allen clas amme · 2015-12-21 · physics ka/00ct214004 5/44 space for rough work / dpps dk;z ds...

A. lkekU; / General :1. ;g iqfLrdk vkidk iz'u&i= gSA bldh eqgj rc rd u rksM+s tc rd fujh{kd ds }kjk bldk funZs'k u fn;k tk;sA

This booklet is your Question Paper. Do not break the seal of this booklet before being instructed to do so by theinvigilator.

2. iz'u&i= dk dksM (CODE) bl i`"B ds Åijh nk;sa dkSus ij Nik gSAThe question paper CODE is printed on the right hand top corner of this sheet.

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;sxkABlank spaces and blank pages are provided in the question paper for your rough work. No additional sheets willbe provided for rough work.

4. dksjs dkxt] fDyi cksMZ] ykWx rkfydk] LykbM :y] dSYdqysVj] dSejk] lsyQksu] istj vkSj fdlh Hkh izdkj ds bysDVªkWfud midj.k dhijh{kk d{k esa vuqefr ugha gSaABlank papers, clipboards, log tables, slide rules, calculators, cameras, cellular phones, pagers and electronicgadgets of any are NOT allowed inside the examination hall.

5. bl iqfLrdk ds fiNys i`"B ij fn, x, LFkku esa viuk uke vkSj QkWeZ uEcj fyf[k,AWrite your name and Form number in the space provided on the back cover of this booklet.

6. mÙkj i=] ,d ; a=&Js.khdj.k ;ksX; i= (ORS) gS tks fd vyx ls fn;s tk;saxsAThe answer sheet, a machine-readable Optical Response Sheet (ORS), is provided separately.

7. vks-vkj-,l-(ORS) ;k bl iqfLrdk esa gsj&Qsj@foÏfr u djsa / DO NOT TAMPER WITH/MUTILATE THE ORS OR THIS BOOKLET.8. bl iqfLrdk dh eqgj rksM+us ds i'pkr d`i;k tk¡p ysa fd blesa 44 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 ds izkjEHk esa fn;s gq, funsZ'kksa dks /;ku ls i<+ sAOn breaking the seal of the booklet check that it contains 44 pages and all the 20 questions in each subject andcorresponding answer choices are legible. Read carefully the instructions printed at the beginning of each section.

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

A candidate has to write his / her answers in the ORS sheet by darkening the appropriate bubble with the help ofBlack ball point pen as the correct answer(s) of the question attempted.

10. ORS ds (i`"B la[;k 1) ij ekaxh xbZ leLr tkudkjh /;ku iwoZd vo'; Hkjsa vkSj vius gLrk{kj djsaAWrite all information and sign in the box provied on part of the ORS (Page No. 1).

C. iz'ui= dk izk:i / Question Paper Formate :bl iz'u&i= ds rhu Hkkx (HkkSfrd foKku] jlk;u foKku vkSj xf.kr) gSaA gj Hkkx ds rhu [kaM gSaAThe question paper consists of 3 parts (Physics, Chemistry and Mathematics). Each part consists of three sections.

11. [kaM–I / SECTION – I(i) Hkkx esa 8 cgqfodYi iz'u gSaA gj iz'u esa pkj fodYi (A), (B), (C) vkSj (D) gSa ftuesa ls ,d ;k vf/kd lgh gSaA

Contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of whichONE or MORE are correct.

(ii) Hkkx esa 4 cgqfodYi iz'u gSaA izR;sd iz'u esa lqesyu lwpha gSA lwfp;ksa ds fy, dksM ds fodYi (A), (B), (C) vkSj (D) gSa ftuesals dsoy ,d lgh gSaAContains 4 multiple choice questions. Each question has matching lists. The codes for the lists have choices(A), (B), (C) and (D) out of which ONLY ONE is correct

12. [kaM–II esa ,d Hkh iz'u ugha gSA / There is no questions in SECTION-II13. [kaM-III es a 4 iz'u gSaA izR;sd iz'u dk mÙkj 000 ls 999 rd (nksuksa 'kkfey) ds chp dk rhu vad dk iw.kk±d gSA

Section-III contains 4 questions The answer to each question is a three digit integer, ranging from 000 to 999(both inclusive)

14. [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 gSASection-IV contains 4 questions The answer to each question is a single digit integer, ranging from 0 to 9(both inclusive)

funs Z'k / INSTRUCTIONS

Path to Success

ALLENCAREER INSTITUTEKOTA (RAJASTHAN)

T M

CLASSROOM CONTACT PROGRAMME(ACADEMIC SESSION 2014-2015)

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

PAPER CODE 0 0 C T 2 1 4 0 0 4

DO N

OT B

REAK

THE

SEA

LS W

ITHO

UT B

EIN

G IN

STRU

CTED

TO

DO S

O BY

THE

INVI

GILA

TOR

\ fuj

h{kd

ds v

uqns'k

ksa ds fc

uk e

qgjsa u

rksM

+s

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.

le; : 3 ?k.Vs egÙke vad : 180Time : 3 Hours Maximum Marks : 180

PATTERN : JEE (Advanced)TEST TYPE : MAJOR

ENTHUSIAST & LEADER COURSE

Date : 08 - 02 - 2015TARGET : JEE (Advanced) 2016

ALL INDIA OPEN TEST #01

isij – 2PAPER – 2

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

Hkkx-1 HkkSfrd foKku I(i) ,d ;k vf/kd lgh fodYi izdkj 03 - 09Part-1 Physics One or More Options Correct Type

I(ii) lqesyu lwpha izdkj 10 - 14Matching List Type

III iw.kk±d eku lgh izdkj (000 ls 999) 15 - 16Integer Value Correct Type (000 to 999)

IV iw.kk±d eku lgh izdkj (0 ls 9) 17 - 19Integer Value Correct Type (0 to 9)

Hkkx-2 jlk;u foKku I(i) ,d ;k vf/kd lgh fodYi izdkj 20 - 22Part-2 Chemistry One or More Options Correct Type




Hkkx-3 xf.kr I(i) ,d ;k vf/kd lgh fodYi izdkj 32 - 34Part-3 Mathematics One or More Options Correct Type




Kota/00CT2140042/44

ALL INDIA OPEN TEST/JEE (Advanced)/08-02-2015/PAPER-2

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 constantpe

9

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


PHYS

ICS

3/44Kota/00CT214004

PART-1 : PHYSICS Hkkx-1 : HkkSfrd foKku

SECTION–I : (i) One or more options correct Type [k.M-I : (i) ,d ;k vf/kd lgh fodYi izdkj

This section contains 8 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 8 cgqfodYi iz'u gSaA izR;sd iz'u esa pkj fodYi (A), (B), (C) vkSj (D) gSa] ftuesa ls ,d ;kvf/kd lgh gSA

1. A charged particle having its charge to mass ratio as b goes in a conical pendulum of length L makingan angle q with vertical and angular velocity w. If a magnetic field B is directed vertically downwards(see figure) :

(A) é ù= w-ê úb w që û

1 gB

L cos(B) Angular momentum of the particle about the point of suspension remains constant.(C) If the direction of B were reversed maintaining same w and L, then q will remain unchanged.(D) Rate of change of angular momentum of the particle about the point of suspension is not a constant

vector.

