PAPER CODE 0 0 C T PAPER-1

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1 Path to Success PAR- : PHYSICS V. Ans. (A,B,C,D) m ibrating in th harmonic nl l l l Distance of antinode from node m 6 Now at t.55 m mean particle is as antinode condition Mean at t string will be straight.. Ans. (C) A m l m B If we obsere motion of particle from point A as frame of A non inertial so angular momentum of rod as well as linear momentum will not consered. Ans. (A,C,D) ALLEN CAREER INSIUE KOA (RAJASHAN) Relatie number of molecule M ES YPE : MAJOR mp ag rms PAPER CODE C DISANCE LEARNING PROGRAMME (ACADEMIC SESSION -5) NURURE ES SERIES / JOIN PACKAGE COURSE ARGE : JEE (Adanced) 6 ALL INDIA JEE-ADVANCED OPEN ES # PAPER- SOLUION Corporate Office : ALLEN CAREER INSIUE, SANKALP, CP-6, Indra Vihar, Kota (Rajasthan) dlp@allen.ac.in V V V V rms mp ag rms R where temperature is in k M R M Date : R pm V rms ( + C) R 7 ( + C) R 7 ANSWER KEY Q A. A,B,C,D C A,C,D A,C,D A,C B,D A,B,C A,D B,D A Q A. 6 5 M. Ans. (A, C, D) 5. Ans. (A,C) Area of PV cure in process > area of PV cure in process Means w process > w process As final olume in both process is same P µ final process < final process 6. Ans. (B,D) 7. Ans. (A,B,C) 8. Ans. (A,D) (klq)l > mg l sin q mgl kl q> q mg k > l 9. Ans. (B, D) At this point particle will lie on the y-ais and moing in horizontal direction.. Ans. (A) M HS-/

2 V. Ans. VP k constant 5% heat as work. dq dw + du HS-/ du dw f nrd f f dq nrd + nrd dq fnrd ncd C FR R as gas is monoatomic é C Rê - ë ( g-) ( -) ù ú û é ù ê ú R Rê - ú êæ 5 ( - - ) ú êç ëè úû k. Ans. 6 Let a be the length of the sting ension at any point is gien by M u ag cos ag a é ë + q- ù û... (i) Where M is the mass of the particle and as such W M M. g \ From Eq. (i), we get : W ( u + ag cos q- ag )...(ii) ag At the highest point mw (gien) and q p \ From Eq. (ii) W mw u ag ag ag é ë - - ù û mag u 5ag... (iii) At the lowest point nw (gien) and q \ From Eq. (ii) W nw u ag ag ag é ë + - ù û nag u + ag...(i) Eq. (iii) Eq. (i) mag nag 5ag ag m + 6 n ALL INDIA OPEN ES/JEE (Adanced)/8--5/PAPER-. Ans. Longitudinal speed ranserse speed y r Ar y r Ar ( - ) p p 6 - p. 68 m. Ans. Net loss k (temperature different between both sphere & surrounding) dq ms - k[ q-q s] dt qs q ò Cdq ( q-qs) q t ò ( ) kdt qs -C éëln q-q s ùû kt é-qsù ê C n ú l ê ú kt ê -qs ú ë û æ - Cln ç kt è Cln() kt C t ln( ) k 5. Ans. When the source (the train) moes towards the obserer, the apparent frequency is f' f -... (i) s When the source is moing away, the apparent frequency is f" f +... (ii) s f' + s \ f" - s Kota/C

