Pits 01 Kgs Final Pcm [Test 10.06.12]

8/11/2019 Pits 01 Kgs Final Pcm [Test 10.06.12]

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erformance Improvement Test Series-1

IIT JEE, 2013

Time: 3 hours Maximum Marks: 270

Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose. You are not allowed to leave the examination hall before end of the test.

INSTRUCTIONS

A. General

1. This booklet is your uestion Paper.

!. This uestion Paper contains "# $uestions. Attempt A%% the $uestions.

&. The uestion Paper contains blank spaces for your rou'h work. (o additional sheets will beprovided for rou'h work.

4. Blank papers, clip boards, log tables, slide rule, calculators, cellular phones, pagers andelectronic gadgets, in any or!, are NOT allo"ed.

5. )rite your (ame and *nrolment (o. in the space provided at the bottom of this sheet.

+. This booklet also contains the answer sheet ,-./. 0illin' in the ,-

2. ,n the ,-3 write in ink your (ame3 *nrolment (o.3 (ame of the 4entre and put your i'naturein the appropriate boxes.

6. *very $uestion has four choices for its answer 7A83 7/83 748 9 7:8.

". ,nly ,(* of them is the ri'ht answer.

1#. ,n the ,-3 for each $uestion number3 darken ,(%Y one bubble correspondin' to what youconsider to be the most appropriate answer3 from amon' the four choices.

4. -arkin' cheme

11. a8 0or each $uestion you will be awarded & marks if you have darkened only one bubblecorrespondin' to the ri'ht answer.

b8 ;n case you have not darkened any bubble you will be awarded # mark for that $uestion.

c8 ;n all other cases you will be awarded


2/28

SECTION I: PHYSICSSECTION I: PHYSICS

1. 0i'ure shows a plank with a block placed on it. Thereexists friction between plank and block but hori=ontalsurface on which plank moves is smooth. A variableforce which 'rows with time is applied on plank as

shown ;t was observed that at time t > t # relativeslippin' between block and plank starts and at t > !t#3 block separates from plank. peed of block atthis instant as observed from frame of plank is

7A8! !m'

!k

towards left 7/8

! !& m'

k

towards left

748! !" m'

!k

towards ri'ht 7:8 (one of these

!. There is a trian'ular wed'e as shown in the fi'ure. Thereare two ima'inary planes that divide the wed'e into three

portions. %et and be the inclination of these two planeswith respect to the hori=ontal as shown in the fi'ure. The

two portion of the wed'e divided by plane ; are havin'

reaction vector 1 3 similarly reaction vector between two

portion divided by plane ;; is !

)hat are the possible an'les between 1 ! and

7A8 # 7/8!

+

748 + 7:8 &. Two particles start movin' at t > # from the same point3 in a 'ravity free space3 with e$ual speeds v

directed3 respectively alon' the ?x and the ?y axes. ;f a third particle starts movin' simultaneously

from the same point alon' a strai'ht line in the x@y plane inclined at an an'le of 1&5 with the x@axissuch that all the three remain collinear for all values of t3 then the speed of the third particle will be

7A8 ! v 7/8 v!

748 !v 7:8 !! v

. The fi'ure shows two rods /4 and ,A hin'ed from / and , respectively.od /4 has a 'roove in which point A of rod ,A can slide smoothly. At a

'iven instant3 the rod /4 has an an'ular velocity #. The speed of end ,Aof the rod A at this instant will be

7A8 #l 7/8 !#l748 l 7:8 (one of these.

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3/28

5. A small body slides from rest alon' two e$ually rou'h circular surfaces in

the vertical plane from A to / throu'h path 1 and ! of e$ual radius if v 1and

v!are the speed of the block at point / via path 718 and 7!83 then

7A8 v1B v! 7/8 v1C v!

748 v1> v! 7:8 canDt be predicted

+. A proEectile is thrown with speed u from ori'in , in a vertical plane asshown in the fi'ure. ;f r is the position vector of the proEectile and v isits velocity vector at any instant t. The distance of the proEectile from theori'in always increases with time durin' its motion. 4hoose the correctstatement.

