leader test series / joint package course · allen-2/42 note : i cor , [email protected]...

READ THE INSTRUCTIONS CAREFULLY / Ïi;k bu funsZ'kks a dks /;ku ls i<+ s aGENERAL / lkekU; %1. This sealed booklet is your Question Paper. Do not break the seal till you are told to do so.

;g eksgjcU/k iqfLrdk vkidk iz'ui= gSA bldh eqgj rc rd u rksM+s tc rd bldk funsZ'k u fn;k tk;sA2. Use the Optical Response sheet (ORS) provided separately for answering the questions.

iz'uksa dk mÙkj nsus ds fy, vyx ls nh x;h vkWfIVdy fjLikal 'khV (vks- vkj- ,l-) (ORS) dk mi;ksx djsaA3. Blank spaces are provided within this booklet for rough work.

dPps dk;Z ds fy, bl iqfLrdk esa [kkyh LFkku fn;s x;s gSaA4. Write your name, form number and sign in the space provided on the back cover of this booklet.

bl iqfLrdk ds fiNys i`"B ij fn, x, LFkku esa viuk uke o QkWeZ uEcj fyf[k, ,oa gLrk{kj cukb;sA5. After breaking the seal of the booklet, verify that the booklet contains 42 pages and that all the

18 questions in each subject and along with the options are legible. If not, contact the invigilator forreplacement of the booklet.bl iqfLrdk dh eqgj rksM+us ds ckn Ïi;k tk¡p ysa fd blesa 42 i`"B gSa vkSj izR;sd fo"k; ds lHkh 18 iz'u vkSj muds mÙkj fodYi Bhdls i<+ s tk ldrs gSaA ;fn ugha] rks iz'ui= dks cnyus ds fy, fujh{kd ls lEidZ djsaA

6. You are allowed to take away the Question Paper at the end of the examination.ijh{kkFkhZ iz'ui= dks ijh{kk dh lekIrh ij ys tk ldrs gSaA

OPTICAL RESPONSE SHEET / vkWfIVdy fjLikal 'khV (vks-vkj-,l-) %7. The ORS will be collected by the invigilator at the end of the examination.

vks- vkj- ,l- dks ijh{kk ds lekiu ij fujh{kd ds }kjk ,d= dj fy;k tk,xkA8. Do not tamper with or mutilate the ORS. Do not use the ORS for rough work.

vks- vkj- ,l- esa gsj&Qsj@foÏfr u djsaA vks-vkj-,l- dk dPps dke ds fy, iz;ksx u djs aA9. Write your name, form number and sign with pen in the space provided for this purpose on the ORS.

Do not write any of these details anywhere else on the ORS. Darken the appropriate bubbleunder each digit of your form number.viuk uke vkSj QkWeZ uEcj vks-vkj-,l- esa fn, x, [kkuksa esa dye ls fy[ksa vkSj vius gLrk{kj djsaA buesa ls dksbZ Hkh fooj.kvks-vkj-,l- es a dgha vkSj u fy[ks aA QkWeZ uEcj ds gj vad ds uhps vuq:i cqycqys dks dkyk djsaA

DARKENING THE BUBBLES ON THE ORS / vks-vkj-,l- ij cqycqyks a dks dkyk djus dh fof/k %10. Use a BLACK BALL POINT PEN to darken the bubbles on the ORS.

vks-vkj-,l- ds cqycqyksa dks dkys ckWy ikWbUV dye ls dkyk djsaA11. Darken the bubble COMPLETELY. / cqycqys dks iw.k ± :i ls dkyk djsaA12. The correct way of darkening a bubble is as : / cqycqys dks dkyk djus dk mi;qDr rjhdk gS : 13. The ORS is machine-gradable. Ensure that the bubbles are darkened in the correct way.

vks-vkj-,l- e'khu tk¡P; gSA lqfuf'pr djsa dh cqycqys lgh fof/k ls dkys fd, x;sa gSaA14. Darken the bubbles ONLY IF you are sure of the answer. There is NO WAY to erase or

"un-darken" a darkened bubble.cqycqys dks rHkh dkyk djsa tc vki mÙkj ds ckjs esa fuf'pr gksA dkys fd, gq, cqycqys dks feVkus vFkok lkQ djus dk dksbZ rjhdk ugha gSA

15. Take g = 10 m/s2 unless otherwise stated. / g = 10 m/s2 iz;qDr djsa] tc rd fd vU; dksbZ eku ugha fn;k x;k gksAQUESTION PAPER FORMAT AND MARKING SCHEME / iz'ui= dk izk:i vkSj vadu ;k stuk %16. The question paper has three parts : Mathematics, Physics and Chemistry.

bl iz'ui= esa rhu Hkkx gSa % xf.kr] HkkSfrd foKku ,oa jlk;u foKkuA17. Each part has two sections as detailed in table page 42.

izR;sd Hkkx esa nks [k.M gSa ftudk fooj.k i`"B 42 ij rkfydk esa fn;k x;k gSA

LEADER TEST SERIES / JOINT PACKAGE COURSETARGET : JEE (Main + Advanced)

DISTANCE LEARNING PROGRAMME

Please see the page 42 of this booklet for rest of the instructions/Ïi;k 'ks"k funs Z'kk sa ds fy, bl iqfLrdk ds i`"B 42 dks i<+ s aA

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LS W

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TO

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SO B

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VIGI

LATO

R / f

ujh{

kd d

s vuqn

s'kksa d

s fcuk

eqgj

sa u r

ksM+s

Test Type : ALL INDIA OPEN TEST (MAJOR) Test Pattern : JEE-AdvancedTime : 3 Hours Maximum Marks : 183PAPER – 1

ALLEN

LTS-2/42

Note : In case of any correction in the test paper, please mail to [email protected] within 2 days along with Paper Code& Your Form No.(uksV % ;fn bl iz'u i= esa dksbZ Correction gks rks Ïi;k Paper Code ,oa vkids Form No. ,oa iw.kZ Test Details ds lkFk 2 fnu ds [email protected] ij mail djsaA)

Space for Rough Work / dPps dk;Z ds fy, LFkku

SOME USEFUL CONSTANTSAtomic No. H = 1, B = 5, C = 6, N = 7, O = 8, F = 9, Al = 13, P = 15, S = 16,

Cl = 17, Br = 35, Xe = 54, Ce = 58,Atomic masses : H = 1, Li = 7, B = 11, C = 12, N = 14, O = 16, F = 19, Na = 23, Mg = 24,

Al = 27, P = 31, S = 32, Cl = 35.5, Ca=40, Fe = 56, Br = 80, I = 127,Xe = 131, Ba=137, Ce = 140,

· Boltzmann constant k = 1.38 × 10–23 JK–1

· Coulomb's law constant pe9

0

1 = 9×104

· Universal gravitational constant G = 6.67259 × 10–11 N–m2 kg–2

· Speed of light in vacuum c = 3 × 108 ms–1

· Stefan–Boltzmann constant s = 5.67 × 10–8 Wm–2–K–4

· Wien's displacement law constant b = 2.89 × 10–3 m–K· Permeability of vacuum µ0 = 4p × 10–7 NA–2

· Permittivity of vacuum Î0 = 20

1

cm· Planck constant h = 6.63 × 10–34 J–s

All India Open Test/Leader Test Series/Joint Package Course/JEE (Advanced)/PAPER-1

ALLEN

LTS-3/42

BEWARE OF NEGATIVE MARKINGHAVE CONTROL ¾® HAVE PATIENCE ¾® HAVE CONFIDENCE Þ 100% SUCCESS

PART-1 : MATHEMATICS Hkkx-1 : xf.krSECTION–I(i) : (Maximum Marks : 28)

[k.M–I(i) : (vf/kdre vad : 28)� This section contains SEVEN questions.� Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these

four option(s) is (are) correct.� For each question, darken the bubble(s) corresponding to all the correct option(s) in the ORS� For each question, marks will be awarded in one of the following categories :

Full Marks : +4 If only the bubble(s) corresponding to all the correct option(s) is (are)darkened.

Partial Marks : +1 For darkening a bubble corresponding to each correct option,Provided NO incorrect option is darkened.

Zero Marks : 0 If none of the bubbles is darkened.Negative Marks : –2 In all other cases.

� for example, if (A), (C) and (D) are all the correct options for a question, darkening all thesethree will result in +4 marks; darkening only (A) and (D) will result in +2 marks; and darkening(A) and (B) will result in –2 marks, as a wrong option is also darkened

� bl [k.M esa lkr iz'u gSa

� izR;sd iz'u ds pkj mÙkj fodYi (A), (B), (C) vkSj (D) gSa ftuesa ls ,d ;k ,d ls vf/kd fodYi lgh gSaA

� izR;sd iz'u ds fy, vks-vkj-,l- ij lkjs lgh mÙkj (mÙkjksa) ds vuq:i cqycqys (cqycqyksa) dks dkyk djsaA

� izR;sd iz'u ds fy, vad fuEufyf[kr ifjfLFkfr;ksa esa ls fdlh ,d ds vuqlkj fn;s tk,axs %

iw.kZ vad : +4 ;fn flQZ lkjs lgh fodYi (fodYiksa) ds vuq:i cqycqys (cqycqyksa) dks dkyk fd;k gSA

vkaf'kd vad : +1 izR;sd lgh fodYi ds vuq:i cqycqys dks dkyk djus ij] ;fn dksbZ xyr fodYi dkyk

ugha fd;k gSA

'kwU; vad : 0 ;fn fdlh Hkh cqycqys dks dkyk ugha fd;k gSA

½.k vad : –2 vU; lHkh ifjfLFkfr;ksa esaA

� mnkgj.k % ;fn ,d iz'u ds lkjs lgh mÙkj fodYi (A), (C) vkSj (D) gSa] rc bu rhuksa ds vuq:i cqycqyksa dks dkys

djus ij +4 vad feysaxsa_ flQZ (A) vkSj (D) ds vuq:i cqycqyksa dks dkys djus ij +2 vad feysaxsa_ rFkk (A) vkSj

(B) ds vuq:i cqycqyksa dks dkys djus ij –2 vad feysaxs D;ksafd ,d xyr fodYi ds vuq:i cqycqys dks Hkh dkyk

fd;k x;k gSA


ALLEN

LTS-4/42


1. Ram throws a fair six-sided die untill six appears for the first time. Independently shyam rollsthe die untill six appears for the first time. Now, which of the following options holds good ?

