LEADER & ENTHUSIAST COURSE PATTERN : JEE (Adaced) …ADV... · TARGET : JEE (M + A) 2014 ATHEMATIC...

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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

36 i`"B gSa vkSj izR;sd fo"k; ds lHkh 20 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 rhu [kaM gSaA

INSTRUCTIONS / lwpuk,¡

SCORE-I : JEE (Advanced) # 01

PATTERN : JEE (Advanced)LEADER & ENTHUSIAST COURSESCORE-I

PAPER CODE

01CT313071 KOTA - 1/36

TM

Path to success KOTA (RAJASTHAN)

Your 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 36 pages and all the 20 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 part consists of threesections.

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

0 1 C T 3 1 3 0 7 1

Date : 08 - 03 - 2014TARGET : JEE (Main + Advanced) 2014

TEST # 03

le; : 3 ?k.Vs egÙke vad : 225Time : 3 Hours Maximum Marks : 225

01CT313071KOTA - 2/36

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

Hkkx-1 xf.kr I(i) dsoy ,d lgh fodYi izdkj 03 - 04Part-1 Mathematics Only One Option Correct Type

I(ii) ,d ;k vf/kd lgh fodYi izdkj 05 - 05One or More Options Correct Type

I(iii) vuqPNsn izdkj 06 - 06Paragraph Type

II eSfVªDl&esy izdkj 07 - 07Matrix Match Type

IV iw.kk±d eku lgh izdkj 08 - 10Integer Value Correct Type

Hkkx-2 HkkSfrd foKku I(i) dsoy ,d lgh fodYi izdkj 11 - 15Part-2 Physics Only One Option Correct Type





Hkkx-3 jlk;u foKku I(i) dsoy ,d lgh fodYi izdkj 27 - 28Part-3 Chemistry Only One Option 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 = 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

TARGET : JEE (Main + Advanced) 2014 08-03-2014TM


LEADER & ENTHUSIAST COURSE(DATE : 08-03-2014)

MATHEMATICSTM


01CT313071 KOTA - 3/36

PART-1 : MATHEMATICS

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

[k.M-I : (i) dsoy ,d lgh fodYi izdkj

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

bl [k.M esa 6 cgqfodYi iz'u gSA izR;sd iz'u esa pkj fodYi (A), (B), (C) vkSj (D) gSa ftuesa ls dsoy ,d

lgh gSA

1. If [ ]n

2 n

x dxn x f(x) C

x x1 x ...

2 n

= - ++ + + +

ò , where f(0) = 0.

then nl im ( f (1 ) )

® ¥ is equal to

(A) 1 (B) 2 (C) 3 (D) does not exist

;fn [ ]n

2 n

x dxn x f(x) C

x x1 x ...

2 n

= - ++ + + +

ò , tgk¡ f(0) = 0.

rc nl i m ( f (1 ) )

® ¥ cjkcj gS

(A) 1 (B) 2 (C) 3 (D) vfLrRo ugha gS

2. The sum of all coefficients in the expansion of (1 + 2z+ 3z2 + ........+ nzn)2 is equal to

(1 + 2z+ 3z2 + ........+ nzn)2 ds izlkj esa xq.kkadksa dk ;ksx gS-(A) S1 (B) Sn (C) Sn2 (D) Sn3

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

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

MATHEMATICSTARGET : JEE (Main + Advanced) 2014(DATE : 08-03-2014)

TM


01CT313071KOTA - 4/36


3.7 3

2x 0

cos x cos xlim

x®

- is equal to

7 3

2x 0

cos x cos xlim

x®

- cjkcj gS

(A) 121

(B) 221

(C) 421

(D) 821

4. If x y y x 5+ + - = , then 2

2

d ydx

is equal to

;fn x y y x 5+ + - = , rc 2

2

d ydx

cjkcj gS

(A) 225

- (B) 225

(C) 25

(D) 25

-

5. Let A = {1, 2, .... ,10} and B = {1, 2, ..... , 5}

f: A ® B is a non-decreasing into function, then number of such functions is

ekuk A = {1, 2, .... ,10} rFkk B = {1, 2, ..... , 5}

f: A ® B váleku vUr{ksZih Qyu gSA rc , sls Qyuksa dh la[;k gSA

(A) 1001 (B) 1876 (C) 205 (D) 875

6.1 1 12 1 3 2 4 3

sin sin sin6 12 20

- - -æ ö æ ö- - -

+ +ç ÷ ç ÷è ø è ø + ......... ¥ is

1 1 12 1 3 2 4 3sin sin sin

6 12 20- - -

æ ö æ ö- - -+ +ç ÷ ç ÷è ø è ø + ......... ¥ gS

(A) 2p

(B) 3p

(C) 4p

(D) 5p


MATHEMATICSTM


01CT313071 KOTA - 5/36


(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

7. ( ){ } ( )n2 1n

nlim 2 1 ( 1)

é ù+ê úë û

®¥+ - (where n Î N and [.], {.} denotes G.I.F. and {.} respective) is equal to :-

(A) – 1 when n is even (B) 0 when n is odd(C) 0 when n is even (D) –1 when n is odd

( ){ } ( )n2 1n

nlim 2 1 ( 1)

é ù+ê úë û

®¥+ - (tgk¡ n Î N rFkk [.], {.} Øe'k% egÙke iw.kk±d Qyu o viw.kk±'k Qyu dks O;Dr

djrk gS :-

(A) – 1 tc n le gS (B) 0 tc n fo"ke gS

(C) 0 tc n le gS (D) –1 tc n fo"ke gS

8. All the natural numbers, sum of whose digits is 8 are arranged in ascending order then the 170thnumber in the list does not contain the digit

lHkh izkÏr la[;k,¡ ftuds vadks dk ;ksx 8 gS dks c<+rs Øe esa O;ofLFkr djus ij 170th la[;k fuEu vad dks xzg.k ugha

djrh gSA

(A) 2 (B) 3 (C) 4 (D) 5

9. f : R ® R is a continuous functions such that

x2

20

f(t)f(x) (1 x ) 1 dt

1 t

æ ö= + +ç ÷+è øò then

f : R ® R ,d lrr~ Qyu gS fd x

22

0

f(t)f(x) (1 x ) 1 dt

1 t

æ ö= + +ç ÷+è øò rc

(A) f(1) = 2e (B) f ' (1) = 4e (C) f(0) = 1 (D) f '(1) = 1

10. The sum of the greatest and the smallest terms in (x + 12)21 is zero then x can take values.

(x + 12)21 esa egÙke ,oa y?kqÙke in dk ;ksx 'kwU; gSA rc x fuEu eku xzg.k dj ldrk gSA

(A) –10 (B) –21 (C) –28 (D) –6


TM


01CT313071KOTA - 6/36

(iii) Paragraph Type

(iii) vuqPNsn izdkj

This section contains 1 paragraph each describing theory, experiment, data etc. Three questionsrelate to paragraphs. Each question of paragraph has only one correct answer among the four choices(A), (B), (C) and (D).

bl [k.M esa fl¼kUrksa] iz;ksxksa vkSj vk¡dM+k sa vkfn dks n'kkZus okyk 1 vuqPNsn gSA vuqPNsn ls lacaf/kr rhu iz'u

gSaA vuqPNsn esa gj iz'u ds pkj fodYi (A), (B), (C) vkSj (D) gSa ftuesa ls dsoy ,d lgh gSA

Paragraph for Questions 11 to 13

iz'u 11 ls 13 ds fy;s vuqPNsn

If a, b, c are distinct non-zero complex numbers such that |a| = |b| = |c|.

;fn a, b, c rhu fHkék v'kwU; lfEeJ la[;k, bl izdkj gS fd |a| = |b| = |c|.

