FIITJEE SOLUTIONS TO JEE (ADVANCED) 2018 PART I: PHYSICS

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1 JEE(DVNED)-08-Paper--PM- Note: For the benefit of the students, specially the aspiring ones, the question of JEE(advanced), 08 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 30 minutes, minutes and 5 minutes respectively. FIITJEE SLUTINS T JEE (DVNED) 08 PRT I: PHYSIS SETIN (Maimum Marks: 4) This section contains SIX (06) questions. Each question has FUR options for correct answer(s). NE R MRE THN NE of these four option(s) is (are) correct option(s). For each question, choose the correct option(s) to answer the question. nswer to each question will be evaluated according to the following marking scheme: Full Marks : +4 If only (all) the correct option(s) is (are) chosen. Partial Marks : +3 If all the four options are correct but NLY three options are chosen. Partial Marks : + If three or more options are correct but NLY two options are chosen, both of which are correct options. Partial Marks : + If two or more options are correct but NLY one option is chosen and it is a correct option. Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered). Negative Marks : - In all other cases. For Eample: If first, third and fourth are the NLY three correct options for a question with second option being an incorrect option; selecting only all the three correct options will result in +4 marks. Selecting only two of the three correct options (e.g. the first and fourth options), without selecting any incorrect option (second option in this case), will result in + marks. Selecting only one of the three correct options (either first or third or fourth option),without selecting any incorrect option (second option in this case), will result in + marks. Selecting any incorrect option(s) (second option in this case), with or without selection of any correct option(s) will result in - marks. *Q. The potential energy of a particle of mass m at a distance r from a fied point is given by V (r) = kr /, where k is a positive constant of appropriate dimensions. This particle is moving in a circular orbit of radius R about the point. If v is the speed of the particle and L is the magnitude of its angular momentum about, which of the following statements is (are) true? () () B, v k R m (B) L mk R (D) kr dv V F kr dr mv k kr v R r m ngular momentum L = mvr = m k R m mk R v L k R m mk R FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fa 65394

2 JEE(DVNED)-08-Paper--PM- *Q. onsider a body of mass.0 kg at rest at the origin at time t = 0. force F t iˆ ˆj is applied on the body, where.0ns and.0 N. The torque acting on the body about the origin at time t =.0 s is. Which of the following statements is (are) true? () Nm 3 (B) The torque is in the direction of the unit vector + ˆk () The velocity of the body at ˆ ˆ t s is v i j ms (D) The magnitude of displacement of the body at t s is 6 m *Q.3, a t iˆ ˆj m/s t v i ˆ t ˆ j m/s 3 r t iˆ t ˆj m 6 r F k k Nm t t so, ˆ t ˆ uniform capillary tube of inner radius r is dipped vertically into a beaker filled with water. The water rises to a height h in the capillary tube above the water surface in the beaker. The surface tension of water is. The angle of contact between water and the wall of the capillary tube is. Ignore the mass of water in the meniscus. Which of the following statements is (are) true? () For a given material of the capillary tube, h decreases with increase in r (B) For a given material of the capillary tube, h is independent of () If this eperiment is performed in a lift going up with a constant acceleration, then h decreases (D) h is proportional to contact angle, cos h ( is density of water) r g eff Q.4 In the figure below, the switches S and S are closed simultaneously at t = 0 and a current starts to flow in the circuit. Both the batteries have the same magnitude of the electromotive force (emf) and the polarities are as indicated in the figure. Ignore mutual inductance between the inductors. The current I in the middle wire reaches its maimum magnitude I ma at time t =. Which of the following statements is (are) true? () Ima V R (B) () L ln (D) R V Ima V 4R L ln R R L R L S S I V FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fa 65394

3 JEE(DVNED)-08-Paper--PM-3 B, D Rt Rt V L V I e e L R R For I ma di 0 dt L t ln and Ima V R 4R Q.5 Two infinitely long straight wires lie in the y-plane along the lines = ±R. The wire located at = +R carries a constant current I and the wire located at = R carries a constant current I. circular loop of radius R is suspended with its centre at (0, 0, 3R ) and in a plane parallel to the y-plane. This loop carries a constant current I in the clockwise direction as seen from above the loop. The current in the wire is taken to be positive if it is in the + ĵ direction. Which of the following statements regarding the magnetic field B is (are) true? () If I I, then B cannot be equal to zero at the origin (0, 0, 0) (B) If I 0, and I 0, then B can be equal to zero at the origin (0, 0, 0) () If I 0, and I 0, then B can be equal to zero at the origin (0, 0, 0) 0I (D) If I I, then the z-component of the magnetic field at the centre of the loop is R, B, D z () If I = I, magnetic field due to infinite wires is equal to zero. So there must be a non zero magnetic field at due to the current carrying I loop. (B) If I 0 & I 0, magnetic field due to straight I lines are along positive z ais and due to loop it is along negative z ais. () If I 0 & I 0 magnetic field due to straight wires are along negative z ais and due to the loop it is also along negative z ais. 0I (D) B ˆ k along z ais R I *Q.6 ne mole of a monatomic ideal gas undergoes a cyclic process as shown in the figure (where V is the volume and T is the temperature). Which of the statements below is (are) true? T I II III () Process I is an isochoric process (B) In process II, gas absorbs heat () In process IV, gas releases heat (D) Processes I and III are not isobaric B,, D Process II is an isothermal epansion Process IV is an isothermal compression In isobaric process, volume is directly proportional to temperature. IV V FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fa 65394

