH > N NH > HN N onjugate acid has two equivalent canonical forms onjugate acid has two equivalent canonical forms FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 65394

16 JEE(ADVANED)-07-Paper--PM-6 Q.6. SETION [Maximum Marks: 8] This section contains SEVEN questions Each question has FOUR options [A], [B], [] 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 : 0 If none of the bubbles is darkened Negative Marks : In all other cases For example, if [A], [] and [D] are all the correct options for a question, darkening all these three will get +4 marks; darkening only [A] and [D] will result in + marks; and darkening [A] and [B] will result in marks, as a wrong option is also darkened. 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 [] loud 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 A, B In adsorption process both H & S is ve. Higher the critical temperature of a gas higher the extent of adsorption. loud is not an emulsion. Brownian motion depends on the size of the particles. *Q.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 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 [] 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 surrounding decreases B, D [B] With increase in temperature, the value of K for endothermic reaction increases because unfavourable change in entropy of the surroundings decreases [D] With increase in temperature, the value of K for exothermic reaction decreases because favourable change in entropy of the surrounding decreases Q.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 [] 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 FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 65394

17 JEE(ADVANED)-07-Paper--PM-7 Q.9. A, B [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 For the following compounds, the correct statement(s) with respect to nucleophilic substitution reactions is(are) Br Br H 3 Br Br I II III IV [A] I and III follow S N mechanism [B] I and II follow S N mechanism [] ompound IV undergoes inversion of configuration [D] The order of reactivity for I, III and IV is: IV > I > III A, B,, D Benzylic and 3 o halides both follow S N mechanism. Benzylic and o halides both follow S N mechanism. Benzylic o halides can undergo inversion of configuration. The order of reactivity would be IV > I > III if both S N and S N are considered suitably for substrates. *Q.30. Among the following, the correct statement(s) is(are) [A] Al( ) 3 has the three-centre two-electron bonds in its dimeric structure [B] BH 3 has the three-centre two-electron bonds in its dimeric structure [] All 3 has the three-centre two-electron bonds in its dimeric structure [D] The Lewis acidity of Bl 3 is greater than that of All 3 A, B, D Both Al( ) 3 and BH 3 has 3c e bonds in the dimeric structure. H 3 Al Al H 3 Bl 3 is stronger Lewis acid than All 3. Q.3. The option(s) with only amphoteric oxides is(are) [A] r O 3, BeO, SnO, SnO [B] r O 3, ro, SnO, PbO [] NO, B O 3, PbO, SnO [D] ZnO, Al O 3, PbO, PbO A, D Amphoteric oxides are ZnO, Al O 3, PbO, PbO, BeO, SnO, SnO, r O 3 NO is a neutral oxide. ro is a basic oxide. B O 3 is an acidic oxide FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 65394

18 JEE(ADVANED)-07-Paper--PM-8 Q.3. ompounds P and R upon ozonolysis produce Q and S, respectively. The molecular formula of Q and S is 8 H 8 O. Q undergoes annizzaro reaction but not haloform reaction, whereas S undergoes haloform reaction but not annizzaro reaction. (i) i) O /H l 3 P ii) Zn/HO Q H O i) O /H l (ii) R ii) Zn/HO S 8H8O The option(s) with suitable combination of P and R, respectively, is(are) [A] H 3 and [B] and H 3 [] H 3 and [D] H 3 H 3 and H 3 A, B H 3 i)o 3 /Hl ii) Zn /HO H 3 O H Q( 8 H 8 O) Undergoes annizzaro but not haloform + H H O i)o 3 /Hl ii) Zn /HO O + H H O Undergoes Haloform but not annizzaro H 3 i)o 3 /Hl ii) Zn/H O H 3 Q O H O + H 3 H Undergoes annizzaro but not haloform i)o 3 /Hl ii) Zn/HO O + S Undergoes Haloform but not annizzaro For () and (D) options no. of carbons are not matching FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 65394

