Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose. 5

Transcription

1 Time : Hrs. : PAPER - Ma. Marks : 8 : 8 Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose. 5 INSTRUCTIONS ( ) A. General. This booklet is your Question Paper. Do not break the seal of the booklet before being instructed to do so by the invigilators.. The question paper CODE is printed on the left hand top corner of this sheet and on the back cover page of this booklet. (CODE). Blank spaces 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 NOT 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 Optical Response Sheet, which is privided separately. The ORS is a doublet of two sheets - upper and lower, having identical layout. The upper sheet is a machine-gradable Objective Response Sheet (ORS) 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- on the back cover page for the correct way of darkening the bubbles for valid answer.) ORS 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- on the back cover page for appropriate way of darkening the bubbles for valid answers. (BUBBLES) Page

2 JEE-ADVANCED-4 DATE: PAPER- CODE- 8. DO NOT TAMPER WITH/MULTIPLE THE ORS OR THIS BOOKLET. (ORS) 9. On breaking the seal of the booklet check that it contains all the 6 questions and corresponding answer choices are legible. Rad carefully the instruction printed at the beginning of each section. 6 B. Filling the right part of the ORS (ORS). The ORS also has a CODE printed on its left and right parts.. Verify that the CODE printed on the ORS (on both the left and right parts) is the same as that on this booklet and put your signature in the Bo designated as R4. R4. IF THE CODES DO NOT MATCH, ASK FOR A CHANGE OF THE BOOKLET/ORS AS APPLICABLE.. Write your Name, Roll No. and the name of centre and sign with pen in the boes provided on the upper sheet of ORS. Do not write any of the 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) (BUBBLE) C. Question Paper Format The question paper consists of three parts (Physics, Chemistry and Mathematics). Each part consists of three sections. 4. Section contains multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE is correct. (A), (B), (C) (D) 5. 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 partcular passage should have only one correct answer among the four given choices (A), (B), (C) and (D). (A), (B), (C) (D) 6. Section contains 4 multiple choice questions. Each question has two lists (list- : P, Q, R and S; List- :,, and 4). The options for the correct match are provided as (A), (B), (C) and (D) out of which ONLY ONE is correct. 4 ( - : P, Q, R S; - :,, 4) (A), (B), (C) (D) D. Marking Scheme 7. For each question in Section, and you will be awarded marks if you darken the bubble corresponding to the correct answer and zero mark if no bubble is darkened. In all other cases, minus one ( ) mark will be awarded., () ( ) Page

3 JEE-ADVANCED-4 DATE: PAPER- CODE- PHYSICS PART - I : PHYSICS SECTION- : (One one options correct Type) This section contains multiple choice questions. Each questions has four choices (A), (B), (C) and (D) out of which Only ONE option is correct : ( ) (A),(B),(C) (D). A glass capillary tube is of the shape of a 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) (truncated cone) h b (surface tension) S, h (g ) S (A) cos( b g ) S (C) cos( b g / ) S (B) cos( b g ) S (D) cos( b g / ) (D) Sol. Using geometry : b Re cos Using Pressure method : P S R c + h g P h S R g e S cos / b g Page

4 JEE-ADVANCED-4 DATE: PAPER- CODE- PHYSICS. A planet of radius R (radius of Earth) has the same mass density as Earth. Scientists dig a well of depth 5 R on it and lower a wire of the same length and of linear mass density kgm 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 6 6 m and the acceleration due to gravity on Earth is ms ) R ( ) 5 R kgm ( 6 6 m ms ) (A) 96 N (B) 8N (C) N (D) 5N (B) Sol. Given, R planet R earth and density, M 4 earth R earth M 4 R Planet planet M M planet earth GM g surface of planet R planet planet GM e..r e GMe R e g surface of earth g depth of planet g surface of planet R where distance from centre of planet T R 4R / 5 d g R g R R 4R / 5 8 N HINDI, R planet R earth, M 4 R earth earth M 4 R Planet planet M M planet earth GM g R planet planet GM e..r e GM e gsurface R e of earth g g R T R 4R / 5 d g R g R R 4R / 5 8 N Page 4

5 JEE-ADVANCED-4 DATE: PAPER- CODE- PHYSICS. Charges Q, Q and 4Q are uniformly distributed in three dielectric solid spheres, 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, and are E E and E respectively, then (dielectric) R/, R R Q, Q 4Q P R, E, E E (A) E > E > E (B) E > E > E (C) E > E > E (D) E > E > E KQ Sol. E R (C) k(q) E R k(4q)r E (R) kq E R kq E R E < E < E 4. 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 ( 9) K X- Cu ( 4) K X- Mo Cu / Mo (A).99 (B).4 (C).5 (D).48 (B) Sol. Using Mosley's law, for K line : a (z b) where b K : a (z b) b cu a(9 a(4 ) ) mo cu mo Page 5

6 JEE-ADVANCED-4 DATE: PAPER- CODE- PHYSICS 5. A wire, which passes through the hole is 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 B (A) always radially outwards (B) always radially inwards (C) radially outwards initially and radially inwards later (D) radially inwards initially and radially outwards later. (A) (radially outwards) (B) (radially inwards) (C) (D) (D) Sol. Using conservation of energy : mgr (- cos ) mv Radial force Equ n : mgcos N mv R N mgcos mv mg ( cos ) R Normal act radially outward on bead if cos > Normal radially inward on bead if cos < Normal on ring is opposite to reaction on bead. Page 6

7 JEE-ADVANCED-4 DATE: PAPER- CODE- PHYSICS HINDI: : mgr (- cos ) mv : mgcos N mv R N mgcos mv mg ( cos ) R cos > cos < 6. A metal surface is illuminated by light of two different wavelengths 48 nm and nm. The maimum speeds of the photoelectrons corresponding to these wavelengths are u and u, respectively. If the ratio u : u : and hc 4 ev nm, the work function of the metal is nearly 48 nm nm (corresponding) (photoelectrons) u u u : u : hc 4 ev nm (A).7 ev (B). ev (C).8 ev (D).5 ev (A) Sol. 48 nm 4 / 48 ev 5 ev nm 4 / ev 4 ev K.E K.E 4 5ev 4ev W W 6 4W S W W W.67 ev.7 ev Page 7

