DO N

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BY T

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Ïi;k bu fun sZ'kks a dks /;ku ls i<+ sA vkidks 5 feuV fo'ks"k :i ls bl dke ds fy, fn;s x;s gS aAPlease read the instructions carefully. You are allotted 5 minutes specifically for this purpose.

A. lkekU; :

1. ;g iqfLrdk vkidk iz'u&i= gSA bldh eqgj rc rd u

rksM+ s tc rd fujh{kd ds }kjk bldk fun Z s'k u fn;k

tk;sA

2. iz'u&i= dk dksM (CODE) bl i`"B ds Åijh ck;sa dkSus ij

Nik gSA

3. dPps dk;Z ds fy, [kkyh i`"B vkSj [kkyh LFkku bl iqfLrdk

esa gh gSaA dPps dk;Z ds fy, dksbZ vfrfjDr dkxt ugha fn;k

tk;sxkA

4. dksjs dkxt] fDyi cksMZ] ykWx rkfydk] LykbM :y] dSYdqysVj]

dSejk] lsyQksu] istj vkSj fdlh izdkj ds bysDVªkWfud midj.k

ijh{kk d{k esa vuqefr ugha gSaA

5. bl iqfLrdk ds fiNys i`"B ij fn, x, LFkku esa viuk uke

vkSj QkWeZ uEcj fyf[k,A

6. mÙkj i=] ,d ;a=&Js.khdj.k ;ksX; i= (ORS) gS tks fd vyx

ls fn;s tk;saxsA

7. vks-vkj-,l-(ORS) ;k bl iqfLrdk esa gsj&Qsj@foÏfr u

djs aA

8. bl iqfLrdk dh eqgj rksM+us ds i'pkr d`i;k tk¡p ysa fd blesa

38 i`"B gSa vkSj izR;sd fo"k; ds lHkh 22 iz'u vkSj muds mÙkj

fodYi Bhd ls i<+ s tk ldrs gSaA lHkh [kaMksa dh 'kq:vkr esa

fn;s gq, funsZ'kksa dks /;ku ls i<+ sA

B. vks-vkj-,l- (ORS) dk Hkjko :

9. ijh{kkFkhZ dks gy fd;s x;s iz'u dk mÙkj ORS mÙkj iqfLrdk

esa lgh LFkku ij dkys ckWy ikbUV dye ls mfpr xksys dks

xgjk djds nsuk gSA

10. ORS ds (i`"B la[;k 1) ij ekaxh xbZ leLr tkudkjh /;ku

iwoZd vo'; Hkjsa vkSj vius gLrk{kj djsaA

C. iz'ui= dk izk:i :

bl iz'u&i= ds rhu Hkkx (xf.kr] HkkSfrd foKku vkSj

jlk;u foKku) gSaA gj Hkkx ds nks [kaM gSaA

INSTRUCTIONS / lwpuk,¡

01CT313083 KOTA - 1/38Your Target is to secure Good Rank in JEE 2014

A. General :

1. The booklet is your Question Paper. Do not break

the seal of this booklet before being instructed to

do so by the invigilator.

2. The question paper CODE is printed on the left hand

top corner of this sheet.

3. Blank spaces and blank pages are provided in the

question paper for your rough work. No additional

sheets will be provided for rough work.4. Blank papers, clipboards, log tables, slide rules,

calculators, cameras, cellular phones, pagers andelectronic gadgets are NOT allowed inside theexamination hall.

5. Write your name and Form number in the space

provided on the back cover of this booklet.

6. The answer sheet, a machine-readable Optical

Response Sheet (ORS), is provided separately.

7. DO NOT TAMPER WITH/MUTILATE THE ORS OR THE

BOOKLET.8. On breaking the seal of the booklet check that it

contains 38 pages and all the 22 questions in eachsubject and corresponding answer choices arelegible. Read carefully the instructions printed at thebeginning of each section.

B. Filling the ORS :9. A candidate has to write his / her answers in the ORS

sheet by darkening the appropriate bubble with thehelp of Black ball point pen as the correct answer(s)of the question attempted.

10. Write all information and sign in the box provied on

part of the ORS (Page No. 1).

C. Question Paper Formate :The question paper consists of 3 parts(Mathematics, Physics and Chemistry). Each partconsists of two sections.

Ïi;k 'ks"k funs Z'kks a ds fy;s bl iqfLrdk ds vfUre i`"B dks i<+ sAPlease read the last page of this booklet for read the instructions

PAPER – 2

SCORE-I : JEE (Advanced) # 01

PATTERN : JEE (Advanced)LEADER & ENTHUSIAST COURSE

PAPER CODETM

Path to success KOTA (RAJASTHAN)0 1 C T 3 1 3 0 8 3

Date : 23 - 04 - 2014TARGET : JEE 2014

SCORE-II : TEST # 03

le; : 3 ?k.Vs egÙke vad : 243Time : 3 Hours Maximum Marks : 243

01CT313083KOTA - 2/38

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PAPER – 2

fo"k; [k.M i`"B la[;kSubject Section Page No.

Hkkx-1 xf.kr I(i) dsoy ,d lgh fodYi izdkj 03 - 06Part-1 Mathematics Only One Option Correct Type

I(ii) ,d ;k vf/kd lgh fodYi izdkj 07 - 08One or More Options Correct Type

I(iii) vuqPNsn izdkj 09 - 10Paragraph Type

IV iw.kk±d eku lgh izdkj 11 - 12Integer Value Correct Type

Hkkx-2 HkkSfrd foKku I(i) dsoy ,d lgh fodYi izdkj 13 - 18Part-2 Physics Only One Option Correct Type

I(ii) ,d ;k vf/kd lgh fodYi izdkj 19 - 22One or More Options Correct Type

I(iii) vuqPNsn izdkj 23 - 25Paragraph Type

IV iw.kk±d eku lgh izdkj 26 - 27Integer Value Correct Type

Hkkx-3 jlk;u foKku I(i) dsoy ,d lgh fodYi izdkj 28 - 30Part-3 Chemistry Only One Option Correct Type

I(ii) ,d ;k vf/kd lgh fodYi izdkj 31 - 32One or More Options Correct Type

I(iii) vuqPNsn izdkj 33 - 34Paragraph Type

IV iw.kk±d eku lgh izdkj 35 - 36Integer Value Correct Type

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 = 58Atomic 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

TARGET : JEE 2014 23-04-2014

LEADER & ENTHUSIAST COURSE(DATE : 23-04-2014)

MATHEMATICSTM

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PAPER – 2

PART-1 : MATHEMATICS

Hkkx-1 : xf.krSECTION–I : (i) Only One option correct Type

[k.M-I : (i) dsoy ,d lgh fodYi izdkjThis section contains 9 multiple choice questions. Each question has four choices (A), (B), (C) and(D) out of which ONLY ONE is correct.

bl [k.M esa 9 cgqfodYi iz'u gSA izR;sd iz'u esa pkj fodYi (A), (B), (C) vkSj (D) gSa ftuesa ls dsoy ,d

lgh gSA1. Given two points A º (–2, 0) and B º (0, 4). The coordinates of point M on the line x = y so that

perimeter of DAMB is least is :(A) (1, 1) (B) (0, 0) (C) (2, 2) (D) (3, 3)

nks fcUnq A º (–2, 0) rFkk B º (0, 4) fn, x, gSA js[kk x = y ij fcUnq M ds funsZ'kkad bl izdkj gS fd f=Hkqt AMB

dh ifjfèk U;wure gS :(A) (1, 1) (B) (0, 0) (C) (2, 2) (D) (3, 3)

2. Evaluate 3

5 4 2

3 5x

x x x xdx, x 0

x x x 1

+ +>

- - -ò

(A) - - - +4 34

2 1x x x C

xx(B) 4 52 1 x x x C+ + + +

(C) - - - +4 3 12 x x x C

x(D) None of these

(Where 'C' is constant of integration)

3

5 4 2

3 5x

x x x xdx , x 0

x x x 1

+ +>

- - -ò Kkr dhft,

(A) - - - +4 34

2 1x x x C

xx(B) 4 52 1 x x x C+ + + +

(C) - - - +4 3 12 x x x C

x(D) buesa ls dksbZ ugha

(tgk¡ 'C' lekdyu vpj gS)

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

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

MATHEMATICSTARGET : JEE 2014(DATE : 23-04-2014)

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01CT313083KOTA - 4/38

PAPER – 2

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

3. If a1, a2, a3 ..... an be in A.P. , (1 + sin 2q) is arithmetic mean of a1, an , (1 – tan q) is arithmetic mean of

1 n 2 n 1 n 1

1 1 1, , .....,

a a a a a a-, (1 + tan q) is arithmetic mean of

1 2 n

1 1 1, ,.....

a a a then general solution of

q is/are :

(A) n

n ,4 2p p

p - (n Î I) (B) n , n4p

p - p (n Î I)

(C) n4p

p - (n Î I) (D) np (n Î I)

;fn a1, a2, a3 ..... an lekUrj Js.kh esa gSa rFkk a1, an dk lekUrj ek/; (1 + sin 2q) gS] 1 n 2 n 1 n 1

1 1 1, , .....,

a a a a a a- dk

lekUrj ekè; (1 – tan q) gS] 1 2 n

1 1 1, ,.....

a a a dk lekUrj ek/; (1 + tan q) gS] rks q ds O;kid gy gksxk@gksaxs :

(A) n

n ,4 2p p

p - (n Î I) (B) n , n4p

p - p (n Î I)

(C) n4p

p - (n Î I) (D) np (n Î I)

4. sin xef : 0, R, if f(x) log ((sinx) 1),

2pæ ö® = +ç ÷

è øthen the minimum value of f(x) is :

(A) loge 2 (B) 1 / e

e1

log 1e

æ öæ ö +ç ÷ç ÷ç ÷è øè ø(C) e

elog (e 1)+ (D) 2

pæ ö® = +ç ÷è ø

sin xef: 0, R, f(x) log ((sinx) 1)