19 JEE(ADVANED)-07-Paper--PM-9 SETION 3 (Maximum Marks: ) This section contains TWO paragraphs Based on each paragraph, there will be TWO questions Each question has FOUR options [A], [B], [] 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 : +3 If only the bubble corresponding to all the correct option is darkened Zero Marks : 0 In all other cases. PARAGRAPH Upon heating KlO 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 HNO 3 gives Y and Z. Q.33 W and X are, respectively [A] O 3 and P 4 O 6 [B] O and P 4 O 6 [] O and P 4 O 0 [D] O 3 and P 4 O 0 Q.34 Y and Z are, respectively [A] N O 3 and H 3 PO 4 [B] N O 5 and HPO 3 [] N O 4 and HPO 3 [D] N O 4 and H 3 PO 3 Solution for the Q. No. 33 to 34 KlO Kl 3O MnO 3 W P 5O P O X P4O 0 4HNO3 4HPO3 NO5 Q.33. Q.34. B PARAGRAPH The reaction of compound P with MgBr(excess) in ( H 5 ) O followed by addition of H O gives Q. The compound Q on treatment with H SO 4 at 0 o gives R. The reaction of R with Ol in the presence of anhydrous All 3 in H l followed by treatment with H O produces compound S [Et in compound P is ethyl group] (H 3 ) 3 O Et Q R S Q. 35 The product S is P [A] H 3 (H 3 ) 3 [B] (H 3 ) 3 O O FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 65394

20 JEE(ADVANED)-07-Paper--PM-0 [] H 3 O [D] HO 3 S (H 3 ) 3 H3 (H 3 ) 3 O O *Q. 36 The reactions, Q to R and R to S, are [A] Dehydration and Friedel-rafts acylation [B] Aromatic sulfonation and Friedel-rafts acylation [] Friedel-rafts alkylation, dehydration and Friedel-rafts acylation [D] Friedel-rafts alkylation and Friedel-rafts acylation Solution for the Q. No. 35 to 36 (H 3 ) 3 O Et i) H3MgBrexcess ii) H5 O, HO (H 3 ) 3 OH H SO P Q (H 3 ) 3 i)h3ol,anhydrawsall 3,5H l ii)ho (H 3 ) 3 Q.35. Q.36. R Intramolecular Fridel-raft alkylation & dehydration A O Fridel-raft alkylation FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 65394

21 JEE(ADVANED)-07-Paper--PM- PART III: MATHEMATIS This section contains SEVEN questions. SETION (Maximum Marks: ) Each question has FOUR options [A], [B], [] 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 : 0 If none of the bubbles is darkened. Negative Marks : In all other cases. Q.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 = [B] 4x y + 5z = 7 [] 4x + y + 5z = 3 [D] 4x + y + 5z = 3 Let direction ratios of normal be n ˆi ˆj kˆ n 4i ˆ ˆj 5kˆ 3 6 equation of plane is 4(x ) (y ) 5(z ) = 0 4x + y + 5z = 3 Q.38 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 [] incentre [D] orthocenter D OP OQ OR OS OR OP OQ OS OP OQ OR OSOQ OR OP OSRQ 0 SP RQ 0 SP RQ Similarly SR QP and SQ PR Hence, S is orthocentre Q.39 If y = y(x) satisfies the differential equation 8 x 9 x dy 4 9 x dx, x > 0 and y(0) = 7, then y(56) = [A] 3 [B] 9 [] 6 [D] 80 FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 65394

22 JEE(ADVANED)-07-Paper--PM- A dy dx 8 x 9 x 4 9 x Let 4 9 x t dx dt 9 x x dt dy t dy dt t y t c y 4 9 x c y0 7 c = 0 y 4 9 x y(56) = 3 Q.40 If f : R R is a twice differentiable function such that f (x) > 0 for all x R, and then [A] f () 0 [B] 0 f [] f [D] f () > f, f () =, D Let h(x) = f(x) x h 0 h h() = 0 for some (0, ) by rolle s theorem f() = as f(x) > 0 f(x) is increasing f() > f() f() > Q.4 How many 3 3 matrices M with entries from {0,, } are there, for which the sum of the diagonal entries of M T M is 5? [A] 6 [B] 98 [] 6 [D] 35 B Let matrix M = [k ij ] Then sum of diagonal entries = ij k 5 k ij FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 65394

23 JEE(ADVANED)-07-Paper--PM-3 where k ij are from {0,, } Total number of matrices = = 98 Q.4 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] 0 [B] 5 [] 5 [D] 6 D There are only 4 even numbers in S Any subset of 5 elements of S will have at least odd number. N + N + N 3 + N 4 + N 5 = 9 5 = 6 Q.43 Three randomly chosen nonnegative integers x, y and z are found to satisfy the equation x + y + z = 0. Then the probability that z is even, is [A] [] [B] 6 [D] 5 B Total number of solutions = = 66 Favourable number of solutions = = 36 P(req) = This section contains SEVEN questions. SETION (Maximum Marks: 8) Each question has FOUR options [A], [B], [] 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 : 0 If none of the bubbles is darkened. Negative Marks : In all other cases. For example, if [A], [] 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. sin( x) Q.44 If g(x) = sin t dt, then sin x [A] g [B] g [] g [D] g Bonus FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 65394

