2014-11-16 (5)

7AIlS-P! -I (Paper-2)-PCM(Sol)-JEE(Advanced)/15

Mat1aematics PART -: IIISECTION-A

L

1.

2.

3.

4.

5.

6.

7.

8.

f-! (8x) _f-l (x). f-1(x)lim-~~-----.'~ = fim-- (Taking 8x = t in first part)x ... 1X) X1/3 X113 x ... oo X1/3 '

(leW1 (x) = t):) lim t 1/3 =.!t_ (f(t)) 2

2-" < 2-" [2" ] :s; 1:) [2-" [2" ]] = 0 and whenever

2-" [2"] = 1[2"]=2"

Also if x E (-2,2), then 1:s;2" < 16

2 2 2 2

J f(x)[x]dx= J f(x)dx = J f(x)x- J f'(x)xdxo 1 ! 1

1 22f(2)- f(l)-- J(t'(x) x + f'(3 - x)(3 - x)):lx2 12f(2) _ f(l) _% (f(2) _ f(l)) = f(l) ~ f(2)

Maximum occurs when figure is cut along diagonal and minimum when we cut it parallel tosmaller side

y = g(x) is inverse of f(x)

f'(x) + f"(x) + f"'(x)g(x) = J1 . dx

f(x) + f'(x) + f"(x) + f"'(x)= x -Injf(x) + f'(x) + f"(x) + f"'(x)1+c=x-3Inlx!+c

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8AITS-PT -I (Paper-2)-PCM(Sol)-JEE(Advanced)/15

9.(x3+3x)-Ix3 _x2 +2x-21 Ix2+x+21

,,; -1x2+x+2 x2+x+2

10.(iX4)(ix5) (!) (!)

lim x.1 x., _ 5' 6 -~=>1=27n_ (n )( n ) ( 1 )( 1 ) 5't;x' t;x9-' 1+1 .10-1

11. (f-'g-' (x)= (gf(x)f'

d ( -, -1 ( )1 1dx f 9 X x.12= -d----dx (gof(x)lx.1

12. Area willbe same as bounded bYY=f(X), y=g(x)1 1

=>J 1x3- ~x = -_, 2

. dA d2A (dA dB)13.-14. -=2(F-A-B)=>-=2 ----dl dl2 dl dl

dB= !(A)dl 2After solvingwe get A =2F le-', B=F(1- le-' - e-' )(F beinglheinilialfoodquanlily)

16. y=g(x)

_ - _ ..- .._ _.._ ..- - ..- y = 1/2-'

(-2.5,0)(-2,0) (-1,0)(-.5,0) (1,0) (1.5,0)

1 1/2 1/2 (1 ) (1 )17. (P) Jln(1-x)lnxdx=2fIn(1-x)lnxdx=2fIn --x In -+x dx.o 0 0 2 2

,/2 ( e) ( e) e e ( 1)=2.[ In sin2" In cos2" sin2"cos2"de Bylakingx= 2"cose

1/.fi=8 f xlnxln(1-x2~x. 0

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LIMITED

I ALL INDIA TEST SERIES"",

JEE (Advanced), 2015PART TEST - I ( PAPER"- 1 & 2 )

(16 November, 2014)

PHrsIcs ,CBEJfIsTRY&MATHEMATICS(OBJEcTIvE)

-ANSWERS & SOLUTIONS

1AITS-PT -I (Paper-1 )-PCM(Sol).JEE(Advanced)/15

FIITJEE .JEEEAdvanced].2015ANSWERS, HINTS & SOLUTIONS

PART TEST-I[Paper. 11

:~IIIII!f]~Irllll'jlll!I!IIIII.III~J'IIIIIIIII!!jllllln'.I.IIJ~:~t~!11~1!lii~lllilllltil~lillll1 A, C

.A, B,e, 0 B,e,0

2. A, e, 0 A, B,e A, B ,e3 B,e A, B,e B , e4 A, e B,e A, B ,e5 A, 0 A, B,0 e,06 A, B,0 A, B,0 e, 07. A, e A, B,0 A, B8. A, B,0 A, B,e,0 A, B ,09. A, e A, B ,0 A, B,01O. A, 0 A, e, 0 B ,e1 2 1 2

2. 5 9 23 2 1 2

4. 1 7 3

.s. 5 7 46. 2 4 5

.

7 3 5 3

8. 7 8 0

9. 2 2 1

10 7 2 4

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I

I

2AilS-PI-I (Paper-1}-PCM(Sol~ EE(Advanced)/15

Physics PART-I

2.

4.

SECTION-AFor different pivoted ends, moment of inertia is different torque is same and work by torque issame

3

3AITS-PT-I (Paper-1 )-PCM(Sol)-JEE(Advanced)/15

5.

6.

F=(10)(10)Z x 3xO.82 8 .

= 5 x 100 x 0.3 = 150 N .

dy dx-=18x-=6xdt dt

dZy dx_ Za =-=6--2m!sY de dt

ax= 0

50 F

Chemistry . PART-II

SECTION-A1. (A) .Radial wave function contains information about n and f.

(B) R('l =:-. 1 [~]31'(2_~) e -,.~('.0) ,[a ao ao

2.High pressure region P

p/Idealgas

3.

Low pressure region P E

V~

At low pressure v, < v;, attractive forces dominate.At high pressure v, > Vi, repulsive forces dominate.

vr PVrZ=-=--v, nRT

d[A] d[B] d[e](A) --=-+-

dt dt dt(B) K[A] = K,[A] + Kz[A)

Where K is overall rate constant.K = K, + Kz

dK dK, dK,or-=-+-

dT dT dT

A -:; [ Ea ] = A -;'; [Ea,] A -* [Ea, ]or e. RT' ,e RT' +,e RT'

Real gas

(Vr ~ real volumev;-+ ideal volume)

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r 4AITS-PT -I (Paper-1 )-PCM(Sol)-JEE(Advanced)/15or K.Ea = K,Ea, + K2Ea2

. K,E. +K,E. x,y, +J

I 5AllS-Pl -I (Paper-1 )-PCM(Sol)-JEE(Advaneed)/15

. (B) K (600 K) = 25x25 = 25p 175. 7

K (900K)= 60x60 =15p 240

Kp (600 K) 25 15Kp(900 K) - 7x15 = 21

K, AH[ 1 1 ](C) In-=- ---K, R T,T,

21 AH[ 1 1]. In 5"=R 600 - 900AH= 1800R in 4.2

8. (B)C(s)+H,O(g)s 1270KN; lCO(g)+H,(g)Water gas

(0) At cathode2W +2e- ~H2(g)At

6Alrs-pr -I (Paper-1I'PCM(Sol~EE(Adyanced)115

(106 + 18y) x 2.5 x 10-3= 0.715

106 + 18y = 286 so that ratio of ~ = 1Y

2.

y = 10

Oxide (A) M20x2 _ w(m). w(o)x - A(m)" A(O)2 0.7745. 0.2255x A(m)' 16

2 0.7745 x 16x = 0.2255 x A(m)

54.95A(m)

3.

4.

2 _ w(m). w(o)5- A(m)' A(o)2 0.4952 0.5048 15.695Y = A(m): 16 A(m)So that '1 = 54.95 = 3.5

x 15.695So that '1 = !..

x 2A(m) = 54.95Metal M is MnOxide A 4 MnO (green colour crystal)Oxide B 4 Mn20, (dark green):. x + y = 9

Y,-=21lr,

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~------ ---------

7AllS-Pl -I (Paper-1 )-PCM(Sol)-JEE(Advanced)/15

5. (i}CCI,+2H,0----+no readion(ii)Except BF3 all boron halide are readily hydrolysed.

6. Pb(OCOCHJ>" Pb02, TlCIO, and 1l20, ad as oxidizingagent.

9. Beryl is cyclic silicate, two oxygen atoms per tetrahedron are shared to fonn closed ring.

Mathematics PART-III

SECTION-A1.f'(x) s0

=>f(x)f'(x)+f'(x)f"(x) sO

d~ ((f(X))2 +(f'(X)n S 0

=>(t2(x))+(f'(X))2 Sf2(0)+(f(0))2 \fx;:,O

=> f(x) =O\fx" 0

2. Period cannot be rational number as solution is unique.

3. Because of symmetry of f(x)& g(x), f(7.) = f(P) &g(7.) = g(P)p p pf f(x)g'(x)dx = I~f(x)g(x) - ff'(x)9(X)dX = -f f'(x)g(x)dx~ ~ ~

4. r(x) = 2x + g'(1} (1)rW=2 ~g'(x) = 2x + f'(2) (3)g"(x) = 2 (4)On putting x = 1, in equation (1) and (3),we get option (A)By taking x = 2, in equation (1) and (3), we get option (B)

5. If(x)-f(y)1 S 2 =>f(x) is bounded and for y = x +2nll, n e Iwe getIf(x)-f(x +2nll)1sO=> f(x) = f(x +2nll)

6. f(x) = 1(f(t)-f(a) I'(a) Jdta (t_a)2 (t-a)

f'(x) f(x)-f\a) f'(a) o(x-a) (x-a)

f(x)-f(a)Using LMVTwe get(x_a)2

f'(a) f"(a)=--(x-a) 2

mJ.' Uc:f.,)"arnB Bena 29

8A1TS-PT-I (Paper-1)-PCM(SoI~EE(Advanced)/15

7. Ar(OADE) 3x => 2'+1> 2' +3x

=> (x -2) +21092(2' +3X) < (x -2)+21092 2'+1=> 2' < 3x

Contradiction h'encesol. lies on2' = 3x , which has two solution

i'l',o)

2.1

LetlxJ=n+f (neN,0';f

9AITS-PT.I (Paper-1 )-PCM(Sol)-JEE(Advanced)/15

2".11(" k k 2)1 ( n 2 ( 1)") 2".1Iim- 3~)-1)-+- = lim (-1)"-+- -- - =2"-n+1 k.1 ~ 3 "~~ 2" 3 2 n+1

4. f2014(X)

1AITS.PT.I (Paper.2)-PCM(Sol)-JEE(Advanced)/15

lllll~~ltl.I't'~II~llIJI~lrjl~~11~lllilll:I;I!lil~liiilllllllll!lilllll~llll'lllll1 A e A2 e A A3 B D B4 e D A5 D B A

.

6 B D D

7 A B A8. e B B9. B D D10 B e B11 A e D12. B e B13. D A B14 A e e15. e D B16 D e A17. e e A18. B B B19 B e D20 . A A e

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2AIlS-PT -I (Paper-2)-PCM(Sol)-JEE(Advanced)/15

Physics PART-I

1.

SECTION-A

Taking origin at 0 and OC as y-axis equation of bridge is"x2

y: 20dy 2x 2x10

So, slope at F = dx : 20 = ~ : 1

At point F,1 mv2

mg~-N:R

2N : mg _ mv and f = mg

~ R ~AtF,

(1 1)3/2R = + = 20~ (metre)110

N- 10m 100m _ 5m- ~ -20~- ~

f _ 10m.- ~:. Net force =mJ~~ + 1~O : 5mJ%N

N

,

2. At the instant shown, 3Xa+ J~+16 = LSO 3VB +. Xo dxo = 0

J16+X~ dtSO, VB : 0.6 mls.

3.

4.

If a ....0, T ....00It mg

and a ....- T ....-. 2' 2

. v2 (5)2 25R: - = ---~----:-

an 25cos37-25cos53 5= 12.5 mAlternative

ivl3Ivxal:5m.

y

ii

x

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3AITS-PT-l (Paper-2)-PCM(Sol)-JEE(Advanc:ed)/15.

