Solution to IIT JEE 2018 (Advanced) : Paper - I

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1 Solution to IIT JEE 018 (Advanced) : Paper - I PART I PHYSICS SECTION 1 (Maimum Marks: 4) This section contains SIX (06) questions. Each question has FOUR options for correct answer(s). ONE OR MORE THAN ONE of these four option(s) is (are) correct option(s). For each question, choose the correct option(s) to answer the question. Answer to each question will be evaluated according to the following marking scheme: Full Mark : +4 If only (all) the correct option(s) is (are) chosen. Partial Marks : +3 If all the four options are correct but ONLY three options are chosen. Partial Marks : + If three or more options are correct but ONLY two options are chosen, both of which are correct options. Partial Marks : +1 If two or more options are correct but ONLY 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 ONLY 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 +1 marks. Selecting any incorrect option(s) (second option in this case), with or without selection of any correct option(s) will result in - marks. 1. The potential energy of a particle of mass m at a distance r from a fied point O 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 O. If v is the speed of the particle and L is the magnitude of its angular momentum about O, which of the following statements is (are) true? (A) v k k R (B) v R (C) L mk R m m (D) L 1. (B), (C) The potential energy of the particle is, kr V = v F = dv F = kr mk R dr At r = R, F = kr To keep the particle in circular motion, this force must be equal to centripetal force. mv So, kr = or v = kr k R R m m The angular momentum of the particle is, r O L = mvr = m k R m 0518/IITEQ18/Paper1/QP&Soln/Pg.1

2 () Vidyalankar : IIT JEE 018 Advanced : Question Paper & Solution. Consider a body of mass 1.0 kg at rest at the origin at time t = 0. A force F ( tˆi ˆj) is applied on the body, where = 1.0 Ns 1 and = 1.0 N. The torque acting on the body about the origin at time t = 1.0 S is Which of the following statements is (are) true? 1 (A) Nm 3 (B) The torque is in the direction of the unit vector ˆk 1 1 (C) The velocity of the body at t = 1 is v (i ˆ j)m ˆ s (D) The magnitude of displacement of the body at t = 1 s is 1 m 6. (A), (C) The force applied on the body is, F tˆ i ˆ j At = 1 N s 1 and = 1 N F t ˆi ˆ mdv j or t ˆi ˆj dt On integrating both the sides, t mv ˆi t ˆj t v ˆi t ˆj [m 1 kg] dr t ˆi t ˆj dt At t = 0, the body is at rest so, v 0 and r 0. On integrating both the sides, 3 t ˆ t r i ˆj 6 At t = 1 s, torque acting on the body is, r F 1 = ˆ 1 ˆ i j ˆi ˆj kˆ kˆ kˆ ˆ ˆ 1 v i j ˆi j ˆ ms At t = 1 s, the magnitude of displacement of the body is, 1 ˆ 1 ˆ 1 1 s r1 r0 i j 0 ˆi ˆ j At t = 1 s, velocity of the body is, s m A 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? (A) 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 (C) If this eperiment is performed in a lift going up with a constant acceleration, then h decreases (D) h is proportional to contact angel 0518/IITEQ18/Paper1/QP&Soln/Pg.

3 IIT JEE 018 Advanced : Question Paper & Solution (Paper I) (3) 3. (A), (C) gh R where R is the radius of meniscus. h = gr r R =, where r is the radius of capillary and is the angle of contact. cos h = cos gr (A) For a given material, is constant. Therefore, h 1 r (B) h depends on as h. (C) If lift is going up with constant acceleration a, g eff = g + a cos h = ; h decreases (g a) r (D) h is proportional to cos. 4. In the figure below, the switches S 1 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 a at time t = T. Which of the following statements is (are) true? (A) I ma V V (B) I ma (C) R 4R L ln (D) R L ln R 4. (B), (D) I I I ma 1 ma I 1 I V R/Lt V (R/L)t I I I1 1 e 1 e R R V R/Lt R/Lt = e e R 0518/IITEQ18/Paper1/QP&Soln/Pg.3

4 (4) Vidyalankar : IIT JEE 018 Advanced : Question Paper & Solution For I ma, d (I) = 0. dt e 1 e R/Lt e R t n L R/L t 1 R/L t t = L n R When I = I ma, t = = L n R V R L I ma = n L R R e e I ma = V 1 1 R 4 = V 4R R L n L R 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 1 and the wire located at = R carries a constant current I. A 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 / 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? (A) If I 1 = I, then B cannot be equal to zero at the origin (0, 0, 0) (B) If I 1 > 0 and I < 0, then B can be equal to zero at the origin (0, 0, 0) (C) If I 1 < 0 and I > 0, then B can be equal to zero at the origin (0, 0, 0) (D) If I 1 = I, then the z-component of the magnetic field at the centre of the loop is 0 I R 5. (A), (B), (D) z I (0, 0, 3 R) y R O R y (A) At the origin, B 0 due to two wires if I 1 = I, hence B net at the origin is equal to B, due to loop, which is non zero. (B) If I 1 > 0 and I < 0, B at the origin due to wires will be along ˆk direction and B due to loop is along ˆk direction, hence B can be zero at the origin. 0518/IITEQ18/Paper1/QP&Soln/Pg.4