7A8 r v # < 7/8 r v # =

748 r v # > 7:8 r v # >

2. Two inclined planes ,A and ,/ of inclinations to the

hori=ontal are and 3 each e$ual to as shown in thefi'ure. A particle is proEected at an an'le of "#with plane,A from point A and its strikes the plane ,/ at point /normally. Then find the speed of proEection in mFs. 7'iven

that ,A > ,/ > !# cm and ' > 1# mFs!87A8 ! 7/8 5748 1# 7:8 !#

6. /lock of mass m > 1# k' is suspended by two sprin's as shown in thedia'ram. Acceleration of block Eust after cuttin' the ri'ht sprin' is

7A85 &

!mFs! 7/8 1# mFs!

748 5& mFs! 7:8 5 mFs!

". Two persons namely am3 and ahim3 who have 'one to picnic.0indin' a beautiful place they rush towards it simultaneously. %et us

take the time when they start simultaneously at t > #. At this instant3the position of am and ahim be x > # and x > ? 6 respectively.am maintains a uniform velocity of ?1# mFs throu'hout the Eourneywhereas ahim starts from rest and continuously accelerates with ? 1

mFs!3 ;f the position@time 'raph for two persons and be depictedas shown in the dia'ram then choose the correct alternative

7A8 ;nitially3 and are at relative rest7/8 The minimum separation between and is A4748 The minimum separation between and is A/7:8 The speed of and continuously increases

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4/28

1#. An ant carryin' a char'e has to crawl alon' the inner surface of a cube of side a . tartin' from onecorner of the cube it has to 'o from the dia'onally opposite corner in least possible time . Assumin'that the ant crawls with a constant speed v 3 find the maximum ma'netic field produced by the ant atthe centre of the cube .

7A8 #

!

5 $v

+ $

7/8 #

!

5 $v

+ $

748 #

!

5 $v

+ $

7:8 (one of these

11. A small block of mass m3 havin' char'e $ is placed on frictionless inclined

plane makin' an an'le with the hori=ontal. There exists a uniform ma'neticfield / parallel to the inclined plane but perpendicular to the len'th of sprin'. ;fm is sli'htly pulled on the inclined in downward direction3 the time period ofoscillation will be 7assume that the block does not leave contact with the plane8

B

7A8 m

!G

7/8 !m

!G

748 $/

!G

7:8 $/

!!G

.

1!. A uniform conductin' rectan'ular loop of sides l 3 b and mass mcarryin' current i is han'in' hori=ontally with the help of two verticalstrin's. There exists a uniform hori=ontal ma'netic field / which isparallel to the lon'er side of loop. The value of tension which is least is

7A8mg Bb

2

7/8

mg Bb

2

+

b

B

l

748mg 2iBb

2

7:8

mg 2Bb

2

+.

1&. ;n fi'ure3 infinite conductin' rin's each havin' current i in the

direction shown are placed concentrically in the same planeas shown in the fi'ure. The radius of rin's are r3 !r3 !!r3 !&r H.

. The ma'netic field at the centre of rin's will be7A8 Iero

7/8 0i

r

748 0i

2r

7:8 0i

3r

.

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5/28

1. A particle of specific char'eq

m= 4Fk' is proEected from the ori'in towards positive x


6/28

1". 4onsider a spherical isolated conductin' body of radius r1havin' a concentric cavity of radius r!with a

point char'e $ at the centre. *lectric field E will have ma'nitude 20

q

4 rfor a point at a distance r

from centre for

7A8 r B # 7/8 10 r r 748 # C r C r! 7:8 only for r B r1

!#. A uniform solid sphere 7A8 is surrounded symmetrically by a uniform thin spherical shell 7/8 of e$ualmass but twice its radius 7i.e. r/> !rA8. The 'ravitational field ' is plotted as a function of the distancefrom the center of the solid sphere 7r8. )hich of the followin' shows the correct plotM

7A8 r

g

rA r BO 7/8 r

g

r A r BO

748O r

g

rA r B 7:8 r

g

r A r BO

!1. A point char'e $ is placed at P7#3 #3 a8. The electric flux throu'h

4A/ due to the electric field of $ is . Then

7A80

q

48 =

7/8

0

q

48 ? 14 and $! > < !4 are placed at A and /respectively as shown in the fi'ure. The distance between $1and $!is cm. A B

q 1

4 ' m

q 2

!5. A line of force emanates from $1makin' an an'le "#N with A/. This line of force7A8 enters $!at an an'le "#N 7/8 enters $!at an an'le +#N

