Sri Chaitanya IIT Academy

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2 8-Jee-Advanced JEE (ADVANCED) 8 PAPER PART I-PYSICS SECTION (Maximum Marks: 4) Question Paper-_Key & Solutions This section contains SIX (6) questions. Each question has FOUR options for correct answer(s). ONE OR MORE TAN 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 : + 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 : + If two or more options are correct but ONLY one option is chosen and it is acorrect option. Zero Marks : 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. Q. 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 Key: (A, B, D) Sol: dk dt t as k mv dk dv mv t dt dt v m vdv tdt t mv t v t m dv a dt m Constant Since F=ma F m m Constant m # 4, Kasetty eights, Ayyappa Society, Madhapur, yderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net Page-

3 8-Jee-Advanced Question Paper-_Key & Solutions Q. 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 velocityu. 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 u D) The tangential (shear) stress on the plate varies linearly with the viscosity of the liquid Key: (A, C, D) h u Sol: viscous force is given by F dv A since h is very small therefore; magnitude of viscous dy force is given by F dv A dy Au F F & F u F, F A h h Q. 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 at the centre O of the spherical shell, as shown in the figure. The permittivity of free space is. Which of the following statements is (are) true? A) The electric flux through the shell is R / 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 # 4, Kasetty eights, Ayyappa Society, Madhapur, yderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net Page-

4 8-Jee-Advanced Key: (A, B) Question Paper-_Key & Solutions P R 6 O z-axis Q Sol: filed due to straight wire is perpendicular to the wire & radially outward. ence Ez length, PQ R sin 6 R according to gauss s law Total flux= qin R E. ds Q.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) Key: (D) # 4, Kasetty eights, Ayyappa Society, Madhapur, yderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net Page-4

5 8-Jee-Advanced B Question Paper-_Key & Solutions A f / 45 f F f Sol: distance of a point A is Let A is the image of A from mirror, for this image v f f v f f f Image of line AB should be perpendicular to the principle axis & image of F will form at infinity, therefore correct image diagram is ' B f-x f h f u h ' A h f f x h f x f or x f-x Q.5 Ina radioactive decay chain, 9Th nucleus decay to 8 Pb nucleus. Let N and N 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 Key: (A, C) Sol: 9 Th is converting into 8 Pb Change in mass number (A) = Number of particle = 5 4 Due to 5 particle, z will be change by unit. Since given change is 8, therefore number of particle is # 4, Kasetty eights, Ayyappa Society, Madhapur, yderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net Page-5

6 8-Jee-Advanced Question Paper-_Key & Solutions Q.6 In an experiment to measure the speed of sound by a resonating air column, a tuning fork of frequency 5z 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 5.7cm and 8.9 cm. Which of the following statements is (are) true? A) The speed of sound determined from this experiment is m/s B) The end correction in this experiment is.9cm C) The wavelength of the sound wave is 66.4cm D) The resonance at 5.7cm corresponds to the fundamental harmonic Key: (A, C) Sol: let n harmonic is corresponding to 5.7 cm & n harmonic is corresponding 8.9 cm. ( ) cm. cm 66.4 cm 6.6 cm 4 Length corresponding to fundamental mode must be close to 4 & 5.7 cm must be an odd multiple of this length cm. therefore 5.7 is rd harmonic If end correction is e, then e 5.7 cm Speed of sound, v f 4 v cm/sec. m/s SECTION (Maximum Marks: 4) This section contains EIGT (8) 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., -., -.,.7, -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 : + If ONLY the correct numerical value is entered as answer. Zero Marks : In all other cases. Q.7 A solid horizontal surface is covered with a thin layer of oil. A rectangular block of mass m.4kg is at rest on this surface. An impulse of.ns is applied to the block at time t/ t so that it starts moving along the x-axis with a velocity vt ve, where v is a constant and 4s. The displacement of the block, in metres, at t is. Take Key: 6. e.7. # 4, Kasetty eights, Ayyappa Society, Madhapur, yderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net Page-6

7 8-Jee-Advanced Question Paper-_Key & Solutions J= m=.4 Sol: J V.5 m/s m t/ V Ve t/ V Ve dx Ve dt t/ x x t/ x e dx V e dt e dx e t/ x V X e e.5( 4)( ) X 54.7 X 6. Q.8 A ball is projected from the ground at an angle of 45 with the horizontal surface. It reaches a maximum height of 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 with the horizontal surface. The maximum height it reaches after the bounce, in meters, is. Key:. u 45 m V u Sol: u sin 45 g u When half of kinetic energy is lost V= u..(i) 4g sin u u..(ii) From (i) & (ii) g 6g on. # 4, Kasetty eights, Ayyappa Society, Madhapur, yderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net Page-7

