Regd. Office : Aakash Tower, Plot No.-4, Sec-11, MLU, Dwarka, New Delhi Ph.: Fax : Answers & Solutions

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1 DATE : 5/05/04 CDE 5 Regd. ffice : Aakash Tower, Plo No.-4, Sec-, MLU, Dwarka, New Delhi-0075 Ph.: Fax : Answers & Soluions Time : hrs. Max. Marks: 80 for JEE (Advanced)-04 PAPER - (Code - 5) INSTRUCTINS Quesion Paper Forma The quesion paper consiss of hree pars (Physics, Chemisry and Mahemaics). Each par consiss of hree secions.. Secion conains 0 muliple choice quesions. Each quesion has four choices (A), (B), (C) and (D) ou of which NLY NE is correc.. Secion conains paragraphs each describing heory, experimen and daa ec. 6 quesions relae o hree paragraphs wih wo quesions on each paragraph. Each quesion peraining o a paricular passage should have only one correc answer among he four given choices (A), (B), (C) and (D).. Secion conains 4 muliple choice quesions. Each quesion has wo liss (Lis-, P, Q, R and S; Lis- :,, and 4). The opions for he correc mach are provided as (A), (B), (C) and (D) ou of which NLY NE is correc. Marking Scheme 4. For each quesion in Secion,, and you will be awarded marks if you darken he bubble corresponding o he correc answer and zero mark if no bubble is darkened. In all oher cases, minus one ( ) mark will be awarded.

2 PART I : PHYSICS SECTIN - : (nly ne pion Correc Type) This secion conains 0 muliple choice quesions. Each quesion has 4 choices (A), (B), (C) and (D) ou of which NLY NE pion is correc.. A glass capillary ube is of he shape of a runcaed cone wih an apex angle so ha is wo ends have cross secions of differen radii. When dipped in waer verically, waer rises in i o a heigh h, where he radius of is cross-secion is b. If he surface ension of waer is S, is densiy is, and is conac angle wih glass is. he value of h will be (g is he acceleraion due o graviy) (A) (B) (C) (D) S cos( ) bg S cos( ) bg S cos( /) bg S cos( /) bg h Answer (D) b h Sb cos b h S cos( ) bg h. A plane of radius R (radius of Earh) has he same mass densiy as Earh. Scieniss dig a well of 0 deph 5 R on i and lower a wire of he same lengh and of linear mass densiy 0 kg m ino i. If he wire is no ouching anywhere, he force applied a he op of he wire by a person holding i in place is (ake he radius of Earh = m and he acceleraion due o graviy on Earh is 0 ms ) (A) 96 N (B) 08 N (C) 0 N (D) 50 N ()

3 Answer (B) R P Re m P = E GM G 4 g R R R 4 g GR gp RP ge gp m/s g R 0 e E Acceleraion due o graviy a a deph x g x x g R P Force on a small segmen of wire a a deph x from surface df ( dx) gx x df gdx R R /5 x F 0 gdx R x x R R/5 R R 9 9 g 5 0 R g g x dx R P N 5. Charges Q, Q and 4Q are uniformly disribued in hree dielecric solid spheres, and of radii R/, R and R respecively, as shown in figure. If magniudes of he elecric fields a poin P a a disance R from he cenre of sphere, and are E, E and E respecively, hen P P P Q R Q R 4Q R R/ R Sphere Sphere Sphere (A) E > E > E (B) E > E > E (C) E > E > E (D) E > E > E Answer (C) E E E KQ R K ( Q ) KQ R R K(4 Q) R KQ 8R R E E E () Q

4 4. If Cu is he wavelengh of K X-ray line of copper (aomic number 9) and Mo is he wavelengh of he K X-ray line of molybdenum (aomic number 4), hen he raio Cu / Mo is close o (A).99 (B).4 (C) 0.50 (D) 0.48 Answer (B) Cu ( ZMo ) 4.4 Mo ( ZCu ) 8 5. A wire, which passes hrough he hole in a small bead, is ben in he form of quarer of a circle. The wire is fixed verically on ground as shown in he figure. The bead is released from near he op of he wire and i slides along he wire wihou fricion. As he bead moves from A o B, he force i applies on he wire is A 90 B (A) Always radially ouwards (B) Always radially inwards (C) Radially ouwards iniially and radially inwards laer (D) Radially inwards iniially and radially ouwards laer Answer (D) 6. A meal surface is illuminaed by ligh of wo differen wavelenghs 48 nm and 0 nm. The maximum speeds of he phooelecrons corresponding o hese wavelenghs are u and u, respecively. If he raio u : u = : and hc = 40 ev nm, he work funcion of he meal is nearly (A).7 ev (B). ev (C).8 ev (D).5 ev Answer (A) hc mu W hc mu W hc W u u hc W 4hc hc 4 W W ( u ) u 4hc hc W W 6 5 = W 0 48 = W W =.7 ev (4)

