PAPER 2 (JEE ADVANCED 2018) PART I : PHYSICS

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1 PAPER (JEE ADVANCED 08) PART I : PHYSICS SECTION (Maximum Marks: 4) This section contains SIX questions. Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four options is correct. For each question, choose the correct option(s)to answer the question. For each question, marks will be awarded in one of the following categories: Full Marks : +4 If only (all) the correct option(s) is (are) chosen. Full Marks : +3 If all the four options are correct but ONLY three options are chosen. Full Marks : + If three or more options are correct but ONLY two options are chosen, both of which are correct options. Full Marks : + 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 Example: 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 + marks. Selecting any incorrect option(s) (second option in this case), with or without selection of any correct option(s) will result in marks.. A particle of mass m is initially at rest at the origin. It is subjected to a force and starts moving along the x-axis. Its kinetic energy K changes with time as dk/dt = t, where is a positive constant of appropriate dimensions. Which of the following statements is (are) true? (A) The force applied on the particle is constant (B) The speed of the particle is proportional to time (C) The distance of the particle from the origin increases linearly with time (D) The force is conservative. Solution: (A, B, D) dk t t K dt t mv v t m dv F m constant dt. Consider a thin square plate floating on a viscous liquid in a large tank. The height h of the liquid in the tank is much less than the width of the tank. The floating plate is pulled horizontally with a constant velocity uo. Which of the following statements is (are) true? (A) The resistive force of liquid on the plate is inversely proportional to h (B) The resistive force of liquid on the plate is independent of the area of the plate (C) The tangential (shear) stress on the floor of the tank increases with uo IIT KALRASHUKLA CIVIL LINES, KAKADEO, SOUTH KANPUR []

2 (D) The tangential (shear) stress on the plate varies linearly with the viscosity of the liquid. Solution: (A, C, D) u o dv Velocity gradient dz h dv Au o F Fv A dz h 3. An infinitely long thin non-conducting wire is parallel to the z-axis and carries a uniform line charge density. It pierces a thin non-conducting spherical shell of radius R in such a way that the arc PQ subtends an angle 0 at the centre O of the spherical shell, as shown in the figure. The permittivity of free space is o. Which of the following statements is (are) true? (A) The electric flux through the shell is 3R / (B) The z-component of the electric field is zero at all the points on the surface of the shell (C) The electric flux through the shell is R / (D) The electric field is normal to the surface of the shell at all points 3. Solution: (A, B) Length of wire inside shell = qin R 3 R 3 o o R sin 60 R 3 E due to wire is parallel to z-axis hence z-component is zero at all points. o o IIT KALRASHUKLA CIVIL LINES, KAKADEO, SOUTH KANPUR []

3 4. A wire is bent in the shape of a right angled triangle and is placed in front of a concave mirror of focal length f, as shown in the figure. Which of the figures shown in the four options qualitatively represent(s) the shape of the image of the bent wire? (These figures are not to scale.) (A) (B) (C) (D) 4. Solution: (B) f When object is at u f f m f u f f The image will be virtual and enlarged. f AB A'B' AB f Image of point C will be at. 5. In a radioactive decay chain, 3 90 Th nucleus decays to Pb nucleus. Let N and N 8 be the number of and particles, respectively, emitted in this decay process. Which of the following statements is (are) true? (A) N 5 (B) N 6 (C) N (D) N 4 IIT KALRASHUKLA CIVIL LINES, KAKADEO, SOUTH KANPUR [3]

4 5. Solution: (A, C) 3 0 No. of particles The atomic number must decrease by 0. But in this case it is decreased by 8 only hence - particles are emitted. 6. In an experiment to measure the speed of sound by a resonating air column, a tuning fork of frequency 500 Hz is used. The length of the air column is varied by changing the level of water in the resonance tube. Two successive resonances are heard at air columns of length 50.7 cm and 83.9 cm. Which of the following statements is (are) true? (A) The speed of sound determined from this experiment is 33 ms (B) The end correction in this experiment is 0.9 cm (C) The wavelength of the sound wave is 66.4 cm (D) The resonance at 50.7 cm corresponds to the fundamental harmonic 6. Solution: (A, C) l e ( n) 4 l e (n 3) 4 hence, ( l l ) ( l l ) cm cm 500 Hz v 33 m/ sec. n = 0 then e = ve (not possible) if n = then e = ve (not possible) = 0.9 IIT KALRASHUKLA CIVIL LINES, KAKADEO, SOUTH KANPUR [4]

5 SECTION (Maximum Marks: 4) This section contains EIGHT 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, 7.30) using the mouse and the onscreen 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. A solid horizontal surface is covered with a thin layer of oil. A rectangular block of mass m = 0.4 kg is at rest on this surface. An impulse of.0 Ns is applied to the block at time t = 0 so that it starts moving along the x-axis with a velocity v(t) = voe t/, where vo is a constant and = 4s. The displacement of the block, in metres, at t = is. Take e = Solution: (6.30) t/ v voe J here vo.5 m / sec. m 0.4 t t/ dx voe dt 0 x.5 e t/ / 0 t/ t x.5 4 e x 0 e 0 ( 0.37) x A ball is projected from the ground at an angle of 45 with the horizontal surface. It reaches a maximum height of 0 m and returns to the ground. Upon hitting the ground for the first time, it loses half of its kinetic energy. Immediately after the bounce, the velocity of the ball makes an angle of 30 with the horizontal surface. The maximum height it reaches after the bounce, in metres, is. 8. Solution: (30.00) o vo sin 45 0m () g o v sin 30 g vo / 0. v / 4 x 0. x x 30m. x () IIT KALRASHUKLA CIVIL LINES, KAKADEO, SOUTH KANPUR [5]