L q

w

B

,d vkosf'kr d.k ftldk vkos'k&æO;eku vuqikr b gS] fp=kuqlkj yEckbZ L okys ,d 'kaDokdkj yksyd esa xfr djrk

gSA ;g ÅèokZ/kj ls q dks.k cukrk gS rFkk bldk dks.kh; osx w gSA ;fn ;gk¡ ,d pqEcdh; {ks= B Å/okZ/kj uhps dh fn'kk

esa dk;Zjr gks rks %&

(A) é ù= w-ê úb w që û

1 gB

L cos

(B) fuyEcu fcUnq ds lkis{k d.k dk dks.kh; laosx fu;r cuk jgrk gSA

(C) ;fn w o L ds eku ogh j[ks tk;sa rFkk B dh fn'kk dks O;qRØfer dj nsa rks q dk eku ifjofrZr ugha gksxkA

(D) fuyEcu fcUnq ds lkis{k d.k ds dks.kh; laosx esa ifjorZu dh nj ,d fu;r lfn'k ugha gSA

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

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

4/44

PHYS

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Kota/00CT214004

2. In the a-decay of a U-238 nucleus the energy released in the decay is Q. The U-238 nucleus was initially

stationary. Which of the following is (are) true?

(A) Ratio of K.E. of a-particle and Thorium nucleus is 117 : 2

(B) Ratio of K.E. of Thorium nucleus and a-particle is 1 : 234

(C) Momentum of a-particle is a234Qm119

(D) Recoil velocity of Thorium nucleus is ´ Th

234Q119 117m

U-238 ukfHkd ds a-fo?kVu esa mRlftZr ÅtkZ dk eku Q gSA U-238 ukfHkd izkjEHk esa fLFkj FkkA lgh dFku@dFkuksa dks

pqfu, %&

(A) a-d.k rFkk Fkksfj;e ukfHkd dh xfrt ÅtkZ dk vuqikr 117 : 2 gSA

(B) Fkksfj;e ukfHkd rFkk a-d.k dh xfrt ÅtkZ dk vuqikr 1 : 234 gSA

(C) a-d.k dk laosx a234Qm119

gSA

(D) Fkksfj;e ukfHkd dk izfrf{kIr osx ´ Th

234Q119 117m gSA



PHYS

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5/44Kota/00CT214004


3. Given a square frame of diagonal length 2r made of insulating wires. There is a short dipole, having dipolemoment P, fixed in the plane of the figure lying at the center of the square, making an angle q as shownin figure. Four identical particles having charges of magnitude q each and alternatively positive andnegative sign are placed at the four corners of the square. Select the correct alternative (s) :-

(A) Electrostatic force on the system of four charges due to dipole is 3

6kPq

r

(B) Electrostatics force on the system of four charges due to dipole is 3

6kPqcos

rq

(C) Net torque on the system of four charges about the centre of the square due to dipole is zero

(D) Net torque on the system of four charges about the centre of the square due to dipole is 23kPq

r

q

r r

r r

P

–q +q

–q+q

fp= esa dqpkyd rkjksa ls cus ,d oxkZdkj Ýse dks n'kkZ;k x;k gS] ftldh fod.kZ yEckbZ 2r gSA ;gk¡ fp= ds ry esa oxZ ds

dsUæ ij f}/kzqo vk?kw.kZ P okyk ,d y?kq f}/kzqo fp=kuqlkj q dks.k ij fo|eku gSA ,dkUrj :i ls /kukRed rFkk ½.kkRed

fpUg~ rFkk izR;sd q ifjek.k okys pkj ,d tSls d.k bl oxZ ds pkjks a dksuksa ij j[k fn, tkrs gSaA lgh dFku@dFkuks a dks

pqfu, %&

(A) f}/kz qo ds dkj.k pkjksa vkos'kksa ds fudk; ij fLFkj oS|qr cy 3

6kPq

r gksxkA

(B) f}/kz qo ds dkj.k pkjksa vkos'kksa ds fudk; ij fLFkj oS|qr cy 3

6kPqcos

rq gksxkA

(C) f}/kzqo ds dkj.k oxZ ds dsUæ ds lkis{k pkjksa vkos'kksa ds fudk; ij dqy cyk?kw.kZ 'kwU; gksxkA

(D) f}/kz qo ds dkj.k oxZ ds dsUæ ds lkis{k pkjksa vkos'kksa ds fudk; ij dqy cyk?kw.kZ 23kPq

r gksxkA

6/44

PHYS

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Kota/00CT214004

4. Figure shows stationary orbit of an hydrogen atom upon the transition of electron from given excited

state to ground state.

(A) Average change in angular momentum is 3h2p

(B) The ratio of de-Broglie wavelengths in final state to initial state is 4

(C) Energy of emitted photon is nearly 12.75 eV.

(D) Ratio of orbital time period in final state to initial state is 43

fp= esa ,d gkbMªkstu ijek.kq dh fLFkj d{kk dks n'kkZ;k x;k gS ftlesa bysDVªkWu nh xbZ mÙksftr voLFkk ls ewy voLFkk esa

laØe.k djrk gS %&

(A) dks.kh; laosx esa vkSlr ifjorZu dk eku 3h2p

gSA

(B) vfUre voLFkk rFkk izkjfEHkd voLFkk esa Mh&czksXyh rjaxnS/; Z dk vuqikr 4 gSA

(C) mRlftZr QksVkWu dh ÅtkZ yxHkx 12.75 eV gSA

(D) vfUre voLFkk rFkk izkjfEHkd voLFkk esa d{kh; vkorZdky dk vuqikr 43 gSA

5. A radioactive sample decays by three modes simultaneously. Half lives corresponding to these modes

are in G.P. and half life of sample is 10 years. When the sample decays by the mode having largest half

life it takes 70 years for the sample to become half. If sample decays exclusively by other modes possible

values of half life is :