5 IIT JEE 018 Advanced : Question Paper & Solution (Paper I) (5) (C) If I 1 < 0 and I > 0, B at the origin due to wires is along ˆk and also due to ring is along ˆk so, B can not be zero. (D) At the centre of the loop B due to wires is along -ais. Hence the z-component of 0 the magnetic field at the center of the loop is I ˆk. R 6. One 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? (A) Process I is an isochoric process (C) In process IV, gas releases heat (B) In process II, gas absorbs heat (D) Processes I and III are not isobaric 6. (B), (C), (D) (A) Volume V is decreasing in process I. (B) U = 0, W > 0 Q > 0 Process II is isothermal epansion. (C) U = 0, W < 0 Q < 0 Process III is isothermal compression. (D) For an isobaric process TV graph must be linear. SECTION II (Maimum Marks:4) This section contains EIGHT (08) questions. The answer to each question is a NUMERICAL VALUE. 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, ) using the mouse and the on-screen virtual numeric keypad in the place designated to enter the answer. Answer to each question will be evaluated according to the following marking scheme: Full Marks : +3 If ONLY the correct numerical value is entered as answer. Zero Marks : 0 In all other cases. 7. Two vectors A and B are defined as A ai ˆ and B a(cos tˆi sint ˆj), where a is a 7. [] constant and = /6 rad s 1. If A B 3 A B at time t = for the first time, the value of, in, seconds, is. t A B a cos 0518/IITEQ18/Paper1/QP&Soln/Pg.5

6 (6) Vidyalankar : IIT JEE 018 Advanced : Question Paper & Solution t A B a sin t t So, a cos 3 a sin tan t 1 3 t 6 t = 3 Now, = 6 rad s 1 t 6 3 t = s 8. Two men are walking along a horizontal straight line in the same direction. The man in front walks at a speed 1.0 m s 1 and the man behind walks at a speed.0 ms -1. A third man is standing at a height 1 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 1430 Hz. The speed of sound in air is 330 m s -1. At the instant, when the moving men are 10 m apart, the stationary man is equidistant from them. The frequency of beats in Hz, heard by the stationary man at this instant, is. 8. [5] C 13 m 13 m 1 m.0 m s m s 1 A 5 m O 5 m B 1 v 330 cos fa f cos v cos 330 cos v 330 cos fb f v cos 330 cos 330 f = cos 13cos 330 = Hz From CAO, cos A 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 ) / 10 s, then the height of the top of the inclined plane, in metres, is. Take g = 10 m s. 0518/IITEQ18/Paper1/QP&Soln/Pg.6

7 IIT JEE 018 Advanced : Question Paper & Solution (Paper I) (7) 9. [0.75] C Acceleration down the inclined plane is h g sin a = 60 k 1 A B R gsin a R gsin a D 3 h 1 g sin 4h 16h 3 tr tr sin 60 sin g sin 3g h 1 g sin 3h 4h 1 td td cos 60 sin 3 g sin g 3 Since tr td 10 16h 4h 3 3g g 10 h 4 3 g 3 10 [g = 10 m s ] h h 0.75m A spring-block system is resting on a frictionless floor as shown in the figure. The spring constant is.0 N m 1 and the mass of the block is.0 kg. Ignore the mass of the spring. Initially the spring is in an unstretched condition. Another block of mass 1.0 kg moving with a speed of.0 m s 1 collides elastically with the first block. The collision is such 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. 10. [.09] After collision 1v 1 + v = 1 v 1 + v 1 =. (i) v v1 e = 1 v v 1 =.. (ii) 4 1 On solving v = ms and v 1 = m s m Time period T = k m s 1 K = N m /IITEQ18/Paper1/QP&Soln/Pg.7

8 (8) Vidyalankar : IIT JEE 018 Advanced : Question Paper & Solution Displacement of 1 st block in time t = T s = s =.09 m 11. Three identical capacitors C 1, C and C 3 have a capacitance of 1.0 F each and they are uncharged initially. They are connected in a circuit as shown in the figure and C 1 is then filled completely with a dielectric material of relative permittivity r. The cell electromotive force (emf) V 0 = 8 V. First the switch S 1 is closed while the switch S is kept open. When the capacitor C 3 is fully charged, S 1 is opened and S is closed simultaneously. When all the capacitors reach equilibrium, the charge on C 3 is found to be 5 C. The value of r =. 11. [1.50] V 0 = 8V 1 F + 8 C 8 C r + 3 C 3 C 1F + 5 C 5 C Applying loop rule, r r F +3 C 3 C r 1. In the y-plane, the region y > 0 has a uniform magnetic field Bkand 1ˆ the region y < 0 has another uniform magnetic field Bk. ˆ A positively charged particle is projected from the origin along the positive y-ais with speed v 0 = m s 1 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 1, the average speed of the particle, in m s 1, along the -ais in the time interval T is. 0518/IITEQ18/Paper1/QP&Soln/Pg.8