6 9. A particle, of mass 0 3 kg and charge.0 C, is initially at rest. At time t = 0, the particle comes under the influence of an electric field E(t) E sin t ˆ i, where E.0 NC and 3 0 rad s speed, in. Consider the effect of only the electrical force on the particle. Then the maximum ms, attained by the particle at subsequent times is. 9. Solution: (.00) F qe qe ˆ o sin t i Max speed will be when t rad. V / qeo dv sin t dt m 0 0 qeo qeo Vmax cos 3 3 m m A moving coil galvanometer has 50 turns and each turn has an area 0 m. The magnetic field produced by the magnet inside the galvanometer is 0.0T. The torsional constant of the 4 suspension wire is 0 Nm rad. When a current flows through the galvanometer, a full scale deflection occurs if the coil rotates by 0. rad. The resistance of the coil of the galvanometer is 50. This galvanometer is to be converted into an ammeter capable of measuring current in the range 0.0 A. For this purpose, a shunt resistance is to be added in parallel to the galvanometer. The value of this shunt resistance, in ohms, is. 0. Solution: (5.56) For moving coil galvanometer, NiAB c i g ig 0.A igg S 5.56 i i 0. 9 g o o. A steel wire of diameter 0.5 mm and Young s modulus 0 Nm carries a load of mass M. The length of the wire with the load is.0m. A vernier scale with 0 divisions is attached to the end of this wire. Next to the steel wire is a reference wire to which a main scale, of least count.0 mm, is attached. The 0 divisions of the vernier scale correspond to 9 divisions of the main scale. Initially, the zero of vernier scale coincides with the zero of main scale. If the load on the steel wire is increased by. kg, the vernier scale division which coincides with a main scale division is. Take g = 0 ms and = 3... Solution: (3.00) MgL.0 Extension ( l ) 3 r Y 3. (0.50 ) 0 4 l 3 0 m 0.3 mm VSR 0.3mm n(0. m) n = 3 IIT KALRASHUKLA CIVIL LINES, KAKADEO, SOUTH KANPUR [6]

7 . One mole of a monatomic ideal gas undergoes an adiabatic expansion in which its volume becomes eight times its initial value. If the initial temperature of the gas is 00 K and the universal gas constant R = 8.0 J mol K, the decrease in its internal energy, in Joule, is.. Solution: (900.00) TV T V constant, here T V /3 /3 00(V) T (8 V) /3 / T 5K /3 (8) 3R U ncvt (75) 900J 3. In a photoelectric experiment a parallel beam of monochromatic light with power of 00 W is incident on a perfectly absorbing cathode of work function 6.5 ev. The frequency of light is just above the threshold frequency so that the photoelectrons are emitted with negligible kinetic energy. Assume that the photoelectron emission efficiency is 00%. A potential difference of 500 V is applied between the cathode and the anode. All the emitted electrons are incident 4 normally on the anode and are absorbed. The anode experiences a force F n 0 N due to the impact of the electrons. The value of n is. Mass of the electron 3 9 me = 9 0 kg and.0 ev =.6 0 J. 3. Solution: (4.00) No. of photons incident per sec. = No. of electrons emitted per sec F nmv n mke n mev F N 4. Consider a hydrogen-like ionized atom with atomic number Z with a single electron. In the emission spectrum of this atom, the photon emitted in the n = to n = transition has energy 74.8 ev higher than the photon emitted in the n = 3 to n = transition. The ionization energy of the hydrogen atom is 3.6 ev. The value of Z is. 4. Solution: (3.00) 3.6Z 36Z ev 3.6Z Z 3 IIT KALRASHUKLA CIVIL LINES, KAKADEO, SOUTH KANPUR [7]

8 SECTION 3 (Maximum Marks : ) This section contains FOUR questions Each question has TWO matching lists: LIST I and LIST II. FOUR options are given representing matching of elements from LIST I and LIST II. ONLY ONE of these four options corresponds to a correct matching. For each question, choose the option corresponding to the correct matching. For each question, marks will be awarded according to the following marking scheme: Full Marks : +3 If ONLY the option corresponding to the correct matching is chosen. Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered). Negative Marks : In all other cases. 5. The electric field E is measured at a point P (0, 0, d) generated due to various charge distributions and the dependence of E on d is found to be different for different charge distributions. List-I contains different relations between E and d. List-II describes different electric charge distributions, along with their locations. Match the functions in List-I with the related charge distributions in List-II. List-I List-II P. E is independent of d. A point charge Q at the origin Q. E d. A small dipole with point charges Q at (0, 0, l) and Q At (0, 0, l). Take l << d R. E d 3. An infinite line charge coincident with the x-axis, with uniform linear charge density S. E 3 d 4. Two infinite wires carrying uniform linear charge density parallel to the x-axis. The one along (y = 0, z = l) has a charge density + and the one along (y = 0, z = l) has a charge density. Take l << d. 5. Infinite plane charge coincident with the xy-plane with uniform surface charge density (A) P 5; Q 3, 4; R ; S (B) P 5; Q 3; R, 4; S (C) P 5; Q 3; R, ; S 4 (D) P 4; Q, 3; R ; S 5 5. Solution: (B) [] kq E d R [] kp E 3 3 d d S [3] E o d d Q 3 [4] ( l) E o (d l ) o (d l ) o d d R 4 [5] E = constant P 5 IIT KALRASHUKLA CIVIL LINES, KAKADEO, SOUTH KANPUR [8]

