FIITJEE SOLUTIONS TO JEE(ADVANCED)-2014

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1 Note: For the benefit of the students, specially the aspiring ones, the question of JEE(advanced), 14 are also given in this booklet. Keeping the interest of students studying in class XI, the questions based on topics from class XI have been marked with *, which can be attempted as a test. For this test the time allocated in Physics, Chemistry & Mathematics are minutes, 4 minutes and 5 minutes respectively. FIITJEE SLUTINS T JEE(ADVANCED)-14 CDE 8 PAPER- Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose. Please read the last page of this booklet for rest of the instructions. P Note: Front and back cover pages are reproduced from actual paper back to back here. Time: Hours Maimum Marks: 18 INSTRUCTINS A. General 1. This booklet is your Question Paper. Do not break the seal of this booklet before being instructed to do so by the invigilators.. The question paper CDE is printed on the left hand top corner of this sheet and on the back cover page of this booklet.. Blank space and blank pages are provided in the question paper for your rough work. No additional sheets will be provided for rough work. 4. Blank papers, clipboards, log tables, slide rules, calculators, cameras, cellular phones, pagers and electronic gadget of any kind are NT allowed inside the eamination hall. 5. Write your name and roll number in the space provided on the back cover of this booklet. 6. Answers to the questions and personal details are to be filled on an ptical Response Sheet, which is provided separately. The RS is a doublet of two sheets upper and lower, having identical layout. The upper sheet is a machine-gradable bjective Response Sheet (RS) which will be collected by the invigilator at the end of the eamination. The upper sheet is designed in such a way that darkening the bubble with a ball point pen will leave an identical impression at the corresponding place on the lower sheet. You will be allowed to take away the lower sheet at the end of the eamination (see Figure-1 on the back cover page for the correct way of darkening the bubbles for valid answers). 7. Use a black ball point pen only to darken the bubbles on the upper original sheet. Apply sufficient pressure so that the impression is created on the lower sheet. See Figure -1 on the back cover page for appropriate way of darkening the bubbles for valid answers. 8. D NT TAMPER WITH / MUTILATE THE RS S THIS BKLET. 9. n breaking the seal of the booklet check that it contains 8 pages and all the 6 questions and corresponding answer choices are legible. Read carefully the instruction printed at the beginning of each section. B. Filling the right part of the RS 1. The RS also has a CDE printed on its left and right parts. 11. Verify that the CDE printed on the RS (on both the left and right parts) is the same as that on the this booklet and put your signature in the Bo designated as R4. 1. IF THE CDES D NT MATCH, ASK FR A CHANGE F THE BKLET / RS AS APPLICABLE. 1. Write your Name, Roll No. and the name of centre and sign with pen in the boes provided on the upper sheet of RS. Do not write any of this anywhere else. Darken the appropriate bubble UNDER each digit of your Roll No. in such way that the impression is created on the bottom sheet. (see eample in Figure on the back cover) C. Question Paper Format The question paper consists of three parts (Physics, Chemistry and Mathematics). Each part consists of two sections. 14. Section 1 contains 1 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which NE is correct. 15. Section contains paragraphs each describing theory, eperiment and data etc. Si questions relate to three paragraphs with two questions on each paragraph. Each question pertaining to a particular passage should have only one correct answer among the four given choices (A), (B), (C) and (D). 16. Section contains 4 multiple choice questions. Each questions has two lists (Lits-1: P, Q, R and S; List-, : 1,,, and 4). The options for the correct match are provided as (A), (B), (C) and (D) out of which NLY one is correct. FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -1116, Ph 4616, , Fa 65194

2 () PART I : PHYSICS SECTIN 1 : (nly ne ption Correct Type) This section contains 1 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which NLY NE option is correct. 1. Charges Q, Q and 4Q are uniformly distributed in three dielectric solid spheres 1, and of radii R/, R and R respectively, as shown in figure. If magnitudes of the electric fields at point P at a distance R from the centre of spheres 1, and are E 1, E and E respectively, then Q P R R/ Q P R 4Q P R R Sphere 1 Sphere Sphere (A) E 1 > E > E (A) E > E 1 > E (C) E > E 1 > E (D) E > E > E 1 *. A glass capillary tube is of the shape of truncated cone with an ape angle so that its two ends have cross sections of different radii. When dipped in water vertically, water rises in it to a height h, where the radius of its cross section is b. If the surface tension of water is S, its density is, and its contact angle with glass is, the value of h will be (g is the acceleration due to gravity) S (A) cos( ) bg (B) (C) (D) S cos( ) bg S cos( / ) bg S cos( / ) bg h. If cu is the wavelength of K X-ray line of copper (atomic number 9) and Mo is the wavelength of the K X-ray line of molybdenum (atomic number 4), then the ratio cu / Mo is close to (A) 1.99 (B).14 (C).5 (D).48 *4. A planet of radius R = 1 1 (radius of Earth) has the same mass density as Earth. Scientists dig a well of depth R 5 on it and lower a wire of the same length and of linear mass density 1 kgm 1 into it. If the wire is not touching anywhere, the force applied at the top of the wire by a person holding it in place is (take the radius of Earth = m and the acceleration due to gravity of Earth is 1 ms ) (A) 96 N (B) 18 N (C) 1 N (D) 15 N FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -1116, Ph 4616, , Fa 65194