5. T= 2mac2mR2RT1=--o:2

m.PT +m(ac + o:R) = -2-

m2RT .i

T+-+T=--2 2mR

5T .i:. 2=2mR

.iT=-

5mRT J2 .i

ac=-= ----2m 5mR2m 10m2R

putting all the values .ae = 4 m/s2.

~T .e

FBD 01 disc FBD 01 particle

6.

7.

8.

9.

.10.

U = 40(xy) + CAs the mid point of rod the particle would be at equilibrium so from conservation of mechanicalenergy between A and the mid point of the rod1 2 a a-mvo = 40-- Where a = 1 m2 22

:. Vo = a~ 40 so Vo = 2 m1sec.2m

I + 2m ="32 x:. x = 1130

Required moment of Inertia = -I x 2 x 3 = .!..30 5

from work energy theorem

1 (2 2)Wmg+WF ="2m v,-v-2Fh = !m(yf - v2)

2v2

where h= ( F)2 g+-. m

v =vJmg-Ff mg+F

= 2mvx6v _ mv2

12(L-10r) L-10r

Concept accelerates so the speed increases linearly w~h time. But Ihe distance fallen increasesas t2 So the average speed occurs al half of time taken to pass Ihe window, which is before ~has covered half ofthe height of Ihe window.

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http://www.flltJ-.com

cc----------------------- ---

4AllS-Pl -I (Paper-2)-PCM(Sol)-JEE(Advanced)/15

11-12. N1 + N2 = Mg (1)I'(N, +N,) =Ma (2)Net torque aboutc=o

Nro Nro~ "(N +N )h+-'-=-'-r, , 2 2.~ N2> N1 . (3)

Solving, N, = ~{g- 2~h}. Contact ~ N1> 0

~ (A)(forQ.ll).a = 1'9 (from equation (1) and (2)

~ (8) (for Q. 12)

13. Corresponding F(x) vs x graph'~ At x = 15, F = 0 ~ a = 0 ~ (1) possible

At x = - 5, F '" 0 ~ a '" 0~ (ii) not possible

Betweenx=10&x=15, F=O ~ a=O~ const. vel. ~ (iii) possible

rol2 00/2N1 -----_ C .--------- N2,,

I hII

II

F(x)

10-10 0 (x

14. Total energy (E) will be conserved.E=K+U

Let's consider x = - 5 ~ v = 0 (,:

~ K=O atx=-5From graph, U = - 5 at x = - 5

~ E=-5J

dX=O)dt

15. Case -1: mvo = (m + 3m)v1 .2 1 . 2

and 2(4m)y- = 2k(M)

and Tma, = kiM)Case 2: mvo = mV1+ 3mv2 and Vo= V2- v,

1 2 1 2 1 2and 23mv2 = 2k(!>e) + 23mvf

Vo.. v,= ..[6'

16.

17.

3 , 1 (v~)-my +-m-2 r 2 4 3

Ratio = ------ =-1 4-m~2 0

(P) Only gravny is working and we know rotot= !>K~ (3)(Q) After falling through any given height, v is same but length of track (i) > length of track (ii)~ (1)(R) rotot=!>K ~ v same (independent of mass)~ (1)(5) Work by gravny same but time (i) > time (ii)~ (2)

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5AllS-Pl -I (Paper-2}-PCM(Sol)-JEE(Advanced)/15

18. (P) forces are conservative so E is constant K will increase upto equilibrium and then decreases(0) Gravnationalforce is conservative, so E is constant U decreases and K increases., 1 1 F2(R) U = mgh =>constant, K = -mv2 = _ma2t2 = _t2

222m, F2

E = U + K = mgh + -t22m

(8) Uo=mgH, AB=vot

J1... = sine =>h = Vosin e tvotU,:, Uo-mgvosinet

1 2K =-mvo2

E = K + U = (Uo+ .!mv~) - mgvosine!2

19. COlM mvo= mv, - 2mv2mr2

COAM mvor= mv,r + -W, 2For no slippingror- v, = V2

v, + v2vo=v,+ ---2

=> 2vo= 3v, + V2

A~B

g~

15 =_ VoAfter solving v1 = 5v0/7, V2= 2vo- "7 Vo 7W= ~(5Vo + V,o) = 4vo

r 7 7 r

KEq,_ = .!m25~ +.! mr216v~

I 2 49 2 2 49r2

25mv~ 8m~ 33 2---+--=-mvo98 98 98

KE '= .!2m v~ = m~, plane 2 49 49

Win;;'" = -(.!m~ _ 33mv~_ mv~)= _14mv~lOll 2 0 98 49 98

. ' 1 16v~ 1mv2, 16~ _ 24mv~K.E. of cyhnderw.r.t plank = '2m-4-9-+'2-2-'-4-9-- -98~

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6AilS-PI-I (Paper-2)-PCM(Sol)-JEE(Advanced)/15

Chemistry PART-II

2.

6.

i.~

13.

SECTION-A2M + 2H,o----+2MOH + H2

n = 2.271dm' = 0.1 moleH, 22.71dm'

80 that number of moles of metal = 0.2 mole.

Average molar mass = 4.6 gm - 23 gm Imole0.2gm/mole

Considering the molar masses of alkali metals orily lithium can come into considerationn(Rb)+ n(Li)= 0.2 molen(Rb)+ M(Rb)+ (0.2 - n(Rb)M(Li)= 4.6n(Rb)x 85.5 + (0.2 - n(Rbx 7 = 4.6n(Rb)= 0.0408 molen(Li)= 0.1592 mole

KE = hv - hvom1(slopej = heV = hv - hv~V = hie Iv - volm2(slope) " hie~ h-=--=em, hie

2Cu+2+ 41-----+Cu212 + 1228,0;2 +~----+2r +8,.2

2Cu+2=122820;' =1280, Cu+' = 8,0;'Mole of 8,0;' = 20 x 1= 20 m.moleMole of Cu+2= 20 m moleMole of Cu = 20 x 10-3x 63.58 gm

Percentage purity = 20 x 1O-~x 63.58 x 100

= 63.58%

14. 51"+2Cu+2----+Cu,12 +1;

13+ 28,0;2 ----+31- + 8.0.'

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Mathematics

7

PART -III

AITS-PT -I (Paper-2)-PCM(Sol)-JEE(Advanced)/15

l

1.

2.

3.

4.

5.

6.

7.

8.

SECTION-A

f-1(8x) _f-,l(x) , f-1(x)lim-~~-~--"- = lim-- (Taking 8x = t in first part)x-+co X1/3 X113 x ....co X1/3

(leW'(x)=t)::=>lim I V3 _1t~~(f(I)) 2

2-X' < 2-X'[2X'] ~ 1::=>[2-x' [2X']] = 0 and whenever

2-x' [2X'] = 1

[2X']= 2x'

Also if x E (-2,2), Ihen 1~ 2x'

8AITS-PT -I (Paper-2)-PCM(Sol).JEE(Advanced)/15

9.(x3 +3x)-Ix3 _x2 +2X-21 Ix2+X+21---------" -1

x2+x+2 x2+x+2

10.(tx4)(tx") (~)(~)

lim x=1 ,=1 _ 5' 6 - ~ =>I = 2 7

n_(t,x')(t,x9-') (I:J10l_l) 5'11. (t-'g-' (x) = (gf(x)f'

d~(f-'g-1(X)I'=12= d 1dx (gof(xI'=1 .

12. Areawill besameas boundedbyy = f(x). Y=g(x)1 1

=>Ilx3-4Jx=--1 . 2

'. I

. dA d2A (dA dB)13.-14. -=2(F-A-B)=>-=2 ----dl dl2 dl dl

dB=~(A)dl 2After solvingwe gel A = 2F le-!. B =F(l- le-' - e-I)(F beingIheinitialfoodquanlily)

16. y=g(x)

_ _.. .. _ .._.._.._.._................ -......................... - y = 1/2

(-2.5,0)( -2,0) (-1,0)(-.5,0) (1,0)(1.5,0)

17. 1 112 112 (1 ) (1 )(P) fln(1-x)lnxd)(=2fIn(1-x)lnxdx=2fIn --x in-+x dx.o 0 0 2 2

,/2 ( e) ( e) e e ( 1)=2 fin sin- In cos- sin-cos-de Bylakingx = -cose'0 2 2.2 2. . 211-/2

=8 f xlnxln(1-x2~x. 0

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18.

19.

20.

9AITS-PT -I (Paper-2)-PCM(Sol)-JEE{Advanced)/15

. (Q) xJ(X)-X2 -1=a(x-l)(x-2)(x-3)

1puttingx = 0, weget a = -6

(R). f'(x)f(x) ;, f(X)4+ 1

~ f'(x)f(x) ;, 1 ~ j f'(x)f(x) dx;, jdx(f(X4 +1 0(f(X4 + 1 0

tan-'(.J3) - tafl-' (1);, 2a

"-;,a24(S) letx + f(x) = g(x),wegetg"((x = g'((x

g' x 9 x~ g(x) = c,ec~,,sof (x) = c,ec" - x

dy ( )-y; = dX' Xi+, - X;,(P) For xy2 = 1(X;..1 - X;= 2X;)

i(Q) For y = e-3,(x;., -x;) = 3"

(R) For y=-cot"'x, X;..,-X; =(-COt"'X;)(l+Xn

Space for Rough Work .

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FIITJEE JEE (Advanced), 2015

INSTRUCTIONS

000000000000000000000000000000000000000000000

PARTTEST-I(16 November 2014)

Paper 1I-Co-d-e I 1502A I

Time Allotted: 3 Hours Maximum Marks: 180 Please read the instructions carefully. You are allotted 5 minutes

specifically for this purpose. You are not allowed to leave the Examination Hall before the end of

the test.

Name of the Candidate

Enrolment No,:

B. Filling of OMR Sheet1. Ensure matching of OMR sheet with the Question paper before you start marking your answers

onOMRsheet.2. On the OMR sheet, darken the appropriate bubble with black pen for each character of your

Enrolment No. and write your Name, Test Centre and other details at the designated places.3. OMR sheet contains alphabets, numerals & special characters for marking answers.

A. General Instructions1. AtlemptALL the questions.Answershaveto bemarkedon the.OMRsheets.2. This question paper contains Three Parts.3. Part-I is Physics.Part-II isChemistryandPart-III is Mathematics.4. Eachpart is furtherdividedintotwo sections:SecUon-A& Sectlon-C5. Rough spaces are provided for rough work inside the question paper. No additional sheets will be

provided for rough work.6. Blank Papers, clip boards, log tables, slide rule, calculator, cellular phones, pagers and electronic

devices, in any form, are not allowed.

C. Marking Scheme For All Three Parts;(I) Sectlon~A (01 - 10) contains 10 multiple choice questions which have one or more than one

correct answer. Each question carries +3 marks for correct answer. There is no negative'marking.

(11) Sectlon..c (01 - 10) contains 10 Numerical based questions .with answers as numerical valuefrom 0 to 9 and each question carries +3 marks for correct answer. There is no negativemarking.

Caution: Question Paper CODE as given above MUST be correctly marked in the answerOMR sheet before attempting the paper. Wrong CODE or no CODE will give wrongresults.

