Paper-1 READ THE INSTRUCTIONS CAREFULLY

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2 Page- 7-Jee-Advanced Paper- READ THE INSTRUCTIONS CAREFULLY Question Paper-_Key & Solutions GENERAL. This sealed booklet is your Question Paper. Do not break the seal till you are told to do so.. The paper CODE is printed on the right hand top corner of this sheet and the right hand top corner of the back cover of this booklet.. Use the Optical Response Sheet (ORS) provided separately for answering the questions 4. The paper CODE is printed on the left part as well as the right part of the ORS. Ensure that both these codes are identical and same as that on the question paper booklet. If not, contact the invigilator for change of ORS. 5. Blank spaces are provided within this booklet for rough work. 6. Write your name, roll number and sign in the space provided on the back cover of this booklet. 7. After breaking the seal of the booklet at 9: am, verify that the booklet contains 6 pages and that all the 54 questions along with the options are legible. If not, contact the invigilator for replacement of the booklet. 8. You are allowed to take away the Question Paper at the end of the eamination. OPTICAL RESPONSE SHEET: 9. The ORS (top sheet) will be provided with an attached Candidate s Sheet (bottom sheet). The Candidate s Sheet is a carbon-less copy of the ORS.. Darken the appropriate bubbles on the ORS by applying sufficient pressure. This will leave an impression at the corresponding place on the Candidate s sheet.. The ORS will be collected by the invigilator at the end of the eamination.. You will be allowed to take away the Candidate s Sheet at the end of the eamination.. Do not tamper with or mutilate the ORS. Do not use the ORS for rough work. 4. Write your name, roll number and code of the eamination center, and sign with pen in the space provided for this purpose on the ORS. Do not write any of these details anywhere else on the ORS. Darken the appropriate bubble under each digit of your roll number. DARKENING THE BUBBLES ON THE ORS 5. Use a BLACK BALL POINT PEN to darken the bubbles on the ORS. 6. Darken the bubble COMPLETELY. 7. The correct way of darkening a bubble is as : 8. The ORS is machine-gradable. Ensure that the bubbles are darkened in the correct way. 9. Darken the bubbles ONYLY IF you are sure of the answer. There is NO WAY to erase or un-darken a darkened bubble. # 4, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net

3 Page- 7-Jee-Advanced PART-:PHYSICS SECTION - (Maimum Marks: 8) Question Paper-_Key & Solutions This Section Contains SEVEN questions. Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four options is(are) Correct. For each question, darken the bubble(s) corresponding to all the correct options(s) in the ORS. For each question, marks will be awarded in one of the following categories: Full Marks : +4 If only the bubble(s) corresponding to all the correct options(s) is (are) darkened. Partial Marks : + For darkening a bubble corresponding to each correct option, provided NO incorrect option is darkened. Zero Marks : If none of the bubbles is darkened. Negative Marks : - In all other cases. For eample, if (A), (C)and (D) are all the correct options for a question, darkening all these three will get +4 marks; darkening only (A) and (D) will get + marks; and darkening (A) and (B) will get - marks; as a wrong option is also darkened. ************************************************************************************************** Q. In the circuit shown, L H, C Fand R k. They are connected in series with an a.c. source V V sin t as shown. Which of the following options is/are correct? L H C F RK V sint A) At the current flowing through the circuit becomes nearly zero B) The frequency at which the current will be phase with the voltage is independent of R C) The current will be in phase with the voltage if rad. s 4 6 D) At rad. s, the circuit behaves like a capacitor A, B ~ O V sin A so no current and option A is correct (A) current in phase with voltage implies resonance so, L C or capacitance and inductance not on resistance. (B) is correct 6 rad 6 6 so (C) is not correct. or circuit behaves as inductive so (D) is not correct. 6 XL XC LC frequency depends on # 4, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net

4 Page-4 7-Jee-Advanced Question Paper-_Key & Solutions Q. For an isosceles prism of angle A and refractive inde, it is found that the angle of minimum deviation m A. Which of the following options is/are correct? i A A) For the angle of incidence B) At minimum deviation, the incident angle i and the refracting angle by r i /, the ray inside the prism is parallel to the base of the prism r at the first refracting surface are related C) For this prism, the emergent ray at the second surface will be tangential to the surface when the angle of incidence at the first surface is A i sin sin A 4cos cos A D) For this prism the refractive inde and the angle of prism A are related as A, B, C A m sin sin A m A& sin A/ sin A/ sin A/ cos A/ or cos A / sin A / cos A / / A / cos / or and or A cos / (D) is not correct. m i i A i i Aand i i A So option A is correct. sini r rand sin r sin sin / cos / cos A/ A sin r A A sin r cos A/ r A / r i / option (B) is correct. Emergent ray tangential to the surface means i 9 r C (critical angle) r A C sini sin r sini sin r cos A / sin AC cos A / sin AcosC cos Asin C cos A / sin A sin C cos A cos A / sin A cos A cos A / sin A cos A 4cos A/ cos A/ cos A # 4, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net