9 IIT JEE 018 Advanced : Question Paper & Solution (Paper I) (9) 1. [] mv y R 1 = qb1 Bk 1ˆ mv V R = 0 = m/s qb B = 4B F V Bk R = ˆ R 1 4 Distance traveled in direction = R 1 + R R1 5R1 = R 1 + m T 1 = qb1 m T1 T = qb 4 T1 T 5T1 Total time = = 8 5R1 R1 Average speed V = 4 t 5T1 T1 8 R = mv qb T = m qb 4R 4V Average speed T Average speed = 13. Sunlight of intensity 1.3 kw m 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. 13. [130] I = 1.3 kw/m f = 0 cm ABF ~ PQF r 1 R 0 10 a 1 A 100 Total energy incident E = P 0 A = P a A P = P 0 a P = = 130 R B A F 0 cm cm P r Q 0518/IITEQ18/Paper1/QP&Soln/Pg.9

10 (10) Vidyalankar : IIT JEE 018 Advanced : Question Paper & Solution 14. Two conducting cylinders of equal length but different radii are connected in series between two heat baths kept at temperatures T 1 = 300 K and T = 100 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 1 and K respectively. If the temperature at the junction of the two cylinders in the steady state is 00 K, then K 1 / K =. 14. [4] T = 00 K k 1 k T 1 = 300 K T = 100 K r = r 1 A = 4A 1 dq k1a1 ka (300 00) (00 100) dt L L k 1 A 1 = k A k1 A 4 k A 1 SECTION III (Maimum Marks:1) This section contains TWO (0) paragraphs. Based on each paragraph, there are TWO (0) questions. Each question has FOUR options. ONLY ONE of these four options is corresponds to the correct answer. For each question, choose the option corresponding to the correct answer. Answer to each question will be evaluated according to the following marking scheme: Full Marks : +3 If ONLY the correct option is chosen. Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered). Negative Marks : 1 In all other cases. Paragraph for Q. No. 15 & 16 PARAGRAPH 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. All the quantities are given in SI units. 15. The relation between [E] and [B] is (A)[E] = [B] [L] [T] (C) [E] = [B] [L] [T] 1 (B) [E] = [B] [L] 1 [T] (D)[E] = [B] [L] 1 [T] /IITEQ18/Paper1/QP&Soln/Pg.10

11 IIT JEE 018 Advanced : Question Paper & Solution (Paper I) (11) 15. (C) F = qvb = qe E = vb [E] = [L] [T] 1 [B] 16. The relation between [ 0 ] and [ 0 ] is (A)[ 0 ] = [ 0 ] [L] [T] (B) [ 0 ] = [ 0 ] [L] [T] (C) [ 0 ] = [ 0 ] 1 [L] [T] (D)[ 0 ] = [ 0 ] 1 [L] [T] 16. (D) = c 0 = [L] [T] [ 0 ] 1 Paragraph for Q. No. 17 & 18 PARAGRAPH A 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, y and z, respectively, then z z = y 1 1 y y y y 1 y The series epansion for 1, to first power in y/y, is 1 (y/y). The relative errors y y in independent variables are always added. So the error in z will be z = z. y The above derivation makes the assumption that / << 1, y/y << 1. Therefore, the higher powers of these quantities are neglected. (1 a) 17. Consider the ratio r to be determined by measuring a dimensionless quantity a. (1 a) If the error in the measurement of a is a (a/a << 1), then what is the error r in a (A) (1 a) (B) 17. (B) r = 1 a 1 a r (1 a) (1 a) r (1 a) (1 a) r a a r 1a 1a r a r (1 a)(1 a) ( a) 1a r = (1 a)(1 a) 1 a a r (1 a) a (1 a) (C) a (1 a ) 1. aa (D) (1 a ) 0518/IITEQ18/Paper1/QP&Soln/Pg.11

12 (1) Vidyalankar : IIT JEE 018 Advanced : Question Paper & Solution 18. In an eperiment the initial number of radioactive nuclei is It is found that nuclei decayed in the first 1.0 s. For << 1, In (1 + ) = up to first power in. The error, in the determination of the decay constant, in s 1, is (A) 0.04 (B) 0.03 (C) 0.0 (D) (C) N = t Ne 0 nn = nn 0 t Differentiate both sides. dn d t N (There is no error in calculation of time t) N 40 = 0.0 Nt (where N is number of atoms remaining) PART II : CHEMISTRY SECTION I (Maimum Marks:4) This section contains SIX (06) questions. Each question has FOUR options for correct answer(s). ONE OR MORE THAN ONE of these four option(s) is (are) correct option(s). For each question, choose the correct option(s) to answer the question. Answer 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 ONLY three options are chosen. Partial Marks : + If three or more options are correct but ONLY two options are chosen, both of which are correct options. Partial Marks : +1 If two or more options are correct but ONLY 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 ONLY 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 +1 marks. Selecting any incorrect option(s) (second option in this case), with or without selection of any correct option(s) will result in - marks. 1. The compound(s) which generate(s) N gas upon thermal decomposition below 300 C is (are) (A) NH 4 NO 3 (B) (NH 4 ) Cr O 7 (C) Ba(N 3 ) (D) Mg 3 N 1. (B), (C) NH 4 NO 3(s) N O (g) + H O (NH 4 ) Cr O 7(s) N (g) + Cr O 3(s) + 4H O Ba(N 3 ) Ba + 3N (g) Mg 3 N doesn t undergo thermal decomposition below 300C. 0518/IITEQ18/Paper1/QP&Soln/Pg.1