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2. '_ttl"'" , t '! ((filii is si(((W( iN" (('WyW" ,",Cwo .. w. C . (tf !_ co. : '" iiWij

Useful Data

PHYSICS

Acceleration due to gravity

Planck constant

Charge of electron

Mass of electron

Permittivity of free space

Density of water

Atmospheric pressure

Gas consta nt

9 = 10 mls2

h = 6.6 x10-34 J-s

e= 1.6x 10-'9C

me = 9.1 x 10-31kg

R =8.314 J K-' mol-'

CHEMISTRY

Gas Constant R = 8.314 J K-' mor'= 0.0821 lit atm K-' mol-'= 1.987", 2 Cal K-' mol-'

Avogadro's Number N. = 6.023 X 1023

Planck's constant h = 6.625 x 10-34J.s= 6.625 x 10-27 erg.s

1 Faraday = 96500 coulomb1 calorie = 4.2 joule1 amu = 1.66 x 10-27 kg1 eV = 1.6 x 10-19 J

Atomic No: H=1, He = 2, Li=3, 8e=4, 8=5, C=6, N=7, 0=8,N=9, Na=11, Mg=12, 5i=14, AI=13, P=15, 5=16,CI=17, Ar=18, K =19, Ca=20, Cr=24, Mn=25,Fe=26, Co=27, Ni=28, Cu = 29, 2n=30, As=33,8r=35, Ag=47, 5n=50, 1=53, Xe=54, 8a=56,Pb=82, U=92.

Atomic masses: H=1, He=4, Li=7, 8e=9, 8=11, C=12, N=14, 0=16,F=19, Na=23, Mg=24, AI = 27, 5i=28, P=31, 5=32,CI=35.5, K=39, Ca=40, Cr=52, Mn=55, Fe=56, Co=59,Ni=58.7, Cu=63.5, 2n=65.4, As=75, 8r=80, Ag=108,5n=118.7, 1=127, Xe=131, 8a=137, Pb=207, U=238.

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3AIlS-PT ~(Paper.1)-PCM.JEE(Advanced)/15

Physics PART-ISECTION-A

(One or More thanOne Options Correct Type)

This section contains 10 multiple choice questions. Each question has four choices (A), (B), (C)and (0) out of which ONE or MORE THAN ONE are correc!.

1. A man is running with a constant acceleration on a plate which is placed on a horizontalsmooth surface. Then choose the correct option(s) -(A) Work done by friction on the man is negative(B) Work done by friction on the man is posttive(C) Work done by friction on the plank is positive(0) work done by friction on the plank is negative

2. The density of a rod gradually changes from one end to the other. It is pivoted at one ofthe end so that it can rotate about a vertical axis through the pivot. A horizontal force F isapplied on the free end in a direction perpendicular to the rod. The quanttties, thatdepend on axis of rotation Onthis situation) are:(A) angular acceleration(B) Total kinetic energy ofthe rod, when the rod completes one revolution(C) Angular momentum when the rod completes one revolution(0) Angular veloctty of rod

3. Two identical blocks are connected by a light spring. The combination issuspended at rest from a string attached to the ceiling, as shown in thefigure below. The string breaks suddenly. Immediately after the stringbreaks.(A) Acceleration of both the blocks would be g downward(B) Acceleration of centre of mass of the combined block system would be

g downward(C) Acceleration of upper block would be 2g downward(0) Acceleration of lower block would be g upward.

~W"T"f5"jf'" '.T ......., .S"... ' ' r."',"',,,,. _' '.Space for Rough work

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4AITS-pr -1(Paper-1)-PCM"EE(Advanced)/15

2m/s-restThe elastic collision between two bodies, A and B can beconsidered using the following model. A and B are free tomove along a common line without friction. Whenseparation between the surfaces is greater than d = 1 m,the interacting force is zero, when their distance less thand, a constant repulsive force F = 6N is present. The massof body A is mA = 1 kg and it is initially at rest. The mass ofbody B is ma = 3 kg and it is approaching towards A with aspeed Vo = 2 m/s. TheJl choose the correct option(s)(A) The common velocity attained by the bodies are 1.5 mls(B) The minimum separation between the bodies is 0.25 m(C) The minimum separation between the bodies is 0.75 m(0) The common velocity attained by the bodies are 2.0 mls

The sum of all the external forces on a system of particles is zero. Which of the followingmust be true for the system of particles?(A) The total mechanical energy is constant.(B) The total potential energy is constant(C) The total kinetic energy is constant(0) The total momenllJm is constant

5.

4.

In the system shown in the figure the mass m moves in acircular path of anguiar amplitude 60' and mass 4 m isstationary. Then choose the correct statement(s)

6.

~ "'~~-'->->->-.",-"60' .': A

B(A) The minimum value of coefficient of friction between the mass 4 m and the surface of

table is 0.5(B) The work done by gravitation force on the block m is positive when it moves from

A~B -(C) The power delivered by the tension when m moves from A to B is non-zero(0) The kinetic energy of m in pos~ion B equals the work done by gravitational force on

the block when the moves from position A to position B.

Space for Rough work

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"

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5AITS-pr ~(Paper-1 )-PCM.JEE(Advanced)/15

7. A weightless rigid rod AB of length e connects two equal masses m one jAparticle is fixed at the end B and the other at the middle of the rod as shownin the figure. The rod can rotate in the vertical plane freely around the hinge CpointA. Choose the correct option(s).

B(A) The minimum horizontal velocity required to be given to the particle B so as to make

the rod go around in the complete vertical circle is ~245ge

(B) The minimum horizontal velocity required to be given to the particle B so as to make

the rod go around in the complete vertical circle is ~249ge

(C) The ratio of compressive force in the rods AC and BC is 2 : 1 when the masses are athighest point.

(D) The ratio of compressive force in the rods AC and BC is 3 : 1 when the masses are athighest point.

.... ;>

L

8AITS-pr -/(Paper.1 )-PCM.JEE(AcIvanced)115

7. The minimum force F needed topush the two 50 kg cylinders up theincline is 200x Newton. The forceacls parallel to the plane and thecoefficients of friction at contactingsu !faces are as JlA = 0.3 betweencylinder A and ground, JlB = 0.25between cylinder B and ground andJlc = 0, between two cylinders. Thecylinder A can rotate about ils axiswithout friction. Find the value of x.

8. A 12 kg disc has an angular velocity of (i) = 20 radls and radius 200 mm. If the brake ABCis applied such.that the magnnude of force P varies wnh time as shown, detennine thetime (in sec) needed to stop the disc. The coefficient of friction at B is Jl = 0.4

p

9. AIlS-PT -1(Paper-1)-PCM.JEE(AdvancedV15

Chemistry PART-IISECTION -A

(One or More than One Options Correct Type)This section contains 10 multiple choice questions. Each question has four choices (A). (6). (C)and (0) out of which ONE or MORE THAN ONE are correc:.:l=--- --------

1. Choose the correct statements among the following regarding hydrogen atom? (Allgraphs given below are indicative in nature, not to the scale)

(A) Radial wave function R(r) gives information about the energy and size of the orbital.(B) Plot of R(2.0) (r) versus r, for 2s electron is

,--+(r - dislance from nucleus)

(C) Plot of Rl,.o)versus r for 2s electron is

,--(D) Plot of function 4xr'R;,o against r (distance from nucleus) is

'--( 11 111%1 : tWWWZ5Wl1WjjWZ55SSW' t ttWW tttS wms: SWWW1155 swsws w wrs ss w ::s sw w %T ::s zw w W WS 1W "

Space for Rough worle

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10AIlS-PT ~(Paper-1)-PCM'-'EE(Advanced)/15

2. The behaviour of a real gas is usually depicted by plolting compressibility factor versus Pat a constant temperature. Which of the following facts are true regarding van der Waal.gas equation?(A) In a region where Z > 1, molecular interaction is repulsive.in nature.(B) In a region where Z < 1, molecular interaction is altractive in nature.(C) In a region where Z 1, molecular interaction neither repulsive non-altractive innature.(0) Z is independent on molecular interaction at all, it vary with variation of pressure only.

\

3./

B

A

(Given that y, > y,)Choose the correct statements regarding above chemical reaction:(A) Overall rate of reaction is equal to sum of formation of Band C.

(B) Overall activation energy ofthe reactionE. = x;y, +x,y, .x,+x2

(C) Half life of reaction is ~., x1 +x2(0) As the temperature decreases rate of reaction of both reactions decreases but ratio

of rate of reaction A-----.C decrease more than'A-----.B.

4. The structure of 503 is a planar triangle. The 5 atom at centre used Sp2orbitals, the S-o bonds in 503 are much shorter than 5 - 0 single bonds. Which ofthe followingstatement (s) is/are correct regarding S03? (Assuming xy plane as molecular plane)(A) 503 has 24 outer electrons like NO; ion. So that 503 !lnd NO; are isostructural and

isoelectronic too.(B) The six atomic orbitals used in 503 for It-bonding are the 2pz orbitals on the three 0

atoms and the 3p, .3d", and 3d" orbitals on S.(C) By combining three 2pz atomic orbitals of oxygen and three 3pz, 3d", and 3d", of

sulphur three It-bonding molecular orbitals and three It-antibonding molecular orbitalsare obtained.

(0) The atomic orbitals of 5 atom used for It-bonding can be 3p" 3d, , and 3d, .x -y z

t.\tWWWWYZWZi:YS'WS j jWWS1WWWWSSiZWWSWWWWSiWjsSWYWSSW'.5WW5W .WWWZ"P':':!i5t:'tNiS itW-Xw.xwwzwwwwww SfWWi ssW'lwsswwZSpace for Rough work

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F"TJEE JEE (Advanced), 2015

INSTRUCTIONS


Please read the instructions carefully. You are allotted 5 minutesspecifically for this purpose.

You are not allowed to leave the Examination Hall before the end ofthe test.

Paper 1I-C-od-e I 1502A I

Maximum Marks: 180

000000000000000OOOOOOOOOOODDOODDDDDDDDDDDDDDD


Time Allotted: 3 Hours

B. Filling of OMR Sheet1. Ensure matching of OMR sheet with the Question paper before you start marking your answers

on OMRsheet2. On the OMR sheet, darken the appropriate bubble with black pen for each character of your


A. General Instructions1. AttemptALL the questions.Answershaveto be markedon the.OMRsheets.2. This question paper contains Three Parts.3. Part-l is Physics,Part-ll is Chemistryand Part.1IIis Mathematics.4. Eachpart is furtherdividedintotwosections:Sectlon-A & Section-C5. Rough spaces are provided for rough work inside the question paper. No additional sheets will be

provided for rough work.6. Blank Papers, clip boards, log tables, slide rule, calculator, cellular phones, pagers and electronic



Enrolment No:.

C. Marking Scheme For All Three Parts,(i) Sectlon-A (01 - 10) contains 10 multiple choice questions which have one or more than one

correct answer. Each question carries +3 marks for correct answer. There is no negative.marking.

(II) Section-C (01 -10) contains 10 Numerical based questions.with answers as numerical valuefrom 0 to 9 and each question carries +3 marks for correct answer. There is no negativemarking.

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2"'ff'X"'X'C"'" ((( .. )' ( C((fWWU"W . -HN' "e ow'""'" w"'("f' 'Xt-F"WiNt (

Useful Data

PHYSICS


Planck constant

Charge of electron

Mass of electron

Permittivity of free space

Density of water


Gas constant

9 = 10 mls2

h = 6.6 x10-34 J-s

e = 1.6 x 1O-19C

me = 9.1 x 10-3' kg'

Pwaler = 103 kg/m3

R= 8.314J K-' mor1

. ,

., .