5 7. A ennis ball is dropped on a horizonal smooh surface. I bounces back o is original posiion afer hiing he surface. The force on he ball during he collision is proporional o he lengh of compression of he ball. Which one of he following skeches describes he variaion of is kineic energy K wih ime mos appropriaely? The figures are only illusraive and no o he scale. K K (A) (B) K K (C) (D) Answer (B) v v K v v K.E. 8. During an experimen wih a mere bridge, he galvanomeer shows a null poin when he jockey is pressed a 40.0 cm using a sandard resisance of 90, as shown in he figure. The leas coun of he scale used in he mere bridge is mm. The unknown resisance is R 90 (A) 60 ± 0.5 (C) 60 ± cm (B) 5 ± 0.56 (D) 5 ± 0. (5)

6 Answer (C) R R x (00 x) For R, (Balanced Wheasone) R R R x x For R, ln R ln x ln(00 x) R x (00 x) R x (00 x) R x x R x (00 x) R R = 0.5 R = 60 ± Parallel rays of ligh of inensiy I = 9 Wm are inciden on a spherical black body kep in surroundings of emperaure 00 K. Take Sefan-Bolzmann consan = Wm K 4 and assume ha he energy exchange wih he surroundings is only hrough radiaion. The final seady sae emperaure of he black body is close o (A) 0 K (B) 660 K (C) 990 K (D) 550 K Answer (A) 4 R.( T T ) 9R T T T (00) (40 8) 0 T = 0 K 0. A poin source S is placed a he boom of a ransparen block of heigh 0 mm and refracive index.7. I is immersed in a lower refracive index liquid as shown in he figure. I is found ha he ligh emerging from he block o he liquid forms a circular brigh spo of diameer.54 mm on he op of he block. The refracive index of he liquid is Liquid Block S (A). (B).0 (C).6 (D).4 (6)

7 Answer (C) 0 mm 0 mm = 0 sin.7 sin0.7.6 =.7 S SECTIN - : Comprehension Type (nly ne pion Correc) This secion conains paragraphs, each describing heory, experimens, daa ec. 6 quesions relaed o he hree paragraphs wih wo quesions on each paragraph. Each quesion has only one correc answer among he four given opions (A), (B), (C) and (D). Paragraph For Quesions and In he figure, a conainer is shown o have a movable (wihou fricion) pison on op. The conainer and he pison are all made of perfecly insulaion maerial allowing no hea ransfer beween ouside and inside he conainer. The conainer is divided ino wo comparmens by a rigid pariion made of a hermally conducing maerial ha allows slow ransfer of hea. The lower comparmen of he conainer is filled wih moles of an ideal monaomic gas a 700 K and he upper comparmen is filled wih moles of an ideal diaomic gas 5 a 400 K. The hea capaciies per mole of an ideal monaomic gas are CV = R, CP = R, and hose for an 5 7 ideal diaomic gas are CV = R, CP = R.. Consider he pariion o be rigidly fixed so ha i does no move. When equilibrium is achieved, he final emperaure of he gases will be (A) 550 K (B) 55 K (C) 5 K (D) 490 K Answer (D) Q = Q R 7R ( T 700) ( T 400) 6T T = 0 T T = 490 K (7)

8 . Now consider he pariion o be free o move wihou fricion so ha he pressure of gases in boh comparmens is he same. The oal work done by he gases ill he ime hey achieve equilibrium will be (A) 50 R (B) 00 R (C) 00 R (D) 00 R Answer (D) nc ( T700) nc (400 T) P P 5R 7R ( T 700) (400 T) 600 5T T T T 55 W = U = nc (55 700) nc (55 400) V V R 5R 75 5 = 55R + 65R W = U = 00 R Paragraph For Quesions and 4 A spray gun is shown in he figure where a pison pushes air ou of a nozzle. A hin ube of uniform cross secion is conneced o he nozzle. The oher end of he ube is in a small liquid conainer. As he pison pushes air hrough he nozzle, he liquid from he conainer rises ino he nozzle and is sprayed ou. For he spray gun shown, he radii of he pison and he nozzle are 0 mm and mm respecively. The upper end of he conainer is open o he amosphere.. If he pison is pushed a a speed of 5 mms, he air comes ou of he nozzle wih a speed of (A) 0. ms (B) ms (C) ms (D) 8 ms Answer (C) A V = A V (0.) VL VL VL V L = 4. If he densiy of air is a and ha of he liquid l, hen for a given pison speed he rae (volume per uni ime) a which he liquid is sprayed will be proporional o (A) a l (B) a l (C) l a (D) l (8)

