test 3 paper 1

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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.

INSTRUCTIONS / lwpuk,¡

NURTURE COURSE : TARGET - JEE (Main + Advanced) 2015

PAPER CODE

01CT113053 KOTA - 1/36

TM

Path to success KOTA (RAJASTHAN)

Your Hard Work Leads to Strong Foundation

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 – 1le; : 3 ?k.Vs egÙke vad : 228Time : 3 Hours Maximum Marks : 228

0 1 C T 1 1 3 0 5 3

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 21 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] Hkk Sfrd foKku vkSj

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

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 21 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.

MAJOR TEST # 03 Date : 16 - 02 - 2014 Pattern : JEE (Advanced)

16-02-2014TARGET : JEE (Main + Advanced) 2015TM


01CT113053KOTA - 2/36

PAPER – 1

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,

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

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

I(ii) ,d ;k vf/kd lgh fodYi izdkj 04 - 05One or More Options Correct Type

I(iii) vuqPNsn izdkj 06 - 06Paragraph Type

III iw.kk±d eku lgh izdkj [000 ls 999] 07 - 08Integer Value Correct Type [000 to 999]

IV iw.kk±d eku lgh izdkj 09 - 10Integer Value Correct Type

Hkkx-2 HkkSfrd foKku I(i) dsoy ,d lgh fodYi izdkj 11 - 12Part-2 Physics Only One Option Correct Type





Hkkx-3 jlk;u foKku I(i) dsoy ,d lgh fodYi izdkj 24 - 25Part-3 Chemistry Only One Option Correct Type





NURTURE COURSE(DATE : 16-02-2014)

MATHEMATICSTM


PAPER – 1

KOTA - 3/3601CT113053

PART-1 : MATHEMATICS Hkkx-1 : xf.kr

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

This section contains 4 multiple choice questions. Each question has four choices (A), (B), (C) and(D) out of which ONLY ONE is correct.bl [k.M esa 4 cgqfodYi iz'u gSA izR;sd iz'u esa pkj fodYi (A), (B), (C) vkSj (D) gSa ftuesa ls dsoy ,dlgh gSA

1. If a,b are roots of x2 + 3x – 15 = 0 and b,g are roots of x2 – 3x + b = 0, then value of (1 – a) (2 – g) is -(A) 9 (B) 11 (C) 8 (D) 10;fn a,b lehdj.k x2 + 3x – 15 = 0 ds ewy rFkk b,g lehdj.k x2 – 3x + b = 0 ds ewy gks] rks (1 – a) (2 – g) dkeku gksxk&(A) 9 (B) 11 (C) 8 (D) 10

2. Let a < b < c be three integers such that a,b,c are in arithmetic progression and a,c,b are in gemoetricprogression. The smallest possible value of 'c' is-(A) –2 (B) 1 (C) 2 (D) 4ekuk a < b < c rhu iw.kk±d bl izdkj gS fd a,b,c lekUrj Js.kh esa rFkk a,c,b xq.kksÙkj Js.kh esa gSA 'c' dk U;wure laHko ekugksxk&(A) –2 (B) 1 (C) 2 (D) 4

3. Sum of all elements in range of ƒ(x) = sgn (sin–1x) + sgn (sinx) + sgn (cosx) + sgn (cos–1x) is(where sgn denotes signum function)-(A) 0 (B) 6 (C) 9 (D) 12ƒ(x) = sgn (sin–1x) + sgn (sinx) + sgn (cosx) + sgn (cos–1x) ds ifjlj esa lHkh vo;oksa dk ;ksxQy gksxk(tgk¡ sgn flXue Qyu dks n'kkZrk gS)-(A) 0 (B) 6 (C) 9 (D) 12

4. Number of solution(s) of the equation [cos x] + {sin x} = 13

, x Î [0,2p] is

(where [.] & { } denotes greatest and fractional part function respectively)-(A) 0 (B) 2 (C) 4 (D) 6

lehdj.k [cos x] + {sin x} = 13

, x Î [0,2p] ds gyksa dh la[;k gksxh&

(tgk¡ [.] rFkk { } Øe'k% egÙke iw.kk±d rFkk fHkUukRed Hkkx Qyu dks n'kkZrk gS)-(A) 0 (B) 2 (C) 4 (D) 6

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) 2015(DATE : 16-02-2014)

TM


PAPER – 1

01CT113053KOTA - 4/36

(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

5. Let A be set of first 25 natural numbers and a,b,c are three distinct elements of set A, then number ofunordered triplets (a,b,c) if-

(A) a + b + c is divisible by 3, is 772 (B) a + b + c is an even number, is 1156

(C) abc is an even number, is 2014 (D) abc is divisible by 3, is 1620

ekuk A, izFke 25 izkd`r la[;kvksa dk leqPp; rFkk a,b,c leqPp; A ds rhu fHkUu vo;o gS] rks vØfer f=dksa (a,b,c)

dh la[;k] ;fn&

(A) a + b + c, 3 ls foHkkftr gS] rks 772 gksxhA (B) a + b + c, ,d le la[;k gS] rks 1156 gksxhA

(C) abc, ,d le la[;k gS] rks 2014 gksxhA (D) abc, 3 ls foHkkftr gS] rks 1620 gksxhA

6. Consider polynomial ƒ(x) = (x – 1) (x – 2) (x – 3).......... (x – 24) (x – 25), then -

(A) coefficient of x24 is 325. (B) coefficient of x23 is 50050.

(C) constant term is 25!. (D) sum of all coefficients is 0.

ekuk cgqin ƒ(x) = (x – 1) (x – 2) (x – 3).......... (x – 24) (x – 25) gks] rks&

(A) x24 dk xq.kkad 325 gksxkA (B) x23 dk xq.kkad 50050 gksxkA

(C) vpj in 25! gksxkA (D) lHkh xq.kkadksa dk ;ksxQy 0 gksxkA



MATHEMATICSTM


PAPER – 1

KOTA - 5/3601CT113053


7. In DABC, a = 2, b = 3, C3p

Ð = , then-

(A) length of internal angle bisector through vertex C is 6 3

5

(B) length of internal angle bisector through vertex C is 3 3

5

(C) length of median through vertex C is 192

(D) length of median through vertex C is 194

f=Hkqt ABC esa a = 2, b = 3, C3p

Ð = gks] rks -

(A) 'kh"kZ C ls xqtjus okys vUr% dks.k v¼Zd dh yEckbZ 6 3

5 gksxhA

(B) 'kh"kZ C ls xqtjus okys vUr% dks.k v¼Zd dh yEckbZ 3 3

5 gksxhA

(C) 'kh"kZ C ls xqtjus okyh ekf/;dk dh yECkkbZ 192

gksxhA

(D) 'kh"kZ C ls xqtjus okyh ekf/;dk dh yECkkbZ 194

gksxhA

8. Function ƒ(x) satisfies the relation ƒ(xy) = ƒ(x).ƒ(y) for all positive numbers x,y and ƒ(2) = 10, then-(ƒ–1(x) represent inverse of ƒ(x))