8 JEE-ADVANCED-4 DATE: PAPER- CODE- PHYSICS 7. 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. (collision) t K (A) K (B) t (C) (D) (B) Sol. K mg t K t : parabolic graph then during collision kinetic energy first decreases to elastic potential energy and then increases. Most appropriate graph is B. Hindi K mg t K t : B Page 8

9 JEE-ADVANCED-4 DATE: PAPER- CODE- PHYSICS 8. During an eperiment with a metre bridge, the galvanometer shows a null point when the joceky 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 meter bridge is mm. The unknown resistance is : 9 4. cm (least count) m.m. (A) 6. 5 (B) (C) 6. 5 (D) 5. Sol. (C) For balanced meter bridge X R X X 6 X R X X X.5. 4 so X (6 +.5) Parallel rays of light of intensity 9 Wm are incident on a spherical balck body kept in surroundings of temperature K. Take Stefan-Biltzmann constant Wm K 4 and assume that the energy echange with the surroundings is only through radiation. Th final steady state temperature of the black body is close to: (black body) K 9 Wm Wm K 4 (A) K (B) 66 K (C) 99 K (D) 55 K (A) Page 9

10 JEE-ADVANCED-4 DATE: PAPER- CODE- PHYSICS Sol. In steady state R (T 4 4 T ) 4 R (T 4 4 T ) 4 T 4 T T T 4 8 T K. A point source S is placed at the bottom of a transparent block of height 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 from the block to the liquid forms a circular bright spot of diameter.54 mm on the top of the block. The refractive inde of the liquid is (S) mm.7.54 mm (spot) (A). (B). (C).6 (D).4 (C) Sol. Sin i C r h r n r n B r h n r r h Page

11 JEE-ADVANCED-4 DATE: PAPER- CODE- PHYSICS SECTION- : Comprehension Type (Only One options 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) (A), (B), (C) (D) Paragraph For Questions to The figure shows a circular loop of radius a with two long parallel wires (numbered 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. The current in the loop is in the counterclockwise direction if seen from above. a (loop) d. When d a but wires are not touching the loop, it is found that the net magnetic filed on the ais of the loop is zero at a height h above the loop. In that case (A) current in wire and wire is the direction PQ and RS, respectively and h a (B) current in wire and wire is the direction PQ and SR, respectively and h a (C) current in wire and wire is the direction PQ and SR, respectively and h. a (D) current in wire and wire is the direction PQ and RS, resepectively and h. a d a h (A) PQ RS h (B) PQ SR h (C) PQ SR h (D) PQ RS h (C) a a. a. a Page

12 JEE-ADVANCED-4 DATE: PAPER- CODE- PHYSICS Sol. B R B due to ring B B due to wire - B B due to wire - In magnitudes B B I r a Resultant of B and B B cos r I a B R 4 r For zero magnetic field at P h a oi a r 4 r.a Hindi B R B B B - B B - B B I r a B B B cos r I a B R 4 r P h a oi a r 4 r.a. Consider d >> a, and the loop is rotated about its diameter parallel to the wires by º 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) d >> a º (torque) ( ) (A) a d (B) a d (C) d a (D) a d (B) Page

13 JEE-ADVANCED-4 DATE: PAPER- CODE- PHYSICS Sol. Magnetic field at mid point of two wires I d Magnetic moment of loop I a Torque on loop M B sin 5 a I d I d I a M B sin 5 a I d Paragraph For Questions to 4 4 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 C V R, CP 5 R, and those for an ideal diatomic gas are C V 5 R, CP 7 R. (monatomic) 7 K (diatomic) 4 K C V R, CP 5 R CV 5 R, CP 7 R Page

14 JEE-ADVANCED-4 DATE: PAPER- CODE- PHYSICS. 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) 5 K (D) 49 K (D) Sol. Let final temperature of gases is T Heat rejected by gas in lower compartment (nc v T) R(7 T) 7 Heat received by gas in above compartment (nc P T) R(T 4) Equating above T 7T 8 T 49 K T (nc v T) R(7 T) (nc P T) R(T 4) T 7T 8 T 49 K 7 4. 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) R (D) R (D) Sol. W + U Q W + U Q Q + Q 7 5 R (T 4) R (7 T) T 6 55 K So W + W. R. (55 4) + R(55 7) + 5R 5 R R Page 4

15 JEE-ADVANCED-4 DATE: PAPER- CODE- PHYSICS Paragraph for Question 5 to 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 mm respectively. The upper end of the container is open to the atmosphere. (nozzle) mm mm (atmosphere) 5. If the piston is pushed at a speed of 5mms, the air comes out of the nozzle with a speed of 5mms (A).ms (B) ms (C) ms (D) 8ms (C) Sol. A V A V A 4 A 4 (5 ) V V m/ s (C) 6. 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) a (B) a (C) a (D) Sol. (A) Pressure at A and B will be same A B P ava P v gh v a v a gh Page 5

16 JEE-ADVANCED-4 DATE: PAPER- CODE- PHYSICS SECTION- : Matching List Tpype (Only One options 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 options (A), (B), (C) and (D) out of which one is correct - 4 (A), (B), (C) (D) 7. 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. 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.. d. m Q. Lift is accelerating vertically down. d >.m with an accelerating less than the gravitational acceleration. R. List is moving vertically up with constant. d <. m 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-, S-4 (C) P-, Q-, R-, S-4 (D) P-, Q-, R-, S-. m, d -I -II -I -II -I -II P.. d. m Q.. d >.m R.. d <. m S. 4. Page 6

17 JEE-ADVANCED-4 DATE: PAPER- CODE- PHYSICS : (A) P-, Q-, R-, S-4 (B) P-, Q-, R-, S-4 (C) P-, Q-, R-, S-4 (D) P-, Q-, R-, S- Sol. (C) Match the column When lift is at rest: (P) g eff > g d 4 h h. m (Q) g eff < g d 4 h h. m (R) g eff g d 4 h h. m (S) g eff o No water leaks out of the jar. (C) P Q R S 4 8. Four charge Q,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- with List-II and select the correct answer using the code given below the lists. List-I List-II P. Q,Q,Q, Q 4, all positive. + Q. Q,Q positive Q,Q 4 negative. R. Q,Q 4 positive Q, Q negative. +y S. Q,Q positive Q, Q 4 negative 4. y Code : (A) P-, Q-, R-4,S- (B) P-4, Q-, R-, S- (C) P-, Q-, R-,S-4 (D) P-4, Q-, R-, S- Page 7