9 Answer (A) P P v a a P P v P = P l l lvl ava vl a l v a a Volume flow rae l Paragraph For Quesions 5 and 6 The figure shows circular loop of radius a wih wo long parallel wires (numbered and ) all in he plane of he paper. The disance of each wire from he cenre of he loop is d. The loop and he wires are carrying he same curren I. The curren in he loop is in he counerclockwise direcion if seen from above. Q d d S Wire a Wire P R 5. When d a bu wires are no ouching he loop, i is found ha he ne magneic field on he axis of he loop is zero a a heigh h above he loop. In ha case (A) Curren in wire and wire is he direcion PQ and RS, respecively and h a (B) Curren in wire and wire is he direcion PQ and SR, respecively and h a (C) Curren in wire and wire is he direcion PQ and SR, respecively and h.a (D) Curren in wire and wire is he direcion PQ and RS, respecively and h.a Answer (C) 0Ia 0Ia ( a h ) ( a h ) a a a h 4 a 4a a h 0a 4a 4h 6a 4h a h h.a 6. Consider d a, and he loop is roaed abou is diameer parallel o he wires by 0 from he posiion shown in he figure. If he currens in he wires are in he opposie direcions, he orque on he loop a is new posiion will be (assume ha he ne field due o he wires is consan over he loop) 0I a 0I a (A) (B) d d 0I a 0I a (C) (D) d d (9)

10 Answer (B) 0I B d 0I I a d 0I a d SECTIN - : Maching Lis Type (nly ne pion Correc) This secion conains 4 quesions, each having wo maching liss. Choice for he correc combinaion of elemens from Lis-I and Lis-II are given as opions (A), (B), (C) and (D), ou of which only one is correc. 7. Four charges Q, Q, Q and Q 4 of same magniude are fixed along he x axis a x = a, a, +a and +a, respecively. A posiive charge q is placed on he posiive y axis a a disance b > 0. Four opions of he signs of hese charges are given in Lis I. The direcion of he forces on he charge q is given in Lis II. Mach Lis I wih Lis II and selec he correc answer using he code given below he liss. q (0, b) Q Q Q Q4 ( a,0) ( a,0) (+ a,0) (+ a,0) Lis-I Lis-II P. Q, Q, Q, Q 4 all posiive. +x Q. Q, Q posiive; Q, Q 4 negaive. x R. Q, Q 4 posiive; Q, Q negaive. +y S. Q, Q posiive; Q, Q 4 negaive 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- Answer (A) Hin: P. No along +y Q. No along +x Q Q Q Q 4 Q Q Q Q 4 R. No along y S. No along x Q Q Q Q 4 Q Q Q Q 4 (0)

11 8. Four combinaions of wo hin lenses are given in Lis I. The radius of curvaure of all curved surfaces is r and he refracive index of all he lenses is.5. Mach lens combinaions in Lis I wih heir focal lengh in Lis II and selec he correc answer using he code given below he liss. Lis-I Lis-II P.. r Q.. r R.. r S. 4. r Code : (A) P-, Q-, R-, S-4 (B) P-, Q-4, R-, S- (C) P-4, Q-, R-, S- (D) P-, Q-, R-, S-4 Answer (B) (P) ( u ) (.5 ) f r r r r r f ' f' f f P f r (Q) 0.5 (.5 ) f r r f r f f r f f f f r r Q 4 ()

12 (R) 0.5 (.5 ) f r f r r f r f r r f f f r r f r R (S) (.5 ) 0.5 f r f r r r r 0.5 (.5 ) f r f r f r f f f r r r f r S 9. A block of mass m = kg anoher mass m = kg, are placed ogeher (see figure) on an inclined plane wih angle of inclinaion. Various values of are given in Lis I. The coefficien of fricion beween he block m and he plane is always zero. The coefficien of saic and dynamic fricion beween he block m and he plane are equal o = 0.. In Lis II expressions for he fricion on block m are given. Mach he correc expression of he fricion in Lis II wih he angles given in Lis I, and choose he correc opion. The acceleraion due o graviy is denoed by g. [Useful informaion : an(5.5 ) 0.; an(.5 ) 0.; an(6.5 ) 0.] m m Lis-I Lis-II P. = 5. m g sin Q. = 0. (m + m )g sin R. = 5. m g cos S. = 0 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- ()

13 Answer (D) m m mgcos ( m m) gsin an m m m 0. So block m sars sliding a >.5. When = 5 and = 0, hen block will be res, so fricion will be ( m m) gsin For = 5 and = 0, block sars sliding. So fricion will be mg cos. So opion (D) is correc. 0. A person in lif is holding a waer jar, which has a small hole a he lower end of is side. When he lif is a res, he waer je coming ou of he hole his he floor of he lif a a disance d of. m from he person. In he following, sae of he lif's moion is given in Lis I and he disance where he waer je his he floor of he lif is given in Lis II. Mach he saemens from Lis I wih hose in Lis II and selec he correc answer using he code given below he liss. Lis-I Lis-II P. Lif is acceleraing verically up. d =. m Q. Lif is acceleraing verically down. d >. m wih an acceleraion less han he graviaional acceleraion R. Lif is moving verically up. d <. m wih consan speed S. Lif is falling freely 4. No waer leaks ou of he 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- Answer (C) a R gh h g H d v H( g+a) x h H( g a) g a d Independen of acceleraion of lif. P-, Q-, R-, S-4 () h d