2;fn gks] rks f(x) dk U;wure eku gksxk :

(A) loge 2 (B) 1 / e

e1

log 1e

æ öæ ö +ç ÷ç ÷ç ÷è øè ø(C) e

elog (e 1)+ (D) 2

LEADER & ENTHUSIAST COURSE(DATE : 23-04-2014)

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01CT313083 KOTA - 5/38

PAPER – 2

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

5. The area bounded by y = f(x), x-axis and the line y =1, where x

1

1f(x) 1 f(t)dt

x= + ò is :

(A) 2(e + 1) (B) 1

2 1e

æ ö-ç ÷è ø

(C) 2(e 1)- (D) None of these

y = f(x), x-v{k rFkk js[kk y =1 }kjk ifjc¼ {ks=Qy] tgk¡ x

1

1f(x) 1 f(t)dt

x= + ò gS] gksxk :

(A) 2(e + 1) (B) 1

2 1e

æ ö-ç ÷è ø

(C) 2(e 1)- (D) buesa ls dksbZ ugha

6. Given A(0, 0) and B(x, y) with x Î (0, 1) and y > 0 let the slope of line AB equals m1 . Point C lies in

the line x = 1 such that slope of BC equals m2 where 0 < m2 < m1. If the area of DABC can be expressed

as (m1 – m2) f(x), then the largest possible value of f(x) is :

(A) 1 (B) 12

(C) 14

(D) 18

fn;k gS A(0, 0) rFkk B(x, y), tgk¡ x Î (0, 1) rFkk y > 0 gSA ekuk js[kk AB dh izo.krk m1 gSA fcUnq C, js[kk x = 1ij

bl izdkj fLFkr gS fd BC dh izo.krk m2, tgk¡ 0 < m2 < m1 gSA ;fn f=Hkqt ABC ds {ks=Qy dks (m1 – m2) f(x) }kjk

O;Dr dj ldrs gks] rks f(x) dk vf/kdre laHko eku gksxk :

(A) 1 (B) 12

(C) 14

(D) 18

7. A straight line cuts the x-axis at point A(1, 0) and y-axis at point B, such that ÐOAB = a (a >4p

),

2pæ öa ¹ç ÷

è ø C is the middle point of AB. if B' is is mirror image of B with respect to the line OC and C' is the

mirror image of C with respect to the line BB', then find the ratio of areas of triangles ABB' and BB'C' :

(A) 1 (B) 12

(C) 2 (D) depends upon a

,d ljy js[kk x-v{k dks fcUnq A(1, 0) rFkk y-v{k dks fcUnq B ij bl izdkj dkVrh gS fd ÐOAB = a (a >4p

),

2pæ öa ¹ç ÷

è ø gS rFkk C, AB dk e/; fcUnq gSA ;fn js[kk OC ds lkis{k fcUnq B dk izfrfcEc B' rFkk js[kk BB' ds lkis{k fcUnq

C dk izfrfcEc C' gks] rks f=Hkqtksa ABB' rFkk BB'C' ds {ks=Qy dk vuqikr gksxk -

(A) 1 (B) 12

(C) 2 (D) a ij vkfJr

MATHEMATICSTARGET : JEE 2014(DATE : 23-04-2014)

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01CT313083KOTA - 6/38

PAPER – 2

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

8.24

20

x (sin 2x cos 2x)dx

(1 sin 2x)cos x

p

-

+ò equals to :

(A) 2

ln216 4p p

- (B) 2

ln216 4p p

+ (C) 2

ln 28 4p p

+ (D) None of these

24

20

x (sin 2x cos 2x)dx

(1 sin 2x)cos x

p

-

+ò dk eku gksxk :

(A) 2

ln216 4p p

- (B) 2

ln216 4p p

+ (C) 2

ln 28 4p p

+ (D) buesa ls dksbZ ugha

9. If A, B, Cur ur ur

are three non-zero vectors, no two of them are parallel if A B+ur ur

is collinear to Cur

, B C+ur ur

is

collinear to Aur

then A B C+ +ur ur ur

is equal to :

(A) Aur

(B) Bur

(C) Cur

(D) None of these

;fn A, B, Cur ur ur

rhu v'kwU; lfn'k] ftlesa dksbZ Hkh nks lekUrj ugha gS ;fn A B+ur ur

, Cur

ds lejS[kh;] B C+ur ur

, Aur

ds

lejS[kh; gks] rks A B C+ +ur ur ur

cjkcj gksxk :

(A) Aur

(B) Bur

(C) Cur

(D) buesa ls dksbZ ugha

LEADER & ENTHUSIAST COURSE(DATE : 23-04-2014)

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PAPER – 2

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

(ii) One or more options correct Type

(ii) ,d ;k vf/kd lgh fodYi izdkjThis section contains 4 multiple choice questions. Each question has four choices (A), (B), (C) and(D) out of which ONE or MORE are correct.

bl [k.M esa 4 cgqfodYi iz'u gSaA izR;sd iz'u esa pkj fodYi (A), (B), (C) vkSj (D) gSa] ftuesa ls ,d ;k

vf/kd lgh gSA

10. If tan6q + 27 tan2q = 3(1 + 11 tan4 q) is satisfied if q is equals to :

(A) 20° (B) 15° (C) 60° (D) 40°

;fn tan6q + 27 tan2q = 3(1 + 11 tan4 q) lUrq"V gksxk] ;fn q cjkcj gks :

(A) 20° (B) 15° (C) 60° (D) 40°

11. Let a = 5p

and A = cos sinsin cos

a aé ùê ú- a aë û

, then B = A + A2 + A3 + A4 is

(A) Singular (B) Non-singular (C) Symmetric (D) det (B) = 1

ekuk a = 5p

rFkk A = cos sinsin cos

a aé ùê ú- a aë û

gks] rks B = A + A2 + A3 + A4 gksxk

(A) vO;qRØe.kh; (B) O;qRØe.kh; (C) lefer (D) det (B) = 1

MATHEMATICSTARGET : JEE 2014(DATE : 23-04-2014)

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PAPER – 2

12. Let 5x , 3

2pæ öÎ pç ÷

è ø then :

(A) ( )1 1 1 7cos sin(cos (cos x) 2 tan (tan x)) x

2- - - p

- = -

(B) ( )1 1 1 9cos sin(cos (cos x) 2 tan (tanx)) x

2- - - p

- = -

(C) ( )1 1 1tan cot(2sin (sin x) tan (tanx)) x2

- - - p+ = +

(D) ( )1 1 1 5tan cot(2sin (sinx) tan (tanx)) x

2- - - p

+ = -

ekuk 5x , 3

2pæ öÎ pç ÷

è ø gks] rks :

(A) ( )1 1 1 7cos sin(cos (cos x) 2 tan (tan x)) x

2- - - p

- = -

(B) ( )1 1 1 9cos sin(cos (cos x) 2 tan (tanx)) x

2- - - p

- = -

(C) ( )1 1 1tan cot(2sin (sin x) tan (tan x)) x2

- - - p+ = +

(D) ( )1 1 1 5tan cot(2sin (sinx) tan (tanx)) x

2- - - p

+ = -

13. The points P, Q and R are taken on the ellipse 2 2

2 2

x y1

a b+ = with eccentric angles q, q + a, q +2a then

area of triangle PQR is :(A) Independent of q (B) Independent of a

(C) Maximum when 23p

a = (D) Maximum when 2p

q=

nh?kZoÙk 2 2

2 2

x y1

a b+ = ij fcUnq P, Q rFkk R ftuds mRØsUæu dks.k q, q + a, q +2a gks] rks f=Hkqt PQR dk {ks=Qy gksxk :

(A) q ls LorU= (B) a ls LorU=

(C) vf/kdre tc 23p

a = (D) vf/kdre tc 2p

q =

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

LEADER & ENTHUSIAST COURSE(DATE : 23-04-2014)

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PAPER – 2

(iii) Paragraph Type (iii) vuqPNsn izdkj

This section contains 2 paragraphs each describing theory, experiment, data etc. Six questions relateto two paragraphs with three questions on each paragraph. Each question of a paragraph has onlyone correct answer among the four choices (A), (B), (C) and (D).bl [k.M esa fl¼kUrksa] iz;ksxksa vkSj vk¡dM+ksa vkfn dks n'kkZus okys 2 vuqPNsn gSA nksuksa vuqPNsn ls lacaf/kr N% iz'ugSa] ftuesa ls gj vuqPNsn ij rhu iz'u gSaA vuqPNsn esa gj iz'u ds pkj fodYi (A), (B), (C) vkSj (D) gSa ftuesals dsoy ,d lgh gSA

Paragraph for Question 14 to 16iz'u 14 ls 16 ds fy;s vuqPNsn

A matrix is a rectangular array of number. The numbers may be real or complex. A matrix with m rowsand n columns is called m × n matrix. If ij m nA (a ) ´= be a matrix. the transpose of A, denoted by A' or byAT. A square matrix is said to be symmetric if A' = A and skew symmetric if A' = –A. If A is any n-rowed squared matrix then a matrix B if exist, such that AB = BA = In is called inverse of A and denotedby B = A–1.,d vkO;wg la[;kvksa dh ,d vk;rkdkj lkj.kh ds :i esa izcU/k gSA la[;k;sa okLrfod ;k dkYifud gks ldrh gSA miafDr;ksa rFkk n LrEHkksa okys vkO;wg dks m × n vkO;wg dgrs gSA vkO;wg ij m nA (a ) ´= ds ifjorZ dks A' ;k AT }kjk n'kkZ;k

tkrk gSA ,d oxZ vkO;wg dks lefer dgk tkrk gS] ;fn A' = A rFkk fo"ke lefer dgk tkrk gS] ;fn A' = –A gSA ;fnA dksbZ n-iafDr oxZ vkO;wg gks rFkk oxZ vkO;wg B bl izdkj fo|eku gks fd AB = BA = In gks] rks bls A dk izfryksedgrs gS rFkk B = A–1 }kjk iznf'kZr gksrk gSA