18 JEE-ADVANCED-4 DATE: PAPER- CODE- PHYSICS Q,Q,Q, Q 4, a, a, +a +a q, + y b > (sign) -I q -II -I -II -I -II P. Q,Q,Q, Q 4,. + Q. Q,Q Q,Q 4. R. Q,Q 4 Q, Q. +y S. Q,Q Q, Q 4 4. y : (A) P-, Q-, R-4,S- (B) P-4, Q-, R-, S- (C) P-, Q-, R-,S-4 (D) P-4, Q-, R-, S- (A) Sol. (P) Component of forces along -ais will vanish. Net force along +ve y-ais - y- (Q) Component of forces along y-ais will vanish. Net force along +ve -ais y- - Page 8

19 JEE-ADVANCED-4 DATE: PAPER- CODE- PHYSICS (R) Component of forces along -ais will vanish. Net force along -ve y-ais. - y- (S) F +q F F 4 F +Q - Q +Q - Q Component of forces along y-ais will vanish. Net force along -ve -ais. y- - (A) P, Q, R 4, S 9. Four combinations of two thin lenses are given in List-I. The radius of curvature of all curved surface is r and the refractive inde of all lenses is.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. -I r (r.i).5 -I -II -I -II List-I -I List-II -II P.. r Q.. r/ R.. r S. 4. r Page 9

20 Code : JEE-ADVANCED-4 DATE: PAPER- CODE- PHYSICS : (A) P-, Q-, R-,S-4 (B) P-, Q-4, R-, S- (C) P-4-,Q-, R-,S- (D) P-, Q-, R-, S-4 (B) Sol. (P) f r r r f r f + f r f eq f eq r (Q) f r f r + f f f r f eq r (R) f r r f r + f f r f eq f eq r (S) feq r r r f eq r (B) P Q 4 R S Page

21 JEE-ADVANCED-4 DATE: PAPER- CODE- PHYSICS. A block of mass m 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 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 epression for the friction on block m 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 ). ; tan (.5 ). ; tan(6.5º.)] m kg m kg ( ) I m m. II m II g [ : tan(5.5 ). ; tan (.5 ). ; tan(6.5º.)] List-I List-II -I -II P. 5. m g sin Q.. (m + m )g sin R. 5. m g cos S. 4. (m + m )g cos Code : (A) P-, Q-, R-,S- (B) P-, Q-, R-,S- (C) P-, Q-, R-,S-4 (D) P-, Q-, R-,S- (D) Page

22 JEE-ADVANCED-4 DATE: PAPER- CODE- PHYSICS Sol. Block will not slip if (m + m ) g sin m g cos sin () cos tan 5.5 (P) 5 friction is static f (m + m )g sin (Q) friction is static f (m + m )g sin (R) 5 friction is kinetic f m g cos (S) friction is kinetic f m g cos (D) P Q R S (m + m ) g sin m g cos sin () cos tan 5.5 (P) 5 f (m + m )g sin (Q) f (m + m )g sin (R) 5 f m g cos (S) f m g cos (D) P Q R S Page

23 JEE-ADVANCED-4 DATE: PAPER- CODE- CHEMISTRY PART II : CHEMISTRY Atomic masses : [H, D, Li 7, C, N 4, O 6, F 9, Na, Mg 4, Al 7, Si 8, P, S, Cl 5.5, K 9, Ca 4, Cr 5, Mn 55, Fe 56, Cu 6.5, Zn 65, As 75, Br 8, Ag 8, I 7, Ba 7, Hg, Pb 7] SECTION : (Only One option correct Type) : ( ) This section contains multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct. (A), (B), (C) (D). Assuming s-p miing is NOT operative, the paramagnetic species among the following is : s-p (operative) (paramagnetic) (species) (A) Be (B) B (C) C (D) N (C) Sol. If s-p miing is not operative, the increasing order of Molecular orbitals will be : s, *s, s, *s, P py pz Considering this Be & B become diamagnetic, so does N. Only C would be paramagnatic with electronic configuration as above s-p s, *s, s, *s, P py pz Be B N C. For the process H O( ) H O(g) at T C and atmosphere pressure, the correct choice is : (A) S system > and S surroundings > (B) S system > and S surroundings < (C) S system < and S surroundings > (D) S system < and S surroundings < T C H O( ) H O(g) (A) S > S > (B) S > S < (C) S < S > (D) S < S < (B) Sol. For H O( ) H O(g) at T ºC, atm equilibrium eists. G, H T S H T S > for system, since evaporation is endothermic ( S) system >, also ( S) surrounding q T surr surr Heat gained by system heat lost by surroundings q surr. < ( S) surr. < Page

24 T º, atm JEE-ADVANCED-4 DATE: PAPER- CODE- H O( ) H O(g) G, H T S H T S >, ( S) >, ( S) q T q < ( S) < CHEMISTRY. 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 : M N M M (rate of disappearance) 8 M (order of the reaction) (A) 4 (B) (C) (D) (B) Sol. M N r K [M] as [M] is doubled, rate increases by a factor of 8. i.e. 8 r K [M] 8 () Sol. M N r K [M] [M] 8 8 r K [M] 8 () 4. For the identification of -naphthol using dye test, it is necesary to use : (A) dichloromethane solution of -naphthol. (B) acidic solution of -naphthol. (C) neutral solution of -naphthol. (D) alkaline solution of -naphthol. - (A) - (B) - (C) - (D) - (D) Sol. In dye test, phenolic OH is converted to, which activates the ring towards EAS. This is possible only in alkaline solution. Hence (D). OH EAS (D) 5. Isomers of heane, based on their branching, can be divided into three distinct classes as shown in the figure. [Figure] The correct order of their boiling point is (A) I > II > III (B) III > II > I (C) II > III > I (D) III > I > II Page 4