5 Page-5 7-Jee-Advanced Question Paper-_Key & Solutions sin A 4cos A/ cos A/ cos A cos A/ cos A/ sin A 4cos A / cos A i sin sin A 4cos A / cos Aoption C is correct Q. A circular insulated copper wire loop is twisted to form two loops of area A and A as shown in the figure. At the point of crossing the wires remain electrically insulated from each other. The entire loop lies in the plane (of the paper). A uniform magnetic field B points into the plane of the paper. At t, the loop starts rotating about the common diameter as ais with a constant angular velocity in the magnetic field. Which of the following options is/are correct? Area A B Area A A) The emf induced in the loop is proportional to the sum of the areas of the two loops B) The rate of change of the flu is maimum when the plane of the loops is perpendicular to plane of the paper C) The net emf induced due to both loops is proportional to cos t D) The amplitude of the maimum net emf induced due to the both loops is equal to the amplitude of maimum emf induced in the smaller loop alone # 4, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net

6 Page-6 7-Jee-Advanced Question Paper-_Key & Solutions B,D emf induced in the loop d BA cos t dt emf induced in the loop d B A cos t dt But these two are in opposite in sense as Area vectors are in opposite direction. net emf induced difference of area so option A is wrong d Rate of change of flu BA cos t or d B A cos t dt dt Will be ma when t 9so option B is correct and option D is currect Q4. A flat plate is moving normal to its plane through as a gas under the action of a constant force F. The gas is kept at a very low pressure. The speed of the plate v is much less than the average speed u of the gas molecules. Which of the following options is/are true? A) At a later time the eternal force F balance the resistance force B) The plate will continue to the move with constant non-zero acceleration, at all times C) The resistive force eperienced by the plate is proportional to v D) The pressure difference between the leading and trailing faces of the plate is proportional to uv ACD S area of plate V F u v u v P f p p p u v u v p p 4Uv p p uv Resistance form uvs US V KV K US (constant) Resistance force v p p s After long time, resistance can balance eternal force. # 4, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net

7 Page-7 7-Jee-Advanced Question Paper-_Key & Solutions Q5. A block of mass M has a circular cut with a frictionless surface as shown. The block rests on the horizontal frictionless surface of a fied table. Initially the right edge of the block is at, in a co-ordinate system fied to the table. A point mass m is released from rest at the topmost point of the path as shown and it slides down. When the mass loses contact with the block, its position is and the velocity is v. At that instant, which of the following options is/are correct? R y m R M gr A) The velocity of the point mass m is: v m M mr B) The component of displacement of the centre of mass of the block M is: M m C) The position of the point mass is mr M m m D) The velocity of the block M is: V gr M A,B ) mv MV v and v are velocities of m and M ) mgr mv M V m mgr mv M V M gr Option (A) V m M Option (B) mr M mr M m But in negative direction hence B is correct. # 4, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net

8 Page-8 7-Jee-Advanced Question Paper-_Key & Solutions Q6. A block M hangs vertically at the bottom end of a uniform rope of constant mass per unit length. The top end of the rope is attached to a fied rigid support at O. A transverse wave pulse (Pulse ) of wavelength is produced at pointo on the rope. The pulse takes time TOA to reach point A. If wave pulse of wavelength is produced at point A (Pulse ) without disturbing the position of M it takes timet AO to reach point O. Which of the following options is/are correct? O Pulse Pulse A M T T A) The time AO OA B) The wavelength of pulse becomes longer when it reaches point A C) The velocity of any pulse along the rope is independent of its frequency and wavelength D) The velocity of the two pulses (Pulse and pulse ) are the same at the midpoint of rope ACD option A is correct as variation of velocity is same w.r.t y T Velocity Q7. A human body has a surface area of approimately m. The normal body temperature is K above the surrounding room temperature T 46Wm 4 T. Take the room temperature to be T K. ForT K, the value of ( where is the Stefan Boltzmann constant). Which of the following options is/are correct? A) If the body temperature rises significantly then the peak in the spectrum of electromagnetic radiation emitted by the body would shift to longer wavelengths B) If the surrounding temperature reduces by a small amount T T, then to maintain the same body temperature the same (living) human being needs to radiate W 4 T T more energy per unit time C) The amount of energy radiated by the body in second is close to 6 joules D) Reducing the eposed surface area of the body (e.g. by curling up) allows humans to maintain the same body temperature while reducing the energy lost by radiation BCD or D (A) is clearly wrong from win s displacement law. W A T 4 T 4 (B) correct:- W A T T T 4 W W 4 AT T 4 (C) net heat radiated is 6J per second. (D) True, by reducing are a net power emitted is reduced while his body will same heat inside so will be easy to maintain temperature. # 4, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net