3 æ ç + s or è 5 æ ç - ès or... (iii) s When the obserer moes with the train, there is no relatie motion between the source and the obserer and hence he will listen the true frequency f. Substituting (iii) in (i), we get or 6. Ans. f Kota/C f 66.6Hz F cos q mg mg Fcos q F sin q (rgr) pr rgr p rgr p mg tan q r p R g PAR- : CHEMISRY V F. Ans.(A, B, C, D) ;. Ans. (A,B,C). Ans.(A, B) ;. Ans. (A, B, D) 5. Ans. (B, C, D) ; 6. Ans. (A) 7. Ans. (B) ; 8. Ans. (B, C, D) 9. Ans. (A, B, C, D) ;. Ans. (A, B, D) q C F ALL INDIA OPEN ES/JEE (Adanced)/8--5/PAPER- 7. Ans. 8. Ans. d dq CAd dt d dq d CA dt ò ò d dq CA ln dt dq l CA ln dt ln/ l ln / / l æ ç è 9. Ans. 5. Ans. Speed just before collision a w a - ç w è Speed just after collision æa w a k a ' m k a 6m Q A. A,B,C,D A,B,C A,B A,B,D B,C,D A B B,C,D A,B,C,D A,B,D Q A SOLUION V. Ans.. Ans. H (g) + I (g) HI (g) at equib. ANSWER KEY after addition of H new equib. HS-/

4 ALL INDIA OPEN ES/JEE (Adanced)/8--5/PAPER- K. Ans. la P æ æ ç - - ç è è ( ). Ans. 5. Ans Ans. 7. Ans. 8. Ans Ans. 9. Ans. 6 PAR- : MAHEMAICS V P increases 5 times then l becomes 5 cm. Ans. (B,C) æ pù It is possible if m Î ç, ú è û. Ans. (A,B) ƒ() ƒ() ƒ(n) " n Î I \ we get atleast solutions in [ 5,5] and also if we get p solution in (,) then we get also p solution in (n,n+), nîi \ we get always l +, lîw solutions.. Ans. (A,B,C) a + b + c a & ab + bc + ca b & abc c...(i)...(ii)...(iii) on soling (i), (ii) & (iii) we get a a a b & b & b & one other solution c c c. Ans. (C) V n a n+ + S n n éë- + ( n+ ) ùû+ é- 8+ ( n-) ù ë û n -5n - V n is minimum at n or n Minimum alue of V n is 9. SOLUION 5. Ans. (A,C,D) (8 ) +.(8 ). + (8 ).() [8 + ] (5 9) / æ + ç - 8 è 8 / ƒ() ƒ () ƒ() [Q ƒ() is continuous and fn] + 8 ( ) ( + ) 6. Ans. (B,C) ,, Gien cure is circle - - ætan a-cot a p centre ç, è length of line L diameter - p æp æ ç + ç tan a- è è ép ù L Îê, pú ë û ANSWER KEY Q A. B,C A,B A,B,C C A,C,D B,C A,B,C,D A,C A,D B,D Q A HS-/ Kota/C

5 7. Ans. (A,B,C,D) Kota/C ìa ï íb ï î g ì/ a ï í b ï î/ g / Gien epression æ æ åç - ç - è a b è æ æ æ ç - ç - ç - a b g è è è a b g é ù 7-6ê a b g ú ab bg ag ë û æ æ æ ç - ç - ç - a b g è è è 8. Ans. (A,C) S º + y y 7 S º + y 8 y + 6 Intersection point of direct common tangent is ( 8, ) equation of line AB is + y 8 Let equation of S is ( + y y ) + l( + y 8) 6 Q AB is diameter l 5 required circle 5 + 5y 76y 8 equation of common chord is 6 + 8y æ8 PQ 9- ç è 5 9. Ans. (A,D) Gien ƒ(ƒ()) aƒ() + b cƒ() + d - d+ b ƒ() c-a (,) P Q...() ALL INDIA OPEN ES/JEE (Adanced)/8--5/PAPER- a+ b gien ƒ()...() c+ d from () & () a d...() ƒ(9) 9 8a + b 9 9c...() ƒ(97) 97 9a + b 97 97c...(5) from (5) () a 58c \ a Ans. (B,D) ƒ(a ) < ƒ(a ) < ƒ(a ) <... < ƒ(a ) it is equialent to distribution of objects to 5 persons such that each person get atleast one object. \ number of such mapings 99 C 9 99 C 5 V. Ans. 8 ƒ (g(5)) sin (sin5) sin (sin(p 5)) p - 5. Ans. ( m/,) Q(,) P(,) Equation of line PQ is chord of circle perpendicular distance < radius m- m - m m m > \ M 5. Ans. 7 y + + y y y y ++ +y +9+y 6 6y (y ) + ( + y ) y and + y + y ( + y) y 7. Ans. 5 b < <...() a p q & b - ac>...() & a b + c >...() from (), () & () we get minimum alue of a 5 HS-5/