748 enters $!at an an'le 5N 7:8 does not enter $!but 'oes off to

!+. The net electric flux will be =ero.7A8 over any surface that encloses a volume includin' A and /3 but havin' very lar'e radius7/8 over any surface that includes A twice and / once748 over any surface that encloses a volume excludin' A and /7:8 only over a surface that encloses =ero volume

!2. The electrostatic potential is =ero at7A8 a point on the line A/ between $1and $!but closer to $!7/8 a point on the line A/ but not between A and /

748 infinitely many point in space7:8 no point in space

#aragraph or 0uestion Nos. )% to -'

An obEect is thrown from a point A on hori=ontal'round as shown Air blows from ri'ht to left so that itimparts a constant acceleration Oa# to obEecttowards left. Answer the followin'

!6. The value of Oa# such that proEectile follows a strai'ht line path is

7A8 ' tan 7/8 ' cot748 ' sin 7:8 ' cos

!". ;f condition in above case is satisfied then the maximum vertical hei'ht attained by obEect abovehori=ontal 'round is

7A8

! !#u sin

'

7/8

! !#u sin

!'

748

! !#!u sin

'

7:8 (one of these

. ;n the above case mentioned3 when obEect falls back on surface3 it will strike the surface

7A8 To the left of point A 7/8 To the ri'ht of point A748 *xactly at point A 7:8 :ata insufficient

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9/28

SECTION II: CHEMISTRYSECTION II: CHEMISTRY

This section contains -' !ultiple choice 2uestions. *ach $uestion has four choices 7A83 7/83 748and 7:8 out of which ON3 ON5 is correct.

1. The product of the followin' reaction is

4Q&

,

&Q , Product+

7A8 ,Q

4Q&,Q

7/8

,Q4Q&

,Q 748

4Q&

,Q 7:8 4Q&

,Q

4Q&

!. The product / in the reaction is! +

!

/ Q4u

& & 52& G ,Q F: ,74Q 8 4,Q A /

7A8

Q&4 4Q&

4Q&

,:

7/8

Q&4

4Q&

Q ,Q

748

Q&4

4Q&

Q ,:

7:8

Q&4

4Q&

: ,Q

&. A few drops of Q&P,were added to a hydrocarbon 4&Q+and heated with ben=ene. ,xy'en'as was then passed throu'h the hot product to form 7A8. 7A8 was then further treated withconc. Q!, to form 7/8. ;dentify the correct combination of final product 7/83 theintermediate3 and oxidation number of oxy'en in 7A8.

O6idation nu!ber +7 Inter!ediate #roduct +B7A8

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10/28

. A compound R:D 746Q1+,8 has followin' observations7a8 ;t reacts with %ucas rea'ent within 5 minutes.7b8 ,n heatin' in presence of acid 'ives a compound S3 which can decolouri=e -n,as

well as /r!F44l.7c8 ,=onolysis (a,;4ompound S Y ! moles of iodoform sodium salt of dibasic acid +

odalime

n@butane

Then :3 S and Y are7A8

,Q

3 3 ,,

7/8

Q,

3 3

,

,

748Q,

3

,Q

3

,

,

7:8

,Q

33

,

,

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11/28

5.

4Q!4Q!(-e&4l

4Q&

,

)olffishner

reduction Product 7P8 -

The structure of 7P8 is

7A8

4Q!4Q!(-e&4l

4Q&7/8

4Q&

,

4Q!

7484Q&

4Q!

7:8

4Q&

,

4Q!

+. ,

4Q&

&

7i8 (aQF dry ether

7 ii 8 4Q ; 7A 8

7maEor8&

7i8 %:A

7ii8 4Q ;7/8

7maEor8

Qence3 A and / are

7A8 ,

Q&

4 4Q&

and

,

4Q&

4Q&

7/8 ,

Q&4 4Q&

,

4Q&

4Q&

and

748 ,

4Q&

Q&4

,

4Q&

4Q&

and

7:8

,

4Q&

4Q&

,

4Q&

4Q&

and

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12/28

2.

,

Q&4 4Q&

,

! &a$ueous G 4, -aEor product 7A8

Product 7A8 in the above reaction is

7A8

,Q

4Q& 7/8

Q,

,

4Q&

748 4Q&

,Q

,

7:8

Q,

4Q&

,

6.