4 JEE(DVNED)-08-Paper--PM-4 SETIN (Maimum Marks: 4) This section contains EIGHT (08) questions. The answer to each question is a NUMERIL VLUE. For each question, enter the correct numerical value (in decimal notation, truncated/rounded-off to the second decimal place; e.g. 6.5, 7.00, -0.33, -.30, 30.7, -7.30) using the mouse and the onscreen virtual numeric keypad in the place designated to enter the answer. nswer to each question will be evaluated according to the following marking scheme: Full Marks : +3 If NLY the correct numerical value is entered as answer. Zero Marks : 0 In all other cases. *Q.7 Two vectors and B are defined as ai ˆ and B a cos t iˆ sin t ˆj, where a is a constant and / 6 rad s -. If B 3 B at time t = for the first time, the value of, in seconds, is..00 t t a cos 3 asin t tan 3 t for the first time t sec 6 -B +B t B B t *Q.8 Two men are walking along a horizontal straight line in the same direction. The man in front walks at a speed.0 ms and the man behind walks at a speed.0 ms. third man is standing at a height m above the same horizontal line such that all three men are in a vertical plane. The two walking men are blowing identical whistles which emit a sound of frequency 430 Hz. The speed of sound in air is 330 ms -. t the instant, when the moving men are 0 m apart, the stationary man is equidistant from them. The frequency of beats in Hz, heard by the stationary man at this instant, is f f cos 330 cos f 0 m/s m/s f = 5 Hz 3330 *Q.9 ring and a disc are initially at rest, side by side, at the top of an inclined plane which makes an angle 60 with the horizontal. They start to roll without slipping at the same instant of time along the shortest path. If the time difference between their reaching the ground is ( 3 ) / 0 s, then the height of the top of the inclined plane, in metres, is. Take g = 0 ms. FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fa 65394

5 JEE(DVNED)-08-Paper--PM h g sin t sin I mr h I t sin g mr 3 h h 0.75m *Q.0 spring-block system is resting on a frictionless floor as shown in the figure. The spring constant is.0 N m and the mass of the block is.0 kg. Ignore the mass of the spring. Initially the spring is in an unstretched condition. nother block of mass.0 kg moving with a speed of.0 m s collides elastically with the first block. The collision is such ms kg kg that the.0 kg block does not hit the wall. The distance, in metres, between the two blocks when the spring returns to its unstretched position for the first time after the collision is..09 For collision : using com = u + v Using e = u + v 4 u m / s v m / s 3 3 time taken for the block to came to the unstretched position of spring for the first time after the collision m = sec k distance between blocks = m.09 m (taking 3.4) 3 Q. Three identical capacitors, and 3 have a capacitance of.0 F each and they are uncharged initially. They are connected in a circuit as shown in the figure and is then filled completely with a dielectric material of relative permittivity r. The cell electromotive force (emf) V 0 = 8V. First the switch S is closed while the switch S is kept open. When the capacitor 3 is fully charged, S is opened and S is closed simultaneously. When all the capacitors reach equilibrium, the charge on 3 is found to be 5. The value of r =. V 0 S S 3.50 fter S is closed, 3 has charge 8. when S opened and S closed, 3 has charge 5. so remaining 3 resides on and 5 V eq r 3 q 3 r.5 FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fa 65394

6 JEE(DVNED)-08-Paper--PM-6 Q. In the y-plane, the region y > 0 has a uniform magnetic field ˆ B k and the region y < 0 has another uniform magnetic field y B ˆ. k positively charged particle is projected from the origin along the positive y-ais with speed v0 m s at t = 0, as shown in the figure. Neglect gravity in this problem. Let t = T be the time when the particle crosses the -ais from below for the first time. If B = 4B, the average speed of the particle, in ms, along the -ais in the time interval T is. v 0 = ms B B.00 vg. speed along -ais total distance travelled along -ais total time taken R r V 0 m / s R r V V 0 0 Q.3 Sunlight of intensity.3 kwm is incident normally on a thin conve lens of focal length 0 cm. Ignore the energy loss of light due to the lens and assume that the lens aperture size is much smaller than its focal length. The average intensity of light, in kw m, at a distance cm from the lens on the other side is The area of over which the light falls satisfy : lens 0 00 so intensity at = 00.3 = 30 kw/m. Lens R r 0 cm cm *Q.4 Two conducting cylinders of equal length but different radii are connected in series between two heat baths kept at temperatures T = 300 K and T = 00 K, as shown in the figure. The radius of the bigger cylinder is twice that of the smaller one and the thermal conductivities of the materials of the smaller and the larger cylinders are K and K respectively. If the temperature at the junction of the two cylinders in the steady state is 00 K, then K / K = In steady state, heat current in both material is same K K L L K K 4 T Insulating material K K T L L FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fa 65394