25 JEE-ADVANCED-4 DATE: PAPER- CODE- CHEMISTRY (Isomers) Sol. (boiling point) (A) I > II > III (B) III > II > I (C) II > III > I (D) III > I > II (B) Greater the etent of branching, lesser is the boiling point of hydrocarbon. Hence III > II > I III > II > I 6. The acidic hydrolysis of ether (X) shown below is fastest when : [Figure] (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. (X) (hydrolysis) (A) (B) (C) (D) X (C) Page 5

26 JEE-ADVANCED-4 DATE: PAPER- CODE- CHEMISTRY Sol. Ph C CO H H O Ph C OH + ROH The reaction proceeds by S N Mechanism : Ph C OR + H Ph COHR Ph C + ROH H H O Ph C OH Ph C OH Greater the electron releasing effect of the attached groups greater is the stability of intermediate carbocation, & faster is the rate of reaction. If two ph groups are replaced by MeO groups, strong +M effect of MeO groups stablize, the carbocation better there by making the reaction faster.. Ph C CR H H O Ph C OH + ROH S N Ph C OR + H Ph COHR Ph C H O Ph C OH H Ph C OH ph MeO MeO +M 7. Hydrogen peroide in its reaction with KIO 4 and NH OH 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 KIO 4 NH OH (A) (reducing agent) (oidising agent) (B) (C) (D) (A) Sol. KlO 4 + H O KlO + H O + O H O acts as a reductant NH OH + H O N + 4H O H O acts as an oidant. Sol. KlO 4 + H O KlO + H O + O H O NH OH + H O N + 4H O H O Page 6

27 JEE-ADVANCED-4 8. The major product in the following reaction is : [Figure] DATE: PAPER- CODE- CHEMISTRY (A) (B) (C) (D) (A) (B) (C) (D) (D) Sol. Cl O CH CH MgBr/ether ºC Cl CH O CH O +Cl Page 7

28 JEE-ADVANCED-4 DATE: PAPER- CODE- CHEMISTRY 9. Under ambient conditions, the total number of gases released as products in the final step of the reaction scheme shown below is : (A) (B) (C) (D) (ambient conditions) Sol. XeF 6 (A) (B) (C) (D) (C) Complete Hydrolysis Page 8

29 JEE-ADVANCED-4 DATE: PAPER- CODE- CHEMISTRY XeF 6. The product formed in the reaction of SOCl with white phosphorous is : SOCl (A) PCl (B) SO Cl (C) SCl (D) POCl (A) Sol. P 4 + 8SOCl 4PCl +4SO S Cl (NCERT Reaction) (NCERT ) SECTION : Comprehension Type (Only One Option Correct) : This section contains paragraphs each describing theory, eperiment, data etc. Si questions relate to 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). (A), (B), (C) (D) Paragraph for questions and X and Y are two volatile liquids with molar weights of g mol and 4 g mol 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 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. L 4 cm Cotton wool soaked in X d Initial formation of the product Cotton wool soaked in Y X Y, g 4 g X Y 4 cm (atmosphere pressure) K X Y X d cm X Y (molecular diameters) (ideal behaviour) Page 9

30 JEE-ADVANCED-4 DATE: PAPER- CODE- CHEMISTRY L 4 cm X d Y. The value of d in cm (shown in the figure), as estimated from Graham s law, is : d cm (A) 8 (B) (C) 6 (D) (C) Sol. According to Grham s law, if all conditions are identical, r M As in this question, all conditions are identical for X and Y, it will be followed Hence d 4 d r r y 4 M M y d 4 d d 48 d d 48 d 6 cm. r M X Y r r y M M y d 4 d 4 d 4 d d 48 d d 48 d 6 cm. Page

31 JEE-ADVANCED-4 DATE: PAPER- CODE- CHEMISTRY. 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. (etimate) d (A) Y X (mean free path) (B) X Y (mean free path) (C) X Y (collision frequency) (D) Y X (collision frequency) (D) Sol. The general formula of mean free path ( ) is RT d N P (d diameter of molecule A p pressure inside the vessel). d & p are same for both gases, ideally their are same. Hence it must be the higher drift speed of X due to which it is seeing more collisions per second, with the inert gas in comparison to gas Y. So X see comparably more resistance from noble gas than Y and hence covers lesser distance than that predicted by Graham's Law. ( ) RT (d d N P A p ). d p X Y X Y Paragraph for questions and 4 Schemes and describe sequential transformation of alkynes M and N. Consider only the major products formed in each step for both the schemes. HO M H. NaNH (ecess). CH CH I ( equivalent). CH I ( equivalent) 4. H, Lindlar's catalyst X Scheme- Y Scheme- Page

32 JEE-ADVANCED-4 DATE: PAPER- CODE- CHEMISTRY 4 M N (sequential transformation) HO M H. NaNH. CH CH. CH I I 4. H, (Lindlar's catalyst) X - Y -. The product X is : X H CO (A) (B) H H CH CH O H (C) (A) H H (D) CH CH O H Sol. Page

33 JEE-ADVANCED-4 DATE: PAPER- CODE- CHEMISTRY 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. Y (A) X (functional isomer) (B) X (geometrical isomer) (C) X (D) X (C) Sol. (Y) can answer iodoform test (but not Tollen's test and it is a functional isomer of (X) (Y) (X) Paragraph For question 5 to 6 An aqueous solution of metal ion M 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: Tetrahedral Tetrahedral Q ecess M Q ecess M R ecess Square planar R ecess Tetrahedral S, stoichiometric amount White precipitate S ecess precipitate dissolves Page

34 JEE-ADVANCED-4 DATE: PAPER- CODE- CHEMISTRY 5 6 M Q R M M S S (Tetrahedral) Q M R (Square planar) (Tetrahedral) Q M R (Tetrahedral) S, (stoichiometric amount) (White precipitate) S (precipitate dissolves) 5. M, Q and R, respectively are : (A) Zn +, KCN and HCl (C) Cd +, KCN and HCl M, Q R, (A) Zn +, KCN HCl (C) Cd +, KCN HCl (B) (B) Ni +, HCl and KCN (D) Co +, HCl and KCN (B) Ni +, HCl KCN (D) Co +, HCl KCN 6. Reagent S is : S (A) K 4 [Fe(CN) 6 ] (B) Na HPO 4 (C) K CrO 4 (D) KOH (D) (5 & 6) Page 4