(A) 28 years (B) 35 years (C) 14 years (D) 17.5 years

,d jsfM;kslfØ; inkFkZ ,d lkFk lEiUu gksus okyh rhu fo/kkvksa }kjk fo?kfVr gksrk gSA bu fo/kkvksa ds laxr v/kZ&vk;q ,d

xq.kksÙkj Js.kh esa gS rFkk inkFkZ dh v/kZ&vk;q 10 o"kZ gSA ;fn bl inkFkZ dk lokZf/kd v/kZ&vk;q okyh fo/kk }kjk fo?kVu gksrk

gS rks bl inkFkZ dks vk/kk gksus esa 70 o"kZ dk le; yxrk gSA ;fn inkFkZ dk vU; fo/kkvksa }kjk fo?kVu gksrk gS rks v/kZ&vk;q

ds laHkkfor eku gksxsa %&

(A) 28 o"kZ (B) 35 o"kZ (C) 14 o"kZ (D) 17.5 o"kZ



PHYS

ICS

7/44Kota/00CT214004

6. A ring of mass m, radius R, cross sectional area A and Young's modulus Y is kept on a smooth cone ofradius 2R and semi vertical angle 45°, as shown in the figure. Assume that the extension in the ring issmall :-

(A) The tension in the ring will be same throughout

(B) The tension in the ring will be independent of the radius of ring.

(C) The extension in the ring will be mgRAY

(D) Elastic potential energy stored in the ring will be p

2 2m g R8 YA

2R

45

ring

æO;eku m, f=T;k R, vuqizLFk dkV {ks=Qy A rFkk ;ax xq.kkad Y okyh ,d oy; dks f=T;k 2R rFkk v/kZ'kh"kZ dks.k 45°okys ,d fpdus 'kadq ij fp=kuqlkj j[kk tkrk gSA ekuk oy; esa gksus okyk foLrkj vYi gS %&

(A) oy; esa ruko loZ= leku gksxkA

(B) oy; esa ruko dk eku bldh f=T;k ij fuHkZj ugha djrkA

(C) oy; esa mRiUu foLrkj mgRAY

gksxkA

(D) oy; esa lafpr izR;kLFk fLFkfrt ÅtkZ p

2 2m g R8 YA

gksxhA


8/44

PHYS

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Kota/00CT214004

7. Following are equations of four waves where x, y, z represent rectangular coordinate system :

(i) y1 = a sin

w

xtn

æ ö-ç ÷è ø (ii) y2 = a cos

w

xtn

æ ö+ç ÷è ø

(iii) z1 = a sin

w

xtn

æ ö-ç ÷è ø (iv) z2 = a cos

w

xtn

æ ö+ç ÷è ø

Which of the following statements is/are CORRECT ?

(A) On superposition of waves (i) and (iii), a travelling wave having amplitude aÖ2 will be formed.

(B) Superposition of waves (ii) and (iii) is not possible.

(C) On superposition of waves (i) and (ii), a transverse stationary wave having maximum amplitude

aÖ2 will be formed.

(D) On superposition of waves (iii) and (iv), a transverse stationary wave will be formed.

pkj rjaxksa dh lehdj.ksa fuEufyf[kr gSa] tgk¡ x, y, z ledksf.kd funsZ'kkad fudk; dks n'kkZrs gSA

(i) y1 = a sin

w

xtn

æ ö-ç ÷è ø (ii) y2 = a cos

w

xtn

æ ö+ç ÷è ø

(iii) z1 = a sin

w

xtn

æ ö-ç ÷è ø (iv) z2 = a cos

w

xtn

æ ö+ç ÷è ø

fuEu esa ls dkSulk@dkSuls dFku lR; gSa\

(A) rjaxksa (i) o (iii) ds v/;kjksi.k ls aÖ2 vk;ke okyh izxkeh rjax cusxhA

(B) rjaxksa (ii) o (iii) dk v/;kjksi.k laHko ugha gSA

(C) rjaxksa (i) o (ii) ds v/;kjksi.k ls aÖ2 vf/kdre vk;ke okyh vuqizLFk vizxkeh rjax cusxhA

(D) rjaxksa (iii) o (iv) ds v/;kjksi.k ls ,d vuqizLFk vizxkeh rjax cusxhA



PHYS

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9/44Kota/00CT214004

8. Inside a uniform sphere of mass M (M is mass of complete sphere) and radius R, a cavity of radius R/3

is made in the sphere as shown :-

(A) Gravitational field inside the cavity is uniform

(B) Gravitational field inside the cavity is non-uniform

(C) The escape velocity of a particle projected from centre of cavity is 88GM45R

(D) The escape velocity of a particle projected from centre of cavity is 20GM

9R

R/3

V

æO;eku M, f=T;k R okys le:i xksys (M iwjs xksys dk æO;eku gS) ds vUnj f=T;k R/3 okyh ,d xqfgdk fp=kuqlkj

cuk;h tkrh gS %&

(A) xqfgdk ds vUnj xq:Rokd"kZ.k {ks= le:i gSA

(B) xqfgdk ds vUnj xq:Rokd"kZ.k {ks= vle:i gSA

(C) xqfgdk ds dsUæ ls iz{ksfir d.k dk iyk;u osx 88GM45R

gSA

(D) xqfgdk ds dsUæ ls iz{ksfir d.k dk iyk;u osx 20GM

9RgSA


10/44

PHYS

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Kota/00CT214004

(ii) Matching List Type (ii) lqesyu lwpha izdkj

This Section contains 4 multiple choice questions. Each question has matching lists. The codes for thelists. have choices (A), (B), (C) and (D) out of which ONLY ONE is correct.bl [k.M esa 4 cgqfodYi iz'u gSaA izR;sd iz'u esa lqesyu lwpha gSA lwfp;ksa ds fy, dksM ds fodYi (A), (B), (C) vkSj (D)gSa ftuesa ls dsoy ,d lgh gSA

9. A non conducting piston divides an adiabatic container into two equal parts. Such that piston is inequilibrium and temperature on both side is also same as T0. Now Ist chamber is heated such that pistonmoves very slowly until volume of IInd chamber is reduced to 1/4th of the initial. In both chambers samemonatomic gas is filled such that the number of moles (n) are same. (24/3 = 2.5)

List-I List-II

(P)st

0

Work done by gas of I chambernRT (1) 27

(Q)st

0

Final temp. of II chamberT (2)

52

(R)st

0

Final temp. of I chamberT (3)

94

(S)st

0

Heat supplied to I chambernRT (4)