7 JEE(DVNED)-08-Paper--PM-7 SETIN 3 (Maimum Marks: ) This section contains TW (0) paragraphs. Based on each paragraph, there are TW (0) questions. Each question has FUR options. NLY NE of these four options corresponds to the correct answer. For each question, choose the option corresponding to the correct answer. nswer to each question will be evaluated according to the following marking scheme: Full Marks : +3 If NLY the correct option is chosen. Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered). Negative Marks : In all other cases. PRGRPH X In electromagnetic theory, the electric and magnetic phenomena are related to each other. Therefore, the dimensions of electric and magnetic quantities must also be related to each other. In the questions below, [E] and [B] stand for dimensions of electric and magnetic fields respectively, while [ 0 ] and [ 0 ] stand for dimensions of the permittivity and permeability of free space respectively. [L] and [T] are dimensions of length and time respectively. ll the quantities are given in SI units. (There are two questions based on PRGRPH X, the question given below is one of them) *Q.5 The relation between [E] and [B] is () E B L T (B) E B L T () E B L T (D) E B L T F E = ; F qvb q E BV; E B L T PRGRPH X In electromagnetic theory, the electric and magnetic phenomena are related to each other. Therefore, the dimensions of electric and magnetic quantities must also be related to each other. In the questions below, [E] and [B] stand for dimensions of electric and magnetic fields respectively, while [ 0 ] and [ 0 ] stand for dimensions of the permittivity and permeability of free space respectively. [L] and [T] are dimensions of length and time respectively. ll the quantities are given in SI units. (There are two questions based on PRGRPH X, the question given below is one of them) *Q.6 The relation between [ 0 ] and [ 0 ] is () L T (B) L T () L T (D) L T 0 0 D L T 0 0 FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fa 65394

8 JEE(DVNED)-08-Paper--PM-8 PRGRPH If the measurement errors in all the independent quantities are known, then it is possible to determine the error in any dependent quantity. This is done by the use of series epansion and truncating the epansion at the first power of the error. For eample, consider the relation z = /y. If the errors in,y and z are Δ, Δy and Δz, respectively, then y z z. y y y y y The series epansion for, to first power in Δy / y, is (Δy / y). The relative errors in independent y variables are always added. So the error in z will be y z z. y The above derivation makes the assumption that Δ /, Δ y /y. Therefore, the higher powers of these quantities are neglected. (There are two questions based on PRGRPH, the question given below is one of them) *Q.7 a onsider the ratio r a to be determined by measuring a dimensionless quantity a. If the error in the measurement of a is Δa ( Δ a /a ), then what is the error Δr in determining r? a () () B a a a r + r = a a a a a (B) (D) a aa a a a a a a a = a a a a a a a a a r r a a alternate : a a a r a a a r a a FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fa 65394

9 JEE(DVNED)-08-Paper--PM-9 PRGRPH If the measurement errors in all the independent quantities are known, then it is possible to determine the error in any dependent quantity. This is done by the use of series epansion and truncating the epansion at the first power of the error. For eample, consider the relation z = /y. If the errors in,y and z are Δ, Δy and Δz, respectively, then y z z. y y y y y The series epansion for, to first power in Δy / y, is (Δy / y). The relative errors in independent y variables are always added. So the error in z will be y z z. y The above derivation makes the assumption that Δ /, Δ y /y. Therefore, the higher powers of these quantities are neglected. (There are two questions based on PRGRPH, the question given below is one of them) *Q.8 In an eperiment the initial number of radioactive nuclei is It is found that 000 ± 40 nuclei decayed in the first.0s. For, ln( + ) = up to first power in. The error Δ, in the determination of the decay constant, in s, is () 0.04 (B) 0.03 () 0.0 (D) 0.0 t Nd N0 e t 000 = 3000 e 0 N N e d t 3 e 0.0 alternate : t N = N0e N N N e 0 t t N N 0 e e N N 0 e e e t t t t t N N0e e N N t 0.0 t FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fa 65394