3 () *5. A tennis ball is dropped on a horizontal smooth surface. It bounces back to its original position after hitting the surface. The force on the ball during the collision is proportional to the length of compression of the ball. Which one of the following sketches describes the variation of its kinetic energy K with time t most appropriately? The figures are only illustrative and not to the scale. (A) (B) K K (C) K t (D) K t t t 6. A metal surface is illuminated by light of two different wavelengths 48 nm and 1 nm. The maimum speeds of the photoelectrons corresponding to these wavelengths are u 1 and u, respectively. If the ratio u 1 : u = : 1 and hc = 14 ev nm, the work function of the metal is nearly (A).7 ev (B). ev (C).8 ev (D).5 ev *7. A wire, which passes through the hole in a small bead, is bent in the form of quarter of a circle. The wire is fied vertically on ground as shown in the figure. The bead is released from near the top of the wire and it slides along the wire without friction. As the bead moves from A to B, the force it applies on the wire is (A) always radially outwards. A (B) always radially inwards. (C) radially outwards initially and radially inwards later. (D) radially inwards initially and radially outwards later. 9 B 8. During an eperiment with a metre bridge, the galvanometer shows a null point when the jockey is pressed at 4. cm using a standard resistance of 9, as shown in the figure. The least count of the scale used in the metre bridge is 1 mm. The unknown resistance is R 9 4.cm (A) 6.15 (B) (C) 6.5 (D) Parallel rays of light of intensity I = 91 Wm are incident on a spherical black body kept in surroundings of temperature K. Take Stefan-Boltzmann constant = Wm K 4 and assume that the energy echange with the surroundings is only through radiation. The final steady state temperature of the black body is close to (A) K (B) 66 K (C) 99 K (D) 155 K 1. A point source S is placed at the bottom of a transparent block of height 1 mm and refractive inde.7. It is immersed in a lower refractive inde liquid as shown in the figure. It is found that the light emerging S from the block to the liquid forms a circular bright spot of diameter mm on the top of the block. The refractive inde of the liquid is (A) 1.1 (B) 1. (C) 1.6 (D) 1.4 Liquid Block FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -1116, Ph 4616, , Fa 65194

4 (4) SECTIN : Comprehension type (nly ne ption Correct) This section contains paragraphs, each describing theory, eperiments, data etc. Si questions relate to the three paragraphs with two questions on each paragraph. Each question has only one correct answer among the four given options (A), (B), (C) and (D). Paragraph for Questions 11 & 1 The figure shows a circular loop of radius a with two long parallel wires (numbered 1 and ) all in the plane of the paper. The distance of each wire from the centre of the loop is d. The loop and the wires are carrying the same current I. The current in the loop is in the counterclockwise direction if seen from above. Q Wire I d d S Wire a P R 11. When d a but wires are not touching the loop, it is found that the net magnetic field on the ais of the loop is zero at a height h above the loop. In that case (A) current in wire 1 and wire is the direction PQ and RS, respectively and h a (B) current in wire 1 and wire is the direction PQ and SR, respectively and h a (C) current in wire 1 and wire is the direction PQ and SR, respectively and h 1.a (D) current in wire 1 and wire is the direction PQ and RS, respectively and h 1.a 1. Consider d >> a, and the loop is rotated about its diameter parallel to the wires by from the position shown in the figure. If the currents in the wires are in the opposite directions, the torque on the loop at its new position will be (assume that the net field due to the wires is constant over the loop) (A) I a d (B) I a d (C) I a Paragraph for Questions 1 & 14 d (D) I a In the figure a container is shown to have a movable (without friction) piston on top. The container and the piston are all made of perfectly insulating material allowing no heat transfer between outside and inside the container. The container is divided into two compartments by a rigid partition made of a thermally conducting material that allows slow transfer of heat. The lower compartment of the container is filled with moles of an ideal monatomic gas at 7 K and the upper compartment is filled with moles of an ideal diatomic gas at 4 K. The heat capacities per mole of an ideal monatomic gas are CV R, CP R, and those for an ideal diatomic gas are CV R, CP R. *1. Consider the partition to be rigidly fied so that it does not move. When equilibrium is achieved, the final temperature of the gases will be (A) 55 K (B) 55 K (C) 51K (D) 49 K *14. Now consider the partition to be free to move without friction so that the pressure of gases in both compartments is the same. Then total work done by the gases till the time they achieve equilibrium will be (A) 5 R (B) R (C) 1 R (D) 1 R d FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -1116, Ph 4616, , Fa 65194