14 PART II : CHEMISTRY SECTIN - : (nly ne pion Correc Type) This secion conains 0 muliple choice quesions. Each quesion has 4 choices (A), (B), (C) and (D) ou of which NLY NE pion is correc.. Isomers of hexane, based on heir branching, can be divided ino hree disinc classes as shown in he figure. [Figure] and and (I) (II) (III) The correc order of heir boiling poin is (A) I > II > III (B) III > II > I (C) II > III > I (D) III > I > II Answer (B) As branching increases boiling poin decreases, so order of boiling poin is III > II > I.. The major produc in he following reacion is Cl. CH MgBr, dry eher, 0 C. aq. acid CH (A) HC CH (B) HC H CH CH CH (C) CH (D) CH Answer (D) Cl CH MgI Cl CH CH CH. The acidic hydrolysis of eher (X) shown below is fases when [Figure] R Acid H + RH (4)

15 (A) ne phenyl group is replaced by a mehyl group (B) ne phenyl group is replaced by a para-mehoxyphenyl group (C) Two phenyl groups are replaced by wo para-mehoxyphenyl groups (D) No srucural change is made o X Answer (C) CH group has +R effec. I increases he sabiliy of he carbocaion. 4. Hydrogen peroxide in is reacion wih KI 4 and NH H respecively, is acing as a (A) Reducing agen, oxidising agen (B) Reducing agen, reducing agen (C) xidising agen, oxidising agen (D) xidising agen, reducing agen Answer (A) H acing as a reducing agen wih KI 4 and H acing as an oxidising agen wih NH H. 5. The produc formed in he reacion of SCl wih whie phosphorous is (A) PCl (B) S Cl (C) SCl (D) PCl Answer (A) P4 8SCl 4PCl 4S SCl (Whie) 6. For he idenificaion of -naphhol using dye es, i is necessary o use (A) Dichloromehane soluion of -naphhol (B) Acidic soluion of -naphhol (C) Neural soluion of -naphhol (D) Alkaline soluion of -naphhol Answer (D) + NCl H H NaH N N 7. For he process H (I) H (g) a T = 00 ºC and amosphere pressure, he correc choice is (A) S sysem > 0 and S surroundings > 0 (B) S sysem > 0 and S surroundings < 0 (C) S sysem < 0 and S surroundings > 0 (D) S sysem < 0 and S surroundings < 0 Answer (B) Given condiions are boiling condiions for waer due o which S oal = 0 S sysem + S surroundings = 0 S sysem = S surroundings For process, S sysem > 0 S surroundings < 0 (5)

16 8. For he elemenary reacion M N, he rae of disappearance of M increases by a facor of 8 upon doubling he concenraion of M. The order of he reacion wih respec o M is (A) 4 (B) (C) (D) Answer (B) r [M] r [M] 8 = () = So, order of reacion is. 9. Under ambien condiions, he oal number of gases released as producs in he final sep of he reacion scheme shown below is XeF 6 Complee Hydrolysis P + her produc Q H /H Slow disproporionaion in H /H Producs (A) 0 (B) (C) (D) Answer (C) XeF 6 Complee Hydrolysis Xe + H F H /H HXe 4 Slow disproporionaion in H /H 4 Xe 6 + Xe(g) + H + (g) 0. Assuming s-p mixing is NT operaive, he paramagneic species among he following is (A) Be (B) B (C) C (D) N Answer (C) Be = s *s s *s B = s *s s *s p z C = s *s s *s p z p x p y N = s *s s *s p z p x p y (6)

17 SECTIN - : Comprehension Type (nly ne pion Correc) This secion conains paragraphs, each describing heory, experimens, daa ec. 6 quesions relaed o he hree paragraphs wih wo quesions on each paragraph. Each quesion has only one correc answer among he four given opions (A), (B), (C) and (D). Paragraph For Quesions and Schemes and describe sequenial ransformaion of alkynes M and N. Consider only he major producs formed in each sep for boh he schemes. H M H. NaNH (excess). CH CH I ( equivalen). CH I ( equivalen) 4. H, Lindlar s caalys X Scheme- N H. NaNH ( equivalen). H Br. H, (mild) 4. H, Pd/C 5. Cr Y Scheme-. The produc X is HC H (A) H H (B) HC H CH C H H (C) H H (D) CH C H H Answer (A) H NaNH + CH CH I CNa Na + ( eq) H Na + C CH I CH H H. The correc saemen wih respec o produc Y is H Lindlar's caalys CH (A) I gives a posiive Tollens es and is a funcional isomer of X (B) I gives a posiive Tollens es and is a geomerical isomer of X (C) I gives a posiive iodoform es and is a funcional isomer of X (D) I gives a posiive iodoform es and is a geomerical isomer of X (7)