24 JEE(ADVANED)-07-Paper--PM-4 g(x) = sin( x) sin x sin t dt g(x) = (cos x) sin (sin x) (cos x) sin (sin x) g 0 cos = 0 g 0 no option is matching. * Q.45 Let and be nonzero real numbers such that (cos cos) + cos cos =. Then which of the following is/are true? [A] tan 3 tan 0 [B] 3 tan tan 0 [] tan 3 tan 0 [D] 3 tan tan 0 A, cos ( + cos) = + cos cos cos cos cos 3cos cos cos cos 3 cos sin sin tan 3tan tan 3 tan. Q.46 If f : R R is a differentiable function such that f (x) > f (x) for all x R, and f (0) =, then [A] f (x) is increasing in (0, ) [B] f (x) is decreasing in (0, ) [] f (x) > e x in (0, ) [D] f (x) < e x in (0, ) A, f(x) f(x) > 0 d dx (f(x). e x ) > 0 f(x). e x is increasing function f(x). e x > f(0) = x > 0 f(x) > e x x > 0 f(x) > f(x) > e x > 0 x > 0. Q.47 Let x x f x cos for x. Then x x [A] lim f x 0 [B] lim f x does not exist [] x x lim f x 0 [D] lim f x does not exist x x FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 65394

25 JEE(ADVANED)-07-Paper--PM-5 A, D h lim f x lim cos 0 x h 0 h h h h lim f x lim cos does not exist x h0 h h cos x cos x sin x Q.48 If f x cos x cos x sin x, then sin x sin x cos x [A] f (x) = 0 at exactly three points in (, ) [B] f (x) = 0 at more than three points in (, ) [] f (x) attains its maximum at x = 0 [D] f (x) attains its minimum at x = 0 B, cos x cos x sin x f x cos x cos x sin x cos 4x cos x sin x sin x cos x Now f(x) = sinx 4sin4x = 0 f(x) = sinx ( + 4 cos x) = 0 Then sin x = 0 or cos x = 4 For sin x = 0; x = 0, / / For cos x = /4 there are four solutions. f(x) = 0 has more than three solutions. Again f(x) = (4 cos x + 6 cos 4x) f(0) < 0 Q.49 If the line x = divides the area of region R = {(x, y) R : x 3 y x, 0 x } into two equal parts, then [A] 0 < [B] < < [] = 0 [D] = 0 B, 3 3 x x dx x x dx x x = 0 f() = (0, ) O (0, 0) (, ) O x = y = f() (, 0) (/, 0) (, ) FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 65394

26 JEE(ADVANED)-07-Paper--PM-6 Q.50 If I = 98 k k k k dx, then x x [A] I > log e 99 [B] I < log e 99 [] I < 49 [D] I > B, D 98 k 98 k 98 k k dx dx dx x x(x ) x k k k k k k n k n k I n k nk k k n 50 I n I n This section contains TWO paragraphs. Based on each paragraph, there are TWO questions. SETION 3 (Maximum Marks: ) Each question has FOUR options [A], [B], [] 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 : 0 In all other cases. PARAGRAPH 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. Q.5 OX OY = [A] sin(p + Q) [B] sin R [] sin(p + R) [D] sin(q + R) A P OZ OY Q OX OX OY OX OY sin R R = sin R = sin(p + Q) FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 65394

27 JEE(ADVANED)-07-Paper--PM-7 Q.5 If the triangle PQR varies, then the minimum value of cos P Q cos Q R cos R P is 5 3 [A] [B] 3 [] 3 [D] 5 3 B (cos P + cos Q + cos R) = OX OY OY OZ OZOX OX OY OZ OX OY OZ 3 PARAGRAPH Let p, q be integers and let, be the roots of the equation, x x = 0, where. For n = 0,,,, let a n = p n + q n. FAT : If a and b are rational numbers and a b 5 0, then a = 0 = b. * Q.53 a = [A] a a 0 [B] a + a 0 [] a + a 0 [D] a + a 0 B As a n + a n a n = 0 So a = a + a 0 * Q.54 If a 4 = 8, then p + q = [A] [B] 4 [] 7 [D] D a 4 = a 0 + 3a a 4 = (q p) (3) + 5p + q = 8 p = q = 4. FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 65394