9 6. A planet of mass M, has two natural satellites with masses m and m. The radii of their circular orbits are R and R respectively. Ignore the gravitational force between the satellites. Define, L, K and T to be, respectively, the orbital speed, angular momentum, kinetic energy and time period of revolution of satellite, and v, L, K and T to be the corresponding quantities of satellite. Given m/m = and R/R = ¼, match the ratios in List-I to the numbers in List-II. List-I List-II v P.. v 8 Q. R. S. L L K K T T (A) P 4; Q, 4; R ; S 3 (C) P ; Q 3; R, ; S 4 6. Solution: (B) GM v R R and (B) P 3; Q ; R 4; S (D) P ; Q 3, 3; R 4; S v R 4 P 3 v R GM L mvr m R m R R L m R.. Q L m R u GM m R R k m R k m R R 4 k mv m. 3/ (R) T R GM 3/ 3/ 3/ T R T R 4 8 S IIT KALRASHUKLA CIVIL LINES, KAKADEO, SOUTH KANPUR [9]

10 7. One mole of a monatomic ideal gas undergoes four thermodynamic processes as shown schematically in the PV-diagram below. Among these four processes, one is isobaric, one is isochoric, one is isothermal and one is adiabatic. Match the processes mentioned in List-I with the corresponding statements in List-II. List-I List-II P. In process I. Word done by the gas is zero Q. In process II. Temperature of the gas remains unchanged R. In process III 3. No heat is exchanged between the gas and its surroundings S. In process IV 4. Work done by the gas is 6 PoVo (A) P 4; Q 3, 4; R ; S (B) P ; Q 3; R ; S 4 (C) P 3; Q 4; R, ; S (D) P 3; Q 4, 3; R ; S 7. Solution: (C) Slope of adiabatic is more than isothermal hence I is adiabatic and (IV) is isothermal. (P) (3) (Q) (4): W = Area under PV curve = 3P0(V0) = 6P0V0 (R) () (S) () 8. In the List-I below, four different paths of a particle are given as functions of time. In these functions, and are positive constants of appropriate dimensions and. In each case, the force acting on the particle is either zero or conservative. In List-II, five physical quantities of the particle are mentioned: p is the linear momentum, L is the angular momentum about the origin, K is the kinetic energy, U is the potential energy and E is the total energy. Match each path in List-I with those quantities in List-II, which are conserved for that path. List-I List-II P. r(t) t ˆi t ˆj. p Q. r(t) cos t ˆi sin t ˆj. L R. r(t) (cost ˆi sin t ˆj) 3. K S. ˆ r(t) t i t ˆj 4. U 5. E (A) P,, 3, 4, 5; Q, 5; R, 3, 4, 5; S 5 (B) P,, 3, 4, 5; Q 3, 5; R, 3, 4, 5; S, 5 (C) P, 3, 4; Q 5; R,, 4; S, 5 (D) P,, 3, 5; Q, 5; R, 3, 4, 5; S, 5 IIT KALRASHUKLA CIVIL LINES, KAKADEO, SOUTH KANPUR [0]

11 8. Solution: (A) (P) r ti ˆ tj ˆ v ˆi ˆ j cons. p const. L m(r v) m tkˆ tkˆ 0 i.e., constant v constant F zero v = constant E = constant. P,, 3, 4, 5 (Q) r cos t ˆi sin t ˆj v sin t ˆi cost ˆj ˆ a cos t i sin t ˆj v = variable p conserve. F a r r F 0 i.e., 0 k variable. Motion periodic and force is conservative E = constant. Q, 5 (R) r a cos t ˆi sin t ˆj v a sin t ˆi cos t ˆj v constant, L constant, p constant R, 3, 4, 5 t (S) r ti ˆ ˆj v ˆi tj ˆ S 5 a ˆ j L constant. END OF PART I : PHYSICS IIT KALRASHUKLA CIVIL LINES, KAKADEO, SOUTH KANPUR []

12 PART II : CHEMISTRY SECTION (Maximum Marks: 4) This section contains SIX questions. Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four options is correct. For each question, choose the correct option(s)to answer the question. For each question, marks will be awarded in one of the following categories: Full Marks : +4 If only (all) the correct option(s) is (are) chosen. Full Marks : +3 If all the four options are correct but ONLY three options are chosen. Full Marks : + If three or more options are correct but ONLY two options are chosen, both of which are correct options. Full Marks : + 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 Example: 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 + marks. Selecting any incorrect option(s) (second option in this case), with or without selection of any correct option(s) will result in marks.. The correct option(s) regarding the complex [Co(en)(NH3)3(HO)] 3+ (en = HNCHCHNH) is (are) (A) It has two geometrical isomers (B) It will have three geometrical isomers if bidentate en is replaced by two cyanide ligands (C) It is paramagnetic (D) It absorbs light at longer wavelength as compared to [Co(en)(NH3)4] 3+. Solution: (A, B, D) 3 (A) [Co(en)(NH ) (H O)] gives Geometrical isomers 3 3 (B) Compound [Co(CN) (NH 3) 3(HO)] give 3 Geometrical isomers (C) [Co(en)(NH ) (H O)] is diamagnetic (D) [Co(en)(NH ) (H O)] absorbs light at longer wavelength as compared to [Co(en)(NH ) ] as HO is weaker ligand NH3 IIT KALRASHUKLA CIVIL LINES, KAKADEO, SOUTH KANPUR []

13 . The correct option(s) to distinguish nitrate salts of Mn + and Cu + taken separately is (are) (A) Mn + shows the characteristic green colour in the flame test (B) Only Cu + shows the formation of precipitate by passing HS in acidic medium (C) Only Mn + shows the formation of precipitate by passing HS in faintly basic medium (D) Cu + /Cu has higher reduction potential than Mn + /Mn (measured under similar condition. Solution: (B, D) (A) Cu + show characteristic green colour in the flame test (B) Only Cu + can give ppt in Acidic medium on passing HS (C) Both Cu + and Mn + show the formation of ppt. by passing HS in faintly basic medium. 0 0 (D) E E as per electrochemical series. Cu /Cu Mn /Mn 3. Aniline reacts with mixed acid (conc. HNO3 and conc. HSO4) at 88 K to give P (5 %), Q (47%) and R (%). The major product(s) of the following reaction sequence is (are) (A) (B) (C) (D) 3. Solution: (D) IIT KALRASHUKLA CIVIL LINES, KAKADEO, SOUTH KANPUR [3]