CHEMISTRY

Gas Constant R = 8.314 J K-' mol-'= 0.0821 Lit atm K-1 mor'= 1.987", 2 Cal K-' mol-'

Avogadro's Number Na = 6.023 x 1023

Planck's constant h = 6.625 x 10-3' J.s= 6.625 x 10-27 erg.s

1 Faraday = 96500 coulomb1 calorie = 4.2 joule1 amu - 1.66 x 10-27 kg1 eV = 1.6 x 10-19 J

Atomic No: H=1, He = 2, Li=3, Be=4, B=5, C=6, N=7, 0=8,N=9, Na=11, Mg=12, 5i=14, AI=13, P=15, 5=16,CI=17, Ar=18, K =19, Ca=20, Cr=24, Mn=25,Fe=26, Co=27, Ni=28, Cu = 29, Zn=30, AS=33,Br=35, Ag=47, 5n=50, 1=53, Xe=54, Ba=56,Pb=82, U=92.

Atomic masses: H=1, He=4, Li=7, Be=9, B=11, C=12, N=14, 0=16,F=19, Na=23, Mg=24, AI = 27, 5i=28, P=31, 5=32,CI=35.5, K=39, Ca=40, Cr=52, Mn=55, Fe=56, Co=59,Ni=58.7, Cu=63.5, Zn=65.4, As=75, Br=80, Ag=108,5n=118.7, 1=127, Xe=131, Ba=137, Pb=207, U=238 .

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3AITS-PT-1(Paper-1)-PCM.JEE(Advanced)/15

Physics PART-ISECTION-A

(One or More than One Options Correct Type)

This section contains 10 multiple choice questions. Each question has four choices (A), (8), (C)and (0) out of which ONE or MORE THAN ONE are correct.

1. A man is running with a constant acceleration on a plate which is placed on a horizontalsmooth surface. Then choose the correct option(s) -(A) Work done by friction on the man is negative(B) Work done by friction on the man is pos~ive(C) Work done by friction on the plank is positive(D) work done by friction on the plank is negative

2. The dens~of a rod gradually changes from one end to the other. It is pivoted at one ofthe end so that ~ can rotate about a vertical axis through the pivot. A horizontal force F isapplied on the free end in a direction perpendicular to the rod. The quantities, thatdepend on axis of rotation Qnthis situation) are:(A) angular acceleration(B) Total kinetic energy of the rod, when the rod completes one revolution(C) Angular momentum when the rod completes one revolution(D) Angular velocity of rod

3. Two identical blocks are connected by a light spring. The combination issuspended at rest from a string attached to the ceiling, as shown in thefigure below. The siring breaks suddenly. Immediately after the stringbreaks.(A) Acceleration of both the blocks would be g downward(B) Acceleration of centre of mass of the combined block system would be

gdownward(C) Acceleration of upper block would be 2g downward(D) Acceleration of lower block would be g upward .

::< , . _-


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------------------4

AITS-PT ~(Paper-1).PCM.JEE(Advanced)f15

2m/s-restThe elastic collision between two bodies, A and B can be

considered using the following model. A and B are free tomove along a common line without friction. Whenseparation between the surfaces is greater than d = 1 m,the interacting force is zero, when their distance less thand, a constant repulsive force F = 6N is present. The massof body A is rnA= 1 kg and ~ is initially at rest. The mass ofbody B is rna = 3 kg and ~ is approaching towards A with aspeed Vo = 2 m/s. The]l choose the correct option(s)(A) The common veloc~ attained by the bodies are 1.5 mls(B) The minimum separation between the bodies is 0.25 m(C) The minimum separation between the bodies is 0.75 m(0) The common veloc~ attained by the bodies are 2.0 mls

The sum of all the external forces on a system of particles is zero. Which of the followingmust be true for the system of particles?(A) The total mechanical energy is constant.(B) The total potential energy is constant(C) The total kinetic energy is constant(0) The total momentum is constant

4.

5.

In the system shown in the figure the mass m moves in acircular path of anguiar amplitude 60 and mass 4 m isstationary. Then choose the correct statement(s)

6. ......:'PiB

(A) The minimum value of coefficient of friction between the mass 4 m and the surface ofta~is05 .

(B) The work done by gravitation force on the block m is positive J'hen it moves fromA to B

(C) The power delivered by the tension when m moves from A to B is non-zero(0) The kinetic energy of m in position B equals the work done by gravitational force on

the block when the moves from position A to position B.


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5AIlS-PT ~(Paper-1).pCM.JEE(Advanced)/15

7. A weightiess rigid rod AB of length f connecls two equal masses m one IAparticle is fixed al the end B and the other at the middle of the rod as shownin the figure. The rod can rotate in the vertical plane freely around lhe hinge Cpoint A. Choose the correcl option(s).

B(A) The minimum horizontal velocity required to be given to the particle B so as to make

the rod go around in the complete vertical circle is J245gf

(B) The minimum horizontal velocity required to be given to the particle B so as to make

the rod go around in the complele vertical circle is J24ggf(C) The ratio of compressive force in lhe rods AC and BC is 2 : 1 when the masses are al

highest point.(0) The ratio of compressive force in lhe rods AC and BC is 3 : 1 when the masses are al

highest point.

l:

..._ ~

L

AITS-PT -1(Paper-1 )-PCM.J EE(Advanced)/15,

6

9. A block of mass m is pulled by a force of constant power P placed on a rough horizontalplane. The friction coefficient between the block and the surface is 11. Then

(A) The maximum velocity of the block during the motion is ~ .. llmg

(B) The maximum velocity of the block during the motion is ._p-211m9

(C) The block's speed is never decreasing and finally becomes constant.(D) The speed of the block first increases to a maximum value and then decreases.

10. Consider a particle at rest which may decay into two (daughter) particles or into three(daughter) particles. Which ofthe following is/are true? (There are no extemal forces)(A) The velocity vectors of the daughter particles must lie in a plane.(B) Given the total kinetic energy of system and the mass of each daughter particle, it is

possible to determine the speed of each daughteJ particle.(C) Given the speed (s) of all but one daughter particle it is possible to deterinine the

speed ofthe remaining particle.(D) The total momentum of the daughter particles is zero.

SECTION -C(One Integer Value Correct Type)

This section contains 10 questions. Each question, when worked out will result in one integerfrom a to 9 (both inclusive).1. A cyclist is going in a straight line at a uniform velocity 18 kmlh. The resistance force can

be expressed as force kv" where k = 0.8 where velocity is in mls and unit of force isnewton. The mass of the cyclist with the bicycle is 100 kg. Neglect the rolling resistanceforce. If the power of the cyclist during the ride is lOx watt, then what is x? .

Space fOf Rough work

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7AITS-PT~(Paper-1)-PCM.JEE(Advanced)/15

o

.ItpIA massive horizontal platform is movinghorizontally with a constant acceleration of 10mis' as shown in the figure. A particle P of massm = 1 kg is kept at rest at the smooth surface as

shown in the figure. The particle is hinged at 0with the help of a massless rod OP of length o.e m. Hinge 0 is fixed on the platform andthe rod can freely rotate about point O. Now the particle P is imparted a velocity in theoppostte direction of the platform's acceleration such that tt is just able to complete thecircular motion about point O. Then the maximum tension appearing in the rod during themotion is 10n. Find the value of n.

2.

3. A hemispherical bowl of radius R = 5 metre is fixed on a horizontal surface its innersurface to be smooth. Three smooth and perfectly inelastic ball of mass 3 mg, 4 kg and 5kg are simultaneously released from the edge of bowl in such way that during thecollision wtth each other, their kinetic energy loss is maximum. If the maximum kineticenergy lost during the collision is 300n Joule. Find the value of n. (take g = 10 m/s')

0--@"B Cl f 1

1m1m

ClA ,

Three identical balls A, Band C each of mass m = 3 kgare connected by strings AB and BC as shown in thefigure. The 'whole system is placed on a smoothhorizontal surface.Now the ball B is given an initial velocity Vo = "3 mis, perpendicular to the strings andalong the horizontal surface. Find the tension (in Newton) in the string just before theballs A and C collide.

4.

5. A 10 kg solid sphere of radius r = 0.8 m is rolling without slippingon a horizontal rough surface with 8 m/s. The force applied bythe right half of the sphere on the left half is 30x New1on. Findthe value of x.

6. A particle moves along a parabolic path y = ex'" in such a way that the x component ofvelocity remains constant and has a value 1/3 mls. Find the magnitude of acceleration (inmls') of the particle. ,


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8AIlS-PT ~(P.aper-1)-PCM.JEE(Advanced)/15

7. The minimum force F needed topush the two 50 kg cylinders up theincline is 200x Newton. The forceacts parallel to the plane and thecoefficients of friction at contactingsurfaces are as llA = 0.3 betweencylinder A and ground, llB = 0.25between cylinder B and ground andllc = 0, between two cylinders. Thecylinder A can rotate about its axiswithout friction. Find the value of x.

8. A 12 kg disc has an angular velocity of OJ = 20 rad/s and radius 200 mm. If the brake ABCis applied such.that the magnttude of force P varies with time as shown, determine thetime (in sec) needed to stop the disc. The coefficient of friction at B is 11 = 0.4

p ~.~~.~.!ml'H.~~~..Jl:I!l> pA;:.~::

5(I) = 20rad

2 tIs)

9. A ball wtth mass m projected horizontally off the end of a table wtth an initial kineticenergy k. At a time t after tt leaves the end of the table it has kinetic energy 3k. The time t

is ~ fk. Find the value ~f n. Neglect air resistance.g'i";:n10. Two persons A and B are standing at the origin and 210 meter away from the origin along

the y-axis, respectively. A starts to move along x axis with acceleration 3.2 m/s2 andthereafter with the same retardation and comes to rest after moving 320 m. B moves with

a constant velocity and catches A in 15 s. The speed ofB is 10n mls. Find the value of n., '. 3 .Space for Rough work

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Chemistry


PART-II

SECTION-A(One or More than One Options Correct Type)

This section contains 10 multiple choice questions. Each question has four choices (A), (8), (C)and (0) out of which ONE or MORE THAN ONE are correct.

1. Choose the correcl stalements among the following regarding hydrogen atom? (Allgraphs given below are indicative in nature, not 10 the scale)

(A) Radial wave function R(r) gives information about the energy and size of Ihe orbital.(B) Piol of R(2,0) (r) versus r, for 2s electr.on is

R,.,

,--(r - distance rrom nucleus)

(C) Plot of R~.,)versus r for 2s eleclron is

'-(0) Plot of function 4"r'R;, against r (distance from nucleus) is, .

4ltr"'R~.o

'-Space for Rough. worlc

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2.

3.

The behaviour of a real gas is usually depicted by plotling compressibility factor versus Pat a constant temperature. Which of the following facts are true regarding van der Waal,gas equation?(A) ,In a region where Z > 1, molecular interaction is repulsive in nature.(B) In a region where Z < 1, molecular interaction is attractive in nature.(C) In a region where Z 1, molecular interaction neither repulsive non-attractive innature.(0) Z is independent on molecular interaction at all, it vary with variation of pressure only.

B

A

(Given that y, > Y2)Choose the correct statements regarding above chemical reaction:(A) Overall rate ofreaction is equalto sum offorrnation of Band C.

(B) Overall activation energy ofthe reactionE. _ x,y, +X2Y2 .x1 +x2

(C) Half life of reaction is ~., X1+X2

(0) As the temperature deCreases rate of reaction of both reactions decreases but ratioof rate of reaction A----4C decrease more than A----4B.

