ALL INDIA TEST SERIES

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1 AITS-FT-II-(Paper-) PCM(Sl)-JEE(Advanced)/5 FIITJEE Students Frm All Prgrams have bagged 4 in Tp 00, 66 in Tp 00 and 74 in Tp 500 All India anks. FIITJEE Perfrmance in JEE (Advanced), 04: 5 FIITJEE Students frm Classrm / Integrated Schl Prgrams & 579 FIITJEE Students frm All Prgrams have qualified in JEE (Advanced), 04. FIITJEE ALL INDIA TEST SEIES ANSWES, HINTS & SLUTINS FULL TEST II (PAPE-) Q. N. PHYSICS CHEMISTY MATHEMATICS. C B D. B C B D D D A A A A A A A A D 7. C C B 8. B, C, D B, C, D A, B, C, D A, B, C, D A, C A, C A, C, D A, B, C, D A, B A, B, C A, D A, B, C, D. A B C. B B C 4. A B D 5. C C C 6. D C B. (A) (r), (B) (q), (A q) (B s) (A) (s) (B) (p) (C) (p), (D) (p) (C r) (D p) (C) (r) (D) (t). (A) (p, s, t), (B) (q, r), (C) (A) (q), (B) (s), (C) (r), (D) (p) (A) (p) (B) (r) (C) (p) (D) (r) (q, r), (D) (p, s, t) JEE(Advanced)-05 7 n = FIITJEE Ltd., FIITJEE Huse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 6594

2 AITS-FT-II-(Paper-) PCM(Sl)-JEE(Advanced)/5 Physics PAT I SECTIN A. Electric field is tangential directin at all pints (E.rˆ 0) and decreasing with increase in r. (r = magnitude f radius vectr f a pint). Since, Electrstatic ptential is zer at all pints n axis, v S, alng axis 0 E r = 0 r. Va Vb Vc because electric field inside cnductr is zer. b rb rb ind ind ind, because ind V (E E ' E E ).dr (E ' E ' ).dr cnductr r utside the cnductr. 4. An infinite wire can be taken as half f clsed lp. Hence Bd 0 i 5. Let us cnsider a strip f thickness dx at a distance x E ' E ' 0 fr all pints inside. x d adx r a a dx eff eff d x a r aln r 6. VA Vind 0 q ind (d asin t) q K q K. ind 0 (dasint).q dqind qacst dt (d asint) 8. Frce applied by hinge n rd may be twards P, away frm P r zer depending upn distance frm pint. Accrdingly linear mmentum f rd + bullet system may decrease increase r remains cnstant respectively. Similarly angular mmentum f system abut Q may decrease, increase r remain cnstant respectively. 9. = m v max = A = 4 y / t 40 v 0m / s y / x 40 A = 0 v m als 0Hz and max a A 80 cm / s FIITJEE Ltd., FIITJEE Huse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 6594

3 AITS-FT-II-(Paper-) PCM(Sl)-JEE(Advanced)/5 0. Frm centre f mass f system ball and centre f tube mves in circular path. M adius f curvature f ball at the time f prjectin f ball is m M Nrmal frce between tube and ball if M = m, at the time f prjectin f ball is mv V cm = 0 m M V 0 V cm N m r. Equivalent circuit is D A C B 4. k k L k. k L 5.60Å k k L K K M L K 5. cut ff cut ff 4 8 hc hc ev V 0 V e SECTIN - B. Using Kirchf s first law at pint P SECTIN C. V = V P KQ KQ V Kq shell 5 KQ Vshell 5 KQ 4 5 KQ 0 9 KQ = 5 5. Using Bernulli s therem v v p gh r h.. () gh Put : V = r = r () in () 5. v u f FIITJEE Ltd., FIITJEE Huse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 6594

4 AITS-FT-II-(Paper-) PCM(Sl)-JEE(Advanced)/5 4 v ( / ) 4 v v = 4 v 4 m u 4 amplitude f image = 4 FIITJEE Ltd., FIITJEE Huse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 6594