8 8-Jee-Advanced Q.9 A particle, of mass kg Question Paper-_Key & Solutions and charge.c, is initially at rest. At time t, the particle comes under the influence of an electric field sin E t E t i, where E. N / C and rad / s. Consider the effect of only the electrical force on the particle. Then the maximum speed, in m/s, attained by the particle at subsequent times is. Key:. Sol: n kg q C t E E sint Force on particle will be At max' F qe qe sin t V a, F qe sint F qe sint dv E q sint dt m qe V m V / dv / cos t qe V m qe m cos cos V sint dt m/s =. 4 Q. A moving coil galvanometer has 5 turns and each turn has an area m. The magnetic field produced by the magnet inside the galvanometer is.t. The torsional constant of the suspension wire is 4 Nm rad. When a current flows through the galvanometer, a full scale deflection occurs if the coil rotates by.rad. The resistance of the coil of the galvanometer is 5. This galvanometer is to be converted into an ammeter capable of measuring current in the range.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.. Key: 5.55 # 4, Kasetty eights, Ayyappa Society, Madhapur, yderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net Page-8

9 8-Jee-Advanced Sol: n= 5 turns A= m 4 Question Paper-_Key & Solutions B=. T m K 4 Q =. rad R 5 I =-.A MB C, M nia A BINA=C I g =.A g For galvanometer, resistance is to be connected to ammeter in shunt. I. R 5 g g I I g S I R I I S g g g. 5. S 5 S Q. A steel wire of diameter.5mm and Young s modulus N / m carries a load of mass M. The length of the wire with the load is.m. A vernier scale with 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.mm, is attached. The 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 m / s and.. Key:. Sol: d.5 mm Fl mgl l Ay d y 4 l Y l m = So rd division of a vernier scale will coincicle with main scale..mm # 4, Kasetty eights, Ayyappa Society, Madhapur, yderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net Page-9

10 8-Jee-Advanced Question Paper-_Key & Solutions Q. One mole of a monatomic ideal gas undergoes an adiabatic expansion in which its volume Key: 9 Sol: becomes eight times its initial value. If the initial temperature of the gas is K and the universal gas constant Vi. V V 8V F R 8.J mol K 5 For adiabatic process { for monotonic process TV T. V V = T V 8, the decrease in its internal energy, in Joule, is T 5k FR v Joule U nc T Q. In a photoelectric experiment a parallel beam of monochromatic light with power of W is Key: 4 incident on a perfectly absorbing cathode of work function 6.eV. The frequency of light is just above the thres hold frequency so that the photoelectrons are emitted with negligible kinetic energy. Assume that the photoelectron emission efficiency is %. A potential difference of 5V is applied between the cathode and the anode. All the emitted electrons are incident normally on the anode and are absorbed. The anode experiences a force electron 4 F n N due to the impact of the electrons. The value of n is. Mass of the me 9 9 kg and.ev.6 J. Sol: power = nhv n= number of protons per second Since KE=, hv 9 n joule n As photon just above threshold frequency difference of 5V. KE f qv P q V P m V m KEmax is zero and they are accelerated by potential Since efficiency is %, number of electrons = number of photons per second As photon is completely absorbed force exerted = nmv # 4, Kasetty eights, Ayyappa Society, Madhapur, yderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net Page-

11 8-Jee-Advanced Question Paper-_Key & Solutions Q.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 74.8eV higher than the photon emitted in the n to n transition. The ionization energy of the hydrogen atom is.6ev. The value of Z is. Key: Sol: E.6 z.6 z E.6 z.6 z E z.6 z z z 9 z SECTION (Maximum Marks: ) This section contains FOUR (4) 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 : + If ONLY the option corresponding to the correct matching is chosen. Zero Marks : If none of the options is chosen (i.e. the question is unanswered). Negative Marks : - In all other cases. Q.5 The electric field E is measured at a point P,, 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. # 4, Kasetty eights, Ayyappa Society, Madhapur, yderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net Page-