13 IIT JEE 018 Advanced : Question Paper & Solution (Paper I) (13). The correct statement(s) regarding the binary transition metal carbonyl compounds is (are) (Atomic numbers: Fe = 6, Ni = 8) (A) Total number of valence shell electrons at metal centre in Fe(CO) 5 or Ni(CO) 4 is 16 (B) These are predominantly low spin in nature (C) Metal-carbon bond strengthens when the oidation state of the metal is lowered (D) The carbonyl C O bond weakens when the oidation state of the metal is increased. (B), (C) Total number of valence shell electrons at metal centre in Fe(CO) 5 or Ni(CO) 4 is 18. Metal carbonyl compounds are predominantly low spin in nature. MetalCarbon bond strengthens when the oidation state of the metal is lowered. 3. Based on the compounds of group 15 elements, the correct statement(s) is (are) (A) Bi O 5 is more basic than N O 5 (B) NF 3 is more covalent than BiF 3 (C) PH 3 boils at lower temperature than NH 3 (D) The N N single bond is stronger than the P P single bond 3. (A), (B), (C) Metal oides are basic in nature while non-metal oides are acidic in nature. Bi O 5 is more basic than N O 5. Non-metals mainly forms covalent bonds while metal forms ionic bonds. NF 3 is more covalent than BiF 3. Boiling point : NH 3 > PH 3 Bond energy P P > N N 4. In the following reaction sequence, the correct structure(s) of X is (are) (A) (B) (C) (D) 4. (B) Me OH PBr NaI NaN 3 3 EtO MeCO HCONMe Me N 3 Enantiomerically Pure 5. The reaction(s) leading to the formation of 1, 3, 5-trimethylbenzene is (are) (A) (B) (C) (D) 0518/IITEQ18/Paper1/QP&Soln/Pg.13

14 (14) Vidyalankar : IIT JEE 018 Advanced : Question Paper & Solution 5. (A), (B), (D) 6. A 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. The correct option (s) is (are) (A) q AC = U BC and W AB = P (V V 1 ) (B) W BC = P (V V 1 ) and q BC = H AC (C) H CA < U CA and q AC = U BC (D) q BC = H AC and H CA > U CA 6. (B), (C) AB = Isothermal process, E or U = 0 AC = Isochoric process, v = U BC = Isobaric process, P = H 0518/IITEQ18/Paper1/QP&Soln/Pg.14

15 IIT JEE 018 Advanced : Question Paper & Solution (Paper I) (15) SECTION II (Maimum Marks:4) This section contains EIGHT (08) questions. The answer to each question is a NUMERICAL VALUE. 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, ) using the mouse and the on-screen virtual numeric keypad in the place designated to enter the answer. Answer to each question will be evaluated according to the following marking scheme: Full Marks : +3 If ONLY the correct numerical value is entered as answer. Zero Marks : 0 In all other cases. 7. Among the species given below, the total number of diamagnetic species is. H atom, NO monomer, O (superoide), dimeric sulphur in vapour phase, MnsO 4, (NH 4 ) [FeCl 4 ], (NH 4 ) [NiCl 4 ], K MnO 4, K CrO 4 7. [1.00] K CrO 4 i.e. Cr +6 d 0 (diamagnetic) Paramagnetic = H-atom, NO (monomer), O (superoide), S (vap), Mn 3 O 4, (NH 4 ) [FeCl 4 ], (NH 4 ) [NiCl 4 ], K MnO The ammonia prepared by treating ammonium sulphate with calcium hydroide is completely used by NiCl.6H O to form a stable coordination compound. Assume that both the reactions are 100% complete. If 1584 g of ammonium sulphate and 95 g of NiCl.6H O are used in the preparation, the combined weight (in grams) of gypsum and the nickelammonia coordination compound thus produced is. (Atomic weights in g mol 1 : H = 1, N = 14, O = 16, S = 3, Cl = 35.5, Ca = 40, Ni = 59) 8. [99.00] (NH 4) SO 4 + Ca(OH) CaSO 4 + NH 3 + H O 1584g = 1 mol 1 mol 4 mol NiCl 6H O + 6NH 3 [Ni(NH 3 ) 6 ] Cl + 6H O 95 g = 4 mol 4 mol 4 mol Total mass = = 99 g 9. Consider an ionic solid MX with NaCl structure. Construct 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) numberof anions The value of in Z is. numberof cations 9. [3.00] MX have NaCl type structure. M + occupies all FCC positions and X occupies all octahedral voids. 0518/IITEQ18/Paper1/QP&Soln/Pg.15

16 (16) Vidyalankar : IIT JEE 018 Advanced : Question Paper & Solution (i) (ii) (iii) (iv) = M + = X The value of numberof anions numberof cations in Z = For the electrochemical cell, Mg(s) Mg + (aq, 1 M) Cu + (aq, 1 M) Cu(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 = K V1, where F is the Faraday constant and R is the gas constant, The value of ln(10) =.30) 10. [10.00] Mg (s) Mg + (aq. 1M) Cu + (aq. 1M) Cu (s) 0 E = RT [Mg E n ] F [Cu ] 0 E = E.70, when [Mg + ] = [Cu + ] Mg (s) Mg + (aq, M) Cu + (aq. 1M) Cu (s) Mg (s) + Cu Mg Cu(aq) (s) (aq) (Z) = M + = X E = RT [Mg ] E0 ne F [Cu ] =.70 ne = n e() ne = A closed tank has two compartments A 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 1). If the old partition is replaced by a new partition which can slide and conduct heat but does NOT allow the gas to leak across (Figure ), the volume (in m 3 ) of the compartment A after the system attains equilibrium is. 0518/IITEQ18/Paper1/QP&Soln/Pg.16