(A) ( )4

r

r 0

ƒ 2 10000=

=å (B) ( )4

r

r 0

ƒ 2 11111=

=å

(C) ( )4

1 r

r 0

ƒ 10 31-

=

=å (D) ( )4

1 r

r 0

ƒ 10 32-

=

=ålHkh /kukRed la[;kvksa x,y ds fy;s Qyu ƒ(x), laca/k ƒ(xy) = ƒ(x).ƒ(y) dks larq"V djrk gS rFkk ƒ(2) = 10 gS] rks

(ƒ–1(x), ƒ(x) ds izfrykse dks n'kkZrk gS) -

(A) ( )4

r

r 0

ƒ 2 10000=

=å (B) ( )4

r

r 0

ƒ 2 11111=

=å

(C) ( )4

1 r

r 0

ƒ 10 31-

=

=å (D) ( )4

1 r

r 0

ƒ 10 32-

=

=å


TM


PAPER – 1

01CT113053KOTA - 6/36


(iii) Paragraph Type (iii) vuqPNsn izdkj

This section contains 2 paragraphs each describing theory, experiment, data etc. Four questions relateto two paragraphs with two questions on each paragraph. Each question of a paragraph has only onecorrect 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 pkj iz'ugSa] ftuesa ls gj vuqPNsn ij nks iz'u gSaA vuqPNsn esa gj iz'u ds pkj fodYi (A), (B), (C) vkSj (D) gSa ftuesals dsoy ,d lgh gSA

Paragraph for Questions 9 and 10 iz'u 9 ,oa 10 ds fy;s vuqPNsn

Let ƒ(x) = |x – 1| + 2|x – 2| + 3|x – 3| and g(x) = 4|x – 4| + 5|x – 5|.

ekuk ƒ(x) = |x – 1| + 2|x – 2| + 3|x – 3| rFkk g(x) = 4|x – 4| + 5|x – 5| gSA

9. Minimum value of ƒ(x) + g(x) is-

ƒ(x) + g(x) dk U;wure eku gksxk&(A) 18 (B) 16 (C) 12 (D) 15

10. Number of solutions of the equation ƒ(x) = g(x).

lehdj.k ƒ(x) = g(x) ds gyksa dh la[;k gksxh&(A) 1 (B) 2 (C) 3 (D) 4

Paragraph for Questions 11 and 12 iz'u 11 ,oa 12 ds fy;s vuqPNsn

If line L º 3x + 4y + 4 = 0 intersect circle S1 º x2 + y2 + 2x – 2y – 2 = 0 at A and B. Circle S2=0intersect circle S1 = 0 orthogonally and touches line L = 0.

;fn js[kk L º 3x + 4y + 4 = 0, o`Ùk S1 º x2 + y2 + 2x – 2y – 2 = 0 dks fcUnq A rFkk B ij izfrPNsn djrh

gSA o`Ùk S2= 0, o`Ùk S1 = 0 dks yEcdks.kh; dkVrk gS rFkk js[kk L = 0 dks Li'kZ djrk gSA

11. (a,b) is intersection point of tangents of circle S1 = 0 at A and B, then the value of (a + b) is-

o`Ùk S1 = 0 ds fcUnq A rFkk B ij [khaph xbZ Li'kZ js[kkvksa dk izfrPNsn fcUnq (a,b) gks] rks (a + b) dk eku gksxk-

(A) 235

- (B) 285

- (C) 235

(D) 285

12. Number of circles S2 = 0 (radius r = 1) is-

o`Ùkksa S2 = 0 (f=T;k r = 1) dh la[;k gksxh-

(A) 1 (B) 2 (C) 3 (D) 4

SECTION –II : Matrix-Match Type [k.M – II : eSfVªDl&esy izdkj

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


MATHEMATICSTM


PAPER – 1

KOTA - 7/3601CT113053


SECTION–III : (Integer Value Correct Type)

[kaM-III : (iw.kk±d eku lgh izdkj)This section contains 5 questions. The answer to each question is a three digit Integer, ranging from000 to 999.

bl [kaM esa 5 iz'u gSaA gj iz'u dk mÙkj rhu v ad dk iw.kk ±d] 000 ls 999 rd ] gSA

1. The number of positive integer divisors of (12)! which is of form 4k + 2 (where k is a whole number) is

(12)! ds 4k + 2 (tgk¡ k ,d iw.k ± la[;k gS) :i ds /kukRed iw.kk±d Hkktdksa dh la[;k gksxh2. Let ABC be a triangle with AB = 5, BC = 4 and AC = 3. Let P and Q

A B

C

PQ

be squares inside ABC with disjoint interiors such that they both have oneside lying on AB. Also, the two squares each have an edge lying on acommon line perpendicular to AB and P has one vertex on AC and Qhas one vertex on BC. If the minimum value of sum of the areas of two

squares can be expressed. as rs

(where r & s are relatively prime), then (r + s) is

ekuk f=Hkqt ABC ftlesa AB = 5, BC = 4 rFkk AC = 3 gSA ekuk f=Hkqt ABC

A B

C

PQ

ds vUnj dh vksj vla;qDr {ks=Qy ds nks fHkUu oxZ P rFkk Q bl izdkj gS fd mudh

,d Hkqtk] AB ij fLFkr gS rFkk nksuksa oxks ±] izR;sd dh ,d Hkqtk AB dh yEcor~

mHk;fu"B js[kk ij fLFkr gS rFkk P dk ,d 'kh"kZ AC ij rFkk Q dk ,d 'kh"kZ

BC ij fLFkr gSA ;fn bu nksuksa oxks ± ds {ks=Qyksa ds ;ksxQy dk U;wure eku rs

(tgk¡ r rFkk s ijLij vHkkT; gS) }kjk O;Dr fd;k x;k gks] rks (r + s) dk eku gksxk


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PAPER – 1

01CT113053KOTA - 8/36


3. Let A º (0,4) and B º (3,8) be two points such that ÐAXB is maximum where X is a point on x-axis.

If x-co-ordinate of X is a 2 b- , then (a4 + b4) is

ekuk A º (0,4) rFkk B º (3,8) nks fcUnq bl izdkj gS fd ÐAXB vf/kdre gS] tgk¡ X, x-v{k ij ,d fcUnq gSA