(A) Probability that Ram gets six as result on third throw is 25216

.

(B) Probability that Shyam gets six as result on or after third throw is 2536

.

(C) Probability that Shyam gets six as a result on or after third throw is 25211

.

(D) Let a and b be the number of throws required by Ram and Shyam respectively to obtain six

as a result and 'P' be the probability that |a – b| < 1, then P is equal to 833

.

jke igyh ckj N% vkus rd ,d fu"i{k ikrh N% Hkqtkvksa okys ikls dks mNkyrk gSA Lora= :i ls ';ke igyh ckj N% vkus

rd ikls dks QSdrk gSA vc fuEu esa ls dkSulk@dkSuls fodYi lgh gksxk@gksaxs\

(A) jke ds }kjk rhljh ckj mNkyus ij ifj.kke N% vkus dh izkf;drk 25

216 gksxhA

(B) ';ke ds }kjk rhljh ckj mNkyus ij ;k mlds ckn ifj.kke N% izkIr gksus dh izkf;drk 2536 gksxhA

(C) ';ke ds }kjk rhljh ckj mNkyus ij ;k mlds ckn ifj.kke N% izkIr gksus dh izkf;drk 25

211 gksxhA

(D) ekuk jke rFkk ';ke ds }kjk vko';d Qsdks dh la[;k Øe'k% a rFkk b gS ftlls ifj.kke N% izkIr gksrk gS rFkk

|a – b| < 1 dh izkf;drk P gks] rks P dk eku 8

33 gksxkA

2. If g : R ® R be a continuous derivable function such that g(0) = 0 and |g'(x)| < |g(x)| " x Î R.The value of g(1) + g'(1) + g(2) + g'(2) can not be -;fn g : R ® R ,d larr vodyuh; Qyu bl izdkj gS fd g(0) = 0 rFkk |g'(x)| < |g(x)| " x Î R gSAg(1) + g'(1) + g(2) + g'(2) dk eku ugha gks ldrk gS -(A) 1 (B) –1 (C) 3 (D) 7


ALLEN

LTS-5/42


3. The vector + + ˆˆ ˆi x j 3k is rotated about its initial point through an angle of -1 11cos14

and its

magnitude is doubled. If the vector in the new position is given by ( )+ - + ˆˆ ˆ4i 4x 2 j 2k . Then

which of the following can't be the value of x ?

lfn'k + + ˆˆ ˆi x j 3k dks blds izkjfEHkd fcUnq ds lkis{k -1 11cos14

ls ?kqekrs gS blds ifjek.k dks nqxquk fd;k tkrk gSA ;fn

u;h fLFkfr esa lfn'k ( )+ - + ˆˆ ˆ4i 4x 2 j 2k }kjk fn;k tkrk gSA rc fuEu esa ls dkSulk x dk eku ugha gks ldrk gS\

(A) -23 (B)

23 (C) -

2017 (D) 2

4. If ( ) ( )

{ }æ ö+ += + + + + + + + + " Î - -ç ÷+ + + +è ø

201822018 2017 20181 2

2018 2017 1 0 2 2018bb bx 2x 2 a x a x .... a x a ... x R 1

x 1 x 1 x 1 x 1,

then (ai, bi Î Constant)

;fn ( ) ( )

{ }æ ö+ += + + + + + + + + " Î - -ç ÷+ + + +è ø

201822018 2017 20181 2

2018 2017 1 0 2 2018bb bx 2x 2 a x a x .... a x a ... x R 1

x 1 x 1 x 1 x 1 gks] rks

(ai, bi Î vpj)

(A) = =

æ ö+ = ç ÷è ø

å å20182018 2018

ii i

i 0 i 1

b 5a22 (B)

=

æ ö= - ç ÷

è øå2018

2018i

i 1

2018b 2

1009

(C) =

æ ö= + ç ÷

è øå2018

2018i

i 1

2018b 2

1009 (D) æ ö

= ç ÷è ø

02018

a1009

5. Define the permutation of the set {1,2,3,...n} to be sortable if upon cancelling appropriate term ofsuch permutation remaining n–1 terms are in increasing order. If ƒ(n) is the number of sortablepermutations of {1,2,3,...n}, then-leqPp; {1,2,3,...n} ds fdlh Øep; dks Js.khc¼ ekuk tkrk gS] ;fn fdlh ,d in dks gVkus ij 'ks"k n–1 in o/kZeku

Øe esa jgsA ;fn ƒ(n) ,sls {1,2,3,...n} Js.khc¼ Øep;ksa dh la[;k gks] rc&

(A) ƒ(5) = 17 (B) ƒ(5) = 16 (C) ƒ(3) = 5 (D) ƒ(4) = 10


ALLEN

LTS-6/42

6. For the function ( ) = - + +2ƒ x 4x 4x 2 x which of the following holds good-

(A) The least value of the function is +3 1

2 2

(B) The least possible value of function is +3 12

(C) The function takes its least value at -=

3 3x6

(D) Function has exactly one point of minima

Qyu ( ) = - + +2ƒ x 4x 4x 2 x ds fy;s fuEu esa ls dkSulk lR; gksxk@gksaxs -

(A) Qyu dk U;wure eku +3 1

2 2 gksxk

(B) Qyu dk U;wure lEHko eku +3 12 gksxk

(C) Qyu] -

=3 3x

6 ij blds U;wure eku dks xzg.k djrk gSA

(D) Qyu dk Bhd ,d fufEu"B fcUnq gksxk

7. If complete set of values of 'a' for which ƒ(x) = loga(4ax – x2) is strictly increasing in é ùê úë û

3 ,22 , then

possible value(s) of a

;fn 'a' ds ekuksa dk iw.kZ leqPp;] ftlds fy;s vUrjky é ùê úë û

3 ,22 esa ƒ(x) = loga(4ax – x2) fujUrj o/kZeku gks] rks a ds

lEHko eku gksaxs

(A) 58 (B)

14 (C) 2 (D) 3



ALLEN

LTS-7/42

SECTION–I(ii) : (Maximum Marks : 18) [k.M–I(ii) : (vf/kdre vad : 18)

� This section contains SIX questions of matching type.� This section contains Two tables (each having 3 columns and 4 rows)� Based on each table, there are THREE questions� Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is

correct� For each question, darken the bubble corresponding to the correct option in the ORS.� For each question, marks will be awarded in one of the following categories :

Full Marks : +3 If only the bubble corresponding to the correct option is darkened.Zero Marks : 0 If none of the bubbles is darkened.Negative Marks : –1 In all other cases

� bl [k.M esa lqesy izdkj ds N% iz'u gSaA� bl [k.M esa nks Vscy gSa (izR;sd Vscy esa 3 dkWye vkSj 4 iafDr;k¡ gSa)� izR;sd Vscy ij vk/kkfjr rhu iz'u gSaA� izR;sd iz'u ds pkj mÙkj fodYi (A), (B), (C) vkSj (D) gSa ftuesa flQZ ,d fodYi lgh gSaA� izR;sd iz'u ds fy, vks- vkj- ,l- ij lgh mÙkj ds vuq:i cqycqys dks dkyk djsaA� izR;sd iz'u ds fy, vad fuEufyf[kr ifjfLFkfr;ksa esa ls fdlh ,d ds vuqlkj fn;s tk,axs %

iw.kZ vad : +3 ;fn flQZ lgh fodYi ds vuq:i cqycqys dks dkyk fd;k gSA'kwU; vad : 0 ;fn fdlh Hkh cqycqys dks dkyk ugha fd;k gSA½.k vad : –1 vU; lHkh ifjfLFkfr;ksa esaAAnswer Q.8,Q.9 and Q.10 by appropriately matching the information given in the threecolumns of the following tableuhps nh x;h Vscy ds rhu dkWyeks a esa miyC/k lwpuk dk mi;qÙkG < ax ls lqesy dj iz'uks a Q.8, Q.9 vkSj Q.10ds mÙkj nhft;sAColumn-1 contains curves and column-2 contains the curves which touches corresponding curvesin column-1. Column-3 contains equation of tangent corresponding point of contact.dkWye-1 esa oØ fo|eku gS rFkk dkWye-2 esa oØ fo|eku gS tks dkWye esa laxr oØks Li'kZ djrs gSA dkWye-3 esa Li'kZfcUnq ds laxr Li'kZ js[kk dk lehdj.k fo|eku gSA

Column-1 / dkWye-1 Column-2 / dkWye-2 Column-3/dkWye-3(I) y2 = 4x (i) 2(x + 1)2 + 4y2 = 2 (P) x = 0

(II) x2 + y2 – 2x = 0 (ii) (x + y – 3)2 = ( )- - +4 2 x y 1 (Q) x – y + 1 =0(III) x2 – y2 + 2x = 0 (iii) (x + 3)2 + y2 = 1 (R) x + 2 = 0(IV) (x + 2)2 + y2 = 4 (iv) y2 + 4(x + 4) = 0 (S) x + 4 = 0

8. Which of the following options is the only CORRECT combination ?fuEu esa ls dkSulk fodYi dsoy lgh la;kstu gS\(A) (I) (i) (Q) (B) (II) (i) (R) (C) (III) (i) (P) (D) (III) (i) (S)

9. Which of the following options is the only CORRECT combination ?fuEu esa ls dkSulk fodYi dsoy lgh la;kstu gS\(A) (IV) (iv) (S) (B) (IV) (iv) (R) (C) (I) (ii) (P) (D) (II) (iii) (R)