14. Let M and N be two 3 × 3 non-singular symmetric matrix such that MN = NM. If PT dentotes thetranspose of a matrix P then M2N2 (MTN)–1 (MN–1)T is equal to:(A) M2 (B) –N2 (C) –M2 (D) MNekuk M rFkk N nks 3 × 3 ds O;qRØe.kh; lefer vkO;wg bl izdkj gS fd MN = NM gSA ;fn PT, vkO;wg P ds ifjorZdks n'kkZrk gS] rks M2N2 (MTN)–1 (MN–1)T cjkcj gksxk :(A) M2 (B) –N2 (C) –M2 (D) MN

15. Let p be an odd prime number and Tp be the following set of 2 × 2 matrices:

Pa b

T A ; a, b, c {0, 1, .....p 1}c a

ì üé ùï ï= = Î -í ýê úï ïë ûî þ

The number of A in Tp such that A is either symmetric or skew symmetric or both and det(A) isdivisible by p:(A) (p–1)2 (B) 2(p–1)2 (C) (p–1)2 + 1 (D) 2p –1ekuk p ,d fo"ke vHkkT; la[;k rFkk Tp fuEu 2 × 2 vkO;wgksa dk leqPp; gS

Pa b

T A ; a, b, c {0, 1, .....p 1}c a

ì üé ùï ï= = Î -í ýê úï ïë ûî þ

vkO;wgksa A dh la[;k] Tp bl izdkj gS fd A ;k rks lefer ;k fo"ke lefer vkO;wg ;k nksuksa gS rFkk det(A), p lsfoHkkftr gS] gksxh :(A) (p–1)2 (B) 2(p–1)2 (C) (p–1)2 + 1 (D) 2p –1

16. Let A and B be two non singular matrices such that (AB)k = AkBk for three consecutive positiveintegral values of k then calculate BA2B–1:(A) A2 (B) B (C) A (D) B2

ekuk A rFkk B nks O;qRØe.kh; vkO;wg gS rFkk k ds rhu Øekxr /kukRed iw.kk±d eku bl izdkj gS fd (AB)k = AkBk

gS] rks BA2B–1 gksxk :(A) A2 (B) B (C) A (D) B2

MATHEMATICSTARGET : JEE 2014(DATE : 23-04-2014)

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01CT313083KOTA - 10/38

PAPER – 2

Paragraph for Question 17 to 19

iz'u 17 ls 19 ds fy;s vuqPNsnLet x and y be two numbers chosen at random from the set {1, 2, 3 .... n} with replacement. Let Qn (p)denotes the probability that xp–1 – yp–1 is divisible by p where p is a prime number.

ekuk x rFkk y iquLFkkZiu ds lkFk leqPp; {1, 2, 3 .... n} esa ls ;knPN;k p;fur nks la[;k;sa gSaA ekuk Qn (p), xp–1 – yp–1

dh p ls foHkkftr gksus dh izkf;drk dks n'kkZrk gS] tgk¡ p ,d vHkkT; la[;k gSA17. Q25 (3) equals :

(A) 0.5648 (B) 0.5678 (C) 0.5698 (D) 0.5628

Q25 (3) cjkcj gksxk :(A) 0.5648 (B) 0.5678 (C) 0.5698 (D) 0.5628

18. If [x] represents greatest integer function then Qn (p) equals :

(A) 2

2

n 2 n1 2

p pn

é ù é ù+ -ê ú ê ú

ë û ë û(B)

2

2

n 1 n1 3

p pn

é ù é ù- -ê ú ê ú

ë û ë û(C)

2

2

n 1 n1

p pn

é ù é ù- +ê ú ê ú

ë û ë û(D) None of these

;fn [x] egÙke iw.kk±d Qyu dks n'kkZrk gS] rks Qn (p) cjkcj gksxk :

(A) 2

2

n 2 n1 2

p pn

é ù é ù+ -ê ú ê ú

ë û ë û(B)

2

2

n 1 n1 3

p pn

é ù é ù- -ê ú ê ú

ë û ë û(C)

2

2

n 1 n1

p pn

é ù é ù- +ê ú ê ú

ë û ë û(D) buesa ls dksbZ ugha

19. nnlim Q (p)®¥

equals:

(A) 2

2 21

p p- + (B)

21

p- (C) 2

21

p- (D) None of these

nnlim Q (p)®¥

cjkcj gksxk :

(A) 2

2 21

p p- + (B)

21

p- (C) 2

21

p- (D) buesa ls dksbZ ugha

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

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

LEADER & ENTHUSIAST COURSE(DATE : 23-04-2014)

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01CT313083 KOTA - 11/38

PAPER – 2

SECTION-IV : (Integer Value Correct Type)

[k.M-IV : (iw.kk±d eku lgh izdkj)This section contains 3 questions. The answer to each question is a single digit Integer, ranging from0 to 9 (both inclusive)

bl [k.M esa 3 iz'u gSaA izR;sd iz'u dk mÙkj 0 ls 9 rd (nksuksa 'kkfey) ds chp dk ,dy vadh ; iw.kk ±d

gSA1. Let a, b Î R be distinct numbers satisfying |a – 1| + |b – 1| = |a| + |b| = |a+1| + |b+1|, then the minimum

value of |a – b| =ekuk fHkUu la[;k;sa a, b Î R, |a – 1| + |b – 1| = |a| + |b| = |a+1| + |b+1| dks lUrq"V djrh gS] rks |a – b| dk U;wureeku gksxk -

2. If the vectors ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆi cj bk , ci j ak, bi aj k- + + - + + - are co-planar vectors |a| £ 1 then maximum value of

2ˆˆ ˆ|ai bj ck|+ + is (where a, b, c are scalars)

;fn lfn'k ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆi cj bk , ci j ak, bi aj k- + + - + + - leryh; gksa rFkk |a| £ 1 gks] rks 2ˆˆ ˆ|ai bj ck|+ + dk vf/kdre eku gksxk

(tgk¡ a, b, c vfn'k gS)

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

MATHEMATICSTARGET : JEE 2014(DATE : 23-04-2014)

TM

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01CT313083KOTA - 12/38

PAPER – 2

3. Let O be the origin and OA and OB be two perpendicular chords of equal length of the circle

2 2(x 1) (y 2) 5- + + = . If m1 and m2 be the slopes of these perpendicular chords then if 2 21 2m m+ is k then

find [k] where [.] represents greatest integer function.

ekuk O ewyfcUnq ,oa OA rFkk OB o`Ùk 2 2(x 1) (y 2) 5- + + = dh leku yEckbZ dh nks yEcor~ thok;sa gSa] ;fn m1 rFkk

m2 bu yEcor~ thokvksa dh izo.krk;sa gSa rFkk 2 21 2m m+ dk eku k gks] rks [k] dk eku Kkr dhft, (tgk¡ [.] egÙke iw.kk±d

Qyu dks n'kkZrk gS)

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

LEADER & ENTHUSIAST COURSE(DATE : 23-04-2014)

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01CT313083 KOTA - 13/38

PAPER – 2

PART-2 : PHYSICS

Hkkx-2 : HkkSfrdhSECTION–I : (i) Only One option correct Type

[k.M-I : (i) dsoy ,d lgh fodYi izdkjThis section contains 9 multiple choice questions. Each question has four choices (A), (B), (C) and(D) out of which ONLY ONE is correct.

bl [k.M esa 9 cgqfodYi iz'u gSA izR;sd iz'u esa pkj fodYi (A), (B), (C) vkSj (D) gSa ftuesa ls dsoy ,d

lgh gSA

1. A triangular wedge of mass M which is free to move, lies at rest on smooth horizontal plane. A man ofmass m initially at rest starts running over wedge with constant speed u relative to wedge. If we doanalysis from ground while man is running without jumping with constant speed over wedge.(A) the power due to normal contact force on man is zero.(B) the friction force acting on man is mg sinq(C) power due to friction on man is zero.(D) power due to normal contact force on wedge is positive.

m

u

M

q

xfr ds fy, Lora= ,d M æO;eku dk f=Hkqtkdkj ost fpdus {kSfrt ry ij fojkekoLFkk esa j[kk gqvk gSA ,d m æO;eku

dk O;fDr] tks izkjEHk esa fojkekoLFkk esa gS] ost ds lkis{k fu;r pky u ls ost ij nkSM+uk izkjEHk djrk gSA ;fn ge /kjkry

ls bl xfr dk fo'ys"k.k djsa tcfd O;fDr ost ds Åij fu;r pky ls fcuk dwns nkSM+rs gq;s tk jgk gks rks %&

(A) O;fDr ij vfHkyEc lEiZd cy ds dkj.k 'kfDr 'kwU; gSA

(B) O;fDr ij dk;Zjr ?k"kZ.k cy mg sinq gSA

(C) O;fDr ij ?k"kZ.k ds dkj.k 'kfDr 'kwU; gSA

(D) ost ij vfHkyEc lEiZd cy ds dkj.k 'kfDr /kukRed gSA

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

PHYSICSTARGET : JEE 2014(DATE : 23-04-2014)

TM

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01CT313083KOTA - 14/38

PAPER – 2

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

2. A uniform thin rod AB is equipped at both ends with the hooks shown and is supported by a frictionlesshorizontal table. Initially the rod is hooked at A to a fixed pin C about which it rotates with a constantangular velocity w1. Suddenly end B of the rod hits pin D and gets hooked to pin D, causing end A tobe released. Determine the magnitude of the angular velocity w2 of the rod in its subsequent rotationabout D. (Assume length and mass of the hook is negligible. Pin C & D are lying on a same horizontalline),d le:i iryh NM + AB ds nksuksa fljkas ij gqd yxs gq;s gS rFkk ;g ?k"kZ.kjfgr {kSfrt Vscy ij j[kh gq;h gS] fp= ns[ksaAizkjEHk esa bl NM + ds fljs A dks ,d fLFkj fiu C ls gqd }kjk tksM+ fn;k tkrk gS] ftlds lkis{k ;g fu;r dks.kh; osx w1

ls ?kw.kZu djrh gSA vpkud NM+ dk fljk B fiu D ls Vdjkrk gS rFkk gqd }kjk fiu D ls tqM + tkrk gS] ftlds dkj.k fljkA NwV tkrk gSA fljs D ds lkis{k NM + dh rnksijkUr ?kw.kZu xfr esa NM + ds dks.kh; osx w2 dk ifjek.k gksxk (gqd dk æO;ekurFkk yEckbZ ux.; ekusa ,oa fiu C rFkk D ,d gh {kSfrt js[kk esa gS) :-