9 Page-9 7-Jee-Advanced SECTION - (Maimum Marks: 5) Question Paper-_Key & Solutions This section Contains FIVE questions. The answer to each question is a SINGLE DIGIT INTEGER ranging from to 9, both inclusive. For each question, darken the bubble corresponding to the correct integer in the ORS. For each question, marks will be awarded in one of the following categories: Full Marks : + If only the bubble corresponding to the correct answer is darkened. Negative Marks : - In all other cases. ************************************************************************************************** Q8. An electron in a hydrogen atom undergoes a transition from an orbit with quantum number ni to another with quantum number n f. Vi and V f are respectively the initial and final potential energies of the electrons. If Vi 6.5 V, then the smallest possible n f is f 5 7. inev n Vi andv f n n Potential energy Vi V f i n f 6.5(given) n n f.5 n i i for ni n 5 (smallest possible value) f f Q9.. A drop of liquid of radius R m having surface tension S Nm divides itself into K identical drops. 4 In this process the total change the surface energy U J. If K then the value of is 6 Surface energyu R S (for big drop) 4 U K4 r S ( for small drop) R R Kr r / K U K4r S 4R S 4 RS / K 4R S 4R S / K K /. / 4R SK 4 K 4 5 / K / K / 6 # 4, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net

10 age- 7-Jee-Advanced Q. A stationary surface emits sound of frequency f 49 Hz # 4, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net Question Paper-_Key & Solutions. The sound is reflected by a large car approaching the source with a speed of ms. The reflected signal is received by the source and superposed with the original. What will be the beat frequency of the resulting signal in Hz? (Given that the speed of sound in air is ms and the car reflects the sound at the frequency it has received). 6 frequency of sound received by car is V f 49 V us V V Reflected sound from the car will have a frequency f f V V V 49 V V Q. 5 V 49 V Beat frequency f f f 49 V V V V 49 V I is an isotope of iodine that decays to an isotope of Xenon with a half-life of 8 days A small amount of a serum laelled with I is injected into the blood of a person. The activity of the amount of I injected was 5.4 Becquerel (Bq). It is known that the injected serum will get distributed uniformly in the blood stream in less than half an hour. After.5 hours.5 ml of blood is drawn from the person s body, and gives an activity of 5 Bq. The total volume of blood in the person s body, in liters is approimately (you may use e for and ln.7 ). I Xe dn N.4 dt i.e. initially. 5.4 N t.4 N e e For.5 ml we get 5 Bq So for V vol. we get 5.5 V N 5 5V.4 t e.5 5 V lit In litres

11 age- 7-Jee-Advanced Question Paper-_Key & Solutions Q. A monochromatic light is travelling in a medium of refractive inde n=.6. It enters a stack of glass layers from the bottom side at an angle. The interfaces of the glass layers are parallel to each other. The refractive indices of different glass layers are monotonically decreasing as n n m n, n is the refractive inde of the m th and m m th m slab and n. m where m (see the figure). The ray is refracted out parallel to the interface between the th m slabs from the right side of the stack. What is the value of m? n mn n m n 8 nsin n m n n n n n n n n sin9 " m 8" ( but n for a material i.e n m n can t be less than ). # 4, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net

12 age- 7-Jee-Advanced SECTION - (Maimum Marks: 8) Question Paper-_Key & Solutions This section Contains SIX questions of matching type. This section Contains TWO tables (each having columns and 4 rows) Based on each table, there are THREE questions Each question has FOUR options [A], [B], [C], and [D]. ONLY ONE of these four options is correct For each question, darken the bubble corresponding to the correct option in the ORS. For each question, marks will be awarded in one of the following categories: Full Marks : + If only the bubble corresponding to the correct option is darkened. Zero Marks : If none of the bubbles is darkened Negative Marks : - In all other cases. ************************************************************************************************** A charged particle (electron or proton) is introduced at the origin (=,y=,z=) with a given initial velocity. A uniform electric field E and a uniform magnetic field B eist everywhere. The velocity, electric field E and magnetic field B are given columns, and, respectively. The quantities E, B are positive in magnitude. Column Column Column (I) Electron with E B (i) E E z (P) B B (II) Electron with E y B (ii) E E y (Q) B B (III) Proton with (iii) E E (R) B B y (IV) Proton with E B (iv) E E (S) B B z Q. In which case would the particle move in a straight line along the negative direction of y-ais (i.e., move along- y )? A) (IV) (ii) (S) B) (II) (iii) (Q) C) (III) (ii) (R) D) (III) (ii) (P) C Y B y proton release X E y Proton released from rest at origin F magnetic q( V y B y) V y qeyt j m # 4, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net