11. If one of the roots of the equation az2 + bz + c = 0 has modulus 1 then a, b, c are in

(A) A.P. (B) G.P. (C) H.P. (D) None

;fn lehdj.k az2 + bz + c = 0 dk ,d ewy ekikad 1 gSA rc a, b, c gS

(A) A.P. (B) G.P. (C) H.P. (D) dksbZ ugha12. If az2 + bz + c = 0 and bz2 + cz + a = 0 each has a root having modulus 1 then the points a, b, c in the

complex plane are(A) collinear (B) vertices of an equilatral triangle(C) vertices of a scalane triangle (D) vertices of a right angled triangle

;fn az2 + bz + c = 0 rFkk bz2 + cz + a = 0 esa izR;sd dk ,d ewy ekikad 1 gSA rc a, b, c lefeJ ry esa gSA(A) lejs[kh; (B) leckgq f=Hkqt ds 'kh"kZ(C) fo"keckgq f=Hkqt ds 'kh"kZ (D) ledks.k f=Hkqt ds 'kh"kZ

13. If az2 + bz + c = 0 and bz2 + cz + a = 0 have a root of modulus 1 then the equations az2 + bz + c = 0 andbz2 + cz + a = 0

(A) cannot have a common root (B) have both roots common

(C) have exactly one common root (D) data insufficient

;fn az2 + bz + c = 0 rFkk bz2 + cz + a = 0 esa izR;sd dk ,d ewy ekikad 1 gSA rc lehdj.k az2 + bz + c = 0 rFkkbz2 + cz + a = 0

(A) dksbZ mHk;fu"B ewy ugha (B) nksuksa ewy mHk;fu"B gS

(C) dsoy ,d mHk;fu"B ewy (D) tkudkjh iw.kZ ugha



MATHEMATICSTM


01CT313071 KOTA - 7/36

SECTION–II : Matrix-Match Type

[k.M - II : eSfVªDl&esy izdkjThis Section contains 1 question. Question has four statements (A, B, C and D) given in Column I andfive statements (P, Q, R, S and T) in Column II. Any given statement in Column I can have correctmatching with ONE or MORE statement(s) given in Column II. For example, if for a given question,statement B matches with the statements given in Q and R, then for the particular question, againststatement B, darken the bubbles corresponding to Q and R in the ORS.

bl [k.M esa 1 iz'u gSA iz'u esa dkWye I esa 4 dFku (A, B, C vkSj D) vkSj dkWye II esa 5 dFku (P, Q, R, S vkSj T) gSaAdkWye I dk dksbZ Hkh dFku dkWye II ds ,d dFku ;k ,d ls vf/kd dFkuksa ls esy [kkrk gSA mnkgj.k ds fy,] fn, gq, iz'uesa ;fn dFku B dFkuks a Q vkSj R ls esy [kkrk gS] rks vksvkj,l (ORS) esa ml iz'u ds fy;s dFku B ds lkeus Q vkSj R lslEcfUèkr cqycqyksa dks dkyk dhft;sA

1. 8 players P1, P2, ........ P8 of equal strength are paired in four pairs in first round, two pairs in secondround and one pair in final randomly, in a knock out tournament. Find the probability that

Column-I Column-II

(A) P1 does not reach the final (P) 34

(B) P1and P2 play against each other during the course of tournament (Q) 14

(C) P1 is the champion, P2 is the runner up (R) 1

420 and P3, P4 reached in the second round

(D)P1 is the champion and P1and P2 did not play against each other (S) 1

14

(T) 1

8408 f[kykfM+;ksa P1, P2, ........ P8 dks pkj ; qXeksa esa cjkcj ls izFke pj.k esa ;qfXer fd;k tkrk gSA nks ; qXe f}rh; pj.k rFkk,d ;qXe vfUre esa rks ,d [ksy esa izkf;drk Kkr djks fd –

Column-I Column-II

(A) P1 vfUre pj.k esa ugha igq¡ps (P) 34

(B) P1 rFkk P2 ,d nwljs ds foi{k esa [ksys (Q) 14

(C) P1 fotsrk] P2 mifotsrk rFkk (R) 1

420 P3, P4 f}rh; pj.k esa igq¡ps

(D) P1 fotsrk rFkk P1, P2 ,d nwljs ds foi{k esa u [ksys (S) 1

14

(T) 1

840Space for Rough Work / dPps dk;Z ds fy, LFkku


TM


01CT313071KOTA - 8/36

SECTION –III : Integer Value Correct Type[k.M – III : iw.kk±d eku lgh izdkj

No question will be asked in section III / [k.M III esa dksb Z iz'u ugha gSASECTION-IV : (Integer Value Correct Type)

[k.M-IV : (iw.kk±d eku lgh izdkj)

This section contains 6 questions. The answer to each question is a single digit Integer, ranging from0 to 9 (both inclusive)

bl [k.M esa 6 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 least value of n for which (n–2)x2 + 8x + n + 4 > sin–1(sin12) + cos–1(cos12) " x Î R,where n Î Nn dk og U;wure eku ftlds fy, (n–2)x2 + 8x + n + 4 > sin–1(sin12) + cos–1(cos12) " x Î R,tgk¡ n Î N gS %

2. Given f(x) = 2

2if | x | 1

| x |a bx if | x | 1

ì >ïíï + £î

where a, b Î R. If f(x) is differentiable at x = 1 then (2a+b) is equal to

fn;k gS f(x) = 2

2if | x | 1

| x |a bx if | x | 1

ì >ïíï + £î

tgk¡ a, b Î R, ;fn f(x), x = 1 ij vodyuh; gS] rks (2a+b) dk eku gS %



MATHEMATICSTM


01CT313071 KOTA - 9/36

3. If 1 1 1

3 a 4 b 5 ca b c

æ ö æ ö æ ö+ = + = +ç ÷ ç ÷ ç ÷è ø è ø è ø when a, b, c are positive numbers and ab + bc + ca = 1 then

2 2 2

2 2 2

1 a 1 b 1 c5

1 a 1 b 1 c

æ ö- - -+ + =ç ÷+ + +è ø

.........

;fn 1 1 1

3 a 4 b 5 ca b c

æ ö æ ö æ ö+ = + = +ç ÷ ç ÷ ç ÷è ø è ø è ø tc a, b, c /kukRed okLrfod la[;k,a gSA rFkk ab + bc + ca = 1 rc

2 2 2

2 2 2

1 a 1 b 1 c5

1 a 1 b 1 c

æ ö- - -+ + =ç ÷+ + +è ø

.........

4. Let M be any point M on the arc AB of circumcircle of right angled triangle ABC and the maximum

value of AM + BM + CM is 29 N , where a = 3, b = 4, c = 5. Then N is ...............

ekuk M ledks.k f=Hkqt ABC ds ifjo`Ùk ij fLFkr pki AB ij dksbZ fcUnq gS rFkk AM + BM + CM dk vf/kdre eku

29 N gSA tgk¡ a = 3, b = 4, c = 5 rc N gS ...............



TM


01CT313071KOTA - 10/36


5. Let f(x) = n 2 n 2

n nn

sin x coslim

sin x cos x

+ +

®¥

++

for 0 £ x £ 2p

, if ( )2

0

Af x dx

B

p p +=ò , then (A+B) is equal to

;fn f(x) = n 2 n 2

n nn

sin x coslim

sin x cos x

+ +

®¥

++

, 0 £ x £ 2p

ds fy, , ;fn ( )2

0

Af x dx

B

p p +=ò , rc (A+B) cjkcj gS

6. f(x) is a countinous function such that f(2x) - f(x) = x " xÎR and f(0) = 1. then 2

0

f(x)dx =ò .....

ekuk fd f(x) ,d lrr~ Qyu bl izdkj gS fd f(2x) - f(x) = x " xÎR vkSj f(0) = 1, rc 2

0

f(x)dx =ò .....


PHYSICSTM


01CT313071 KOTA - 11/36

PART-2 : PHYSICS

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

[k.M-I : (i) dsoy ,d lgh fodYi izdkjThis section contains 6 multiple choice questions. Each question has four choices (A), (B), (C) and(D) out of which ONLY ONE is correct.


lgh gSA

1. A cylindrical tube filled with water (mw = 4/3) is closed at its both ends by two silvered plano-convex

lenses as shown in the figure. Refractive indices of lenses L1 and L

2 are 2.0 and 1.5 while their radii of

curvature are 5 cm and 9 cm respectively. A point object is placed some where at a point O on the axis

of cylindrical tube. It is found that the object and image coincide with each other after every reflection.