5 (5) Paragraph for Questions 15 & 16 A spray gun is shown in the figure where a piston pushes air out of a nozzle. A thin tube of uniform cross section is connected to the nozzle. The other end of the tube is in a small liquid container. As the piston pushes air through the nozzle, the liquid from the container rises into the nozzle and is sprayed out. For the spray gun shown, the radii of the piston and the nozzle are mm and 1 mm respectively. The upper end of the container is open to the atmosphere. *15. If the piston is pushed at a speed of 5 mms 1, the air comes out of the nozzle with a speed of (A).1 ms 1 (B) 1 ms 1 (C) ms 1 (D) 8 ms 1 *16. If the density of air is a and that of the liquid, then for a given piston speed the rate (volume per unit time) at which the liquid is sprayed will be proportional to (A) a (B) a (C) SECTIN : Match List Type (nly ne ption Correct) a (D) This section contains four questions, each having two matching lists. Choices for the correct combination of elements from List-I and List-II are given as option (A), (B), (C) and (D) out of which one is correct. *17. A person in a lift is holding a water jar, which has a small hole at the lower end of its side. When the lift is at rest, the water jet coming out of the hole hits the floor of the lift at a distance d of 1. m from the person. In the following, state of the lift s motion is given in List I and the distance where the water jet hits the floor of the lift is given in List II. Match the statements from List I with those in List II and select the correct answer using the code given below the lists. List I List II P. Lift is accelerating vertically up. 1. d = 1. m Q. Lift is accelerating vertically down. d > 1. m with an acceleration less than the gravitational acceleration. R. Lift is moving vertically up with. d < 1. m constant speed. S. Lift is falling freely. 4. No water leaks out of the jar Code: (A) P-, Q-, R-, S-4 (B) P-, Q-, R-1, S-4 (C) P-1, Q-1, R-1, S-4 (D) P-, Q-, R-1, S Four charges Q 1, Q, Q and Q 4 of same magnitude are fied along the ais at = -a, a, +a and +a, respectively. A positive charge q is placed on the positive y ais at a distance b >. Four options of the signs of these charges are given in List I. The direction of the forces on the charge q is given in List II. Match List I with List II and select the correct answer using the code given below the lists. List I P. Q 1, Q, Q Q 4 all positive 1. + Q. Q 1, Q positive; Q, Q 4 negative. R. Q 1, Q 4 positive ; Q, Q negative. +y S. Q 1, Q positive; Q, Q 4 negative 4. y List II Code: (A) P-, Q-1, R-4, S- (B) P-4, Q-, R-, S-1 (C) P-, Q-1, R-, S-4 (D) P-4, Q-, R-1, S- (+, b) q Q 1 Q Q Q 4 (a, ) (a, ) (+a, ) (+a, ) FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -1116, Ph 4616, , Fa 65194

6 (6) 19. Four combinations of two thin lenses are given in List I. The radius of curvature of all curved surfaces is r and the refractive inde of all the lenses is 1.5. Match lens combinations in List I with their focal length in List II and select the correct answer using the code given below the lists. List I List II P. 1. r Q.. r/ R.. r S. 4. r Code: (A) P-1, Q-, R-, S-4 (B) P-, Q-4, R-, S-1 (C) P-4, Q-1, R-, S- (D) P-, Q-1, R-, S-4 *. A block of mass m 1 = 1 kg another mass m = kg, are placed together (see figure) on an inclined plane with angle of inclination. Various values of are given in List I. The coefficient of friction between the block m 1 and the plane is always zero. The coefficient of static and dynamic friction between the block m and the plane are equal to =.. In List II epressions for the friction on the block m are given. Match the correct epression of the friction in List II with the angles given in List I, and choose the correct option. The acceleration due to gravity is denoted by g. [Useful information: tan (5.5).1; tan (11.5).; tan (16.5).] m 1 m List I List II P. = 5 1. m g sin Q. = 1. (m 1 +m ) g sin R. = 15. m g cos S. = 4. (m 1 + m )g cos Code: (A) P-1, Q-1, R-1, S- (B) P-, Q-, R-, S- (C) P-, Q-, R-, S-4 (D) P-, Q-, R-, S- FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -1116, Ph 4616, , Fa 65194

7 (7) PART II : CHEMISTRY SECTIN 1 : (nly ne ption Correct Type) This section contains 1 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which NLY NE option is correct. *1. Assuming s p miing is NT operative, the paramagnetic species among the following is (A) Be (B) B (C) C (D) N *. For the process H H g at T = 1C and 1 atmosphere pressure, the correct choice is (A) Ssystem and Ssurrounding (B) Ssystem and Ssurrounding (C) Ssystem and Ssurrounding (D) Ssystem and Ssurrounding *. For the elementary reaction M N, the rate of disappearance of M increases by a factor of 8 upon doubling the concentration of M. The order of the reaction with respect to M is (A) 4 (B) (C) (D) 1 4. For the identification of -naphthol using dye test, it is necessary to use (A) dichloromethane solution of -naphthol. (B) acidic solution of -naphthol. (C) neutral solution of -naphthol. (D) alkaline solution of -naphthol. *5. Isomers of heane, based on their branching, can be divided into three distinct classes as shown in the figure. [Figure] I and II and III The correct order of their boiling point is (A) I > II > III (C) II > III > I (B) III > II > I (D) III > I > II 6. The major product in the following reaction is [Figure] Cl (A) 1. CH MgBr, dry ether, C CH. aqueous acid (B) H (C) H C CH (D) H C CH CH CH CH CH FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -1116, Ph 4616, , Fa 65194