;fn X dk x- funsZ'kkad a 2 b- gks] rks (a4 + b4) dk eku gksxk4. The number of rectangles whose edges lie completely on the grid lines of following figure is

vk;rksa dh la[;k] ftldh Hkqtk;sa iw.k ± :i ls fp=kuqlkj xzhM js[kkvksa ij fLFkr gS ] gksxh

5. Three pairs of real numbers (x1,y1), (x2,y2) and (x3,y3) satisfy both the equations x3 – 3xy2 = 2005 and

y3 – 3x2y = 2004. Then the value ( )( )( )1 2 3

1 1 2 2 3 3

y y y2 y x y x y x- - -

is

okLrfod la[;kvks a ds rhu ; qXe (x1,y1), (x2,y2) rFkk (x3,y3) nk suks a lehdj.kks a x3 – 3xy2 = 2005 rFkk

y3 – 3x2y = 2004 dks larq"V djrs gSA rc ( ) ( )( )1 2 3

1 1 2 2 3 3

y y y2 y x y x y x- - -

dk eku gksxk


MATHEMATICSTM


PAPER – 1

KOTA - 9/3601CT113053


SECTION-IV : (Integer Value Correct Type)

[k.M-IV : (iw.kk±d eku lgh izdkj)This section contains 4 questions. The answer to each question is a single digit Integer, ranging from0 to 9 (both inclusive)

bl [k.M esa 4 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. In a DABC (with usual notations), if a cosB – b cosA = 35

c, then tan Atan B

is

f=Hkqt ABC esa (lkekU; ladsrksa ds lkFk)] ;fn a cosB – b cosA = 35

c gks] rks tan Atan B

dk eku gksxk

2. Let ƒ(x) be an odd function defined on R such that ƒ(x) = x2 when x > 0. If ƒ(x + a) > 2ƒ(x) holds"xÎ[a,a + 2], then the least integral value of a is

ekuk ƒ(x), R ij ifjHkkf"kr ,d fo"ke Qyu bl izdkj gS fd ƒ(x) = x2 gS] tc x > 0 gSA ;fn ƒ(x + a) > 2ƒ(x)

"xÎ[a,a + 2] lR; gks] rks a dk U;wure iw.kk±d eku gksxk


TM


PAPER – 1

01CT113053KOTA - 10/36


3. If the area of a right angled triangle is numerically equal to twice its perimeter, then the number of suchtriangles with integral sides is

;fn ledks.k f=Hkqt dk {ks=Qy vkafdd :i ls bldh ifjf/k ds nksxqusa ds cjkcj gks] rks , sls f=Hkqtksa dh la[;k ftudhHkqtk;sa iw.kk±d gks] gksxh

4. Let ƒ(x) = ax + b, with a,b real numbers, ƒ1(x) = ƒ(x) & ƒn+1(x) = ƒ(ƒn(x)); n Î N.

If ƒ7(x) = 128x + 381, then (a + b) is

ekuk ƒ(x) = ax + b, ftlesa a,b okLrfod la[;k;sa rFkk ƒ1(x) = ƒ(x) ,oa ƒn+1(x) = ƒ(ƒn(x)); n Î N gSA

;fn ƒ7(x) = 128x + 381 gks] rks (a + b) dk eku gksxk


PHYSICSTM


PAPER – 1

KOTA - 11/3601CT113053

PART-2 : PHYSICS

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

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

This section contains 4 multiple choice questions. Each question has four choices (A), (B), (C) and

(D) out of which ONLY ONE is correct.

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

lgh gSA

1. The temperature at which the speed of sound wave in helium gas is same as that in hydrogen gas

at 27°C, is :-

fdl rkieku ij ghfy;e xSl esa /ofu rjax dh pky] 27°C ij gkbMªk stu xSl esa /ofu rjax dh pky ds leku gksxh %&

(A) 504° C (B) 45° C (C) 327°C (D) 231° C

2. A string of linear mass density 0.8 kg/m is stretched to a tension of 500 N. The mean power required to

maintain a travelling wave of amplitude of 10 mm and wavelength 0.5 m is :-

jSf[kd æO;eku ?kuRo 0.8 kg/m okyh ,d jLlh dks 500 N ruko ds v/khu foLrkfjr fd;k x;k gSA bl jLlh ij

10 mm vk;ke rFkk 0.5 m rjaxnS/; Z okyh ,d izxkeh rjax dks cuk;s j[kus ds fy, vko';d ek/; 'kfä gksxh :-

(A) 70 W (B) 85.3 W (C) 98.7 W (D) 110 W


PHYSICSTARGET : JEE (Main + Advanced) 2015(DATE : 16-02-2014)

TM


PAPER – 1

01CT113053KOTA - 12/36

3. A spherical metal ball of radius 'r' is lying at the bottom of a stationary container containing liquid of density

r as shown in the figure. The force exerted on the upper hemispherical portion of the sphere due to pressure

(P0 = atmospheric pressure) is :-

,d 'r' f=T;k dh xksykdkj /kkfRod xsan fp=kuqlkj r ?kuRo ds æo ls Hkjs ,d fLFkj ik= ds isansa ij j[kh gqbZ gSA nkc ds dkj.k

xksys ds Åijh v/kZxksykdkj Hkkx ij yxus okyk cy gksxk (P0 = ok;qe.Myh; nkc) :-

r4r

(A) [ ]2

0

r3P 7r g

3p

+ r (B) [ ]2

0

r3P 7r g

2p

+ r (C) [ ]20r 3P 7r gp + r (D) [ ]2

02 r 3P 7r gp + r

4. A uniform disc is lying on a board of same mass. Equal and opposite forces have been applied on the disc

and the board as shown in figure. The horizontal surface, on which the board is placed, is frictionless. No

slipping takes place between the disc and the board. Initially the whole system was in the state of rest. If

at any instant the centre of the disc has a velocity V with respect to earth then its angular velocity will be

equal to :-

,d le:i pdrh leku æO;eku ds cksMZ ij j[kh gqbZ gSA bl pdrh rFkk cksMZ ij leku rFkk foijhr cyksa dks fp=kuqlkj

yxk;k tkrk gSA ;g cksMZ ftl {kSfrt lrg ij j[kk gqvk gS] og ,d ?k"kZ.kjfgr lrg gSA ;gk¡ pdrh rFkk cksMZ ds e/; dksbZ

fQlyu ugha gkssrhA izkjEHk esa ;g lEiw.kZ fudk; fojkekoLFkk esa FkkA ;fn fdlh {k.k pdrh ds dsUæ dk /kjkry ds lkis{k

osx V gks rks bldk dks.kh; osx gksxk %&

F

F

(A) VR

(B) V

2R(C)

