ALL INDIA TEST SERIES

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1 AITS-FT-IV-(Pper-)-PCM(S)-JEE(Advnced)/4 From Clssroom/Integrted School Progrms 7 in Top 0, 3 in Top 00, 54 in Top 300, 06 in Top 500 All Indi Rnks & 34 Students from Clssroom /Integrted School Progrms & 373 Students from All Progrms hve been Awrded Rnk in JEE (Advnced), 03 FIITJEE JEE(Advnced), 04 ALL INDIA TEST SERIES ANSWERS, HINTS & SOLUTIONS FULL TEST IV (Pper-) Q. No. PHYSICS CHEMISTRY MATHEMATICS. C A B. C B C 3. C B B 4. D C B 5. B A B 6. A C D 7. D B D 8. B B D 9. A, B, C, D A, B A, C, D 0. A, C A, B A, B, C, D. A, C, D A, D A, B. A, C A, B, C A, B, D 3. C A B 4. C D B 5. A A C 6. A B B 7. B A C 8. A A A.. (A) (q),(b) (p, q, s, t), (C) (p, q, s, t), (D) (r, s) (A) (q, r), (B) (s), (C) (p, s, t), (D) (p, q, r, t) A (p, q) B (p, q) C (q) D (r, t) A (p, q, s, t) B (p, r, s) C (q, s) D (q, s, t) (A) (p), (B) (p, r) (C) (r), (D) (p, r) (A) (q), (B) (p) (C) (t), (D) (r) FIITJEE Ltd., FIITJEE House, 9-A, Klu Sri, Srvpriy Vihr, New Delhi -006, Ph , , F 65394

2 AITS-FT-IV-(Pper-)-PCM(S)-JEE(Advnced)/4 Physics PART I d. Prticle vel = 0 n cos nt dt m prticle vel = 0 n putting cos term = wve vel = for 0 n = 4n 0 n T. mv mg sin 60 mv0 mg sin30 3 v 0 = m/s 4. = LC eq = 00 T = /50 sec 5. sin C = / cos C = ˆ ˆ n.p = E n = 3.4 ev En n E 3.6 ev n = Angulr momentum = nh h h 34.0 Js 7. v = (A ) = 3 v = 3 = rd/sec = sin(t + ), v = cos (t + ) At t = 0, = & v = +ve = sin = n + () n 6 6 = 5 6 v t = /6 = + ve nd v t = 5/6 = ve So, = /6 = sin (t + ) = [sin t cos + cost sin] = ( 3 sin t + cos t) FIITJEE Ltd., FIITJEE House, 9-A, Klu Sri, Srvpriy Vihr, New Delhi -006, Ph , , F 65394

3 3 AITS-FT-IV-(Pper-)-PCM(S)-JEE(Advnced)/4 8. For polytropic process R U Q R R For polytomic gs, = 4/3 U Q 3 This is m t = 3/ 9. Use the concept of cpcitor. 0. Are in (v L -t)grph = LI (i f 0) = 0 i f = 5A 3-5. For long time cpcitor gets full chrged nd chrge on ech cpcitor must be sme. Q Q 0 3 Q = 4 C 0 50 rd/s LC 6 0 Q(t) = 4 sin t C Q(t) = (4 sin t) C 500t = n n t = n n E m = Q C 0 = Joule = 0.44 mj 6. Q = Q AB + Q BC + Q CA Q = 5U 0 + 3U 0 + 0U0 ln W AB = Q AB U AB = 5U 0 (3U 0 ) = U 0 8. Process AB U = constnt P RT nd U t M P = const Process BC isochoric Process CA isotherml FIITJEE Ltd., FIITJEE House, 9-A, Klu Sri, Srvpriy Vihr, New Delhi -006, Ph , , F 65394

4 AITS-FT-IV-(Pper-)-PCM(S)-JEE(Advnced)/4 4 SECTION - B. I = 6m 36 m4 + m = 4 m for p, q, s & t v0 p v ˆ j, 4m = mv 0 ˆk + mv 0 ˆk v = 0 ˆk 9 6 q v 0, 4m = mv 0 ˆk + mv 0 ˆk v = 0 ˆk 4 v s v 0 ˆj, 4m = mv 0 ˆk + mv 0 ( ˆk ) = 0 3 4v t v 0 ˆj, 4m = 4mv 0 ˆk + mv 0 ( ˆk ) v = 0 ˆk 9 r v 0, (8 m + m +m4 ) = 4mv 0 ˆk + mv 0 ˆk 6v0 = ˆk 5 FIITJEE Ltd., FIITJEE House, 9-A, Klu Sri, Srvpriy Vihr, New Delhi -006, Ph , , F 65394