10. Which of the following options is the only CORRECT combination ?fuEu esa ls dkSulk fodYi dsoy lgh la;kstu gS\

(A) (I) (iii) (R) (B) (II) (i) (Q) (C) (I) (ii) (Q) (D) (I) (ii) (R)


ALLEN

LTS-8/42

Answer Q.11,Q.12 and Q.13 by appropriately matching the information given in thethree columns of the following tableuhps nh x;h Vscy ds rhu dkWyeks a esa miyC/k lwpuk dk mi;qÙkG < ax ls lqesy dj iz'uks a Q.11, Q.12 vkSjQ.13 ds mÙkj nhft;sA

Column-1 and column-2 represents loci of variable point on argand plane and column-3 representscorresponding minimum distance between thendkWye-1 rFkk dkWye-2 vkxZ.M lery eas pj fcUnq dk fcUnqiFk O;Dr fd;k x;k gS rFkk dkWye-3 esa buds e/; U;wurenwjh dks O;Dr fd;k x;k gSA

Column-1 / dkWye-1 Column-2 / dkWye-2 Column-3 / dkWye-3

(I) |z + 2| + |z + 4| = 4 (i) |z + 10| + |z + 8| = 4 (P) 12

(II) |z + 13| = –11 – Re(z) (ii) + + + =22 2z 8z z 8z 2 z (Q) 1

(III) |z + 2| = 1 (iii) |z + 12| = |z + 11| (R) 2(IV) |z + 10| = 1 (iv) |z| = |z + 1| (S) 4

11. Which of the following options is the only INCORRECT combination ?fuEu esa ls dkSulk fodYi dsoy lgh la;kstu ugha gS\

(A) (I) (i) (R) (B) (I) (ii) (Q) (C) (I) (iv) (P) (D) (I) (iii) (S)12. Which of the following options is the only INCORRECT combination ?

fuEu esa ls dkSulk fodYi dsoy lgh la;kstu ugha gS\

(A) (III) (i) (S) (B) (III) (ii) (Q) (C) (III) (iv) (P) (D) (III) (iii) (R)13. Which of the following options is the only CORRECT combination ?

fuEu esa ls dkSulk fodYi dsoy lgh la;kstu gS\(A) (II) (i) (P) (B) (II) (iii) (R) (C) (IV) (iii) (P) (D) (IV) (iii) (Q)

SECTION –II / [k.M – II & SECTION –III / [k.M – III Matrix-Match Type / eSfVªDl&esy izdkj Integer Value Correct Type / iw.kk±d eku lgh izdkj

No question will be asked in section II and III / [k.M II ,oa III esa dksb Z iz'u ugha gSA



ALLEN

LTS-9/42

SECTION–IV : (Maximum Marks : 15) [k.M–IV : (vf/kdre vad : 15)

� This section contains FIVE questions.� The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 to 9, both inclusive.� For each question, darken the bubble corresponding to the correct integer in the ORS.� For each question, marks will be awarded in one of the following categories :

Full Marks : +3 If only the bubble corresponding to the correct answer is darkened.Zero Marks : 0 In all other cases.

� bl [k.M esa ik¡p iz'u gSa

� izR;sd iz'u dk mÙkj 0 ls 9 rd (nksuksa 'kkfey) ds chp dk ,d ,dy vadh ; iw.kk±d gSA� izR;sd iz'u esa] vks-vkj-,l- ij lgh iw.kk±d ds vuq:i cqycqys dks dkyk djsaA


iw.kZ vad : +3 ;fn flQZ lgh mÙkj ds vuq:i cqycqys dks dkyk fd;k gSA

'kwU; vad : 0 vU; lHkh ifjfLFkfr;ksa esaA

1. Consider = + +r ˆˆ ˆr ai bj ck . If a,b and g be the projection of rr along + - +ˆ ˆ ˆ ˆ2i j 2i j,

5 5 and k respectively

such that |a| = |b| = |g| and =rr 1 , then number of such ordered triplets (a, b, c) is

ekuk = + +r ˆˆ ˆr ai bj ck gSA ;fn

+ - +ˆ ˆ ˆ ˆ2i j 2i j,5 5

rFkk k ds vuqfn'k rr ds iz{ksi Øe'k% a, b rFkk g blizdkj gS

|a| = |b| = |g| rFkk =rr 1 gks] rks ,sls Øfer f=dks (a, b, c) dh la[;k gksxh



ALLEN

LTS-10/42

2. ƒ : R ® R

( )ì <ïï= íï ³ïî

10 , x2ƒ x11 , x2

. If the value of ( )p =ò1

0

pƒ sin x dxq (where p and q) are relatively prime, then

the value of p + q is

ƒ : R ® R

( )ì <ïï= íï ³ïî

10 , x2ƒ x11 , x2

gSA ;fn ( )p =ò1

0

pƒ sin x dxq (tgk¡ p rFkk q ijLij vHkkT; gS) gks] rks p + q dk eku gksxk

3. If ( )

( )+

l®

-=

òlog 1 x 1

y

0x 0

1 tan 2y dy1lim

sin x e, then 'l' is

;fn ( )

( )+

l®

-=

òlog 1 x 1

y

0x 0

1 tan 2y dy1lim

sin x e gks] rks 'l' dk eku gksxk



ALLEN

LTS-11/42

4. Let A and B be 3 × 3 square matrices such that AB = 9I where -æ ö

ç ÷= -ç ÷ç ÷- lè ø

0 1 1A 4 3 4

3 3. If B33 = 9A23,

then the value of 2 2A BTr82

æ ö++ lç ÷

è ø, is

(where xij represents element of ith row and jth column of square matrix and Tr(x) representstrace of square matrix x).

ekuk A rFk k B, dksfV 3 × 3 dk oxZ vkO;wg bl izdkj gS fd AB = 9I tgk ¡ -æ ö

ç ÷= -ç ÷ç ÷- lè ø

0 1 1A 4 3 4

3 3 g SA

;fn B33 = 9A23 gks] rks 2 2A BTr82

æ ö++ lç ÷

è ø dk eku gksxkA

(tgk¡ xij, oxZ vkO;wg dh ioha iafDr rFkk josa LrEHk ds vo;o dks n'kkZrk gS rFkk Tr(x) oxZ vkO;wg x ds vuqjs[k dks n'kkZrk gS)

5. Let |z| = 1 & the minimum value of the expression |z + 1| + |1 – z + z2| be m then the valueof 2m2 isekuk |z| = 1 rFkk O;atd |z + 1| + |1 – z + z2| dk U;wure eku m gS] rks 2m2 dk eku gksxk



ALLEN

LTS-12/42

PART-2 : PHYSICS Hkkx-2 : HkkSfrd foKkuSECTION–I(i) : (Maximum Marks : 28)







� bl [k.M esa lkr iz'u gSa� izR;sd iz'u ds pkj mÙkj fodYi (A), (B), (C) vkSj (D) gSa ftuesa ls ,d ;k ,d ls vf/kd fodYi lgh gSaA� izR;sd iz'u ds fy, vks-vkj-,l- ij lkjs lgh mÙkj (mÙkjksa) ds vuq:i cqycqys (cqycqyksa) dks dkyk djsaA� izR;sd iz'u ds fy, vad fuEufyf[kr ifjfLFkfr;ksa esa ls fdlh ,d ds vuqlkj fn;s tk,axs %

iw.kZ vad : +4 ;fn flQZ lkjs lgh fodYi (fodYiksa) ds vuq:i cqycqys (cqycqyksa) dks dkyk fd;k gSAvkaf'kd vad : +1 izR;sd lgh fodYi ds vuq:i cqycqys dks dkyk djus ij] ;fn dksbZ xyr fodYi dkyk

ugha fd;k gSA'kwU; vad : 0 ;fn fdlh Hkh cqycqys dks dkyk ugha fd;k gSA½.k vad : –2 vU; lHkh ifjfLFkfr;ksa esaA

� mnkgj.k % ;fn ,d iz'u ds lkjs lgh mÙkj fodYi (A), (C) vkSj (D) gSa] rc bu rhuksa ds vuq:i cqycqyksa dks dkysdjus ij +4 vad feysaxsa_ flQZ (A) vkSj (D) ds vuq:i cqycqyksa dks dkys djus ij +2 vad feysaxsa_ rFkk (A) vkSj(B) ds vuq:i cqycqyksa dks dkys djus ij –2 vad feysaxs D;ksafd ,d xyr fodYi ds vuq:i cqycqys dks Hkh dkykfd;k x;k gSA

1. The electric quadrupole moment of a nuclear charge distribution which is of the shape of an

ellipsoid is given by ( )2 22Zep b a5

= -

Total nuclear charge = Ze

Average nuclear radius 3 20R a b= ; b = R0 + dR0 ; b is semi major axis & a is semi minor axis.