2VR

(D) V

3R



PHYSICSTM


PAPER – 1

KOTA - 13/3601CT113053





vf/kd lgh gSA

5. A large rock is tied to a balloon filled with air. Both are placed in a lake. As the balloon sinks :

(A) The air pressure inside the balloon increases

(B) The average density of the balloon + air + rock increases

(C) The magnitude of the net force on the balloon + air + rock increases

(D) The magnitude of the net force on the balloon + air + rock decreases

,d cM+s iRFkj dks ok;q ls Hkjs xqCckjs ds lkFk cka/k fn;k tkrk gSA nksuksa dks ,d >hy esa j[k nsrs gSaA xqCckjs ds uhps Mwcus ij

(A) xqCckjs ds vanj ok;q nkc c<+ tkrk gSA

(B) xqCckjs + ok;q + iRFkj dk vkSlr ?kuRo c<+ tkrk gSA

(C) xqCckjs + ok;q + iRFkj ij ifj.kkeh cy dk ifjek.k c<+rk gSA

(D) xqCCkkjs + ok;q + iRFkj ij ifj.kkeh cy dk ifjek.k ?kVrk gSA

6. Which of the following processes must violate the first law of thermodynamics ? (There may be more than

one answer) :-

fuEu esa ls dkSuls izØe fuf'pr :i ls Å"ekxfrdh ds izFke fu;e dk mYya?ku djrs gSa (blds ,d ls vf/kd mÙkj

gks ldrs gSa) :-

(A) W > 0, Q < 0, and DEint

= 0 (B) W > 0, Q < 0 and DEint

> 0

(C) W > 0, Q < 0 and DEint

< 0 (D) W < 0, Q > 0, and DEint

< 0


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PAPER – 1

01CT113053KOTA - 14/36


7. A rod CD of length L and mass M is placed horizontally on a frictionless horizontal surface as shown.A second identical rod AB which is also placed horizontally (perpendicular to CD) on the same horizontalsurface is moving along the surface with a velocity v in a direction perpendicular to rod CD and its endB strikes the rod CD at end C and sticks to it rigidly. Then :-

(A) velocity of centre of mass of the system just after impact is v4

(B) the w (angular speed) of system just after collision is 3v5L

(C) velocity of centre of mass of the system just after impact is v2

(D) the w (angular speed) of system just after collision is 5v3L

A L B

v

C

L

D

yEckbZ L rFkk æO;eku M okyh ,d NM + CD fp=kuqlkj ?k"kZ.kjfgr {kSfrt lrg ij {kSfrt :i ls j[kh gqbZ gSA ,d nwljhblds tSlh NM+ AB blh {kSfrt lrg ij CD ds yEcor~ {kSfrt :i ls j[kh gqbZ gS rFkk ;g CD dh yEcor~ fn'kk esa lrgds vuqfn'k v osx ls xfr'khy gSA bldk fljk B NM+ CD ds C fljs ls Vdjkdj blls n`<+rkiwoZd fpid tkrk gSA rc %&

(A) VDdj ds rqjUr i'pkr~ fudk; ds æO;eku dsUæ dk osx v4

gSA

(B) VDdj ds rqjUr i'pkr~ fudk; dh dks.kh; pky w dk eku 3v5L

gksxkA

(C) VDdj ds rqjUr i'pkr~ fudk; ds æO;eku dsUæ dk osx v2

gksxkA

(D) VDdj ds rqjUr i'pkr~ fudk; dh dks.kh; pky w dk eku 5v3L

gksxkA


PHYSICSTM


PAPER – 1

KOTA - 15/3601CT113053


8. A particle on a spring executes simple harmonic motion. If the mass of the particle and the amplitude are

both doubled then :-

(A) The period of oscillation will change by a factor of 2

(B) The maximum speed of the particle will change by a factor of 2

(C) The magnitude of the maximum acceleration of the particle will change by a factor of 1 (remains

unchanged)

(D) The magnitude of the maximum acceleration of the particle will change by a factor of 2

,d d.k fLizax ij ljy vkoZr xfr djrk gSA ;fn d.k ds æO;eku rFkk vk;ke nksuks a dks nqxuk dj fn;k tk, rks %&

(A) nksyu dk vkorZdky 2 ds xq.kt esa ifjofrZr gks tk,xkA

(B) d.k dh vf/kdre pky 2 ds xq.kt esa ifjofrZr gks tk,xhA

(C) d.k ds vf/kdre Roj.k dk ifjek.k ogh jgsxkA

(D) d.k ds vf/kdre Roj.k dk ifjek.k 2 ds xq.kt esa ifjofrZr gks tk,xkA


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PAPER – 1

01CT113053KOTA - 16/36


(iii) Paragraph Type

(iii) vuqPNsn izdkj

This section contains 2 paragraphs each describing theory, experiment, data etc. Four questions relate

to two paragraphs with two questions on each paragraph. Each question of a paragraph has only onecorrect 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 pkj iz'u

gSa] ftuesa ls gj vuqPNsn ij nks 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 9 and 10

iz'u 9 ,oa 10 ds fy;s vuqPNsn

An elevator is moving in vertical direction such that its velocity varies with time as shown in figure. Its

upward direction is taken as positive. A block of mass 60 kg is placed inside lift on a weight-machine

and also attach to a spring balance as shown in figure. Reading of weight machine and spring balance

are respectively 40 kg and 20 kg when lift was at rest. Lift start moving at t = 0 (g = 10 m/s2)

,d fy¶V Å/okZ/kj fn'kk esa bl izdkj xfr'khy gS fd bldk osx le; ds lkFk fp=kuqlkj ifjofrZr gksrk gSA bldh Åij

dh vksj fn'kk dks /kukRed fy;k x;k gSA ,d 60 kg æO;eku ds CykWd dks fy¶V ds vUnj j[kh ,d Hkkjekih e'khu ij

j[k nsrs gSa ftlls ,d fLizax rqyk Hkh fp=kuqlkj tqM+h gqbZ gSA tc fy¶V fojkekoLFkk esa gS rks Hkkjekih e'khu rFkk fLizax rqyk

ds ikB~;kad Øe'k% 40 kg rFkk 20 kg gSaA ;g fy¶V t = 0 ij xfr djuk izkjEHk dj nsrh gSA (g = 10 m/s2)

60kg 2 4 6

810 12 13 t(sec.)