4. The structure of S03 is a planar triangle. The S atom at centre used Sp2orbitals, the S-o bonds in S03 are much shorter than S - 0 single bonds. Which of the followingstatement (s) is/are correct regarding S03? (Assuming xy plane,as molecular plane)(A) S03 has 24 outer electrons like NO; ion. So that S03 ,md NO; are isostructural and

isoelectronic too.(B) The six atomic orbitals used in S03 for x-bonding are the 2pz orbitals on the three 0

atoms and the 3p"3d,,, and 3dyzorbitals on S.(C) By combining three 2Pz atomic orbitals of oxygen and three 3pz, 3d", and 3dyz, of

sulphur three x-bonding molecular orbitals and three It-antibonding molecular orbitalsare obtained.

(0) The atomic orbitals of S atom used for x-bonding can be 3pz, 3d, ,and 3d , .l( -y Z

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-11

AITS-PT-1(Paper-1)-PCM.JEE (Advanced)/15

5. Choose the correct statement(s) among the following statements(A) Ionization energy of H2 molecule is greater than ionization energy of H-atom.(B) In resonating structure of ozone, electron pair migrates from non-boding orMal of

oxygen to antibonding ,,-orbital of other oxygen.(C) Order of bond strength

O-H>C-C>C-H>H-H(D) Thermal stability of alkali metal sulphate increases down the group.

6. A sealed vessel of a volume 5 dm3 was filled with ethane at a temperature 300 K and 1bar pressure was sealed. The vessel was then heated and pressure was measured atdistinct temperature as followed? (1 bar = 100 KPJ

T(K) Pressure measured (KPa)300 100 KPa600 225 KPa900 360 KPa

Due to thermal decomposition of ethane (C2H, (g) ... C2H, + H2(g)J, measured pressurewas higher than calculated pressure.Which of the fOllowing statements give correct information regarding abovedecomposition?(A) Ratio of degree of dissociation of ethane at 600 K and 900 K is 5/8.(B) Ratio of equilibrium constant (Kp) at 600 K and 900 K is 512. .(C) Mean value of L'.H(heat of reaction) in the temperature range of 600.K to 900 K is

(-1800 Rfn4.2)(D) Reaction is shifted forward when temperature rises and shifted reversed when

pressure is elevated.

7. Choose the correct statement(s) among the fOllowing?(A) Ionic character of halogen acid (HX) decreases in the group.

HF> HCI > HBr > HI(B) Boiling point of covalent hydrides of elements N, 0 and F are abnormally high.(C) Dipole moment of NF3 is greater than NH3.(D) Bond angles of NO; , N02 and NO; are in the order of NO; < N02 < NO;

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12AITS-PT -l(Paper-1 )-PCM.JEE(Advanced)/15

8. Oihydrogen can be prepared both in laboratory and industry which of the followingstatement(s) is/are correct regarding hydrogen preparation?(A) Highly pure hydrogen is obtained by electrolyzing warm aqueous Ba(OHh solution'

between nic~el electrode ..(B) Reaction of steam on hydrocarbons or coke at high temperature.(C) Impure H2(g) is purified by passing through AgN03 solution; (CH3COOhPb, KOH

solution, anhydrous CaCb in orderly manner.(0) Hydrogen gas obtained as by-product in manufacture of H202 by electrolysis of

sulphate solution at high current density.

9. There are certain facts regarding water, which of the following are correct?(A) Water free from soluble salts of calcium and magnesium is called soft water.(B) A water sample contain 122 PPM HCO,. The hardness of water in terms of CaC03

is 100 PPM.(C) At atmospheric pressure ice crystallizes in cubic form, but at very low temperature ij

condenses to hexagonal form.(0) Hydrating tendency of water is due to high dielectric constant.

10. Consider the following statement regarding Maxwell's distribution of velocities., Thecorrect statement(s) is/are?(A) As temperature increases, the peak (maxima) of the curve is shifted towards right,side.(B) At temperature increases, the most probable velocity of molecules increases but

fracti,on of molecules having maximum velocity decreases.(C) The area under the curve at all temperature is same, because it represents the total

number of gaseous molecules.(0) At same temperature a light gas has broader distribution of molecular speed than a

heavier gas.

Space for rough worle

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13AITS-PT~(Paper.1)-PCM.JEE(Advanced)/15

SECTION :-C(One Integer Value CorrectType)

This section contains 10 questions. Each question, when worked out will result in one integerfrom 0 to 9 (both inclusive). .

1. A sample of crystalline soda (A) w~h a mass 2.574 g was allowed to react with an excessof hydrochloric acid and 204.3 cm3 of gas was liberated at 1 bar pressure and oDe. .

Another sample of different crystalline soda (B) w~h a mass of 0.715 g was decomposedby50 cm3 of 0.2 N sulphuric acid. After total decomposition of soda B, the excess of thesulphuric acid was neutralized which required 50 cm3 of 0.1 N sodium hydroxide solution(by titration of methyl orange)What is the ratio of water of crystallization in crystalline soda (A) and crystalline soda (6).

2. Oxide (A) of a certain metal contains 22.55% of oxygen by mass. Oxide (A) adopts rocksalt structure and is a component of fertilizers and food add~ives. It forms green crystals.Another oxide (B) of same metal contains 50.48% of oxygen by mass. Oxide (B) is darkgreen. It has similar structure as pyrosulphate and pyrophosphate.What is the sum of oxidation states of metal ions in oxide (A) and oxide (6)?

3. The number of revolutions made by an electron in one second in H-atom in its nO.orb~ iseight times of number of revolutions made by electron in one second in 2

ndorb~ of H.

atom, then n is .

4. A gas in a closed flexible container is slowly cooled from 273C to 27~OC. The ratio of

final volume of the gas to its initial volume is xly. Assume ideal behaviour of gas.What is the value of x + y?


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I

14A1lS-Pl~(Paper-1)-PCM.JEE(Advanced)/15

5. Howmany numberof the following reaction(s) are feasible:3Ga- (aq)~2Ga(s)+Ga-'(aq)

B(OH), +H,O~[B(OH).J (aq) +H' (aq)

CCI. +2H,0~CO,(g)+4HCI(aq)BCI, +3H,0~H,BO, +3HCIBF, +3H,O~H,BO, +3HF

B,H. +6H,0~2H,BO, +6H,

B,H, +HI~B,H,I+H,(g)B,H. +2KOH+2H,O~2KBO, +6H,C(s)+H,O(g)

15AITS.PT ~(Paper.1 ).PCM.JEE(Advanced)/15

,I

Mathematics PART-IIISECTION -A

(One or More than One O~tions Correct Type)This section contains 10 multiple choice questions. Eachquestionhas four choices (A), (8), (C)and (0) out of which ONEor MORETHAN ONEare correct. .

1. If f(x) is a'non.increasing function such that f(x)+f"(x);"O VXER&f(O)=f'(O)=Othen f(x) is not strictiydecreasing in(A) [-4,-2) (B) [-2,2)(C) [3,4) (D) [2,4)

2. If f:R->[2,4),f(x) be periodic function such that the equation f(x)=1-cosKx has auniquesolution then periodof f (x) can beWx ~e(C) .J3 (D) 1

p .

3. If Y= f(x) and y = g(x) are symmetrical aboutthe line x = a.; J3 , then f f(x )g'(x)dx isa

equal top p

(A) ff'(x)g(x)dx (B) -ff'(x)g(x)dxa a

1p 1p(C) 2 f( f(x)g'(x) -f'(x)g(x ))dx (D) 2f(f(X)9'(x) + f'( x)g(x ))dx

a a

4. Let f(x)=x' +xg'(1)+g"(2) andoption/s is/are correct?(A) f'(1) = 4+f'(2)(C) g"(2)+f"(3) = 4

g(x) = x' + xf'(2) + f"(3) then which of the following

(B) g'(2)=8+g'(1)(D) g'(2) = 6 +g'(1)

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16AITS-PT ~(Paper-1)-PCM.JEE(Advanced)/15

5.

6.

7.

If f:R-+R isafunctionsuchlhallf(x)-f(y)[slsinx-siny[\;Ix,yeR,lhen f(x) is(A) surjective (6) one-one(C) many one (D) periodic

If 1([f(I)-f\a))_ (f'~a)))dl has no critical points,then'f(x) can bea (I-a) 1 a .

(A) x +sinx (6) x3 +3x +2(C) x2 +sinx (D) x2 -sil)x

If A be the area boundedby f(x) = sinx, x = O,x= ~ and x-axis, Ihen which of followingx 2

is/are truex

(A) 1 y

(6) a

17AITS-PT~(Paper-1 )-PCM..JEE(Advanced)l15

SECTION -C(One Integer Value Correct Type)

This section contains 10 questions. Each question, when worked out will result in one integerfrom 0 to 9 (both inclusive).

1. The number of solution of equation (x -2)+21092 (2" +3x) = 2' is __

2. The value of !iTolxI2(1+2+3+ ......[I~I])is equal to ([.)isGJF.) __3.

2",1 I n 0 k 21The value of Um--13~) -1) ..+ - is equal to. n~n+1 0=1 ""- 3

4. Let fix) = 30 - 2x - x' and fn (x) = fofof(n'mes)....f(x), then find the number of positiveintegral values of x which satisfies 1,014 (x) Delhi .JJOOI6, Ph 46106000. 26669493, Fax 265J3942. wclIrit.: Il1U1U1Jfltjee.eom

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18AIlS -Pl ~(Pape r-1)-PCM.J EE(AdvancedI/15

6.

7.

y=f(x)is differentiable function and g(x)=f(x-x2).lf y=g(x) has local maxima at

x = ~ but the absolute maximumexists at some other points, then minimum number of

solution of g' (x) = 0 is__

~fU-Y.+ifU-F]8. If f(x) is differentiable function such that f"(x) < O\;lxE R, then ,-0 n 1n "1 n n2jf(x)dxo .

is equal to __ (where [.) is GIF)

9. Let f(x)=x3+3x+2 and x=e be a point such that

values of a,b E R, then numberof such point are__

I

f'(e);t f(b)-f:(a)b-10. Let A. denotes the area of rectangleformed by taking two of tts vertices on x-axis and

remaining two on curve f(x) = e-2'~2,.and B. denotes the maximum value of A. for

x ER If t,B. =P, then 16109.(~)isequalto __ .Space for rough work

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FIITJEE JEE(Main), 2015

INSTRUCTIONS


C. Marking Scheme For All Three Parts.(I) Section~A (01 to 30) contains 30 multiple choice questions which have only one correct answer,

Each question carries +4 marks for correct answer and - 1 mark for wrong answer.

Maximum Marks: 360You are allotted 5 minutes

ICodel1501A I

DDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDD


Please read the instructions carefully.specifically for this purpose.You are not allowed to leave the Examination Hall before the end ofthe test.

Time Allotted: 3 Hours

A. General Instructions1, AttemptALL the questions.Answershaveto bemarkedon the OMRsheets.2. This question paper contains'Three Parts.3. Part-l is Physics:Part-ll is Chemistryand Part-III is Mathematics.4. Each part has only one section: Sectlon.A.5. Rough spaces are provided for rough work inside the question paper. No additional sheets will be

provided for rough work.6. Blank Papers. clip boards, log tables, slide' rule, calculator, cellular phones, pagers and electronic


B. Filling of OMR Sheet1. Ensure matching of OMR sheet with the Question paper before you start marking 'your answers

on OMRsheel -2. On the OMR sheet, darken the ap"propriate bubble with- black pen for each character of your


Enrolment No.