14 4. The Fischer presentation of D-glucose is given below. The correct structure(s) of -L-glucopyranose is (are) (A) (B) (C) (D) 4. Solution: (D) IIT KALRASHUKLA CIVIL LINES, KAKADEO, SOUTH KANPUR [4]

15 5. For a first order reaction A(g) B(g) + C(g) at constant volume and 300 K, the total pressure at the beginning (t = 0) and at time t are P0 and Pt, respectively. Initially, only A is present with concentration [A]0, and t/3 is the time required for the partial pressure of A to reach /3 rd of its initial value. The correct option(s) is (are) (Assume that all these gases behave as ideal gases) (A) (B) (C) (D) 5. Solution: (A, D) A B C P0 P p p p 0 P0 + p = Pt Pt P0 p 3P0 Pt P0 p = P0 k = l n t (3P P ) 0 t ln(3p0 Pt) = ln P0 kt 3P0 k l n t P t /3 /3 0 n 3 l k IIT KALRASHUKLA CIVIL LINES, KAKADEO, SOUTH KANPUR [5]

16 6. For a reaction, A P, the plots of [A] and [P] with time at temperature T and T are given below. If T >T, the correct statement(s) is (are) (Assume ΔH Ɵ and ΔS Ɵ are independent of temperature and ratio of lnk at T to lnk at T is T greater than. Here H, S, G and K are enthalpy, entropy, Gibbs energy and equilibrium T constant, respectively.) (A) ΔH Ɵ < 0, ΔS Ɵ < 0 (B) ΔG Ɵ < 0, ΔH Ɵ > 0 (C) ΔG Ɵ < 0, ΔS Ɵ < 0 (D) ΔG Ɵ < 0, ΔS Ɵ > 0 6. Solution: (A, C) [P] Keq = K K [A when T > T ΔHƟ < 0 ln K T ln K T T ln K > T ln K G = RT lnk 0 0 G G H T S H T S (T T) S 0 < 0 S 0 < 0 Keq > So G 0 < 0 IIT KALRASHUKLA CIVIL LINES, KAKADEO, SOUTH KANPUR [6]

17 SECTION (Maximum Marks: 4) This section contains EIGHT 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, 7.30) using the mouse and the onscreen 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. The total number of compounds having at least one bridging oxo group among the molecules given below is. NO3, NO5, P4O6, P4O7, H4PO5, H5P3O0, HSO3, HSO5 7. Solution: (5) NO3 O N N O O NO5 P4O6 P4O7 H4PO5 H5P3O0 HSO3 HSO5 HO P O P OH OH OH O O O H O P O P O P OH OH OH OH S H O S O H O S O HO S S OH O IIT KALRASHUKLA CIVIL LINES, KAKADEO, SOUTH KANPUR [7]

18 8. Galena (an ore) is partially oxidized by passing air through it at high temperature. After some time, the passage of air is stopped, but the heating is continued in a closed furnace such that the contents undergo self-reduction. The weight (in kg) of Pb produced per kg of O consumed is. (Atomic weights in g mol : O = 6, S = 3, Pb = 07) 8. Solution: (6.47) PbS + 3O PbO + SO PbO + PbS 3Pb + SO 3 3 gm of O produces = 3 07 gm of Pb. kg of O produces = = 6.47 kg 9. To measure the quantity of MnCl dissolved in an aqueous solution, it was completely converted to KMnO4 using the reaction, MnCl + KSO8 + HO KMnO4 + HSO4 + HCl (equation not balanced). Few drops of concentrated HCl were added to this solution and gently warmed. Further, oxalic acid (5 mg) was added in portions till the colour of the permanganate ion disappeared. The quantity of MnCl (in mg) present in the initial solution is. (Atomic weights in g mol : Mn = 55, Cl = 35.5) 9. Solution: (6) MnO4 5CO4 6H Mn 0CO 8HO 5 moles of oxalic acid react with = moles of KMnO = 0 3 Moles of KMnO4 = moles of MnCl Moles of MnCl = gm = 6 mg 0. For the given compound X, the total number of optically active stereoisomers is. 0. Solution: (7) IIT KALRASHUKLA CIVIL LINES, KAKADEO, SOUTH KANPUR [8]

19 . In the following reaction sequence, the amount of D (in g) formed from 0 moles of acetophenone is. (Atomic weights in g mol : H =, C =, N = 4, O = 6, Br = 80. The yield (%) corresponding to the product in each step is given in the parenthesis). Solution: (495) Moles of D formed =.5 Mass of D formed = gram.. The surface of copper gets tarnished by the formation of copper oxide. N gas was passed to prevent the oxide formation during heating of copper at 50 K. However, the N gas contains mole % of water vapour as impurity. The water vapour oxidises copper as per the reaction given below: Cu(s) + HO(g) CuO(s) + H(g) ph is the minimum partial pressure of H (in bar) needed to prevent the oxidation at 50 K. The value of ln (p ) is. H IIT KALRASHUKLA CIVIL LINES, KAKADEO, SOUTH KANPUR [9]