352

,d vpkyd fiLVu Å"ek:¼ ik= dks nks cjkcj Hkkxksa eas foHkkftr djrk gSA fiLVu lkE;koLFkk esa gS rFkk nksuksa Hkkxksa dkrkieku leku ,oa T0 ds cjkcj gSA vc izFke d{k dks bl izdkj xeZ fd;k tkrk gS fd fiLVu /khjs&/khjs bruh xfr djrkgS fd f}rh; d{k dk vk;ru izkjfEHkd eku dk ,d pkSFkkbZ gks tk,A nksuksa d{kksa esa leku ,d ijekf.od xSl Hkjh tkrh gSrFkk v.kqvksa dh la[;k n Hkh leku gSA (24/3 = 2.5)

lwph&I lwph&II

(P)0nRT

izFke d{k dh xlS }kjk fd;k x;k dk;Z(1) 27

(Q)0T

f}rh; d{k dk vfUre rkieku(2)

52

(R)0T

izFke d{k dk vfUre rkieku(3)

94

(S)0nRT

izFke d{k dks nh xbZ Å"ek(4)

352

Codes :P Q R S

(A) 3 2 4 1

(B) 2 4 1 3

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


PHYS

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11/44Kota/00CT214004

10. A train is moving with speed u. Engine of train blows a long whistle. The sound of whistle is composedof components varying from f

1 to f

2 as shown in figure. List-I shows the distribution of the sound

intensity of the whistle as observed by different observers. Match List-I with List-II.,d Vªsu u pky ls xfr'khy gSA Vªsu dk batu yEch lhVh ctkrk gSA lhVh dh /ofu fp=kuqlkj f

1 ls f

2 rd ifjofrZr gksus

okys ?kVdksa ls feydj cuh gSA lwph-I esa fofHkUu izs{kdksa }kjk izsf{kr lhVh dh /ofu rhozrk ds forj.k dks n'kkZ;k x;k gSAlwph&I dk lwph-II ls feyku dhft,A

f1 f2 frequency

Intensity

List-I/lwph&I List-II/lwph&II

(P)

f1 f2 frequency

Intensity

(1) Observer in train

izs{kd Vªsu esa gSA

(Q)

f1 f2 frequency

Intensity(2) Observer on ground close to track and infront of train

izs{kd /kjkry ij Vªsd ds lehi rFkk Vªsu ds lkeus gSA

(R)

f1 f2 frequency

Intensity

(3) Observer on ground close to track and behind train

izs{kd /kjkry ij Vªsd ds lehi rFkk Vªsu ds ihNs gSA

(S)

f1 f2 frequency

Intensity

(4) Observer on ground for which engine approachesand then recedes

izs{kd /kjkry ij gS rFkk batu izs{kd dh vksj vkrk gS rFkk fQj

blls nwj tkrk gSA

Codes :P Q R S

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

12/44

PHYS

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Kota/00CT214004

11. A spherical ball just able to complete a vertical circular track as shown in figure. List-I gives variouspositions of ball as shown in figure. Match list-I with list-II.

,d xksykdkj xsan fp=kuqlkj fdlh Å/okZ/kj o`Ùkkdkj iFk dks iw.kZ djus esa l{ke gSA lwph&I esa xsan dh fofHkUu fLFkfr;k¡

n'kkZ;h xbZ gSA budk lwph&II ls feyku dhft, %&

BD

C

A

List-I/lwph&I List-II/lwph&II(P) A (1) Friction acts vertically down

?k"kZ.k Å/okZ/kj uhps dh vksj dk;Zjr gSA(Q) B (2) Friction acts vertically up

?k"kZ.k Å/okZ/kj Åij dh vksj dk;Zjr gSA(R) C (3) No normal reaction

dksbZ vfHkyEc izfrfØ;k ugha yxrhA(S) D (4) No friction force

dksbZ ?k"kZ.k cy ugha yxrkACodes :

P Q R S(A) 3 2 1, 4 2(B) 2 3 1 4(C) 4 2 3,4 2(D) 3 2 4 1



PHYS

ICS

13/44Kota/00CT214004

12. List-I describes various arrangements to obtain interference pattern on screen.List-I List-II

(P)S

S1

S2

O

Screen

d (1) Shape of fringes is strictly linear.

S1, S

2 are slits.

d = 0.5 l

(Q)S

S1

S2

O

Screen

d

(2) First order maxima will be seen.

S1, S

2 are slits.

d = 100.5 l

(R)S

S1

S2

O

Screen

d

(3) Shape of fringes is hyperbolic

S1, S

2 are pin holes.

d = 0.5 l

(S) O

Screen

S2S1

d

(4) Shape of fringes is circular.

S1, S

2 are coherent point sources

with zero initial phase difference.d = 2.5 l

Codes :P Q R S

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

14/44

PHYS

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Kota/00CT214004

lwph-I esa insZ ij O;frdj.k izfr:i izkIr djus ds fy, fofHkUu O;oLFkk,sa n'kkZ;h xbZ gSAlwph&I lwph&II

(P)S

S1

S2

O

Screen

d (1) fÝatks a dh vkd`fr fuf'pr :i ls js[kh; gSA

S1, S

2 fLyVsa gSaA

d = 0.5 l

(Q)S

S1

S2

O

Screen

d

(2) izFke Øe dk mfPp"B fn[kkbZ nsxkA

S1, S

2 fLyVsa gSaA

d = 100.5 l

(R)S

S1

S2

O

Screen

d

(3) fÝatks a dh vkd`fr vfrijoy;kdkj gSA

S1, S

2 lwph fNæ gSA

d = 0.5 l

(S)O

Screen

S2S1

d

(4) fÝatks a dh vkd`fr o`Ùkkdkj gSA

S1, S

2 dyk laxr fcUnq L=ksr gSa ftuesa

izkjfEHkd dykUrj dk eku 'kwU; gSAd = 2.5 l

Codes :P Q R S

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


PHYS

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15/44Kota/00CT214004

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

SECTION–III : (Integer Value Correct Type) [kaM-III : (iw.kk±d eku lgh izdkj)

This section contains 4 questions. The answer to each question is a three digit Integer, ranging from000 to 999.

bl [kaM esa 4 iz'u gSaA gj iz'u dk mÙkj rhu vad dk iw.kk ±d] 000 ls 999 rd ] gSA

1. ABCD is a square frame made from different wires of same length and each having different uniformresistance per unit length. Resistances of wires forming sides AB, BC, CD and DA are 100W, 400 W,500 W and 200 W respectively. An ideal cell is connected across B and D. A straight conducting wirecontaining a resistance and a galvanometer in series starts rotating about pivoted point A from initial

position as shown with uniform angular velocity 360

=pw rad/sec. One end of the straight wire (rotating)

is pivoted at A and other end always in sliding contact with a side of the square. The time (in second)after start when galvanometer shows zero deflection is.

fp= esa ABCD leku yEckbZ okys fofHkUu rkjksa ls cuk ,d oxkZdkj Ýse gSA izR;sd rkj dk izfr bdkbZ yEckbZ dk le:i

izfrjks/k vyx&vyx gSA Hkqtk AB, BC, CD rFkk DA dks fufeZr djus okys rkjksa dk izfrjks/k Øe'k% 100W, 400 W,