Mark the CORRECT statement(s) under condition dR0 << R0 :

ijoy;t dh vkdfr okys fdlh ukfHkdh; vkos'k forj.k dk fo|qr prq/kzqZoh vk?kw.kZ ( )2 22Zep b a5

= - }kjk fn;k tkrkgSA dqy ukfHkdh; vkos'k = ZevkSlr ukfHkdh; f=T;k 3 2

0R a b= ; b = R0 + dR0 gS] tgk¡ b v¼Znh?kZ v{k rFkk a v¼Zy?kq v{k gSA;fn dR0 << R0 gks rks lgh dFku@dFkuksa dks pqfu;sA

(A) 2 2 00

0

Ra R 1R

æ öd@ -ç ÷

è ø(B) 2 2 2 0

00

Rb a 3RR

æ öd- @ ç ÷

è ø

(C) 2 00

0

R6Zep R5 R

æ öd= ç ÷

è ø(D)

2 00

0

R2Zep R5 R

æ öd= ç ÷

è ø

ALLEN

LTS-13/42


2. In figure (a) and (b) shown the mass M is very large and exerts a gravitational force on m1 andm2. Gravitational force between m1 and m2 is very small as compared to that of M. Net force netF

r

on m1 in accelerating reference frame of m2 is called tidal force. If m1 is to be at rest in referenceframe of m2 an additional force equal to tidal force is to be applied on m1. Mark the correctstatement(s) using appropriate approximation :

(A) Tidal force on m1 in figure (a) is given by 13

GMm xR

(B) Tidal force on m1 in figure (a) is given by 13

2GMm xR

(C) Tidal force on m1 in figure (b) is given by 13

GMm xR

(D) Tidal force on m1 in figure (b) is given by 13

2GMm xR

m2 m1 M

Rxx << R

Figure (a)

x

m1

m2 Rq M

Figure (b)

x<<Rcos 1qsinq x

R

iznf'kZr fp= (a) rFkk (b) esa æO;eku M cgqr vf/kd gS rFkk æO;eku m1 o m2 ij ,d xq:Rokd"kZ.k cy yxkrk gSA m1 o

m2 ds e/; xq:Rokd"kZ.k cy M dh rqyuk esa vR;Yi gSA m2 ds Rofjr funsZ'k ra= esa m1 ij dqy cy netFr

Tokjh; cy

dgykrk gSA ;fn m1 dks m2 ds funsZ'k ra= esa fojkekoLFkk esa j[kuk gks rks Tokjh; cy ds cjkcj ,d vfrfjDr cy m1 ij

yxkuk gksxkA mfpr vo/kkj.kk ds mi;ksx }kjk lgh dFku@dFkuksa dks pqfu;s%&

(A) fp= (a) esa m1 ij Tokjh; cy 13

GMm xR gSA

(B) fp= (a) esa m1 ij Tokjh; cy 13

2GMm xR gSA

(C) fp= (b) esa m1 ij Tokjh; cy 13

GMm xR gSA

(D) fp= (b) esa m1 ij Tokjh; cy 13

2GMm xR gSA


ALLEN

LTS-14/42


3. A spherical ball of radius R is made by joining two hemispherical parts. The two parts havedensity r and 2r. When placed in a water tank, the ball floats while remaining completelysubmerged. Neglect viscous forces. Mark the correct statement(s) :-

(A) Density of water is r0, then r = 023r .

(B) Position of centre of mass of ball is at distance of R8 from O.

(C) Moment of inertia of ball about COM of ball perpendicular to the plane of paper is 5123 R160

rp

(D) Time period of small angular oscillations of the ball about its equilibrium position is given

by 123R240g

p

O

water

r

2r

nks v¼Zxksykdkj Hkkxksa dks tksM+dj R f=T;k dh xksykdkj xsan cuk;h tkrh gSA bu nksuksa Hkkxksa ds ?kuRo r o 2r gSA bUgsa ty

ls Hkjs VSad esa j[kus ij ;g xsan iw.kZr;k Mwch gqbZ rSjus yxrh gSA ';ku cyksa dks ux.; ekusA lgh dFku@dFkuksa dks pqfu;s %&

(A) ;fn ty dk ?kuRo r0 gks rks r = 023r

gksxkA

(B) xsan ds æO;eku dsUæ dh fLFkfr O ls R8 nwjh ij gksxhA

(C) dkxt ds ry ds yEcor~ xsan ds æO;eku dsUæ ds lkis{k xsan dk tM+Ro vk?kw.kZ 5123 R160

rp gksxkA

(D) xsan ds vYi dks.kh; nksyuksa dk bldh lkE;koLFkk fLFkfr ds lkis{k vkorZdky 123R240g

p gksxkA


ALLEN

LTS-15/42

4. Two sound waves, travelling in same direction can be represented as

( ) ( )11 1

xy 0.02mm sin 400 rads t330ms

--

é ùæ ö= p -ç ÷ê úè øë û

And

( ) ( )12 1


--


The wave superimpose.

(A) Distance between two nearest points where an intensity maximum is recorded simultaneouslyis 165 m.

(B) Distance between two nearest points where an intensity maximum is recorded simultaneouslyis 330 m.

(C) The time gap between two successive intensity maxima at a given point is 0.5 sec.

(D) The time gap between two successive intensity maxima at a given point is 1 sec.

leku fn'kk esa xfr'khy nks /ofu rjaxs

( ) ( )11 1


--


,oa

( ) ( )12 1


--


}kjk nh tkrh gSA ; s rjaxsa v/;kjksfir gksrh gSA

(A) nks fudVre fcUnq tgk¡ ,d rhozrk mfPp"B ,dlkFk lalwfpr gksrk gS] ds e/; nwjh 165 m gSA

(B) nks fudVre fcUnq tgk¡ ,d rhozrk mfPp"B ,dlkFk lalwfpr gksrk gS] ds e/; nwjh 330 m gSA

(C) fdlh fn;s x;s fcUnq ij nks Øekxr rhozrk mfPp"Bksa ds e/; le;kUrjky 0.5 sec gSA

(D) fdlh fn;s x;s fcUnq ij nks Øekxr rhozrk mfPp"Bksa ds e/; le;kUrjky 1 sec gSA



ALLEN

LTS-16/42

5. In the arrangement shown in the figure the cylinder is insulated from the surrounding. Bothsides same diatomic gas is trapped by two insulting massless pistons with the help of an idealspring. The natural length of the spring is equal to the length of the cylinder. Initial state of thegases are as shown in the figure.

(A) The value of energy stored in the spring is P0V0

(B) The value of energy stored in the spring is 2P0V0

(C) Now the gases are heated slowly, such that their temperature becomes three times to theirinitial temperature. The total heat given to the system is 12P0V0

(D) Now the gases are heated slowly, such that their temperature becomes three times to theirinitial temperature. The total heat given to the system is 6P0V0

vaccumnPV

0

0

n

P0

V0

iznf'kZr O;oLFkk esa csyu ifjos'k ls foyfxr gSA blds nksuksa vksj leku f}ijekf.od xSl dks ,d vkn'kZ fLizax dh lgk;rk

ls nks dqpkyd æO;ekughu fiLVuksa }kjk ifjc¼ fd;k x;k gSA fLizax dh ewy yEckbZ] csyu dh yEckbZ ds leku gSA

fp= esa xSlksa dh izkjfEHkd voLFkk n'kkZ;h x;h gSA

(A) fLizax esa lafpr ÅtkZ dk eku P0V0 gSA

(B) fLizax esa lafpr ÅtkZ dk eku 2P0V0 gSA

(C) vc xSlksa dks /khjs&/khjs bl izdkj xeZ fd;k tkrk gS fd budk rkieku] buds izkjfEHkd rkieku dk rhu xquk gks tkrk gSA

fudk; dks nh x;h dqy Å"ek 12P0V0 gSA

(D) vc xSlksa dks /khjs&/khjs bl izdkj xeZ fd;k tkrk gS fd budk rkieku] buds izkjfEHkd rkieku dk rhu xquk gks tkrk gSA

fudk; dks nh x;h dqy Å"ek 6P0V0 gSA



ALLEN

LTS-17/42

6. Two identical dumbbells (having point masses at end) of length l are connected by an axlesituated at the midpoint of their lengths and are placed on a frictionless horizontal plane invertical position. A small impulse is imparted to each body at the bottom in the opposite directionsso that both dumbbells will rotate vertically and reach the horizontal position.(A) The velocity of the mass which is initially at the top, relative to the one on the plane on the

same dumbbell at the moment when the dumbbells reach the horizontal position is 2gl

(B) The velocity of the mass which is initially at the top, relative to the one on the plane on the

same dumbbell at the moment when the dumbbells reach the horizontal position is 2 gl

(C) Centre of mass of system initially at rest will not move horizontally.(D) Now we assume that friction exists between the plane and the bodies of the dumbbells. The

angle a made by dumbbells relative to the horizontal when the system can start moving byitself is given by tan a = µ. The coefficient of friction µ is known.

l yEckbZ okys nks ,dtSls MEcyksa (ftuds fljksa ij fcUnq æO;eku fo|eku gS) dks mudh yEckb;ksa ds e/; fcUnq ij fLFkr,Dlsy }kjk vkil esa tksM+dj bUgsa ?k"kZ.kjfgr {kSfrt ry ij Å/okZ/kj fLFkfr esa j[kk tkrk gSA izR;sd fi.M ds iSans ij ,dvYi vkosx foijhr fn'kkvksa esa bl izdkj yxk;s tkrs gSa fd nksuksa MEcy Å/okZ/kj :i ls ?kw.kZu djrs gS rFkk {kSfrt fLFkfr esaigq¡prs gSaA(A) tc MEcy {kSfrt fLFkfr esa igq¡prs gSa rks izkjEHk esa 'kh"kZ ij fLFkr æO;eku dk] leku MEcy ij ry ij fLFkr nwljs

æO;eku ds lkis{k osx 2gl gksxkA

(B) tc MEcy {kSfrt fLFkfr esa igq¡prs gSa rks izkjEHk esa 'kh"kZ ij fLFkr æO;eku dk] leku MEcy ij ry ij fLFkr nwljs

æO;eku ds lkis{k osx 2 gl gksxkA

(C) izkjEHk esa fojke esa fLFkr fudk; dk æO;eku dsUæ {kSfrt :i ls xfr ugha djsxkA(D) vc ekuk fd ry rFkk MEcyksa ds e/; ?k"kZ.k fo|eku gSA tc fudk; Lo;a xfr djuk izkjEHk dj ldrk gS] ml {k.k

MEcyksa }kjk {kSfrt ds lkis{k cuk;k x;k dks.k tan a = µ gksrk gS tgk¡ ?k"kZ.k xq.kkad µ Kkr gSA



ALLEN

LTS-18/42

7. A circuit consists of two capacitors, a 24V battery and an AC source connected as shown infigure. The AC voltage is given by e = (20 cos 120pt)V, where t is in second.