5 5 AITS-FT-IV-(Pper-)-PCM(S)-JEE(Advnced)/4 Chemistry PART II SECTION - A. Due to non vilbility of d-orbitls, boron is unble to epnd its octet. Therefore it cnnot etend its covlency more thn BCO 3 + HCl BCl + H O + CO ZnS + HCl ZnCl + H S 4. Ag CO 3 55 gm Ag CO 3 43 g of Ag 4 Ag + CO + O.76 g Ag CO g NO in iron comple hs + oidtion number Thus, 6. It is clled Zinc blend 7. From left to right in period cidic chrcter increses due to increse in electronegtivity. H C H C NH CH Roult s lw for idel solutions cn be represented in the bove two given wys.. Hund s rule.. Addition of Br in CCl 4 is nti ddition. 3. In the originl compound Z toms form CCP X toms present in tetrhedrl void. rx rz 00 Y toms present in octhedrl void FIITJEE Ltd., FIITJEE House, 9-A, Klu Sri, Srvpriy Vihr, New Delhi -006, Ph , , F 65394

6 AITS-FT-IV-(Pper-)-PCM(S)-JEE(Advnced)/4 6 ry rz 00 Assuming, ll voids re occupied, formul of the originl compound = X 8 Y 4 Z 4. On removing toms long one body digonl, We lose Z toms, X toms, Y tom (body centre) The new formul becomes X 6 Y 3 Z Simplest formul X 8 Y 4 Z 5 4. On removing toms long nother body digonl, we lose Z toms X toms The new formul becomes = X 4 Y 3 Z 3.5 Simplest formul = X 8 Y 6 Z 7 Bse 6. ph = pk + log Slt [Bse] = Let milli moles of (NH 4 ) SO 4 [NH 4 + ] = ; 500 ph = log 0.0 / = log = [Slt] = [NH 4 + ] Moles of (NH 4 ) SO 4 dded = Solution (I) CH3COON HCl CH3COOH NCl No.of moles CH3COOH Volume of solution K H C C K.C K C i.e. ph = log K pk Solution (II) CH3COON CH3COOH FIITJEE Ltd., FIITJEE House, 9-A, Klu Sri, Srvpriy Vihr, New Delhi -006, Ph , , F 65394

7 7 AITS-FT-IV-(Pper-)-PCM(S)-JEE(Advnced)/4 It is buffer. ph = pk + log CH 3COON CH3COOH ph = pk + log ph = pk ph ph 8. HA BOH BA HO t t eq (.) 0.. Meq of HA = Meq of BOH t end point A = =.66 Meq of HA left =.66. Now, ph = pk + log Slt Acid left. 5 = log K + log.66. K = FIITJEE Ltd., FIITJEE House, 9-A, Klu Sri, Srvpriy Vihr, New Delhi -006, Ph , , F 65394

8 AITS-FT-IV-(Pper-)-PCM(S)-JEE(Advnced)/4 8 Mthemtics PART III SECTION A. We cn ssume tht OP nd OR re is nd y is respectively. Let OP =, then r(sq. OPQR) = coordintes of M nd N re, nd, respectively 3 r OMN 8 8 k 3 6 k = 6 R O y N P Q M. Eqution of the two circles be ( r) + (y r) = r i.e. + y r ry + r = 0 where r = r nd r. Condition of orthogonlity gives r r r r r r 4r r r r Circle psses through (, b) + b r rb + r = 0 i.e. r r( + b) + + b = 0 r + r = ( + b) nd r r = + b 4( + b ) = 4( + b) ( + b ) i.e. 4b + b = 0 3. f() is decresing function nd for mjor is to be is f(k + k + 5) > f(k + ) k + k + 5 < k + k ( 3, ) 4. f() = sin + 4 cos ( + ) sin sin + cos ( + ) = sin + cos ( + ) + cos ( + ) cos ( ) cos ( + ) = sin + cos ( + ) + cos sin cos ( + ) = cos f f cos sin 4 n n n 5. Consider the function 0 n f... n n n n Then f(0) = 0 nd f() = 0 Hence f() = 0 hs t lest one solution in (0, ) 6. fog e t I e d dt t (where e t ) = t tn t + C = A + B = + ( ) = 0 e tn e C fog tn fog C FIITJEE Ltd., FIITJEE House, 9-A, Klu Sri, Srvpriy Vihr, New Delhi -006, Ph , , F 65394