13 age- 7-Jee-Advanced V y is antiparallel to B y F magnetic Because of Electrostatic force, if the proton moves along ve y-direction. Q4. In which case will the particle move in a straight line with constant velocity? A) (II) (iii) (S) B) (III) (iii) (P) C) (IV) (i) (S) D) (III) (ii) (R) A Y Question Paper-_Key & Solutions V E FM F E X When Fm FE, E the electron moves with constant velocity along y-ais. qv B qe q B qe B F net y V y remains constant. 5. In which case will the particle describe a helical path with ais along the positive z direction? A) (II) (ii) (R) B) (III) (iii) (P) C) (IV) (i) (S) D) (IV) (ii) (R) C Y E V X B Z The proton describes helical path with increasing pitch, with ais of heli along z-ais and plane of heli is X, Y An ideal gas is undergoing a cyclic thermodynamic process in different ways as shown in the corresponding P-V diagrams in column of the table. Consider only the path from state to state. W denotes the corresponding work done on the system. The equations and plots in the table have standard notations as used in thermodynamic processes. Here is the ration of heat capacities at constant pressure and constant volume. The number of moles in the gas is n. Column Column Column (I) W PV PV (i) Isothermal (P) P V # 4, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net

14 age-4 7-Jee-Advanced (II) W PV PV (ii) Isochoric (Q) Question Paper-_Key & Solutions P (III) W (iii) Isobaric P V V (IV) W (iv) Adiabatic nrt ln( ) V (R) P V (S) V Q6. Which one of the following options correctly represents a thermodynamic process that is used as a correction ` in the determination of the speed of sound in an ideal gas? A) (IV) (ii) (R) B) (I) (ii) (Q) C) (I) (iv) (Q) D) (III) (iv) (R) C V sound RT M as the sound wave propagates, the air in a chamber undergoes compression & rare fraction, & Undergo a adiabatic process. So curves are steeper than isothermal dp P dv Adi V dp P dv Iso V Graph Q satisfies eqn() Q7. Which of the following options is the only correct representation of a process in which U Q P V? A) (II) (iii) (S) B) (II) (iii) (P) C) (III) (iii) (P) D) (II) (iv) (R) B U Q P V U PV Q As U W Q The Process represents, Isobaric process gas W P V P v v Pv Pv graph p satisfies isobaricprocess Q8. Which one of the following options is the correct combination? A) (II) (iv) (P) B) (III) (ii) (S) C) (II) (iv) (R) D) (IV) (ii) (S) B Work done in isochoric process is zero W as V Graph S represents isochoric processes # 4, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net

15 age-5 7-Jee-Advanced PART-:CHEMISTRY SECTION - (Maimum Marks: 8) Question Paper-_Key & Solutions This Section Contains SEVEN questions. Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four options is(are) Correct. For each question, darken the bubble(s) corresponding to all the correct options(s) in the ORS. For each question, marks will be awarded in one of the following categories: Full Marks : +4 If only the bubble(s) corresponding to all the correct options(s) is (are) darkened. Partial Marks : + For darkening a bubble corresponding to each correct option, provided NO incorrect option is darkened. Zero Marks : If none of the bubbles is darkened. Negative Marks : - In all other cases. For eample, if (A), (C)and (D) are all the correct options for a question, darkening all these three will get +4 marks; darkening only (A) and (D) will get + marks; and darkening (A) and (B) will get - marks; as a wrong option is also darkened. ************************************************************************************************** Q9. The colour of the X molecules of group 7 elements changes gradually from yellow to violet down the group. This is due to A) The physical state of X at room temperature changes from gas to solid down the group B) Decrease in HOMO-LUMO gap down the group C) Decrease in * O* gap down the group D) Decrease ionization energy down the group BC IF Consider, X as F F MoT config is / s */s s *s *p z = p p y * p *p HOMO y p * z LuMo if gap between HOMO & LUMO Due to absorbed energy decreases complimentary colour is seen light energy is seen both B & C are thus applicable Q. Addition of ecess aqueous ammonia to a pink coloured aqueous solution of MCl.6HO (X) and NH 4Cl gives an octahedral comple Y in the presence of air. In aqueous solution, comple Y behaves as : electrolyte. The reaction of X with ecess HCl at room temperature results in the formation of a blue coloured comple Z. The calculated spin only magnetic moment of X and Z is.87 B.M., whereas it is zero for comple Y. A) The hybridization of the central metal ion in Y is d sp B) When X and Z are in equilibrium at C, the colour of the solution is pink C) Z is a tetrahedral comple D) Addition of silver nitrate to Y gives only two equivalent of silver chloride ABC # 4, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net

16 age-6 7-Jee-Advanced Question Paper-_Key & Solutions X Co HO Cl 6 pink sp d Y Co NH Cl 6 d sp Z CoCl blue sp 4 Option (B) is correct CBSE book equilibrium chapter. Hence A is correct Z is tetrahedral, C is correct Q. An ideal gas is epanded from,, among the following is (are) p V T to p, V, T under different conditions. The correct statement(s) A) If the epansion is carried out freely, it is simultaneously both isothermal as well as adiabatic B) The work done by the gas is less when it is epanded reversibly from V to V under adiabatic conditions as compared to that when epanded reversibly from V to V under isothermal conditions C) The work done on the gas is maimum when it is compressed irreversibly from p, V to, constant pressure p p V against D) The change in internal energy of the gas is (i) zero, if it is epanded reversibly with, and (ii) positive, if it is epanded reversibly under adiabatic conditions with T T T T ABC in free epansion the heat supplied in zero (z=) And work done is zero in both isothermal and adiabatic process But T T P isothermal Adiabatic process V In adiabatic process PV cons tan t is always greater than so, opp. Pressure is less for same ep is Case of adiabatic process compare to isothermal so work done is less. # 4, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net