AC

B

B'D

A'w2

w1

(A) 3w1 (B) w1/2 (C) w1 (D) none of these

3. A light ray travelling in air is incident vertically on one face of a right angled prism with a refractive

index m = 3 as shown in the figure and the ray follows the path shown in the figure. If q = 60° and the

base of the prism is horizontal and silvered, what is the angel f made by emergent ray with the normalto the right face of the prism?

ok;q esa xfr'khy ,d izdk'k fdj.k m = 3 viorZukad okys ,d ledks.k fizTe ds ,d Qyd ij Å/okZ/kj :i ls

fxjrh gS rFkk ;g fdj.k fp=kuqlkj iFk dk vuqlj.k djrh gSA ;fn q = 60° gks rFkk fizTe dk vk/kkj {kSfrt o jtfrr gksrks fizTe ds nka;s Qyd ij cus vfHkyEc ls fuxZr fdj.k }kjk cuk;k x;k dks.k f gksxk%&

q

f Emergent rayIncident

ray

(A) 0° (B) 30° (C) sin-11

3

æ öç ÷è ø

(D) none of these

LEADER & ENTHUSIAST COURSE(DATE : 23-04-2014)

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01CT313083 KOTA - 15/38

PAPER – 2

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

4. A vessel with a hole in its bottom is fastened on a cart. The mass of the vessel and the cart is M and thearea of the vessel base is A. What force F should the cart be pulled with so that maximum amount ofwater remains in the vessel? The dimensions of the vessel are shown in the figure. Assume there is nofriction between cart and ground and vessel is fixed with cart. (r is density of water)

,d ik= ds isnsa esa fNnz cuk gqvk gS rFkk ;g ik= ,d xkM+h ij j[kk gqvk gSA bl ik= rFkk xkM+h dk æO;eku M rFkk ik=

ds vk/kkj dk {ks=Qy A gSA bl xkM+h dks fdl cy F ls [khapk tk;s rkfd ik= esa ty dh vf/kdre ek=k cuh jgs\ bl

ik= dk vkdkj fp= esa n'kkZ;k x;k gSA ekuk fd xkM+h rFkk /kjkry ds e/; ?k"kZ.k fo|eku ugh gS rFkk ;g ik= xkM+h ij fLFkj

gSA (r ty dk ?kuRo gS)

l

c b

F

µ=0

(A) bcA b

M g2 c

æ ö+ rç ÷è øl

(B) bcA b

2M g2

æ ö+ rç ÷è øl l

(C) bcA 2b

2M g2

æ ö+ rç ÷è øl l

(D) none of these

5. The radius and surface tension of a spherical soap bubble be r and T respectively. Find the chargeuniformly distributed over the outer surface of the bubble, is required to double its radius. (Given thatatmospheric pressure is P

0 and inside temperature of the bubble during expansion remains constant.)

,d xksykdkj lkcqu ds cqycqys dh f=T;k rFkk i`"B ruko Øe'k% r o T gSA bl cqycqys dh f=T;k dks nqxquh djus ds fy;s

cqycqys dh ckgjh lrg ij fdruk vkos'k ,dleku :i ls forjhr gksuk pkfg,\ (ekuk ok;qe.Myh; nkc P0 gS rFkk izlkj

ds nkSjku cqycqys ds vUnj dk rkieku fu;r cuk jgrk gS)(A) 8pr[e

0r (7P

0r + 12T)]1/2 (B) 4pr[e

0r (7P

0r + 12T)]1/2

(C) pr[e0r (7P

0r + 12T)]1/2 (D) 8pr[e

0r (7P

0r + 12T)]1/3

PHYSICSTARGET : JEE 2014(DATE : 23-04-2014)

TM

Path to success KOTA (RAJASTHAN)

01CT313083KOTA - 16/38

PAPER – 2

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

6. A wire carrying a current I is bent into the shape of an exponential spiral, r = eq, from q = 0 toq = 2p as shown in figure. To complete a loop, the ends of the spiral are connected by a straight wirealong the x-axis. Find the magnitude of B at the origin.

,d rkj esa I /kkjk izokfgr gks jgh gS rFkk bls q = 0 ls q = 2p rd ,d pj?kkrkadh lfiZykdkj iFk r = eq esa fp=kuqlkj eksM+k

tkrk gSA ,d ywi iwjk djus ds fy;s bl iFk ds fljksa dks x-v{k ds vuqfn'k ,d lh/ks rkj }kjk tksM+ nsrs gSA ewy fcUnq ij

B dk ifjek.k Kkr dhft,Ay

r xI

I

I

q

(A) 20I

1 e2 2

- pm é ù-ë ûp(B)

20I 2 1 e4

- pm é ù-ë ûp(C)

20I1 e

2- pm é ù-ë ûp

(D) 20I

1 e4

- pm é ù-ë ûp

7. Mark the INCORRECT statement :-(A) The saturation behavior of nuclear forces, implies that the nuclear forces act over a very small

range.(B) At very short distances much smaller than the range the nuclear forces become repulsive that is why

nuclear volume being proportional to square of the total number of nucleons.(C) At very short distances much smaller than the range the nuclear forces become repulsive therefore

nuclear density is constant.(D) The saturation behavior of nuclear forces, implies that a particular nucleon interacts only with its

nearest neighbors.

xyr dFku pqfu;s %&

(A) ukfHkdh; cyks dk lar`Ir O;ogkj ;g n'kkZrk gS fd ukfHkdh; cy ,d cgqr vYi ijkl rd gh dk;Z djrs gSA

(B) ijkl dh rqyuk esa cgqr vYi nwfj;ksa ij ukfHkdh; cy izfrd"khZ gks tkrs gSA ;gh dkj.k gS fd ukfHkdh; vk;ru

U;wfDy;kWuksa dh dqy la[;k ds oxZ ds lekuqikrh gksrk gSA

(C) ijkl dh rqyuk esa cgqr vYi nwfj;ksa ij ukfHkdh; cy izfrd"khZ gks tkrs gS blfy;s ukfHkdh; ?kuRo fu;r gksrk gSA

(D) ukfHkdh; cyksa dk larIr O;ogkj ;g n'kkZrk gS fd ,d U;wfDy;kWu dsoy blds lehiLFk iM+kSfl;ks ds lkFk gh vU;ksU;

fØ;k djrk gSA

LEADER & ENTHUSIAST COURSE(DATE : 23-04-2014)

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01CT313083 KOTA - 17/38

PAPER – 2

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

8. In the below figure a dielectric is released inside the capacitor at t = 0 from the end of the capacitor. Thenwhich of the following current (i) Vs time(t) graph is correct in one complete oscillation. If clockwisecurrent is taken as positive and anticlockwise as negative. (a > c). Neglect electrical and mechanicalresistance.

iznf'kZr fp= esa ,d ijkoS|qr dks t = 0 ij la/kkfj= ds fljs ls la/kkfj= ds vUnj fojkekoLFkk ls NksM+k tkrk gSA ,d iw.kZ

nksyu ds fy;s /kkjk (i) rFkk le; (t) ds e/; lgh vkjs[k pqfu;sA ;gk¡ nf{k.kkorhZ /kkjk dks /kukRed ,oa okekorhZ /kkjk dks

½.kkRed ekuk x;k gS] (a > c) gS ,oa fo|qr rFkk ;kaf=d izfrjks/k dks ux.; ekuasA

V

a ´ b c b

k

a c

(A)

i

t (B)

i

t

(C)

i

t (D)

i

t \

PHYSICSTARGET : JEE 2014(DATE : 23-04-2014)

TM

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01CT313083KOTA - 18/38

PAPER – 2

9. Three capacitor of capacitance C each are connected in series as shown in the figure. Initially switch Sis open. Now capacitors are charged by a battery of emf V by connecting between terminal A and B.After long time battery is disconnected and inductor of inductance L is connected between A and B attime t = 0 so that an oscillatory circuit is formed. Now at an instant t0 switch S is closed, then find theamplitude of charge oscillations of the remaining capacitors.