8 (8) 7. Under ambient conditions, the total number of gases released as products in the final step of the reaction scheme shown below is Complete Hydrolysis XeF P other product 6 H / H Q slow disproportionation in H / H products (A) (B) 1 (C) (D) 8. The product formed in the reaction of SCl with white phosphorous is (A) PCl (B) S Cl (C) SCl (D) PCl *9. Hydrogen peroide in its reaction with KI 4 and NH H respectively, is acting as a (A) reducing agent, oidising agent (B) reducing agent, reducing agent (C) oidising agent, oidising agent (D) oidising agent, reducing agent. The acidic hydrolysis of ether (X) shown below is fastest when [Figure] R acid H RH (A) one phenyl group is replaced by a methyl group. (B) one phenyl group is replaced by a para-methoyphenyl group. (C) two phenyl groups are replaced by two para-methoyphenyl groups. (D) no structural change is made to X. SECTIN : Comprehension type (nly ne ption Correct) This section contains paragraphs, each describing theory, eperiments, data etc. Si questions relate to the three paragraphs with two questions on each paragraph. Each question has only one correct answer among the four given options (A), (B), (C) and (D). Paragraph for Questions 1 & X and Y are two volatile liquids with molar weights of 1 g mol 1 and 4 g mol 1 respectively. Two cotton plugs, one soaked in X and the other soaked in Y, are simultaneously placed at the ends of a tube of length L = 4 cm, as shown in the figure. The tube is filled with an inert gas at 1 atmosphere pressure and a temperature of K. Vapours of X and Y react to form a product which is first observed at a distance d cm from the plug soaked in X. Take X and Y to have equal molecular diameters and assume ideal behaviour for the inert gas and the two vapours. FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -1116, Ph 4616, , Fa 65194

9 (9) L = 4 cm Cotton wool soaked in X d Initial formation of the product Cotton wool soaked in Y *1. The value of d in cm (shown in the figure), as estimated from Graham s law, is (A) 8 (B) 1 (C) 16 (D) *. The eperimental value of d is found to be smaller than the estimate obtained using Graham s law. This is due to (A) larger mean free path for X as compared to that of Y. (B) larger mean free path for Y as compared to that of X. (C) increased collision frequency of Y with the inert gas as compared to that of X with the inert gas. (D) increased collision frequency of X with the inert gas as compared to that of Y with the inert gas. Paragraph for Questions & 4 Schemes 1 and describe sequential transformation of alkynes M and N. Consider only the major products formed in each step for both schemes. 1. NaNH ecess. CHCHI 1 equivalent H X. CH I 1 equivalent H M 4. H, Lindlar 's catalyst Scheme 1 N H 1. NaNH equivalent. H Br Y.H, mild 4. H, Pd/C 5. Cr Scheme. The product X is (A) H C (B) H H (C) CH CH H H (D) H C H H H H CH CH 4. The correct statement with respect to product Y is (A) It gives a positive Tollens test and is a functional isomer of X. (B) It gives a positive Tollens test and is a geometrical isomer of X. (C) It gives a positive iodoform test and is a functional isomer of X. (D) It gives a positive iodoform test and is a geometrical isomer of X. FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -1116, Ph 4616, , Fa 65194

10 (1) Paragraph for Questions 5 & 6 An aqueous solution of metal ion M1 reacts separately with reagents Q and R in ecess to give tetrahedral and square planar complees, respectively. An aqueous solution of another metal ion M always forms tetrahedral complees with these reagents. Aqueous solution of M on reaction with reagent S gives white precipitate which dissolves in ecess of S. The reactions are summarized in the scheme given below. SCHEME: Q ecess R ecess Tetrahedral M1 Square planar Tetrahedral Q R ecess M ecess Tetrahedral S,stoichiometricamount Whiteprecipitate S ecess precipitate dissolves 5. M1, Q and R, respectively are (A) Zn (C) Cd,KCN and HCl,KCN and HCl (B) (D) Ni Co, HCl and KCN,HCl andkcn 6. Reagent S is K Fe CN (C) KCr 4 (A) 4 6 (B) NaHP 4 (D) KH SECTIN : Match List Type (nly ne ption Correct) This section contains four questions, each having two matching lists. Choices for the correct combination of elements from List-I and List-II are given as option (A), (B), (C) and (D) out of which one is correct. 7. Match each coordination compound in List-I with an appropriate pair of characteristics from List-II and select the correct answer using the code given below the lists {en = H NCH CH NH ; atomic numbers: Ti =, Cr = 4; Co= 7; Pt = 78) List I List - II P. [Cr(NH ) 4 Cl ]Cl 1. Paramagnetic and ehibits ionization isomerism Q. [Ti(H ) 5 Cl](N ). Diamagnetic and ehibits cis-trans isomerism R. [Pt(en)(NH )Cl]N. Paramagnetic and ehibits cis-trans isomerism S. [Co(NH ) 4 (N ) ]N 4. Diamagnetic and ehibits ionization isomerism Code: P Q R S (A) 4 1 (B) 1 4 (C) 1 4 (D) 1 4 FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -1116, Ph 4616, , Fa 65194

11 (11) *8. Match the orbital overlap figures shown in List-I with the description given in List-II and select the correct answer using the code given below the lists. List I List - II P. 1. p d antibonding Q.. d d bonding R.. p d bonding S. 4. d d antibonding Code: P Q R S (A) 1 4 (B) 4 1 (C) 1 4 (D) 4 1 FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -1116, Ph 4616, , Fa 65194

12 (1) 9. Different possible thermal decomposition pathways for peroyesters are shown below. Match each pathway from List I with an appropriate structure from List II and select the correct answer using the code given below the lists. P C R R ' Q C R R ' R X carbonyl compound R R' (Peroyester) R RC R ' R X ' carbonyl compound C S RC R ' R R ' C List I P. Pathway P 1. List II C 6 H 5 CH Q. Pathway Q. C 6 H 5 R. Pathway R. CH CH CH C 6 H 5 CH CH CH C6H5 S. Pathway S 4. Code: P Q R S (A) 1 4 (B) 4 1 (C) 4 1 (D) 1 4 C 6 H 5 C 6 H 5 CH CH FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -1116, Ph 4616, , Fa 65194