10 JEE(DVNED)-08-Paper--PM-0 PRT II: HEMISTRY SETIN (Maimum Marks: 4) This section contains SIX (06) questions. Each question has FUR options for correct answer(s). NE R MRE THN NE of these four option(s) is (are) correct option(s). For each question, choose the correct option(s) to answer the question. nswer to each question will be evaluated according to the following marking scheme: Full Marks : +4 If only (all) the correct option(s) is (are) chosen. Partial Marks : +3 If all the four options are correct but NLY three options are chosen. Partial Marks : + If three or more options are correct but NLY two options are chosen, both of which are correct options. Partial Marks : + If two or more options are correct but NLY one option is chosen and it is a correct option. Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered). Negative Marks : In all other cases. For Eample: If first, third and fourth are the NLY three correct options for a question with second option being an incorrect option; selecting only all the three correct options will result in +4 marks. Selecting only two of the three correct options (e.g. the first and fourth options), without selecting any incorrect option (second option in this case), will result in + marks. Selecting only one of the three correct options (either first or third or fourth option),without selecting any incorrect option (second option in this case), will result in + marks. Selecting any incorrect option(s) (second option in this case), with or without selection of any correct option(s) will result in - marks. Q. The compound(s) which generate(s) N gas upon thermal decomposition below 300 is (are) () NH 4 N 3 (B) (NH 4 ) r 7 () Ba(N 3 ) (D) Mg 3 N B, (a) NH N N H (N can further decompose to N and at temperature above 300 o ) 4 3 (b) NH r N r 4H (c) Ba N 3N Ba 3 (d) Mg 3 N does not decompose at any temperature. Q. The correct statement(s) regarding the binary transition metal carbonyl compounds is (are) (tomic numbers: Fe = 6, Ni = 8) () Total number of valence shell electrons at metal centre in Fe() 5 or Ni() 4 is 6 (B) These are predominantly low spin in nature () Metal carbon bond strengthens when the oidation state of the metal is lowered (D) The carbonyl bond weakens when the oidation state of the metal is increased B, (a) Electronic configuration of central metal atom in both cases is [r] 3d 0 4s 4p 6 outermost shell and 8 valence electrons respectively). (b) Low spin comple because is a strong field ligand. (c) Metal-carbon bond strengthens when the oidation state of metal is lowered. (d) The carbonyl bond becomes stronger when the oidation state is increased. (8 electrons in Q.3 B ased on the compounds of group 5 elements, the correct statement(s) is (are) () Bi 5 is more basic than N 5 (B) NF 3 is more covalent than BiF 3 () PH 3 boils at lower temperature than NH 3 (D) The N N single bond is stronger than the P P single bond FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fa 65394

11 JEE(DVNED)-08-Paper--PM-, B, Bi 5 is more basic than N 5. NF 3 is more covalent than BiF 3. PH 3 boils at lower temperature than NH 3. N N single bond is weaker than P P single bond. Q.4 In the following reaction sequence, the correct structure(s) of X is (are) Me N ) PBr 3, Et 3 X ) NaI, Me 3) NaN, HNMe 3 Me H () enantiomerically pure Me (B) H () Me H (D) Me H B Me H PBr 3,Et SN Me Br NaI, Me SN Me I NaN3 DMF,SN Me N 3 Q.5 The reaction(s) leading to the formation of,3,5-trimethylbenzene is(are) () onc. HS4 (B) Me H () ) Br, NaH ) H3 3) sodalime, (D) H H heated iron tube 873 K H Zn/Hg, Hl, B, D H 3 onc. H S 4 H 3 H 3, 3, 5 trimethylbenzene H 3 H H heated iron tube K H 3 H 3 FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fa 65394

12 JEE(DVNED)-08-Paper--PM- Br,NaH H3 3 Sodamide, H H 3 H H Zn/Hg, Hl H 3 H 3 * Q.6 reversible cyclic process for an ideal gas is shown below. Here, P, V, and T are pressure, volume and temperature, respectively. The thermodynamic parameters q, w, H and U are heat, work, enthalpy and internal energy, respectively. Volume (V) (P, V,T ) B(P, V, T ) (P, V,T ) Temperature (T) The correct option(s) is (are) () q = U B and w B = P (V V ) () H < U and q = U B (B) w B = P (V V ) and q B = H (D) q B = H and H > U B, H H H 0 B B H 0 (Temperature is constant) B H H () B We know that qp H (In path B, P = constant) Hence qb HB From Eq. () q H B (B) qb P V V P V V () H np T T np T T U n T T n T T V V s, P > V H U So, SETIN (Maimum Marks: 4) This section contains EIGHT (08) questions. The answer to each question is a NUMERIL VLUE. For each question, enter the correct numerical value (in decimal notation, truncated/rounded-off to the second decimal place; e.g. 6.5, 7.00, 0.33,.30, 30.7, 7.30) using the mouse and the on-screen virtual numeric keypad in the place designated to enter the answer. nswer to each question will be evaluated according to the following marking scheme: Full Marks : +3 If NLY the correct numerical value is entered as answer. Zero Marks : 0 In all other cases. FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fa 65394