The length of the cylindrical tube is :-

ty (mw = 4/3) ls Hkjh ,d csyukdkj uyh ds nksuksa fljksa dks fp=kuqlkj nks leryksÙky (jtfrr) ysalksa }kjk can dj fn;k

x;k gSA ysalks L1 rFkk L

2 ds viorZukad Øe'k% 2.0 rFkk 1.5 gSa o mudh oØrk f=T;k,a Øe'k% 5 cm o 9 cm gSaA bl

csyukdkj uyh dh v{k ij fdlh fcUnq O ij ,d fcUnq fcEc dks j[kk x;k gSA ;g izsf{kr gksrk gS fd fcEc rFkk bldk

izfrfcEc izR;sd ijkorZu ds i'pkr~ ,d nwljs ds laikrh gSA csyukdkj uyh dh yEckbZ gS

m 2m 1 mw

OL1L2

(A) 82 cm (B) 92 cm (C) 72 cm (D) 18 cm


PHYSICSTARGET : JEE (Main + Advanced) 2014(DATE : 08-03-2014)

TM


01CT313071KOTA - 12/36

2. In the circuit shown in figure, A is a sliding contact which can move over a smooth rod PQ.

Resistance per unit length of the rod PQ is 1 ohm/m. Initially slider is just left to the point P and

circuit is in the steady state. At t = 0 slider starts moving with constant velocity v = 5 m/s towards

right. Current in the circuit at t = 2 sec is :-

(A) 1 amp (B) less than 0.5 amp

(C) more than 1 amp (D) in between 0.5 to 1.0 amp

R=20WL=2H

30V

A VP Q

iznf'kZr ifjiFk esa A ,d lihZ laidZ(sliding contact) gS tks fd fpduh NM+ PQ ij xfr dj ldrk gSA NM+ PQ dh

izfr ,dkad yEckbZ dk izfrjks/k 1 W/m gSA izkjEHk esa LykbMj fcUnq P ds Bhd cka;h vksj gS rFkk ifjiFk LFkk;h voLFkk esa

gSA t = 0 ij LykbMj v = 5 m/s ds fu;r osx ls nka;h vksj xfr djuk izkjEHk dj nsrk gSA t = 2 sec ij ifjiFk esa /kkjk

gksxh

(A) 1 ,fEi;j (B) 0.5 ,fEi;j ls de

(C) 1 ,fEi;j ls vf/kd (D) 0.5 ls 1.0 ,fEi;j ds chp



PHYSICSTM


01CT313071 KOTA - 13/36

3. An electron and a proton each travel with equal speeds around circular orbits in the same uniform

magnetic field, as shown in the diagram (not to scale). The field is into the page on the diagram. Because

the electron is less massive than the proton and because the electron is negatively charged and the proton

is positively charged :-

(A) the electron travels clockwise around the smaller circle and the proton travels counterclockwise

around the larger circle

(B) the electron travels counterclockwise around the smaller circle and the proton travels clockwise

around the larger circle

(C) the electron travels clockwise around the larger circle and the proton travels counterclockwise around

the smaller circle

(D) the electron travels counterclockwise around the larger circle and the proton travels clockwise around

the smaller circle

ÄB

,d bysDVªkWu rFkk ,d izksVkWu ,d gh le:i pqEcdh; {ks= esa o`Ùkkdkj d{kkvksa ds pkjksa vksj leku pkyksa ls xfr djrs gSa

tSlk fd fp= esa n'kkZ;k x;k gS (;g iSekus ij ugha gS)A ;g {ks= fp= ij dkxt esa vUnj dh vksj fo|eku gSA bysDVªkWu ds

izksVkWu ls gYdk gksus ds dkj.k ,oa bysDVªkWu ds ½.kkosf'kr o izksVkWu ds /kukosf'kr gksus ds dkj.k %&

(A) bysDVªkWu NksVh d{kk esa nf{k.kkorhZ rFkk izksVkWu cM+h d{kk esa okekorhZ ?kw.kZu djrk gSA

(B) bysDVªkWu NksVh d{kk esa okekoehZ rFkk izksVkWu cM+h d{kk esa nf{k.kkorhZ ?kw.kZu djrk gSA

(C) bysDVªkWu cM+h d{kk esa nf{k.kkorhZ rFkk izksVkWu NksVh d{kk esa okekorhZ ?kw.kZu djrk gSA

(D) bysDVªkWu cM+h d{kk esa okekoehZ rFkk izksVkWu NksVh d{kk esa nf{k.kkorhZ ?kw.kZu djrk gSA



TM


01CT313071KOTA - 14/36

4. The diagram shows a modified meter bridge, which is used for measuring two unknown resistances atthe same time. When only the first galvanometer is used, for obtaining the balance point, it is found atpoint C. Now the first galvanometer is removed and the second galvanometer is used, which givesbalance point at D. Using the details given in the diagram, find out the value of R

1 and R

2 :-

fp= esa ,d la'kksf/kr ehVj lsrq n'kkZ;k x;k gS ftldk mi;ksx ,d gh le; ij nks vKkr izfrjks/kksa ds ekiu gsrq fd;k tkrk

gSA tc larqyu fcUnq izkIr djus ds fy, izFke xsYosuksehVj dks iz;qä fd;k tkrk gS rks ;g fcUnq C ij izkIr gksrk gSA vc

izFke xsYosuksehVj dks gVkdj f}rh; xsYosuksehVj dks iz;qä djus ij larqyu fcUnq D ij izkIr gksrk gSA R1 o R

2 ds eku gksxsa %&

GG

R2R1R

BDCA

AB = L; AC = L/4 and AD = 2L/3

(A) R1 = 5R/3, R

2 = 4R/3 (B) R

1 = 4R/3, R

2 = R

(C) R1 = 4R/3, R

2 = 5R/3 (D) R

1 = R, R

2 = R/3

5. In the shown figure (1) and (2), capacitors are in the steady state. Charging batteries are removed andswitches S

1 and S

2 are closed at time t = 0. The plot of lnI (I is the current in the resistor) in against time

t in the resistor R1 and R

2 are shown by the graphs 1 and 2 respectively in the figure (3). Choose the

INCORRECT option :-(A) If e

1 = 2e

2, R

1 must be equal to 2R

2(B) If R

1 = R

2, C

1 must be less than C

2

(C) If C1 = C

2, R

1 must be less than R

2(D) R

1C

1 is equal to R

2C

2

(1)

e1

R1 S1

C1

(2)

e2

R2 S2

C2

(3) (1)

(2)lnI

t

iznf'kZr fp= (1) rFkk (2) esa la/kkfj= LFkk;h voLFkk esa gSA vkos'ku cSVfj;ksa dks gVkdj le; t = 0 ij fLop S1 o S

2 dks

can dj nssrs gSaA izfrjks/k R1 o R

2 esa izokfgr /kkjk I ds fy, lnI rFkk le; t ds e/; vkjs[k fp= (3) esa Øe'k% vkjs[k 1 rFkk

2 }kjk n'kkZ;s x;s gSaA xyr fodYi pqfu, %&

(A) ;fn e1 = 2e

2 gks rks R

1 fuf'pr :i ls 2R

2 ds rqY; gksxkA

(B) ;fn R1 = R

2 gks rks C

1 fuf'pr :i ls C

2 ls de gksxkA

(C) ;fn C1 = C

2 gks rks R

1 fuf'pr :i ls R

2 ls de gksxkA

(D) R1C

1 dk eku R

2C

2 ds rqY; gSA



PHYSICSTM


01CT313071 KOTA - 15/36

6. Three medium with refractive index n1, n

2 and n

3 (n

1 > n

2 > n

3 > 1) are as shown in Figure. The two

beams are parallel to each other, with a beam-1 passes only through the medium I and III, and beam-2-through the medium II and III. Determine the angle between the beams in the medium III.