17 11. [.] IIT JEE 018 Advanced : Question Paper & Solution (Paper I) (17) A Old Partition 1 m 3, 5 bar 400 K 3 m 3, 1 bar 300 K Fig. 1 n A = No. of moles in vessel A = n B = No. of moles in vessel B = R R R R.(1).() New Partition (Conducting) A B Fig. Pressure and temperature on the both sides are must be same. A = No. of moles in vessel A = P.(3) RT P(4) n B = No. of moles in vessel B =.(4) RT Using equation 1,, 3 and 4, we get = Liquids A and B form ideal solution over the entire range of composition. At temperature T, equimolar binary solution of liquids A and B has vapour pressure 45 Torr. At the same temperature, a new solution of A and B having mole fractions A and B, respectively, has vapour pressure of.5 Torr. The value of A / B in the new solution is. (given that the vapour pressure of pure liquid A is 0 Torr at temperature T) 1. [19.00] The vapour pressure of pure liquid A, PA 0 torr The vapour pressure of pure liquid B, P B =? P Total = PA XAP B XB = 0 P B 90 = 0 PB PB 70 torr 0 For new solution, P Total =.5 = P X P X A A B B.5 = (1 ) = 0.95 = X A X B = 1 = 0.05 X X 0.05 A B 0518/IITEQ18/Paper1/QP&Soln/Pg.17

18 (18) Vidyalankar : IIT JEE 018 Advanced : Question Paper & Solution 13. The solubility of a salt of weak acid (AB) at ph 3 is Y 10 3 mol L 1. The value of Y is. (Given that the value of solubility product of AB (K sp ) = and the value of ionization constant of HB (K a ) = ) 13. [4.47] Solubility = [n ] Ksp 1 Ka 10 Y Y Y 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 NaCl in these solvents. NaCl completely dissociates in both the solvents. On 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. 14. [0.05] ( T ) i (K ) m b X NaCl b X ( T ) i (K ) m b Y NaCl b Y (K b) X (K b) Y On addition of equal no. of moles of a non-volatile solute, S in equal amount (in kg) of these solvent. SoluteS undergo dimerisation in solventx and Y Van t Hoff factor = i = 1 ( = degree of ionization) 0.7 i Y = i X =? ( T b) X3( T b) Y i (K ) m i (K ) m X b X Y b Y X 1 (K B) Xm (0.65) (K b) Ym X = /IITEQ18/Paper1/QP&Soln/Pg.18

19 IIT JEE 018 Advanced : Question Paper & Solution (Paper I) (19) SECTION III (Maimum Marks:1) This section contains TWO (0) paragraphs. Based on each paragraph, there are TWO (0) questions. Each question has FOUR options. ONLY ONE of these four options corresponds to the correct answer. For each question, choose the option corresponding to the correct answer. Answer to each question will be evaluated according to the following marking scheme: Full Marks : +3 If ONLY the correct option is chosen. Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered). Negative Marks : 1 In all other cases. Paragraph for Q. No. 15 & 16 PARAGRAPH X Treatment of benzene with CO/HCl in the presence of anhydrous AlCl 3 /CuCl followed by reaction with Ac O/NaOAc gives compound X as the major product. Compound X upon reaction with Br /Na CO 3, followed by heating at 473 K with moist KOH furnishes Y as the major product. Reaction of X with H /Pd-C, followed by H 3 PO 4 treatment gives Z as the major product. 15. The compound Y is (A) (B) (C) (D) 16. The compound Z is (A) (B) (C) (D) 15. (C), 16. (A) O C H CH CH COOH CO, HCl anhy. AlCl 3/CuCl AcO/NaOAc CH CH COOH Br, NaCO3 473K moistkoh (X) Cinnamic Acid C C H (X) H / Pd C CH CH COOH (Y) H3PO 4 (Z) O 0518/IITEQ18/Paper1/QP&Soln/Pg.19

20 (0) Vidyalankar : IIT JEE 018 Advanced : Question Paper & Solution Paragraph for Q. No. 17 & 18 PARAGRAPH A An organic acid P (C 11 H 1 O ) can easily be oidized to a dibasic acid which reacts with ethyleneglycol 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. 1) H /PdC ) NH 3/ 1) HPdC 1) HCl 3) Br /NaOH ) SOCl ) Mg/Et O 4) CHCl 3, KOH, 3) MeMgBr, CdCl 3) CO (dryice) /PdC 4) NaBH4 4) H3O S P Q R 17. The compound R is (A) (B) (C) (D) 18. The compound S is (A) (B) (C) (D) 17. (A), 18. (B) Organic Acid (C 11 H 1 O ) [O] dibasic acid ethylene glycol polymer (dacron) 0518/IITEQ18/Paper1/QP&Soln/Pg.0