12 8-Jee-Advanced Question Paper-_Key & Solutions LIST- LIST-II P. E is independent of d. A point charge Q at the origin Q. E d,,l and -. A small dipole with point charges Q at Q, At,, l R. S. Key: (B) take l d E. An infinite line charge coincident with the x-axis, d with uniform linear charge density E 4. Two infinite wires carrying uniform linear charge d density parallel to the x- axis.the one along y, z l has a charge density and the one along y, 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, 4; R ; S B) P 5; Q ; R, 4; S C) P 5; Q ; R, ; S 4 D) P 4; Q, ; R ; S 5 KQ Sol: (i) E E d d (ii) dipole E kp d cos E (iii) for line charge K K (iv) E d l d l dr t V iˆ ˆj dt k E d (V) Electric filed due to sheet for dipole d E d d l d l Kl d l E d V is independent of r # 4, Kasetty eights, Ayyappa Society, Madhapur, yderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net Page-

13 8-Jee-Advanced Question Paper-_Key & Solutions Q.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 v, 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 m corresponding quantities of satellite.given m and R R the numbers in List-II. 4, match the ratios in List-I to LIST I v P. v L Q. L K R. K T S. T A) P 4; Q ; R ; S B) P ; Q ; R 4; S C) P ; Q ; R ; S 4 D) P ; Q ; R 4; S Key: (B) LIST II M m m m R R m R R 4 Sol: GMm R mv R V GM GM, V R R V V R V 4 (P) V R # 4, Kasetty eights, Ayyappa Society, Madhapur, yderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net Page-

14 8-Jee-Advanced L mv R (Q) L mvr L m v R 4 Question Paper-_Key & Solutions (R) K K m v 8 mv K mv (S) T R / V T R v R v T v R R v 4 8 Q.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- with the corresponding statements in List-II. LIST I LIST II P. In process I. Work done by the gas is zero Q. In process II. Temperature of the gas remains unchanged R. In process III. No heat is exchanged between the gas and its surroundings S. In process IV 4. Work done by the gas is 6PV A) P 4; Q ; R ; S B) P ; Q ; R ; S 4 C) P ; Q 4; R ; S D) P ; Q 4; R ; S Key: (C) Sol: process-i is an adiabatic process Q U W Q W U Volume of gas is decreasing W U Temperature of gas increases # 4, Kasetty eights, Ayyappa Society, Madhapur, yderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net Page-4

15 8-Jee-Advanced No heat is exchanged between the gas and surrounding. Question Paper-_Key & Solutions Process II is an isobaric process (pressure remain constant) W PV P V V 6PV Process-III is an isochoric process ( volume remain constant) Q U W Q U Process-IV is an isothermal process (temperature remains constant) Q U W U Q.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 r t t i t j. p P. r t t i t j. L Q. cos sin R. r t cost i sint j S.. K r t t i t j 4. U 5. E A) P,,, 4, 5; Q, 5; R,, 4, 5; S 5 B) P,,, 4, 5; Q, 5; R,, 4, 5; S, 5 C) P,, 4; Q 5; R,, 4; S, 5 D) P,,, 5; Q, 5; R,, 4, 5; S, 5 Key: (A) # 4, Kasetty eights, Ayyappa Society, Madhapur, yderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net Page-5

16 8-Jee-Advanced r t ti tj Sol: (P) ˆ ˆ dv a dt k P mv (Remain constant) dr t V iˆ ˆj {Constant} dt U x U y mv {Remain constant} F iˆ iˆ U Constant E= K+ U dl r F L =constant dt (Q) r costiˆ sin t ˆj dv a cost i ˆ sint ˆ j dt a r U r U depends on r hence it will change with time Total energy remain constant because force is central dr ( t) (R) r ( t) costiˆ sint ˆj dt dr ( t) v( t) sintiˆ cost ˆj cost i dt ˆ sin t ˆ j (S) ˆ r ti t ˆ dv j a ˆ j Constant F ma Constant dt Question Paper-_Key & Solutions t U F. dr m ˆj. iˆ tj ˆ dt E k U m [Remain constant] # 4, Kasetty eights, Ayyappa Society, Madhapur, yderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net Page-6