13 (1) 4. Match the four starting materials (P, Q, R, S) given in List I with the corresponding reaction schemes (I, II, III, IV) provided in List II and select the correct answer using the code given below the lists. List I List - II P. H H 1. Scheme I H Q. i KMn 4, H, heat ii H, H iii SCl iv NH? C7H6N. Scheme II i Sn /HCl iich CCl iii conc. H S iv HN v dil. H S, heat vi H 4 4? C6H6N R. N H. Scheme III i red hot iron, 87 K ii fu min g HN, HS 4, heat iii H S.NH iv NaN, H S v hydrolysis 4? C6H5N S. N 4. Scheme IV i conc. HS 4, 6 C ii conc. HN, conc. H S iii dil. H S, heat 4 4? C6H5N4 CH Code: P Q R S (A) 1 4 (B) 1 4 (C) 4 1 (D) 4 1 PART III : MATHEMATICS SECTIN 1 : (nly ne ption Correct Type) This section contains 1 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which NLY NE option is correct. 41. Three boys and two girls stand in a queue. The probability, that the number of boys ahead of every girl is at least one more than the number of girls ahead of her, is (A) 1 (B) 1 (C) (D) 4 *4. In a triangle the sum of two sides is and the product of the same two sides is y. If c = y, where c is the third side of the triangle, then the ratio of the in-radius to the circum-radius of the triangle is y y (A) (B) c c c (C) y 4 c (D) y 4c c FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -1116, Ph 4616, , Fa 65194

14 (14) *4. Si cards and si envelopes are numbered 1,,, 4, 5, 6 and cards are to be placed in envelopes so that each envelope contains eactly one card and no card is placed in the envelope bearing the same number and moreover the card numbered 1 is always placed in envelope numbered. Then the number of ways it can be done is (A) 64 (B) 65 (C) 5 (D) 67 *44. The common tangents to the circle + y = and the parabola y = 8 touch the circle at the points P, Q and the parabola at the points R, S. Then the area of the quadrilateral PQRS is (A) (B) 6 (C) 9 (D) 15 *45. The quadratic equation p() = with real coefficients has purely imaginary roots. Then the equation p(p()) = has (A) only purely imaginary roots (B) all real roots (C) two real and two purely imaginary roots (D) neither real nor purely imaginary roots 46. The following integral cosec log 1 / 17 d is equal to /4 16 u u u u (A) e e du (B) log 1 log 1 17 e e du 17 u u u u (C) e e du (D) log 1 16 e e du 47. The function y = f () is the solution of the differential equation f () =. Then (A) (C) f d is (B) 4 (D) dy y d in (1, 1) satisfying 48. Let f : [, ] R be a function which is continuous on [, ] and is differentiable on (, ) with f() = 1. Let F f t dt for [, ]. If F() = f () for all (, ), then F() equals (A) e 1 (B) e 4 1 (C) e 1 (D) e 4 *49. Coefficient of 11 in the epansion of (1 + ) 4 (1 + ) 7 (1 + 4 ) 1 is (A) 151 (B) 116 (C) 111 (D) 11 FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -1116, Ph 4616, , Fa 65194

15 (15) *5. For (, ), the equation sin + sin sin = has (A) infinitely many solutions (B) three solutions (C) one solution (D) no solution SECTIN : Comprehension Type (nly ne ption Correct) This section contains paragraph, each describing theory, eperiments, data etc. Si questions relate to the three paragraphs with two questions on each paragraph. Each question has only one correct answer among the four given options (A), (B), (C) and (D). Paragraph For Questions 51 and 5 Bo 1 contains three cards bearing numbers 1,, ; bo contains five cards bearing numbers 1,,, 4, 5 ; and bo contains seven cards bearing numbers 1,,, 4, 5, 6, 7. A card is drawn from each of the boes. Let i be the number on the card drawn from the i th bo, i = 1,,. 51. The probability that is odd, is (A) 9 15 (C) (B) 5 15 (D) 1 5. The probability that 1,, are in an arithmetic progression, is 9 (A) (B) (C) 11 7 (D) Paragraph For Questions 5 and 54 Let a, r, s, t be non-zero real numbers. Let P(at, at), Q, R(ar, ar) and S(as, as) be distinct points on the parabola y = 4a. Suppose that PQ is the focal chord and lines QR and PK are parallel, where K is the point (a, ). *5. The value of r is (A) (C) 1 t 1 t (B) (D) t 1 t t 1 t *54. If st = 1, then the tangent at P and the normal at S to the parabola meet at a point whose ordinate is (A) t 1 a t 1 (B) t t (C) 1 a t t (D) a t t FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -1116, Ph 4616, , Fa 65194

16 (16) 1h a Given that for each a (, 1), lim 1 h h function g(a) is differentiable on (, 1). Paragraph For Questions 55 and 56 a1 t t dt eists. Let this limit be g(a). In addition, it is given that the 55. The value of (A) (C) 1 g is (B) (D) The value of (A) (C) 1 g is (B) (D) SECTIN : Matching List Type (nly ne ption Correct) This section contains four questions, each having two matching list. Choices for the correct combination of elements from List-I and List-II are given as options (A), (B), (C) and (D), out of which NE is correct. 57. Match the following: (P) List I The number of polynomials f () with non-negative integer coefficients of degree, satisfying f() = and 1 f d 1, is (Q) The number of points in the interval 1, 1 at which f() = sin( ) + cos( ) attains its maimum value, is (R) (S) 1 e 1/ 1/ 1/ d equals 1 cos log d 1 equals 1 cos log d 1 Codes: P Q R S (A) 4 1 (B) 4 1 (C) 1 4 (D) 1 4 (1) 8 () () 4 (4) List II FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -1116, Ph 4616, , Fa 65194