10m/s

–10m/s

V


PHYSICSTM


PAPER – 1

KOTA - 17/3601CT113053


9. What will be the maximum reading of the weight machine :-

Hkkjekih e'khu dk vf/kdre ikB~;kad gksxk %&

(A) 60 kg (B) 70 kg (C) 80 kg (D) None

10. The reading of the spring balance fluctuate with time as (till 12 sec.)

12 lsd.M rd fLizax rqyk dk ikB~;kad le; ds lkFk fdl izdkj ifjofrZr gksxk %&

(A) 2 4 68 10

12 t(sec.)

Reading

(B) 2 4

6 810 12

t(sec.)

Reading

(C) 12 t(sec.)

Reading

(D) 24 8

12 t(sec.)

Reading


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PAPER – 1

01CT113053KOTA - 18/36



THERMODYNAMICS BY CLASSICAL MECHANICS (fpjlaor~ ;kaf=dh }kjk Å"ekxfrdh)The pressure of a gas on the container can be assumed to be the result of force exerted by the gas moleculeon the container when they collide elastically & changes their momentum. This can be viewed as, if Nnumber of balls collide with the wall/sec, then F = 2NmV

v

m

m

v

Now consider a situation, where the space is separated into two parts by a semipermeable membrane(allows N2 to pass but does not allow O

2 to pass) with temperature T

1 & T

2 respectively as

shown in diagram

N + O2 2

T1 T2

Only N2

T < T2 1

Semipermeable membrane

The velocity of molecule at any temperature is given by 8RT

VM

=p

, R is a constant, M is the molecular

mass of the gas. If n1 molecules of O

2 & n

2 molecules of N

2 collides with the semipermeable membrane

per second, then (Assume equal number of moles of N2 in both chambers)

fdlh ik= esa Hkjh xSl }kjk ik= ij yxk;k x;k nkc] xSl ds v.kqvksa }kjk izR;kLFk :i ls Vdjkus rFkk buds laosx esa ifjorZu

gksus ds dkj.k bu v.kqvksa }kjk ik= dh nhokjksa ij yxk;s x;s cy dk ifj.kke ekuk tk ldrk gSA bls le>us ds fy, ekuk

N xsansa nhokj ls izfr lsd.M Vdjkrh gS rc F = 2NmV

v

m

m

v

vc ,d fLFkfr ij fopkj djrs gSa] tgk¡ fdlh LFkku dks ,d v¼ZikjxE; f>Yyh (tks N2 dks xqtjus nsrh gS ijUrq O

2 blesa

ls gksdj ugha xqtj ikrh) }kjk T1 o T

2 rkieku okys nks Hkkxksa esa ck¡Vk x;k gS] fp= ns[ksaA


PHYSICSTM


PAPER – 1

KOTA - 19/3601CT113053

N + O2 2

T1 T2

Only N2

T < T2 1

Semipermeable membrane

fdlh rkieku ij v.kq dk osx 8RT

VM

=p

}kjk fn;k tkrk gS] tgk¡ R ,d fu;rkad gS rFkk M xSl dk vk.kfod æO;eku

gSA ;fn O2 ds n

1 v.kq rFkk N

2 ds n

2 v.kq izfr lsd.M bl f>Yyh ls Vdjkrs gksa rks (nksuksa d{kksa esa N

2 ds v.kqvksa dh leku

la[;k ekusa)11. The average force exerted by O

2 molecule on the membrane is :-

f>Yyh ij O2 v.kqvksa }kjk vkjksfir vkSlr cy gksxk %&

(A) 2

2

11 O

O

8RTn M

Mp(B)

2

2

11 O

O

8RT2n M

Mp(C)

2

2

21 O

O

8RTn M

Mp(D)

2

2

21 O

O

8RT2n M

Mp

12. The direction of net force on the membrane due to (N2 + O

2) will be :-

(A) Towards left (B) Towards right (C) Zero (D) Can't be determined

(N2 + O

2) ds dkj.k f>Yyh ij yxus okys ifj.kkeh cy dh fn'kk gksxh %&

(A) ck¡;h vksj (B) nk¡;h vksj (C) 'kwU; (D) Kkr ugha fd;k tk ldrkA



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PAPER – 1

01CT113053KOTA - 20/36





bl [kaM esa 5 iz'u gSaA gj iz'u dk mÙkj rhu v ad dk iw.kk ±d] 000 ls 999 rd] gSA1. A glass sphere of volume 7 litres contains air at 27°C and is attached to a pipe full of mercury as shown

in the figure. Initially the mercury is level with the bottom of the sphere in both arms of the tube and theoutside pressure is 700 mmHg. The air in the sphere is then heated so that the mercury level is raised by35 mm in the outer arm. Neglecting the cross-sectional area of the pipe, find the final temperature (in K)of air in the sphere.,d 7 yhVj vk;ru okys dk¡p ds xksys esa 27°C ij ok;q Hkjh gS rFkk bls fp=kuqlkj ikjs ls iw.kZr;k Hkjs ,d ikbi ds lkFktksM+ fn;k tkrk gSA izkjEHk esa uyh dh nksuksa Hkqtkvksa esa ikjs dk Lrj xksys ds isans rd gS rFkk ckgjh nkc 700 mmHg gSA vcxksys esa Hkjh ok;q dks xeZ fd;k tkrk gS rkfd ckgjh Hkqtk esa ikjs dk Lrj 35 mm c<+ tk,A ikbi ds vuqizLFk dkV {ks=Qydks ux.; ekurs gq, xksys esa ok;q dk vfUre rkieku (K esa) Kkr dhft,A

V = 7l

Hg2. An experiment is performed to determine the value of gravitational acceleration 'g' on planet Vulcan. Two

equal masses M hang at rest from the ends of a string on each side of a frictionless pulley (see figure). Amass m = 0.05 M is placed on the left-hand mass. After the heavier side has moved down by h = 1 m thesmall mass m is removed. The system continues to move for next 2 s, covering a distance of H = 1.2 m.Find the value of g (in cm/sec2) from these data.fdlh xzg Vulcan ij xq:Roh; Roj.k 'g' ds ekiu ds fy, ,d iz;ksx fd;k tkrk gSA nks leku æO;ekuksa M dks fp=kuqlkj?k"kZ.kjfgr f?kjuh ij ls xqtj jgh jLlh ds nksuksa fljk sa ls fojkekoLFkk esa yVdk;k x;k gSA vc ck¡;s æO;eku ij ,dm = 0.05 M vfrfjä æO;eku j[k fn;k tkrk gSA Hkkjh okys fljs ds h = 1 m uhps tkus ds i'pkr~ bl NksVs æO;eku mdks gVk nsrs gSaA ;g fudk; vxys 2 s rd yxkrkj xfr djrs gq, H = 1.2 m nwjh r; dj ysrk gSA bu vk¡dM+ksa ds vk/kkjij g dk eku (cm/sec2 esa) Kkr dhft,A