(A) 1 1

1 2

sin sinsin sin

n n- -æ ö æ öq q

-ç ÷ ç ÷è ø è ø

(B) 1 13 3

1 2

n sin n sinsin sin



(C) 1 11 2

3 3

n sin n sinsin sin



(D) None of these

q

I

2n = 1

II

III

n2n1

n3

1

viorZukad n1, n

2 rFkk n

3 (n

1 > n

2 > n

3 > 1) okys rhu ek/;e fp= esa n'kkZ;s x;s gSaA nksuksa iqat ,d&nwljs ds lekUrj gSa

rFkk iqat–1 dsoy ek/;e I rFkk III ls xqtj ldrk gS vkSj iqat-2 ek/;e II rFkk III ls xqtj ldrk gSA ek/;e III esa iqatksa

ds e/; dks.k gksxk %&

(A) 1 1

1 2

sin sinsin sin



(B) 1 13 3

1 2

n sin n sinsin sin



(C) 1 11 2

3 3

n sin n sinsin sin



(D) buesa ls dksbZ ugha



TM


01CT313071KOTA - 16/36


(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.


vf/kd lgh gSA

7. If an endoergic nuclear reaction is brought about by bombarding a stationary nucleus with a projectile

then INCORRECT statement(s) is/are:-

(A) kinetic energy of the projectile must be equal to the magnitude of Q-value of the reaction

(B) kinetic energy of the projectile must be less than the magnitude of the Q-value of the reaction

(C) the kinetic-energy of the projectile must be more than the magnitude of Q-value of the reaction

(D) the momentum will not be conserved in such a nuclear reaction

;fn fdlh fLFkj ukfHkd ij iz{ksI; ds Vdjkus ds QyLo:i ,d Å"ek{ksih ukfHkdh; vfHkfØ;k lEiUu gksrh gks rks fuEu

esa ls xyr dFku@dFkuksa dks pqfu;s %&

(A) bl iz{ksI; dh xfrt ÅtkZ fuf'pr rkSj ij vfHkfØ;k ds Q-eku ds ifjek.k ds cjkcj gksxhA

(B) bl iz{ksI; dh xfrt ÅtkZ fuf'pr rkSj ij vfHkfØ;k ds Q-eku ds ifjek.k ls de gksxhA

(C) bl iz{ksI; dh xfrt ÅtkZ fuf'pr rkSj ij vfHkfØ;k ds Q-eku ds ifjek.k ls vf/kd gksxhA

(D) ,slh ukfHkdh; vfHkfØ;k esa laosx lajf{kr ugha gksrkA



PHYSICSTM


01CT313071 KOTA - 17/36

8. In the figure, AB is a given real object and its image is A'B'. Comment on the nature of the lens used to

make the image and the position of the lens.

(A) The lens used can be converging

(B) The lens used can be diverging

(C) The lens could be located between object and the image

(D) The lens could be located to the right of the image

B'

A'A

B

iznf'kZr fp= esa AB ,d okLrfod fcEc gS rFkk A'B' bldk izfrfcEc gSA izfrfcEc ds fuekZ.k ds fy, iz;qä ysUl dh

izd`fr rFkk bldh fLFkfr ds ckjs esa lgh dFku@dFkuksa dks pqfu, %&

(A) iz;qä ysUl vfHklkjh gks ldrk gSA

(B) iz;qä ysUl vilkjh gks ldrk gSA

(C) ;g ysUl fcEc rFkk izfrfcEc ds e/; fLFkr gks ldrk gSA

(D) ;g ysUl izfrfcEc ds nk¡;h vksj fLFkr gks ldrk gSA



TM


01CT313071KOTA - 18/36

9. A very small circular loop of radius a is initially coplanar and concentric with a much larger circular loopof radius b (>>a). A constant current I is passed in the large loop which is kept fixed in space and thesmall loop is rotated with angular velocity w about a diameter. The resistance of the small loop is R & itsself inductance is negligible. The current in the larger loop is clockwise. (Here q is to be measured fromaxis of large ring)

(A) The current in the small loop as a function of time is2

0a µ I sin t2bR

p w w

(B) The current in the small loop as a function of time is 2

0a µ I cos t2bR

p w w

(C) Torque that must be exerted on the small loop to rotate it, is

220b µ I sin t

R 2b

æ öp wwç ÷è ø

(D) Torque that must be exerted on the small loop to rotate it, is

220a µ Isin t

R 2b

æ öp wwç ÷è ø

w

a b

I

,d a f=T;k dk cgqr NksVk o`Ùkkdkj ywi izkjEHk esa f=T;k b (>>a) okys ,d cgqr cM+s oÙ̀kkdkj ywi ds leryh; rFkk

ladsUæh; gSA ,d fu;r /kkjk I dks cM+s ywi ls xqtkjk tkrk gS tks lef"V esa fLFkj fd;k x;k gS rFkk NksVs ywi dks O;kl ds lkis{k

dks.kh; osx w ls ?kw.kZu djk;k tkrk gSaA NksVs ywi dk izfrjks/k R gS o bldk LoizsjdRo ux.; gSA cM+s ywi esa /kkjk nf{k.kkorhZ

gS (;gk¡ q dk ekiu cM+s ywi dh v{k ls fd;k x;k gSA )

(A) le; ds Qyu ds :i esa NksVs ywi esa /kkjk dk eku 2

0a µ I sin t2bR

p w w gksxkA

(B) le; ds Qyu ds :i esa NksVs ywi esa /kkjk dk eku 2

0a µ I cos t2bR

p w wgksxkA

(C) NksVs ywi dks ?kw.kZu djkus ds fy, bl ij yxk;s x;s cyk?kw.kZ dk eku

220b µ I sin t

R 2b

æ öp wwç ÷è ø

gksuk pkfg,A

(D) NksVs ywi dks ?kw.kZu djkus ds fy, bl ij yxk;s x;s cyk?kw.kZ dk eku

220a µ Isin t

R 2b

æ öp wwç ÷è ø

gksuk pkfg,A



PHYSICSTM


01CT313071 KOTA - 19/36

10. Positive charge Q is distributed uniformly throughout an insulating sphere of radius R, centered at theorigin. A particle with positive charge Q is placed at x = 2 R on the x axis.

(A) The electric potential at x = R2

is 0

49Q96 Rp Î

(B) The electric field at x = R2

is 20

Q72 RpÎ

(C) Electric field at x = R2

is 20

17Q72 RpÎ

(D) Electric potential at x = R2

is 0

3Q2 RpÎ

,d R f=T;k ds dqpkyd xksys] ftldk dsUæ ewy fcUnq ij gS] ij /kukos'k Q dks ,dleku :i ls forfjr fd;k tkrk gSA

,d Q /kukos'k okys d.k dks x-v{k ij x = 2R ij j[kk tkrk gS %&

(A) x = R2

ij fo|qr foHko 0

49Q96 Rp Î gSA

(B) x = R2

ij fo|qr {ks= 20

Q72 RpÎ gSA

(C) x = R2

ij fo|qr {ks= 20

17Q72 RpÎ gSA

(D) x = R2

ij fo|qr foHko 0

3Q2 RpÎ gSA



TM


01CT313071KOTA - 20/36

(iii) Paragraph Type

(iii) vuqPNsn izdkj

This section contains 1 paragraph each describing theory, experiment, data etc. Three questionsrelate to paragraphs. Each question of paragraph has only one correct answer among the four choices

(A), (B), (C) and (D).

bl [k.M esa fl¼kUrksa] iz;ksxksa vkSj vk¡dM+k sa vkfn dks n'kkZus okyk 1 vuqPNsn gSA vuqPNsn ls lacaf/kr rhu iz'u

gSaA vuqPNsn esa gj iz'u ds pkj fodYi (A), (B), (C) vkSj (D) gSa ftuesa ls dsoy ,d lgh gSA