17 8-Jee-Advanced Question Paper-_Key & Solutions PART-:CEMISTRY SECTION - (Maximum Marks: 4) Each question has FOUR options for correct answer(s). ONE OR MORE TAN 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 : + 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 : + If two or more options are correct but ONLY one option is chosen and it is acorrect option. Zero Marks : 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) (N ) ( O)] + :- Key. ABD Sol. (en = NC C N ) 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) (N ) 4 ] + A) [Co(en)(N ) ( O)] + complex is type of [M(AA)b c] have two G.I. B) If (en) is replaced by two cynide ligand, complex will be type of [Ma b c] and have G.I. C) [Co(en)(N ) ( O)] + have d 6 configuration (t 6 g) on central metal with SFL therefore it is dimagnetic in nature. D)Complex [Co(en)(N ) ( O)] + have lesser CFSE ( O ) value than [Co(en)(N ) 4 ] + therefore complex [Co(en)(N ) ( O)] + absorbs longer wavelength for d d transition.. The correct option(s) to distinguish nitrate salts of Mn + and Cu + taken separately is (are) :- Key. BD A) Mn + shows the characteristic green colour in the flame test B) Only Cu + shows the formation of precipitate by passing S in acidic medium C) Only Mn + shows the formation of precipitate by passing S in faintly basic medium D) Cu + /Cu has higher reduction potential than Mn + /Mn (measured under similar conditions) # 4, Kasetty eights, Ayyappa Society, Madhapur, yderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net Page-7

18 8-Jee-Advanced Question Paper-_Key & Solutions Sol. A) Cu + and Mn + both gives green colour in flame test and cannot distinguished. B) Cu + belongs to group-ii of cationic radical will gives ppt. of CuS in acidic medium. C) Cu + and Mn + both form ppt. in basic medium. D) Cu + /Cu = +.4 V (SRP) Mn + /Mn =.8 V (SRP). Aniline reacts with mixed acid (conc. NO and conc. SO 4 ) at 88 K to give P (5%), Q (47%) and R (%). The major product(s) the following reaction sequence is (are) :- ) Ac O, pyridine ) Sn/ Cl ) Br, CCO ) Br / Oexcess R S major product s ) ) O NaNO, Cl/7798K 4) NaNO 4), Cl/7798K PO 5) EtO, Br Br Br Br Br Br A) Br Br B) Br Br C) Br Br D) Br Br Br Key. D Sol. NO / SO 4 ACO Py Br / CCO Br NaNO Cl Br Br 5 C PO Br Br # 4, Kasetty eights, Ayyappa Society, Madhapur, yderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net Page-8

19 8-Jee-Advanced 4. The Fischer presentation of D-glucose is given below. CO O Question Paper-_Key & Solutions O O O CO D-glucose The correct structure(s) of -glucopyranose is (are) :- O CO O O O CO O O O A) O B) O O CO O O O CO O O O O O O C) Key. D Sol. O C=O CO D-glucose O O O O O O O O O Dglucopyranose D) O O CO O O O O Lglucopyranose 5. For a first order reaction A(g) B(g) + C(g) at constant volume and K, the total pressure at the beginning (t = ) and at time t are P and P t, respectively. Initially, only A is present with concentration [A], and t / is the time required for the partial pressure of A to reach / rd of its initial value. The correct option(s) is (are) :- (Assume that all these gases behave as ideal gases) # 4, Kasetty eights, Ayyappa Society, Madhapur, yderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net Page-9

20 8-Jee-Advanced Question Paper-_Key & Solutions ln P P t t A) Time B) A ln P P t Rate constant Key. AD C) Time Sol. A B C t P t t P P P P P P Pt P K ln ln t P P t P P P P P ln ln ln t P Pt t D) K Kt P P P t A P and t ln ln cons tan t K P / K Rate constant does not depends on concentration 6. For a reaction, A P, the plots of [A] and [P] with time at temperatures T and T are given below. Key. AC If T > T, the correct statement(s) is (are) (Assume and S are independent of temperature and ratio of lnk at T to lnk at T is T greaterthan T constant, respectively.). ere, S, G and K are enthalpy, entropy, Gibbs energy and equilibrium A), S B) G, C) G, S D) G, S # 4, Kasetty eights, Ayyappa Society, Madhapur, yderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net Page-

21 8-Jee-Advanced Sol. A P given T T Question Paper-_Key & Solutions ln ln K K T T G G T ln k T ln k T S T S TS TS S SECTION (Maximum Marks: 4) This section contains EIGT (8) 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., -., -.,.7, -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 : + If ONLY the correct numerical value is entered as answer. Zero Marks : In all other cases. 7. The total number of compounds having at least one bridging oxo group among the molecules Ans. 5 or 6 given below is. N O, N O, P O, P O, P O, PO, S O, S O Sol: # 4, Kasetty eights, Ayyappa Society, Madhapur, yderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net Page-