20 (Given: total pressure = bar, R (universal gas constant) = 8 J K mol, ln(0) =.3. Cu(s) and CuO(s) are mutually immiscible. At 50 K: Cu(s) + ½ O(g) CuO(s); ΔG Ɵ = 78,000 J mol H(g) + ½ O(g) HO(g); ΔG Ɵ =,78,000 J mol ; G is the Gibbs energy). Solution: ( 4.6) Cu + HO(g) CuO(s) + H(g) 0 Rxn G 78 78KJ / mole 0 Rxn G 00KJ / mole 0 G G RT l n Q 850 p 0 00 l n 000 p p H O bar 00 H H O ln ph l n l n p 0 log0 H = Consider the following reversible reaction, A(g) + B(g) AB(g). The activation energy of the backward reaction exceeds that of the forward reaction by RT (in J mol ). If the pre-exponential factor of the forward reaction is 4 times that of the reverse reaction, the absolute value of ΔG Ɵ (in J mol ) for the reaction at 300 K is. (Given; ln() = 0.7, RT = 500 J mol at 300 K and G is the Gibbs energy) 3. Solution: (8500) Eaf + Eab = RT = H A f A b Keq = 4 Af e H/RT A b Keq = 4e G 0 = RT ln Keq = 500 [ln 4 + ln e] = = 8500 J/mole IIT KALRASHUKLA CIVIL LINES, KAKADEO, SOUTH KANPUR [0]

21 4. Consider an electrochemical cell: A(s) A n+ (aq, M) B n+ (aq, M) B(s). The value of ΔH Ɵ for the cell reaction is twice that of ΔG Ɵ at 300 K. If the emf of the cell is zero, the ΔS Ɵ (in J K mol ) of the cell reaction per mole of B formed at 300 K is. (Given: ln() = 0.7, R (universal gas constant) = 8.3 J K mol. H, S and G are enthalpy, entropy and Gibbs energy, respectively.) 4. Solution: (.6) n 0 RT [A ] Ecell Ecell l n n nf [B ] 0 RT Ecell l n 4 nf nfe RT l n 0 cell G H T S RT ln 0 S T S J / K mol 0 IIT KALRASHUKLA CIVIL LINES, KAKADEO, SOUTH KANPUR []

22 SECTION 3 (Maximum Marks : ) This section contains FOUR questions Each question has TWO matching lists: LIST I and LIST II. FOUR options are given representing matching of elements from LIST I and LIST II. ONLY ONE of these four options corresponds to a correct matching. For each question, choose the option corresponding to the correct matching. For each question, marks will be awarded according to the following marking scheme: Full Marks : +3 If ONLY the option corresponding to the correct matching is chosen. Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered). Negative Marks : In all other cases. 5. Match each set of hybrid orbitals from LIST I with complex(es) given in LIST II. LIST-I LIST-II P. dsp. [FeF6] 4 Q. sp 3. [Ti(HO)3Cl3] R. sp 3 d 3. [Cr(NH3)6] 3+ S. d sp 3 4. [FeCl4] 5. Ni(CO)4 6. [Ni(CN)4] The correct option is (A) P 5; Q 4,6; R,3; S (B) P 5, 6; Q 4; R 3; S, (C) P 6; Q 4,5; R ; S,3 (D) P 4, 6; Q 5, 6; R,; S 3 5. Solution: (C) () FeF 3d 6 WFL Hybridization sp 3 d 4 4 T (H O) Cl 3d WFL d sp 3 () 3 3 (3) Cr(NH ) (4) FeCl 4 3d 3 SFL d sp 3 3d 6 WFL sp 3 (5) Ni(CO)4 3d 0 sp 3 (6) Ni(CN) 4 3d 8 SFL dsp 6. The desired product X can be prepared by reacting the major product of the reactions in LIST-I with one or more appropriate reagents in LIST-II. (given, order of migratory aptitude: aryl > alkyl > hydrogen) IIT KALRASHUKLA CIVIL LINES, KAKADEO, SOUTH KANPUR []

23 LIST-I LIST-II P.. l, NaOH Q.. [Ag(NH3)OH R. 3. Fehling solution S. 4. HCHO, NaOH 5. NaOBr The correct option is (A) P ; Q,3; R,4; S,4 (B) P,5; Q 3,4; R 4,5; S 3 (C) P,5; Q 3,4; R 5; S,4 (D) P,5; Q,3; R,5; S,3 6. Solution: (D) (P) (Q) (R) (S) IIT KALRASHUKLA CIVIL LINES, KAKADEO, SOUTH KANPUR [3]

24 7. LIST-I contains reactions and LIST-II contains major products. LIST-I P.. LIST-II Q.. R. 3. S Match each reaction in LIST-I with one or more products in LIST-II and choose the correct option. (A) P,5; Q ; R 3; S 4 (B) P,4; Q ; R 4; S 3 (C) P,4; Q,; R 3,4; S 4 (D) P 4,5; Q 4; R 4; S 3,4 7. Solution: (B) (P) (Q) (R) (S) IIT KALRASHUKLA CIVIL LINES, KAKADEO, SOUTH KANPUR [4]