/r

Ph

alc.G,Q

U Product

;dentify the product.7A8

,Q

Ph

7/8

Ph

748

Ph

7:8

,QPh

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13/28

". ;n the reaction se$uence

4+Q5< 4Q > 4Q @,;;4

@4Q&&

!

7i84Q -'/rJSK

7ii8Q , F Q+7maEor product8

S will be

7A8

4+Q5

4Q& 4Q&

,Q

7/8

Q54+ 4Q&

,Q

4Q&

7484+Q5

4Q& 4Q&

,

7:8

4+Q5

4Q&

4Q&

,Q

4Q&

#7SS7857ro!aticity

$uestions %0 to %%:Aromatic compounds are more stable than their non@aromatic counterparts. There are differentways to measure the so@called aromatic stabili=ation ener'y. The followin' experiment wasperformed to compare the stabili=ation ener'y of ben=ene with that of naphthalene 7Aromaticstabili=ation ener'y of naphthalene > +1 kcalFmol and that of ben=ene > &+ kcalFmol8.

40&0&4

(4 4(

bG

40&

40&

4(

4(

40&0&4

(4 4(

nG

40&

40&

4(

4(

1#. The relation between band nat same temperature is 7band nare e$uilibrium constants87A8 bC n 7/8 bB n748 b> n 7:8 :ata insufficient

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14/28

11.

!/r P

P can be

7A8 /r

/r

7/8 /r /r

748 /r

/r

7:8 (one of the above

1!.

4Q&

,

4,,4!Q5

& !

7i8 (aQF TQ0 7i8 dil. G,Q

7ii8 4Q 4,4l 7ii8 Q ,FQ 37A8 7/8+

The product 7/8 in the above se$uence of reaction is

7A8

4Q&

,

4,,4!Q5

4,4Q&

7/8

4Q&

,

4,,Q

4,4Q&

748

4Q&

,

4Q&

, 7:8

4Q&

,

4,,Q

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15/28

1&. A certain metal A has fcc structure with unit cell len'th &.1+ o

A3 the si=e of lar'est atom that

can be accommodated into the interstices of the metal without chan'in' the position of metalatoms is

7A8 o

#.5& A 7/8 o

#."&5 A 748 o

#.1" A 7:8 (one of the above

1. ;n diamond3 the lattice is fcc with 4 atoms occupyin' lattice points as well as some of the

voids. )hich of the followin' statement7s8 isFare correct about the diamond structureM7;8 ;n the diamond lattice3 half of the tetrahedral voids are occupied.7;;8 Percenta'e space occupied 7packin' efficiency8 is 2V.7;;;8 ;n the diamond lattice all octahedral voids are occupied.7A8 ,nly ; 7/8 ,nly ;;; 748 ;3 ;; and ;;; 7:8 ,nly ;;

15. ;n a fcc crystal3 which of the followin' shaded plane contains the followin' arran'ement ofatomsM

7A8 7/8

748 7:8

1+. ;n an ionic solid3 /x


16/28

16. uppose you have a collection of 'lass marbles arran'ed in hexa'onal close packin'. ;f the

density of the 'lass is &!.+ ' F cm 3 what would be the avera'e density of matter in the

collection

7A8 &!.& ' F cm 7/8 &!.+ ' F cm 748 &1."!5 'Fcm 7:8 &&.5 'F cm

1". ;f all 0e!?ions are replaced by r? ions in #.1 mole of 0e#.",3 then number of vacancies

created are7A8 #.#&5 (A 7/8 #.#1 (A 748 #.#2 (A 7:8 #.1 (A7(Ais Avo'adro number.8

!#. /aTi,& crystalli=es in perovskite structure. This structure may be described as a cubiclattice3 with barium ions occupyin' the corners3 oxide ions at the face centres and titaniumions in one of the octahedral holes. )hich of the followin' is correct about /aTi, &crystallatticeM7A8 Ti?occupies the body centred octahedral hole.7/8 Ti? occupies the ed'e centred octahedral hole.748 Ti? can occupy any of the octahedral holes.7:8 1F!nd of the octahedral holes are occupied.