22 8-Jee-Advanced Question Paper-_Key & Solutions 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 furnance such that the contents undergo self-reduction. The weight (in kg) of Pb produced per kg of O consumed is. Ans (Atomic weights in g mol : O = 6, S =, Pb = 7) Sol: PbS O Pb SO mol 7 gm Mol of Pb=mol of O mol mass of Pb 7 g 7 kg 6.47kg 9. To measure the quantity of MnCl dissolved in an aqueous solution, it was completely converted Ans. 6 to KMnO 4 using the reaction, MnCl + K S O 8 + O KMnO 4 + SO 4 + Cl (equation not balanced). Few drops of concentrated Cl were added to this solution and gently warmed. Further, oxalic acid(5 g) was added in portions till the colour of the permanganate ion disappeard. The quantity of MnCl (in mg) present in the initial solution is. (Atomic weights in g mol : Mn = 55, Cl = 5.5) MnCl K S O O KMnO SO Cl Sol: amole amole 4 4 C O MnO CO m of C O m of MnO eq 4 eq 4.5 / 9 a 5 a mg. For the given compound X, the total number of optically active stereoisomers is. Ans. 7 # 4, Kasetty eights, Ayyappa Society, Madhapur, yderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net Page-

23 8-Jee-Advanced Question Paper-_Key & Solutions. In the following reaction sequence, the amount of D (in g) formed from moles of acetophenone is. (A tomic weight in g mol : =, C =, N = 4, O = 6, Br = 8. The yield (%) corresponding to the product in each step is given in the parenthesis) Ans. 495 Sol:. 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 5 K. owever, the N gas contains mole % of water vapour as impurity. The water vapour oxidizes copper as per the reaction given below: Cu s Og Cu O s g p is the minimum partial pressure of (in bar) needed to prevent the oxidation at 5K. The value of In P is (Given : total pressure= bar, R(universal gas constant)= Ans Sol: Cu O s are mutually immiscible. At 5K: Cu s / O g Cu O s ; G 78, J mol / ;.78, 8JK Cu s and mol, In ()=.. g O g O g G Jmol ; G is the Gibbs energy) # 4, Kasetty eights, Ayyappa Society, Madhapur, yderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net Page-

24 8-Jee-Advanced Question Paper-_Key & Solutions. Consider the following reversible reaction, The activation energy of the backward reaction Ans. -85 Sol: 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 G (in the reaction at K is mol ) for n n 4. Consider an electrochemical cell: A(s) A aq, M B aq,m B s. The value of for the cell reaction is twice that of G at K. If the emf of the cell is zero, the S (in JK mol ) of the cell reaction per mole of B formed at K is Ans. -.6 Sol: # 4, Kasetty eights, Ayyappa Society, Madhapur, yderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net Page-4

25 8-Jee-Advanced Question Paper-_Key & Solutions SECTION (Maximum Marks: ) This section contains FOUR (4) 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 : + If ONLY the option corresponding to the correct matching is chosen. Zero Marks : 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. Q. R. S. dsp. FeF 4 6 sp. Ti O Cl sp d. Cr N 6 d sp 4. FeCl 4 Ans. C Sol: The correct option is 5. NiCO NiCN A) P 5; Q 4, 6; R,; S B) P 5,6; Q 4,6; R ; S, C) P 6; Q 4,5; R ; S, D) P 4,6; Q 5,6; R, ; S # 4, Kasetty eights, Ayyappa Society, Madhapur, yderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net Page-5

26 8-Jee-Advanced Question Paper-_Key & Solutions 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: ary>alky>hydrogen) O Ph O Ph Me X Ans. D # 4, Kasetty eights, Ayyappa Society, Madhapur, yderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net Page-6

27 8-Jee-Advanced 7. LIST-I contains reactions and LIST-II contains major products. LIST-I LIST-II Question Paper-_Key & Solutions P.. Q.. R.. S Match each reaction in LIST-I with one or more product in LIST-II and choose the correct option. A) P,5; Q ; R ; S 4 B) P,4; Q ; R 4; S C) P,4; Q,; R,4; S 4 D) P 4,5; Q 4; R 4; S,4 Ans. B Sol: P. Q. R. S. 8. Dilution process of different aqueous solutions; with water, are given in LIST-I. The effects of dilution of the solutions on are given in LIST-II (Note: Degree of dissociation of weak acid and weak base is <<; degree of hydrolysis of salt <<; represents the concentration of ions) # 4, Kasetty eights, Ayyappa Society, Madhapur, yderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net Page-7