Paragraph for Questions 11 to 13

iz'u 11 ls 13 ds fy;s vuqPNsn

As shown in figure, a thin spherical shell of radius R has a fixed charge +q distributed uniformly over its

surface. A small circular section (radius r << R) of charge is removed from the surface.

iznf'kZr fp= esa R f=T;k okys ,d irys xksykdkj dks'k ij fu;r vkos'k +q bldh lrg ij ,dleku :i ls forjhr gSA

bldh lrg ls vkos'k dk ,d NksVk o`Ùkkdkj [k.M (f=T;k r << R) vyx dj fn;k tkrk gSA

R

w+z

–z

P

11. The electric field just inside and just outside the sphere at the hole is :-

bl fNæ ij xksys ds Bhd vUnj rFkk Bhd ckgj fo|qr {ks= gksxk :-

(A) 20

q4 Rpe (B) 0 (C) 2

0

q2 Rpe (D) 2

0

q8 Rpe



PHYSICSTM


01CT313071 KOTA - 21/36

FOR THE NEXT QUESTION NO. (12) : Have following setup, the cut section is replaced and the

sphere is set in motion rotating with constant angular velocity w = w0 about the z-axis.

vxys iz'u (12) ds fy, % bl O;oLFkk esa FkksM+k la'kks/ku fd;k tkrk gS] ftlesa dVs gq, [k.M dks okil j[kdj xksys

dks z-v{k ds lkis{k fu;r dks.kh; osx w = w0 ls ?kw.kZu djk;k tkrk gSA

12. The line integral of the magnetic field ( Bdz¥

-¥ò ) along the z-axis is :-

z-v{k ds vuqfn'k pqEcdh; {ks= dk js[kh; lekdy ( Bdz¥

-¥ò ) gksxk :-

(A) 0 0q4m w

p(B) 0 0q

2m w

p(C) 0 0q

3m w

p(D) 0 02q

3m wp

FOR THE NEXT QUESTION NO. (13) : Now the sphere's angular velocity increases linearly with

time w = w0 + kt.

vxys iz'u (13) ds fy, % vc xksys ds dks.kh; osx dks le; ds lkFk w = w0 + kt ds vuqlkj jSf[kd :i ls c<+k;k tkrk

gSA

13. The line integral of the electric field (PE.dò lÑ ) around the circular path P located at the centre of the

sphere is (Assume that the normal to the plane containing the path is along the +z axis and that the radius

rp << R) :-

xksys ds dsUæ ij fLFkr o`Ùkh; iFk P ds pkjksa vksj fo|qr {ks= dk js[kh; lekdy (PE.dò lÑ ) gksxk (;g iFk ftl ry esa gSS

ekuk ml ry ij vfHkyEc + z v{k ds vuqfn'k gS rFkk f=T;k rp << R gS) %-

(A) 2

0 Pkr q2R

m(B)

20 Pkr q3R

m(C)

20 Pkr q6R

m(D)

20 Pkr q4R

m



TM


01CT313071KOTA - 22/36



bl [k.M esa 1 iz'u gSA iz'u esa dkWye I esa 4 dFku (A, B, C vkSj D) vkSj dkWye II esa 5 dFku (P, Q, R, S vkSj T) gSaA

dkWye I dk dksbZ Hkh dFku dkWye II ds ,d dFku ;k ,d ls vf/kd dFkuksa ls esy [kkrk gSA mnkgj.k ds fy,] fn, gq, iz'u

esa ;fn dFku B dFkuks a Q vkSj R ls esy [kkrk gS] rks vksvkj,l (ORS) esa ml iz'u ds fy;s dFku B ds lkeus Q vkSj R ls

lEcfUèkr cqycqyksa dks dkyk dhft;sA1. Column-I shows position vector as function of time for a particle. Match the characteristic of the motion

with their description in column-II.Column-I Column-II

(A) A particle has its position as a function (P) Angular momentum of the system

of time given by ( )0 0r t r v t= +r r r

where about the origin is conserved.

0rr

and 0vr

are non-zero constants.

(B) A particle has its position vector as (Q) The particle may be charged andfunction of time given by moving in a uniform electric and magnetic

( ) ( )20 0

ˆ= +rr t v t c t k + ( ) ( )0 0

ˆ ˆcos sina t i a t jw w+ field, no other force acts on it

where 0ar

and 0vr

, c0 are non-zero

constant. (angle between 0vr

and 0ar

is 90° < q < 180°)(C) A particle is moving in a plane with (R) Projection of particle on x or y axis is

its position vector given by simple harmonic motion.

( ) ( ) ( )0 0ˆ ˆsin cosr t a t i b t jw w= +

r

where a0, b

0 (a

0 ¹ b

0) and w are

positive constant.(D) A particle is moving in a plane (S) Particle may be moving under the influence

with position vector of force obeying inverse square force law

( ) ( ) ( )0 0ˆ ˆcos sinr t a t i a t jw w= +

r

where a0 and w are positive constant.

(T) Area swept by position vector of particlein equal time interval is equal.



PHYSICSTM


01CT313071 KOTA - 23/36

LrEHk-I esa ,d d.k ds fy, le; ds Qyu ds :i esa fLFkfr lfn'k dks n'kkZ;k x;k gSA budk LrEHk-II esa fn, x, dFkuksa ls

feyku fdft,A

LrEHk-I LrEHk-II

(A) ,d d.k dh fLFkfr le; ds Qyu ds :i esa (P) ewy fcUnq ds lkis{k fudk; dk dks.kh; laosx lajf{kr gSA

( )0 0r t r v t= +r r r

}kjk fy[kh tkrh gS] tgk¡

0rr

rFkk 0vr

v'kwU; vpj gSaA

(B) ,d d.k dk le; ds Qyu ds :i esa fLFkfr lfn'k (Q) ;g d.k vkosf'kr gks ldrk gS rFkk le:i fo|qr o

( ) ( )20 0

ˆ= +rr t v t c t k + ( ) ( )0 0

ˆ ˆcos sina t i a t jw w+ pqEcdh; {ks= esa xfr'khy gS ,oa bl ij dksbZ vU; cy

}kjk fn;k tkrk gS] tgk¡ 0ar

, 0vr

rFkk c0 v'kwU; fu;rkad dk;Zjr ugha gSA

gS ( 0vr

o 0ar

ds e/; dks.k 90° < q < 180° gS)

(C) ,d d.k fdlh ry esa xfr'khy gS ftldk fLFkfr lfn'k (R) x ;k y v{k ij d.k dk iz{ksi ljy vkorZ xfr gSA

( ) ( ) ( )0 0ˆ ˆsin cosr t a t i b t jw w= +

r }kjk fn;k tkrk

gS] tgk¡ a0, b

0 (a

0 ¹ b

0) rFkk w /kukRed fu;rkad gSaA

(D) ,d d.k fdlh ry esa xfr'khy gS ftldk fLFkfr lfn'k (S) d.k O;qRØe oxZ cy fu;e dh ikyuk djus okys

( ) ( ) ( )0 0ˆ ˆcos sinr t a t i a t jw w= +

r }kjk fn;k tkrk cy ds v/khu xfr'khy gks ldrk gSA

gS] tgk¡ a0 rFkk w /kukRed fu;rkad gSA

(T) d.k ds fLFkfr lfn'k }kjk leku le; vUrjky esa

leku {ks=Qy r; fd;k tkrk gSA


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



TM


01CT313071KOTA - 24/36

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

This section contains 6 questions. The answer to each question is a single digit Integer, ranging from

0 to 9 (both inclusive)

bl [k.M esa 6 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 Ka X-ray of molybdenum has wavelength 71 pm. If the energy of a molybdenum atom with a

K electron knocked out is 23.32 keV, what will be the energy (in keV) of this atom when an L electron

is knocked out? Approximate your answer to nearest integer (Take hc = 1242 eV-nm)

eksfyCMsue dh Ka X-fdj.k dh rjaxnS/; Z 71 pm gksrh gSA ;fn eksfyCMsue ijek.kq ftldk K bysDVªkWu ckgj fudy x;k

gks] dh ÅtkZ 23.32 keV gks rks ,d L bysDVªkWu ds ckgj fudyus ij bl ijek.kq dh ÅtkZ (keV esa) D;k gksxh\ viuk mÙkj

fudVre iw.kk±d esa nsaA (hc = 1242 eV-nm)

2. Figure shows a cylindrical conducting rod of diameter d and length l >> d is uniformly charged in vacuum

such that electric field near its surface and far from its ends is E0. Electric field at r >> l on the axis of

the cylinder is 0

23E d

4Nr

l

. Then find the value of N.