M

Mm



PHYSICSTM


PAPER – 1

KOTA - 21/3601CT113053

3. A conical flask of mass 10 kg and base area as 103 cm2 is floating in liquid of specific gravity 1.2,as shown in the figure. The force (in N) that liquid exerts on curved surface of conical flask will be(Given g = 10 m/s2)

æO;eku 10 kg rFkk vk/kkj {ks=Qy 103 cm2 okyk ,d 'kaDokdkj ¶ykLd fp=kuqlkj 1.2 fof'k"V xq:Ro okys æo esa rSj

jgk gSA bl ¶ykLd dh oØh; lrg ij æo }kjk yxk;k x;k cy (N esa) Kkr dhft,A(g = 10 m/s2)

q10cm

4. A certain mass (2 mole) of ideal gas with specific heat at constant volume (CV) given by 2R, is cooled at

constant atmospheric pressure P0 = 105 Pa. As a result its volume decreases from V

0 = 1 m3 to V

f = 0

1V

2.

The amount of heat lost by the gas in this process is given by a kJ. Fill a in OMR sheet.

2 eksy vkn'kZ xSl] ftldh fu;r vk;ru ij fof'k"V Å"ek (CV) = 2R gS] dks fu;r ok;qe.Myh; nkc P

0 = 105 Pa ij

B.Mk fd;k tkrk gSA blds ifj.kkeLo:i bldk vk;ru V0 = 1 m3 ls V

f = 0

1V

2 rd ?kV tkrk gSA bl izfØ;k esa xSl

}kjk Å"ek gkfu a kJ gks rks a Kkr dhft,A5. Water is filled in a rectangular vessel of dimension 4m × 3m × 2m. Now the container starts to move with

uniform acceleration, a = 1.25 m/s2 at t = 0. The volume of liquid in vessel is 1.8m3. Net horizontal forceacting on the liquid by the container is given by 10a newton. Fill the value of a.vkdkj 4m × 3m × 2m okys ,d vk;rkdkj ik= esa ty Hkjk tkrk gSA vc t = 0 ij ;g ik= a = 1.25 m/s2 le:i Roj.kls xfr djuk izkjEHk dj nsrk gSA ik= esa æo dk vk;ru 1.8m3 gSA ;fn ik= }kjk æo ij yxk;k x;k ifj.kkeh {kSfrt cy10a U;wVu gks rks a Kkr dhft,A

a4m

2m



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PAPER – 1

01CT113053KOTA - 22/36


[k.M-IV : (iw.kk±d eku lgh izdkj)

This section contains 4 questions. The answer to each question is a single digit Integer, ranging from

0 to 9 (both inclusive)

bl [k.M esa 4 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. A tube in the shape of a rectangle with rounded corners is placed in a vertical plane, as shown in figure.

Two ball bearings are introduced at the upper right-hand corner. One travels by path AB and the other

by path CD to arrive at the lower left-hand corner. If the time taken by them is tAB

and tCD

respectively,

find the value of tAB

/tCD

.

fp= esa iznf'kZr ,d vk;rkdkj uyh Å/okZ/kj ry esa j[kh gqbZ gS rFkk blds fljs fp=kuqlkj eqM+s gq, gSaA nks fc;fjax xsanksa dks

Åijh nk¡;s fljs ls izfo"V djk;k tkrk gSA buesa ls ,d iFk AB ls rFkk nwljh iFk CD ls gksrs gq, fupys ck¡;s fljs rd tkrh

gSA ;fn buds }kjk fy;k x;k le; Øe'k% tAB

,oa tCD

gks rks tAB

/tCD

dk eku Kkr dhft,A

15°

A

D

C

B

2. A long string of mass per unit length 0.2 kg m–1 is stretched to a tension of 500 N. The power required

to maintain a travelling wave of amplitude 9 mm and wavelength 0.5 m is P. The string is joined to another

string of mass per unit length 0.8 kg m–1. The amplitude (in mm) of transmitted wave.

,d yEch jLlh ftldk izfr bdkbZ yEckbZ dk æO;eku 0.2 kg m–1 gS] dks 500 N ruko ds v/khu f[kapk tkrk gSA bl jLlh

ij 9 mm vk;ke rFkk 0.5 m rjaxnS/; Z okyh izxkeh rjax dks cuk;s j[kus ds fy, vko';d 'kfä P gSA bl jLlh dks

0.8 kg m–1 bdkbZ yEckbZ æO;eku okyh ,d vU; jLlh ds lkFk tksM + fn;k tkrk gSA ikjxfer rjax dk vk;ke (mm esa)

Kkr dhft,A



PHYSICSTM


PAPER – 1

KOTA - 23/3601CT113053

3. Two trains A and B are travelling in opposite directions along straight parallel tracks at the same speed

v = 60 km/h. A light air plane crosses above them. A person on train A sees it cross at right angles, while

a person on train B sees it cross the track at an angle q = 30°. If angle at which the airplane crosses the

tracks as seen from the ground is tan–12æ ö

ç ÷aè ø then find the value of a.

nks Vªsusa A rFkk B lh/ks lekUrj iFkksa ij foijhr fn'kk esa v = 60 km/h dh leku pky ls xfr'khy gSaA ,d gYdk ok;q;ku

buds Åij ls gksdj xqtjrk gSA Vªsu A esa cSBk ,d O;fä bls ledks.k ij xqtjrs gq, ns[krk gS tcfd Vªsu B esa cSBk O;fä

bls q = 30° dks.k ij iFk dks ikj djrs gq, ns[krk gSA ;fn /kjkry ls ns[kus ij ;g ok;q;ku iFkks a dks tan–12æ ö

ç ÷aè ø dks.k

ij ikj djrs gq, fn[kkbZ nsrk gks rks a Kkr dhft,A

4. A wire breaks down under a stress of 105 Pa. If the density of the wire is 2 × 103 kg/m3, find the

minimum length (in m) of the wire which will break under its own weight.