35 JEE-ADVANCED-4 DATE: PAPER- CODE- CHEMISTRY Reaction with (S) indicates amphoteric nature of M. Amongst the options mentioned for (S) in Q.6, only KOH can give a compleing agent OH ; which is its answer. [Note : M may be Zn +, which (a) is amphoteric (b) has cordination number 4 and (c) always from tetrahedral complees. It may be noted that Be + will also qualify with these characterstics] Let us consider the possiblities of M. M should be able to form square planner comple (dsp hybridisation) as well as tetrahedral (sp ). This rules out Zn + & Cd + [ d configuration will not allow dsp hybridization] Ni + is most suitable Ni + (aq.) + 4CN (ecess) Ni ( CN) 4 (square planner) Ni + (aq.) + Cl (ecess) NiCl 4 (tetrahedral) [Note : Co + + 6CN 4 (ecess) Co ( CN) 6, an octahedral compele] M ( ) ( ) Q ( ) R ( ) Q ( ) M ( ) S R ( ) (S) (S) M 6 (S) KOH OH [ : M, Zn + (a) (b) 4 (c) Be + M M (dsp ) (sp ) Zn + Cd + [ d dsp ] Ni + Ni + ( ) + 4CN ( ) Ni ( CN) 4 ( ) Ni + ( ) + Cl ( ) NiCl 4 ( ) [ : Co + + 6CN 4 ( ) Co ( CN) 6, ] Page 5

36 JEE-ADVANCED-4 DATE: PAPER- CODE- CHEMISTRY SECTION : Matching List Type (One One Option 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. 4 (A), (B), (C) (D), 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; Cp 7; Pt 78} List-I List-II P. [Cr(NH ) 4 Cl )Cl]. Paramagnetic and ehibits ionisation isomerism Q. [Ti(H O) 5 Cl](NO ). Diamagentic and ehibits cis-trans isomerism R. [Pt(en)(NH )Cl]NO. Paramagentic and ehibits cis-trans isomerism S. [Co(NH ) 4 (NO ) ]NO 4. Diamagentic and ehibits ionisation isomerism Code : P Q R S (A) 4 (B) 4 (C) 4 (D) 4 -I (coordination compound) -II {en H NCH CH NH ; : Ti ; Cr 4; Cp 7; Pt 78} -I -II P. [Cr(NH ) 4 Cl )Cl]. (Paramagnetic) (ionisation isomerism) Q. [Ti(H O) 5 Cl](NO ). (Diamagentic) (cis-trans) R. [Pt(en)(NH )Cl]NO. S. [Co(NH ) 4 (NO ) ]NO 4. : P Q R S (A) 4 (B) 4 (C) 4 (D) 4 (B) Sol. (P) [Cr(NH ) 4 Cl ]Cl Cr + is d. It is paramagnetic and it shows cis-trans isomerism. (Q) [Ti(H O) 5 Cl](NO ) Ti + is d. It is paramagnetic and it show ionisation isomerism. (R) [Pt(en)(NH )Cl]NO Pt + is d 8. But this comple is square planar and all electron are paired. So it is diamagnatic. It ehibit ionisation isomerism. (S) [Co(NH ) 4 (NO ) ]NO Co + is d 6. Since ligands are strong, so electron are paired. it is diamagnetic. It ehibit cis-trans isomerism. is (B). Page 6

37 JEE-ADVANCED-4 DATE: PAPER- CODE- CHEMISTRY Sol. (P) [Cr(NH ) 4 Cl ]Cl Cr + d (Q) [Ti(H O) 5 Cl](NO ) Ti + d (R) [Pt(en)(NH )Cl]NO Pt + d 8 (S) [Co(NH ) 4 (NO ) ]NO Co + d 6 (B) 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.. p d antibonding Q.. d d bonding R.. p d bonding S. 4. d d antibonding Code : P Q R S (A) 4 (B) 4 (C) 4 (D) 4 -I (orbital overlap) -II -I -II P.. p d (antibonding) Q.. d d (bonding) R.. p d (bonding) S. 4. d d (antibonding) : P Q R S (A) 4 (B) 4 (C) 4 (D) 4 Page 7

38 JEE-ADVANCED-4 DATE: PAPER- CODE- CHEMISTRY (C) Sol. It is d d aial overlap in same phase, so d d bonding. It is p & d lateral overlap in same phase, so it is p d bonding. It is p and d lateral overlap in opposite phase, so it is p d antibonding. It is d d aial overlap in opposite phase, so it is d d antibonding.. d d d d p d p d p d p d d d d d 9. Different possible thermal decomposition pathways for peroyesters are shwon 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. List-I List-II P. Pathway P. Q. Pathway Q. R. Pathway R. Page 8

39 JEE-ADVANCED-4 DATE: PAPER- CODE- CHEMISTRY S. Pathway S 4. Code : P Q R S (A) 4 (B) 4 (C) 4 (D) 4 (thermal decomposition) -I -II -I -II P. P. Q. Q. R. R. S. S 4. : P Q R S (A) 4 (B) 4 (C) 4 (D) 4 (A) Page 9

40 JEE-ADVANCED-4 DATE: PAPER- CODE- CHEMISTRY Sol. This is an ecellent question, probably the best in the paper. Pyrolysis of peroyesters is not conventionally taught anywhere for IIT-JEE preparation, not even in any standard organic chemistry course. But the question is throughly logical. It can be answered after a bit of reflection by anyone with good basics of reaction mechanism. Four pathways for reactions are given : Note that first homolytic fission is epected to give + R'O. This is usually unstable (remember kolbe's electrolysis) and decomposes to R + CO. But break. Hence in () & (), which have Ph CH C O O R O is stable, since it is a difficult bond to groups, Ph CH. will be formed ( benzyl radical is stable) so P or Q are possible pothways for them. For Ph C O O R ( & 4), R & S pathways will be more likely. : P or Q : R or S : P or Q 4 : R or S Now consider R'O. If it is CH O, It cannot splite. CH CH But Ph CH C CH Ph CH + C can occur. Hence carbonyl compound will be formed CH O O with & 4, but not with &. : P or S : Q or R : P or S 4 : Q or R combining these options gives : P, : S, : Q, 4 : R I cannot but help commenting that such questions are those which make IIT-JEE what it is. A good motivation for students to keep thinking while studying. O Sol. IIT-JEE Page 4