13 JEE(DVNED)-08-Paper--PM-3 Q.7 mong the species given below, the total number of diamagnetic species is. H atom, N monomer, (superoide), dimeric sulphur in vapour phase, Mn 3 4, (NH 4 ) [Fel 4 ], (NH 4 ) [Nil 4 ], K Mn 4, K r 4 Paramagnetic: H, N monomer (NH 4 ) [Fel 4 ], (NH 4 ) [Nil 4 ], K Mn 4 Diamagnetic: K r 4 superoide, S (Vapour), [Mn 3 4 is mied oide of Mn + and Mn +3 ], Q.8 The ammonia prepared by treating ammonium sulphate with calcium hydroide is completely used by Nil.6H to form a stable coordination compound. ssume that both the reactions are 00% complete. If 584 g of ammonium sulphate and 95 g of Nil.6H are used in the preparation, the combined weight (in grams) of gypsum and the nickel-ammonia coordination compound thus produced is. (tomic weights in g mol : H =, N = 4, = 6, S = 3, l = 35.5, a = 40, Ni = 59) 99 a H NH 4 S NH 4 3 as 4.H 584 Gypsum moles 3 4 Weight of gypsum formed = 7 = 064 g Nil.6H 6NH3 NiNH3 l 6 moles Mass of [Ni(NH 3 ) 6 ] l formed = 4 3 = 98 g Total weight = = 99 g. Q.9 onsider an ionic solid MX with Nal structure. onstruct a new structure (Z) whose unit cell is constructed from the unit cell of MX following the sequential instructions given below. Neglect the charge balance. (i) Remove all the anions (X) ecept the central one (ii) Replace all the face centered cations (M) by anions (X) (iii) Remove all the corner cations (M) (iv) Replace the central anion (X) with cation (M) number of anions The value of in Z is. number of cations 3 X = ctahedral void M + = F point M + X - (i) = (ii) (iii) 3 + = 4 (iv) = 3 Hence anion 3 3 cation Q.0 F or the electrochemical cell, Mg(s) Mg + (aq, M) u + (aq, M) u(s) the standard emf of the cell is.70 V at 300 K. When the concentration of Mg + is changed to M, the cell potential changes to.67 V at 300 K. The value of is. (given, F R = 500 K V, where F is the Faraday constant and R is the gas constant, ln(0) =.30) FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fa 65394

14 JEE(DVNED)-08-Paper--PM-4 0 ase I: 0 RT Ecell Ecell n F =.7 0 =.7 V ase II: 0 RT Ecell Ecell n F R =.7 n F R n F 0.03 F n R = 0 * Q. closed tank has two compartments and B, both filled with oygen (assumed to be ideal gas). The partition separating the two compartments is fied and is a perfect heat insulator (Figure ). If the old partition is replaced by a new partition which can slide and conduct heat but does NT allow the gas to leak across (Figure ), the volume (in m 3 ) of the compartment after the system attains equilibrium is. m 3, 5 bar, 400 K 3 m 3, bar, 300 K B Figure B Figure. s in fig, the system attains equilibrium, so, P = P B and T = T B P V PB VB Rn T n T R V V n B B B n n B R 3 nb 300R Due to sliding of piston vol. of will be increased by and that of B will be decreased by. V = + V B = 3 FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fa 65394

15 JEE(DVNED)-08-Paper--PM R R = 9 Hence volume of container will be 0 V. 9 9 Q. Liquids and B form ideal solution over the entire range of composition. t temperature T, equimolar binary solution of liquids and B has vapour pressure 45 Torr. t the same temperature, a new solution of and B having mole fractions and B, respectively, has vapour pressure of.5 Torr. The value of / B in the new solution is. (given that the vapour pressure of pure liquid is 0 Torr at temperature T) 9, B 0 0 PT P PB (Given P 0 ) P P () 0 0 B 0 PB P P B B B 0 * Q.3 The solubility of a salt of weak acid (B) at ph 3 is Y 0 3 mol L. The value of Y is. (Given that the value of solubility product of B (K sp ) = 0 0 and the value of ionization constant of HB (K a ) = 0 8 ) 4.47 B B S S 0 0 S S () B H HB S S FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fa 65394

16 JEE(DVNED)-08-Paper--PM-6 5 S 0 () Multiply equation () and (). S. = 0-5 From Eq. () 0 S S S S S = y = 4.47 Q.4 The plot given below shows P T curves (where P is the pressure and T is the temperature) for two solvents X and Y and isomolal solutions of Nal in these solvents. Nal completely dissociates in both the solvents pressure (mmhg). solvent X. solution of Nal in solvent X 3. solvent Y 4. solution of Nal in solvent Y Temperature (K) n addition of equal number of moles of a non-volatile solute S in equal amount (in kg) of these solvents, the elevation of boiling point of solvent X is three times that of solvent Y. Solute S is known to undergo dimerization in these solvents. If the degree of dimerization is 0.7 in solvent Y, the degree of dimerization in solvent X is K m b y b K m K K b by T K b bm T K b 3 T 0.7 b y K T K b y b ym b b y S S i i FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fa 65394

17 JEE(DVNED)-08-Paper--PM SETIN 3 (Maimum Marks: ) This section contains TW (0) paragraphs. Based on each paragraph, there are TW (0) questions. Each question has FUR options. NLY NE of these four options corresponds to the correct answer. For each question, choose the option corresponding to the correct answer. nswer to each question will be evaluated according to the following marking scheme: Full Marks : +3 If NLY the correct option is chosen. Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered). Negative Marks : In all other cases. PRGRPH X Treatment of benzene with /Hl in the presence of anhydrous ll 3 /ul followed by reaction with c /Nac gives compound X as the major product. ompound X upon reaction with Br /Na 3, followed by heating at 473 K with moist KH furnishes Y as the major product. Reaction of X with H /Pd-, followed by H 3 P 4 treatment gives Z as the major product. (There are two questions based on PRGRPH X, the question given below is one of them) Q.5 The compound Y is () Br (B) H Br () (D) H Br Br Br H H H H i /Hl ii ll 3 /ul / Nac Br / Na3 X H Br + H H H Br Moist KH/ 473 K H Y FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fa 65394