500 W rFkk 200 W gSA B o D ds e/; ,d vkn'kZ lsy tksM+k tkrk gSA ,d lh/kk pkyd rkj ftlesa ,d izfrjks/k rFkk ,d

xsYosuksehVj Js.khØe esa yxs gq, gSa] fp=kuqlkj bldh izkjfEHkd fLFkfr ls dhyd fcUnq A ds lkis{k ,dleku dks.kh; osx

360=

pw rad/sec ls ?kw.kZu djuk izkjEHk djrk gSA ?kw.kZu'khy lh/ks rkj dk ,d fljk fcUnq A ij dhydhr gS rFkk nwljk

fljk lnSo oxZ dh ,d Hkqtk ds lkFk pyk;eku lEidZ esa jgrk gSA izkjEHk ls fdrus le; (lsd.M esa) i'pkr~ xsYosuksehVj

'kwU; fo{ksi n'kkZ;sxkA

G

E

B A

D C

w

Initial position


16/44

PHYS

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Kota/00CT214004

2. The experimenteer collected 640g wet snow (mixture of ice & water) from street at a temperature of0°C, placed it in a freezer and started measuring its temperature at regular intervals, recording data(the first record was made immediately after the start of the experiment). Subsequently, however, the filehas been corrupted, so it could only read the temperature values corresponding to the tenth and eleventhrecords : –0.5°C and –4°C, respectively. Determine from these data (Assuming constant power loss) themass of water (in gm) in wet snow initially. Specific heat of ice 0.5 Cal/g°C, latent heat of ice 80 Cal/g.

,d iz;ksxdrkZ us lM+d ls 0°C rkieku okyh 640g Hkhxh gqbZ cQZ (cQZ rFkk ty dk feJ.k) mBkdj bls ,d fÝtj esa

j[k fn;k rFkk fu;fer varjky ij blds rkieku dks ntZ djuk izkjEHk dj fn;kA lcls izFke izs{k.k iz;ksx izkjEHk djus ds

rqjUr i'pkr~ fy;k x;k FkkA blh Øe esa fdlh xyrh ds dkj.k iz;ksxdrkZ dh Qkby [kjkc gks tkrh gS rFkk og dsoy nlosa

rFkk X;kjgosa izs{k.k ds laxr rkieku gh i<+ ikrk gS tks fd Øe'k% –0.5°C rFkk –4°C gSaA fu;r 'kfä âkl ekurs gq, buvk¡dM+ksa ds vk/kkj ij izkjEHk esa Hkhxh gqbZ cQZ esa ty dk æO;eku (gm esa) Kkr dhft,A cQZ dh fof'k"V Å"ek

0.5 Cal/g°C rFkk cQZ dh xqIr Å"ek 80 Cal/g yhft,A3. An x-ray tube is working at a potential difference of 20 kV. The potential difference is decreased to

10kV. It is found that difference of the wavelength of Ka x-ray and the most energetic continuous x-raydecreases by four times. Find the atomic number of the target element.

,d x-fdj.k uyh 20 kV foHkokUrj ij dk;Z djrh gSA foHkokUrj dk eku 10kV rd ?kVk fn;k tkrk gSA ;g ik;k x;k

fd Ka x-fdj.k dh rjaxnS/; Z rFkk lokZf/kd mftZr lrr x-fdj.k dh rjaxnS/; Z esa vUrj pkj xquk ?kV tkrk gSA y{; rRo

dk ijek.kqØekad Kkr dhft,A4. The final image I of the object O shown in the arrangement is formed at point 20 cm below a thin

equi-concave lens, which is at a depth of 65 cm from principal axis. Calculate the radius of curvature incm of lens kept at "A".

fp= esa iznf'kZr O;oLFkk esa fcEc O dk vfUre izfrfcEc I ,d irys leryksory ysUl ls 20 cm uhps fLFkr fcUnq ij curkgS] tks eq[; v{k ls 65 cm uhps fLFkr gSA "A" ij j[ks ysUl dh oØrk f=T;k (cm esa) Kkr dhft,A

A

20cm

m=32

65cm

45°

f=30cm

O

36cm 1m



PHYS

ICS

17/44Kota/00CT214004

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. The diameter of a wire of length 100 cm is measured with the help of a screw gauge. The main scale

reading is 1mm and circular scale reading is 25. Pitch of the screw gauge is 1 mm and the total number

of divisions on the circular scale is 100. The wire is used in an experiment for determination of Young's

modulus of the wire by Searle's method. The following data are available : elongation in the wire

Dl = 0.125 cm under the tension of 50 N, least count for measuring normal length of wire is

0.01 cm and for elongation in the wire is 0.001 cm. The maximum error in calculating the value of

Young's modulus (Y), assuming that the force is measured very accurately, is 8 n

%10´

, where n is very

nearly an integer, then fill the value of n in OMR sheet.

,d 100 cm yEcs rkj dk O;kl] LØwxst dh lgk;rk ls ekik tkrk gSA eq[; iSekus dk ikB~;kad 1mm o o`Ùkkdkj iSekus

dk ikB~;kad 25 gSA LØwxst dk pwM+h vUrjky 1 mm o o`Ùkkdkj iSekus ij Hkkxksa dh dqy la[;k 100 gSA bl rkj dk mi;ksx

lyZ fof/k }kjk rkj ds ;ax izR;kLFkrk xq.kkad dks Kkr djus gsrq fd;s x;s iz;ksx esa fd;k tkrk gSA izkIr vkadM+s fuEu izdkj

gS ; 50N ruko ds vUrxZr rkj esa foLrkj Dl = 0.125 cm, rkj dh lkekU; yEckbZ ds ekiu ds fy;s vYirekad

0.01 cm o rkj esa foLrkj ds fy;s 0.001 cm gSA ;fn cy dks cgqr T;knk ;FkkFkZrk ds lkFk ekik x;k gks rks ; ax izR;kLFkrk

xq.kkad (Y) dh x.kuk esa vf/kdre =qfV 8 n

%10´

izkIr gksrh gS] tgk¡ n yxHkx ,d iw.kk±d gS rks n Kkr djsaA


18/44

PHYS

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Kota/00CT214004

2. A horizontal conducting cylindrical hollow pipe of radius R = 54 mm and length L = 100 cm (R<<L)has a small hole P at its top, at the middle of the length as shown in the figure. Drops of mass m = 231mgand charge q = 1nC are falling into the hole from point A, at height 2R measured from the axis of thecylinder. Assume that the charge in the fallen drop gets uniformly distributed over the surface of thecylinder and charge distributed on cylinder remains uniform throughout. If the number of drops that willbe able to enter the cylinder is given as n = x × 10y in scientific notation. Find the value of (x + y).