17 age-7 7-Jee-Advanced compression Question Paper-_Key & Solutions PV P PV, P PV V Epansion PV V Work done in compression is more than epansion or (Area under curve is more) w irr w rev comp Ep opp., pressure is more in compression P V In case of adiabatic epansion And temperature also. E et Q. For a solution formed by miing liquids L and M, the vapour pressure of L plotted against the mole fraction of M in solution is shown in the following figure. Here L and M in the solution. The correct statement(s) applicable to this system is (are) represent mole fractions of L and M, respectively, Z p L M A) Attractive intermolecular interactions between L-L in pure liquid L and M-M in pure liquid M are stronger than those between L-M when mied in solution B) The point Z represents vapour pressure of pure liquid M and Raoult s law is obeyed when L C) The point Z represents vapour pressure of pure liquid M and Raoult s law is obeyed from L D) The point Z represents vapour pressure of pure liquid L and Raoult s law is obeyed when L AD to L pl Q actual given behaviour Z epected ideal behaviour X m M X M X X L L # 4, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net

18 age-8 7-Jee-Advanced As actual is above epected ideal +ve deviation L L attn > L M attn Also m mattn > L M attm X only L is present P P, P Pt. z L L L M Question Paper-_Key & Z P L P L From pt Q to Z the graph (actual) is very similar to epected ideal behaviour where X it mm ideal behaviour. Q. The IUPAC name(s) of the following compound is (are) L HC Cl A) -chloro-4-methybenzene B) 4-chlorotoluene C) -methy-4-chlorobenzene D) 4-methylchlorobenzene AB Cl CH Alphabetical order is the priority order C of chloro has greater priority over W of methyl -chloro -4-methyl benzene A 4- Chlorotoluene B According to NCERT BOOK Q4. The correct statement(s) for the following addition reactions is (are) (i) (ii) HC H HC H H CH Br CHCl CH Br CHCl H Mand N Oand P # 4, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net

19 age-9 7-Jee-Advanced A) O and P are identical molecules B) Bromination proceeds through trans-addition in both the reactions C) (M and O) and (N and P) are two pairs of enantiomers D) (M and O) and (N and P) are two pairs of diastereomers BD Question Paper-_Key & Solutions CH CH () i Br/ CHCl H H OH OH HO HO H H CH CH M N CH CH ( ii) Br CHCl H HO OH H HO H H OH CH CH Bromination is anti addition. M & N are identical (M& O) & (N& P) two pairs of diastereomere O P Q5. The correct statement(s) about the ooacids, HClO 4 and HClO, is(are) A) The conjugate base of HClO 4 is weaker base than HO B) The central atom in both HClO 4 and HClO is sp hybridized C) HClO 4 is formed in the reaction between Cl and HO D) HClO 4 is more acidic than HClO because of the resonance stabilization of its anion ABD HClO 4 and HClO O O Cl O O is resonance stabilized, so charge density is less, it is a weak base than mo # 4, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net

20 age- 7-Jee-Advanced O sp Cl O O O ClO 4 Cl sp OH Question Paper-_Key & Solutions so HClO 4 is strong acid due to resonance stabilization of anion SECTION - (Maimum Marks: 5) This section Contains FIVE questions. The answer to each question is a SINGLE DIGIT INTEGER ranging from to 9, both inclusive. For each question, darken the bubble corresponding to the correct integer in the ORS. For each question, marks will be awarded in one of the following categories: Full Marks : + If only the bubble corresponding to the correct answer is darkened. Negative Marks : - In all other cases. ************************************************************************************************** 6. The conductance of a.5 M aqueous solution of a weak monobasic acid was determined by using a conductivity cell consisting of platinized Pt electrodes. The distance between the electrode is cm with an area 7 of cross section of cm. The conductance of this solution was found to be 5 S. The ph of the solution is 4. The value of limiting molar conductivity m of this weak monobasic acid in aqueous solution is Z S cm mol. The value of Z is 6 K Gcell constant k 5 cm S 6 cm S cm m 4 C.5M ph 4 4 HT C 5 M m m 4 m 6 /5 6S cm mol The sum of the number of lone pairs of electrons on each central atom in the following species is TeBr, BrF, SNF 6, and XeF (Atomic numbers: N = 7, F = 9, S = 6, Br = 5, Te = 5, Xe = 54) 6 # 4, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net