(A) Charge on capacitor C2 as function of time Q2 = (30µC) cos (120pt) + 36µC

(B) Steady state current is 33.9mA sin (120pt + p)

(C) Maximum energy stored in capacitor C1 & C2 is 36 mJ

(D) Minimum energy stored in capacitors C1 and C2 is 36µJ

C1 3µF C2 1.5µF

24V

e0=20V

iznf'kZr ifjiFk esa nks la/kkfj=] ,d 24V cSVjh rFkk ,d AC L=ksr tqM+s gq, gSaA AC oksYVrk dk eku e = (20 cos 120pt)V}kjk fn;k tkrk gS tgk¡ t lsd.M esa gS %&

(A) le; ds Qyu ds :i esa la/kkfj= C2 ij vkos'k Q2 = (30µC) cos (120pt) + 36µC gSA

(B) LFkk;h voLFkk /kkjk dk eku 33.9mA sin (120pt + p) gSA

(C) la/kkfj= C1 o C2 esa lafpr vf/kdre ÅtkZ 36 mJ gSA

(D) la/kkfj= C1 o C2 esa lafpr U;wure ÅtkZ 36µJ gSA



ALLEN

LTS-19/42






iw.kZ vad : +3 ;fn flQZ lgh fodYi ds vuq:i cqycqys dks dkyk fd;k gSA'kwU; vad : 0 ;fn fdlh Hkh cqycqys dks dkyk ugha fd;k gSA½.k vad : –1 vU; lHkh ifjfLFkfr;ksa esaAAnswer Q.8, Q.9 and Q.10 by appropriately matching the information given in the threecolumns of the following table.

uhps nh x;h Vscy ds rhu dkWyeksa esa miyC/k lwpuk dk mi;qÙkG < ax ls lqesy dj iz'uks a Q.8, Q.9 vkSj Q.10 dsmÙkj nhft;sAColumn-1 gives the external magnetic field and a black box present in the field. Column-2 givesthe component of magnetic field at any point P in the plane of paper and Column-3 representsforce on charged particle when projected from point P in the plane of paper at the same instant.Match them correctly if the system given in question is kept in the black box.dkWye-1 esa ,d ckg~; pqEcdh; {ks= rFkk bl {ks= esa fLFkr ,d CySd ckWDl dks n'kkZ;k x;k gSA dkWye-2 esa dkxt ds ryesa fLFkr fdlh fcUnq P ij pqEcdh; {ks= dk ?kVd fn;k x;k gS rFkk dkWye-3 esa ,d vkosf'kr d.k ij ml {k.k cy dksn'kkZ;k x;k gS tc bls dkxt ds ry esa fcUnq P ls iz{ksfir fd;k tkrk gSA ;fn iz'u esa fn;k x;k fudk; CySd ckWDl esa j[kkgks rks budk feyku dhft;sA

Y

XZ



ALLEN

LTS-20/42

Column–1 / dkWye–1 Column-2 / dkWye–2 Column-3 / dkWye–3

(I) Blackbox

Bext(i) Bx = 0 (P) Fx = 0

(II) Blackbox

Bext

(ii) By = 0 (Q) Fy = 0

(III)Blackbox

Idealtoroid

(iii) Bz ¹ 0 (R) Fz = 0

(IV)Blackbox

Bext

(iv) Bx ¹ 0, By ¹ 0 (S) Fnet = 0

8. If the black box contains an ideal current carrying solenoid with its axis parallel to x-axis :-;fn CySd ckWDl esa ,d vkn'kZ /kkjkokgh ifjukfydk bl izdkj j[kh gks fd bldh v{k] x–v{k ds lekUrj gS %&(A) (I) (iv) (S) (B) (I) (i) (S) (C) (III) (iii) (Q) (D) (IV) (iii) (S)

9. If the black box contains an ideal current carrying solenoid with its axis parallel to y-axis :-;fn CySd ckWDl esa ,d vkn'kZ /kkjkokgh ifjukfydk bl izdkj j[kh gks fd bldh v{k] y–v{k ds lekUrj gS %&(A) (I) (i) (S) (B) (II) (i) (S) (C) (III) (iii) (R) (D) (IV) (iii) (S)

10. If the black box contains a current carrying ring with its axis parallel to x-axis :-;fn CySd ckWDl esa ,d /kkjkokgh oy; bl izdkj j[kh gks fd bldh v{k] x–v{k ds lekUrj gS %&(A) (I) (iv) (S) (B) (II) (iii) (P) (C) (III) (iii) (S) (D) (IV) (ii) (S)



ALLEN

LTS-21/42

Answer Q.11, Q.12 and Q.13 by appropriately matching the information given in thethree columns of the following table.In the shown DC–RC circuit galvanometer of resistance ‘R’ is connected as shown. (Initially att = 0 all switchs are open & capacitors are uncharged)

G

E

RS3

R3

R2

R4R1

C1 S2

E

E S1

R

C2

Condition of switch, resistance & capacitance is given in column (1), (2), (3) respectively.Column–1 Column-2 Column-3

(I) S1 closed, S2 & S3 open (i) R1 = 2R, R2 = 2R (P) C1 = C2 = CR3 = R, R4 = R

(II) S1 open, S2 & S3 closed (ii) R2 = R, R4 = 3R (Q) C1 = C2 = 2CR3 = 2R, R1 = 6R

(III) S1, S2 closed S3 open (iii) R1 = R2 = R3 = R4 = R (R) C1 = C, C2 = 2C(IV) S1, S3 open, S2 closed (iv) R1 = R2=2R, R4=R3=R (S) C1 = 2C, C2 = C

11. Initially both capacitor is uncharged which of the following case galvanometer shown no deflectionat t = 0 (just after switches is closed)(A) (IV) (ii) (S) (B) (I) (i) (R) (C) (I) (ii) (S) (D) (IV) (iii) (R)

12. In which of the following case galvanometer shown no deflection at (t = ¥) (steady state)(A) (II) (iv) (P) (B) (IV) (i) (S) (C) (II) (ii) (S) (D) (IV) (ii) (P)

13. In which of the following case final charge on capacitor C2 is zero :-(A) (III) (iv) (R) (B) (III) (iii) (Q) (C) (II) (ii) (Q) (D) (IV) (ii) (R)



ALLEN

LTS-22/42

uhps nh x;h Vscy ds rhu dkWyeks a esa miyC/k lwpuk dk mi;qÙkG < ax ls lqesy dj iz'uks a Q.11, Q.12 vkSj Q.13 dsmÙkj nhft;sAiznf'kZr DC–RC ifjiFk esa izfrjks/k ‘R’ okys xsYosuksehVj dks fp=kuqlkj tksM+k x;k gSA (izkjEHk esa t = 0 ij lHkh fLop [kqysgS rFkk la/kkfj= vukosf'kr gS)

G

E

RS3

R3

R2

R4R1

C1 S2

E

E S1

R

C2

dkWye (1), (2), (3) esa Øe'k% fLop dh fLFkfr] izfrjks/k rFkk /kkfjrk dks n'kkZ;k x;k gSA

dkWye–1 dkWye-2 dkWye-3(I) S1 can, S2 o S3 [kqys gS (i) R1 = 2R, R2 = 2R (P) C1 = C2 = C

R3 = R, R4 = R(II) S1 [kqyk gS, S2 o S3 can gS (ii) R2 = R, R4 = 3R (Q) C1 = C2 = 2C

R3 = 2R, R1 = 6R(III) S1, S2 can gS] S3 [kqyk gS (iii) R1 = R2 = R3 = R4 = R (R) C1 = C, C2 = 2C(IV) S1, S3 [kqys gS] S2 can gS (iv) R1 = R2=2R, R4=R3=R (S) C1 = 2C, C2 = C

11. izkjEHk esa nksuksa la/kkfj= vukosf'kr gSA fuEu esa ls fdl izdj.k esa xsYosuksehVj t = 0 (fLop can djus ds Bhd i'pkr~) ij dksbZfo{ksi ugh n'kkZ;sxk\(A) (IV) (ii) (S) (B) (I) (i) (R) (C) (I) (ii) (S) (D) (IV) (iii) (R)

12. fuEu esa ls fdl izdj.k esa xsYosuksehVj (t = ¥) ij (LFkk;h voLFkk) dksbZ fo{ksi ugha n'kkZ;sxk\(A) (II) (iv) (P) (B) (IV) (i) (S) (C) (II) (ii) (S) (D) (IV) (ii) (P)

13. fuEu esa ls fdl izdj.k esa la/kkfj= C2 ij vafre vkos'k 'kwU; gS\(A) (III) (iv) (R) (B) (III) (iii) (Q) (C) (II) (ii) (Q) (D) (IV) (ii) (R)





ALLEN

LTS-23/42




� bl [k.M esa ik¡p iz'u gSa� izR;sd iz'u dk mÙkj 0 ls 9 rd (nksuksa 'kkfey) ds chp dk ,d ,dy vadh; iw.kk±d gSA� izR;sd iz'u esa] vks-vkj-,l- ij lgh iw.kk±d ds vuq:i cqycqys dks dkyk djsaA� izR;sd iz'u ds fy, vad fuEufyf[kr ifjfLFkfr;ksa esa ls fdlh ,d ds vuqlkj fn;s tk,axs %

iw.kZ vad : +3 ;fn flQZ lgh mÙkj ds vuq:i cqycqys dks dkyk fd;k gSA'kwU; vad : 0 vU; lHkh ifjfLFkfr;ksa esaA