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http://www.JI.tJ-.com

,f'.!!S;PT -1.p?~';;!o~of;Jo/!,?!,~l'.~,?,"'''"'' ",,"",,",,",,"" "".. ,,""'"'~o'O'O'Oo,,"".. '"C,,' """"" ."'""'""""'"''0'''' "'" "",,""""',""'''' "0,,""

Useful Data

,------------------.---------,PHYSICS


Planck constant

Charge of electron

Mass of electron

Permiltivijy of free space

Density of water


Gas constant

g=10mls2

h = 6,6 x10-34 J-s

e=1.6x1Q-'9C

me = 9.1 x 10-31kg

R = 8.314 J K-' mol-'

.

CHEMISTRY

Gas Constant R = 8.314 J K-' mol-'= 0.0821 Lit atm K-' mor'= 1.987" 2 Cal K-' mor'

Avogadro's Number Na = 6.023 X 1023

Planck's constant h = 6,625 x 10-34J.s= 6.625 x 10-27erg.s

1 Faraday = 96500 coulomb1 calorie = 4.2 joule1 amu = 1.66x 10-27kg1 eV - 1.6 x 10-19JAtomic No: H=1, He = 2, Li=3, Be=4, B=5, C=6, N=7, 0=8,

N=9, Na=11, Mg=12, 81=14, AI=13, P=15, 5=16,CI=.17, Ar=18, K =19, Ca=20, Cr=24, Mn=25,Fe=26, Co=27, Ni=28, Cu = 29, Zn=30, As=33,Br=35, Ag=47, 5n=50, 1=53, Xe=54 , Ba=56,Pb=82, U=92.

Atomic masses: H=1, He=4, Li=7, Be=9, 8=11, C=12, N=14, 0=16,F=19, Na=23, Mg=24, AI = 27, 51=28, P=31, 5=32,CI=35.5, K=39, Ca=40, Cr=52, Mn=55, Fe=56, Co=59,Ni=58.7, Cu=63.5, Zn=65.4, As=75, Br=80, Ag=108,5n=118,7, 1=127, Xe=131, Ba=137, Pb=207, U=238,

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3AITS-PT ~-PCM.JEE(Main)f15

Physics PART-I

SECTION -ASingle Correct Choice Type

This section contains 30 multiple choice questions. Each question has four choices (A), (6), (C)arid (0) out of which ONLY ONE is correct.

1. If velocity of the particle is given by v = .JX, where x denotes the position of the particle.Initially particle was at x = 4, then which of the following is incorrect?(A) At t = 2 sec, the position of the particle is at x = 9.(6) Particle acceleration at t = 2 sec is 1 m/s"(C) Particle acceleration is ~ mls2 throughout the motion

(0) Particle will never go in negative direction from its starting position.

2.. A stone is projected vertically upward at t = 0 second from ground and collision between.the ground and the stone is peliectly elastic. The net displacement of stone between t =o second to t = T seconds is zero. The time of flight of stone is T. Pick up theINCORRECT statement.(A) From time t = T/4 second to t = 3T/4 second, the average velocity is zero.(6) The change in velocity from time t = 0 to t = T14 second is same as change in velocity. from t = T/8 second to t = 3T/8 second.(C) The distance travelled from t = 0 to t = T/4 second is larger than the distance

travelled from t = T/4second to t = 3T/4 second.(0) The distance traveled from t = TI2 second to t = 3T12 second is half the distance

traveled from t = TI2 second to t = T second.

3. A stone is projected from a horizontal plane. it attainsmaximum height 'H' and strikes a stationary smooth walland falls on the ground vertically below the maximumheight. Assume the collision to be elastic, the height of thepoint on the wall where ball strikes is:(A) HI2(6) H/4(C) 3H/4(0) none ofthese


R

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AITS-PT-1-PCM.JEE(Main)/154

4. There is a stationary field of force F = (ay) i (i is untt vector in x-direction and y isdistance along the y-axis.)(A)This force is conservative in nature.(8) This force is non-conservative in nature.(C) Potential energy function is U = axy.(D) None of these.

5. A particle free to move along the x-axis has potential energy given by U(x)=k[1-exp(-lt)for all values of x, where k is a positive constant of appropriate dimension. Then(A) At any point away from origin particle is in unstable equilibrium.(8) For any finite non-zero value of 'x' there is a force directed away from origin.(C) If its total mechanical energy is 1

5AIlS-PT ~-PCM.JEE(Main)/15

8. A block of mass m slides down a rough inclined plane of inclination 0 wtth horizontal wtthzero initial velocity. The coefficient of friction between the block and the plane is I' and0> tan-'(I'). Rate of work done by the force of friction at time t is(A) I'mg2t sinO (B) mg2t(sinO -I' cos 0)(C) I'mg2t cos O(sin 0 -I' cos 0) (0) I'mg2tcosO

9. A car is driven on a straight road. The maximum speed, the car can altain is 24 mls andthe maximum deceleration that can be applied to the car is 4 m/s2 The car starts fromrest and stops after moving a distance of 1032 m. If the minimum time in which tt cantravel the above distance is 56 sec, the maximum acceleration that the car achieves is(A) 1.2 m/s2 (B) 3.6 m/s2

(C) 6 mls2 (0) 12 mls2

10. The co-efficient of friction (both static and kinetic), between theinclined plane and the block A of mass 1 kg is l/-J3 whereasthere is no friction between block B (1 kg) and the inclinedplane. The system starts from rest. When the system covers adistance s = 30 m on the inclined plane, the cord between Aand B is burnt off. The veloctty of A when tt covers a furtherdistance of 20 m on the inclined plane is(A) Zero (B) ~120m/s

(C) ~150m/s (0) ~300m/s

11. A block of mass M, moving with a velocity Va on a -+ v,horizontal table strikes a spring of spring constant k at t = 0and begins to experience a frictional force. The co-efficientof friction is variable and is given by I' = bx, where x is thecompression in the spring. The loss in mechanical energywhen the block first comes momentarily to rest is

(A) gbMv6 (B) M9bv62k 2(k +gb)

(C) Zero (0) M29bv6

2(k+Mgb)


k

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6AITS-PT-I.PCM .JEE(Main;;,)l1;,;5;"",,,,,,,,!!!!!!!!~!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!,,,,,

(A) Zero

3 2( 1)(C)-mu 1--2 2./2

12.

13.

14.

15.

Consider 3 identical balls, each having mass of 0.1 Kg. The ball A has an innial velocityof 10,J3m/s. It then collides simultaneously wnh balls 8 and C, whose centres are onaline perpendicular to the initial velocity of the ball A. They are innially in contact with eachother and are at rest. The ball A is aimed directly at the contact point of 8 and C. After thecollision, if the ball A comes to rest, then(A) Collision is elastic(8) collision is inelastic and final kinetic energy of the system -Initial kinetic energy of the

system = 5 J(C) Collision is inelastic and final kinetic energy of the system - Initial kinetic energy of

the system = - 5 J(D) Collision is elastic, but the final kinetic energy of the system is less than the innial

kinetic energy of the system. .

A ball of mass '2m' moving wtth velocity ul collides with another ball of mass 'm' moving

with velocity -2ul. After the collision, mass 2m moves wnh velocity (1- J). Then thechange in kinetic energy ofthe system (both balls) is

3(8) "2mu2

(D) ~mu2(1+ 2~)A small object of mass m moves in a circular orbn under an attractive central forcekr3 (i.e.F = _kr3f). The radius of the orbit is aD.Take the potential energy to be zero at theorigin i.e. r = O.The total mechanical energy of the object is

(A) kao4 (8) ~ kao4

, 41 4 ('0) 1k 4~-~ -~2 4

A particle of mass m, moving in a circle of radius R, has a velocity v = vo (1+ e-Y') .The

power delivered to the particle by the force at t = 0 is(A) mvo2y (8) 2mvo

2y(C) -2mvo2y (D) -mvo

2y.. or

"' ff 1 1"'15'".. ', .. , 0'...'..."'......'.1 "w _Space for Rough worle

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16. A mass m = OAkg is attached to a vertical rod by means of twostrings of equal length I = 2 m. The system rotates about the axis ofthe rod with '" = 4 rad/sec. as shown. If T, is the tension in the upperstring and T2 is the tension in the lower string, then(A) 1; = T2(8) T, = lOAN, T2 = 2AN

(C) 1; =3N, T2 =lON(0) 1;= T2+4N

A

B

17. A solid sphere of radius R is lying on a horizontal smooth surface. The sphere is hit by a

stick, held horizontally at a height 3R above the surface. As a result, the sphere moves2

forward with linea'r velocity Vo and also rotates about its centre of mass ,with angularvelocity "'0, such that

5 15(A) "'0 = 14R VO (8) "'0 = 4R VO

vo 5(C) "'0 =R (0) "'0 = 4R vo

18. The velocity of a particle moving along x-axis is v=a.,JX, where a. is a constant. At t = 0,the particle is at x = O. Its acceleration at time t is

2(A) ~ (8) --a.

2 2(C) .JO. ' (0) -.JX

t t

19. A ball is thrown up with an initial velocity 'v'. Air friction acts on It against its direction ofvelocity. If It takes time t, to reach the top and time t, to fall down. Then(A)t, > h (8)t, = h(C) t, < t2 (0) Not determinable


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A1TS-PT -/-PCM .JEE(Main}l158

20. A stone A of mass rnA = 0.1 kg is thrown vertically upwards with initial speed ofUA= 5 m/s from the ground. Simultaneously, another stone of mass rna = 0.2 kg isprojected from the same place with initial velocity ua = 10 m/s at an angle of 30 to thehorizontal. Then which ofthe following statement is correct?(A) A will ~se higher than B(B) A will take less time to Jeach to maximum compared to B.(C) Both A and B take same time to fall back to the ground.(0) When both finally strike the ground, the kinetic energy of A is twice that ~f B.

21. A plane is flying at a speed of 720 kmh-1 wKh respect to air. The wind is blowing at aspeed of 54 kmh-1 from west to east. With respect to ground, the plane is found to bemoving northwards. In which direction is the plane heading?

(A) North-West at angle sin-1 ( 430)to north.

(B) North-East at angle tan-1 (:o)t0 west.

(C) North-East at angle sin-' ( :0 )to north

(0) North-East at angle tan-' (:0 )to east.

22. A block of mass m slides down an inclined plane which makes an angle e with thehorizontal. The coefficient of friction between the block and the plane is ~. The forceexerted by the block on the plane is

(A) mgcosS (B) (~~2+1)mgcose .

(C) ~mgcose (0) ~mgcose. ~~2 +1


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9.AJTS-PT~-PCM.JEE(Main)f15

(B) 2hgsine.cose

(D) ~VgC;;;2 e

23. A wedge of height h is released from rest with a particle P placed on ~as shown. The wedge slides down an incline which makes an angle ewith the horizontal. All surfaces are frictionless. P will reach thesurface of the incline in time.