6 5. Ans. 5 O r 9 B A B º 9 C C r r 9 B 9 C B 5 C in DAO B r O AB r...() sin B cos B AC r in DAO B...() sin C cosc from () & () AB.AC rr sin B.sin C cos BcosC RsinC.RsinB rr sinb.sinc æ 5 rr R ç 5 èsin 6. Ans. Let i sina i Gien 9 å i 9 å i i sin a -sin a ç è æ i i 9 9 å å sin a sin i i i 9 å ( a ) ( a ) sin æ 9 i 9 i å i i i ç å è i ma 7. Ans. (tan sin) + (cot cos) + 5 (sin- sincos+ cos) (cos- sincos+ sin) + cos sin i æ ( cos+ sin- sincos) ç + ècos sin sin + cos sin cos or tan- Case-I : tan- principal solutions ALL INDIA OPEN ES/JEE (Adanced)/8--5/PAPER- Case-II : sin + cos sin cos (sin + cos) (sin + cos) \ sin + cos + (rej.) or - sin + cos - æ p sinç + - è principal sol. \ otal solutions 8. Ans. 5 On subtract gien equations [(tan ) ( coty) ] + 5 [(tan ) ( coty)] [tan + coty ] [(tan ) + ( coty) + (tan )( coty) + 5] it is possible if tan + coty p p \ sinz + cosz z - p, -,,,p 9. Ans. ( ) ( ) + ( ) CéD.C C.D.ù ë û [6 +.] 8 D(n) means dearrangement of n objects.. Ans. Let r p + \ r n c p n p n p mp \ n P...() + m r m C p.( ) m p ( ) p ( n 6 ) p m P...() n - from () & () n (m+) r \ m reject for minimum alue m, then n 6 HS-6/ Kota/C

7 Path to Success PAR- : PHYSICS. Ans. (C) l w l l cm / sec k p p w A - ( ) - p 6 - p p cm / sec. Ans. (C) By mechanical energy conseration m m - mgl -mglcos q é ù g [ cos ] ê - - q ë ú l û g ( cos ) 8 l - q é 8gl ( -cosq) ù ê ú êë... (i) úû m At point A - mg l m 5mg - mg gl... (ii) l By soling equation (i) & (ii) gl ALLEN CAREER INSIUE KOA (RAJASHAN) ( - q) 8gl cos M ES YPE : MAJOR PAPER CODE C DISANCE LEARNING PROGRAMME (ACADEMIC SESSION -5) NURURE ES SERIES / JOIN PACKAGE COURSE ARGE : JEE (Adanced) 6 ALL INDIA JEE-ADVANCED OPEN ES # PAPER- SOLUION 8 ( -cos q) -cos q 8 - cosq q. Ans. (A). Ans. (B) Corporate Office : ALLEN CAREER INSIUE, SANKALP, CP-6, Indra Vihar, Kota (Rajasthan) dlp@allen.ac.in r r O q A dm rdv dm rpr dr I ( r p ) B R 5 r 5 ò r dr r r dr I ( p) òr 5r dr dz r dr dz 5 p r dz 5ò Area under cure I p 5 æ ç rr 5 è 5 Date : pr R ò r p r dr let r 5 z r r 5 pr 5 ANSWER KEY Q A. C C A B B A A D A B Q A. C A A D C B A C B B HS-7/ z