çR;sd C /kkfjrk okys rhu la/kkfj=ksa dks fp=kuqlkj Js.khØe esa tksM+k x;k gSA izkjEHk esa fLop S [kqyk gqvk gSA vc fljkas A

rFkk B ds e/; V fo|qr okgd cy okyh cSVjh dks tksM+dj la/kkfj=ksa dks vkosf'kr fd;k tkrk gSA ,d yEcs le; i'pkr~

cSVjh dks gVkdj le; t = 0 ij L izsjdRo okyh ,d izsjd dq.Myh dks A o B ds e/; tksM+ nsrs gS rkfd ,d nksyuh

ifjiFk cu tk;sA vc {k.k t0 ij fLop S dks can dj fn;k tkrk gSA 'ks"k la/kkfj=ksa ds vkos'k nksyuksa dk vk;ke gksxk%&

AC C

S

B

CL

(A)

20

3cos t

LC1CV 1

6 3

æ öæ öç ÷ç ÷ç ÷ç ÷è ø-ç ÷ç ÷ç ÷è ø

(B)

20

3cos t

LCCV 11

3 5 3

æ öæ öç ÷ç ÷ç ÷ç ÷è ø-ç ÷ç ÷ç ÷è ø

(C)

20

3cos t

LC1CV 1

4 3

æ öæ öç ÷ç ÷ç ÷ç ÷è ø-ç ÷ç ÷ç ÷è ø

(D) CV3

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

LEADER & ENTHUSIAST COURSE(DATE : 23-04-2014)

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01CT313083 KOTA - 19/38

PAPER – 2

(ii) One or more options correct Type

(ii) ,d ;k vf/kd lgh fodYi izdkjThis section contains 4 multiple choice questions. Each question has four choices (A), (B), (C) and(D) out of which ONE or MORE are correct.

bl [k.M esa 4 cgqfodYi iz'u gSaA izR;sd iz'u esa pkj fodYi (A), (B), (C) vkSj (D) gSa] ftuesa ls ,d ;k

vf/kd lgh gSA

10. Three particles, each of mass m are placed at the points (x1,y1,z1), (x2,y2,z2) and (x3,y3,z3) on theinner surface of a paraboloid of revolution obtained by rotating the parabola, x2 = 4ay about they-axis. Neglect the mass of the paraboloid. (y-axis is along the vertical)(A) The moment of inertia of the system about the axis of the paraboloid is I = 4ma (y1 + y2 +y3).(B) If potential energy at O is taken to be zero, the potential energy of the system is

mg (y1 + y2 + y3).

(C) If the particle at (x1,y1,z1) slides down the smooth surface, its speed at O is 1gy2(D) If the paraboloid spins about OY with an angular speed w, the kinetic energy of the system

will be ma (y1 + y2 +y3)w2.

y

xO

1

23

,d ijoy; x2 = 4ay ; y-v{k ds ifjr% ?kw.kZu dj jgk gSA blds dkj.k cus ijoy;t dh vkarfjd lrg ij mæO;eku okys rhu d.k fcUnqvksa (x1,y1,z1), (x2,y2,z2) rFkk (x3,y3,z3) ij fLFkr gSA bl ijoy;t dk æO;eku ux.;ekusa rFkk y-v{k ÅèokZ/kj ds vuqfn'k gS %&(A) ijoy;t dh v{k ds ifjr% fudk; dk tM+Ro vk?kw.kZ I = 4ma (y1 + y2 +y3) gSA(B) ;fn O ij fLFkfrt ÅtkZ 'kwU; ekuh tk;s rks fudk; dh fLFkfrt ÅtkZ mg (y1 + y2 + y3) gksxhA

(C) ;fn (x1,y1,z1) ij fLFkr d.k fpduh lrg ij uhps dh vksj xfr djrk gS rks O ij bldh pky 1gy2 gksxhA

(D) ;fn ijoy;t OY ds ifjr% dks.kh; pky w ls pØ.k djs rks fudk; dh xfrt ÅtkZ ma (y1 + y2 +y3)w2

gksxhA

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

PHYSICSTARGET : JEE 2014(DATE : 23-04-2014)

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01CT313083KOTA - 20/38

PAPER – 2

11. If a graph of Binding energy (B/A) per nucleon versus mass number (A) looks like as shown in figure.Then using the curve choose CORRECT option(s).(A) Fusion of two nuclei with mass numbers 30 and 45 will have Q value 99 MeV(B) Fission of nuclei with mass number 80 into two nuclei of equal mass number will have Q value

256 MeV(C) Fission of nuclei with mass number 150 into two nuclei of equal mass number will release energy(D) Fission of nuclei with mass number 80 into two nuclei of equal mass number will release energy

2 4 6 8

50 100 150 200 A

(MeV) B/A

izfr U;wfDy;kWu ca/ku ÅtkZ (B/A) rFkk æO;eku la[;k (A) ds e/; vkjs[k fp= esa iznf'kZr gSA lgh dFku@dFkuksa dks

pqfu;s %&

(A) æO;eku la[;k 30 rFkk 45 okys nks ukfHkdkas ds lay;u ds fy;s Q eku 99 MeV gksxkA

(B) æO;eku la[;k 80 okys ukfHkd ds leku æO;eku la[;k okys nks ukfHkdkas esa fo[k.Mu dh izfØ;k ds fy;s Q eku

256 MeV gksxkA

(C) æO;eku la[;k 150 okys ukfHkd ds leku æO;eku la[;k okys nks ukfHkdkas esa fo[k.Mu ds QyLo:i ÅtkZ mRlftZr

gksrh gSA

(D) æO;eku la[;k 80 okys ukfHkd ds leku æO;eku la[;k okys nks ukfHkdkas esa fo[k.Mu ds QyLo:i ÅtkZ mRlftZr

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

LEADER & ENTHUSIAST COURSE(DATE : 23-04-2014)

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01CT313083 KOTA - 21/38

PAPER – 2

12. n moles of a monoatomic gas undergoes a cyclic process ABCDA as shown. ProcessesAB ® Isobaric, BC ® adiabatic, CD ® isochoric, DA ® Isothermal. The maximum temperature andminimum temperature in the cycle are 4T0 and T0 respectively. Then(A) TB> TC> TD

(B) heat is released by the gas in the process CD.(C) heat is supplied to the gas in the process AB(D) total heat supplied to the gas is zero

V

P

A B

C

D

n eksy ,d ijekf.od xSl fp=kuqlkj pØh; çØe ABCDA ls xqtjrh gSA izØe AB ® lenkch;, BC ® :¼ks"e,

CD ® levk;rfud, DA ® lerkih; gSA bl pØ esa vf/kdre rFkk U;wure rkieku Øe'k% 4T0 rFkk T0 gSA rc %&(A) TB> TC> TD

(B) izØe CD esa xSl }kjk Å"ek mRlftZr dh tkrh gSA

(C) izØe AB esa xSl dks Å"ek nh tkrh gSA

(D) xSl dks nh x;h dqy Å"ek 'kwU; gSA

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

PHYSICSTARGET : JEE 2014(DATE : 23-04-2014)

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01CT313083KOTA - 22/38

PAPER – 2

13. The current velocity of a river is directly proportional to the perpendicular distance from its bank and isits maximum ‘v

0’ in the middle. Near the banks the velocity is zero. A boat is moving in the river such

that its velocity u relative to the river flow is constant and perpendicular.(A) The horizontal drift through which the boat crossing the river will be carried away by the current

0v c2u

(B) The horizontal drift through which the boat crossing the river will be carried away by the current

0v c4u

(C) Time taken by the boat to cross the river will be cu

(D) Time taken by the boat to cross the river will be c

2u

y

x O

c v0

,d unh dk /kkjk izokg dk osx blds fdukjs ls yEcor~ nwjh ds lh/ks lekuqikrh gS rFkk unh ds e/; esa ;g vfèkdre ‘v0’

gks tkrk gSA fdukjkas ds lehi osx 'kwU; gSA ,d uko unh esa bl izdkj xfr'khy gS fd unh izokg ds lkis{k bldk osx uyEcor~ rFkk fu;r jgrk gS%&

(A) /kkjk ds dkj.k uko }kjk unh esa r; fd;k x;k {kSfrt viokg 0v c2u

gSA

(B) /kkjk ds dkj.k uko }kjk unh esa r; fd;k x;k {kSfrt viokg 0v c4u

gSA

(C) uko dks unh ikj djus esa cu

le; yxsxkA

(D) uko dks unh ikj djus esa c

2u le; yxsxkA

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

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(iii) Paragraph Type (iii) vuqPNsn izdkj

This section contains 2 paragraphs each describing theory, experiment, data etc. Six questions relateto two paragraphs with three questions on each paragraph. Each question of a paragraph has onlyone correct answer among the four choices (A), (B), (C) and (D).bl [k.M esa fl¼kUrksa] iz;ksxksa vkSj vk¡dM+ksa vkfn dks n'kkZus okys 2 vuqPNsn gSA nksuksa vuqPNsn ls lacaf/kr N% iz'ugSa] ftuesa ls gj vuqPNsn ij rhu iz'u gSaA vuqPNsn esa gj iz'u ds pkj fodYi (A), (B), (C) vkSj (D) gSa ftuesals dsoy ,d lgh gSA

Paragraph for Question 14 to 16iz'u 14 ls 16 ds fy;s vuqPNsn

A free uniform disc of mass m = 1 kg and radius R = 1 m is rotating on smooth horizontal floor withangular velocity wo = 10 rad/s. A particle of mass 2 kg is released from height 5 m above disc. Particlehits disc at distance R/2 from centre of disc. Collision of particle with disc is inelastic and particle sticksto disc after collision due to friction. Disc does not bounce from ground during collision of particle withdisc.