(A) ~ 2hgsin2e

(C)) 2h .gtanS

p

th~

e

24. The value ofthe following expression 1.(] x k) + ].(1x k) +k.(] x 1) is~O ~1(C)- 1 (D)3

2S. If A;41-2]+6k & B;I-2]-3k,theangle,which A+B makes with x-axis, is

(A) cos" ( b)'ISO

(C) cos"( ~.). 'ISO

(B) cos" ( ~)'ISO

(D) cos" ( .E.. )$0

26. Vectors A and B are given by A = 101-21] and vector B = sl +S]. Find the time whenA is perpendicular to BW~ ~~(C)2s (D) Ss

27. Two particles having posttion vectors r; ; (31+ S]) metres and [2 ; (-SI - 3]) metres aremoving with veloctties vi; (41+3])m/s and v2 ;(CtI+7])m/s. If they collide after 2

seconds; the value of 'Ct' isW2 ~4~6 ~8

",n""w",w",,, ."'55"''''",,"U'''" ""-"Z""""ss""w",ws""w",ws",,w,,,sW"'C',,"W"'ws,,", .. "" .s.. ",ex"l1""S5"Wi,",W",, ... "'Y"5S"X"WS""W"'SiW,",W"'W w",.w",w,,".,,,s s ,,,,nsw,,,q,,,s """T"" -C ... "'Y'"",c",sw"w,,,sW"'W,",i.s ", .. ""..,."'65' '5" :"6wSpace for Rough work

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AITS-PT-(.PCM .JEE(Main)/1512

6. Which ofthe following nitrate on heating decomposes to produce nitrite?(A) LiNOa (B) NaNOa(C) Mg(NOa), (D) Ca(N03)"

7. If )'0 and A are the threshold wavelength and wavelength of incident light, then thevelocity of photo electron ejected from the metal surface is

(A) /2h (Ao_).) (B) 1-2h-C-(l.-o_-;.-)Vm Vm ,(C) 2hC(Ao -A)' (D) 12h(..!-_~)

m A1,0 ~ m Ao ).

8. The correct order of stability of alkali metal compounds is?(A) LiH> NaH > KH (B) Li,C03> Na,C03 > K,C03(C) K02 > Rb02 > Cs02 (D) LiOH > NaOH > KOH

9. The radius of which of the following orbit is same as 'the radius of 2'" Bohr's orbit ofhydroaen atom.(A) 41!i' orbit of He+ (B) 20d orM of Li2+~~o~~~ ~~o~~B~

10. ' The incorrect statement among the following is(A) CO reduces aqueous PdCI2 solution to Pd.(B) CO reduces 1205 to 12,(C) CO when heated with Ni forms NiO.(D) CO reacts with CI, to form phosgene.


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13AITS-PT ~-PCM.JEE(Main)/15

11. Whichof the followingstatement is incorrect?(A) Calciumcarbidewhen heatedwith N2 producescalciumcyanamide.(B) Water gas is a mixtureof CO and H2.(C) Producergas is a mixtureof CO2 and N2.(D) SiCI. on hydrolysisproducessilicic acid.

12. PCI5(g}at a certaintemperaturedissociatesas:Pq (g).----"PC~ (g) +CI2 (g)If the degree at dissociation of PCI5 at a certain temperature is 80%. then the vapourdensityof PCI5at that temperaturewill (Atwt. of P = 31 and CI = 35.5)(A) 115.83 (B) 57.92(C) 1.287. (D) 2.572

13. Which ofthe followingpair of moleculeshave identicalshape?(A) BrFsXeOF. (B) XeO. SF.(C) CIF" XeO, (D) 502, CO2

14. Hydrogen atoms in the ground state are excited by monochromatic radiation ofwavelength ),. The resulting spectrum consists of maximum 10 different lines, Thewavelength ), is?

(A) 937.3 A(C) 102.5.6A

(B) 875.2 Ao

(D) 949.6A

15. AgNO,(s} is slowly added to a solutionwhich 0.1 Min Ct' and 0.1 Min Br-. The % ofBr- ions precipitatedwhenAgCI ion starts precipitatingis?Ksp(AgCI}= 1 x 10.10 and K", (AgBr)= 1 x 10.13.(A) 0.1 (B) 0.01(C) 99.9 (D) 99.99

"iiSfWWWWS 55 (WHo/X Sf'1 _. itW11WsiwW'WZWWiWl'WK"S1j'W:!WW ssWrlSWl1W-YWlwwwwswswwZ'$ "Wsw W-WXWi WWWWWitWWSWiWWj


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16AITS-PT-1-PCM.JEE(Main)/15

26. The increasingorder of bondangle in H,O, OF" CI,O and ICI, is(A) H,O < OF, < CI,O< ICI, (B) OF, < H.,O< CI,O < ICI,(e) OF, < H,O < ICI, < CI,O (D) ICI, < CI,O < H,O < OF,

27. Inwhich ofthe followingprocessnitrogenis oxidized?(A) NO; -4NO (B) NH; -4N,

(C) NO; -4NH; (D) NO, -4NO,

28. 10 ml of 1~H,sO, is added to 40 ml of 1~ NH,OH solution. The pH of the resulting

solution is?(pKoNH,OH = 4.76)W4~ ~9~(C) 9.24 (D) 4.76

29. The general formula of pyrosilicateis(A) SiO;'

(C) Si,o;'

(B) (Si03)~"~.

(D) Si,07'

l

30. The equilibriumconstantKpfor the reaction:H, (g) +CO, (g)~H,O(g) +CO(g)is 16. Initially 0.4 mole of H, and 0.4 mole of CO, are present in a 5 L flask. Theequilibriumconcentrationof CO,(g) is:(A) 0.08M (B) 0.016M(C) 0.32M (D) 0:064M

Space for Rough wor/(

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17AlTS..pT ~.PCM.JEE(Main)115

Mat1&ematics PART-III, ,

SECTION-A, Single Correct Choice Type

This section contains30 multiple choice questions. Each question has four choices (A), (8), (C)and (0) out of whichONLY ONE is correct.

'1. ~If lim (((X +a)(x + /3)(x +y))L X) = liml' aX+ /3x+ y")i (a,/3,y > 0) and a +/3+y.= 6th~nx-+

AITS-PT -/-PCM .JEE(Main)/1518

(A) 2x =siny(1+2x)

(C) 2x+siny(1+cx2)=0

. (X2+4)(d)2II x=secS-cosS and y=sec"S-cos"S(nEN) then 2 .Y... is

y +4 dx

(A) (n_1)2 +1 -{8) n2(C) 2n2 (0) 3n-2

If I: {1,2,3,4}---->{1,2,3,4}y = I(x) be a lunction such that

C< E{1,2,3,4} then total number01such functionsare(A) 81 (8) 36(C) 54 (0) noneolthese

Solution olthe differentialequation dy + .!tany = -;'tanysiny isdx x x

(8) 2x =siny(1+cx2)

(0) noneofthese

6.

8.

7.

9. II I(x + y) + I(y+ z) + I(z + x) ;, 31(x+2y +3z) lor allx ER, then

(A) I(X);,O'tXER (8) I(X)SO'tXER(C) I (x) is constant (0) I (x) is periodicw~hperiod 1

10. The values 01a forwhich l(x)=2e2'-4ae'-(4a-5)x is always increasing(A)a>1 (8)as1(C) a > 2 (0) a < 2

l

11. Let I = [sinx,tanxl lor the small positivevalues 01x, consider the statements(1) x EI(2) x lies left olthe pointwhich divides the interval I in 1: 2(3) x lies right olthe pointwhich divide"sthe interval I in 1: 3(4) x lies left olthe pointwhich dividesthe interval I in 1: 3(A) 1 and 2 only (8) 1,2,4(C) 1,2, 3 , (0) 1,4 onlykit J')\ \ "')'P % Space for rough work

Cc.o)7r d'1. , n.~n~G.$e(t9.fn

r19

AITS-PT~-PCM.JEE(Main)l15

sinx

12. f(x)= f (1-t+2t3)dthasin the interval [0,2,,)(A) . "d' . 3"a maximaat 4" an a minimaat "4

(C) . 5" d .. 7"a maXImaat "4 an a minimaat "4

(B) . t 3" d .. t 7"a maximaa "4an a minimaa "4

(D) noneofthese

13. If f: N~ N, f(x) = (x + 1)' +x - [J(x +1)' + (x +1)]2,then f(x) will be ([.] is GIF)(A) bijectivefunction (B)manyone function(C) periodicfunction . (D) into function

14. Let f'(X)=sinx and!J tltX)dX=f(K)-f(O),then K isx 20x 1-x

w" OO!2(e) 1 (D)~

2

15.

16.

The numberof real roots of ~5X2- 6x - 6 - ~5X2- 6x - 7 = 1 is/are(A) 1 ...-{5)2~3 ~4

16 r+1 dxThe value of L f ( )( .) is equal to,=2, 2r-x 2r+2-xW~(~ ~~(~(C) tan-'m ~n( 2~)

~. _. _15"_.. ~_ ..._"fS_jj _. _._l$_S"_1$_.. _. _. __ ~ ~ ._'1 ._. __ ._5

20AllSoPl -l-PCM .JEE(MainV15

(B) x2(D) .JX

(B) symmetric(D) equivalence

17.

18.

19.

f: [0,00)---+[0,00), f (X) is a function such that area ,bounded between the curve, x-axis

and x =t is J of the area bounded by triangle having (0,0), (t,O)&(t,f(t)) as itsvertices. f(x) can be

(A) x(C) x3

Let X be a non empty set and P(X) be the set of all subsets of X. For A,B e P(X) ARB if

and only if A v B = X then the relation R is(A) reflexive(C) transnive

I , 'dIf qx)=e,-I') for all neN and fo(x)=x then -{fo(x) } is equal todx

, d(A) fo(x)'dx{fo.' (x) } (B) f,(x).fo_, (x)

(C) ~ (x).fo_, (x) ... .f2 (x).f, (x)fo (x) (D) none ofthese

20. If maximum value of the function y = _11_1

+ 11

I is ~5' then ns value at point of1+x 1+x-a

local minima is

(A) ! (B) 13

~~ ~O3

< ,

21. An idealized hourglass is made by jOining two identical circular cones, each with baseradius 12 cm and height 18 cm, at their vertices. A tiny hole allows water to drop from theupper cone into the lower one. At a certain instant, the depth of water in the upper cone is6 em which is decreasing at a rate of 0.2 cm/min, and water in the lower cone is 10 cmdeep. The rate of increase of depth of water in lower cone 'is

(A) 1~5 cm/min (B) 890cm/min

(C) :6cm / min (D) none ofthese. Space for rough work

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21AIlS-PI -1-PCM.JEE(Main)/15

22. lim(__ 1_);" ei isequaltoX~O 1+tanx

-1

(A) e2

(C) e

1

(B) e'+1

(D) e2

23. Thedomainof the functionf(x)"= J1-~2- J3-sin2x is(A) [2n1t-~,2nn+~] (B) [n1t-~,n+~]

(C) R (D) [n1t-~,n1t+i]

24. The rangeof functionf(x) = cos' (sinx) +sin' (cosx) is

(A) [1+sin'~2+cos'1J (B) [sin' 1,1+cos' 1J

(C) (1+sin21,2~coS'1) (0)[cos'1.1+sin21J

25. Graph of y = f(x) is shown in the figure; which of thefollowingis true (where[ 1 standsforgreatestintegerx)(A) Y= [t(x)] is discontinu~usfor twovaluesof x

(8) Y=[ -f(x)] is discontinuousfortwovaluesof x

(C) f(x) = [f(x)] hasonesolution'

(0) f(x) =[ ~f(x)] hasonlyonesolution

v

26. Iff: rationalnumbers-Heai numbers.f(x) = sinx. then f is(A)oneto one.into (B)Manyto one.into(C)Bijective (D)Notamapping


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22AlTS-PT-l-PCM .JEE(Maln)/15

(B) symmetric(D) symmetric and transRive

27.

28.

29.

30.