17 (17) 58. Match the following: List I (P) Let y() = cos(cos 1 ), [1, 1],. Then 1 d y dy 1 equals y d d (Q) Let A 1, A,.., A n (n > ) be the vertices of a regular *(R) (S) polygon of n sides with its centre at the origin. Let a k be the position vector of the point A k, k = 1,,.. n. If n1 n1 a a, then the minimum k 1 ak a k1 = k1 k k 1 value of n is If the normal from the point P(h, 1) on the ellipse y 1 is perpendicular to the line + y = 8, then the 6 value of h is Number of positive solutions satisfying the equation tan tan tan is Codes: P Q R S (A) 4 1 (B) 4 1 (C) 4 1 (D) 4 1 (1) 1 () () 8 (4) 9 List II 59. Let f 1 : R R, f : [, ) R, f : R R and f 4 : R [, ) be defined by if ; ; sin if f f f and f if 1 e if 4 f f1 if f f1 1 if List I List II (P) f 4 is (1) onto but not one-one (Q) f is () neither continuous nor one-one (R) f o f 1 is () differentiable but not one-one (S) f is (4) continuous and one-one Codes: P Q R S (A) 1 4 (B) 1 4 (C) 1 4 (D) 1 4 FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -1116, Ph 4616, , Fa 65194

18 (18) k k *6. Let z k = cos i sin ; k = 1,,., List I (P) For each z k there eists a z j such z k. z j = 1 (1) True (Q) (R) (S) There eists a k {1,,.., 9} such that z 1. z = z k has no solution z in the set of comple numbers 1 z 1 z... 1 z equals 9 k 1 cos 1 equals k 1 () False () 1 (4) List II Codes: P Q R S (A) 1 4 (B) 1 4 (C) 1 4 (D) 1 4 FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -1116, Ph 4616, , Fa 65194

19 (19) ANSWERS PAPER- [Code 8] JEE(ADVANCED) 14 ANSWERS PART I : PHYSICS 1. C. D. B 4. B 5. B 6. A 7. D 8. C 9. A 1. C 11. C 1. B 1. D 14. D 15. C 16. A 17. C 18. A 19. B. D PART II : CHEMISTRY 1. C. B. B 4. D 5. B 6. D 7. C 8. A 9. A. C 1. C. D. A 4. C 5. B 6. D 7. B 8. C 9. A 4. C PART III : MATHEMATICS 41. A 4. B 4. C 44. D 45. D 46. A 47. B 48. B 49. C 5. D 51. B 5. C 5. D 54. B 55. A 56. D 57. D 58. A 59. D 6. C FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -1116, Ph 4616, , Fa 65194

20 () 1. For point outside dielectric sphere E Q 4 r r For point inside dielectric sphere E = Es R Eact Ratio E 1 : E : E = : 4 : 1 PART I : PHYSICS. If R be the meniscus radius R cos ( + /) = b Ecess pressure on concave side of meniscus = S R S S hg cos R b S h cos bg + (/) R b /. Cu Mo ZMo 1 ZCu 1 4. Inside planet r 4 gi gs Gr R Force to keep the wire at rest (F) = weight of wire R dr Gr G R 5 4R /5 Me Here, = density of earth = 4 R e Also, R e R 1 ; putting all values, F = 18 N d KE dt hc hc dv mv dt 1 1 u =.7 ev u 7. Initially bead is applying radially inward normal force. During motion at an instant, N =, after that N will act radially outward. FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -1116, Ph 4616, , Fa 65194

21 (1) 8. R 9 1 R 6 dr 1 d R ()(1 ) 1 dr.1 6 (4)(6) =.5 Alternatively dr.1.1 R 4 6 dr =.5 9. Rate of radiation energy lost by the sphere = Rate of radiation energy incident on it 4 4 4r T () 91 r T 111 K 1. r 5.77 tan C h 1 C sin C b Liquid C r C h 11. The net magnetic field at the given point will be zero if. B B wires loop I a Ia / a h a h (a h ) h 1. a The direction of magnetic field at the given point due to the loop is normally out of the plane. Therefore, the net magnetic field due the both wires should be into the plane. For this current in wire I should be along PQ and that in wire RS should be along SR. 1. I MBsin Ia sin I a d d 1. Heat given by lower compartment = R (7 T)...(i) 7 Heat obtained by upper compartment = R (T 4)...(ii) equating (i) and (ii) (7 T) = 7 (T 4) 1 T = 7 T 8 49 = 1 T T = 49 K at eq. TK 4 K 7 K 14. Heat given by lower compartment = 5 R (7 T)...(i) FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -1116, Ph 4616, , Fa 65194