41 JEE-ADVANCED-4 DATE: PAPER- CODE- CHEMISTRY + R'O ( ) R + CO () () Ph CH C O O R O O Ph CH ( ) P Q Ph C O O R ( 4), R S : P Q : R S : P Q 4 : R S R'O CH O CH CH Ph CH C CH Ph CH + C 4 CH O O : P S : Q R : P S 4 : Q R : P, : S, : Q, 4 : R IIT-JEE 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.. Scheme I Page 4

42 JEE-ADVANCED-4 DATE: PAPER- CODE- CHEMISTRY Q.. Scheme II R.. Scheme III S. 4. Scheme IV Code : P Q R S (A) 4 (B) 4 (C) 4 (D) 4 -I (P, Q, R, S) -II (Scheme) (I, II, III, IV) -I -II P.. I Page 4

43 JEE-ADVANCED-4 DATE: PAPER- CODE- CHEMISTRY Q.. II R.. III S. 4. IV : P Q R S (A) 4 (B) 4 (C) 4 (D) 4 (C) Sol. P : Page 4

44 P JEE-ADVANCED-4 DATE: PAPER- CODE- CHEMISTRY Q : Q 4 R : R S : S Hence the answer is (C) ( (C) ) Note : Verifying any two can easily give you the answer complete details are given for reference & understanding. v ki d sfunsz' k (reference) Page 44

45 JEE-ADVANCED-4 DATE: PAPER- CODE- MATHS PART- III - MATHEMATICS SECTION : (Only One Option Correct Type) : ( ) This section contains multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE option is correct. (A),(B),(C) (D) 4. 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 (quadratic equation) p() p(p()) (A) (B) (C) (D) (D) Sol. p() will be of the form a + c. Since it has purely imaginary roots only. Since p() is zero at imaginary values while a + c takes real value only at real '', no root is real. Also p(p()) p() is purely imaginary a + c purely imaginary Hence can not be purely imaginary since will be negative in that case and a + c will be real. Thus.(D) is correct. Hindi p(), a + c p() a + c '' p(p()) p() a + c a + c (D) 4. 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) (B) (C) (D) 4 (probability), (A) (B) (C) (D) 4 (A) Page 45

46 JEE-ADVANCED-4 DATE: PAPER- CODE- MATHS Sol. Hindi Boys & Girls... () B () B () B (4) Girl can't occupy 4 th position. Either girls can occupy of,, position or they can both be a position () or (). Hence total number of ways in which girls can be seated is C!! + C!! Number of ways in which B & A can be seated 5! Hence required prob. 6 5!.... () B () B () B (4) 4 th,, () (). C!! + C!! B A 5! 6 5!. 4. Si cards and si envelopes are numbered,,, 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 is always placed in envelope numbered. Then the number of ways it can be done is (A) 64 (B) 65 (C) 5 (D) 67,,, 4, 5, 6 (A) 64 (B) 65 (C) 5 (D) 67 (C) Sol. Cards Envelopes If '' goes in '' then it is dearrangement of 4 things which can be done in 4!!! 4! 9 ways. If '' doen't go in, it is dearrangement of 5 things which can be done in 44 ways. Hence total 5 ways. Page 46

47 JEE-ADVANCED-4 DATE: PAPER- CODE- MATHS Hindi '', '' 4 4!!! 4! 9 '',, 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 (A) y ( c) (B) y c( c) (C) y 4( c) (D) y 4c( y c y, c (in-radius) (circum-radius) (ratio) c) (A) y ( c) (B) y c( c) (C) y 4( c) (D) y 4c( c) Sol. (B) a + b ab y c y (a + b) c ab a + b + ab c a + b c ab c A b a b c ab 7 B a C cosc C r 4 R s abc 4 a b sin C 4 a b c abc ab 4c c y 4c c Page 47

48 JEE-ADVANCED-4 DATE: PAPER- CODE- MATHS 45. 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) 5 + y (parabola) y 8 (common tangents) P, Q R, S (quadrilateral) PQRS (A) (B) 6 (C) 9 (D) 5 (D) Sol. y m + m If it is tangent to + y Then, P S (,) m m 4 m ( m ) m ±. Q R Hence equation of tangent is y + & y. Chord of contact PQ is Chord of contanct RS is y. 4 ( ) Hence co-ordinates of P, Q, R, S are (, ) ; (, ) ; (, 4) & (, 4) Area of trapezium is (PQ + RS) Height () 5 Hindi y m + m + y, P S (,) m m 4 m ( m ) m ±. Q R y + & y PQ, RS y. 4 ( ) P, Q, R, S (, ) ; (, ) ; (, 4) & (,4) (PQ + RS) () 5 Page 48

49 JEE-ADVANCED-4 DATE: PAPER- CODE- MATHS 4 dy y 46. The function y f() is the solution of the differential equation d in (, ) satisfying f(). Then f ()d is y f() (differential equation) dy d y 4 (, ) f() f ()d (A) (B) (B) 4 (C) 6 4 (D) 6 Sol. I.F. e d d e n e n( ) e y 4 d c 5 y c 5, y c y d d sin d cos d sin cos d cos ( cos )d sin sin 4 Page 49

50 JEE-ADVANCED-4 DATE: PAPER- CODE- MATHS 47. Let f : [, ] R be a function which is continuous on [, ] and is differentiable on (, ) with f(). Let F () f( t )dt for [, ]. If F'() f'() for all (, ), then F() equals f : [, ] R [, ] (continuous) (, ) (differentiable) Sol. f() [, ] F'() f'() F() (A) e (B) e 4 (C) e (D) e 4 (B) f'() f() F () f( t )dt (, ) f '() f() n(f()) + c, f() c n(f()) f() e F() f() + c F() e + c F() c f() e f() e Coefficient of in the epansion of ( + ) 4 ( + ) 7 ( + 4 ) is ( + ) 4 ( + ) 7 ( + 4 ) (epansion) (coefficient) (A) 5 (B) 6 (C) (D) (C) ( ) ( ) ( ) Sol. Coefficent of 4 ( ) ( ) 4 Coefficent of ( 8 ) 4 ( + 4 ) 8 ( + ) 7 ( ) 4 ( 4 8 ) ( + 4 ) 8 ( ) ( ) 4 ( ) ( + 4 ) 8 ( ) 4 Coefficent of 8 ( ) ( ) ( ) 4 ( ) ( ) 4 Coefficent of t 4 (7 + 56t + 5t + 68t 4 ) ( t) C C C Page 5