21 IIT JEE 018 Advanced : Question Paper & Solution (Paper I) (1) COOH COOH + terephthalic acid alkene O3 Zn/HO ethylene glycol H H O CH CH OH dacron (P) aliphatic ketone + other products. O CH CH O C C O O n O C OH H /PdC O C OH SOCl O C Cl (P) Me Mg Br, CdCl Cl OH O HCl Me Me NaBH 4 Me (Q) Mg Et O CO (dryice) H 3 O + COOH Me (R) (P) O C OH H /PdC O OH NH 3/ O NH NC NH Br /NaOH CHCl 3, KOH H /PdC NH CH 3 (S) 0518/IITEQ18/Paper1/QP&Soln/Pg.1

22 () Vidyalankar : IIT JEE 018 Advanced : Question Paper & Solution PART III MATHEMATICS SECTION 1 (Maimum Marks:4) This section contains SIX (06) questions. Each question has FOUR options for correct answer(s). ONE OR MORE THAN ONE of these four option(s) is (are) correct option(s). For each question, choose the correct option(s) to answer the question. Answer 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 ONLY three options are chosen. Partial Marks : + If three or more options are correct but ONLY two options are chosen, both of which are correct options. Partial Marks : +1 If two or more options are correct but ONLY 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 ONLY 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 +1 marks. Selecting any incorrect option(s) (second option in this case), with or without selection of any correct option(s) will result in - marks. 1. For a nonzero comple number z, let arg (z) denote the principal argument with < arg (z). Then, which of the following statement (s) is (are) FALSE? (A) arg (1 i) =, where i = 1 4 (B) The function f : R (, ], defined by f(t) = arg (1 + it) for all t R, is continuous at all points of R, where i = 1 z1 (C) For any two nonzero comple numbers z 1 and z, arg ( arg(z 1 ) + arg(z ) is an z integer multiple of. (D) For any three given distinct comple numbers, z 1, z and z 3, the locus of the point (zz 1) (zz 3) z satisfying the condition arg =, lies on a straight line (zz 3) (zz 1) 1. (A) (B) (D) 3 (A) arg (1 i) = 4 (B) f(t) is discontinuous at t = 0 z1 (C) arg z = arg(z 1) arg (z ) + k Hence, it is true. z1 z Z3 Z (D) arg = arg z3 z Z1 Z is locus of a point lying on circle. 0518/IITEQ18/Paper1/QP&Soln/Pg.

23 IIT JEE 018 Advanced : Question Paper & Solution (Paper I) (3). In a triangle PQR, let PQR = 30 and the sides PQ and QR have lengths 10 3 and 10, respectively. Then, which of the following statement (s) is (are) TRUE? (A) QPR = 45 (B) The area of the triangle PQR is 5 3 and QRP = 10 (C) The radius of the incircle of the triangle PQR is (D) The area of the circumcircle of the triangle PQR is 100. (B),(C),(D) (10 3) 10 (PR) cos (30) = PR = 10 QPR = 30 and QRP = 10 Area = sin(30) = 5 3 sq.units r = S = units Area of circumcircle = R abc = 4 = 100 Sq units 10 3 P Q R 3. Let P 1 : + y z = 3 and P : + y + z = be two planes. Then, which of the following statement (s) is (are) TRUE? (A) The line of intersection of P 1 and P has direction ratios 1,, y z (B) The line is perpendicular to the line of intersection of P 1 and P (C) The acute angle between P 1 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 1 and P, then the distance of the point (, 1, 1) from the plane P 3 is 3 3. (C), (D) Let a, b, c be D.C s of line of intersect Then a + b c = 0 a + b + c = 0 a b = c a : b : c = 1 : 1 : 1 Hence, (A, B) are incorrect. 1 (C) cos () = = 1 = 60 (D) Eq. of required plane is 1( 4) 1(y ) + 1(z + ) = 0 y + z = 0 Distance of (, 1, 1) from y + z = 0 is 1 1 = /IITEQ18/Paper1/QP&Soln/Pg.3

24 (4) Vidyalankar : IIT JEE 018 Advanced : Question Paper & Solution 4. For every twice differentiable function f : R [, ] with (f(0)) + (f(0)) = 85, which of the following statement (s) is (are) TRUE? (A) There eist r, s R, where r < s, such that f is oneone on the open interval (r, s) (B) There eists 0 (4, 0) such that f( 0 ) 1 (C) lim f () = 1 (D) There eists a (4, 4) such that f(a) + f(a) = 0 and f(a) 0 4. (A), (B), (D) (A) f : R [, ] (f(0)) + (f (0)) = 85 (i) As co-domain [, ] and from (1) we can say function is not constant. f() is increasing (or decreasing) in some small interval. There eist r, s R where r < s, such that f is 11 on the (r, s) (B) Apply LMVT in [4, 0] f (0) f ( 4) f [ 0 ] =, 4 0 (4, 0) f (0) f ( 4) f ( 0 ) = 4 1 (C) Let f() = sin 85 f(0) = 0 So, f () = cos ( 85 ) ; 85 f (0) = 85 f() = sin 85 = does not emits lim lim Option (C) not correct. (D) Let g() = f () + (f ()) g() = f() f () + f () f ''() = f () [f() + f ''()] Now from option (B) f ( 0 ) 1 in (4, 0) LMVT for (0, 4) f () f (0) 4 ' f '( 0) 1 g( 0 ) 5 0 (4, 0) ' 0 g ( ) 5 ' 0 (0, 4) = ' f ' ( 0) As g(0) = (f(0)) + (f (0)) = 85 g() has ma in (4, 4) g() = 0 for some (4, 4) g() = f () [f() + f ''()] 0 = f () [f() + f ''()] As f () 0 f() + f ''() = 0 for some (4, 4). 5. Let f : R R and g : R R be two nonconstant differentiable functions. If f() = (f () g()) e g'() for all R, and f(1) = g() = 1, then which of the following statement (s) is (are) TRUE? (A) f() < 1 log e (B) f() > 1 log e (C) g(1) > 1 log e (D) g(1) < 1 log e 5. (B), (C) 0518/IITEQ18/Paper1/QP&Soln/Pg.4