!1. )hich one of the followin' is true for chottky defect in crystalsM7A8 chottky defects increases the density of the crystal.

7/8 chottky defect is found in crystal havin' lar'er

r

+

ratio.

748 chottky defects are non thermodynamic defects.7:8 chottky defect is shown by hi'hly ionic compounds havin' hi'h coordination number.

!!. The closest distance between two atoms amon' the followin' unit cells3 assumin' sameed'e len'th RaD in each unit cell7A8 :iamond unit cell 7/8 Primitive cubic cell 748 /44 unit cell 7:8 044 unit cell

!&. +."# 'm of metal carbonate were dissolved in +# ml of ! ( Q4l. The excess acid wasneutrali=ed by !# ml of 1 ( (a,Q. )hat is the e$uivalent wei'ht of metalM7A8 # 7/8 !# 748 1" 7:8 &"

!. x ml of ,!at (TP reduces 1## ml #.1 ( !4r!,2in presence of Q!,. The value of x is7A8 !! ml 7/8 5+ ml 748 6 ml 7:8 11! ml

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17/28

!5. & ! ! !(Q , ( Q ,+ +

;f molecular wei'hts of (Q&and (!are -1and -!respectively3 and their e$uivalent wei'htsare *1and *!respectively3 then 7*1< *!8 is e$ual to

7A8 1 !!- -

+

7/8 ( )1 !- - 748 ( )1 !&- - 7:8 ( )1 !- &-

!+. A metal oxide has #V oxy'en. The percenta'e of chlorine in its chloride is7A8 "#V 7/8 66.25V 748 2.2V 7:8 &5V

!2. Gaseous ben=ene reacts with hydro'en 'as in presence of a nickel catalyst to form 'aseous

cyclohexane accordin' to the reaction3 + + ! + 1!4 Q 7'8 &Q 7'8 4 Q 7'8.+ A mixture of 4+Q+andexcess Q!has a pressure of +# mm of Q' in an unknown volume. After the 'as is passedover a nickel catalyst and all the ben=ene converted to cyclohexane3 the pressure of the 'asis mm of Q' in the same volume at the same temperature. The fraction of 4 +Q+ 7byvolume8 in the ori'inal mixture is

7A8 1

&7/8

1

748

1

57:8

1

+

!6. ,ne mole of a hydrocarbon havin' molecular wei'ht 26 is subEected to o=onolysis and theQ!,!formed as a result is not removed from the reaction vessel. The product obtained3 onfurther treatment with 4a!?ions3 'ave &6 ' of calcium oxalate. The hydrocarbon is

7A8 & &4Q 4 4 4 4 4Q 7/8 ! !4Q 4 4Q 4Q 4 4Q

748 ! &4Q 4 4Q 4 4 4Q 7:8 ben=ene

!". ;f 1m ' of a metal A displaces !m ' of another metal / from its salt solution and if their

e$uivalent masses are 1 !* and * respectively3 then the e$uivalent mass of A can be

expressed as

7A8

! !

11

m *

* m

= 7/8 1 !

1!

m *

* m

= 748 1 !

1!

m m

* *

= 7:8 1 !

1!

m *

* m

=

. ,Q 7L1ml 3 x(8 is mixed with (a,Q 7L!ml 3 y(8 9 treated with #.1 ( Q4l. 1## ml of Q4l isre$uired for complete reaction3 if L1L!> 1! 9 xy > !1. )hat fraction of Q4l is neutrali=edby (a,Q.7A8 #.5# 7/8 #.!5 748 #.&& 7:8 #.+2

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18/28

SECTION III: MATHEMATICSSECTION III: MATHEMATICS

This section contains -' !ultiple choice 2uestions. *ach $uestion has four choices 7A83 7/83 748and 7:8 out of which ON3 ON5 is correct.

1. ;f fractional part of1

xand x!for some x ( )!3 & are e$ual then the value of x1 1

1

sin x cos x is

7A8 x 1

3 1!

7/8 x 1

13!

748 x 1 1

13 3 1! !

7:8 x J x J!3 5K3 then

( ) ( )( )5

1

!

f x f x dx+ is7A8 # 7/8 748 5 7:8 !1

. %et f7x8 > x ? sinx3 then ( )1

1

a

f x dx is e$ual to3 where a > 1 1x #1 1

lim sec secx x

3 7where

J.K denotes the 'reatest inte'er function87A8

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19/28

5. ;f the derivative of f7x8 w.r. t. x is

( )xf

xsin!