f=T;k R = 54 mm rFkk yEckbZ L = 100 cm (R<<L) okys ,d {kSfrt pkyd csyukdkj [kks[kys ikbi dh Åijh lrgij bldh yEckbZ ds e/; eas fp=kuqlkj ,d NksVk fNæ P cuk gqvk gSA csyu dh v{k ls 2R Å¡pkbZ ij fLFkr fcUnq A lsbl fNæ esa m = 231mg æO;eku rFkk vkos'k q = 1nC okyh cwUnsa fxj jgh gSA ekukfd fxjus okyh cwan esa fo|eku vkos'kcsyu dh lrg ij ,dleku :i ls forfjr gks tkrk gS rFkk csyu ij forfjr vkos'k loZ= le:i cuk jgrk gSA ;fn csyuesa izos'k djus okyh cwUnksa dh la[;k oSKkfud fu:i.k esa n = x × 10y gks rks (x + y) Kkr dhft,A

2R

R

L

A

3. Two identical capacitors having plate separation 1mm are connected parallel to each other across pointsA and B as shown in figure. Total charge of 4 µC is imparted to the system by connecting a batteryacross A and B and battery is removed. Now first plate of first capacitor and second plate of secondcapacitor starts moving with constant velocity 3 m/s towards left. Find the magnitude of current (in mA)flowing in the loop initially.

nks ,dtSls la/kkfj=ksa esa IysVksa ds e/; nwjh 1 mm gSA bUgsa fcUnqvksa A rFkk B ds e/; ,d&nwljs ds lekUrj tksM + nsrs gSa] fp=ns[ksaA vc A o B ds chp ,d cSVjh dks tksM+dj fudk; dks dqy 4 µC vkos'k fn;k tkrk gS rFkk vc cSVjh dks gVk ysrsgSaA vc izFke laèkkfj= dh izFke IysV rFkk f}rh; la/kkfj= dh f}rh; IysV fu;r osx 3 m/s

ls ck¡;h vksj xfr djuk izkjEHk

dj nsrh gSA izkjEHk esa ywi esa izokfgr /kkjk dk ifjek.k (mA esa) Kkr dhft,A

BA

2

1

u0

u0



PHYS

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19/44Kota/00CT214004

4. A superconducting round ring of radius a = 20 cm and inductance L = 0.1 H was located in a uniform

magnetic field of induction B = 10 T. The plane of ring was parallel to the vector B, and the current in

the ring was equal to zero. Then the ring was turned through 90° so that its plane became perpendicular

to the field. The work (in J) performed during the turn is (Assume p2 = 10).

f=T;k a = 20 cm rFkk izsjdRo L = 0.1 H okyh ,d vfrpkyd xksykdkj oy; B = 10 T izsj.k okys le:i pqEcdh;

{ks= esa fo|eku gSA oy; dk ry lfn'k B ds lekUrj gS rFkk oy; esa /kkjk dk eku 'kwU; gSA vc oy; dks 90° ij ?kqek

fn;k tkrk gS rkfd bldk ry {ks= ds yEcor~ gks tk,A oy; dks ?kqekus ds nkSjku fd;k x;k dk;Z (J esa) Kkr dhft,A

(ekuk p2 = 10).


20/44

CHEM

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YALL INDIA OPEN TEST/JEE (Advanced)/08-02-2015/PAPER-2

Kota/00CT214004

PART-2 : CHEMISTRY Hkkx-2 : jlk;u foKku


This section contains 8 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 8 cgqfodYi iz'u gSaA izR;sd iz'u esa pkj fodYi (A), (B), (C) vkSj (D) gSa] ftuesa ls ,d ;kvf/kd lgh gSA

1. If electron of hydrogen atom is replaced by another particle of same charge but of double mass then-(A) Radii of orbits will increase (B) Ionisation energy will increase(C) Velocity of new particle will be more (D) Energy gap between two levels will be doubled

;fn gkbMªkstu ijek.kq dk bysDVªkWu] leku vkos'k ysfdu nqxqus nzO;eku ds nwljs d.k }kjk foLFkkfir fd;k x;k gS] rc&(A) d{kk dh f=T;k c<sxh (B) vk;uu ÅtkZ c<sxh

(C) u;s d.k dk osx vf/kd gksxk (D) nks Lrjks ds e/; ÅtkZ vUrjky nqxquk gksxk2. Which of the following is correct for 0.1 M BOH solution (Kb = 10–5)

(A) pH of solution is 11(B) OH– concentration is 10–3 mol/L(C) it's salt with HCl (i.e. BCl) form the acidic solution in water(D) Phenolphthalein indicator can be used during the titration of BOH with HCl

0.1 M BOH foy;u ds fy, fuEu esa ls dkSulk dFku lgh gS (Kb = 10–5)

(A) foy;u dh pH = 11 gS

(B) OH– dh lkUærk = 10–3 mol/L gS

(C) HCl ds lkFk bldk yo.k (vFkkZr~ BCl) ty esa vEyh; O;ogkj iznf'kZr djsxk(D) HCl ds lkFk BOH ds vuqekiu ds nkSjku fQuks¶Fksyhu lwpd dk mi;ksx fd;k tk ldrk gS

3. An ionic crystalline solid AB having cubic unit cell, may have following arrangement(s) -(A) B– in FCC and A+ occupies all tetrahedral voids.(B) B– in CCP and A+ in alternate tetrahedral voids.(C) B– at each corner and each face center and A+ in all octahedral voids.(D) B– at each corner and A+ at each edge center.

,d vk;fud fØLVyh; Bksl AB tks ?kuh; bdkbZ lsy j[krk gS] esa fuEu O;oLFkk,sa gks ldrh gS-

(A) B– , FCC cukrk gS rFkk A+ ] lHkh prq"Qydh; fjfDr;ksa dks ?ksjrs gSA

(B) CCP esa B– rFkk ,dkUrj prq"Qydh; fjfDr;ks a esa A+ gSaA

(C) B– ] izR;sd dksus ij o izR;sd Qyd ds dsUnz ij rFkk A+] lHkh v"VQydh; fjfDr;ksa esa gSaA(D) izR;sd dksus ij B– rFkk izR;sd fdukjs ds dsUnz ij A+ gSaA



CHEM

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21/44Kota/00CT214004

4. The cohesive energy is the force holding the atoms or ions together in solid (this is the same in magnitude

but the oppsite in sign to the enthalpy of atomization which is the energy required to break up the solid

into gaseous atom) select the CORRECT statements :

(A) Li > Na > K > Rb > Cs (Magnitude of cohesive energy)