1. A point object is located at a distance of 100 cm from a screen. A lens of focal length 23 cmmounted on a movable frictionless stand is kept between the source and the screen. The stand isattached to a spring of natural length 50 cm and spring constant 800 N/m as shown. Mass of thestand with lens is 2 kg. How much minimum impulse P (in kg m/s) should be imparted to thestand so that a real image of the object is formed on the screen four times during a completeoscillation. All these are separated by same time interval. (Neglect width of the stand.),d fcUnqor~ oLrq insZ ls 100 cm dh nwjh ij fLFkr gSA ,d pyk;eku ?k"kZ.kghu LVS.M ij LFkkfir fd;k x;k 23cm Qksdlnwjh dk ,d ySUl] lzksr o insZ ds e/; j[kk x;k gSA LVS.M dks n'kkZ;s vuqlkj 50 cm ewy yEckbZ rFkk 800N/m fLizaxfu;rkad dh ,d fLizax ls la;qDr fd;k x;k gSA ySUl lfgr LVS.M dk nzO;eku 2 kg gSA LVS.M dks fdruk U;wure vkosxP (kg m/s esa) iznku fd;k tk;s rkfd ,d iw.kZ nksyu ds nkSjku insZ ij oLrq dk pkj xquk okLrfod izfrfcEc cus\ ; s lHkhleku le;kUrjky ls i`Fkd fd;s x;s gSA (LVS.M dh pkSM+kbZ dks ux.; ekusaA)

P

Screen

50 cm100 cm

ObjectLens



ALLEN

LTS-24/42

2. There is a bicycle on a bicycle stand. We know that its rear wheel's moment of inertia around itsaxis is I = 0.5 kg·m2 and its radius is r = 30 cm. We speed up the wheel by repeatedly hitting itwith a hand tangentially on its circumference in the direction of its rotation. The speed of eachhit in the rest reference frame is v = 5 m·s–1, the mass of the hand is m = 1.5kg, the hand loses allits velocity in the reference frame of the point of impact after the impact (the hand is movingaround an elbow, but we neglect rotation of the forearm). If velocity of a point on the perimeter of

the wheel after 10 hits be 10

nv 1æ öæ öbç ÷= a - ç ÷ç ÷gè øè ø

then value of |a +2b –g|

lkbfdy LVS.M ij ,d lkbfdy [kM+h gqbZ gSA ge tkurs gSa fd blds fiNys ifg;s dk bldh v{k ds lkis{k tM+Ro vk?kw.kZ

I = 0.5 kg·m2 rFkk bldh f=T;k r = 30 cm gSA ifg;s dh ifj/kh ij blds ?kw.kZu dh fn'kk esa Li'kZjs[kh; :i ls gkFk ls

ckj&ckj izgkj dj bldh pky dks c<+k;k tkrk gSA fojke funsZ'k ra= esa izR;sd izgkj dh pky v = 5 m·s–1 gS rFkk gkFk dk

æO;eku m = 1.5kg gSA VDdj ds ckn la?kV~V fcUnq ds funsZ'k ra= esa gkFk vius laiw.kZ osx dks [kks nsrk gS (gkFk bldh dksguh

ds pkjksa vksj xfr'khy gS ijUrq ;gk¡ ck¡g ds ?kw.kZu dks ux.; ekuk x;k gSA ;fn 10 ckj izgkj djus ds ckn ifg;s ds ifjeki ij

fLFkr fdlh fcUnq dk osx 10

nv 1æ öæ öbç ÷= a - ç ÷ç ÷gè øè ø

gks rks |a +2b –g| dk eku Kkr dhft;sA



ALLEN

LTS-25/42

3. Due to its higher density, cold water stays close to the bottom of a rectangular vessel which isfilled up to the height of h = 30 cm. We assume that the density of water in the vessels growslinearly with increasing depth – at the water level, the density is equal to r1 = 996kg·m–3, while thedensity rb at the bottom of the vessel is unknown. Determine this density using the fact, that ahomogeneous rod with density rr = 997 kg·m–3 and length h immersed in the water and fixed byone of its ends at the water level makes an angle of j = 60° with the vertical. Fill value of x inOMR if rb = 996 + x

B.Ms ty dk mPp ?kuRo gksus ds dkj.k ;g h = 30 cm Å¡pkbZ rd Hkjs fdlh vk;rkdkj ik= ds iSans ds utnhd Bgjk jgrk

gSA ekuk fd ik= esa ty dk ?kuRo c<+rh gqbZ xgjkbZ ds lkFk jSf[kd :i ls c<+rk gSA ty Lrj ij ?kuRo

r1 = 996kg·m–3 rFkk ik= ds iSans ij ?kuRo rb vKkr gSA ty esa h yEckbZ rd Mwch rFkk ?kuRo rr = 997 kg·m–3 okyh

,d le:i NM+ dk ,d fljk ty Lrj ij fLFkj fd;k gqvk gS rFkk ;g Å/okZ/kj ls j = 60° dks.k cukrh gSA bl rF; dk

mi;ksx djrs gq;s bl ?kuRo dk eku Kkr dhft;sA ;fn rb = 996 + x gks rks x dk eku Kkr dhft;sA



ALLEN

LTS-26/42

4. Figure shows a soap film formed between two square figures made of a uniform wire. The biggersquare is held while keeping it in a horizontal plane and the smaller square is slowly allowed todrop vertically. It reaches an equilibrium state after dropping a height h. Let surface tension ofsoap = T. Mass per unit length of wire = l. Acceleration due to gravity = g. Given that

2 2 2

n gah2 4T g

l=

- l; find the integer value n.

fp= esa ,d le:i rkj ls cuh nks oxkZdkj vkdfr;ksa ds e/; lkcqu dh ,d fQYe n'kkZ;h x;h gSA cM+s oxZ dks {kSfrt ryesa jksddj j[kk x;k gS tcfd NksVk okyk oxZ /khjs&/khjs Å/okZ/kj :i ls fxjus fn;k tkrk gSA ;g h Å¡pkbZ rd fxjus ds cknlkE;koLFkk fLFkfr esa igq¡prk gSA ekuk lkcqu dk i`"Bruko T, rkj dh izfr bdkbZ yEckbZ dk æO;eku l] xq:Roh; Roj.k g gSA

;fn 2 2 2

n gah2 4T g

l=

- l gks rks n dk iw.kk±d eku Kkr dhft;sA

h

a

4a



ALLEN

LTS-27/42

5. In the figure shown S1 and S2 are two identical coherent sources of sound, separated by d.A receiver moves along line XY which is parallel to S1S2 to detect intensity of sound at variouspoints on the line. Distance of line XY from S1S2 is D(>>d). Distance of first intensity maxima online XY measured from O is y. Find percentage change in value of y if the temperature of airincreases by 1%. If your answer is X % fill value of 6X.

iznf'kZr fp= esa S1 rFkk S2 ,d&nwljs ls d nwjh ij j[ks nks ,dtSls dyklaca¼ /ofu L=ksr gSA ,d xzkgh js[kk XY, tks fd S1S2

ds lekUrj gS] ds vuqfn'k bl js[kk ij fofHkUu fcUnqvksa ij /ofu dh rhozrk dks lalwfpr djus ds fy;s xfr djrk gSA XY js[kk

dh S1S2 ls nwjh D(>>d) gSA js[kk XY ij O ls ekis x;s izFke rhozrk mfPp"B dh nwjh y gSA y ds eku esa izfr'kr ifjorZu

Kkr dhft;s ;fn ok;q dk rkieku 1% c<+rk gSA ;fn vkidk mÙkj X % gks rks 6X dk eku Kkr dhft;sA

S2

S1

d

X

Y

O

D



ALLEN

LTS-28/42

PART-3 : CHEMISTRY Hkkx-3 : jlk;u foKkuSECTION–I(i) : (Maximum Marks : 28)







� bl [k.M esa lkr iz'u gSa

� izR;sd iz'u ds pkj mÙkj fodYi (A), (B), (C) vkSj (D) gSa ftuesa ls ,d ;k ,d ls vf/kd fodYi lgh gSaA

� izR;sd iz'u ds fy, vks-vkj-,l- ij lkjs lgh mÙkj (mÙkjksa) ds vuq:i cqycqys (cqycqyksa) dks dkyk djsaA


iw.kZ vad : +4 ;fn flQZ lkjs lgh fodYi (fodYiksa) ds vuq:i cqycqys (cqycqyksa) dks dkyk fd;k gSA

vkaf'kd vad : +1 izR;sd lgh fodYi ds vuq:i cqycqys dks dkyk djus ij] ;fn dksbZ xyr fodYi dkyk

ugha fd;k gSA

'kwU; vad : 0 ;fn fdlh Hkh cqycqys dks dkyk ugha fd;k gSA

½.k vad : –2 vU; lHkh ifjfLFkfr;ksa esaA

� mnkgj.k % ;fn ,d iz'u ds lkjs lgh mÙkj fodYi (A), (C) vkSj (D) gSa] rc bu rhuksa ds vuq:i cqycqyksa dks dkys

djus ij +4 vad feysaxsa_ flQZ (A) vkSj (D) ds vuq:i cqycqyksa dks dkys djus ij +2 vad feysaxsa_ rFkk (A) vkSj

(B) ds vuq:i cqycqyksa dks dkys djus ij –2 vad feysaxs D;ksafd ,d xyr fodYi ds vuq:i cqycqys dks Hkh dkyk

fd;k x;k gSA


ALLEN

LTS-29/42


1. One mole of an ideal monoatomic gas is taken round the cyclic process PQRP as shown below infigureThe correct statement(s) about the given cyclic process is/are :(A) Along proces QR ; the temperature first increases ; reaches a maximum value and then

decreases(B) Along process QR ; the temperature first decreases ; reaches a minimum value and then

increases

(C) The net heat exchanged between system and surrounding during process RP is 0 05 P V2

(D) The net heat rejected during process QR is 0 01 P V2

R

2V0V0

P

Q3P0

P0

,d eksy vkn'kZ ,dy ijekf.o; xSl dks uhps fn[kk;s x;s pØh; izØe PQRP es fy;k x;k gS

fn;s x;s pØh; izØe ds ckjs esa lgh dFku gS@gSa\

(A) izØe QR ds vuqfn'k igys rki c<+rk gS ,d vf/kdre eku rd igq¡prk gS vkSj fQj ?kVrk gSA