æO;eku m = 1 kg rFkk f=T;k R = 1 m okyh ,d Lora= le:i pdrh fpduh {kSfrt lrg ij dks.kh; osx

wo = 10 rad/s ls ?kw.kZu dj jgh gSA bl pdrh ls 5 m Å¡pkbZ ls 2 kg æO;eku ds d.k dks NksM+k tkrk gSA d.k pdrh ds

dsUnz ls R/2 nwjh ij pdrh ls Vdjkrk gSA d.k dh pdrh ls VDdj vizR;kLFk gksrh gS rFkk ?k"kZ.k ds dkj.k VDdj ds ckn

d.k pdrh ls fpid tkrk gSA d.k dh pdrh ls VDdj ds nkSjku pdrh /kjkry ij ls mNyrh ugha gSA

R/2

h

m(side view)(Top view)

w0

14. The angular velocity of disc immediately after collision is

VDdj ds rqjUr i'pkr~ pdrh dk dks.kh; osx gksxk%&(A) 5 rads (B) 10 rad/s (C) 7.5 rad/s (D) 12.5 rad/s

15. The friction force acting on particle after impact is

VDdj ds i'pkr~ d.k ij dk;Zjr ?k"kZ.k cy gksxk%&(A) 25/4 N (B) 75/4 N (C) 50 N (D) 50/3 N

16. The impulse on particle due to disc during impact is approximately

VDdj ds nkSjku pdrh ds dkj.k d.k ij yxus okyk vkosx yxHkx gksxk%&(A) 20.2 Ns (B) 10.1 Ns (C) 10 Ns (D) 15 Ns

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

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Paragraph for Question 17 to 19

iz'u 17 ls 19 ds fy;s vuqPNsnIn the figure shown a uniform conducting rod of mass m and length l is suspended in vertical plane bytwo conducting springs of spring constant K. Upper end of springs are connected to each other bycapacitor of capacitance C. A uniform horizontal magnetic field (B0) perpendicular to plane of springsexists in space. Initially rod is in equilibrium but if rod is pulled down and released, it performs SHM.(Assume resistance of springs and rod are negligible also neglect the self inductance of both the spring)

iznf'kZr fp= esa æO;eku m rFkk yEckbZ l okyh ,d le:i pkyd NM+ dks Å/okZ/kj ry esa fLizax fu;rkad K okyh nks

pkyd fLizaxksa dh lgk;rk ls yVdk;k x;k gSA fLizaxkas ds Åijh fljs dks ,d&nwljs ls C /kkfjrk okys ,d la/kkfj= }kjk

vkil esa tksM+ fn;k tkrk gSA ;gk¡ fLizaxkas ds ry ds yEcor~ ,d le:i {kSfrt pqEcdh; {ks= (B0) fo|eku gSA izkjEHk esa

NM+ lkE;koLFkk esa gS ijUrq ;fn NM+ dks uhps dh vksj [khapdj NksM+ fn;k tk;s rks ;g ljy vkorZ xfr djus yxrh gSA

(fLizazxkas rFkk NM+ ds izfrjks/k dks ux.; ekuas rFkk nksuksa fLizaxkas ds LoizsjdRo dks Hkh ux.; ekus)

KK

ml

C B

non conducting roof

17. Find time period of oscillation of rod.

NM+ ds nksyu dk vkorZdky Kkr dhft,A

(A) m

2K

p (B) 2 2B C

2K

pl

(C) 2 2m B C

K+

pl

(D) 2 2B C m

22K

+p

l

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18. Choose CORRECT option from following :

(A) Electrical energy stored in capacitor is maximum when rod is at its lower extreme position

(B) Electrical energy stored in capacitor is maximum when rod is at its mean position

(C) Current in rod is maximum at mean position of rod

(D) If magnetic field is switched off then mean position of rod will change.

lgh fodYi pqfu;s%&

(A) tc NM+ bldh fuEure fLFkfr ij gksrh gS rks la/kkfj= esa laxzfgr fo|qr ÅtkZ vf/kdre gksrh gSA

(B) tc NM+ bldh ek/; fLFkfr ij gksrh gS rks la/kkfj= esa laxzfgr fo|qr ÅtkZ vf/kdre gksrh gSA

(C) NM+ dh ek/; fLFkfr ij NM+ esa /kkjk vf/kdre gksrh gSA

(D) ;fn pqEcdh; {ks= can dj fn;k tk;s rks NM+ dh ek/; fLFkfr ifjofrZr gks tk;sxhA

19. Find the ratio of current (i) to acceleration of rod at any instant (other then equilibrium) :

lkE;koLFkk ds vfrfjDr vU; fdlh {k.k ij NM+ esa /kkjk (i) rFkk mlds Roj.k dk vuqikr gksxk%&

(A) BlC (B) BCl

(C) B2l

2C (D) None of these

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

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

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SECTION-IV : (Integer Value Correct Type)

[k.M-IV : (iw.kk±d eku lgh izdkj)This section contains 3 questions. The answer to each question is a single digit Integer, ranging from0 to 9 (both inclusive)

bl [k.M esa 3 iz'u gSaA izR;sd iz'u dk mÙkj 0 ls 9 rd (nksuksa 'kkfey) ds chp dk ,dy vadh ; iw.kk ±d

gSA

1. The electrostatic flux through a rectangular area of 27

m47

lying in a plane 2x + 3y + 6z = 10 due to

an electric field ˆ ˆ ˆ(8i 6 j 10k)+ + V/m will be in (V-m)

fdlh lery 2x + 3y + 6z = 10 esa j[ks 27m

47{ks=Qy okys ,d vk;r ls fo|qr {ks= ˆ ˆ ˆ(8i 6 j 10k)+ + V/m ds dkj.k

fuxZr fLFkjoS|qr ¶yDl dk eku (V-m) esa gksxkA2. The limbs of a glass U-tube are lowered into vessels A and B as shown in the figure. Some air is

pumped out through a valve at C, placed at the top of tube and then the valve is closed. The liquid in theleft hand limb then rises to a height h1 and in the right hand one to height h2. The density of the liquid in

limb B is 1k

(in gm/cc) if water is present in limb A. Given h1 = 10 cm and h2 = 20 cm. Calculate the

value of K.

dk¡p dh ,d U-uyh dh Hkqtkvksa dks fp=kuqlkj ik=ksa A rFkk B esa Mqcks;k tkrk gSA bl uyh ds 'kh"kZ Hkkx C ij yxs ,dokYo dh lgk;rk ls dqN ok;q ckgj fudky nh tkrh gS rFkk blds ckn okYo dks can dj nsrs gSaA vc ck¡;h Hkqtk esa Hkjk gqvkæo h1 Å¡pkbZ rd rFkk nk¡;h Hkqtk esa Hkjk æo h2 Å¡pkbZ rd p<+rk gSA ;fn Hkqtk A esa ty Hkjk gqvk gks rks Hkqtk B esa æo dk

?kuRo 1k

(gm/cc esa) izkIr gksrk gSA K dk eku Kkr dhft, (fn;k gS h1 = 10 cm rFkk h2 = 20 cm)

h2h1

A B

C

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3. Image distance v vs object distance u curve for two biconvex lens with same radii of curvatures is

shown in the figure. If refractive index of lens 1 is 52

, find refractive index of lens 2.

leku oØrk f=T;kvksa okys nks f}&mÙky ysUlksa ds fy, izfrfcEc nwjh v rFkk fcEc nwjh u ds e/; vkjs[k fp= esa n'kkZ;s

x;s gSaA ;fn ysUl 1 dk viorZukad 52

] gks rks ysUl 2 dk viorZukad Kkr dhft,A

30 cm

20 cm

(1)

(2)

|v|

|u| 45°

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

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PART-3 : CHEMISTRY Hkkx-3 : jlk;u

SECTION–I : (i) Only One option correct Type [k.M-I : (i) dsoy ,d lgh fodYi izdkj

This section contains 9 multiple choice questions. Each question has four choices (A), (B), (C) and(D) out of which ONLY ONE is correct.bl [k.M esa 9 cgqfodYi iz'u gSA izR;sd iz'u esa pkj fodYi (A), (B), (C) vkSj (D) gSa ftuesa ls dsoy ,d

lgh gSA1. In the electrolysis of CuCl

2 soluton using Cu electrodes, the weight of Cu increased by 2gram at cathode.

At the anode -(A) 0.2 mole of Cu2+ will go into solution (B) Cl

2 is produced

(C) No loss in weight (D) 2 gram of copper goes into solution as Cu2+

CuCl2 foy;u ds oS|qr vi?kVu esa Cu bysDVªksMks dk mi;ksx fd;k x;kA dSFkksM ij Cu dk Hkkj 2 xzke c< tkrk gS

,uksM ij -

(A) Cu2+ ds 0.2 eksy foy;u esa pys tk,sxs (B) Cl2 fufeZr gksxh

(C) Hkkj esa gkfu ugha gksxh (D) Cu2+ ds :i esa 2 xzke dkWij foy;u esa ?kqy tk;sxk2. Which of the following is the correct order of C = C bond lengths among these compounds ?

I. CH3O – CH = CH – NO

2II. CH

2 = CH – NO

2

III. CH2 = CH – Cl IV. CH

2 = CH

2

(A) I > II > III > IV (B) IV > III > II > I (C) I > III > II > IV (D) II > III > I > IV

fuEu ;kSfxdksa esa ls C = C cU/k yEckbZ;ksa dk lgh Øe dkSulk gS ?I. CH

3O – CH = CH – NO

2II. CH

2 = CH – NO

2

III. CH2 = CH – Cl IV. CH

2 = CH

2

(A) I > II > III > IV (B) IV > III > II > I (C) I > III > II > IV (D) II > III > I > IV3. Which pairs of the salts should have identical solubilities in methanol ?

yo.kksa ds dkSuls ;qXe dh esFksukWy esa ?kqyu'khyrk leku gksuh pkfg, ?

(A) I & IV (B) I & III (C) II & III (D) II & IV

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

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

4. Salt -A ¾¾¾ ®¾ testLayer If Reddish brown layers comes first then(A) Br– absent (B) I– absent (C) Cl– Present (D) I– present

yo.k -A ¾¾¾¾®ijr ijh{k.k ;fn yky Hkwjh ijr igys curh gS rks

(A) Br– vuqifLFkr gS (B) I– vuqifLFkr gS (C) Cl– mifLFkr gS (D) I– mifLFkr gS5. Which of the followings is correct statement ?

(A) Fe(CO)5 has trigonal bipyramidal geometry.

(B) [NiCl4]2– is a diamagnetic in nature.

(C) [Ni(CN)4]2– is paramagnetic with 2.8 BM magnetic moment.

(D) Sidgwick EAN rule is followed by K3[Fe(CN)

6] complex.

fuEu esa ls dkSulk dFku lgh gS ?