51 JEE-ADVANCED-4 DATE: PAPER- CODE- MATHS ( ) ( ) ( ) Hindi 4 ( ) ( ) 4 8 ) 4 ( + 4 ) 8 ( + ) 7 ( ) 4 ( 4 8 ) ( + 4 ) 8 ( ) ( ) 4 ( ) ( + 4 ) 8 ( ) 4 ( ) ( ) ( ) 4 8 ( ) ( ) 4 (7 + 56t + 5t + 68t 4 ) ( t) 4 t C C C Alterantive : + y + 4z (, y, z) (,, ) 4 C 7 C C (,, ) 4 C 7 C (,, ) 4 C 7 C C (4,, ) 7 C coefficient of For (, ), the equation sin + sin sin has (A) infinitely many solutions (B) three solutions (C) one solution (D) no solution (, ) sin + sin sin (A) (infinitely many) (B) (three) (C) (one) (D) (no solution) (D) Sol. sin + sin sin. sin ( + cos + 4 sin ). (4 sin + cos ) 4 cos + cos sin sin 9 cos 4. sin L.H.S. 9 4 R.H.S.. No solution. Page 5

52 JEE-ADVANCED-4 DATE: PAPER- CODE- MATHS 5. The following integral ( cosec) 7 d is equal to 4 (integral) ( cosec) 7 d? 4 (A) log( ) u u 6 (e e ) du (B) log( ) u (e e ) u 7 du (C) (A) log( ) log( u u 7 ( e e ) du (D) ) (e u e ) u 6 du Sol. (cosec) 7 d 4 Put n tan / t tan sin e t e t e t t t cosec e e t t 6 (e e ).dt n( ) t t 6 (e e ).dt n( ) since(e t + e t ) 6 is an even function (e t + e t ) 6 a a Hence n( ) t t 6 (e e ) dt Page 5

53 JEE-ADVANCED-4 DATE: PAPER- CODE- MATHS SECTION : Comprehension Type (Only One Option 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). (A), (B), (C) (D) Paragraph For Questions 5 and 5 ( 5 5 ) Bo contains three cards bearing numbers,, ; bo contains five cards bearing numbers,,, 4, 5; and bo contains seven cards bearing numbers,,, 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,,.,,,,, 4, 5,,, 4, 5, 6, 7 i (i th bo) i (i,, ) 5. The probability that + + is odd, is (A) 5 5 (B) 5 57 (C) 5 (D) Sol. (B) + + is odd if all three are odd or are even & one is odd (OOO) or (OEE) or (EOE) or (EEO) The probability that,, are in an arithmetic progression, is,, (arithmetic progression) 9 (A) 5 (B) 5 (C) 5 7 (D) 5 (C) Sol. + If & both are odd 4 8 ways & both are even ways Total ways Total ( ) triplets are 5 7 P 5 Page 5

54 JEE-ADVANCED-4 DATE: PAPER- CODE- MATHS Hindi ( ) 5 7 P 5 Paragraph For Questions 5 and 54 ( 5 54 ) Let a, r, s, t be nonzero real numbers. Let P(at, at), Q, R (ar, ar) and (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, ) a, r, s, t (nonzero real numbers) P(at, at), Q, R (ar, ar) (as, as) y 4a PQ (focal chord ) QR PK K (a, ) 5. The value of r is r Sol. (A) t (D) m PK m QR t (B) (C) t t (D) t t at at a ' at ar a(t') ar t t t' (t') r r t tr t rt t + r, tt t tr t + r rt tr + r(t ) + t + t t t 4 t t 4 t t t t t t r t It is not possible as the R & Q will be one same. r t R Q or r t t Page 54

55 JEE-ADVANCED-4 DATE: PAPER- CODE- MATHS 54. If st, then the tangent at P and the normal at S to the parabola meet at a point whose ordinate is st P S (normal) (ordinate) (A) (t ) t (B) a(t ) (B) Sol. Tangent at P is ty + at Normal at S is y + s as + as P ty + at S y + s as + as a ty + a + t ty a + a t ty + at t y at 4 + at + a t a(t ) (C) t a(t ) (D) t y a t t Paragraph For Questions 55 and 56 ( ) Given that for each a (, ) h lim h a h t ( t) a dt eists. Let this limit be g(a). In addition, it is given that the function g(a) is differentiable on (, ). a (, ) h lim h a h t ( t) a dt g(a) (interval) (, ) g(a) 55. The value of g is g (A) (B) (C) (D) 4 (A) Page 55

56 JEE-ADVANCED-4 DATE: PAPER- CODE- MATHS Sol. g '(a) h h a a t ( t) dt a h h a a a a t ( t) dt t t dt g(a) constant g(a) g(a) h Lt h h t t dt h h t dt 4 sin t h h sin (t ) h h sin ( h) sin (h ) 56. The value of g is g (A) (B) (C) (D) (D) Sol. g(a) h lim h h t a ( t) a dt g( a) h h ( h a) lim t ( t) dt ( a) h h a a lim t ( t) dt h h h h a a lim ( t) ( ( t)) dt b by f() d a b a f(a b ) d h lim h h ( t) a t a dt g( a) g(a) g' ( a) g'(a) at a, g ' g ' g'. Page 56

57 JEE-ADVANCED-4 DATE: PAPER- CODE- MATHS SECTION : Matching List Type (Only One Option 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 options (A),(B),(C) and (D), out of which ONE is correct. 4 (A), (B), (C) (D), 57. List I List II P. The number of polynomials f() with non-negative integer. 8 coefficients of degree, satisfying f() and f()d Q. The number of points in the interval, at which. f() sin( ) + cos( ) attains its maimum value, is, is R. ( e ) d equals. 4 / / S. / - I cos log cos log d d equals 4. - II P. (non-negative integer) (polynomials),. 8 f(), (degree) f() f()d Q.,. f() sin( ) + cos( ) R. ( e ) d. 4 / / S. / cos log cos log d d 4. Page 57