5 5 AITS-FT-II-(Paper-) PCM(Sl)-JEE(Advanced)/5 Chemistry PAT II SECTIN A. Ag AgC s NaC aq. KBr aq. AgBr s Ags Ande : Ags C AgC s e Cathde : AgBr s e Ags Br aq. C aq. AgBr s Br aq. Ag C s Br cell cell C E E lg At eqm Br cell C 0 E lg 0 E lg cell C Br Br 0.059lg EC E C Ag/ AgCl/Cl Ag /Ag A E E lg K spagbr sp E E lg Ag/AgC /C Ag K sp AgBr Br K AgBr sp lg lg C K sp Ag C Br C 8 0. N N N ln ln 4 6 N 6 t Ne t e t ln ln ln t ln 4 6 t = 4.8 hurs K 7 Ag C FIITJEE Ltd., FIITJEE Huse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 6594

6 AITS-FT-II-(Paper-) PCM(Sl)-JEE(Advanced)/5 6. CH + CH CH H C CH CH 4. r 0.44 fr ctahedral vid fitted with ur red spherical balln. r r 0.44 r 0.44 Initial r radius = 0.6 Final r radius = 0.44 Initial r Finalr The number f maxima in the radial charge density curve is (n l) Fr d electrn, n = and d = Hence, n l = = There will be nly ne maximum in the curve fr d electrn. 6. In Perkin reactin, the carbanin attacks the carbnyl carbn f armatic aldehyde t prduce finally the unsaturated acid The electrn withdrawing grups increase reactivity f aldehydes in this reactin by increasing the electrn deficiency f carbnyl atm. Thus N increases the reactivity and the grups like N(CH ) and CH decreases the reactivity. 7. The rder f ligand strength in the spectrchemical series F H NH N. A strng ligand causes a larger degree f splitting resulting in high value f E (energy). Therefre crrespnding hc lw value f E Prtnatin ccurs n the H which can prduce mre stable carbcatin and while rearrangement phenyl has higher migratry aptitude ver -substituted phenyl. FIITJEE Ltd., FIITJEE Huse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 6594

7 7 AITS-FT-II-(Paper-) PCM(Sl)-JEE(Advanced)/5 Slutins fr Q. N. t Let the n. f mles f H, and He are x, y and z respectively x + y + z =.0 Pressure exerted by mles f gaseus mixture is 50 atm. After the first electric spark decrease in pressure = 50.5 = 7.5 atm decrease in n. f mles f gaseus mixture is.5 But frm the given infrmatin limiting reagent in first step is frm H g g H H and reacted are in the rati f : 0.5 mle f shuld be reacted with.0 mle f H N. f mle f in the mixture (y) = 0.5 When again is passed, pressure increased frm.5 t 5.0 atm Change in pressure =.5 atm N. f mles f added = mle 50 Nw H will be cmpletely reacted N. f mles H actually left after first electric spark = x H H x x Nw change in pressure in 5 0 = 5 atm change in n. f mles is x x 0.6 x.4 n. f mles f H =.4 n. f mles f He =.0 ( ) = 0. mle fractin f H =.4.0 = Pssible structure after test (a) wuld be that f a r a alchl and nt that f alchl. 5. nly nn-reslvable cmpund is a alchl ut f the given chices. 6. xidatin f cmpund with Cr in pyridine gives a cmpund, that gives Tllen s test. This implies that it is a primary alchl. SECTIN-B. Silver re is disslved by cmplex frmatin using CN in presence f air fllwed by precipitatin with zinc. Lead Invlves rasting f PbS fllwed by self reductin. Irn Invlves calcinatin fllwed by carbnmnxide reductin. Magnesium Invlves calcinatin (f MgCl.6H in presence f dry HCl gas) fllwed by fused salt electrlysis. SECTIN C FIITJEE Ltd., FIITJEE Huse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 6594