21 age- 7-Jee-Advanced F F F Te F F F sp d lonepair Question Paper-_Key & Solutions F N Br F S Xe lonepair F F F lonepair zero-lone pair Total 6 lone pair 8. Among the following, the number of aromatic compounds(s) is 5 Aromatic Aromatic ions Aromatic Aromatic Compounds Aromatic * It is clearly mentioned in the question word compounds then answer is only # 4, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net

22 age- 7-Jee-Advanced Question Paper-_Key & Solutions 9. A crystalline solid of a pure substance has a face-centred cubic structure with a cell edge of 4 pm. If the density of the substance in the crystal is The value of N is FCC: Z=4 Z M d a N 8. O 4 ( Mol. Wt) (4) NO 4 mol. mass 4 N no. of moles No.of atoms N 56 8g cm N 4 N 56, then the number of atoms present in 56 g of the crystal is. Among H, He, Li, Be, B, C, N, O, and F the number of diamagnetic species is (Among numbers: H, He, Li, Be 4, B 5, C 6, N 7, O 8, 9 5 or 6 H diamagnetic He Li Be paramagnetic Diamagnetic Does not eist B Paramagnetic due to s p miing C diamagnetic due to s p miing N diamagnetic (4 e ) O paramagnetic (7 e ) F diamagnetic F ) 4 N. SECTION - (Maimum Marks: 8) This section Contains SIX questions of matching type. This section Contains TWO tables (each having columns and 4 rows) Based on each table, there are THREE questions Each question has FOUR options [A], [B], [C], and [D]. ONLY ONE of these four options is correct For each question, darken the bubble corresponding to the correct option in the ORS. For each question, marks will be awarded in one of the following categories: Full Marks : + If only the bubble corresponding to the correct option is darkened. Zero Marks : If none of the bubbles is darkened Negative Marks : - In all other cases. ************************************************************************************************** # 4, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net

23 age- 7-Jee-Advanced Question Paper-_Key & Solutions Answer Q.,Q. Q by appropriately matching the information given in the three columns of the following table. The wave function is a mathematical function whose value depends upon spherical polar coordinates (r, n,, l ml ), of the electron and characterized by the quantum numbers n, l and m l. Here r is distance from nucleus, is colatitude and is azimuth. In the mathematical functions given in the Table, Z is atomic number and a is Bohr radius. Column Column Column (I) Is orbital (i) Zr Z a n,, l m e l a (P),, n l m r (II) S orbital (ii) one radial node (Q) Probability density at nucleus a r / a (III) P z orbital (IV) (iii) 5 Zr Z a n,, l m re l a COS (R) Probability density is maimum at nucleus d z orbital (iv) y-plane is a nodal plane (S) Energy needed to ecite electron from n= state to n= 4 state is 7 times the energy needed to ecite electron from n= state to n=6 state Q. For He ion, the only INCORRECT combination is A) (I) (i) (R) B) (II) (ii) (Q) C) (I) (i) (S) D) (I) (iii) (R) D s radial nodes, no angular component ( no cos ) s radial node,, no angular component ( no cos ) p z radial node, angular component is present ( cos ) z d radial node, angular component is present ( cos ) a)s (i) R s orbital radial node probability ma@nucleas correct b)s radial node of correct c) s (i) s correct s ( i) s correct ( 4) 6 8 ( 6) 6 # 4, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net

24 age-4 7-Jee-Advanced Question Paper-_Key & Solutions d) s no angular component cos should not be present wrong. Q. For the given orbital in column I, the only CORRECT combination for any hydrogen-like species is A) (I) (ii) (S) B) (IV) (iv) (R) C) (III) (iii) (P) D) (II) (ii) (P) D s radial nodes, no angular component ( no cos ) s radial node,, no angular component ( no cos ) p z radial node, angular component is present ( cos ) z d radial node, angular component is present ( cos ) a) s radial node wrong as (ii),(i) radial node b) d dz R is wrong point z c)for pzgraph p iswrong d) correct Q. For hydrogen atom, the only CORRECT combination is A) (II) (i) (Q) B) (I) (iv) (R) C) (I) (i) (P) D) (I) (i) (S) D s radial nodes, no angular component ( no cos ) s radial node,, no angular component ( no cos ) p z radial node, angular component is present ( cos ) z d radial node, angular component is present ( cos ) a) s radial node / coloumn() (i) no radial node wrong b) s no node plane wrong c)s no radial node coloumn () p radial node wrong d) correct Answer Q.4,Q.5 and Q.6 by appropriately matching the information given in the three columns of the following table. Columns, and, contain starting materials, reaction conditions, and type of reactions, respectively. Column- Column- Column- (I) Toluene (i) NaOH / Br (P) Condensation (II) Acetophenone (ii) Br / hv (Q) Carboylation (III) Benzaldehyde (iii) CH CO O CH COOK (R) Substitution / (IV) Phenol (iv) NaOH / CO (S) Haloform Q4. The only CORRECT combination in which the reaction proceeds through radical mechanism is A) (II) (iii) (R) B) (III) (ii) (P) C) (IV) ii) (Q) D) (I) (ii) (R) D # 4, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net