9 9 AITS-FT-IV-(Pper-)-PCM(S)-JEE(Advnced)/4 7. Are = = n tn d n d tn d 5 n n 5 tn 4 8. Let + y = v 9. I f 4 5 n Let g 4 5 n, then 0 < g < n n n n n n n n I f C 4 C 4 5 C 4 5 C n n n n n n n n g C 4 C 4 5 C 4 5 C n n n n I f g C 4 C = even integer 0 < f + g < f + g = 0 f = g thus I is n odd integer f = g = 4 5 (I + f)( f) = (I + f) g = 3 0. (A) P(E ) = P(R R R) = (B) P(E ) = 3P(B R R) = PR R R (C) P(E ) = P(R R R/R R R B B B) = PR R R PB B B But PB B B P(E 3 ) (D) P(E 4 ) P B B B 5 5 n. z i Re (z) = z lm (z) Let z = + iy, then + iy i = + iy y i.e. + (y ) = ( y) + y i.e. = y i.e. y = o 3 4 o Since,, 3, 4 re linerly independent = = 0, + = 0, + + = 0, + = 0 i.e. =, =, + + = 0, + = 0 i.e. =,,, i.e i 3 e 3 3 FIITJEE Ltd., FIITJEE House, 9-A, Klu Sri, Srvpriy Vihr, New Delhi -006, Ph , , F 65394

10 AITS-FT-IV-(Pper-)-PCM(S)-JEE(Advnced)/ Since i is rel is purely imginry 5. i e i i i 6. Eqution of the second plne is + y 3z + 5 = 0 ( ) + 3, + ( 4)( 3) > 0 origin lies in obtuse ngle ( + 3( ) )( + ( ) ) = ( 6 + 7)( ) > 0 P lies in obtuse ngle < 0 Origin lies in cute ngle Also ( + ( ) 3() + 5)( ) = ( )(0) < 0 P lies in obtuse ngle < 0 Origin lies in cute ngle Further ( )( ) > 0 The point P lies in cute ngle SECTION B sinb b c b. (A) Since cos A, we hve sinc bc c or b + c = b or c = Hence c = nd so the ABC is isosceles (B) cos A (sin B sin C) + (sin B sin C) = 0 or cos A (sin B sin C) + sin (B C) cos (B + C) = 0 or cos A (sin B sin C) cos A sin (B C) = 0 either cos A = 0 A = 90º or (sin B sin C) (sin B cos C cos B sin C) = 0 b c c b b C b c b c or (b c) (b c ) = 0 (b c)[ (b + c)] = 0 b c = 0 or b = c Isosceles (C) Combine first nd thirl nd put the vlue of cos B c b b b c b c bc or 4b + c + b = + b b + c + A = 90º FIITJEE Ltd., FIITJEE House, 9-A, Klu Sri, Srvpriy Vihr, New Delhi -006, Ph , , F 65394

11 AITS-FT-IV-(Pper-)-PCM(S)-JEE(Advnced)/4 sin A B k sin A sin B (D) by sine formul sina B k sin A sin B sina B sina BsinA B or sinc sin A sin B or sinc sin A B 0 sinc sin A sin B Either sin (A B) = 0 A = B i.e. is isosceles or sin A + sin B = sin C or + b = c is right ngled. (A) (3k ) + k 3k 0 [ (k )][ (k + )] 0 k + k+ + f() 0 f( ) 0 Hence k + nd k i.e. k 0 nd k k [0, ] + m = (B) (AM GM) (C) b b c c d d ( + )( + b)( + c)( + d) 6 bcd 6 Minimum vlue = (0, ) (D) f() = nd g() is inverse of f() f() = 5 g(5) = g' f f ' 3 3 g' f 3 3 g' 5 6 FIITJEE Ltd., FIITJEE House, 9-A, Klu Sri, Srvpriy Vihr, New Delhi -006, Ph , , F 65394

MATHEMATICS PART A. 1. ABC is a triangle, right angled at A. The resultant of the forces acting along AB, AC

MATHEMATICS PART A. 1. ABC is a triangle, right angled at A. The resultant of the forces acting along AB, AC FIITJEE Solutions to AIEEE MATHEMATICS PART A. ABC is tringle, right ngled t A. The resultnt of the forces cting long AB, AC with mgnitudes AB nd respectively is the force long AD, where D is the AC foot