25 8. Dilution processes of different aqueous solutions, with water, are given in LIST-I. The effects of dilution of the solutions on [H + ] are given in LIST-II. (Note: Degree of dissociation () of weak acid and weak base is << ; degree of hydrolysis of salt <<; [H + ] represents the concentration of H + ions) LIST-I LIST-II P. (0 ml of 0. M NaOH + 0 ml of 0. M acetic acid) diluted to 60 ml Q. (0 ml of 0. M NaOH + 0 ml of 0. M acetic acid) diluted to 80 ml R. (0 ml of 0. M HCl + 0 ml of 0. M ammonia solution) diluted to 80 ml S. 0 ml saturated solution of Ni(OH) in equilibrium with excess solid Ni(OH) is diluted to 0 ml (solid Ni(OH) is still present after dilution).. the value of [H + ] does not change on dilution. the value of [H + ] changes to half of its initial value on dilution 3. the value of [H + ] changes to two times of its initial value on dilution 4. the value of [H + ] changes to times of its initial value on dilution 5. the value of [H + ] changes to times of its initial value on dilution Match each process given in LIST-I with one or more effect(s) in LIST-II. The correct option is (A) P 4; Q ; R 3; S (B) P 4; Q 3; R ; S 3 (C) P ; Q 4; R 5; S 3 (D) P ; Q 5; R 4; S 8. Solution: (D) (P) NaOH CH3COOH CH3COONa HO 0 From buffer so no change in ph on dilution NaOH CH3COOH CH3COONa HO (Q) 0 0 Initially [CH 3COONa], finaly[ch3coona] Kw [OH ] ck H [H ck f i / 80 [H ] / 40 [H ] (R) Initially [NH Cl],finalyNH Cl [H ] ck H f i / 40 [H ] / 80 [H ] H Ni(OH)(s) Ni + + OH It s a solubility equilibrium so no change in ph on dilution. END OF PART II : CHEMISTRY IIT KALRASHUKLA CIVIL LINES, KAKADEO, SOUTH KANPUR [5]

26 PART III : MATHEMATICS SECTION (Maximum Marks: 4) This section contains SIX questions. Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four options is correct. For each question, choose the correct option(s)to answer the question. For each question, marks will be awarded in one of the following categories: Full Marks : +4 If only (all) the correct option(s) is (are) chosen. Full Marks : +3 If all the four options are correct but ONLY three options are chosen. Full Marks : + If three or more options are correct but ONLY two options are chosen, both of which are correct options. Full Marks : + 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 Example: 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 + marks. Selecting any incorrect option(s) (second option in this case), with or without selection of any correct option(s) will result in marks.. For any positive integer n, define f n : (0, ) as f (x) n for all x (0, ) n tan j (x j)(x j) (Here, the inverse trigonometric function tan x assumes values in,.) Then which of the following statement(s) is (are) true? 5 (A) j j tan f (0) 55 0 ' (B) j j j f (0) sec f (0) 0 n n lim sec f (x) (C) For any fixed positive integer n, lim tan f (x) x (D) For any fixed positive integer n,. Solution: (D) (x j) (x j) n n x f n (x) tan tan (x j) tan (x j) j (x j)(x j) j tan (x n) tan (x) tan x(x n) n n IIT KALRASHUKLA CIVIL LINES, KAKADEO, SOUTH KANPUR [6]

27 tan f n (x) (A) tan f n (0) n x nx n tan f j(0) j 55 [0 is not in domain] j j 6 (False) (B) x = 0 is not in domain (False) (C) n lim tan f n (x) lim 0 x x x nx (False) (D) lim sec f (x) lim( tan f (x)) 0 (True) x n x n. Let T be the line passing through the points P(, 7) and Q(, 5). Let F be the set of all pairs of circles (S, S) such that T is tangent to S at P and tangent to S at Q, and also such that S and S touch each other at a point, say, M. Let E be the set representing the locus of M as the pair (S, S) varies in F. Let the set of all straight line segments joining a pair of distinct points of E and passing through the point R(, ) be F. Let E be the set of the mid-points of the line segmetns in the set F. Then, which of the following statement(s) is (are) True? 4 7 (A) The point (, 7) lies in E (B) The point, does not lie in E 5 5 (C) The point. Solution: (D), lies in E (D) The point 3 0, does not lie in E RP = RM and RM = RQ RP = RM = RQ R is M.P. of PQ Locus of M is a circle with PQ as diameter. Locus of M: E: (x + ) (x ) + (y 7) (y + 5) = 0 (x ) E: x + y y 39 = 0 (x ) E : Locus of mid points of chords of E which pass through (, ) let mid-point be (h, k) Then equation of chord is T = S : hx + ky (y + k) = h + k = qk It passes through (, ) h + k k = h + k k E : x + y x y + = 0 (A) LHS = = 0 = RHS (False) (B) LHS = RHS (False) (C) LHS = RHS 4 (False) IIT KALRASHUKLA CIVIL LINES, KAKADEO, SOUTH KANPUR [7]

28 9 (D) LHS = RHS (True) 4 Note: If we allow point circle then option A can also be correct b 3. Let S be the set of all column matrices b such that b, b, b3 and the system of equations b 3 (in real variables) x y 5z b x 4y 3z b x y z b 3 has at least one solution. Then, which of the following system(s) (in real variables) has (have) at b least one solution for each b S? b 3 (A) x + y + 3z = b, 4y + 5z = b and x + y + 6z = b3 (B) x + y + 3z = b, 5x + y + 6z = b and x y 3z = b3 (C) x + y 5z = b, x 4y + 0z = b and x y + 5z = b3 (D) x + y + 5z = b, x + 3z = b and x + 4y 5z = b3 3. Solution: (A, C, D) x y 5z b () x 4y 3z b () x y z b 3 (3) R R R R system has infinite solutions [No solutions not possible] Let a() + b() = 3 b a =, a 4b + = 0, 5a + 3b =, ab + bb = b3 7 a, b and b 7b 3b (A) [unique solution for any value of b, b, b3] (B) R R R b 7b 36x 5y z 3b3 [No solution] IIT KALRASHUKLA CIVIL LINES, KAKADEO, SOUTH KANPUR [8]

29 (C) (D) b 7b 3x 6y 65z 3b3 [Infinite solution] Consider two straight lines, each of which is tangent to both the circle parabola y x y and the 4x. Let these lines intersect at the point Q. Consider the ellipse whose center is at the origin O(0, 0) and whose semi-major axis is OQ. If the length of the minor axis of this ellipse is, then which of the following statement(s) is (are) TRUE? (A) For the ellipse, the eccentricity is and the length of the latus rectum is (B) For the ellipse, the eccentricity is and the length of the latus rectum is (C) The area of the region bounded by the ellipse between the lines x and x = is ( ) 4 (D) The area of the region bounded by the ellipse between the lines x and x = is ( ) 6 4. Solution: (A, C) Any tangent to y = 4x is It is also tangent to x / m y mx m y m m m 4 m m 0 m m Equation of tangents are y x, y x Q, 0 Equation of (A) & (B) x y / e IIT KALRASHUKLA CIVIL LINES, KAKADEO, SOUTH KANPUR [9]