8 AITS-FT-II-(Paper-) PCM(Sl)-JEE(Advanced)/5 8. CCH H C P = Br.9. n n ml = 0.06 ml. M 60 n hydrazinehydrate = ml = ml. 50 n eqki = 4 N 0.48L = 0.9 n eqhydrazine reacted with KI 0 Br n hydrazine left after reactin with M n+ = 0.9 ml = ml 4 n hydrazine reacted with M n+ = ( ) ml = ml n M 0.06 n n hydrazine 4. F T T T T F F F T T F T T T (IX) A is a buffer slutin; ph f A = pk a = pk w pk a lgc = 8 B is an aqueus slutin f SacNa; ph f B = 5. Ttal wrk dne = Area f circle A (anticlckwise) Area f circle B (clckwise) = 40 6 = 4 unit heat flws int the system alng the path ADB, = 5 + = 4 unit 6. CCl CH(H) + NaH CHCl + HC Na + + H Chlral hydrate (A) CHCl + h CCl + HCl (A) (B) Cl =C + HC H 5 Cl (phsgene) H 5 C C C H 5 + CH MgBr C H 5 =C + HCl C H 5 (diethyl carbnate) (C) H 5 C C C H 5 CH C H 5 H 5 C C CH CH MgCl H CH C CH CH CH MgCl H + CH C CH FIITJEE Ltd., FIITJEE Huse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 6594

9 9 AITS-FT-II-(Paper-) PCM(Sl)-JEE(Advanced)/5 Mathematics PAT III SECTIN A. Pssibilities: (i) three digit ccur twice = 5 6! C 900 ways!!! (ii) tw digit ccur three times = 00 ways (iii) ne digit ccur fur times and ther twice = 00 ways (iv) all digits are same = 5 Ttal pssibilities = 405. z i = represent a circle with centre (0, ), radius unit and z i + z + i = represent an ellipse they tuch each ther at (0, 4). Let n n n r r0 f n C f(9) = 0 =.. Ttal divisrs are = 8 00 x a0 ax ax... a00x cs x cs x... cs x 00 eplacing x by x and adding x x x a0 ax... a00x cs cs Put x = a k = i0 i 5. The desired cnditin is t have the circle N inside the cnvex regin f the parabla y = x clearly a, therwise circle will crss x-axis als fr circle t be inside the parabla the number f pints f intersectin must be less than r equal t Equatin y + (y a) =, has ne r less than ne slutin D 0 a I N = e e x lg sin x dx x lg sin x dx 0 0 = x lge dx x lge sin x dx = D N D 0 0 I I 7. T = A B T I A B FIITJEE Ltd., FIITJEE Huse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 6594

10 AITS-FT-II-(Paper-) PCM(Sl)-JEE(Advanced)/ Let ABCDPQS be a cube, the numbers a, b, c, d, e, f be written n the faces ade + abe + abf + adf + cde + bce + bcf + cdf = 004 (e + f)(a + c)(b + d) = 67 Pssible cmbinatin are 4 67, 6 67, 4, 50 T = a + b + c + d + e + f = 74, 75, 9, r If x 0, x + (a 5)x + = 0 x < 0, x + (a + )x + = 0 Equatin has exactly three distinct slutin if () Either (a 5) > 4 and (a + ) = 4 r (a 5) = 4 and (a + ) > 4 0. DC = HEA = GFB sina cs A sinccsccs A sin A sinb sinc First tw equality gives, tan A = 4 sin C cs C Frm first and last, cs A = sin C ( + 4 cs C) cs A = sin C ( + 4 cs C) sinc r. Scre f 5 can be btained as 5 r (, 4) r (,, ) r (,,, ) P (scring 5) = similarly P (scring 6) = P (scring 7) = P (scring ) is impssible event.-. Let M = d MDC ~ AMB AK + K = A = D = DL + L k x sin + (kx cs d) = x sin + (x cs + d) (k )x = xd (k + ) cs as k + > 0 (k )x = d cs Nw, AB CD = (AK LD) = (k )x sin = d sin If AMB = 60º, = 0º A D x kx L M K C x kx B If AMD = 60º, = 60º in either case AB CD = sin a, c, b are in H.P. c a b Line x y 0 passes thrugh a fixed pint a b c S centre f ellipse is, Which is als centre f auxiliary circle p, q radius f auxiliary circle =, FIITJEE Ltd., FIITJEE Huse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 6594