(B) Li > Na > K > Rb > Cs (Magnitude of enthalpy of atomization)

(C) Li < Na < K < Rb < Cs (softness)

(D) Li > Na > K > Rb > Cs (strength of bond in their metal lattice)

llatu ÅtkZ og cy gS tks Bksl esa ijek.kqvksa ;k vk;uksa dks ,d lkFk ck¡/ksa j[krk gS (;g ifjek.k esa ijekf.o;dj.k

dh ,UFkSYih] tks Bksl dks xSlh; ijek.kqvksa esa rksM+us ds fy, vko';d ÅtkZ gS] ds leku ysfdu fpUg esa foijhr gSA

lgh dFku pqfu, :

(A) Li > Na > K > Rb > Cs (llatu ÅtkZ dk ifjek.k)

(B) Li > Na > K > Rb > Cs (ijekf.o;dj.k dh ,UFkSYih dk ifjek.k)

(C) Li < Na < K < Rb < Cs (eqyk;eiu)

(D) Li > Na > K > Rb > Cs (muds /kkrq tkyd esa ca/k dh lkeF;Z )

5. Select the CORRECT statement when lower the position of a metal's line in the Ellingham diagram

(A) The greater is the stability of it's oxide for example the line for Al (oxidation of Al) is found to be

below that for Fe (formation of Fe2O3)

(B) The greater the gap between any two lines the greater the effectiveness of reducing properties

corresponding to the lower line's metal

(C) The intersection of two line implies an oxidation reduction equilibrium

(D) At the point of intersection the free energy change for redox reaction is zero involving metal oxide

and metal from that two lines, whatever applicable

lgh dFku pqfu, tc ,fy?kae fp= esa '/kkrq js[kk' dh fLFkfr uhps gks

(A) blds vkWDlkbM dk LFkkf;Ro vf/kd gksxkA mnkgj.k] Al (Al dk vkWDlhdj.k) ds fy, /kkrq js[kk dh fLFkfr

Fe (Fe2O3 dk fuekZ.k) ls uhps izsf{kr dh xbZ gS

(B) fdUgh nks js[kkvksa ds e/; vUrj vf/kd gks rks uhps dh js[kk ls lEcfU/kr /kkrq ds vipk;d xq.k vf/kd izHkkoh

gksxsa

(C) nks js[kkvksa dk dVko fcUnq ,d vkWDlhdj.k&vip;u lkE; dks iznf'kZr djrk gS

(D) jsMkWDl vfHkfØ;k ds fy, dVko fcUnq ij eqä ÅtkZ ifjorZu 'kwU; gS] mu nksuks a js[kkvksa ls lEcfU/kr /kkrq rFkk

/kkrq vkWDlkbMksa dks lfEefyr djrs gq;s] tks Hkh ykxw gks


22/44

CHEM

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6. The bond order in O2

+ is the same as in :

O2

+ esa ca/k Øe] fuEu esa ls fdlds leku gS :

(A) N2̄

(B) N2

+(C) CO (D) NO+

7. Pentyne-2 can be obtained in by

(A) Butyne-1 & NaNH2 followed by CH3Br

(B) Propyne-1 & NaNH2 followed by CH3–CH2–I

(C) Ethyne & NaNH2 followed by CH3–I & CH3–CH2–I respectively

(D) Pentyne-1 heated with alc.KOH

isUVkbu-2 fdlds }kjk izkIr dh tk ldrh gS&

(A) C;wVkbu-1 rFkk NaNH2 ds ckn CH3Br }kjk

(B) izksikbu-1 rFkk NaNH2 ds ckn CH3–CH2–I }kjk

(C) ,sFkkbu rFkk NaNH2 ds ckn Øe'k% CH3–I rFkk CH3–CH2–I }kjk

(D) isUVkbu-1 dks ,sYdksgkWfyd KOH ds lkFk xeZ djds

8. Correct order is/are

(A) CH2 = CH2 > HC º CH for reaction with HBr

(B) CH2 = CH2 > HC º CH for reaction with O3

(C) CH2 = CH2 < HC º CH for reaction with H2

(D) CH2 = CH2 < HC º CH for reaction with O3

lgh Øe gS@gS&

(A) HBr ds lkFk vfHkfØ;k ds fy;s CH2 = CH2 > HC º CH

(B) O3 ds lkFk vfHkfØ;k ds fy;s CH2 = CH2 > HC º CH

(C) H2 ds lkFk vfHkfØ;k ds fy;s CH2 = CH2 < HC º CH

(D) O3 ds lkFk vfHkfØ;k ds fy;s CH2 = CH2 < HC º CH


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23/44Kota/00CT214004



This Section contains 4 multiple choice questions. Each question has matching lists. The codes for the

lists. have choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

bl [k.M esa 4 cgqfodYi iz'u gSaA izR;sd iz'u esa lqesyu lwpha gSA lwfp;ksa ds fy, dksM ds fodYi (A), (B), (C) vkSj (D)

gSa ftuesa ls dsoy ,d lgh gSA

9. List-I List-II(P) Conductivity (k) increases (1) Depends on cell constant as well as

concentration

(Q) Molar conductance (lm) increases (2) Independent of concentration as well as cell

constant

(R) Conductance (G) (3) With decreases in concentration but

independent of cell constant

(S) Molar conductance ( m¥l ) at infinite (4) With increase in concentration but independent

dilution. of cell constant

lwph-I lwph-II

(P) pkydrk (k) c<rh gS (1) lSy fu;rkad ds lkFk&lkFk lkUnzrk ij fuHkZj djrk gS

(Q) eksyj pkydRo (lm) c<+rk gS (2) lSy fu;rkad ds lkFk&lkFk lkUnzrk ls Lora= gksrk gS

(R) pkydRo (G) (3) lkUnzrk esa deh ds lkFk------ijUrq lSy fu;rkad ls Lora=

gksrk gS

(S) vuUr ruqrk ij eksyj pkydRo ( m¥l ) (4) lkUnzrk esa o`f¼ ds lkFk------ijUrq lSy fu;rkad ls Lora=

gksrk gS

Codes :P Q R S

(A) 4 1 3 2

(B) 4 3 2 1

(C) 3 4 1 2

(D) 4 3 1 2

24/44

CHEM

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Kota/00CT214004

10. List-I List-II(P) [Ni(gly)

3]¯ (1) Low spin

(Q) [Fe(NH3)

6]+2 (2) High spin

(R) Na[Pt BrCl(NO2) (NH

3)] (3) Optical isomerism

(S) [Co Br2Cl

2(SCN)

2]–3 (4) Geometrical isomerism

lwph-I lwph-II

(P) [Ni(gly)3]¯ (1) fuEu pØ.k

(Q) [Fe(NH3)