25 age-5 7-Jee-Advanced CH CH Br h Br / Question Paper-_Key & Solutions Free radical substitution Ans: (I) (II) (R) Q5. For the synthesis of benzoic acid, the only CORRECT combination is A) (III) (iv) (R) B) (IV) (ii) (P) C) (II) (i) (S) D) (I) (iv) (Q) C C CH NaoH / Br afterh COOH + CHBr Acetophenone Haloform rn S Q6. The only CORRECT combination that gives two different carboylic acids is A) (IV) (iii) (Q) B) (I) (i) (S) C) (III) (iii) (P) D) (II) (iv) (R) C (III) CH O After acidification ACO CH CH COOH ACOH CH COOK Cis & trans Condensation R n P (III)(iii)(P) # 4, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net

26 age-6 7-Jee-Advanced PART-:MATHS SECTION - (Maimum Marks: 8) Question Paper-_Key & Solutions This Section Contains SEVEN questions. Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four options is(are) Correct. For each question, darken the bubble(s) corresponding to all the correct options(s) in the ORS. For each question, marks will be awarded in one of the following categories: Full Marks : +4 If only the bubble(s) corresponding to all the correct options(s) is (are) darkened. Partial Marks : + For darkening a bubble corresponding to each correct option, provided NO incorrect option is darkened. Zero Marks : If none of the bubbles is darkened. Negative Marks : - In all other cases. For eample, if (A), (C)and (D) are all the correct options for a question, darkening all these three will get +4 marks; darkening only (A) and (D) will get + marks; and darkening (A) and (B) will get - marks; as a wrong option is also darkened. ************************************************************************************************** Q7. Let a, b, and y be real numbers such that aband y. If the comple number z iy satisfies az b Im y z (S) of? A) C), then which of the following is (are) possible value y B) y y D) AB az b az b z z z z a zz az bz b a zz y bz az b z zz z a b z z zz z z z or z z az bz bz az z z z z z y z y y # 4, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net

27 age-7 7-Jee-Advanced Q8. Lef f : R, Key Question Paper-_Key & Solutions be a continuous function. Then which of the following functions (S) has (have) the value zero at some point in the interval,? 9 A) f f tsin t dt B) f C) f tcos t dt D) : BC Sol : f is ve ; f t sin dtis ve f f t dt is ve sin 9 B) f f f f ve f for atleast one, C) f f t cot dt f f t cost dt is - ve f f ve f f f tcost dt is ve f for atleast one, D) f e f f t sect dt i f e f t sect dt ve f e f for sin, y Q9. If y is tangent to the hyperbola a 6 right angled triangle? e f t sin t dt A) a,4, B) a,4, C) a,4, D) a,8, BCD y Tangent with slope y 4a 6 Comparing a 7 7 4a 7 y a 6 y a 4 6 a a, 4, is Pythagorean triplet, then which of the following CANNOT be sides of a # 4, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net

28 age-8 7-Jee-Advanced Q4. Let X and Y be two events such that P X, P X / Y and / A) P X Y B) PY C) / BC P X Y D) P X Y 5 P X Y P X. PY / X 5 5 P X Y /5 4 PY P X / Y / 5 PX, PX / Y and PY / X P X / Y PY P X Y P Y P X Y P Y 4 /5 /5 /5 4 /5 4 / P X Y Q4. Which of the following is (are) NOT the square of a A) B) C) D) AC Determent value should not be negative A, C 5 Question Paper-_Key & Solutions P Y X. Then; 5 matri with real entries? For options B I and for option (D) be the greatest integer less than or equals to.then, at which of the following points(s) the functions Q4. Let cos f is discontinuous? A) B) C) D) BCD f cos # 4, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net

29 age-9 7-Jee-Advanced f ( ) f ( ) f () f( ) cos dis continuous f( ) cos( ) f( ) cos f ( ) cos4 dis continuous f() cos 4 f( ) cos dis continuous f( ) cos Q4. If chord, which is not a tangent, of the parabola y 6 has the equation y p, and midpointhk,, the which of the following is (are) possible values (S) of ph, and k? A) p, h, k 4 B) p 5, h 4, k C) p, h, k D) p, h, k 4 Key : A Sol : S S ky 8 h k 6h ky h k 8 8 ky k h Comparing with y p p k 4 4p k 8h 8h4 p 6 h p 4 k k 8 h Question Paper-_Key & Solutions SECTION - (Maimum Marks: 5) This section Contains FIVE questions. The answer to each question is a SINGLE DIGIT INTEGER ranging from to 9, both inclusive. For each question, darken the bubble corresponding to the correct integer in the ORS. For each question, marks will be awarded in one of the following categories: Full Marks : + If only the bubble corresponding to the correct answer is darkened. Negative Marks : - In all other cases. ************************************************************************************************** Q44. The sides of a right angled triangle are in the Arithmetic progression. If the Triangle has area 4, then what is the length of its smallest side? 6 Let the sides a-d, a, a+d ( a d) a ( a d ) 4ad a a 4d Sides are d, 4d, 5d # 4, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net