O;kl d rFkk yEckbZ l >> d okyh ,d csyukdkj pkyd NM+ dks fuokZr esa ,d leku :i ls bl izdkj vkosf'kr fd;k

tkrk gS fd bldh lrg ds utnhd rFkk blds fljksa ls nwj fo|qr {ks= E0 izkIr gksrk gSA csyu dh v{k ij r >> l ij fo|qr

{ks= 0

23E d

4Nr

l

gks rks N dk eku Kkr djsaA

l

d

r



PHYSICSTM


01CT313071 KOTA - 25/36

3. Figure shows a mirrored equiconvex lens placed at 12 cm below water. Find position of object abovethe surface of water (in cm) such that its image coincides with itself.fp= esa ,d niZ.kuqek leksÙky ysal n'kkZ;k x;k gS tks fd ty lrg ls 12 cm uhps fLFkr gSA ty lrg ls fdlh fcEcdks fdruk Åij (cm esa) j[kk tk;s rkfd bldk izfrfcEc] Lo;a fcEc ds laikrh gk s\

mg=3/2R=30cm

12cm

m air=1

\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ \\\\\\\\\\\\\\\

O

mW=4/3

4. An infinitely long cylinder of radius R is made of an unusual exotic material with refractive index –1(Figure). The center of the cylinder O lies along the y-axis. A narrow laser beam is directed along the ydirection from the lower plate. The laser source is at a horizontal distance x from the diameter in the ydirection. The angle of incidence is 10°. If total deviation suffered by ray is d (in degree) as it comes out,

write answer as 10d

.

,d vuUr yEck R f=T;k dk csyu] ,d vlkekU; inkFkZ ftldk viorZukad (–1) gS] ls cuk gqvk gS] fp= ns[ksaA csyudk dsUnz O, y-v{k ds vuqfn'k fLFkr gSA fupyh IysV ls ,d iryk ystj iq at y-fn'kk esa vkifrr gksrk gSA ;g ystj òksry-fn'kk esa O;kl ls {kSfrt nwjh x ij fLFkr gSA vkiru dks.k 10° gSA ;fn fdj.k ds ckgj fudyus ij fdj.k esa mRiUu dqy

fopyu d (fMxzh) gks rks 10d

dk eku Kkr dhft;sA

qrqi

O

y

x incident rayemergent ray



TM


01CT313071KOTA - 26/36

5. A disc of radius r is made of a material of negligible resistance and can rotate about a horizontal shaft. Asmaller disc of radius r is fixed onto the same shaft and has a massless cord wrapped around it, which isattached to a small object of mass m as shown. Two ends of a resistor of resistance R are connected tothe perimeter of the disc and to the shaft by wiping contacts. The system is then placed into a uniformhorizontal magnetic field B and mass m is released. Find the constant angular velocity (in rad/s) with which

the disc will rotate after a certain time. If your answer is w give value of 20w

. Data : r = 10 cm, r = 2 cm,

R = 0.01 W, B = 0.2 T, m = 50 g.,d r f=T;k dh pdrh ux.; izfrjks/k okys inkFkZ ls cuh gS rFkk ;g ,d {kSfrt 'kkW¶V ds lkis{k ?kw.kZu dj ldrh gSA ,dr f=T;k okyh NksVh pdrh dks bl 'kkW¶V ij j[kdj ,d æO;ekughu jLlh blds pkjksa vksj yisV nh tkrh gS tks fp=kuqlkjm æO;eku ds ,d NksVs fi.M ls tqM+h gqbZ gSA izfrjks/k R okys ,d izfrjks/kd ds nksuksa fljksa dks pdrh ds ifjeki rFkk 'kkW¶Vls tksM+ nsrs gSaA bl fudk; dks vc le:i {kSfrt pqEcdh; {ks= B esa j[kdj æO;eku m dks fojkekoLFkk ls NksM+k tkrk gSA,d fuf'pr le; ds i'pkr~ ;g pdrh ftl fu;r dks.kh; osx ls ?kw.kZu djsxh mldk eku (rad/s esa) Kkr dhft,A ;fn

vkidk mÙkj w gks rks 20w

dk eku fyf[k, ,oa r = 10 cm, r = 2 cm, R = 0.01 W, B = 0.2 T, m = 50 g yhft,A

rmB

R

r

B

6. A long cylinder with uniformly charged surface and cross-sectional radius a = 1.0 cm moves with aconstant velocity v = 10 m/ s along its axis. An electric field strength at the surface of the cylinder isequal to E = 0.9 kV/cm. If the resulting convection current, that is, the current caused by mechanicaltransfer of a charge is a µA. Then fill 10a. (Î

0 = 8.85 × 10–12 C2/N-m2)

,dleku :i ls vkosf'kr i`"B rFkk a = 1.0 cm vuqizLFk dkV f=T;k okyk ,d yEck csyu bldh v{k ds vuqfn'k

v = 10 m/ s ds fu;r osx ls xfr djrk gSA bl csyu ds i`"B ij fo|qr {ks= lkeF;Z dk eku E = 0.9 kV/cm gSA ;fn

,d vkos'k ds ;kaf=d LFkkukUrj.k ds dkj.k mRiUu /kkjk tks fd ifj.kkeh laogu /kkjk gS] dk eku a µA gks rks 10a Kkr

dhft,A (Î0 = 8.85 × 10–12 C2/N-m2)



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PART-3 : CHEMISTRY

Hkkx-3 : jlk;uSECTION–I : (i) Only One option correct Type

[k.M-I : (i) dsoy ,d lgh fodYi izdkjThis section contains 6 multiple choice questions. Each question has four choices (A), (B), (C) and(D) out of which ONLY ONE is correct.


lgh gSA

1. A quantity of aluminium has a mass of 54.0 g. What is the mass of the same number of magnesiumatoms? (At. wt. Al = 27, Mg=24)

esXuhf'k;e ds fdru nzO;eku esa mrus gh ijek.kq mifLFkr gksxsa ftrus dh 54.0 g ,Y;wfefu;e esa ijek.kqvksa dh la[;kmifLFkr gSa ? (At. wt. Al = 27, Mg=24)

(A) 12 g (B) 24 g (C) 48 g (D) 97 g

2. + CH3 –CH=CH

2

D¾¾¾¾ ®¾ 43POH

A +¾¾¾¾ ®¾ D

OH).2(

,O.)1(

3

2 B + C

2 2 4NH OH/ H conc.H SOC D E+

D¾¾¾¾® ¾¾¾¾®

Product C reacts with NaOH/I2 but does not show any reaction with ammoniacal silver nitrate.