28 8-Jee-Advanced LIST-I P. ( ml of. M NaO+mL of. M acetic acid) diluted to 6Ml Q. (ml of. M NaO+mL of. M acetic acid) diluted to 8mL Q. (ml of. M Cl+mL of. M ammonia solution) diluted to 8Ml S. ml saturated solution of Ni O in equilibrium with excess solid Ni O is diluted to ml (solid Ni O is still Question Paper-_Key & Solutions LIST-II. the vale of does not change on dilution. the value of change to half of its initial value on dilution. the value of changes to two times of its initial value on dilution 4. the value of changes to times of its initial value on dilution present after dilution) 5. the value of 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 ; S B) P 4; Q ; R ; S C) P ; Q 4; R 5; S D) P ; Q 5; R 4; S Ans. D Sol: S. Because of dilution solubility does not change so [ + ] = constant # 4, Kasetty eights, Ayyappa Society, Madhapur, yderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net Page-8

29 8-Jee-Advanced PART-:MATEMATICS SECTION - (Maximum Marks: 4) Question Paper-_Key & Solutions Each question has FOUR options for correct answer(s). ONE OR MORE TAN 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 : + 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 : + If two or more options are correct but ONLY one option is chosen and it is acorrect option. Zero Marks : 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 Q. For any positive integer n, define f :, as f x Key: D all x, (ere, the inverse trigonometric function n tan n x assume values in Then, which of the following statement(s) is (are) TRUE? 5 A) f j tan 55 f j B) f j ' sec x n lim sec f x C) For any fixed positive integer n, lim tan fn x D) For any fixed positive integer n, n x n tan j,.) x j x j for Sol: f n x x j x j x j x j n tan j n tan tan fn x x j x j j tan f tan n x x n x tan n tan tan tan tan f n f x x n x x x n x fn x x x n n tan x nx # 4, Kasetty eights, Ayyappa Society, Madhapur, yderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net Page-9

30 8-Jee-Advanced fn x fn x sec tan n limsec fn x lim x x Question Paper-_Key & Solutions x nx. 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 tangents 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 segments 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, 5 5 does NOT lie in E C) The point, lies in E D) The point, does NOT lie in E Key. D Sol: AP AQ AM Locus of M is a circle having PQ as its diameter : 7 5 and x ence, E x x y y Locus of B (midpoint) is a circle having RC as its diameter E : x x y Now, after checking the options, we get (D) # 4, Kasetty eights, Ayyappa Society, Madhapur, yderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net Page-

31 8-Jee-Advanced Question Paper-_Key & Solutions b. Let S be the of all column matrices b such that b, b, b and the system of equations (in b real variables) as at least one solution. Then, which of the following system(s) (in real b variables) has (have) at least one solution of each b S? b A) x y z b,4y 5z b and x y 6z b B) x y z b,5x y 6z b and x y z b C) x y 5 z b,x 4y z b and x y 5z b D) x y 5 z b,x z b and x 4y 5z b Key. ACD Sol. We find D= & since no pair of planes are parallel, so there are infinite number of solution s. Let P P P P 7P P b 7b b A) D unique solution for any b, b, b B) D but P 7P P C) D Also b b, b b D) D 4. Consider two straight lines, each of which is tangent to both the circle x y and the parabola y 4x. Let these lines intersect at the point Q. Consider the ellipse whose center is at the origin O(, ) and whose semi-major axis is OQ. If the length of the minor axis of this ellipse is, then the 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 4 D) The area of the region bounded by the ellipse between the lines 6 Key. AD x and x is x and x is # 4, Kasetty eights, Ayyappa Society, Madhapur, yderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net Page-

32 8-Jee-Advanced Sol. Question Paper-_Key & Solutions Let equation of common tangent is m y mx m m 4 m m m Equation of common tangents are y x and y x Point Q is, x y Equation of ellipse is / b A) e and LR a C) Area x.. 9 sin / x dx x x Correct answer are (A) and (D) 5. Let s, t, r be the non-zero complex numbers and L be the set of solutions Key. ACD z x iy x, y, i of the equation sz t z r, where z x iy.then, which of the following statements(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 # 4, Kasetty eights, Ayyappa Society, Madhapur, yderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net Page-

33 8-Jee-Advanced Sol. Given sz tz r.() On taking conjugate s z tz r..() From () and () eliminating z z s t rt rs A) If s t then z has unique value Question Paper-_Key & Solutions B) If s t then rt rs may or may not be zero so L may be empty set C) locus of z is null set or singleton set or a line in all cases it will intersect given circle at most two points D) In this case locus of z is a line so L has infinite elements sin sin f x t f t x 6. Let f :, be a twice differentiable function such that lim sin t t x for all x,. If f, then which of the following statement (s) is (are) TRUE? 6 4 x A) f B) f x x for all x, C) There exists, such that f ' D) f " f Key. BCD f xsin t f tsin x Sol. lim sin x tx t x By using L opital f xcost f tsin x lim sin x tx f x cos x f x sin x sin x f x x f x cos x sin x f x f x d x c sin x sin x Put x & f 6 6 c f x xsin x sin A) f 4 4 x # 4, Kasetty eights, Ayyappa Society, Madhapur, yderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net Page-