22 () 7 Heat obtained by upper compartment = R (T 4)...(ii) By equating (i) and (ii) 5(7 T) 7(T 4) 5T = 7T 8 6 = 1 T T = 55K Work done by lower gas = nrt = 5 R Work done by upper gas = nrt = +5 R Net work done 1 R 15. By A 1 V 1 = A V () 5 (1) V V For given V a a V ava V m/s 17. In P, Q, R no horizontal velocity is imparted to falling water, so d remains same. In S, since its free fall, a eff = Liquid won t fall with respect to lift. 18. P: By Q 1 and Q 4, Q and Q F is in +y Q: By Q 1 and Q 4, Q and Q F is in +ve. R: By Q 1 and Q 4, F is in +ve y (+, b) By Q and Q, F is in ve y But later has more magnitude, since its closer to (, b). Therefore net force is in y S: By Q 1 and Q 4, F is in +ve and by Q and Q, F is in, but later is more in magnitude, since its closer to Q 1 Q Q Q 4 (a, ) (a, ) (+a, ) (+a, ) (, b). Therefore net force is in ve f R1 R Use (P) f = R f = R f f f eq eq R ; feq f R R R f = R (Q) ; f eq R feq R R R (R) ; f eq = - R f R R R (S) eq ; f eq = R f R R R eq FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -1116, Ph 4616, , Fa 65194

23 (). Condition for not sliding, f ma > (m 1 + m ) g sin N > (m 1 + m ) g sin. m g cos sin 6 tan 1/5 tan. tan for P, Q f = (m 1 + m ) g sin For R and S F = f ma = m g sin m 1 g cos m g cos m 1 m (m 1 + m ) g cos (m 1 + m ) g sin PART II : CHEMISTRY 1. Assuming that no s-p miing takes place A Be 1s, *1s, s, *s diamagnetic p z py B B 1s, *1s, s, *s, p, (diamagnetic ) 1 * p p * z 1 * p z y py C C 1s, *1s, s, * s, p,,, p (paramagnetic) * p p * z * p z y py D N 1s, *1s, s, *s, p,,, p (diamagnetic). At 1 C and 1 atmosphere pressure H H g Stotal and Ssystem Ssurrounding Ssystem and Ssurrounding is at equilibrium. For equilibrium. n r1 1 M n n r 8 M 4. N=N-Ph H Ph N N Cl alkaline solution H 5. III > II > I More the branching in an alkane, lesser will be the surface area, lesser will be the boiling point. 6. H MgX Cl CHMgBr, dry ether, C CH Cl CH CH aqueous acid CH CH FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -1116, Ph 4616, , Fa 65194

24 (4) 7. XeF6 H Xe HF H HXe 4 H / H (disproportionation) Xe Xe H 4 6 s g g 8. P4 s 8SCl 4PCl 4S g SCl g 9. KI4 H KI H NHH H HN 4H. When two phenyl groups are replaced by two para methoy group, carbocation formed will be more stable 1. rx d 4 ry 4 d 1 d = 48 d d = 48 d 16cm. As the collision frequency increases then molecular speed decreases than the epected. Solution for the Q. No. to 4. H M NaNH H C CHCHI H C CHI H, Lindlar's catalyst H X H N C NaNH 1eq NaNH H 1 equivalent Br H H H mild H, Pd/C Cr H X and Y are functional isomers of each other and Y gives iodoform test. Y FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -1116, Ph 4616, , Fa 65194

25 (5) Solution for the Q. No. 5 to 6. HClQ KCNR 4 ecess ecess M1 4 Tetrahedral square planar NiCl Ni Ni CN HClQ KCNR ZnCl4 Zn Zn CN ecess M ecess 4 Tetrahedral Tetrahedral KH S KH Zn H Zn H ecess 4 White ppt. Soluble 7. (P) [Cr(NH ) 4 Cl ]Cl Paramagnetic and ehibits cis-trans isomerism (Q) [Ti(H ) 5 Cl](N ) Paramagnetic and ehibits ionization isomerism (R) [Pt(en)(NH )Cl]N Diamagnetic and ehibits ionization isomerism (S) [Co(NH ) 4 (N ) ]N Diamagnetic and ehibits cis-trans isomerism 8. P. d d bonding Q. p d bonding R. p d antibonding S. d d antibonding 9. (P) 1; (Q) ; (R) 4; (S) (P) C6H5 CH C CH H 5 C 6 H C CH (Q) CH C H CH C Ph CH C CH CH CH H 5 C 6 H C 6 5 CH C 6 H 5 Ph CH CH C CH FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -1116, Ph 4616, , Fa 65194

26 (6) (R) C 6 H 5 CH CH C 6 H 5 C H C CH C CH Ph CH C Ph CH 6 5 C H 6 5 -C (S) C 6 H 5 C6H5 C CH CH 6 5 C C H 4. (P) N NH red hot iron87k CH CH fuming HN /HS4 HS.NH selective reduction NaN HS4 N N N + HS4 -. H Hydrolysis (Q) H N H N H H Conc.HS4 6 C N N dil.hs4 N H H H H S H S H (R) N NH NHCCH NHCCH NHCCH Sn/HCl CHCCl conc.hs 4 HN N S H S H dil. HS4 NH NHCCH N H N (S) N N N N N KMn 4 H/ H SCl NH CH C CH CCl CNH FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -1116, Ph 4616, , Fa 65194

27 (7) PART III : MATHEMATICS 41. Either a girl will start the sequence or will be at second position and will not acquire the last position as well. C 1 C1 Required probability = = 1 5 C. 4. = a + b y = ab c = y a b c 1 cos 1 ab C = R = abc 4, r = s r 4 R s abc r y R c c 1 4 ab sin c y c. 4. Number of required ways = 5! 4 4! 4 C! 4 C! 1 = Area = ( 1, 1) (, 4) 45. P () = a + b with a, b of same sign. P(P()) = a(a + b) + b If R or i R R P () R P(P()) Hence real or purely imaginary number can not satisfy P(P()) =. ( 1, 1) (, 4) cosec d FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -1116, Ph 4616, , Fa 65194