11 AITS-FT-II-(Paper-) PCM(Sl)-JEE(Advanced)/5 Length f majr axis = 4 als it segment f fcal chrd are, 4 8 b b 8 Directr circle x y 4 SECTIN B. (A) 0, b, c are in H.P. 0 c b This can be written as (40 b)( c + 0) = 800 we must have 0 < b < 40 r 0 < 40 b < 0, cnsider factrisatin f 800 in which ne term is less that 0 (40 b)(c + 0) = 800 = 800 = 400 = 4 00 = 5 60 = 8 00 = 0 80 = 6 50 Thus pair (b, c) = (9, 780), (8, 80), (6, 80), (5, 40), (, 80), (0, 60), (4, 0) but b divides c nly 5 triples satisfy required requirement (B) Slutins f m + n + p = 0, are 0 C (C) ab + b a + = 0 (a + )(b ) = (a +, b ) {(, ), (, ), (, ), (, )} (a, b) {(0, ), (, ), (, 0), (, ) (D) Let z = x + iy then (x + iy) 0 = x iy z 0 = z z ( z 9 ) = 0 z = r 0 z = 0 rdered pair (0, 0) Let z 0 then (x + iy) 0 (x + iy) = x + y z = z = z = 0, which are rts. (A) f(x) = x 4 + ax + bx + cx + d,,, are real rts f equatin f(x) = 0 f(x) = (x )(x )(x )(x ) f(i) = ( + )( + )( + )( + ) = this is pssible if = = = = 0 a = b = c = d = 0 (B) x 5 = 0 (x )(x 4 + x + x + x + ) = x x x x x x x x x x x x x x x x x x x x x x x x x x x x = = = x x x x x x 4 6 x x x (C) cs x 0 cs x and cs x cs x f(x) = 0 x x f x FIITJEE Ltd., FIITJEE Huse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 6594

12 AITS-FT-II-(Paper-) PCM(Sl)-JEE(Advanced)/5 (D) Let AD = x, BAD = In ABD, BD sin sin = 60º 4 x BD Using csine rule cs 60º x 5 BD a, b. f(9) = f(4 + 5) = f(4.5) = f(0) = f(6 + 4) = f(6.4) = f(64) = f(8.8) = f(8 + 8) = f(6) = f(4.4) = f(4 + 4) = f(8) SECTIN C. g(x) = f(x) f(x).. () g(x) = f(x) f(x).. () F(x) = (f(x)f(x) + g(x)g(x))(f ( x) + g ( x)) + (f (x) + g (x))( f( x)f( x) g( x)g( x)).. () Using () and () in equatin (), we get F(x) = 0 F(x) is cnstant functin F(0) = = F(). f(x) = tan (tan x) 5 f r f f f f 4 f 5 = = 5 5 r a = 5, b = 5 4. Minimum area will be in case f square with vertices,,,,, and, Minimum area = 8s r r 5. rx r x r0 r x... x x... x = x... x 99 x x x Equating cefficient f x 00, x 0,.., x 00 respectively we get 00 = 0 C 0, 0 = 0 C 0,.., 00 = 0 C 0 00 A... r00 r When 00 is divided by 5 we get remainder 6. Take B as the rigin, BC and BA as x and y axis and z axis perpendicular t the plane f ABC BA 6j ˆ, CD 6 ˆi ˆj kˆ n BA CD = k ˆ i ˆ i ˆ k ˆ n The minimum distance d = BC n A x D y 0, 6, 0 A D,, B(0) C 6, 0, 0 x B C FIITJEE Ltd., FIITJEE Huse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fax 6594