25 f () IIT JEE 018 Advanced : Question Paper & Solution (Paper I) (5) e f () = g() g() e e f() f () e g() g() = 0 f () g() e f '() d e g'() d = 0 e f() + e g() = K As given f(1) = g() = 1 e f(1) + e g(1) = K = e f() + e g() e 1 + e g(1) = e f() + e 1 e f() + e g(1) = e So, e f() < e e g(1) < e f () < n 1 g(1) < n 1 f() > 1 n g(1) > 1 n 6. Let f : [0, ) R be a continuous function such that f() = 1 + [0, ). Then, which of the following statement (s) is (are) TRUE? (A) The curve y = f() passes through the point (1, ) (B) The curve y = f() passes through the point (, 1) (C) The area of the region {(, y) [0, 1] R : f() y (D) The area of the region {(, y) [0, 1] R : f() y 6. (B), (C) e f() = (1 )e + 0 e t f (t) dt diff. e f () e f() = (1 ) e e + e f() f () f() = 3 + I.F. = e Soln. y e = C + e (3) d = C + e d 3 e d e e = C e 1 1 t for all } is 4 1 } is 4 0 e f (t)dt y = C e y = Ce + 1 As f(0) = 1 1 = C + 1 C = 0 y = 1 f() = 1 It passes (, 1) (B) A = = = 4 4 (C) (0, 1) (1, 0) 0518/IITEQ18/Paper1/QP&Soln/Pg.5

26 (6) Vidyalankar : IIT JEE 018 Advanced : Question Paper & Solution SECTION II (Maimum Marks:4) This section contains EIGHT (08) questions. The answer to each question is a NUMERICAL VALUE. 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, ) using the mouse and the on-screen virtual numeric keypad in the place designated to enter the answer. Answer to each question will be evaluated according to the following marking scheme: Full Marks : +3 If ONLY the correct numerical value is entered as answer. Zero Marks : 0 In all other cases. 1 1 log (log9) log The value of (log 9) 7 7. [8] log(log 9) 1 log 9 = log log 9 is. 7 log log = 4 = 8 log The number of 5 digit numbers which are divisible by 4, with digits from the set {1,, 3, 4, 5} and the repetition of digits is allowed, is. 8. [65] Divisible by 4 means last two digits 1, 4, 3, 44, 5 1 So, Noof ways Total ways = = 5 4 = Let X be the set consisting of the first 018 terms of the arithmetic progression 1, 6, 11,.., and Y be set consisting of the first 018 terms of the arithmetic progression 9, 16, 3,.., Then, the number of elements in the set X Y is. 9. [3748] X = 1, 6, 11, 16, 1, 6, 31, 36, 41, 46, 51, 56, 61, 66, 71, 76, 81, 86, Y = 9, 16, 3, 30, 37, 44, 51, 58, 65, 7, 79, 86, 1418 X Y = 16, 51, 86, 11, (A.P. d = 35, a = 16) So, n(x Y) = t So, 16 + (t 1) t 88.7 t = 88 n(x Y) = n(x) + n(y) n(x Y) = = /IITEQ18/Paper1/QP&Soln/Pg.6

27 IIT JEE 018 Advanced : Question Paper & Solution (Paper I) (7) 10. The number of real solutions of the equation i 1 i 1 i 1 i sin = cos ( ) i1 i1 i1 i1 1 1 lying in the interval, is. (Here, the inverse trigonometric functions sin 1 and cos 1 assume values in, and [0, ], respectively.) 10. [] i1 = = i1 = i1 i 1 = i i1 = 1 i = i1 = 1 ( ) = 1 1 i i i = As given sin 1 i1 1 cos i1 i1 i1 i1 i i i1 i i1 i1 i1 i (1 )( ) (1 ) ( ) = 0 = 0 Let f() = f(0) < 0, f () = D < 0 f () is increasing So, only two solutions one root lies between 1 0, 0518/IITEQ18/Paper1/QP&Soln/Pg.7

28 (8) Vidyalankar : IIT JEE 018 Advanced : Question Paper & Solution 11. For each positive integer n, let y n = 1 ((n 1) (n )...(n n)) n. n 1 For R, let [] be the greatest integer less than or equal to. If value of [L] is. 11. [1] n y n = 1 log n 1 n... nn n n n n n lim = n 1 r log1 n n n y n = r1 n n lim y n n = n 1 r lim log1 n n 1 0 r1 log (1 ) d 1 n (L) = log (1 ) d 0 1 = log n 1 = log 1 + [ n 0] 4 = n e lim y n = L, then the n L = 4 e [L] = 1 1. Let a and b be two unit vectors such that a. b 0 For some, y R, let c ay b (a b). If c = and the vector c is inclined at the same angle a to both a and b, then the value of 8 cos a is. 1. [3] a = b = 1 a b = 0, c = a c = cos b c = cos c = a + y b + (a b) c a =, c b = y So, = y = cos dot product with c 4 = (a c) + y(b c) + a b c 4 = + y + [a b c] 4 = = 4 4 Solving = or = /IITEQ18/Paper1/QP&Soln/Pg.8 [a b c] = = a a a b a c b a b b b c c a c b c c = 4 [abc] = 4 defined if <