58 P Q R S (A) 4 (B) 4 (C) 4 (D) 4 (D) Sol. (P) Let f() a + b, a, b W (as f() ) a bc JEE-ADVANCED-4 DATE: PAPER- CODE- MATHS a b a + b 6 (a, b) (, ), (, ) Number of such polynomials (Q) f() sin n + if f() is maimum (R) n + 4 for n, [, ] e e e d d a a f()d f() f( ) d a e d d e e 8 (S) / / cos n d Hence P, Q, R, S 4 (D) (as it is an odd function) Hindi. (P) f() a + b, a, b W ( f() ) a bc a b a + b 6 (a, b) (, ), (, ) (Q) f() sin n + f() Page 58

59 JEE-ADVANCED-4 DATE: PAPER- CODE- MATHS (R) n + 4 n, [, ] e e e d d a a f()d f() f( ) d a e d d e e 8 (S) / / cos n d P, Q, R, S 4 ( ) 58. List I List II P. Let y() cos( cos ), [, ], ±. Then. y() d y() d dy() d equals Q. Let A, A,..., A n (n > ) be the vertices of a regular polygon of n. sides with its centre at the origin. Let a k be the position vector of the point A k, k,,..., n. If the minimum value of n is n k n k ak ) (ak.ak ) k ( a, then y R. If the normal from the point P(h, ) on the ellipse is. 8 6 perpendicular to the line + y 8, then the value of h is S. Number of positive solutions satisfying the equation 4. 9 tan tan 4 tan is Page 59

60 JEE-ADVANCED-4 DATE: PAPER- CODE- MATHS - I - II P. y() cos( cos ), [, ], ±,. y() d y() d dy() d Q. A, A,..., A n (n > ) n (regular polygon). (vertices) a k A k, k,,..., n (position vector) n k n k ak ) (ak.ak ) k ( a n y R. (ellipse) P(h, ) y 8 h 4 S. tan tan tan 4. 9 P Q R S (A) 4 (B) 4 (C) 4 (D) 4 (A) Sol. (P) y 4 where cos dy d d y d dy d ( ). 4 + ( ) 6 7 9(4 ) 9y Hence d y dy y d d 9 (Q) a a a a... an a n a. a a. a... a n.an Let a a..... a n (as centre is origin) More over angle between consecutive a i 's is n Page 6

61 JEE-ADVANCED-4 DATE: PAPER- CODE- MATHS Hence given equation reduces to (n ) sin n (n ) cos n tan n n 4 n 8 (R) Equation of normal 6 y h a b y Equation of normal is a b y slope 6 h (as it is perpendicular to z + y ) h (S) tan + tan 4 + tan (as ) ( ) ( + ), is rejected Hindi. (P) y 4 cos dy d d y d dy d ( ). 4 + ( ) 6 7 9(4 ) 9y d y dy y d d 9 (Q) a a a a... an an a. a a. a... a n.an a a..... a n ( ) a i 's n Page 6

62 JEE-ADVANCED-4 DATE: PAPER- CODE- MATHS (n ) sin n (n ) cos n tan n n 4 n 8 6 y (R) h a b y Equation of normal is a b y 6 h ( z + y ) h (S) tan + tan 4 + tan ( ) ( ) ( + ), is rejected 59. Let f : R R, f : [, ) R,f : R R and f 4 : R [, ) be defined by f () e if if, ; f () ; f () and sin if, if f 4 () f(f()) if, f(f()) if List I List II P. f 4 is. onto but not one-one Q. f is. neither continuous nor one-one R. f o f is. differentiable but not one-one S. f is 4. continuous and one-one Page 6

63 JEE-ADVANCED-4 DATE: PAPER- CODE- MATHS f : R R, f : [, ) R,f : R R f 4 : R [, ) f () f () ; f () f 4 (), e ; sin, f (f ()), f (f ()) - I - II P. f 4. (onto) (one-one) Q. f. (continuous) R. f o f. (differentiable) S. f 4. (continuous) P Q R S (A) 4 (B) 4 (C) 4 (D) 4 (D) Sol. f (f ()) (f ()) e f 4 () e f () f () f () f () 4 e Page 6

64 JEE-ADVANCED-4 DATE: PAPER- CODE- MATHS f 4 () is many-one onto, continuous and non-derivable f () is many-one, into, continuous and derivable f () is one-one, into, differentiable f 4 () f () f () Hence R so (D) p, q, R, S 4 6. Let z k cos k k isin ; k,, List I List II P. For each z k there eists a z j such that z k. z j. True Q. There eists a k {,,...,9} such that z.. z z k has. False no solution z in the set of comple numbers. R. 9 z z... z equals. S. 9 k k cos equals 4. k k z k cos isin ; k,, I -II P. z k z j z k. z j. Q. {,,...,9} k z.. z z k. z (comple numbers) R. z z... z9. S. 9 k k cos 4. Page 64

65 JEE-ADVANCED-4 DATE: PAPER- CODE- MATHS P Q R S (A) 4 (B) 4 (C) 4 (D) 4 (C) Sol. (P) z k z j z j z k Hence for each k {,,,...., 9} there eists z j such that z k. z j True (Q) z. z z k z z k for k,, 4,..., 9 & (R) z for k False z, z,...., z 9 are roots of the equation z other then unity, hence z z + z z9 (z z )(z z ).... (z z 9 ) Substituting z, we get z z.... z9 (S) 9 k ( ) k cos {sum of real parts of roots of z ecept } (as + z + z z 9 ) Re zk Hindi. (P) z k z j z j z k k {,,,...., 9} z j z k. z j (Q) z. z z k z z k k,, 4,..., 9 & z, k (R) z, z,...., z 9 z (S) z z + z z9 (z z )(z z ).... (z z 9 ) z 9 k z z.... z9 k cos { z } ( ) ( + z + z z 9 ) Re zk Page 65

66 JEE-ADVANCED-4 DATE: PAPER- CODE- 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 and darkening another bubble Answer will not be evaluated no marks, no negative marks a Attempt to Erase after darkening Figure- : Correct way of bubbling for valid answer and a few eamples of invalid answer. Any other form of partial marking such as ticking or crossing the bubble will be considered invalid. - : Figure- : Correct way of Bubbling your Roll Number on the ORS. (Eample Roll Number : 545) - : ( ORS.) ( : 545) 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 Page 66