(B) izØe QR ds vuqfn'k igys rki ?kVrk gS ,d U;wure eku rd igq¡prk gS vkSj fQj c<+rk gSA

(C) izØe RP ds nkSjku ra= rFkk ifjos'k ds e/; dqy m"ek fofue; 0 05 P V2 gSA

(D) izØe QR ds nkSjku fu"dkflr dqy m"ek 0 01 P V2 gSA


ALLEN

LTS-30/42


2. In a voltmeter, mass of a metal deposited in 30 seconds in 200 gms. Analyse the current v/s timegraph shown below and identify the correct statement(s):(A) Electrochemical equivalent of the metal is 100 g/coulomb(B) 33.33 grams got discharged in the first 10 seconds.(C) A constant current of 66.66 mA would also discharge approximately the same amount in

same time(D) 100 gms of metal got discharged in the first 15 seconds

i

100 mA

10sec 20sec 30sec

t

oksYVehVj esa 30 lSd.M esa fu{ksfir /kkrq dk æO;eku 200 gms gSA uhps fn[kk;s x;s /kkjk v/s le; ds vkjs[k dk fo'ys"k.kdhft;s rFkk lgh dFkuksa dk p;u dhft;s :(A) /kkrq dk oS|qrjklk;fud rqY;kad 100 g/dwykWe gSA

(B) igys 10 lSd.M esa 33.33 xzke fu{ksfir gq;h

(C) 66.66 mA dh fu;r /kkjk ls leku le; es yxHkx leku ek=k es fu{ksfir gksxhA

(D) igys 15 lSd.M es 100 gms /kkrq fu{ksfir gq;hA

3. In which of the following pair both have O-N-O bond angle is/are higher than 120° ?

fuEu esa ls dkSuls ;qXe dh nksuksa Lih'kht esa O-N-O ca/k dks.k 120° ls vf/kd gS ?

(A) N2O4 & NO2 (B) NO2 & N2O5 (C) 3NO- & N2O5 (D) N2O5 & N2O4


ALLEN

LTS-31/42

4. Select correct for the following given two types of double chain silicates anionSiliconOxygen

(I)

(II)

(A) each Si in both is bonded with four 'O' atoms(B) oxidation number of each Si = +4 in both(C) both have tetrahedron with three shared 'O' corners(D) in both all the tetrahedron have three shared 'O' corners

nh xbZ fuEu nks izdkj ds Mcy Ükaa[kyk flfydsV ½.kk;u ds fy, lgh fodYi pqfu,SiliconOxygen

(I)

(II)

(A) nksuksa esa] izR;sd Si pkj 'O' ijek.kqvksa ls caf/kr gS

(B) nksuksa esa] izR;sd Si dh vkWDlhdj.k la[;k = +4 gS

(C) nksuksa esa prq"Qyd rhu 'O' dksuksa ds lkFk lkf>r gS

(D) nksuksa esa lHkh prq"Qyd rhu 'O' dksuksa ds lkFk lkf>r gSa



ALLEN

LTS-32/42

5. Participation of 'd' orbital in hybridisation of x, y & z arex, y rFkk z ds ladj.k esa] fuEu esa ls dkSuls fodYi ds 'd' d{kd Hkkx ysrs gSa\XeF6 + 1H2O ¾® x + 2HFXeF6 + 2H2O ¾® y + 4HFXeF6 + 3H2O ¾® z + 6HF

(A) x = -2 2 2x y zd ,d ; y = 2zd , z = None

(B) x = -2 2 xyx yd ,d ; y = 2zd , z = dxy

(C) x = 2 zxzd , d ; y = -2 2x yd , z = dxy

(D) x = -2 2 2x y zd , d ; y = dxy , z = 2zd

6. Select correct statement(s) out of following for compound given below :

OBr(x)

Br(y)

Br(z)(A) Br(y) is the most reactive Br towards SN2 reaction in this compound.(B) Br(z) is the most reaction Br towards SN1 reaction in this compound.(C) 16 stereoisomers are theoretically possible for this compound.(D) This compound is capable of showing geometrical isomerismuhps fn;s x;s ;kSfxd ds fy;sa lgh dFkuksa dk p;u dhft;s\

OBr(x)

Br(y)

Br(z)(A) bl ;kSfxd esa SN2 vfHkfØ;k ds izfr Br(y) lokZf/kd fØ;k'khy Br gSA

(B) bl ;kSfxd esa SN1 vfHkfØ;k ds izfr Br(z) lokZf/kd fØ;k'khy Br gSA

(C) bl ;kSfxd ds fy;s lS¼kfUrd :i ls 16 f=foe leko;oh lEHko gSA

(D) ;g ;kSfxd T;kfefr; leko;ork iznf'kZr dj ldrk gSA



ALLEN

LTS-33/42

7. Select correct statement(s) out of following ?

(A) O

is more reactive than NH

towards reaction with PhMgBr

(B) O2Cl (aq)hn

¾¾¾¾® O

Cl (major product)

(C) 2Conc. H SO2 4(Catalytic) (Major)

(D) D-glucose + D-mannose + D-fructose 2PhNHNH¾¾¾¾® only one osazone

fuEu es ls lgh dFkuks dk p;u dhft;s\

(A) O

, PhMgBr ds lkFk vfHkfØ;k ds izfr NH

dh rqyuk es vf/kd fØ;k'khy gSA

(B) O2Cl (aq)hn

¾¾¾¾® O

Cl (eq[; mRikn)

(C) 2Conc. H SO2 4(Catalytic) (eq[;)

(D) D-Xywdksl + D-esuksl + D-ÝDVksl 2PhNHNH¾¾¾¾® dsoy ,d vkslktksu



ALLEN

LTS-34/42






iw.kZ vad : +3 ;fn flQZ lgh fodYi ds vuq:i cqycqys dks dkyk fd;k gSA'kwU; vad : 0 ;fn fdlh Hkh cqycqys dks dkyk ugha fd;k gSA½.k vad : –1 vU; lHkh ifjfLFkfr;ksa esaAAnswer Q.8, Q.9 and Q.10 by appropriately matching the information given in the threecolumns of the following table.

Column-1 Column-2 Column-2(A) + (B) pH at equivalence point Potentiometric titration

curve

(I)

(II)

(III)

(IV)

NaOH + HCl

NH OH + HCl4

NaOH + CH COOH3

NH OH + CH COOH4 3

(i) 7 ; no hydrolysis

(ii) 8.5 ; anionic hydrolysis

(iii) 7 ; cationic & anionic hydrolysis

(iv) 5.5 ; cationic hydrolysis

Vequivalence

Vequivalence

Vequivalence

Vequivalence

V addedA

V addedA

V addedA

V addedA

®

®

®

®

®®

®®

Gsolution

Gsolution

Gsolution

Gsolution

(a)

(b)

(c)

(d)


ALLEN

LTS-35/42

0.1 M, 100 ml of reagent (B) is taken & 100 ml of 0.1M A is added to it. The volume of finalsolution is made 1 litre by adding water. Assuming (A) is added to (B) dropwise,(K(a)(CH3COOH) = 10–5 M ; K(b)(NH4OH) = 10–5 M ; Kw(H2O) = 10–14 M2 , log2 = 0.3

8. Identify which of the following combinations represent a correct combination of all three columnentries :(A) (I); (i); (a) (B) (III); (ii); (c) (C) (IV); (iii); (d) (D) (II); (iv); (b)

9. Identify which of the following combinations represents an incorrect combination of all threecolumn entries

(A) (I); (i); (b) (B) (IV); (iii); (d) (C) (I); (iii); (b) (D) (III); (ii); (a)

10. Identify which of the following combinations represents an incorrect combination of all threecolumn entries

(A) (III); (ii); (a) (B) (II); (iv); (c) (C) (I); (ii); (a) (D) (IV); (iii); (d)



ALLEN

LTS-36/42

uhps nh x;h Vscy ds rhu dkWyeks a esa miyC/k lwpuk dk mi;qÙkG < ax ls lqesy dj iz'uks a Q.8, Q.9 vkSj Q.10ds mÙkj nhft;sA

(A) + (B) rqY;kad fcUnq ij pH foHkoekih vuqekiu oØ

(I)

(II)

(III)

(IV)

NaOH + HCl

NH OH + HCl4

NaOH + CH COOH3

NH OH + CH COOH4 3

(i) 7 ; ty vi?kVu ugh

(ii) 8.5 ; ½.kk;fud ty vi?kVu

(iii) 7 ; èkuk;fud rFkk + ½.kk;fud ty vi?kVu

(iv) 5.5 ; èkuk;fud ty vi?kVu

Vequivalence

Vequivalence

Vequivalence

Vequivalence

V addedA

V addedA

V addedA

V addedA

®

®

®

®

®®

®®

Gsolution

Gsolution

Gsolution

Gsolution

(a)

(b)

(c)

(d)

0.1 M, 100 ml vfHkdeZd (B) fy;k x;k gS rFkk blesa 100ml, 0.1M-A feyk;k tkrk gSA ty feykdj vfUre foy;udk vk;ru ,d yhVj fd;k x;k gSA ekus dh (A) dks (B) esa cwan&cwan djds feyk;k tkrk gS

(K(a)(CH3COOH) = 10–5 M ; K(b)(NH4OH) = 10–5 M ; Kw(H2O) = 10–14 M2 , log2 = 0.3

8. fuEu es ls dkSulk la;ksx lHkh rhuks dkWyeks dh izfof"V;ks ds lgha la;ksx dks iznf'kZr djrk gSA

(A) (I); (i); (a) (B) (III); (ii); (c) (C) (IV); (iii); (d) (D) (II); (iv); (b)

9. fuEu es ls dkSulk la;ksx lHkh rhuks dkWyeks dh izfof"V;ks ds xyr la;ksx dks iznf'kZr djrk gSA

(A) (I); (i); (b) (B) (IV); (iii); (d) (C) (I); (iii); (b) (D) (III); (ii); (a)