18 JEE(DVNED)-08-Paper--PM-8 PRGRPH X Treatment of benzene with /Hl in the presence of anhydrous ll 3 /ul followed by reaction with c /Nac gives compound X as the major product. ompound X upon reaction with Br /Na 3, followed by heating at 473 K with moist KH furnishes Y as the major product. Reaction of X with H /Pd-, followed by H 3 P 4 treatment gives Z as the major product. (There are two questions based on PRGRPH X, the question given below is one of them) Q.6 The compound Z is () (B) H () (D) H H H H /Pd H H H H3P4 ES H H PRGRPH n organic acid P ( H ) can easily be oidized to a dibasic acid which reacts with ethylene glycol to produce a polymer dacron. Upon ozonolysis, P gives an aliphatic ketone as one of the products. P undergoes the following reaction sequences to furnish R via Q. The compound P also undergoes another set of reactions to produce S. ) H /Pd ) NH / ) H /Pd ) Hl 3 3) Br /NaH ) Sl ) Mg/Et S P Q R 4) Hl 3, KH, 3) MeMgBr, dl 3) (dry ice) 5) H /Pd 4) NaBH4 4) H3 (There are two questions based on PRGRH, the question given below is one of them) Q.7 The compound R is H () H (B) () H (D) H FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fa 65394

19 JEE(DVNED)-08-Paper--PM-9 H H l H /Pd Sl H H 3 P H 3 Mgl Me H H H H 3 H 3 H Me H H H H 3 H 3 MeMgBr / dl Me i Hl ii Mg/Et NaBH 4 H H H 3 H 3 / H3 H H H 3 H 3 Q H H H 3 H 3 H Me H H H H 3 R H 3 PRGRPH n organic acid P ( H ) can easily be oidized to a dibasic acid which reacts with ethylene glycol to produce a polymer dacron. Upon ozonolysis, P gives an aliphatic ketone as one of the products. P undergoes the following reaction sequences to furnish R via Q. The compound P also undergoes another set of reactions to produce S. ) H /Pd ) NH / ) H /Pd ) Hl 3 3) Br /NaH ) Sl ) Mg/Et S P Q R 4) Hl 3, KH, 3) MeMgBr, dl 3) (dry ice) 5) H /Pd 4) NaBH4 4) H3 (There are two questions based on PRGRPH, the question given below is one of them) FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fa 65394

20 JEE(DVNED)-08-Paper--PM-0 Q.8 The compound S is () (B) NH HN H () NH (D) N B H NH NH i H /Pd ii NH 3, Br / NaH H H 3 P H 3 H H H 3 H 3 NH-H 3 H H H 3 H 3 Hl 3, KH, N H / Pd H H H 3 H 3 S H H H 3 H 3 FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fa 65394

21 JEE(DVNED)-08-Paper--PM- PRT III: MTHEMTIS This section contains SIX (06) questions. SETIN (Maimum Marks: 4) Each question has FUR options for correct answer(s). NE R MRE THN NE of these four option(s) is (are) correct option(s). For each question, choose the correct option(s) to answer the question. nswer to each question will be evaluated according to the following marking scheme: Full Marks : + 4 If only (all) the correct option(s) is (are) chosen. Partial Marks : + 3 If all the four options are correct but NLY three options are chosen. Partial Marks : + If three or more options are correct but NLY two options are chosen, both of which are correct options. Partial Marks : + If two or more options are correct but NLY one option is chosen and it is a correct option. Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered). Negative Marks : In all other cases. For Eample: If first, third and fourth are the NLY three correct options for a question with second option being an incorrect option; selecting only all the three correct options will result in +4 marks. Selecting only two of the three correct options (e.g. the first and fourth options), without selecting any incorrect option (second option in this case), will result in + marks. Selecting only one of the three correct options (either first or third or fourth option), without selecting any incorrect option (second option in this case), will result in + marks. Selecting any incorrect option(s) (second option in this case), with or without selection of any correct option(s) will result in marks. *Q. For a non-zero comple number z, let arg(z) denote the principal argument with < arg(z). Then, which of the following statement(s) is (are) FLSE? () arg( i) =, where i 4 (B) The function f : R (, ], defined by f(t) = arg( + it) for all t R, is continuous at all points of R, where i z () For any two non-zero comple numbers z and z, arg arg z arg z is an integer multiple z of (D) For any three given distinct comple numbers z, z and z 3, the locus of the point z satisfying the condition arg z z z z3 z z z z 3, lies on a straight line, B, D 3 () rg( i) = 4 tan t if t 0 (B) rg( + it) = tan t if t 0 Not continuous at t = 0 z () rg arg z arg z = 0 z FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fa 65394