(A) Fe(CO)5 dh f=dks.kh; f}fijkfeMh; T;kfefr gksrh gS

(B) [NiCl4]2– dh izd`fr izfrpqEcdh; gksrh gS

(C) [Ni(CN)4]2– , 2.8 BM pqEcdh; vk?kw.kZ ds lkFk vuqpqEcdh; gS

(D) K3[Fe(CN)

6] ladqy }kjk fltfod EAN fu;e dk vuqlj.k gksrk gS

6. Which compound will not give the Tollen's test.

dkSulk ;kSfxd VkWySal ijh{k.k ugha nsxk :

(A)

CH OH2CH OH2 CH OH2

HH H

H

HH H

H

HH H

HH

H

OHOH

OH

OH OH OH

OHOH

O

O

O

(B)

CH OH2CH OH2 CH OH2

HH

H

HH H

HH

H H

HH

H

OHOH

OH

OH OH OH

OHOH

O

O O

(C)

CH OH2CH OH2 CH OH2

CH OH2HH

H

HH

H

H

HHH HOHOH OHOH

OHOHOH

O

O

O

(D)

CH OH2CH OH2 CH OH2

HH H

H

HH H

H

HH

HHH

H

OHOH

OH

OH OH

OH

OHOH

O

O

O

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

7. The first ionisation energy of Na, NO, Xe and O2 follows the order Na < NO < Xe = O2 .O2 reacts withpowerful oxidising agent (PtF6) to yield O2

+[PtF6]–. If PtF6 is allowed to react with other mentionedspecies then the product is/are(A) Only Na+[PtF6]– (B) Only NO+[PtF6]–

(C) Only Xe[PtF6]– (D) Na+[PtF6]–, NO+[PtF6]–, Xe+ [PtF6]–

Na, NO, Xe rFkk O2 dh izFke vk;uu ÅtkZ Na < NO < Xe = O2 Øe dk vuqlj.k djrh gSA O2 izcy

vkWDlhdkjd (PtF6) ds lkFk vfHkfØ;k djds O2+[PtF6]– nsrk gSA ;fn PtF6 vU; nh xbZ Lih'kht ds lkFk vfHkfØ;k

djrk gS rks mRikn gSa @ gS

(A) dsoy Na+[PtF6]– (B) dsoy NO+[PtF6]–

(C) dsoy Xe[PtF6]– (D) Na+[PtF6]–, NO+[PtF6]–, Xe+ [PtF6]–

8. The product formed in the following reaction is

fuEu vfHkfØ;k esa cuus okys mRikn gS

CH – CH – CH2 2 2 CH = CH2 ¾¾¾ ®¾NBS

(A) CH – CH – CH2 2 CH = CH2

Br

(B) CH – CH – CH2 2 CH = CH2

|Br

(C) CH – CH – CH2 2 CH = CH2

|Br

(D) CH – CH – CH2 2 2 CH – CH Br2|Br

9. Two liquids A and B are mixed at temperature T in a certain ratio to form an ideal solution. It is foundthat the partial vapour pressure of A i.e. P

A is equal to P

B, the vapour pressure of B for the liquid mixture.

What is the total vapour pressure of the liquid mixture in terms of oAP and o

BP (the vapour pressure of

the pure liquids A and B at the temprature) ?nks æoksa A rFkk B dks ,d fuf'pr vuqikr esa T rki ij fefJr dj ,d vkn'kZ foy;u cuk;k x;kA ;g ik;k x;kfd æo feJ.k ds fy, A dk vkaf'kd ok"inkc] tks P

A gS] B ds vkaf'kd nkc P

B ds cjkcj gSA o

AP rFkk oBP ds inksa

es a æo feJ.k dk dqy ok"inkc D;k gksxk (rki ij 'kq¼ æoksa A rFkk B dk ok"inkc gS) ?

(A) o oA B

o oA B

P PP P+ (B)

oA

o oA B

PP P+ (C)

o oA B

o oA B

2P PP P+ (D)

oB

o oA B

2PP P+

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

(ii) One or more options correct Type (ii) ,d ;k vf/kd lgh fodYi izdkj

This section contains 4 multiple choice questions. Each question has four choices (A), (B), (C) and(D) out of which ONE or MORE are correct.

bl [k.M esa 4 cgqfodYi iz'u gSaA izR;sd iz'u esa pkj fodYi (A), (B), (C) vkSj (D) gSa] ftuesa ls ,d ;k

vf/kd lgh gSA

10. CH CH CH –Cl22 2AlCl3

(catalytic amount) P + HCl(mf C H )10 12

P can not be (P ugha gks ldrk) :

(A) (B) (C) (D)

11. Which of the following will not change if the choice of the zero potential energy is changed in Bohrmodel -

(A) Total energy (B) Potential energy (C) Kinetic energy (D) None of these

;fn cksgj ekWMy esa 'kwU; fLFkfrt ÅtkZ esa ifjorZu gksrk gS rks fuEu esa ls dkSulk ifjofrZr ugha gksxk -

(A) dqy ÅtkZ (B) fLFkfrt ÅtkZ (C) xfrt ÅtkZ (D) buesa ls dksbZ ugha

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12. The product in the following reaction is CH3CHBrF + NaOCH

3 ¾®¾

fuEu vfHkfØ;k esa mRikn gS CH3CHBrF + NaOCH

3 ¾®¾

(A) CH3CHBrOCH

3(B) CH

3CHFOCH

3

(C) CH3CH(OCH

3)

2(D) Optically active (izdkf'kd lfØ;)

13. The product obtained in the following reaction

Br

HBr

H

KIacetone¾¾¾¾® Product.

(A) Decolourise bromine water (B) On ozonolysis gives single product

(C) Shows geometrical isomerism (D) Shows optical isomerism

fuEu vfHkfØ;k esa izkIr mRikn

Br

HBr

H

KIacetone¾¾¾¾® mRikn

(A) czksehu ty dks jaxghu dj nsrk gS (B) vkstksuh vi?kVu djkus ij ,d mRikn nsrk gS

(C) T;kfefr; leko;ork iznf'kZr djrk gS (D) izdkf'kd leko;ork iznf'kZr djrk gS

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(iii) Paragraph Type (iii) vuqPNsn izdkj

This section contains 2 paragraphs each describing theory, experiment, data etc. Six questions relateto two paragraphs with three questions on each paragraph. Each question of a paragraph has onlyone correct answer among the four choices (A), (B), (C) and (D).bl [k.M esa fl¼kUrksa] iz;ksxksa vkSj vk¡dM+ksa vkfn dks n'kkZus okys 2 vuqPNsn gSA nksuksa vuqPNsn ls lacaf/kr N% iz'ugSa] ftuesa ls gj vuqPNsn ij rhu iz'u gSaA vuqPNsn esa gj iz'u ds pkj fodYi (A), (B), (C) vkSj (D) gSa ftuesals dsoy ,d lgh gSA

Paragraph for Question 14 to 16iz'u 14 ls 16 ds fy;s vuqPNsn

A crystalline solid consist of a large number of small units, called crystals, each of which possesses adefinite geometric shape bounded by plane faces. The crystals of a given substance produced under adefinite set of conditions are always of the same shape.From external appearance we come across a varietyof crystal forms in many shapes. But on the basis of length of axes and the axial angles, it has been possibleto classify the various crystal forms into only seven fundamental systems. These are:Cubic, Orthorhombic, Tetragonal, Monoclinic,Triclinic,Hexagonal, Rhombohedral.The crystals belonging to any one system may differ in shape, in size, etc., but their axial ratios as alsothe axial angles will always be the same. The intercepts on the three axes are denoted by a, b and c andthe axial angles by a, b and g.,d fØLVyh; Bksl] NksVh&NksVh bdkbZ;ksa] ftUgs fØLVy dgrs gS] dh cgqr cM+h la[;k ls cus gksrs gS] ; s lHkh ,d fuf'prT;kfefr; vkd`fr esa lery Qyd }kjk cU/kh gksrh gSA fn;s x;s inkFkZ ds fØLVy] fLFkfr;ksa ds ,d fuf'pr leqPp; esafufeZr gskrs gS] tks lnSo leku vkdfr ds gksrs gSA cká ifjorZuksa ls ge dbZ vkdfr;ksa esa fofHkUu fØLVyh; :iksa dks cukldrs gSA ijUrq v{kksa dh yEckbZ rFkk v{kh; dks.kksa ds vk/kkj ij dsoy lkr ewyHkwr ra=ksa esa fofHkUu fØLVyh; :iksa dkoxhZdj.k lEHko gSA;sa gS, ?kuh;, fo"keyEck{k, prq"dks.kh;, ,durk{k, f=urk{k, "kV~dks.kh;] f=leurk{k (Rhombohedral)fdlh ,d ra= ls lEcfU/kr fØLVy dks vkdfr] vkdkj vkfn esa foHksfnr fd;k tk ldrk gSA ijUrq buds v{kh; vuqikrds lkFk v{kh; dks.k Hkh lnSo leku gksxsaA rhuksa v{kksa ij izfrPNsnks dks a, b rFkk c }kjk rFkk v{kh; dks.kks dks a, b rFkk g}kjk vafdr fd;k tkrk gSA

14. A tetragonal system is -,d prq"dks.kh; ra= esa gksrk gS –(A) a ¹ b ¹ c (B) a = b = g = 90º (C) a = b = c (D) a = b = 90º ¹ g

15. In a multi layered close-packed structure(A) there are twice as many tetrahedral holes as there are close-packed atoms(B) there are as many tetrahedral holes as there are closed packed atoms(C) there are twice as many octahedral holes as there are close-packed atoms(D) there are as many tetrahedral holes as there are octahedral holes,d cgqijrh; ca/k ladqyu lajpuk esa(A) ftrus cUn ladqfyr ijek.kq gksrs gS mlds nqxus prq"dks.kh; gkWy gksrs gS(B) ftrus cUn ladqyhr ijek.kq gksrs gS mrus gh prq"dks.kh; gkWy gksrs gS(C) ftrus cUn ladqfyr ijek.kq gksrs gS mlds nqxus v"VQydh; gkWy gksrs gS(D) ftrus v"VQydh; gkWy gksrs gS mrus gh prq"dks.kh; gkWy gksrs gS