18 Answer (C) H. NaNH ( eq). H Br +. H (mild) 4. H, Pd/C 5. Cr Paragraph For Quesions and 4 An aqueous soluion of meal ion M reacs separaely wih reagens Q and R in excess o give erahedral and square planar complexes, respecively. An aqueous soluion of anoher meal ion M always forms erahedral complexes wih hese reagens. Aqueous soluion of M on reacion wih reagen S gives whie precipiae which dissolves in excess of S. The reacions are summarized in he scheme given below : Scheme : Terahedral Q Excess M R Excess Square planar Terahedral Q Excess M R Excess Terahedral S, soichiomeric amoun. M, Q and R, respecively are (A) Zn +, KCN and HCl (C) Cd +, KCN and HCl Answer (B) Ni + + HCl [NiCl 4 ] Ni + + KCN [Ni(CN) 4 ] 4. Reagen S is Whie precipiae R Excess (B) Ni +, HCl and KCN (D) Co +, HCl and KCN Precipiae dissolves (A) K 4 [Fe(CN) 6 ] (B) Na HP 4 (C) K Cr 4 (D) KH Answer (D) H 4 whie pp Soluble Zn H Zn(H) [Zn(H) ] Paragraph For Quesions 5 and 6 X and Y are wo volaile liquids wih molar weighs of 0 g mol and 40 g mol respecively. Two coon plugs, one soaked in X and he oher soaked in Y, are simulaneously placed a he ends of a ube of lengh L = 4 cm, as shown in he figure. The ube is filled wih an iner gas a amosphere pressure and a emperaure of 00 K. Vapours of X and Y reac o form a produc which is firs observed a a disance d cm from he plug soaked in X. Take X and Y o have equal molecular diameers and assume ideal behaviour for he iner gas and he wo vapours. L = 4 cm Coon wool soaked in X d Iniial formaion of he produc 5. The value of d in cm (shown in he figure), as esimaed from Graham's law, is (A) 8 (B) (C) 6 (D) 0 (8) Coon wool soaked in Y

19 Answer (C) x 40 4 x 0 x 4 x x = 6 6. The experimenal value of d is found o be smaller han he esimae obained using Graham's law. This is due o (A) Larger mean free pah for X as compared o ha of Y (B) Larger mean free pah for Y as compared o ha of X (C) Increased collision frequency of Y wih he iner gas as compared o ha of X wih he iner gas (D) Increased collision frequency of X wih he iner gas as compared o ha of Y wih he iner gas Answer (D) Increased collision frequency of X wih he iner gas as compared o ha of Y wih he iner gas. Therefore, he experimenal value of d is found o be smaller han he esimae obained using Graham's law. SECTIN - : Maching Lis Type (nly ne pion Correc) This secion conains 4 quesions, each having wo maching liss. Choice for he correc combinaion of elemens from Lis-I and Lis-II are given as opions (A), (B), (C) and (D), ou of which only one is correc. 7. Differen possible hermal decomposiion pahways for peroxyesers are shown below. Mach each pahway from Lis I wih an appropriae srucure from Lis II and selec he correc answer using he code given below he liss. P C R+ R R R Q C R+ R R+ X+ carbonyl compound (Peroxyeser) R RC + R C R+ X + carbonyl compound Lis-I RC + R P. Pahway P. Q. Pahway Q. S C Lis-II CHCH 6 5 CH 6 5 R. Pahway R. CHCH 6 5 S. Pahway S 4. CH 6 5 (9) R+ R CH CH CH CH CHC6H 5 CH CH CH 6 5

20 Code : P Q R S (A) 4 (B) 4 (C) 4 (D) 4 Answer (A) () C6H5CH C CH C H CH + C + CH 6 5 () CHCH C 6 5 CH C (4) CH 6 5 C CH () CH C 6 5 CH CH CH C C + C6H5CH + CH C CH C H 6 5 CH C H 6 5 CH CH C H C + C CH 6 5 CH CH CH 6 5 C + CH 8. Mach he four saring maerials (P, Q, R, S) given in Lis I wih he corresponding reacion schemes (I, II, III, IV) provided in Lis II and selec he correc answer using he code given below he liss. Lis-I Lis-II P. H H. Scheme I (i) KMn 4, H, hea (ii) H, H? (iii) SCl (iv) NH CHN 7 6 H Scheme II Q. H. (i) Sn/HCl (ii) CHCCl (iii) conc. HS4 (iv) HN (v) dil. HS 4, hea (vi) H? CHN 6 6 N Scheme III R.. (i) red ho iron, 87 K (ii) fuming HN, HS 4, hea (iii) HS.NH (iv) NaN, HS 4 (v) hydrolysis? CHN 6 5 N Scheme IV S. CH 4. (i) conc. HS 4, 60ºC (ii) conc. HN, conc. HS 4 (iii) dil. HS 4, hea? CHN (0)