22 JEE(DVNED)-08-Paper--PM- (D) z z z z rg rg z z z z 3 3 z, z, z, z 3 form a cyclic quadrilateral Hence locus of z is circle *Q. In a triangle PQR, let PQR = 30 and the sides PQ and QR have lengths 0 3 and 0, respectively. Then, which of the following statement(s) is (are) TRUE? () QPR = 45 (B) The area of the triangle PQR is 5 3 and QRP = 0 () The radius of the incircle of the triangle PQR is (D) The area of the circumcircle of the triangle PQR is 00 B,, D () sin sin 50 cos 3 sin 3sin tan 3 = 30º Q P º 50º 0 R (B) () sin30º = 5 3, QRP = 0º r 5 3 s (D) rea = R = / = Q.3 Let P : + y z = 3 and P : + y + z = be two planes. Then, which of the following statement(s) is (are) TRUE? () The line of intersection of P and P has direction ratios,, 3 4 3y z (B) The line is perpendicular to the line of intersection of P and P () The acute angle between P and P is 60 (D) If P 3 is the plane passing through the point (4,, ) and perpendicular to the line of intersection of P and P, then the distance of the point (,, ) from the plane P 3 is 3, D () Direction ratios of line of intersection are given by i ˆ ˆ j k ˆ ˆ i j ˆ k ˆ = 3i ˆ 3j ˆ 3kˆ dr (,, ) a b ˆi ˆj kˆ 3iˆ 3j ˆ 3kˆ (B) () ngle between P, P cos = cos 60º 6 6 (D) P 3 : y + z = 0 Distance of (,, ) from the plane = 3 FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fa 65394

23 JEE(DVNED)-08-Paper--PM-3 Q.4 For every twice differentiable function f : R [, ] with (f(0)) + (f (0)) = 85, which of the following statement(s) is (are) TRUE? () There eist r, s R, where r < s, such that f is one-one on the open interval (r, s) (B) There eists 0 (4, 0) such that f ( 0 ) () lim f (D) There eist (4, 4) such that f() + f () = 0 and f () 0, B, D L.M.V.T. in [ 4, 0] f 0 f 4 f 0 4 f 0 f 4 f 0 4 If f() periodic then lim f Similarly f( ) for some (0, 4) g() = (f()) + (f()) g( 0 ) 5, g( ) 5 g(0) = 85 it has a local maimum having value 85 Say g() = 0, g() 0 f()f() + f()f() = 0 f()(f() + f()) = 0 as f() 0 Q.5 Let f : R R and g : R R be two non-constant differentiable functions. If f () = (e (f()g()) ) g() for all R, and f() = g() =, then which of the following statement(s) is (are) TRUE? () f() < log e (B) f() > log e () g() > log e (D) g() < log e B, f() = e f() g() g() given f() = g() = f g f e d e g d f g d e d e f g e e put = = e e g g = e f() e = g e e g() > log & f() > e loge Q.6 Let f : [0, ) R be a continuous function such that t f e f t dt for all [0, ). Then, which of the following statement(s) is (are) TRUE? () The curve y = f() passes through the point (, ) (B) The curve y = f() passes through the point (, ) 0 FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fa 65394

24 JEE(DVNED)-08-Paper--PM-4 () The area of the region {(, y) [0, ] R : f() y (D) The area of the region {(, y) [0, ] R : f() y } is } is 4 4 B, f() = + e t e f t dt 0 f() = + e t e f t dt + f() 0 f() = + f() + + f() f() f() = 3 e e 3e f e 4 put = 0, f(0) = = 0 f e e e f() = which passes through (, ) area =. 4 4 SETIN (Maimum Marks: 4) This section contains EIGHT (08) questions. The answer to each question is a NUMERIL VLUE. For each question, enter the correct numerical value (in decimal notation, truncated/rounded off to the second decimal place; e.g. 6.5, 7.00, 0.33,.30, 30.7, 7.30) using the mouse and the on-screen virtual numeric keypad in the place designated to enter the answer. nswer to each question will be evaluated according to the following marking scheme: Full Marks : +3 If NLY the correct numerical value is entered as answer. Zero Marks : 0 In all other cases. *Q.7 The value of log 9 7 log log 9 log 4 7 is. 8 loglog 9 log 9 7 log7 4 = / 4 8 *Q.8 The number of 5 digit numbers which are divisible by 4, with digits from the set {,, 3, 4, 5} and the repetition of digits is allowed, is. 65 Divisible by 4 last digits divisible by 4 ends in, 4, 3, 44 or = 65 *Q.9 Let X be the set consisting of the first 08 terms of the arithmetic progression, 6,,.., and Y be the set consisting of the first 08 terms of arithmetic progression 9, 6, 3,... Then, the number of elements in the set X Y is n(x Y) = n(x) + n(y) n(x Y), 6,,. 08 term FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fa 65394