34 8-Jee-Advanced B) f x xsin x x x as sin x x, xsin x x x f x x x, 6 C) f x sin x xcos x 4 tan there exist, for which f x x x Question Paper-_Key & Solutions f f x cos x xsin x D) f f, f f SECTION (Maximum Marks: 4) This section contains EIGT (8) 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., -., -.,.7, -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 : + If ONLY the correct numerical value is entered as answer. Zero Marks : In all other cases. 7. The value of the integral Key. Sol. dx x x 6 dx /4 /4 6 x x 6 x x dx Put t x x dt dt 6 4 x x / / I 6/4 t t 8. Let P be a matrix of order such that all the entries in P are from the set,,. Then the maximum possible value of the determinant of P is. # 4, Kasetty eights, Ayyappa Society, Madhapur, yderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net Page-4

35 8-Jee-Advanced Key. 4 a a a Sol: b b b a b c a b c a b c a b c a b c a b c c c c Now if x and y the can be maximum 6 But it is not possible as x each term of x and y each term of y i a b c and i i i i Which is contradiction. x a b c i i i So now next possibility is 4 y Question Paper-_Key & Solutions 9. Let X be a set with exactly 5 elements and Y be a set with exactly 7 elements. If is the Key. 9 number of one- one functions form X to Y and is the number of onto functions from Y to X, 5! is. then the value of Sol. n X 5 ny 7 Number of one-one function 7 C5 5! Number of onto function Y to X,,,,,,,, 7! 7! ! 5! C. C 5! 4 C 5!!4!!! C C !. Let f : Key..4 be a differentiable function with dy dx f. If y f x equation 5y5y, then the value of lim f x x satisfies the differential is. # 4, Kasetty eights, Ayyappa Society, Madhapur, yderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net Page-5

36 8-Jee-Advanced dy Sol. 5y 4 dx dy So, dx 5y 4 y Integrating, ln 5 5 x c y 5 5 Question Paper-_Key & Solutions 5y ln x c 5y Now, f c as 5y ence 5y let x x x e 5 f 5 f Now, RS = let f x. Let f : x 5 x let e x x let 5x x x be a differentiable function with f O and satisfying the equation ' ' f x y f x f y f x f y for all x, y. Then, then value of Key. P x, y : f x y f x f ' y f ' x f y x. y R Sol. P, : f f f ' f ' f f ' P x, : f x f x. f ' f ' x. f f x f x f x f ' x f x ' f f x e x ln 4 loge f 4 is. Let P be a point in the first octant, whose image Q in the plane x y (that is, the line segment PQ is perpendicular to the plane x y and the mid-point of PQ lies in the plane x y ) 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 Key. 8 Sol. Let P,, Q,, & R,, # 4, Kasetty eights, Ayyappa Society, Madhapur, yderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net Page-6

37 8-Jee-Advanced Now, PQ iˆ ˆj iˆ ˆj iˆ ˆj Also, mid point of PQ lies on the plane 6 Now, distance of point P from X-axis 5 Question Paper-_Key & Solutions 5 6 as as 4 ence PR = 8. Consider the cube in the first octant with sides OP, OQ and OR of length, along the x-axis, y- Key..5 axis and z-axis, respectively, where O(,, ) 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 p q r t is, then the value of Sol. p SP,, i ˆ ˆj k ˆ q SQ,, i ˆ ˆj k ˆ r SR,, i ˆ ˆj k ˆ t ST,, i ˆ ˆj k ˆ iˆ ˆj kˆ iˆ ˆj kˆ p qr t 4 4 ˆ ˆ ˆ ˆ ˆ k i j i j 6 4. Let X C C C... C Key. 646, where r,,,... coefficients. Then, the value of n Sol: r n X r. C ; n r n n n X n. Cn r. C r X. 4 4 r 9 C9 = 646 n r 4 X is n n X n. Cr. C r X n C n n. n; C r denote binomial 9 X. C 9 # 4, Kasetty eights, Ayyappa Society, Madhapur, yderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net Page-7