10. fuEu es ls dkSulk la;ksx lHkh rhuks dkWyeks dh izfof"V;ks ds xyr la;ksx dks iznf'kZr djrk gSA

(A) (III); (ii); (a) (B) (II); (iv); (c) (C) (I); (ii); (a) (D) (IV); (iii); (d)



ALLEN

LTS-37/42

Answer Q.11, Q.12 and Q.13 by appropriately matching the information given in thethree columns of the following table. Column-1 Column-2 Column-3

(I)

Me

MeCH CH CH Cl2 2 2

AlCl3(A) Only one organic (P) Carbocation is formed

product is formed as intermediate

(II)

NO2

(1) NaOH

(2) H

F

(B) Product is optically (Q) Carbanion is formed

active as intermediate

(III)Red P (limited)

Br (leq.)2CH CH COOH3 2 (C) This process involves (R) Free radical is

nucleophite substitution formed asreaction at one of the intermediatecarbon

(IV) CH3

COOH

H

NH2NaNO + HCl2 (D) This process involves (S) Nitrene is formed as

electrophle substitioun intermediatereaction at one of thecarbon

11. Which of the following option is correctly matched :(A) (I),(D),(P) (B) (II), (D), (Q) (C) (III), (A), (R) (D) (IV), (B), (S)

12. Select incorrect match :(A) (I), (D), (P) (B) (II), (A), (Q) (C) (III), (B), (R) (D) None of these

13. Product of (III) can be converted into product of (IV) by :(A) NaNO2 + HCl (B) aq. KOH followed by HÅ

(C) NH3 (D) AgNO2



ALLEN

LTS-38/42

uhps nh x;h Vscy ds rhu dkWyeks a esa miyC/k lwpuk dk mi;qÙkG < ax ls lqesy dj iz'uks a Q.11, Q.12 vkSj Q.13ds mÙkj nhft;sA

dkWye-1 dkWye-2 dkWye-3

(I)

Me

MeCH CH CH Cl2 2 2

AlCl3(A) dsoy ,d dkcZfud (P) dkcZ/kuk;u e/;orhZ ds :i

mRikn curk gSA es curk gSA

(II)

NO2

(1) NaOH

(2) H

F

(B) mRikn izdkf'kd (Q) dkcZ½.kk;u e/;orhZ ds

lfØ; gSA :i es curk gSA

(III)Red P (limited)

Br (leq.)2CH CH COOH3 2 (C) bl izØe es fdlh ,d (R) eqDr ewyd e/;orhZ

dkcZu ij ukfHkdLusgh ds :i esa curk gSAizfrLFkkiu vfHkfØ;klfEefyr gSA

(IV) CH3

COOH

H

NH2NaNO + HCl2

(D) bl izØe es fdlh ,d (S) ukbVªhu e/;orhZ ds

dkcZu ij bysDVªkWuLusgh :i es curk gSAizfrLFkkiu vfHkfØ;klfEefyr gSA

11. fuEu es ls dkSulk fodYi lgh :i ls lqesfyr gS %(A) (I),(D),(P) (B) (II), (D), (Q) (C) (III), (A), (R) (D) (IV), (B), (S)

12. xyr lqesyu dk p;u dhft;s %

(A) (I), (D), (P) (B) (II), (A), (Q) (C) (III), (B), (R) (D) buesa ls dksbZ ugha

13. fuEu es ls fdlds }kjk (III) ds mRikn dks (IV) ds mRikn es :ikarfjr fd;k tk ldrk gS\

(A) NaNO2 + HCl (B) tfy; KOH ds i'pkr HÅ }kjk(C) NH3 (D) AgNO2



ALLEN

LTS-39/42






� bl [k.M esa ik¡p iz'u gSa� izR;sd iz'u dk mÙkj 0 ls 9 rd (nksuksa 'kkfey) ds chp dk ,d ,dy vadh; iw.kk±d gSA

� izR;sd iz'u esa] vks-vkj-,l- ij lgh iw.kk±d ds vuq:i cqycqys dks dkyk djsaA


iw.kZ vad : +3 ;fn flQZ lgh mÙkj ds vuq:i cqycqys dks dkyk fd;k gSA

'kwU; vad : 0 vU; lHkh ifjfLFkfr;ksa esaA1. Number of negatively charged colloids among the following is -

(i) Blood (ii) Ferric hydroxide sol (iii) Gold sol(iv) Clay (v) Acidic dye (vi) Haemoglobin(vii) Basic dye (viii) Starch (ix) As2S3 solfuEu es ls ½.kkRed vkosf'kr dksykWbM dh la[;k crkb;s\

(i) jDr (ii) Qsfjd gkbMªkWDlkbM lkWy (iii) xksYM lkWy

(iv) e`nk (v) vEyh; jatd (vi) gheksXyksfcu

(vii) {kkjh; jatd (viii) LVkpZ (ix) As2S3 lkWy2. Find the number of complex/complex ion

Which have atleast three isomeric complex with same molecular formula.

fuEu esa ls , sls ladqyksa@ladqy vk;uksa dh la[;k crkbZ;s tks leku vkf.od lw= ds lkFk de ls de rhu leko;oh ladqy

j[krs gSa(a) [CoCl2Br2]2– (b) K2[Pt(CN)2(NO2)2] (c) [Co(C2O4)3]3–

(d) [Pt(en)2Cl2]Cl2 (e) [Pt(NH3)3Cl]Br (f) [Pt(NH3)4][PtCl4]



ALLEN

LTS-40/42


3. Borax (Na2B4O7.10H2O) has ratio of s-bond & lone pair is

cksjsDl (Na2B4O7.10H2O) esa mifLFkr s-ca/k rFkk ,dkadh bysDVªkWu ; qXe dk vuqikr gS\

4. Find the number of compound(s) from the following has needed more than three water moleculesfor complete hydrolysis of one molecule of given species?

PCl3, H4P2O8, PCl5, P4O10, IF7, IF5, H2S2O7

fuEu esa ls ,sls ;kSfxdksa dh la[;k crkbZ;s ftuds ,d v.kq ds iw.kZ tyvi?kVu ds fy, rhu ls vf/kd ty ds v.kqvksa dh

vko';drk gksrh gS ?

PCl3, H4P2O8, PCl5, P4O10, IF7, IF5, H2S2O7

5. All the compounds which gives positive test with Bredy's reagent (2,4–DNP) and are havingmolecular mass 100 are treated separately with NaBH4. Product of how many of these can beused effectively to separate racemic mixture of 2-methylbutanoic acid.

vkf.od æO;eku 100 j[kus okys lHkh ;kSfxd tks czsMh vfHkdeZd (2,4–DNP) ds lkFk /kukRed ijh{k.k nsrs gS] dksNaBH4 ds lkFk i`Fkd&i`Fkd mipkfjr fd;k x;k gSA bues ls fdrus mRiknks dk izHkkoh :i ls 2-esfFkyC;wVsuksbZd vEy

ds jslsfed feJ.k dks i`Fkd djus ds fy;s iz;ksx fd;k tk ldrk gSA


ALLEN

LTS-41/42



ALLEN

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eSaus lHkh funs Z'kks a dks i<+ fy;k gS vkSj eS mudkvo'; ikyu d:¡xk@d:¡xhA

____________________________Signature of the Candidate / ijh{kkFkhZ ds gLrk{kj

I have verified the identity, name and Formnumber of the candidate, and that questionpaper and ORS codes are the same.eSaus ijh{kkFkhZ dk ifjp;] uke vkSj QkWeZ uEcj dks iwjh rjgtk¡p fy;k gS ,oa iz'u i= vkSj vks- vkj- ,l- dksM nksuks a leku gSaA

____________________________Signature of the Invigilator / fujh{kd ds gLrk{kj

NAME OF THE CANDIDATE / ijh{kkFkhZ dk uke .............................................................................FORM NO. / QkWeZ uEcj .............................................

Your Target is to secure Good Rank in JEE 2018LTS-42/42

Que. No. Category-wise Marks for Each Question / oxkZuqlkj izR;sd iz'u ds vad MaximumSection Type of Full Marks Partial Marks Zero Marks Negative Marks of the

Que. iw.kZ vad vkaf'kd vad 'kwU; vad Marks section[k.M iz'u dk iz'uksa ½.k vad [k.M esa

izdkj dh vf/kdre vadla[;k

+4 +1 0 –2One or more If only the bubble(s) For darkening a bubble If none In all

correct corresponding corresponding to each of the otheroption(s) to all the correct correct option, provided bubbles is cases

I(i) ,dy ;k ,d ls 7 option(s) is(are) NO incorrect option darkened vU; lHkh 28vf/kd lgh darkened darkened ;fn fdlh Hkh ifjfLFkfr;ksa

fodYi ;fn flQZ lkjs lgh fodYi izR;sd lgh fodYi ds vuq:i cqycqys dks esa(fodYiksa) ds vuq:Ik cqycqys dks dkyk djus ij] dkyk ughacqycqys (cqycqyksa) dks ;fn dksbZ xyr fodYi dkyk fd;k gS

dkyk fd;k x;k gS ugha fd;k gSMatching +3 0 –1Lists Type If only the bubble If none In all

(Single corresponding to of the other correct option) the correct option bubbles is cases

I(ii) lqpha lqesyu 6 is darkened — darkened vU; lHkh 18izdkj ;fn flQZ lgh fodYi ds ;fn fdlh Hkh ifjfLFkfr;ksa

(,dy lgh vuq:Ik cqycqys dks cqycqys dks dkyk esafodYi) dkyk fd;k gS ugha fd;k gS

+3 0Single digit If only the bubble In all

Integer corresponding other(0-9) to correct answer cases

IV ,dy vadh; 5 is darkened — vU; lHkh — 15iw.kk±d ;fn flQZ lgh mÙkj ds ifjfLFkfr;ksa(0-9) vuq:Ik cqycqys dks esa

dkyk fd;k gS


leader test series / joint package course · allen-2/42 note : i cor , [email protected]...

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