,d rkj 105 Pa izfrcy yxkus ij VwV tkrk gSA ;fn rkj dk ?kuRo 2 × 103 kg/m3gks rks rkj dh og U;wure yEckbZ

(m esa) Kkr dhft;s tks vius Lo;a ds Hkkj ds dkj.k VwV tk;sxhA


CHEMISTRYTARGET : JEE (Main + Advanced) 2015(DATE : 16-02-2014)

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PAPER – 1

01CT113053KOTA - 24/36

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

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

lgh gSA1. For the given endothermic reaction, A(g) 2B(g)

The variation in concentration due to different changes is plotted.Neglect the slope of change in concentration when system approaches equilibrium. Assume thatchanges are carried out very fast.

effcet-I effcet-II effcet-III

A

B

Time

Con

cent

ratio

n

Give the correct order of initials T (true) of F (false) for following statements.(a) Effect-I is decrease in temperature at constant volume(b) Effect-II is decrease in total equilibrium pressure by changing volume.(c) Effect-III is addition of B only at constant volume(A) FTT (B) TFT (C) TFF (D) TTTnh x;h Å"ek'kks"kh vfHkfØ;k ds fy,,

A(g) �� 2B(g)fofHkUu ifjorZuksa ds dkj.k lkaærk esa ifjorZu dks vkjsf[kr fd;k x;kAtc ra= lkE; ij igqaprk gS rks lkaærk es a ifjorZu dh <+ky ux.; gSA ekuk fd ifjorZu rhoz gk srk gSA

le;

lkUæ

rk

-II -III -IV

fuEu dFkuksa ds fy, T (lR;) rFkk F (vlR;) ds lgh Øe dks (izkjEHk ls) fyf[k,sA(a) izHkko- I fu;r vk;ru ij rki esa deh djrk gS(b) izHkko- II vk;ru ifjorZu }kjk dqy lkE; nkc esa deh djrk gSA(b) izHkko - III fu;r vk;ru ij dsoy B dk ;ksx djrk gS(A) FTT (B) TFT (C) TFF (D) TTT



CHEMISTRYTM


PAPER – 1

KOTA - 25/3601CT113053

2. A2Ox is oxidised to AO3– by MnO4

– in acidic medium. If 1.5 × 10-3 mole of A2Ox requires 40 ml of 0.03

M-KMnO4 solution in acidic medium. Which of the following statement(s) is/are correct?

(A) The value of "x" = 1

(B) The value of "x" = 3

(C) Empirical formula of oxide is AO3.

(D) Empirical formula of oxide is A2O.

vEYkh; ek/;e esa A2Ox dks AO3– esa MnO4

– }kjk vkWDlhd̀r fd;k tkrk gS] ;fn vEyh; ek/;e esa 1.5 × 10-3 eksyA2Ox dks 0.03 M-KMnO4 foy;u ds 40 ml dh vko';drk gksrh gS] rks fuEu esa ls dkSulk dFku lgh gS ?

(A) "x" dk eku = 1 gS(B) "x" dk eku = 3 gS(C) vkWDlkbM dk ewykuwikrh lw= AO3 gS(D) vkWDlkbM dk ewykuwikrh lw= A2O gS

3. Find the ions is not having X–O–X type of linkage

fuEu esa og vk;u crkb;s ftlesa X–O–X izdkj dk ca/ku mifLFkr ugha gSa]

(A) S2O32– (B) (P3O9)

3– (C) Si2O76– (D) H2P2O5

2–

4. Least stable resonating structure of acrolein (Prop-2-enal) is -

,Øksyhu (izksi&2&busy) dh U;wure LFkk;h vuquknh lajpuk gS&

(A) O (B) O+

–(C)

O+

–(D) O+

–



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PAPER – 1

01CT113053KOTA - 26/36





vf/kd lgh gSA

5. Which of the following is the nodal plane of dxy orbital ?

fuEu esa ls dkSulk dxy d{kd dk uksMy ry gS ?

(A) XY (B) YZ (C) ZX (D) All (lHkh)

6. Find the correct ionic mobility order from the following options-

fuEu esa vk;fud xfr'khyrk dk lgh Øe gS&

(A) Be2+ < Li+ (B) Mg2+ < Sr2+ (C) Fe2+ > Fe3+ (D) Br– < I–

7. Choose the correct order for ionisation energy -

vk;uu ÅtkZ dk lgh Øe pqfu,s&

(A) Tl > Al (1st IE) (B) Pb > Sn (1st IE) (C) Cu > Zn (2nd IE) (D) Ca > K (2nd IE)

8. Compound(s) which have lesser pKa than water

;kSfxd ftlds pKa dk eku ty dh rqyuk esa de gS&

(A) COOH (B) H–CºC–H (C) OH (D) OH


CHEMISTRYTM


PAPER – 1

KOTA - 27/3601CT113053


(iii) Paragraph Type (iii) vuqPNsn izdkj

This section contains 2 paragraphs each describing theory, experiment, data etc. Four questions relateto two paragraphs with two questions on each paragraph. Each question of a paragraph has only onecorrect 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 gaSA nksuksa vuqPNsn ls lacaf/kr pkj iz'ugSa] ftuesa ls gj vuqPNsn ij nks iz'u gSaA vuqPNsn esa gj iz'u ds pkj fodYi (A), (B), (C) vkSj (D) gSa ftuesals dsoy ,d lgh gSA



In a H-like species there are two energy levels A and B above the ground state having principal quantum

numbers of n1 and n2 respectively. A sample of this H-like species has all atoms/ions in excited levels A

or B only and none in any other energy level. Energy of level B is greater than that of level A and a total

of 15 different lines are emitted from this sample on returning to ground state out of which 6 lines are

emitted due to electronic transitions between the levels n1 and n2 only. Also energy difference between

levels n2 and n1, En2

– En1 = 4.53 eV.

,d H-ds leku Lih'kht esa nks ÅtkZ Lrjksa A rFkk B, tks vk| voLFkk ls Åij gS] dh eq[; Dok.Ve la[;k,sa Øe'k%n1 rFkk n2 gSA bl H-ds leku Lih'kht ds ,d uewus esa lHkh ijek.kq@vk;u dsoy A ;k B mÙksftr voLFkk esa gSvkSj vU; fdlh Hkh ÅtkZ Lrj esa ugha gSA Lrj B dh ÅtkZ Lrj A dh ÅtkZ ls vf/kd gS vkSj vk| voLFkk esa ykSVusij bl uewus ls mRlftZr dqy 15 fHkUu js[kkvks a esa ls 6 js[kk,sa dsoy n1 rFkk n2 Lrjksa ds e/; bysDVªksuh; laØe.kds dkj.k gksrh gS n2 rFkk n1Lrjks a ds e/; ÅtkZ vUrj En2

– En1 = 4.53 eV gSA

9. How many lines will be emitted by this sample in Balmer series ?

ckej Js.kh esa bl uewus }kjk fdruh js[kk,sa mRlftZr gksxhA(A) 6 (B) 5 (C) 4 (D) 3