21 Code : P Q R X (A) 4 (B) 4 (C) 4 (D) 4 Answer (C) N Scheme I (S) CH N Scheme II (R) Scheme III H C C H (P) H Scheme IV H 9. Mach each coordinaion compound in Lis-I wih an appropriae pair of characerisics from Lis-II and selec he correc answer using he code given below he liss. {en = H NCH CH NH ; aomic numbers : Ti =, Cr = 4; Co = 7; P = 78} Lis-I Lis-II P. [Cr(NH ) 4 Cl ]Cl. Paramagneic and exhibis ionisaion isomerism Q. [Ti(H ) 5 Cl](N ). Diamagneic and exhibis cis-rans isomerism R. [P(en)(NH )Cl]N. Paramagneic and exhibis cis-rans isomerism S. [Co(NH ) 4 (N ) ]N 4. Diamagneic and exhibis ionisaion isomerism Code : P Q R S (A) 4 (B) 4 (C) 4 (D) 4 ()

22 Answer (B) Magneic characer Isomerism P. [Cr(NH ) 4 Cl ]Cl Paramagneic cis/rans Q. [Ti(H ) 5 Cl](N ) Paramagneic ionizaion R. [P(en)(NH )Cl]N Diamagneic ionizaion S. [Co(NH ) 4 (N ) ]N Diamagneic cis/rans 40. Mach he orbial overlap figures shown in Lis-I wih he descripion given in Lis-II and selec he correc answer using he code given below he liss. Lis-I Lis-II P.. p d anibonding Q.. d d bonding R.. p d bonding S. 4. d d anibonding Code : P Q R S (A) 4 (B) 4 (C) 4 (D) 4 Answer (C) P. d d ( bonding) Q. p d ( bonding) R. p d ( anibonding) S. d d ( anibonding) ()

23 PART III : MATHEMATICS SECTIN - : (nly ne pion Correc Type) This secion conains 0 muliple choice quesions. Each quesion has 4 choices (A), (B), (C) and (D) ou of which NLY NE pion is correc. 4. Coefficien of x in he expansion of ( + x ) 4 ( + x ) 7 ( + x 4 ) is (A) 05 (B) 06 (C) (D) 0 Answer (C) Power of Coefficien of x x x x C C C C C C 4 7 C C C C C C Le f : [0, ] R be a funcion which is coninuous on [0, ] and is differeniable on (0, ) wih f(0) =. Le x ( ) ( ) 0 F x f d for x [0, ]. If F(x) = f(x) for all x (0, ), hen F() equals (A) e (B) e 4 (C) e (D) e 4 Answer (B) F(x) = f(x).x = f(x) f'( x) x fx ( ) f( x) ke x Given f(0) = f( x) e x So x x F( x) e dx 0 x e So F() = e 4 ()

24 4. The funcion y = f(x) is he soluion of he differenial equaion dy xy x x dx x x 4 in (, ) saisfying f(0) = 0. Then f ( x ) dx is (A) (C) (B) 4 (D) Answer (B) dy x x x y dx x x 4 I.F. x dx x x e So 5 4 x y x ( x x) dx x c 5 Given f(0) = 0 c = 0 So y 5 x x 5 x f( x) 5 x x ( ) 5 x dx I f x dx dx x Le x = sin I sin d 0 ( cos ) d sin x 44. The following inegral (A) 7 (cosec ) x dx is equal o 4 log( ) u u 6 ( e e ) du (B) 0 log( ) u u 7 ( e e ) du 0 (C) log( ) u u 7 ( ) 0 e e du (D) (4) log( ) u u 6 ( ) 0 e e du

25 Answer (A) 7 I (cosec x) dx 4 cosec x co x cosec cosec x co x xdx 4 Le cosecx + cox = e u cosecxdx = du 6 0 ln( ) u u 6 u u 6 I ( e e ) du ( e e ) du ln( ) The quadraic equaion p(x) = 0 wih real coefficiens has purely imaginary roos. Then he equaion p(p(x)) = 0 has (A) nly purely imaginary roos (B) All real roos (C) Two real and wo purely imaginary roos (D) Neiher real nor purely imaginary roos Answer (D) Le p(x) = x + a (a > 0) ( roos are purely imaginary) p(p(x)) = (x + a) + a (a R) x 4 + a(x ) + a + a = 0 x = a 4a 4a 4a = a ai x = a ai x iy 46. For x (0, ), he equaion sinx + sinx sinx = has (A) Infiniely many soluions (B) Three soluions (C) ne soluion (D) No soluion Answer (D) sinx + sinx sinx = cosx( sinx) + 4sinxcosx = sinx[cosx cosx] = sinx(cosx (cos x )) = sinx( + cosx cos x) = sinx cos x Possible only when sinx = (i) and cos x 0 cos x From (i) and (ii) No soluion. (ii) (5)