Let S be a set of all the points in a plane, and A,B E S. If relation ARB =:> AB ,; 1 , then

relation R is(A) reflexive(C) reflexive and symmetric

A point P moves inside a square of area 4 square unRs such that R is nearer to point ofintersection of Rsdiagonal than any of the vertex. Area of the region traced by P isW4 ~2~3 ~1

If f(x) satisfies x +If(x)1 =2f(x), then f-'(x) satisfies

(A) 3x+lf-1(x)I=2f-1(x) (B~ x +If-' (x)j = 2f-' (x)

(C) f-l(x)-lxJ=2x (0)3X-lf-1(x)I=2f-1(x)

f (x) = [ 2 x[ +1] ] is discontinuous at; (where [ ]' stands for greatest integer function)x + Ix +1 .(A) one point (B) two points(C) no point (D) infinite many points

Space for rough worlc......wr "if"ZP' ssw:

L _

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I

'I

I

lEII E)LIMITED

ALL INDIA TEST SERIES-

JEE(Main), 2015

PART TEST ~ I

(15 November, 2014)

(OBJEcTIVE)

-,... . .. . . .ANSWERS & SOLUTIONS

1AITS-PT-I-PCM (Sol)-JEE (Main)/15

.JEE[Main].20 15

!$~ill ::1] l;~~j~$il :ltii~~~t!i:ij!jj:j~!:!:t!:!:~!QHgll$II~.:,t;::::m:::::}: d :,::'::::,:-:: .:::,::'::::,::.::;::.:::;::.:::'.:'::::,-:'::::,::'::':::::::':::"::;::~::::::::':-:':i :;::::\::{:i\::::;;: ..M4::tHe,MATIC$]Ji}}.1 B e A2. 0 B B3. e B B4. B A A5. 0 e 06. e B B7. 0 e B8. e A B9. A 0 e1o. e e B11 0 e e12. e B B13. B A A14. B 0 A15. e e B16. B A 017. 0 e e18. A B B19. e A A20. e e e21 A A B22. B e B23. A . e e24. e 0 025. B B 026. 0 B A27. 0 B e28. 0 e B29. A e 030. A B . . B

.i ANSWERS, HINTS & SOLUTIONSPART TEST-I

(Main]

FIITJEE

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Physics PART-I

SECTION -'-A

1.

2.

3.

2 dvV =x and 2V-;1dx

S = 0 = uT - .!gT22

. . T ; 2u, the time to retum to ITS initial position.9 .

Distance travelled fromt= ~ t~ 3; is Is(32T)-s(~)1

;u(3n-ig(32Tr -uG)+igGr=igT2,

But,jsG)-s(T)j;ig(H

Hence, option (D) is wrong.

The equation of trajectory of the stone is :gx2

y=xtane 2 22u cos e

3u2 sin2 eThe x-coordinate of the ball is ---4 .9P tt' th" . . r 3Hu mg IS mto equa lon, y; 4'

u2 sin2eWhere H=

2g

.-

4. F = ay]The work done along the path ABC ;t work doneAlong ADC; where the points are given by:A (0, 0), B (1, 0), C (1, 1), D (0,1).

f F.di';tf F.di'ABC ADC

D

A

,,,,,,,

,,,

B

C (1, 1)

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1_'_-


5. For small X,

U(X)= k(1-e-")

" k[H1-x2)]=kx2

6. The mean .position is at a distance Xo = mg I k below the striking poin\. Applying energy

conservation, mg (h +~) = !mv' + !kx~,2 2'We get V, the speed at the equilibrium position.

7. Suppose that they collide in lime \. -Then:

Vot-(!gSine)e +!(gsine)t2 =22 2 .

t = 2lVo

The quantity to be calculated is (mgsine). (gt). Vo = 2mg2lsin' e

8. Cleariy, mg sine > ~mg coseTherefore, rate ofwor1l done byFriction = ~mgcosexv

= ~mgcose .g(sine - ~cose). t

9. The v-t graph 01 the car is drawn:The total time of deceleration = 6s,

1 .displacement ='2x6x24= 72 m.

The displacementfor the rest olthe time1 .

= 960 m; '224 x t,+24 x (56-6 -t,) = 960.

Coherent, is the period of acceleration.:. t, =20s

Thus, acceleration = 24 =1.2 mts'20

tV

24m/s

10. The velocity of A, when it covers 30 m on the incline is :V2 =2. (2gsin30o-~.gcos300) .30

2or V' = 150or V = .J150 mts

After the cord is burnt off, the net lorce on A is zero, so the velocity remains constant thereafter.

11. If the block comes to rest momentarily after moving a distance x, then application of the wor1lenergy theorem gives,1 1 1 . 1'2MV; = '2M9bx' + '2kx2 and the loss IS: '2M9bx'

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AITS-PT-I-PCM (Sol)-JEE (Main)/154

12. m x 10../3 = 2mv cos 30Where m = 0.1 kgThe loss in energy is

1 {;; 1-m(10,,3)' -2x -my' = 5J2 2

restB

10../3~ _-8

5AITS-PT-I-PCM (Sol).JEE (Main)/15

Ao

20. Since both stones have the same vertical velocny: they travel for the same time, reach the samevertical height and retum at the same instant.

21. The plane's heading is given by AB which can Bbe found from the velocny triangle OAB.

22. The normal reaction is mg cos e, the friction : ~ mg cose,Their resultant is the force exerted by the block on the incline.

23. The motion of P is vertically downward; ns acceleration in the vertical direction is,

a = (gsine)sine = gsin2 e.

If t is the time taken to reach the plane, h = ~ae< 2

24. TheeJqlression =1.1+].(-}) +k.(-k)=-l

25. Wewme: A.B=IAIIBlcose and then compute cose

26. Since, A is perpendicular B, A. B = 0This gives t.

27. In order that they collide after 2s:1',+\;,t = f., +\;2t

Where t = 2. Subsmuting, we get Ct.

28. m,x, + 2mx2 = (3m)x=, 1(F)2where x =~ - t. om 2 3m29. Taking torques about the point of

. 2r ( a )asuspenSIOn, (A.1trg).- = A.-g -1t 2 2 aA.-g

2

30. Momentum: 2v = {2msF(t)dt;o .

Work done = ~.2'V22

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AITS.PT-'-PCM (Sol)-JEE (Main)/15

Chemistry

1. A(g)_B(S)+2C(g)1112= 20 min

No. of half lives = 60 = 320

200 .Pressure of A after 60 min -

2"

6

PART-II

SECTION-A

2.

3.

4.

200=-=25mm2'Pressure of C after 60 min = 175 x 2 = 350 mmTolal pressure = PA + Pc = 25 + 350 = 375 mm.

NH,HS(s)==,NH, (g) +H,S(g)0.2 +P P

P'o'a'= 0.2 + P + P = 0.2 + 2P0.2 + 2P= 0.42P = 0.2, P = 0.1Kp = Pm" x P"",= (0.2 + 0.1) x 0.1= 0.3 x 0.1= 0.03

Trisilyl amine is less basic Ihan lrimelhyl amine.

A_B

a 0a-x x

k = 0.025 M s"x=kt= 0.025 x 40 = 1 M:.a-x=2-1= 1 M

5. p". P" back bonding in BF3.

6. 2NaN03_2NaNO,+O,

4LiN03_2Li,O+ 4N02 +02

2Mg(NO,), _2MgO+ 4NO, +0,

2Ca(NO,), _2CaO+4NO, +0,

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7.

8.

9.

10.

11.

12.

13.

14.

15.


The correct order ofstabimy isLi,C03 < Na2C03 < K2CO,Ko, < Rb02 < Cs02LiOH < NaOH < KOH

Radius of 2nd orbn ,of H = aDx ~ = aDx 4 = 4aD, Z 1

Radius of 4th orbn of Be3+= aDx (4]' = 4aD

(A) PdCI2 + H20 + CO~Pd + CO2 + 2HCI.

(B) 1,0, +5CO--->5C02 +12,

(C) Ni+4CO~Ni(CO),

(D) CO + CI2--->COC~ (phosgene)

Producer gas is a mixlure of CO and N2.

MW of PCI5 = 208.5D = 208.5 = 104 252 .

80ct=-=0.8100

D-d. 0.=---.. d(n-1)

08_104.25-d. d(2-1)

d = 57.92

BrF5 and XeOF. have square pyramidal shape.

n(n-1) -102

n=5

..!. =R(.!- __1 ]A. l' 52

oA. = 949.6A

1x 10-10[Ag+] required for ppt. of AgCI = ---

0.1= 1 x 1Cr9 M[Brl remaining when AgCI starts precipnating = 1x 10~:3 10-'

. 1x10

% Br"remaining unprecipnated _1x10'" x100=0.10.1

% Br" precipnated = 100 - 0.1 = 99.9.

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16.

17.

18.

19.

20.

21.

22.

23.

24.

25.

Add~ion of inert gas at constant pressure shifts the 'equilibrium in that direction which has morenumber of gaseous moles.

In [Fe (H,O), NOJSO, ,oxidation state of Fe is +1.

[H'] =K x [CH3COOHI

[CH3COO']

10" =2xl0" )CH3COOHI[CH3COO' ]

[CH COO.] = 2xl0-5 x 0.1 - 0.023 10'"

Since V = 1 I~reNo. of moles added = 0.02

NaCI is less soluble in 020 than in H20.

N>O>Be>B

Na2[Na4(P03)sl is used to remove hardness. n is called calgon.

[OH'] = 3x 10" ,~2xl0" =2xl0.3 M

pOH = -log (2 x 10'') = 2.7pH= 14-2.7= 11.3.

K = 2.30310 [AI,t g [AI

2.303 = 2.30310g_1_1 [AI

1rog[A( 1, [AI =0.1

Rate = K [Al= 2.303 x 0.1 = O. 2303 M min.1.

2AI(S) + 2NaOH +6H,o----.2Na[ AI(OH),J +3H,

[SiClsr- does not exist because of large size of cr .26. 0/~

H H(104.5')

o/~CI CI

(111)

[C'-i-CT(Linear)1800

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r9

AITS-PT -1-PCM (Sol)-JEE (Main)/15

(+5) (+2)27. NO;~NO

(-,) (0)

NW,~N2(.5j (-,)

NO;~NW,(..) (..)N02~N02

28.

29.

2NH,OH + H,SO,

40x"'!'-=4 10x"'!'-=110 10

4-2=2 0

[NW,]pOH=pK +Iog---

[NH.oH)

=4.76+log~=4.762.

pH= 14- 4.76=9.24

Formula pyrosilicate is SiP;' .

~(NH,), SO,+2H,o

o 01

30.H2(g) +CO,(g)~H,O(g) +CO(g)

~A~~~A-~ x xKp=Kc= 16

[H,O)[CO)Kc=~~~~[H2)[CO,1

16= . Xxx(OA-x)(OA-x)

x= 0.32 .

[C02) = OA~0.32

= 0.016

FaTJElIiLtd., FrrI'.JBBHouse, 29-A, Kalu Sand, SanJt%prlya Vihar, New Delhi -11ooJ6. Ph 46J06Ooo, 215869493, Tax 265J3942rDlf1nlu: wwwJllt,j_.eom

10AilS-PI -I-PCM (Sol)-JEE (Main)/15

Mathematics PART -III

SECTION-A

1.

2.

3.

4.

5.

6.

7.

8.

9..

. (((1+a./X)(1+~/X)(1+Y/X))1I3 -1) a.+~+ylim =----x_ 1/x 3

I. (a.x + ~x + l' )'/X Iim,(a'+p~;"-3J (A )113

= 1m ---- = eH = a.pyx->O 3

~a.=~=y

If (Xo, Yo) is point of contact of tangent,

1 -sin(Xo +2yo) 1~y = =--

1+2sin(Xo +2yo) 3

sin(Xo +2yo) = 1~ Y= O,sinXo = 1

If X" Y&f(x)

2014-11-16 (5)

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