16. For an Ionic solid of the general formula AB and coordination number 6, the value of the radius ratiowill be(A) less than 0.225 (B) in between 0.225 and 0.414(C) between 0.414 and 0.732 (D) greater than 0.732lkekU; lw= AB ds ,d vk;fud Bksl ftldk leUo; la[;k 6 gS] ds fy,] f=fT;; vuqikr ds eku gksxsa(A) 0.225 ls de (B) 0.225 rFkk 0.414 ds e/;(C) 0.414 rFkk 0.732 ds e/; (D) 0.732 ls vf/kd

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Paragraph for Question 17 to 19iz'u 17 ls 19 ds fy;s vuqPNsn

17. Which of the following statement is correct about S?(A) It will not give yellow/orange or red colour with 2, 4-DNP.(B) It will give red colour with Cerric Ammonia Nitrate(C) It will not give Fehling test.(D) It will give Bielstien testS ds lUnHkZ esa fuEu esa ls dkSulk dFku lgh gS ?(A) ;g 2, 4-DNP ds lkFk ihyk@ukjaxh ;k yky jax ugha nsrk gS(B) ;g lsfjd veksfu;e ukbVªsV ds lkFk yky jax nsrk gS(C) ;g Qsgfyax ifj{k.k ugha nsxk(D) ;g fCyLVhu ifj{k.k nsxk

18. Which of the following statement is correct about V?(A) It will not give mustard oil reaction.(B) It will not give carbylamine test or isocyanide test(C) It is less basic than aniline(D) It will give base soluble product with Hinsburg reagentfuEu esa ls dkSulk dFku V ds lUnHkZ esa lgh gS ?(A) ;g eLVMZ vkW;y vfHkfØ;k ugh nsxk(B) ;g dkchZy ,ehu ifj{k.k ;k vkblkslk;ukbM ifj{k.k ugha nsxk(C) ;g ,fufyu dh rqyuk esa U;wu {kkjh; gksrk gS(D) ;g ghal cxZ vfHkdeZd ds lkFk {kkj esa ?kqyu'khy mRikn nsxk

19. Which of the following reaction will be useful for conversion of R into S.(A) Clemmenson's reduction (B) Oppenauer oxidation(C) Meerwin Pondorff Verely reduction (D) Tischenko reactionR ls S esa :ikUrj.k ds fy, fuEu esa ls dkSulh vfHkfØ;k mi;ksxh gksxh(A) DyhesUlu vip;u (B) vksihukWj vkWDlhdj.k(C) ehjfou iwUMªksQ osjyh vip;u (D) fV'kadks vfHkfØ;k

SECTION –II / [k.M – II & SECTION –III / [k.M – IIIMatrix-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

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

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SECTION-IV : (Integer Value Correct Type) [k.M-IV : (iw.kk±d eku lgh izdkj)

This section contains 3 questions. The answer to each question is a single digit Integer, ranging from0 to 9 (both inclusive)

bl [k.M esa 3 iz'u gSaA izR;sd iz'u dk mÙkj 0 ls 9 rd (nksuksa 'kkfey) ds chp dk ,dy vadh; iw.kk ±d

gSA

1. On passing electricity through nitrobenzene solution, it is converted into azobenzene. Calculate themass of azobenzene (in gm) if same quantity of electricity produces oxygen just sufficient to burn 96 gmof fullerene (C60).

Fill your answer as sum of digits (excluding decimal places) till you get the single digit answer.

ukbVªkscsUthu foy;u esa ls oS|qr /kkjk izokfgr djus ij] ;g ,stkscsUthu esa :ikUrfjr gks tkrk gSA ;fn oS|qr /kkjk dh leku

ek=k ls Qyqjhu (C60) ds 96 xzke dks i;kZIr tykus ij] vkWDlhtu dk fuekZ.k gksrk gS] rks ,stkscsUthu ds æO;eku dh x.kuk(xzke esa) dhft,A

vius mÙkj ds vadksa dks (n'keyo LFkku dks NksM+dj) rc rd ;ksx dhft, tc rd vkidks bdkbZ vad izkIr u gks tk,A

2. ZnSO4.xH2O on heating loses y % of water and isomorphous with green vitriol. Find the number ofwater of crystallisation in ZnSO4.xH2O.

ZnSO4.xH2O dks xeZ djus ij ty ds y % dh gkWfu gk srh gS rFkk gjk df'k'k (Green vitriol) ds lkFk

le vkd`frd (isomorphous) cukrk gSA ZnSO4.xH2O esa fØLVyhdj.k esa ty dh la[;k crkbZ;saA

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

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

3. Two liquids 'A' (Molecular mass = 20) and 'B' (Molecular mass = 40) are partially

MN

miscible. When 1 mol of A and 3 mol of B are shaken together and allowed to settle,

two layer 'M' and 'N' are formed as shown in diagram. Layer 'M' contains 0.2 mole

fraction of 'A' and layer 'N' contains 0.6 mole fraction of A. Calculate the ratio of

masses of layer M to layer N.

nks æo 'A' (vkf.od Hkkj = 20) rFkk 'B' (vkf.od Hkkj = 40) vkaf'kd feJ.kh; gSA tc 1 eksy

MN

A rFkk 3 eksy B dks ,d lkFk feykdj fLFkj gksus rd j[kk tkrk gS rc fp=kuqlkj nks ijrs 'M' rFkk

'N' fufeZr gksrh gS ijr 'M' esa 'A' dk eksy izHkkt 0.2 eksy rFkk ijr 'N' esa A dk eksy izHkkt 0.6

eksy gS ijr M ls ijr N ds æO;ekuksa ds vuqikr dh x.kuk dhft,A

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

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Name of the Candidate / ijh{kkFkhZ dk uke

I have read all the instructions and shallabide by them.eSus a lHkh vuqns'kks a dks i<+ fy;k gS vkSj eSa mudkvo'; ikyu d:¡xkA

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

Form Number / QkWeZ la[;k

I have verified all the information filledin by the Candidate.ijh{kkFkh Z }kjk Hkjh xbZ tkudkjh dks eSusa tk¡pfy;k gSA

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

PAPER – 2

11. [kaM–I(i) Hkkx esa 9 cgqfodYi iz'u gSaA gj iz'u esa pkj fodYi

(A), (B), (C) vkSj (D) gSa ftuesa ls ,d lgh gSaA(ii) Hkkx esa 4 cgqfodYi iz'u gSaA gj iz'u esa pkj fodYi

(A), (B), (C) vk Sj (D) g S a ftues a l s ,d ; kvf/kd lgh gSaA

(iii) Hkkx esa fl¼kUrksa] iz;ksxksa vkSj vk¡dM+ksa vkfn dks n'kkZus okys2 vuqPNsn gSA nksuksa vuqPNsnksa ls lacfU/kr N% iz'u gSaAftuesa ls gj vuqPNsn ij rhu iz'u gSaA fdlh Hkh vuqPNsnesa gj iz'u ds pkj fodYi (A), (B), (C) vkSj (D) gSa ftuesals dsoy ,d lgh gSaA

12. [kaM–II & III esa ,d Hkh iz'u ugha gSA13. [kaM-IV es a 3 iz'u gSaA izR;sd iz'u dk mÙkj 0 ls 9 rd

(nksuksa 'kkfey) ds chp dk ,dy vadh; iw.kk±d gSA

D. vadu ;kstuk :14. [kaM-I (i & ii) ds gj iz'u esa dsoy lgh mÙkjksa (mÙkj) okys

lHkh cqycqyksa (cqycqys) dks dkyk djus ij 3 vad vkSj dksbZHkh cqycqyk dkyk ugha djus ij 'kwU; (0) vad iznku fd;ktk;sxkA vU; lHkh fLFkfr;ksa esa ½.kkRed ,d (–1) vad iznkufd;k tk;sxkA

15. [kaM-I (iii) ds gj iz'u esa dsoy lgh mÙkjksa (mÙkj) okys lHkhcqycqyksa (cqycqys) dks dkyk djus ij 4 vad vkSj dksbZ Hkhcqycqyk dkyk ugha djus ij 'kwU; (0) vad iznku fd;k tk;sxkAvU; lHkh fLFkfr;ksa esa ½.kkRed ,d (–1) vad iznku fd;ktk;sxkA

16. [k aM-IV es a gj iz'u es a dsoy lgh mÙkj okys cqycqys(BUBBLE) dks dkyk djus ij 6 vad vkSj dksbZ Hkh cqycqykdkyk ugha djus ij 'kwU; (0) vad iznku fd;k tk;sxk bl [ akMds iz'uksa esa xyr mÙkj nsus ij dksbZ ½.kkRed vad ugha fn;stk;saxsaA

17. g = 10 m/s2 iz;qDr djsa] tc rd fd vU; dksbZ eku ughafn;k x;k gksA

11. SECTION – I(i) Contains 9 multiple choice questions. Each

question has four choices (A), (B), (C) and (D)out of which ONLY ONE is correct.

(ii) Contains 4 multiple choice questions. Eachquestion has four choices (A), (B), (C) and (D)out of which ONE or MORE are correct.

(iii) Contains 2 paragraphs each describing theory,experiment, date etc. Six questions relate totwo paragraphs with three questions on eachparagraph. Each question of a paragraph hasONLY ONE correct answer among the fourchoices (A), (B), (C) and (D)

12. There is no questions in SECTION-II & III13. Section-IV contains 3 questions The answer to each

question is a single digit integer, ranging from0 to 9 (both inclusive)

D. Marking scheme :14. For each question in Section-I (i & ii), you will be

awarded 3 marks if you darken all the bubble(s)corresponding to only the correct answer(s) andzero mark if no bubbles are darkened. In all othercases minus one (–1) mark will be awarded

15. For each question in Section-I (iii), you will beawarded 4 marks if you darken all the bubble(s)corresponding to only the correct answer(s) andzero mark if no bubbles are darkened. In all othercases minus one (–1) mark will be awarded

16. For each question in Section-IV, you will be awarded6 marks if you darken the bubble corresponding tothe correct answer and zero mark if no bubbles aredarkened No negative marks will be awarded forincorrect answers in this section.

17. Take g = 10 m/s2 unless otherwise stated.