30 age- 7-Jee-Advanced Area : 6d 4 d = Smallest side = 6 Q45. For a real number if the system Of linear equations, has infinitely many solutions, then = But gives no solution y z = Question Paper-_Key & Solutions Q46. Words of length are formed using the letters A, B, C D,E,F,G,H,I,J Let be the number of such words where no letter is repeated ; and let y be the number of such words Where eactly one letter is repeated twice and no other letter is repeated. Then 9 y = 5! 5! 9! y C 45!!! y 45! 5 9 9! Q47. For how many Values of p, the circle y 4y p and the coordinate aes have eactly three common points? # 4, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net

31 age- 7-Jee-Advanced Question Paper-_Key & Solutions Q48 Let f : R R be a differentiable function such that f()=, f( )= and f ()=. If g()= f ( t)cosect cot t cos ectf ( t) dt for (, ] then Lim g( ) f = sin g f cosec f Let g Lt Lt f Lt cosn f sin SECTION - (Maimum Marks: 8) This section Contains SIX questions of matching type. This section Contains TWO tables (each having columns and 4 rows) Based on each table, there are THREE questions Each question has FOUR options [A], [B], [C], and [D]. ONLY ONE of these four options is correct For each question, darken the bubble corresponding to the correct option in the ORS. For each question, marks will be awarded in one of the following categories: Full Marks : + If only the bubble corresponding to the correct option is darkened. Zero Marks : If none of the bubbles is darkened Negative Marks : - In all other cases. ************************************************************************************************** Answer Q.49,Q.5 and Q.5 by appropriately matching the information given in the three columns of the following tables Column Column Column I. II. III. IV. y a (i) a y a (ii) y my m a y m a m 4a (iii) y= m a y a (iv) y= m a m a m a a p., m m Q. R. S. ma a, m m am a m, a m am a m, a m # 4, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net

32 age- 7-Jee-Advanced Q49. The Tangent to a suitable conic (Column) at, option is the only CORRECT combination? A) IV iiis B)II iii R C) IV ivs D) II ivr Question Paper-_Key & Solutions is found to be +y=4, then Which of the following D y 4 y If a 4, lieson II, a a 4 y 4. Satisfies D is correct 4 Q5. If a tangent to a suitable conic (Column) is found to be y 8 of the following option is the only CORRECT combination? and it s point of contact is A) IIIi P B)I iiq C)II ivr D) III iiq A y. 8 a y m matches A is correct m 8,6 lieson6 4 a.8 a 8 Q5. Fora, If a Tangent is drawn to a suitable conic (Column ) at the point of contact 8,6 then which,, then which of the following options is the only CORRECT combination for obtaining its equation? A) II iiq B) I ip C)I iiq D) III ip C I y : tangent at, I, ii, Q is C is correct is y y # 4, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net

33 age- 7-Jee-Advanced Let f log log,, e e Column contains information about zeros of f, f and f. Column contains information about the limiting behavior of f, f and f Column contains information about increasing/decreasing nature of (I) Question Paper-_Key & Solutions f and f. at infinity. Column Column Column f for some, e (II) f for some, e (III) f for some (IV), f for some, e (I) lim f (P) f is increasing in, (II) lim f (Q) f is decreasing in ee, (III) lim f (R) (IV) lim f (S) f is increasing in f I decreasing in, ee, Q5. Which of the following options is the only CORRECT combination? A) I ii R B) IV i S C) III iv P D) II iii S D f log lg f log log f f f e e e, f, f e e Let f Let f Let f f, for, f isincrea sin g f, for e, e f is decrea sin g # 4, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net

34 age-4 7-Jee-Advanced f, for, f is decrea sin g f, for e, e f is decrea sin g Question Paper-_Key & Solutions Q5. Which of the following options I the only CORRECT combination? A) I i P B) II ii Q C) III iii R D) IV iv S B f log lg f log log f f f e e e, f, f e e Let f Let f Let f Let f Let f Let f f, for, f isincrea sin g f, for e, e f is decrea sin g f, for, f is decrea sin g f, for e, e f is decrea sin g Q54. Which of the following options is the only INCORRECT combinations? A) II iii P B) I iii P C) III i R D) II iv Q C f log lg f log log f # 4, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net

35 age-5 7-Jee-Advanced f f e e e, f, f e e Let f Let f Let f Let f Let f Let f f, for, f isincrea sin g f, for e, e f is decrea sin g f, for, f is decrea sin g f, for e, e f is decrea sin g Question Paper-_Key & Solutions **************** # 4, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad 58 www. srichaitanya.net, iconcohyd@srichaitanyacollege.net