38 8-Jee-Advanced Question Paper-_Key & Solutions SECTION (Maximum Marks: ) This section contains FOUR (4) 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 : + If ONLY the option corresponding to the correct matching is chosen. Zero Marks : If none of the options is chosen (i.e. the question is unanswered). Negative Marks : - In all other cases. x 5. Let E X : x and and x :sin log x E X e is a real number x. ( ere, the inverse trigonometric functionsin xassumes values in,. x Let f : E be the function defined by f x log e x and g : E be the function x defined by g x sin log. x LIST-I LIST-II e P. The range of f is.,, e e Q. The range of g contains. (, ) R. The domain of f contains., S. The domain of g is 4.,, The correct options is: A) P-4;Q-;R-;S- B) P-;Q-;R-6;S-5 C) P-4;Q-;R-;S-6 D) P-4;Q-;R-6;S-5 Key. A Sol. : x E x e 5., e e 6.,, e E : x,, : x E n x # 4, Kasetty eights, Ayyappa Society, Madhapur, yderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net Page-8

39 8-Jee-Advanced x e e x x Now x e e x e x Question Paper-_Key & Solutions x,, e x also e x e x e x e x,, e e So E :,, e e x as Range of x is R Range of f is,, R or Range of g is, \ or,, Now P 4, Q, R, S ence a is correct 6. In a high school, a committee has to be formed form a group of 6 boys M, M, M, M 4, M 5, M 6 and 5girls G, G, G, G4, G 5. i) Letbe the total number of ways in which the committee can be formed such that the committee has 5 members, having exactly 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 be the number of ways in which the committee can be formed such that the committee has 5 numbers, 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 lest girls and such that both Mand G are NOT in the committee together. LIST-I LIST-II P. The value ofis.6 Q. The value of is. 89 R. The value of is. 9 S. The value of 4 is # 4, Kasetty eights, Ayyappa Society, Madhapur, yderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net Page-9

40 8-Jee-Advanced Question Paper-_Key & Solutions The correct options is: A) P-4;Q-6;R-;S- B) P-;Q-4;R-;S- C) P-4;Q-6;R-5;S- D) P-4;Q-;R-;S- Key. C Sol. 65 () so P () ! So Q () = 8 So R (4) 89 4 So S 7. x y Let :, wherea b, be a hyperbola in the xy-plane whose conjugate axis LM a b subtends an angle of 6 at one of its vertices N, Let the area of the triangle LMN be 4 LIST-I LIST-II P. The length of the conjugate axis of is. 8 Q. The eccentricity of is. 4 R. The distance between the foci of is. S. The length of the latus rectum of is 4. 4 Key. B The correct options is: A) P-4;Q-;R-;S- B) P-4;Q-;R-;S- C) P-4;Q-;R-;S- D) P-;Q-4;R-;S- # 4, Kasetty eights, Ayyappa Society, Madhapur, yderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net Page-4

41 8-Jee-Advanced Question Paper-_Key & Solutions Sol. b tan a a b Now are of 4 b LMN.. b. b b & a P. Length of conjugate axis = b = 4 So P 4 Q. Eccentricity e So Q R. distance between foci = ae 8 So R b 4 S. Length of latus rectum a So S 8. Let f :, f :,, f, e and f : be functions defied by 4 (i) f x sin e x sin x if x (ii) f x tan x, where the inverse trigonometric function tan x assumes if x values in, (iii) f x sinloge x, where for t, t denotes the greatest integer less than or equal to t, x sin if x x (iv) f4 x if x # 4, Kasetty eights, Ayyappa Society, Madhapur, yderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net Page-4

42 8-Jee-Advanced Question Paper-_Key & Solutions LIST-I LIST-II P. the function f is. NOT continuous at x = Q. The function f is. Continuous at x = and NOT differentiable at x = R. The function f is. Differentiable at x = and is derivative is NOT continuous at x = S. The function f4 is 4. Differentiable at x = and its derivative is continuous at x = The correct options is: A) P-;Q-;R-;S-4 B) P-4;Q-;R-;S- C) P-4;Q-;R-;S- D) P-;Q-;R-4;S- Key. D Sol. i f x sin e x ' f x e e x x e at x x does not exist x x cos.. ' f So. P sin x, x ii f x tan x x sin x x lim x x tan x f x does not continuous at x = So Q iii f x n x sin / x e n x sin n x x f So R 4 x 4 So S sin, iv f x x x, x # 4, Kasetty eights, Ayyappa Society, Madhapur, yderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net Page-4