26 47. In a riangle he sum of wo sides is x and he produc of he same wo sides is y. If x c = y, where c is he hird side of he riangle, hen he raio of he in radius o he circum-radius of he riangle is y y (A) (B) xx ( c) ( cxc) y y (C) (D) 4 xx ( c) 4( cxc) Answer (B) a + b = x, ab = y, x c = y Now, (a + b) c = ab a + b c = ab a b c ab cosc c = 0 c c R sinc c r ( sc) an x c r c R c ( ) r x c r ( x c ) ( ) x c R c c ( x c)( x c) cx ( c) ( x c ) cx ( c) y cx ( c) 48. Three boys and wo girls sand in a queue. The probabiliy, ha he number of boys ahead of every girl is a leas one more han he number of girls ahead of her, is (A) (C) Answer (A) n(s) = 5! Case I : G 4! = 48 (B) (D) 4 Case II : G! = n(e) = PE ( ) 5! (6)

27 49. The common angens o he circle x + y = and he parabola y = 8x ouch he circle a he poins P, Q and he parabola a he poins R, S. Then he area of he quadrilaeral PQRS is (A) (B) 6 (C) 9 (D) 5 Answer (D) x + y = Equaion of angen o circle x + y = is y mx m Equaion of angen o y = 8x is y mx m ( m ) m 4 ( m ) m m ( + m ) = m 4 + m = 0 (m + )(m ) = 0 m = y = x + y = x (, ) (, ) (, 4) (, 4) Area = = 5 sq. uni 50. Six cards and six envelopes are numbered,,, 4, 5, 6 and cards are o be placed in envelopes so ha each envelope conains exacly one card and no card is placed in he envelope bearing he same number and moreover he card numbered is always placed in envelope numbered. Then he number of ways i can be done is (A) 64 (B) 65 (C) 5 (D) 67 Answer (C) 6!!!! 4! 5! 6! (7)

28 SECTIN - : Comprehension Type (nly ne pion Correc) This secion conains paragraphs, each describing heory, experimens, daa ec. 6 quesions relaed o he hree paragraphs wih wo quesions on each paragraph. Each quesion has only one correc answer among he four given opions (A), (B), (C) and (D). Paragraph For Quesions 5 and 5 Le a, r, s, be nonzero real numbers. Le P(a, a), Q, R(ar, ar) and S(as, as) be disinc poins on he parabola y = 4ax. Suppose ha PQ is he focal chord and lines QR and PK are parallel, where K is he poin (a, 0). 5. The value of r is (A) (B) + (C) Answer (D) (D) (, ) P a a S a, a ( ar, ar) R ( a, 0) K(a, 0) Le co-ordinaes of Q be (a, a ) Since PQ is focal chord = a, Q a = a a Co-ordinaes of Q, a 0 Slope of PK = = a a (8)

29 Slope of RQ a ar + r + = = a ar r = r Slope of PK = Slope of RQ = r r = r = 5. If s =, hen he angen a P and he normal a S o he parabola mee a a poin whose ordinae is (A) ( + ) (B) a ( + ) (C) a ( + ) (D) a ( + ) Answer (B) Since s = s = a a Co-ordinaes of S =, Equaion of angen a P y = (x + a ) x = y a...(i) Slope of normal a S s = Equaion of normal a a y = x a y + a = x a x = y+ a+...(ii) (9)

30 From (i) and (ii) a y. a = y + a + a y = a + + a y = a + a ( + ) y = y = Given ha for each a (0, ), a + Paragraph For Quesions 5 and 54 + h 0 h a a lim ( ) d h exiss. Le his limi be g(a). In addiion, i is given ha he funcion g(a) is differeniable on (0, ). 5. The value of g is (A) (B) (C) π (D) π 4 Answer (A) g(a) = g ( h) a + h 0 h ( h) / + h 0 h ( h) a lim ( ) d = lim ( ) = lim + h 0 h / d d ( ) lim sin = + h 0 h 0 h h = lim [sin (( h) )) sin (h )] + π π = + = π (0)

31 54. The value of g' is (A) π (C) π (B) (D) 0 Answer (D) g(a) = h 0 h a a lim ( ) d + h h a a g '( a) lim ( ) ln d h0 h h h0 h g' lim ( ) ln d g ' 0 Paragraph For Quesions 55 and 56 Box conains hree cards bearing numbers,, ; box conains five cards bearing numbers,,, 4, 5; and box conains seven cards bearing numbers,,, 4, 5, 6, 7. A card is drawn from each of he boxes. Le x i be number on he card drawn from he i h box, i =,,. 55. The probabiliy ha x + x + x is odd, is 9 (A) (C) 05 Answer (B) B B 4 5 (B) (D) () 5 05 B x + x + x = odd Case I : ne odd ( ) + (4 ) + ( ) Two even = = 9 Case II : odd 4 = 4 = = 5 Toal cases = 5 7 = 05 Probabiliy = 5 05