Product E is

+ CH3 –CH=CH

2

D¾¾¾¾ ®¾ 43POH

A +¾¾¾¾ ®¾ D

OH).2(

,O.)1(

3

2 B + C

2 2 4NH OH/ H conc.H SOC D E+

D¾¾¾¾® ¾¾¾¾®

mRikn C, NaOH/I2 ds lkFk vfHkfØ;k djrk gS ijUrq veksfuÏr flYoj ukbVªsV ds lkFk fØ;k ugha djrk gS] mRikn

E gS :-

(A) (CH3)

2=NOH (B) CH

3CH

2CONH

2

(C) CH3CONHCH

3(D) CH

3COCH

3


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3. The monomer that can undergo radical, cationic and anionic polymerisation with equal ease:

fuEufyf[kr esa ls dkSulk ,dyd ewyd] /kuk;fud rFkk ½.kvk;fud cgqydhdj.k leku nj ls djrs gS :

(A) Me–C=CH2

Me

(B) Ph–CH=CH2

(C) CH2 = CH–CH3 (D) CH2=CH–CN

4. Electrolytic reduction is commmercially employed for the extraction on which of the following metals?

fuEufyf[kr esa ls fdl /kkrq ds fu"d"kZ.k ds fy, oS|qr vi?kVuh; vip;u dks O;kolkf;d Lrj ij iz;qDr fd;k tkrkgS ?(A) Al (B) Cu (C) Ag (D) Fe

5. Which of the following ion does not dissolve in excess of aqueous ammonia

fuEufyf[kr esa ls dkSulk vk;u tyh; veksfu;k ds vkf/kD; esa foys; ugha gksrk gSA(A) Ag+ (B) Cu2+ (C) Al3+ (D) Zn2+

6. Pick the incorrect statement (s) for [Fe(H2O)

5(NO)]SO

4 (spin only magnetic moment = 15BM )

(A) It is formed when NaNO3 is treated with saturated solution of FeSO

4 in presence

of conc. H2SO

4

(B) Oxiation state of Fe is +2

(C) It is an outer-orbital complex

(D) Fe is sp3d2 hybrid.

[Fe(H2O)

5(NO)]SO

4 (pØ.k pqEcdh; vk?kw.kZ = 15BM ) ds fy, vlR; dFku dk p;u dhft, %

(A) bldk fuekZ.k NaNO3 dks lkUnz H

2SO

4 dh mifLFkfr esa FeSO

4 ds lar̀Ir foy;u ds lkFk fØ;k }kjk gksrk gS

(B) vk;ju dh vkWDlhdj.k voLFkk +2 gSA(C) ;g ,d ckg~; d{kd ladqy gSA(D) Fe, sp3d2 ladfjr gSA


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(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.


vf/kd lgh gSA

7. Given that : DGf° (CuO) = –30.4 kcal/mole

DGf° (Cu

2O) = –34.98 kca/mole T = 298 K

Now on the basis of above data which of the following predictions will be most appropriate under thestandard conditions.

(A) The reaction between Cu2O and O

2 would be spontaneous

(B) The reaction between CuO and O2 would be spontaneous

(C) The reaction between CuO and O2 would be non spontaneous

(D) Finely divided form of CuO kept in excess O2 would be converted to a mixture of CuO and Cu

2O

(having more of Cu2O)

fn;k gqvk gS % DGf° (CuO) = –30.4 kcal/mole

DGf° (Cu

2O) = –34.98 kca/mole T = 298 K

mijksDr vk¡dM+ksa ds vk/kkj ij ekud voLFkkvksa ij fuEu esa ls dkSulk vuqeku lokZf/kd lgh gS :-

(A) Cu2O rFkk O

2 ds e/; vfHkfØ;k Lor% gksxh

(B) CuO rFkk O2 ds e/; vfHkfØ;k Lor% gksxh

(C) CuO rFkk O2 ds e/; vfHkfØ;k vLork gksxh

(D) CuO dk lw{e pw.kZ O2 ds vkf/kD; esa j[kus ij CuO rFkk Cu

2O ds feJ.k (Cu

2O dk vkf/kD; gksrk gS) esa

ifjofrZr gksxk


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8. The possible product(s) of the following reaction is/arefuEu vfHkfØ;k esa lEHkkfor mRikn gS@gSa

COOH

COOH

COOH

COOH

4 10P OD¾¾¾®

(A)

O

OO

O

O

O

(B)

HOOC

O

O

O

COOH

(C)

COOH

COOH

O

O

O(D)

O

O

O

O

O

O

9. In which of the following reaction a precipitate is formed ?

(A) BaCl2 + H

2SO

4 (dilute) (B) AgNO

3 + HCl (dilute)

(C) SrCO3 + H

2SO

4 (dilute) (D) BaCl

2 + Ca(OH)

2

fuEu esa ls dkSulh vfHkfØ;k esa vo{ksi curk gS ?

(A) BaCl2 + H

2SO

4 (ruq) (B) AgNO

3 + HCl (ruq)

(C) SrCO3 + H

2SO

4 (ruq) (D) BaCl

2 + Ca(OH)

2

10. Pick the paramagnetic species

vuqpqEcdh; iztkfr;ksa dks pqfu,(A) N

2O (B) NO

2(C) NO (D)N

2O

4


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(iii) Paragraph Type (iii) vuqPNsn izdkj

This section contains 1 paragraph each describing theory, experiment, data etc. Three questionsrelate to paragraphs. Each question of paragraph has only one 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 okyk 1 vuqPNsn gSA vuqPNsn ls lacaf/kr rhu iz'u

gSaA vuqPNsn esa gj iz'u ds pkj fodYi (A), (B), (C) vkSj (D) gSa ftuesa ls dsoy ,d lgh gSAParagraph for Questions 11 to 13

iz'u 11 ls 13 ds fy;s vuqPNsnA hydrogen-like atom (atomic number Z) is in a higher excited state of quantum number n. Thisexcited atom can make a transition to the first excited state by successively emitting two photons ofenergies 10.20 eV and 17.00 eV respectively. Alternatively, the atom from the same excited state canmake a transition to the second excited state by successively emitting two photons of energy 4.25 eVand 5.95 eV respectively.gkbMªkstu ds leku ijek.kq (ijek.kq Øekad Z) mPpÙkj mÙksftr voLFkk ftldh DokaVe la[;k n gS] esa mifLFkr gSA ;gmÙksftr ijek.kq Øekxr nks QksVksuksa ftudh ÅtkZ Øe'k % 10.20 eV rFkk 17.00 eV gS dk mRltZu dj izFke mÙksftrvoLFkk esa LFkkukUrfjr gks ldrk gSA nwljs fodYi ds :i esa leku mÙksftr voLFkk ls ijek.kq dk LFkkukUrj.k nks ØekxrQksVksuks ftudh ÅtkZ Øe'k % 4.25 eV rFkk 5.95 eV dk mRltZu dj f}rh ; mÙksftr voLFkk esa LFkkukUrfjr gksldrk gSA

11. The principle energy level from which transition takes place to first and second excited state :og eq[; ÅtkZ Lrj ftlls izFke o f}rh; mÙksftr voLFkk esa LFkkukUrj.k gksrk gS %(A) n = 4 (B) n = 6 (C) n = 5 (D) n = 7

12. The hydrogen like species which corresponds to the above transition :gkbMªkstu ds leku iztkfr tks fd mi;qZDr LFkkukUrj.k ls lEcfU/kr gS :(A) He+ (B) Be3+ (C) Li2+ (D) None

13. If in the above sample ions, de-excitation takes place from the given transition state upto ground statethen how many spectral lines will be observed in visible range :mi;qZDr lsEiy vk;uksa esa fomÙkstu nh gqbZ laØe.k voLFkk ls fuEuÙke ÅtkZ Lrj esa gksrk gS rks n`' ; ijkl esa fdruhLisDVªe js[kk,sa izsf{kr gksxh :(A) 1 (B) 2 (C) 3 (D) 4


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bl [k.M esa 1 iz'u gSA iz'u esa dkWye I esa 4 dFku (A, B, C vkSj D) vkSj dkWye II esa 5 dFku (P, Q, R, S vkSj T) gSaA

dkWye I dk dksbZ Hkh dFku dkWye II ds ,d dFku ;k ,d ls vf/kd dFkuksa ls esy [kkrk gSA mnkgj.k ds fy,] fn, gq, iz'u

esa ;fn dFku B dFkuks a Q vkSj R ls esy [kkrk gS] rks vksvkj,l (ORS) esa ml iz'u ds fy;s dFku B ds lkeus Q vkSj R ls

lEcfUèkr cqycqyksa dks dkyk dhft;sA1. Match compounds of Column-I with their tendency to react with some reagent given in Column-II

