Q. No. PHYSICS CHEMISTRY MATHEMATICS
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1 AITS-CRT-I (Paper-)-PCM(Sol)-JEE(Advaced)/5 FIITJEE Studets From All Programs have bagged 4 i Top 00, 66 i Top 00 ad 74 i Top 500 All Idia Raks. FIITJEE Performace i JEE (Advaced), 04: 5 FIITJEE Studets from Classroom / Itegrated School Programs & 579 FIITJEE Studets from All Programs have qualified i JEE (Advaced), 04. FIITJEE ALL INDIA TEST SERIES ANSWERS, INTS & SLUTINS CRT I (Paper-) Q. No. PYSICS CEMISTRY MATEMATICS. A, D A,, C, D. C, D A,, C A,, C. A,, D A, C C, D 4. A, C, C A,, C, D 5. A, A, C A, 6. A, D A,, C, D 7. C, D, D 8. A, C A, D A, D 9. A, D, D A,, C 0., D A, A, D JEE(Advaced) FIITJEE Ltd., FIITJEE ouse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fa 6594
2 AITS-CRT-I (Paper-)-PCM(Sol)-JEE(Advaced)/5 Physics PART I SECTIN A. Usig cocept of relative velocity we ca solve v (0, 0) v A v (d, ) R mi v vj ˆ. Use v u as Ad impulse = chage i mometum gh 4. Use v I where, Mr 5. Use work eergy theorem 6. uoyat force F = V e g eff If lift accelerates g eff = g + a v v0 7. Use, app v v vs v = speed of soud 8. for isothermal process, PV = costat For adiabatic process, PV = costat 9. Electric field at cetre = 0 So, charge q will produce same electric field as iduced charge o sphere at cetre so A & D R d q 0. Usig coservatio liear momet ad coservatio mechaical eergy.. For traslatio positio of A will be.. formula for rod ad are both are k ˆ k si si i cos cos ˆj d d So, E 0 ad b = 0 SECTIN C FIITJEE Ltd., FIITJEE ouse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fa 6594
3 AITS-CRT-I (Paper-)-PCM(Sol)-JEE(Advaced)/5 6. work doe = P 0 V 0 P V P V U Q P V k = f = f C 0 fw f 000 z C 0 00 C v f fw z C 0 00 Frequecy ad width = f f = 00 z 0 m/s 0 m/s S 8. Fet Fpseudo Fuoyat V 0 v0 v 0 F v 0 v et a IL 50 A 00 IC 50 I I I 0 source L C m/s FIITJEE Ltd., FIITJEE ouse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fa 6594
4 AITS-CRT-I (Paper-)-PCM(Sol)-JEE(Advaced)/5 Chemistry 4 PART II SECTIN A. Dacro is a copolymer of ethylee glycol ad terephthalic acid. A A 4 A 4 A 4 Z Z Z Z A C D - emissio gives isodiapher.. From kietic theory of gases K. E. T M N Na C W 000 W 000 M C. m V m V 4 Na C W 000 N 000 N C. E V E V factor = -factio = o E redo chage will be egative for o-spotaeous reactio Ni aq Cu s Ni s Cu aq o o o Ni / Ni Cu / Cu E E E V ve, o spo ta eous Cu s aq Cu aq g o o o / Cu /Cu E E E V ve, o spo taeous 6. Rig Ep 7. i C Mg S 4 / CC C C C ii C C C C C C C 8. Compouds have all c atom sp hybridised are coplaar of N =.5, paramagetic N + =.0, diamagetic FIITJEE Ltd., FIITJEE ouse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fa 6594
5 5 AITS-CRT-I (Paper-)-PCM(Sol)-JEE(Advaced)/5 0. Cr, d cofiguratio shows same umber paired e ad EAN but stregth of all ligads are differet so does ot slow same colour ad splittig. SECTIN C. [Mabcd] type square plaar comple has G. I.. N Cl N Cl N N N N N S Cl Cl S C Cl C Cl. Cl Cl Cl Cl 5. Aromatic compouds meso Eatiomers, graphite N,,,,, N N N A layer of graphite (graphee) cotais ifiite lattice of fused aromatic rigs. All the valecies are satisfied (ecept at the edges) ad o bods are eeded betwee layers. (Plaar layered structure of fused aromatic rigs) M 5 mol S gm of S4 000 ml of solutio gm solutio wt. of solvet = = 65 gm moles of solute 5 m 8 mol/ kg wt. of solvet i kg P s,s,p,s p For s = 0, m = 0, e = 6 + For p, =, m = - 0 FIITJEE Ltd., FIITJEE ouse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fa 6594