10. The hydrogen like species for which data is given in above passage is -

gkbMªkstu ds leku fdl Lih'kht ds fy, mijksDr vuqPNsn esa vk¡dM+s fn;s x;s gSA(A) He+ (B) Li+2 (C) Be+3 (D) B+5


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CH –CH2 2

H

OH

(e)

(d)

SH(c)

H(b)

H N2

(a)

11. Correct order of basic strength-

{kkjh; lkeF;Z dk lgh Øe gS&

(A) e > b > c (B) b > e > c (C) a > d > c (D) c > e > b

12. Correct order of acidic strength -

vEyh; lkeF;Z dk lgh Øe gS&

(A) a > d > e (B) a > d > c (C) c > d > a (D) a > c > d




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bl [kaM esa 5 iz'u gSaA gj iz'u dk mÙkj rhu vad dk iw.kk ±d] 000 ls 999 rd ] gSA1. 10 ml of CO is mixed with 25 ml air (20% O2 by volume) in a container at 1 atm. Find final volume

(in ml) of container at 1 atm after complete combustion. (Assume that temperature remain constant).,d ik= esa 1 atm ij 10 ml CO dks 25 ml ok;q (vk;ru dk 20% O2 gS) ds lkFk fefJr fd;k x;kA lEiw.kZ nguds i'pkr ,d 1 atm ij ik= dk vk;ru (ml esaa) Kkr dhft,A (eku fyft,s rki fu;r jgrk gSa ).

2. In the ionic reaction, (vk;fud lehdj.k esa)xBrO3

– + yCr3+ + zH2O ® pBr2 + qHCrO4– + rH+ {x, y, z, p, q, r Î N}

Find the value of x + y + z : (x + y + z dk eku Kkr dhft,)3. Find the number of isoelectronic species from the following having 14 electrons.

fuEu esa ls , slh lebysDVªkWfu; Lih'kht dh la[;k Kkr dhft, ftuesa 14 bysDVªkWu mifLFkr gSaAMg2+ , Na+, N3– , S2– , K+, CN– , N2 , NO+, PH3, P

+


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4. Number of carbanions which are more stable than 2 3CH CH-1

fuEu esa ls ,sls dkcZ½.kk;uksa dh la[;k crkb;s tks 2 3CH CH-1

dh rqyuk esa vf/kd LFkk;h gSA

(i) 3CH

1

(ii) 3CF1

(iii) CH(CH )3 2

1(iv) CCl3

1

(v) 2 3CH CCl-1

(vi) 2 2 3CH CH CH- -1

(vii) 2CH NÅ

-1

(CH3)3 (viii) C(CH )3 3

1

5. Possible number of compounds with IUPAC name P1-bromo -P2-methyl propanoic acid. Where Pi

represents position of side chains/substituents.

IUPAC uke] P1-czk seks-P2-esfFky izk sisuk sbd vEy] j[kus okys ;kSfxdksa dh lEHkkfor la[;k D;k gS tgk¡ Pi

ik'oZ Ja`[kyk @ izfrLFkkidksa dh fLFkfr dks izn£'kr djrk gSA


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[k.M-IV : (iw.kk±d eku lgh izdkj)This section contains 4 questions. The answer to each question is a single digit Integer, ranging from0 to 9 (both inclusive)

bl [k.M esa 4 iz'u gSaA izR;sd iz'u dk mÙkj 0 ls 9 rd (nksuksa 'kkfey) ds chp dk ,dy vadh; iw.kk ±d

gSA1. 1 mol N2 and 3 moles of H2 combines in a 24.63 L cylinder maintained at 27°C according to the equation

N2(g) + 3H2(g) 2NH3(g)

find moles of NH3 produced if pressure inside the cylinder is 3 atm. at equilibrium.fuEu vfHkfØ;k ds vuqlkj 27ºC ij fu;af=r 24.63 yhVj ds ,d csyukdkj ik= esa 1 mol N2 rFkk 3 mol H2 dksfeyk;k x;k gSA

N2(g) + 3H2(g) 2NH3(g)

;fn lkE; ij flys.Mj ds vUnj dk nkc 3atm gks] rks cuus okyh NH3 ds eksyksa dks Kkr dhft,A2. Find the species having lone pair to bond pair ratio equals to 1

fuEu esa , slh Lih'kht dh la[;k crkb;s ftuesa ,dkadh bysDVªkWu ; qXe dk] cU/k ; qXe ls vuqikr 1 gSACO2 , BF4

– , CCl4 , SF2 , XeO4 , SO3, SeO42–



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3. Find the number of electron having (n × l + m) = 3 for Kr-atom (At. Number : Z = 36)

Kr-ijek.kq (ijek.kq Øekad% Z = 36) ds fy, ,sls bysDVªkWuksa dh la[;k Kkr dhft, ftuds fy, (n × l + m) = 3,

gSA

4. How many of the following are soluble in aq.NaOH

fuEu esa ls fdrus ;kSfxd tyh; NaOH esa foys; gSA

(a)

OH(b)

OH(c)

NH2

NO2

(d)

CH3

(e)

COOH

(f) (g)

SO3H

(h)



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NURTURE COURSE 16-02-2014TM


KOTA - 35/3601CT113053

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16-02-2014TARGET : JEE (Main + Advanced) 2015TM


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Kota | Chandigarh | Ahmedabad 01CT113053KOTA - 36/36

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

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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

11. [kaM–I(i) Hkkx esa 4 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 okys2 vuqPNsn gSA nksuksa vuqPNsnksa ls lacfU/kr pkj iz'u gSaAftuesa ls gj vuqPNsn ij nks 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 es a ,d Hkh iz'u ugha gSA13. [kaM-III es a 5 iz'u gSaA izR;sd iz'u dk mÙkj 000 ls 999 rd

(nksuksa 'kkfey) ds chp dk rhu vadksa dk iw.kk±d gSA14. [kaM-IV es a 4 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) 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

17. [kaM-III ds gj iz'u esa dsoy lgh mÙkjksa (mÙkj) okys lHkh cqycqyksa(cqycqys) dks dkyk djus ij 4 vad vkSj dksbZ Hkh cqycqykdkyk ugha djus ij 'kwU; (0) vad iznku fd;k tk;sxkA vU;lHkh fLFkfr;ksa esa ½.kkRed ,d (–1) vad iznku fd;k tk;sxkA

18. [kaM-IV 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

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

11. SECTION – I(i) Contains 4 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. Four questions relate totwo paragraphs with two 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-II13. Section-III contains 5 questions The answer to each

question is a three digit integer, ranging from000 to 999 (both inclusive)

14. Section-IV contains 4 questions The answer to eachquestion 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 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

17. For each question in Section-III, you will be awarded4 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

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

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

test 3 paper 1

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