30 / LR. (C) & (D) Required value = x dx / x sin x x 4 / 5. Let s, t, r be non-zero complex numbers and L be the set of solutions z = x + iy (x, y, i ) of the equation sz tz r 0, where z x iy. Then, which of the following statement(s) is (are) TRUE? (A) If L has exactly one element, then s t (B) If s = t, then L has infinitely many elements (C) The number of elements in L {z : z i 5} is at most (D) If L has more than one element, then L has infinitely many elements 5. Solution: (A, C, D) z x iy sz tz r 0 s z tz r 0 Eliminating z, we get s t z sr tr 0 z s t tr rs (A) If s t, then z has unique value. (True) (B) If s t, then rs tr may or may not be zero. (False) (C) (D) Locus of z is null set, singleton set or a straight line. In any case maximum number of point of intersections with a circle z i 5 are atmost. (Ture) If L has more than one element, then it is a line. Hence L has infinitely many elements. (True) 6. Let f : (0, ) be a twice differentiable function such that f (x) sint f(t) sinx lim sin x for all x (0, ) tx t x If f, then which of the following statement(s) is (are) TRUE? 6 (A) f x (B) f x x for all x (0, ) 6 (C) There exists a (0, ) such that f '(a) 0 (D) f '' f 0 IIT KALRASHUKLA CIVIL LINES, KAKADEO, SOUTH KANPUR [30]

31 6. Solution: (B, C, D) f (x) sint f(t) sinx Lim sin x tx t x f (x) cos x f '(x)sin x sin x ' f (x) sin x f (x) x c (Integrating) sin x Put x c 0 6 f (x) x sin x (A) (B) f x sin x x 6 4 x x sin x x 6 4 x f (x) x 6 f '(x) x cos x sin x f '(x) 0 tan x x (C) (D) f f '' 0 IIT KALRASHUKLA CIVIL LINES, KAKADEO, SOUTH KANPUR [3]

32 SECTION (Maximum Marks: 4) This section contains EIGHT 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, 7.30) using the mouse and the onscreen 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. The value of the integral 7. Solution: () I / 0 3 x (x) x 3/ dx / x Put t dx dt x ( x) I /3 3 dt 3/ t 3 dx is. 0 6 (x) ( x) t 3 /3 8. Let P be a matrix of order 3 3 such that all the entries in P are from the set {, 0, }. Then, the maximum possible value of the determinant of P is. 8. Solution: (4) a b c a b c a b c a b c a b c a b c a b c a b c p (say) a b c Now, P 3 and q 3 Maximum value of can be 6 [provided P = 3, q = 3] P =3 abc3 = a3bc = ab3c = a a a3 b b b3 c c c3 = q = 3 a b3 c = a b c3 = a3 b c=,,.,, = Hence p =3, q = 3 is not possible. Hence maximum value of can be (possible) IIT KALRASHUKLA CIVIL LINES, KAKADEO, SOUTH KANPUR [3]

33 9. Let X be a set with exactly 5 elements and Y be a set with exactly 7 elements. If is the number of one-one functions from X to Y and is the number of onto functions from Y to X, then the value of is. 5! 9. Solution: (9) n(x) 5, n(x) 7 No. of one-one function from x y is 7 P5 = = 7 C No. of onto function from y x is ( ) C C3 C Let f : be a differentiable function with f (0) 0. If y = f(x) satisfies the differential equation dy ( 5 y)(5 y ) dx then the value of lim f(x) x is. 0. Solution: (0.4) dy 5y 5y dx dy (5 y )(5 y ) dx dy dx 4 5y 5y 5y ln 0x c 5y f (0) 0 c 0 5y e 5y 0x 5f (x) lim lim e x 5f (x) x lim f (x) 5 x. Let f : 0x be a differential function with f(0) = and satisfying the equation f (x y) f(x) f'(y) f'(x) f(y) for all x, y. Then, the value of loge(f(4)) is.. Solution: () f (x y) f(x) f '(y) f '(x)f(y) Put x = y = 0 f (0) f(0) f '(0) f '(0) IIT KALRASHUKLA CIVIL LINES, KAKADEO, SOUTH KANPUR [33]

34 Put y = 0 f(x) f (x) f '(x) f (x) f '(x) x/ f (x) e c. f (0) c 0 f (x) e x/ ln f (4) lne. Let P be a point in the first octant, whose image Q in the plane x + y = 3 (that is, the line segment PQ is perpendicular to the plane x+ y =3 and the mid-point of PQ lies in the plane x + y = 3) lies on the z-axis. Let the distance of P from the x-axis be 5. If R is the image of P in the xy-plane, then the length of PR is.. Solution: (8) Let P(h, k, l ), Q(0, 0, l ) and R (h, k, l ) PQ ˆi ˆj h k Mid-point of PQ lies on x + y = 3 h k 3 h k 6 h k 3 Distance of P from X-axis = 5 k l 5 k l 5 l 6 PR l 8 3. Consider the cube in the first octant with sides OP, OQ and OR of length, along the x-axis, y-axis and z-axis, respectively, where O(0, 0, 0) is the origin. Let S,, be the centre of the cube and T be the vertex of the cube opposite to the origin O such that S lies on the diagonal OT. If p SP, q SQ, r SR and t ST, then the value of (p q) (r t) is. 3. Solution: (0.5) ˆ ˆ ˆ p SP i j k ˆi ˆj p q q ˆi ˆj kˆ ˆ ˆ ˆ r i j k ˆi ˆj r t ˆ ˆ ˆ t i j k ˆk p q r t IIT KALRASHUKLA CIVIL LINES, KAKADEO, SOUTH KANPUR [34]