1 !3 then f7x8 is a periodic function with period

7A8 7/8 ! 748 F! 7:8 none ofthese.

+. ;f lo'&7lo'!x8 ? lo'1F&7lo'1F!y8 > 1 and xy!

> "3 then7A8 x > !"2 7/8 x > 2!"

748 y > 1

&&7:8 1F1#

2. ;f 'raph of xy > 1 is reflected in y > !x to 'ive the 'raph 1!x!? rxy ? sy!? t > #3 then7A8 r > 13 s > 1!3 t > !5 7/8 r > 1!3 t > 1748 r > !5 7:8 r ? s > x!? y!< xy3 then -1.-!is

7A8 6 7/8 "

748 + 7:8 2

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20/28

11.

+

1n

n n

11ln1lim is e$ual to

7A8 # 7/8 1748 e 7:8 none of these

1!. %et f be a continuous and differentiable function in 7x13 x!8. ;f f7x8 . f7x8 x ( )( )1 f x and

( )( )1

!

x xlim f x 1

+= and ( )( )

!

!

x x

1lim f x

!= . Then minimum value of ! !1 !x x is 7where J.K denotes

the 'reatest inte'er function87A8 & 7/8 1748 ! 7:8

1&. %et y > k be a line that intersects the curve y > f7x8 atleast at two points. ;f

( ) ( )

x !!!

! x

xf t dt t f t dt

!= + 3 then

7A8 k 7 Pn < 17x < n83 where P#7x8 > x&? 5x!< 22x < 63

then the coefficient of x in P1#7x8 is7A8 < 5# 7/8 !1"#748 1!"6 7:8 !"61

15. ;f 3 7where C 8 are the points of discontinuity of the function '7x8 > f7f7f7x8883 where

f7x8 > 11 x

3 then the points of discontinuity of '7x8 is

7A8 x > #3 1 7/8 x > 1 only748 x > # only 7:8 x > #3 1

1+. ;f k >! !

1

x 1lim sec

ln x x 1

exist3 then the minimum value of JK is 7where J.K denotes the

'reatest inte'er function87A8 ! 7/8 +748 1 7:8

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21/28

12. ;f f7x8 > 7a!< &a ? !8 ( )! !x xcos sin a 1 x cos1

+ + does not posses critical point then

the number of inte'ral values of RaD are7A8 7/8 !748 & 7:8 1

16. ;f f7x8 > 7x < 181##7x < !8!7""87x < &8&7"68H 7x < n81##3 let k >( )

( )

f 1#1

f 1#1

3 then

k"2

5#

is

7A8 7/8 !748 & 7:8 1

1". ;f !p!< &$!? p$ < p > # and a variable line px ? $y > 1 always touches a parabola whoseaxis is parallel to x@axis. Then the e$uation of the parabola is7A8 7y < 8!> !7x < !8 7/8 7y < &8!> 1!7x < 18748 7y < 8!> 1!7x < !8 7:8 7y < !8!> !7x < 8!

!#. ;f m13 m!be the roots of the e$uation x!? 7 & ? !8x ? & < 1 > # and the area of the

trian'le formed by the lines y > m1x3 y > m!x and y > ! is && ? 3 then >7A8 # 7/8 11

748 11 7:8 1!1

!1. A continuous3 even periodic function f with period 6 is such that f7#8 > #3 f718 > < !3 f7!8 > 13f7&8> !3 f78 > &3 then the value of tan


22/28

#aragraph or 0uestion Nos. ) to )1

To the circle x!? y!> two tan'ents are drawn from P73 #83 which touches the circle at T1and T!and a rhombus PT1PT!is completed.

!5. 4ircum centre of the trian'le PT1T!is at

7A8 7!3 #8 7/8 7!3 #8

748&

3 #!

7:8 none of these

!+. atio of the area of trian'le PT1Pto that the PT1T!is7A8 ! 1 7/8 1 !