8 5. Ans. (B) HS-8/ ( -) ( + ) R AR I I A æ m - m æ - ç m + m ç + è è æ- I R I ç + è I R I I R I (at I R I 6. Ans. (A) 7. Ans. (A) F ædv ma ç è D æ V æmw L F µwl ç ç èh/ è H F mw L H 8. Ans. (D) Here sources are incohesent so I resultant I + I + I +... I I S æi S 6 log ç è I æ5i S log ç è I Sole (i) & (ii) 6 / 6- I I S I 5 I take log both side 6 - log.6 S...(i)...(ii) ALL INDIA OPEN ES/JEE (Adanced)/8--5/PAPER db 9. Ans. (A) s t s s t t t æ s s t + t ç + è ag s+ s s+ s t s + t s + diided by s in left so s + s s + s diided by in both side s s + + y + + y y + -- y - y + y y -y - y + y y s (gien) Kota/C

9 ALL INDIA OPEN ES/JEE (Adanced)/8--5/PAPER- + y y -y - y at A PV P V C > A DU C A (DQ) CA < Dw C A < y y h ABCA PV /8 Z D Q +DQ AB BC æ ç è / PV /8 h ADEA Q EA. Ans. (B) p p p p h p w Q ABCA net supplied E q A B D C V V V æp (A B) ç è const. DU > DQ > tanq P P P const. P P wd ABCA Area wd ADEA Area V P -P 8 æ P P ç è 8 nd Q AB nc V d æpv f f - PV i i ç è R æ R ç PV PV - ç R ç è.pv æ9 PV PV Q BC nc P d 5R ç - è R 5 æ5 ç è 8 Q EA DU + Dw PV R æ PV PV ç - è nr + æp V V P ç + è æ PV ç + + è 8 PV 8 h /+ 5/8 h /8 h h 7. Ans. (C) æ 9 At 'C' PV ç P V PV è B C L D Kota/C HS-9/

10 HS-/ Just after collision of first ball with dumb-bell Velocity of COM of dumb-bell V Ball B & C performe SHM in COM frame of dumbell. Velocity of ball D will be V if C coilied with elocity. Velocity of C will be if it is moing towards right with in COM frame. \ Minimum time after which C will be moing with \ min (right) in COM frame is m m t p p, k k L. Ans. (A) m t p k From momentum conseration, finally linear momentum of dumb-bell cm. Ans. (A). Ans. (D) 5. Ans. (C) w m ær ç w è p mr æ m ç p èr Rp 6. Ans. (B) y A A ædl æp DR ç ç è l è pr R p ALL INDIA OPEN ES/JEE (Adanced)/8--5/PAPER- D R R ya m D R pya 7. Ans. (A) () ma wl E ma w l m R Rp ya (w pf & f l ) l m wl m w l l 8 8 () E A A ( ) () ma w l E A A 8 6 (w pr & f l ) m wl+ m wl l l l + l l l m A wl 6 æl ma w l A ç 6 è m w (w pr & f l ) () E A A 8 m w l m w l (w pr & f l ) l l 8. Ans. (C) (P) (Q) w Angular momentum conseration ml ml wl - w 6 Angular momentum conseration Ml wl Ml- w 6 Kota/C

11 (R) (S) Angular momentum conseration Ml wl Ml- w 6 Angular momentum conseration M - w 9. Ans. (B) (P) (Q) l Ml wl A B e e.8 emissie power es area of A & B is same, so heat energy radiated will be more for which hae more alue of e. E l B A l on increasing temperature peak of graph shift f left. so B > A ALL INDIA OPEN ES/JEE (Adanced)/8--5/PAPER- (R) B hae more area under cure so emissie power is more but heat energy of A & B cannot be predicted untiel area of A & B is known. A B Direction & heat flow from A to B so A > B (S) () A B () At steady state dq dt is same for all but direction of heat flow from to so A > B. Ans. (B) r b < r l motion of ball is opposite to g eff PAR- : CHEMISRY. Ans. (C) w w / 58.5 M(NaCl) ;M(KCl) 7.5 V V M (NaCl) > M(KCl). Ans. (C). Ans. (D). Ans. (B) 5. Ans. (B) 6. Ans. (A) 7. Ans. (C) 8. Ans. (C) 9. Ans. (B) Kota/C. Ans. (A). Ans. (B) Q En > En n > n ( n -n)( n - n + ) 6 & n 6 & n.. Ans. (D). Ans. (B). Ans. (D) 5. Ans. (B) 6. Ans. (C) 7. Ans. (B) 8. Ans. (C) 9. Ans. (A). Ans. (A) ANSWER KEY Q A. C C D B B A C C B A Q A. B D B D B C B C A A SOLUION n (n - ) 5 HS-/