32 56. The probabiliy ha x, x, x are in an arihmeic progression, is (A) 9 05 (C) 05 Answer (C) (B) (D) x = x + x x and x mus be even number. boh x {, } x and x odd x {,, 5, 7} also x + x {, 4, 6, 8, 0} x x (,, 5, 7) (,, 5, 7) 8 cases x and x boh even x = x = (, 4, 6) Toal = cases Required probabiliy = 05 cases SECTIN - : Maching Lis Type (nly ne pion Correc) This secion conains 4 quesions, each having wo maching liss. Choice for he correc combinaion of elemens from Lis-I and Lis-II are given as opions (A), (B), (C) and (D), ou of which one is correc. k k 57. Le zk cos i sin ; k =,,..., Lis-I () Lis-II P. For each z k here exiss a z j such ha z k. z j =. True Q. There exiss a k {,,..., 9} such ha z.z = z k has. False R. no soluion z in he se of complex numbers z z... z9 equals 0. 9 k S. k cos equals 4. 0 P Q R S (A) 4 (B) 4 (C) 4 (D) 4

33 Answer (C) z k = (P) z k.z j = π π cos + isin = e 0 0 k iπk 0 e π i ( k + j ) 0 = π ( k+ j) π ( k+ j) cos = and sin = k + j = 0n and k + j = 5m; so rue (Q) z k z = = e z π i ( k ) 0 So, if k {,,...9}, z has soluion; i.e., false (Q) (R) z 0 = 0 (z z )(z z )...(z z 9 ) = + z + z z 9 (S) So, z z... z 9 = 0 i.e., (R) π 4π 8π cos + cos cos π π + 8 cos 0 0 = = 58. Lis-I Lis-II P. The number of polynomials f(x) wih non-negaive ineger. 8 coefficiens of degree, saisfying f(0) = 0 and 0 fxdx ( ), is Q. The number of poins in he inerval [, ] a. which f(x) = sin(x ) + cos(x ) aains is maximum value, is R. S. x dx equals. 4 ( x e ) 0 x cos x log dx x equals 4. 0 x cos x log dx x P Q R S (A) 4 (B) 4 (C) 4 (D) 4 ()

34 Answer (D) (P) Le f(x) = a 0 x + a x + a ; a 0, a, a > 0 Now, a = 0 and f ( x ) dx 0 ax 0 ax a0 a 0 a 0 + a = 6 i.e., for inegral soluion eiher a0, a 0 or a0 0, a i.e., f(x) = x or f(x) = x P () (Q) f( x ) a max 9 x, 4 4 x, So, number of poins = 4 Q () (R) I x e x dx again, I x e x dx Adding, I x dx { x } 6 I = 8 R () (S) As x cosx log x is an odd funcion So, x cos x log 0 dx x S (4) (4)

35 59. Lis-I Lis-II P. Le y(x) = cos( cos x), x [,], x. Then. d y( x) dy( x) ( x ) x yx ( ) dx dx equals Q. Le A, A,..., A n (n > ) be he verices of a regular. polygon of n sides wih is cenre a he origin. Le a k be he posiion vecor of he poin A k, n n k =,,..., n. If ( a a ) ( a. a ), k k k k k k hen he minimum value of n is R. If he normal from he poin P(h, ) on he ellipse. 8 x y 6 is perpendicular o he line x + y = 8, hen he value of h is S. Number of posiive soluions saisfying he equaion 4. 9 an an an x 4x x P Q R S (A) 4 (B) 4 (C) 4 (D) 4 Answer (A) is (P) dy sin(cos x) dx x dy ( x ) 9( cos ( cos x)) 9( y ) dx dy d y dy dy ( x ) ( ) 8 x y dx dx dx dx d y ( ) dy x x 9 y dx dx (Q) ( n) R sin ( n) R cos n n an n n = 8 (5)

36 (R) As, equaion of normal is 6x y cos sin Given is slope = an Also (i) passes hrough (h, ) (i) So, 6h h = (S) an 4 x x an x ( )(4 ) x x 6x x 8x 6x x 4x x x 7x 6 = 0 x, Bu x + > 0 and 4x + > 0 So, soluion are x = (where x 0) 60. Le f : R R, f : [0, ) R, f : R R and f 4 : R [0, ) be defined by x if x 0, f ( x) x e if x 0; f (x) = x ; and f (x) = sin x if x 0, x if x 0 f( f( x)) if x 0, f4 ( x) f( f( x)) if x 0. Lis-I Lis-II P. f 4 is. no bu no one-one Q. f is. Neiher coninuous nor one-one R. f of is. Differeniable bu no one-one S. f is 4. Coninuous and one-one P Q R S (A) 4 (B) 4 (C) 4 (D) 4 (6)

37 Answer (D) (P) f 4 (x) = f[ f( x)], x < 0 f[ f( x)], x 0 no bu no one-one = x x < e x, 0, x 0 Now f(x) is no differeniable a x = 0, no one-one bu coninuous. (P) (Q) Now, f = sin x, x < 0 x, x 0 Differeniable bu no one-one. (Q) (R) f of = x x < e x, 0, x 0 (R) (S) f : [0, ] R, f (x) = x (S) 4 (7)