748 & ! 7:8 none of these

!2. ;f P is taken to be at 7h3 #8 such that Plies on the circle3 the area of the rhombus is7A8 + & 7/8 ! &

748 & & 7:8 none of these

#aragraph or 0uestion Nos. )% to -'

%et ( ) ( ) ( )2 30 1 2 3' x a a x a x a x and f x ' x 3= + + + = f7x8 has its non@=ero local minimum and

maximum values at & and & respectively. ;f a&the domain of the function ( )2

1 1

2

xh x sin

x

+=

.

!6. The value of a1? a!is e$ual to

7A8 7/8 !2 7:8 !2

!". The value of a#is7A8 e$ual to 5# 7/8 'reater than 5748 less than 5 7:8 less than 5#

. f71#8 is defined for7A8 a#B 6 7/8 a#C 6 a#> 6 7:8 none of these

!ace for rou"h #ork



#ITS()'$)(#C*(+$%$&.'())


23/28


IIT JEE, 2013

7NS95RS,,


1. B !. 7 &. B . B

5. 7 +. C 2. 7 6. :

". C 1#. 7 11. 7 1!. C

1&. : 1. B 15. : 1+. :

12. : 16. C 1". C !#. :

!1. B !!. C !&. 4 !. B

!5. / !+. C !2. 4 !6. B

!". B . C


1. B !. C &. : . 7

5. C +. B 2. / 6. C

". B 1#. A 11. / 1!. C

1&. : 1. 7 15. C 1+. C

12. C 16. C 1". 7 !#. 7

!1. : !!. 7 !&. : !. :

!5. 7 !+. C !2. : !6. :

!". B . 7

SESECTION III: MATHEMATICSCTION III: MATHEMATICS

1. B !. C &. : . B

5. 7 +. B 2. C 6. C

". B 1#. C 11. 7 1!. B

1&. C 1. C 15. : 1+. C

12. B 16. 7 1". C !#. B

!1. : !!. B !&. C !. B

!5. 7 !+. : !2. 7 !6. C

!". B . 7



#ITS()'$)(#C*(+$%$&.'()-

#aper Code

$%$&.'


24/28


SOLUTION


1. t#can be found by e$uatin' pseudo force on block to static friction

*$n. for block w.r.t. plank

dv ktm m m'

dt &m !

=

7after relative slippin' starts8

#

#

!tv

# t

kt m'm dv m dt

&m !

=

! !& m'v

k

= 7towards left8

!. 1 is vertical and ! is also vertical

&. At any time t3 ,A > ,/ > vt. 0or third particle ,4 > ut. 0rom similartrian'les A4: and A/,3

ut !

vt>

utvt!

vt

implifyin'3 we 'et u >v

!

. ,A> !/4

2. ince > A, > /, > !# cm

A/ > ! !d d !d dcos1!#+ A/ > &d H.7i8(ow A/ is the ran'e of the proEectile from A to /.

&d >!u sin!

'

&d >!u &

!'

Puttin' d > #.! mu > ! mFs

11. There will be no effect of ma'netic force on timeperiod because the ma'netic force will beperpendicular to the inclined plane.

q , m

1!. Takin' moments about point / to be =ero.

1( ) ib B mg2

+ = l

l l

1

mg 2ibB

2

= .

B

m g

1

2

A



#ITS()'$)(#C*(+$%$&.'()4


25/28

1&. 0 0 0

2

i i iB )))))))

2r 2(2r ! 2(2 r!

= +

0

2 3

i 1 1 1B 1 )))))

2r 2 2 2

= + +

0i 1B12r

12

=

0iB3r

= .

1. Time period2 m

qB

=

2 1

2

= =

s.

Thus particle will beat point P after1

t12

= s. P

v

y

x

3 0 *

v 10+',-30i -i.30j/= +

3 1 v 10 i j & 3 i j2 2

= + = +

rmFs.

!1. The flux is 'reater in the nearer half of the s$uare ,A4/.

!6. The 'iven condition is possible when net acceleration to direction of proEection is =ero

#

#

a sin 'cos

a 'cot

=

=

!". Zse ! !v u !as = alon' direction of proEection. ;t follow strai'ht path up3 then down back to fall to same point A.


5. )olff


26/28

1!.

4Q&

,

4,,4!Q5

(aQFTQ0 4Q&

,

4,,4!Q5

&4Q 4,4l 4Q&

,

4,4Q&

4,,4!Q5

7A8

7i8 dil. G,Q 7ii8 Q

?