28 (8) Let e u + e u = cosec, = u = ln 4 1, = u = cosec + cot = e u and cosec cot = e u e cot (e u e u ) d = cosec cot d u u u u 17 e e e e du cosec cot u u = ln 1 16 e e du ln 1 u u = 16 e e du u e u dy y d 1 1 This is a linear differential equation 1 d ln 1 1 I.F. = e e 1 solution is y d y 1 d = or 4 f () = c = 5 f () 1 5 Now, / / 5 5 f d d (Using property) / / / 1 = / d = sin cos d (Taking = sin ) 1 cos / / sin = sin d 4 = c 48. F() = F() = f() = f() f() = e f() = F() = c e ( f() = 1) e d F() = e 1 ( F() = ) F() = e = 11 FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -1116, Ph 4616, , Fa 65194

29 (9) Possibilities are (, 1, ); (1,, ); (, 1, 1); (4, 1, ). Required coefficients = ( 4 C 7 C 1 1 C ) + ( 4 C 1 7 C 1 C ) + ( 4 C 7 C 1 1 C 1 ) + ( 4 C 4 7 C 1 1) = (1 7 66) + (4 5 1) + (6 7 1) + (1 7) = = sin + sin sin = sin + 4 sin cos sin + 4 sin = sin [ + 4 cos + 4(1 cos )] = sin [ (4 cos 4 cos + 1) + 1] = sin [ ( cos 1) ] = sin = 1 and cos 1 = and which is not possible at same time Hence, no solution 51. Case I : ne odd, even Total number of ways = = 9. Case II: All odd Number of ways = 4 = 4 Favourable ways = Required probability = Here = = even Hence number of favourable ways = C 1 4 C + 1 C 1 C 1 = Slope (QR) = Slope (PK) a ar at t at a a ar t 1 r t t t 1 r t t 1 r t 54. Tangent at P: ty = + at or y = at t a a Normal at S: y t t t a a Solving, y = at + t t y = 1h 1 h h a t 1 t 1/ g lim t 1 t dt 1/ 55. FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -1116, Ph 4616, , Fa 65194

30 () 1 1 dt dt = t t 1 1 t 4 = sin 1 1 sin 1 ( 1) =. = sin 1 1 t We have g (a) = g (1 a) and g is differentiable 1 Hence g = (P) f () = a + b, f d 1 a + b = 6 (a, b) (, ) and (, ). (Q) f () = cos 4 For maimum value, n 4 = n (R) (S) 9 =, as, d d 8 1 e 1 e. 1/ 1/ 1 cos ln d 1 as it is an odd function. 58. (P) y = cos ( cos 1 ) 1 sin cos y = y sin cos 1 y 1 y cos cos. 1 1 y + (1 ) y = 9y 1 1 y y 9 y. (Q) (a k a k+1 ) = a k a k+1 = r sin n r cos n n 1 n 1 a k a k 1 a k a k 1 k 1 k 1 sin r n 1 cos n n r (n 1) FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -1116, Ph 4616, , Fa 65194

31 (1) (R) (S) tan 8 1 n = n 4k 1 n = 8. h 1 1, h = 6 y Tangent at (, 1) is 1 + y = tan tan tan tan tan = =,., 59. f (f 1 ) = e, f 4 : R [, ) f f1, f 4 () = f f1 1,, = e 1, y =f 1 () y = f () y = f () y =f (f 1 ()) y = f 4 () 6. (P) z k is 1 th root of unity z k will also be 1 th root of unity. Take z j as z k. zk (Q) z 1 take z =, we can always find z. z 1 (R) z 1 1 = (z 1) (z z 1 ) (z z 9 ) (z z 1 ) (z z ) (z z 9 ) = 1 + z + z + + z 9 z comple number. FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -1116, Ph 4616, , Fa 65194

32 () Put z = 1 (1 z 1 ) (1 z ) (1 z 9 ) = 1. (S) 1 + z 1 + z + + z 9 = Re (1) + Re (z 1 ) + + Re (z 9 ) = Re (z 1 ) + Re (z ) + + Re (z 9 ) = 1. 9 k 1 cos. 1 k1 FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -1116, Ph 4616, , Fa 65194

33 () D. Marking Scheme 17. For each question in Section 1, and you will be awarded marks if you darken only the bubble corresponding to the correct answer and zero mark if no bubble is darkened. In all other cases, minus one (1) mark will be awarded. Appropriate way of darkening the bubble for your answer to be evaluated: a The one and the only one acceptable a a a a Part darkening Darkening the rim Cancelling after darkening Answer will not be evaluatedno marks, no negative marks a Erasing after darkening Figure-1 : Correct way of bubbling for a valid answer and a few eamples of invalid answers. Any other form of partial marking such as ticking or crossing the bubble will be invalid Figure- : Correct Way of Bubbling your Roll Number on the RS. (Eample Roll Number : 5451) Name of the Candidate Roll Number I have read all instructions and shall abide by them. I have verified all the information filled by the candidate. Signature of the Candidate Signature of the invigilator FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -1116, Ph 4616, , Fa 65194