6 AITS-CRT-I (Paper-)-PCM(Sol)-JEE(Advaced)/ e e Total e = = 9 8. PCl5 PCl Cl a 0 0 aa a a 0.50 P Kp Kp P 0.75 P K P P = K P 9. S S S 0. MCN 6 + 6( ) = = + M Ar d 4S 4 0 tg eg As CN is strog ligad hece due to splittig it will from low spi comple. FIITJEE Ltd., FIITJEE ouse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fa 6594
7 7 AITS-CRT-I (Paper-)-PCM(Sol)-JEE(Advaced)/5 Mathematics PART III Ay poit o y = is P t, t Normal at P is y t t t SECTIN A AC PC APC C ar br r a b A C s s a s s b s s c s cos cos cos bc ac ab abc A P 4. P(A A ) = 4, P(A ) = 4 P(A ) =. Thus P(A A ) = P(A ) P(A ) 5. z 4 z 4 0 which is a circle + y 4 y = 0 z i z i + yy ( + ) (y + y ) = y 4 y It passes through (0, 0), so, the locus of (, y ) is + y y = 0. So, z = + iy, lies o this circle for which the poits (, 0) ad (0, ) are etremities of a diameter. Also (0, 0) ad (, ) represet etremities of aother diameter 6. I e d cosl d e e 7. f(t) = t t + t + f(t) > 0 ma.{f(t): 0 t } = f() FIITJEE Ltd., FIITJEE ouse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fa 6594
8 AITS-CRT-I (Paper-)-PCM(Sol)-JEE(Advaced)/ Put si si t 9. E = + C + C = C + C +.. Addig ad subtractig, we get E + = ( + ) = 4, E = ( ) = Addig ad subtractig agai, we get E = 4 + E = d y y dy d y y dy d y y dy Put t SECTIN C. cos 4 4 A 7 p 4 7 ad cm CM A c M p 7 c C c. + = ( + i)( i) b =, a = c Number of ways of choosig a, b, c = 0 = 0 k =. Uit vector ˆ perpedicular to A ad C C A 7 ˆi 4j ˆ 5 kˆ ˆ C A Shortest distace = A ˆi ˆj kˆ As C A = 4. Replacig by Agai, replacig by, we obtai A ˆ f f FIITJEE Ltd., FIITJEE ouse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fa 6594
9 9 AITS-CRT-I (Paper-)-PCM(Sol)-JEE(Advaced)/5 5. We have a = 6 + = a a a ( ) = { + cos(a + b + c)} solutio eists whe L..S. = R.. S. = 0 = ad cos(a + b + c) = Whe =, = a + b + c =, a b + c = b = ad a + c = 00 i 7. 0i f i i 00 l f i 0 i l i i i i 00 i 0 f ' f i 8. For ay quadrilateral, a + b + c > d. Now, (a b) + (b c) + (c a) 0 a + b + c (ab + bc + ca) (a + b + c ) (ab + bc + ca) (a + b + c ) a + b + c + (ab + bc + ca) (a + b + c ) (a + b + c) (a + b + c ) > d (a b c ). d a b c The miimum value of d is 9. As P(, ( + ) ) lie o y solvig y with L, we get, 5.. () ad with lie L, we get P 5, 6.. () Equatio () ad (), we get 5, 5 L = 0 P P 5, 6 y + = 0 L = 0 0. f() = cos a cos cos + a 0 R si (cos + a) 0 a cos FIITJEE Ltd., FIITJEE ouse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , , Fa 6594
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AITS-PT-I (Paper-)-PCM-JEE(Advaced)/8 FIITJEE JEE(Advaced)-8 ANSWERS, HINTS & SOLUTIONS PART TEST I PAPER- ANSWERS KEY Q. No. PHYSICS Q. No. CHEMISTRY Q. No. MATHEMATICS ALL INDIA TEST SERIES. D. B 7.
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