4Q& 4Q

,

4,4Q&

4,,Q

7 3 keto acid8

!4,

4Q&

,

4Q&

,

7/8

1. ;n diamond3 all the carbon atoms are tetrahedrally bonded to one another so octahedral voids cannotbe occupied. Also3 the lattice points and the holes are occupied by carbon atoms. As a result thepackin' efficiency is less than 2V.

12. r

#.1 #.2&!r

+

= 3 4o. (o. and +

r#.!!5 #.13

r

+

= 4o. (o <

Qence 748

16. uppose vol. of the hexa'on &xcm=%et us assume that entire hexa'on is occupied by 'lass marbles without leavin' voids.

The wei'ht of halo'en !.+ x 'm=

/ut we know only 2.#5V space is occupied in a hexa'on.

o actual wei'ht of hexa'on!.+ x 2.#5

1##

=

:ensity of hexa'on!.+ x 2.#5

1## x

=

&1."! 'Fcm=

1". ,n the basis of solution of . (o. &20e!?> #.#2 mole in #.1 mol of 0e #.",o to balance char'e3 some of the place has been left vacant as 0e!?is replaced by r!?.#.#2 [ ! > x [

#.#2 !x #.#&5

= =

Lacant place > #.#2 < #.#&5 > #.#&5 (AQence3 7A8.

!!. d4

dt=

:urin' meltin' of ice temperature remains constant. Therefore3 4 tends to infinity

!&. *$uivalent of Q4l taken > +# [ ! [ 1# !# [ 1 [ 1# 1## [ 1# &"



#ITS()'$)(#C*(+$%$&.'()/


27/28

!. ! ! 2 ! ! ! ! & !G 4r , Q , &, G , 4r 7, 8 Q ,+ + + +1## ml #.1 ( !4r!,2> #." 'm!" 'm !4r!,2> & [ !!## ml ,!at (TP #." 'm > 11! ml of ,!

!2. P1? P!> +# mm Q' H718

After the reaction3 pressure of Q!> P!< &P1Pressure of 4+Q1!> P1P!< &P1? P1> mm H7!8

olvin' 718 and 7!83 we 'et fraction of ben=ene1

+=

. me$. of ,Q ? me$ of (a,Q > me$ of Q4l

xL1?yL!> 1#

!y ! !L

y L 1#!

+ =

L!y > 53 xL1>5

SESECTION III: MATHEMATICSCTION III: MATHEMATICS

5. Given f 7x8 >( )xf

!

1xsin! +

f 7x8.f7x8 >!

x!cos

!

1xsin! !=+

( ) dxx!cos!

1dxxf8x7f =

or( )[ ]

x!sinc

!

xf !

=+

f7x8 > ( ) c!x!sin!

1

)hich is a periodic function with period .Qence 7A8 is the correct answer

6. ( )

h

xtanhxtan 11 +> sin!y < hsin!y.

!

y!sin? H.

(ow3 takin' limits of both sides as h #3 we 'et

ysin

x1

1 !!=

+

x!> cosec!y < 1 > cot!y

x > coty 3 y > cot@17x8

11.

+

1n

n n

11lo'1lim >

+

+

n

11

n

11

lo'1lim

n

n

>

++

+

n

11lo'

n

11lo'1lim

n

n

>

+

++

n

nn n

11lo'lim

n

11lo'1lim > 1 ? # < lo'e > 1 # .

!!. KxJlimx >

[ ]x

xxlimx



#ITS()'$)(#C*(+$%$&.'()1


28/28

> [ ]x

xlimx

. xlimx

> xlimx

[ ]

=

1

x

xlimasx

KxJ

1nn

x e

1nxxlim

++

>x

1nn

x

e

1nxxlim

++

>

( ) ....

\1n

x

\n

x....

\!

xx1

1nxxlim

1nn!

1nn

x

++

+++++

+++

>

( ) ( ) ....

\!n

x

\1n

x

\n

1....

x!

1

x

1

x

1

x

1

x

n1

lim!

!n1nn

n

x

++

++

+++++

++

>#

1

\n

1#.....##

##1 =

=+++++

++.

#ITS()'$)(#C*(+$%$&.'()%

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Pits 01 Kgs Final Pcm [Test 10.06.12]