Column I Column II

(A) HCHO (P) Reacts with [Ag(NH3)2]+

(B) CH3CHO (Q) Reacts with Fehling solution

(C) CH3COCH3 (R) Makes yellow precipitate of CHI3 with NaOH/I2

(D) HCOOH (S) Makes 2 different molecules of cyanohydrin

(T) Reacts with 2, 4 DNP molecule

LrEHk-I esa fn;s x;s ;kSfxdksa dks LrEHk-II esa fn;s x;s dqN vfHkdedksZ ls mudh fØ;k'khyrk ds vk/kkj ij feykudhft, %

LrEHk-I LrEHk-II

(A) HCHO (P) [Ag(NH3)2]+ ds lkFk fØ;k djrk gS

(B) CH3CHO (Q) Qsgfyax foy;u ds lkFk fØ;k djrk gS

(C) CH3COCH3 (R) NaOH/I2 ds lkFk ihyk vo{ksi cukrk gSA(D) HCOOH (S) lk;uksgkbfMªu ds nks fHkék v.kq cukrk gSA

(T) 2, 4 DNP v.kq ds lkFk fØ;k djrk gSA


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


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SECTION-IV : (Integer Value Correct Type)

[k.M-IV : (iw.kk±d eku lgh izdkj)This section contains 6 questions. The answer to each question is a single digit Integer, ranging from0 to 9 (both inclusive)

bl [k.M esa 6 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. For the reaction A ¾®¾ products, the following data is given for a particular run.time (min.) : 0 5 15 35

[A]M : 1 1/2 1/4 1/8

Determine the order of the reaction.

A ¾®¾ mRikn] vfHkfØ;k esa fy, fdlh fof'k"V izk;ksfxd izs{k.k ds fuEu vk¡dM+s fn;s x;s gSle; (feuV) : 0 5 15 35

[A]M : 1 1/2 1/4 1/8

vfHkfØ;k dh dksfV Kkr dhft,A

2. The total residual activity in a patient after receiving an injection containing 99Tc must not be more

than 0.01 mCi, after 36.0 hours. If sample injected have the initial activity 0.64mCi. The half-life of 99Tc

would be (in hours)99Tc dk batsD'ku fdlh ejht dks yxkus ij vo'kksf"kr fØ;k'khyrk 36 ?k.Vs ckn 0.01 mCi ls vf/kd izkIr ugha gqbZ;fn lsEiy batsD'ku dh izkjfEHkd fØ;k'khyrk 0.64 mCi gS] rks 99Tc dh v¼Zvk;q (?k.Vks esa) D;k gksxh\

3. When a graph is plotted between log x/m and log p, it is straight line with an angle 45° and intercept0.3010 on y-axis. If initial pressure is 0.3 atm, then 6 × 10–y gm of gas is adsorbed per gm of adsorbent,then find value of y ?

tc log x/m rFkk log p ds e/; xzkQ [khapk tkrk gS rks ,d ljy js[kk izkIr gksrh gS tks x-v{k ls 45° dk dks.k cukrhgS rFkk y-v{k ij 0.3010 var[k.M dkVrh gSA ;fn izkjfEHkd nkc 0.3 atm gks] vf/k'kks"kd ds izfr xzke }kjk6 × 10–y gm xSl vf/k'kksf"kr gksrh gS] rks y dk eku Kkr djks ?


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4. For the following reaction

CDiamond

+ O2 CO

2(g) ; DH = – 97.0 kcal

CGraphite

+ O2 CO

2(g) ; DH = – 94.0 kcal

The heat required to change 24 g of Cgraphite

Cdiamond

, in kcal would be

fuEu vfHkfØ;k ds fy,

CDiamond

+ O2 CO

2(g) ; DH = – 97.0 kcal

CGraphite

+ O2 CO

2(g) ; DH = – 94.0 kcal

24 xzke Cgraphite

Cdiamond

, esa ifjorZu ds fy, vko';d Å"ek kcal esa gksxhA

5. The number of carbon atoms whose type of hybridization changed in the given reaction is/arenh xbZ vfHkfØ;k esa dkcZu ijek.kqvks a dh la[;k gS@gSa ftudh ladj.k voLFkk esa ifjorZu gksrk gSA

Cl (excess)2

Sunlight Product

6. Find the number of species from the following which undergoes redox change on treatment withwater.

(i) NO2 (ii) ClO2 (iii) XeF2 (iv) XeF4

(v) XeF6 (vi) N2O4 (vii) SiF4 (viii) Cl2O7

(ix) BF3 (x) F2

nh xbZ iztkfr;ksa esa ls ty ds lkFk mikip; vfHkfØ;k nsus okyh iztkfr;ksa dh la[;k Kkr dhft,A

(i) NO2 (ii) ClO2 (iii) XeF2 (iv) XeF4

(v) XeF6 (vi) N2O4 (vii) SiF4 (viii) Cl2O7

(ix) BF3 (x) F2

LEADER & ENTHUSIAST COURSE 08-03-2014TM


KOTA - 35/3601CT313071


11. [kaM–I(i) Hkkx esa 6 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 ftue s a l s ,d ; kvf/kd lgh gSaA

(iii) Hkkx esa fl¼kUrksa] iz;ksxksa vkSj vk¡dM+ksa vkfn dks n'kkZus okyk1 vuqPNsn gSA vuqPNsn ls lacfU/kr rhu iz'u gSaA vuqPNsnij rhu iz'u gSaA vuqPNsn esa gj iz'u ds pkj fodYi (A),(B), (C) vkSj (D) gSa ftuesa ls dsoy ,d lgh gSaA

12. [kaM–II esa 1 iz'u eSfVªDl lqesy izdkj dk gSA13. [kaM–III es a ,d Hkh iz'u ugha gSA14. [kaM-IV es a 6 iz'u gSaA izR;sd iz'u dk mÙkj 0 ls 9 rd

(nksuksa 'kkfey) ds chp dk ,dy vadh; iw.kk±d gSAD. vadu ;kstuk :15. [kaM-I (i & iii) 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

16. [kaM-I(ii) esa gj iz'u esa dsoy lgh mÙkj okys cqycqys(BUBBLE) dks dkyk djus ij 4 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. [k.M–II ds iz'u ds fy, , flQZ mfpr mÙkj okys cqËs@cqËksadks dkyk fd;k gqvk gS rks izR;sd iafä ds fy, vkidks 2 vadfn;s tk;s axsA vr% bl [k.M ds izR;sd iz'u ds vf/kdre8 vad gSA bl [k.M esa xyr mÙkj @ mÙkjks a ds fy, dksbZ½.kkRed vad ugha fn;s tk;s axsA

18. [k aM-IV es a gj iz'u es a dsoy lgh mÙkj okys cqycqys(BUBBLE) dks dkyk djus ij 4 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

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

11. SECTION – I(i) Contains 6 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 1 paragraph each describing theory,experiment, date etc. Three questions relate toparagraph. Each question of paragraph hasONLY ONE correct answer among the fourchoices (A), (B), (C) and (D)

12. SECTION – II contains 1 question of Matrix MatchType.

13. There is no questions in SECTION-III14. Section-IV contains 6 questions The answer to each

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

D. Marking scheme :15. For each question in Section-I (i & iii), 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

16. For each question in Section-I (ii), you will beawarded 4 marks if you darken the bubblecorresponding to the correct answer and zero markif no bubbles are darkened No negative marks willbe awarded for incorrect answers in this section.

17. For question in Section – II, you will be awarded2 marks for each row in which you have darkenedthe bubble(s) corresponding to the correct answer.Thus, each question in this section carries amaximum of 8 marks. There is no negative markingfor incorrect answer(s) for this section.

18. For each question in Section-IV, you will be awarded4 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.

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

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Corporate OfficeALLEN Career Institute,

“SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan)-324005, Trin : +91 - 744 - 2436001 Fax : +91-744-2435003,E-Mail: [email protected] Website: www.allen.ac.in

TM

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

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