29 IIT JEE 018 Advanced : Question Paper & Solution (Paper I) (9) But, as > ( cos ) = 3 8 cos = Let a, b, c be three nonzero real numbers such that the equation 3 a cos + b sin = c,,, has two distinct real roots a and with a + =. Then, the value of 3 b is. a 13. [0.5] 3 a cos + b sin = c 3 cos + b a sin = c a As, are roots 3 cos + b a sin = c (1) a 3 cos + b a sin = c () a (1) () 3(cos cos ) + b (sin sin ) = 0 a 3sin sin + b cos sin a = 0 Given + = sin + b 3 sin a 1 = b b a a = 1 = A farmer F 1 has a land in the shape of a triangle with vertices at P(0, 0), Q (1, 1) 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 > 1). If the area of the region taken away by the farmer F by the farmer F is eatly 30% of the area of PQR, then the value of n is. 14. [4] 1 n ( ) d = n1 1 n n 1 = n 1 10 n + 1 = 5 n = 4 = 3 10 P (0, 0) y = n Q(1, 1) R(,0) 0518/IITEQ18/Paper1/QP&Soln/Pg.9

30 (30) Vidyalankar : IIT JEE 018 Advanced : Question Paper & Solution SECTION III (Maimum Marks:1) This section contains TWO (0) paragraphs. Based on each paragraph, there are TWO (0) questions. Each question has FOUR options. ONLY ONE of these four options corresponds to the correct answer. For each question, choose the option corresponding to the correct answer. Answer to each question will be evaluated according to the following marking scheme: Full Marks : +3 If ONLY the correct option is chosen. Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered). Negative Marks : 1 In all other cases. Paragraph for Q. No. 15 & 16 PARAGRAPH X Let 5 be the circle in the y-plane defined by the equation + y = Let E 1, E and F 1 F be the chords of S passing through the point P 0 (1, 1) and parallel to the -ais and the y-ais, respectively. Let G 1 G be the chord of S passing through P 0 and having slope 1. Let the tangents to S at E 1 and E meet at E 3, the tangents to S at F 1 and F meet at F 3, and the tangents to S at G 1 and G meet at G 3. Then, the points E 3, F 3, and G 3 lie on the curve (A) + y = 4 (B) ( 4) + (y 4) = 16 (C) ( 4) (y 4) = 4 (D) y = (A) E 3 G 1 G 3 E 1 F 1 E P 0 (1,1) G (,0) F 3 F E 1 ( 3,1), E ( 3,1), F 1 (1, 3), F (1, 3), G 1 (0, ), G (, 0) Tangent at E ( 3,1) is 3 + y = 4 Solving (0, 4) = E 3 Similarly (4, 0) = F 3 (, ) = G 3 E 3, F 3, G 3 satisfy option (A) E 1 ( 3,1) is 3 + y = /IITEQ18/Paper1/QP&Soln/Pg.30

31 IIT JEE 018 Advanced : Question Paper & Solution (Paper I) (31) 16. 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 (A)( + y) = 3y (B) /3 + y /3 = 4/3 (C) + y = y (D) + y = y 16. (D) Tangent cos + y sin = M,0 cos, N 0, sin Let P(h, k) mid point of M and N 0 0 P(h, k) = cos, sin cos = 1 h, cos + sin = = 1 h k h + k = h k Locus + y = y sin = 1 k Paragraph for Q. No. 17 & 18 PARAGRAPH A There are five students S 1, S, S 3, S 4 and S 5 in a music class and for them there are five seats R 1, 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 = 1,, 3, 4, 5. But, on the eamination day, the five students are randomly allotted the five seats. 17. The probability that, on the eamination day, the student S 1 gets the previously allotted seat R 1, and NONE of the remaining students gets the seat previously allotted to him/her is (A) 3 (B) 1 (C) 7 (D) (A) n(s) = 5! = 10 event A = the student S 1 gets the previously allotted seat R 1 and None of the remaining students gets the seat previously allotted n(a) = D 4 = 4! 1 1!! 3! 4! = 9 9 P(A) = 10 = For i = 1,, 3, 4, let T i denote the event that the students S i and S i+1 do NOT sit adjacent to each other on the day of the eamination. Then, the probability of the event T 1 T T 3 T 4 is (A) 1 15 (B) 1 10 (C) 7 60 (D) /IITEQ18/Paper1/QP&Soln/Pg.31

32 (3) Vidyalankar : IIT JEE 018 Advanced : Question Paper & Solution 18. (C) Totaln(T1 T T3 T 4) P(T 1 T T 4 ) = n(s) 5! { n (T ) n (T T ) n (T T T )...} 5! = C1 4!! ( C1 3!! C1!! 3!) ( C1!! C1!!) = { } = = = = /IITEQ18/Paper1/QP&Soln/Pg.3