35 Let X C C 3 C C0, where 0 C r, r {,,...,0} denote binomial coefficients. The, the value of 4. Solution: (646) X 430 is C C 9 C x C 3x C 3... C ( x) x x x Comparing coefficient of x on both sides ( x) x C C 3 C 3... coeff. of x in 0 x x 0 C9 9 0 x IIT KALRASHUKLA CIVIL LINES, KAKADEO, SOUTH KANPUR [35]

36 SECTION 3 (Maximum Marks : ) This section contains FOUR questions Each question has TWO matching lists: LIST I and LIST II. FOUR options are given representing matching of elements from LIST I and LIST II. ONLY ONE of these four options corresponds to a correct matching. For each question, choose the option corresponding to the correct matching. For each question, marks will be awarded according to the following marking scheme: Full Marks : +3 If ONLY the option corresponding to the correct matching is chosen. Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered). Negative Marks : In all other cases. x 5. Let E x R; x and 0 x and x E x E ; sin loge is a real number x. (Here, the inverse trigonometric function sin x assumes values in,.) x Let f: E be the function defined by f(x) loge x and g : E be the function defined by g(x) x log e sin x List-I List-II P. The range of f is. e,, e e Q. The range of g contains. (0, ) R. The domain of f contains 3., S. The domain of g is 4. (, 0) (0, ) 5. e, e 6. e (, 0), e The correct option is: (A) P 4; Q ; R ; S (B) P 3; Q 3; R 6; S 5 (C) P 4; Q ; R ; S 6 (D) P 4; Q 3; R 6; S 5 5. Solution: (A) x E x R : x, 0 (, 0) (, ) x x x e E : ln e x,, x e x e e x Range of is R {} x Range of f is R {0} IIT KALRASHUKLA CIVIL LINES, KAKADEO, SOUTH KANPUR [36]

37 Range of g is, {0} P 4, Q, R, S 6. In a high school, a committee has to be formed form a group of 6 boys M, M, M3, M4, M5, M6 and 5 girls G, G, G3, G4, G5. (i) Let be the total number of ways in which the committee can be formed such that the committee has 5 members, having exactly 3 boys and girls. (ii) Let be the total number of ways in which the committee can be formed such that the committee has at least members, and having an equal number of boys and girls. (iii) Let 3 be the total number of ways in which the committee can be formed such that the committee has 5 members, at least of them being girls. (iv) Let 4 be the total number of ways in which the committee can be formed such that the committee has 4 members, having at least girls and such that both M and G are NOT in the committee together. List-I List-II P. The value of is. 36 Q. The value of is. 89 R. The value of 3 is 3. 9 S. The value of 4 is The correct option is: (A) P 4; Q 6; R ; S (B) P ; Q 4; R ; S 3 (C) P 4; Q 6; R 5; S (D) P 4; Q ; R 3; S 6. Solution: (C) 6 5 C. C Let H: 3 C C C C C C C C C C C C C C C C C C C C C C C C C C C C P 4, Q 6, R 5, S x a y, where a b 0, be a hyperbola in the xy-plane whose conjugate axis LM b subtends an angle of 60 o at one of its vertices N. Let the area of the triangle LMN be 4 3. List I List-II P. The length of the conjugate axis of H is. 8 Q. The eccentricity of H is. 4 3 R. The distance between the foci of H is 3. 3 S. The length of the latus rectum of H is 4. 4 The correct option is: IIT KALRASHUKLA CIVIL LINES, KAKADEO, SOUTH KANPUR [37]

38 (A) P 4; Q ; R ; S 3 (B) P 4; Q 3; R ; S (C) P 4; Q ; R 3; S (D) P 3; Q 4; R ; S 7. Solution: (B) b a o tan 30 a b 3 ar(lmn) 3b 4 3 b and a 3 e 3 3 Now, P 4, Q 3, R, S 8. Let f:, f :,, f 3 :, e and f 4 : be functions defined by (i) f x (x) sin e (ii) sinx if x 0 f (x) tan x, where the inverse trigonometric function tan x assumes if x 0 values in,. f (x) sin(log (x )), where for t. [t] denotes the greatest integer less than or (iii) 3 e equal to. (iv) x sin if x 0 f 4(x) x 0 if x 0 List-I List-II P. The function f is. Not continuous at x = 0 Q. The function f is. Continuous at x = 0 and not differentiable at x = 0 R. The function f3 is 3. Differentiable at x = 0 and its derivative is NOT continuous at x = 0 S. The function f4 is 4. Differentiable at x = 0 and its derivative is continuous at x = 0 The correct option is: (A) P ; Q 3; R ; S 4 (B) P 4; Q ; R ; S 3 (C) P 4; Q ; R ; S 3 (D) P ; Q ; R 4; S 3 IIT KALRASHUKLA CIVIL LINES, KAKADEO, SOUTH KANPUR [38]

39 8. Solution: (D) f (x) sin e x x x cos e. e x f '(x) DNE at (x 0) x e sin x, x 0 f (x) tan x not continuous at x = 0 0, x 0 f (x) sin ln(x ) 0 3 x e 0 ln(x ) 0 sin ln(x ) f 3(x) 0 x sin, x 0 f 4(x) x 0, x 0 P ; Q ; R 4; S 3 END OF PART III : MATHEMATICS IIT KALRASHUKLA CIVIL LINES, KAKADEO, SOUTH KANPUR [39]