12 PAR- : MAHEMAICS. Ans. (A) HS-/ m / w / m w ( C. C ) + ( C. C ) + ( C. C ) + ( C. C ) Ans. (A) y tan + cot y min 7 6 tan tan tan cot. Ans. (D) [] + {} [] {} soling,. Ans. (C) ³ ƒ() k æ ƒç k è k æ ƒ() ƒç k è k k k 6 é ù ƒ() 79 ê ú ë û 5. Ans. (C) Let sin q p p - q sin - æ sin p ç q+ p +q è p p p -p p - +q q SOLUION ALL INDIA OPEN ES/JEE (Adanced)/8--5/PAPER- p p - q - 6. Ans. (C) Let w be circumcirle of P,Q,R,S. Let r i & O i be the radius & centre respectiely of circles w i, i,,. Now O lies on radical ais of w & w and O lies on radical ais of w & w. equating powers O O - r O O -r O O - r O O -r subtracting O O - r O O -r Power of O with respect to w & w same Locus of O is radical ais of w & w i.e. on + y - - tan ( ) tan ( ) 7. Ans. (B) a+b + a+b- tan 8 ƒ() ƒ () has solutions. 8. Ans. (B) Obiously one-one into. 9. Ans. (C) Required number of ways 6 ( + + ) (Principle of inclusion & eeclusion) ANSWER KEY Q A. A A D C C C B B C A Q A. B B C A D B A C B A Kota/C

13 9 q. Ans. (A) sin cos + sin + cos + sin + cos sin cos (Not possible) Paragraph for Question & ( + y) + (y + ) + ( + y) -(y + ) E ì ï + + ³ + E í ïî E ma ( + y, y + ) E > + y E > y + y; y y y + ; y + ³ + y æ æ E ³ ç + + ç y+ - è è E³- Equality holds for y. Ans. (B). Ans. (B). Ans. (C) Paragraph for Question & ALL INDIA OPEN ES/JEE (Adanced)/8--5/PAPER- 9 P(6) P(sin ) P(cos ) sin cos 5. Ans. (D) 6. Ans. (B) 7. Ans. (A) For (P,Q) - - tan(cos ) O p p or p - cos p - p cos or - - D ƒ : é ù é ù ê-, - È, ú ê ú ë û ë û R ƒ : [, ] - For (R,S) - ƒ() cos sin tan ( cos ) ( ( )) Kota/C r Q. Ans. (A) d P d cos q r Required locus is perpendicular bisector of PQ Paragraph for Question 5 & 6 P() C( ) + 8 C( ) + 8 C 9 C P() ( - ) + æ - - ƒ() cos sin ç è ƒ() - ƒ(ƒ()) - ƒ () HS-/

14 Now, ( -) ± Domain ƒ(ƒ()) < or ³ æ ù æ é ç -,- úèç -, È ê, ÇD è è û ë 8. Ans. (C) - ì ü + ( + ) å n n k n P n (-) í- ý k î k + þ k n + - k n+ (n+ ) (-) C - (-). k+ n+ - k n+ n n n- k n+ n-k å k å k k - - n n k n å - + ( ) Cn+- k n+ k + (-) - n+ 9. Ans. (B) n ƒ ALL INDIA OPEN ES/JEE (Adanced)/8--5/PAPER- tana tanb y. y - + y + cosq y sinq + y + (cosq + sinq). Ans. (A) a n (n + ) n a n+ n (n + ) an an+ - n n n+ ( ) a n n a- ç n+ è æ + n+ æn+ a 5 n n + - ç è n+ Now check P(,y) O (,) a y b A(,) HS-/ Kota/C