25 JEE(DVNED)-08-Paper--PM-5 *Q.0 T n = + (n )5 = 5n 4 U K = 9 + (K )7 = 7K + for common terms 5n 4 = 7K + n = 7K 6 08, K K =, 7,. r + (r )5 440 r ma = 88 number of common term = 88 = n(xy) n(xy) = = The number of real solutions of the equation i i i i sin cos i i i i lying in the interval, is. (Here, the inverse trigonometric functions sin and cos assume values in, respectively.) and [0, ], sin / cos sin cos sin cos = sin 0 = 0 or = = 0 increasing function f(0) =, f(/) > 0 one root between 0, total number of solutions = Q. For each positive integer n, let y / n n n n... n n. n For R, let [] be the greatest integer less than or equal to. If. n lim y n L, then the value of [L] is FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fa 65394

26 JEE(DVNED)-08-Paper--PM-6 n n r ln(l) = lim ln n n r n 0 ln d ln ln = ln ( ln ) = ln(4/e) L = 4/e [L] = 0 0 Q. Let a and b be two unit vectors such that a b 0. For some, y R, let c a yb a b and the vector c is inclined at the same angle to both a and b, then the value of 8cos is. 3 c a cos = c b y cos = y c + y + = 4 8 cos = 3 ( a, b, a b all are to each other & are unit vectors). If c *Q.3 Let a, b, c be three non-zero real numbers such that the equation 3a cos b sin c,,, has two distinct real roots and with + = 3. Then, the value of b a is a cos bsin c 3a cos bsin c 3 3 sin 3 3 3a cos b cos sin c 3 3 a 3b cos a b sin c 3 3 3b a cos a 3b sin 0 b a Q.4 farmer F has a land in the shape of a triangle with vertices at P(0, 0), Q(, ) and R(, 0). From this land, a neighbouring farmer F takes away the region which lies between the side PQ and a curve of the form y = n (n > ). If the area of the region taken away by the farmer F is eactly 30% of the area of PQR, then the value of n is. FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fa 65394

27 JEE(DVNED)-08-Paper--PM-7 4 rea (PQR) = sq. unit n 3 = d 0 0 y ais n 5 n = 4. P(0, 0) Q(, ) y = n ais R(, 0) SETIN 3 (Maimum Marks: ) This section contains TW (0) paragraphs. Based on each paragraph, there are TW (0) questions. Each question has FUR options. NLY NE of these four options corresponds to the correct answer. For each question, choose the option corresponding to the correct answer. nswer to each question will be evaluated according to the following marking scheme: Full Marks : +3 If NLY the correct option is chosen. Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered). Negative Marks : In all other cases. PRGRPH X Let S be the circle in the -y plane defined by the equation + y = 4. (There are two questions based on PRGRPH X, the question given below is one of them) *Q.5 Let E E and F F be the chords of S passing through the point P 0 (, ) and parallel to the -ais and the y- ais, respectively. Let G G be the chord of S passing through P 0 and having slope. Let the tangents to S at E and E meet at E 3, the tangents to S at F and F meet at F 3, and the tangents to S at G and G meet at G 3. Then, the points E 3, F 3, and G 3 lie on the curve () + y = 4 (B) ( 4) + (y 4) = 6 () ( 4) (y 4) = 4 (D) y = 4 learly they lie on + y = 4 y ais (0, 4) a F a 3 (, ) E y= P 0 E ais = a F 3 (4, 0) F FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fa 65394

28 JEE(DVNED)-08-Paper--PM-8 PRGRPH X Let S be the circle in the -y plane defined by the equation + y = 4. (There are two questions based on PRGRPH X, the question given below is one of them) *Q.6 Let P be a point on the circle S with both coordinates being positive. Let the tangent to S at P intersect the coordinate aes at the points M and N. Then, the mid-point of the line segment MN must lie on the curve () ( + y) = 3y (B) /3 + y /3 = 4/3 () + y = y (D) + y = y D y yy and are same h k 4 4, y h k y = y h k y ais N(0, k) P ais M(h, 0) F PRGRPH There are five students S, S, S 3, S 4 and S 5 in a music class and for them there are five seats R, R, R 3, R 4 and R 5 arranged in a row, where initially the seat R i is allotted to the student S i, i =,, 3, 4, 5. But, on the eamination day, the five students are randomly allotted the five seats. (There are two questions based on PRGRPH, the question given below is one of them) Q.7 The probability that, on the eamination day, the student S gets the previously allotted seat R, and NNE of the remaining students gets the seat previously allotted to him/her is () 3 (B) 40 8 () 7 (D) 40 5 D4 9 3 P 5! 0 40 PRGRPH There are five students S, S, S 3, S 4 and S 5 in a music class and for them there are five seats R, R, R 3, R 4 and R 5 arranged in a row, where initially the seat R i is allotted to the student S i, i =,, 3, 4, 5. But, on the eamination day, the five students are randomly allotted the five seats. (There are two questions based on PRGRPH, the question given below is one of them) FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fa 65394

29 JEE(DVNED)-08-Paper--PM-9 Q.8 For i =,, 3, 4, let T i denote the event that the students S i and S i+ do NT sit adjacent to each other on the day of the eamination. Then, the probability of the event T T T 3 T 4 is () (B) 5 0 () 7 (D) 60 5 S S S S S S S S S S S S S S S S S S S S S S S S S S S S S S S S S S S Same number of ways in reverse order n E 7 7 PE n S 5! 60 FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fa 65394