6]+2 (2) mPp pØ.k

(R) Na[Pt BrCl(NO2) (NH

3)] (3) izdkf'kd leko;ork

(S) [Co Br2Cl

2(SCN)

2]–3 (4) T;kferh; leko;ork

Codes :P Q R S

(A) 1, 3, 4 2 3, 4 1, 3, 4

(B) 3, 4 1 1, 4 2, 3, 4

(C) 3, 4 2 1, 4 2, 3, 4

(D) 3, 4 2 4 2, 3, 4



CHEM

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25/44Kota/00CT214004

11. Match each of the reactions given in List-I with the corresponding product(s) given in List-II

List-I List-II

(P) Ag + dil.HNO3

(1) NO

(Q) Sn + dil.HNO3

(2) NO2

(R) Fe + conc.HNO3

(3) NH4NO

3

(S) Zn + dil.HNO3

(4) N2O

lwph-I esa nh xbZ izR;sd vfHkfØ;k lwph-II esa fn;s x;s lEcfU/kr mRiknks a ds lkFk lqesfyr dhft,A

lwph-I lwph-II

(P) Ag + ruq.HNO3

(1) NO

(Q) Sn + ruq.HNO3

(2) NO2

(R) Fe + lkUnz.HNO3

(3) NH4NO

3

(S) Zn + ruq.HNO3

(4) N2O

Codes :P Q R S

(A) 1 3 2 4

(B) 1 4 2 3

(C) 3 4 1 2

(D) 4 3 2 1


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12. List - I List - II

(Can be oxidised by)

(P) CH – C3

O

H(1) Tollen's reagent

(Q) H – C – H

O

(2) Cu / 300º C

(R) CH3 – CH2 – OH (3) NaOI

(S) Ph – CH = O (4) Ceric ammonium nitrate

lwph- I lwph- II

(fdlds }kjk vkWDlhd`r fd;s tk ldrs gSa)

(P) CH – C3

O

H(1) VkWysUl vfHkdeZd

(Q) H – C – H

O

(2) Cu / 300º C

(R) CH3 – CH2 – OH (3) NaOI

(S) Ph – CH = O (4) lsfjd veksfu;eukbVªsV

Codes :P Q R S

(A) 1, 3 1 2, 4 1

(B) 1 1 2 1

(C) 1,3 1 2,3,4 1

(D) 1,3,4 1,3 2 1,2





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SECTION–III : (Integer Value Correct Type)

[kaM-III : (iw.kk±d eku lgh izdkj)This section contains 4 questions. The answer to each question is a three digit Integer, ranging from000 to 999.


1. Critical temperature and critical pressure of a real gas is 27°C & 75 atm respectively. Then what is criticalvolume of 1 mole of real gas (in mL) - (R = 0.08 atm-L/mol-K)

,d okLrfod xSl dk Øakfrd rki rFkk Økafrd nkc Øe'k% 27°C rFkk 75 atm gSA rc 1 mole okLrfod xSl dkØkafrd vk;ru D;k gksxk (mL esa) - (R = 0.08 atm-L/mol-K)

2. Consider the structure of Al2(CH3)6 compound and find the value of A + B + C

A = Maximum number of atoms in one plane

B = Total number of 2C – 2e¯ bonds

C = Total number of sp3 hybridised atom

Al2(CH3)6 dh lajpuk ij fopkj dhft, rFkk A + B + C dk eku Kkr dhft,

A = ,sls ijek.kqvksa dh vf/kdre la[;k tks ,d ry esa mifLFkr gS

B = 2C – 2e¯ ca/kksa dh dqy la[;k

C = sp3 ladfjr ijek.kqvksa dh dqy la[;k

3. A cyclic hexapeptide (m. wt. = 488) on complete hydrolysis gives Glycine (mol wt. = 75), alanine,phenyl alanine & valine. Glycine contributes 37.8 % of total hydrolysed product. Maximum number ofpossible valine unit which may present in cyclic hexapeptide.

,d pØh; gsDlkisIVkbM (v.kqHkkj = 488)] iw.kZ ty vi?kVu djkus ij Xykbflu (v.kqHkkj = 75)] , sykfuu] Qsfuy

,sykfuu rFkk osfyu nsrk gSaA Xykbflu dqy ty vi?kfVr mRikn dk 37.8 % Hkkx gksrh gS pØh; gsDlkisIVkbM esa mifLFkr

gks ldus okyh osfyu bdkbZ dh vf/kdre lEHkkfor la[;k gSA


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4. Number of reaction correctly matched with their product

,slh vfHkfØ;kvksa dh la[;k tks buds mRiknksa ds lkFk lgh :i ls lqesfyr gksrh gSA

(i)C – Cl

O

H – Pd2BaSO4

CO

H (ii) N H2 4H O

(1 eq.)2 2

(iii)

O

O

N H2 4H O2 2

(iv) H (1 eq)2

Pd

(v)

O

O

OH Zn–HgHCl

OH

(vi)O

C

O

NaEtOH

CH –OH2+

OH

(vii) CH3 – C º C – CH3 H –Pd2BaSO4

C = CH C3

H H

CH3



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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. For the given reaction

A(g) + 2B(g) ® 3C(g)

Find the equilibrium concentration of B (mol / L) in 1/8 litre container if two moles of A and four moles

of B are taken initially.

Given (Kc = 2 × 10–3)

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

nh xbZ vfHkfØ;k ds fy,

A(g) + 2B(g) ® 3C(g)

1/8 yhVj ds ik= esa B (mol / L) dh lkE; lkUærk crkb;sA ;fn A ds nks eksy rFkk B ds pkj eksy izkjEHk esa fy;s

x;s gksA

fn;k gS (Kc = 2 × 10–3)

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

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2. Use the following data to answer the question below :

H2

Ni DH = –30 kcal mol–1 ;

excess H2

Ni DH = –180 kcal mol–1

Calculate the magnitude of resonance energy (in kcal/mole) for


fuEu vk¡dM+ksa dk mi;ksx dj uhpsa fn;s x, iz'u dk mÙkj nhft,A

H2

Ni DH = –30 kcal mol–1 ;

excess H2

Ni DH = –180 kcal mol–1

ds fy, vuqukn ÅtkZ (kcal/mole esa) ds ifjek.k dh x.kuk dhft,

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


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3. Find the number of basic radicals among the following cations, which can form soluble complex on

adding excess of KCN

Pb+2, Ag+, Fe+2, Fe+3, Cu+2, Cd+2.

fuEu /kuk;uksa esa ls ,sls {kkjh; ewydksa dh la[;k crkbZ;s tks KCN dk vkf/kD; feykus ij foys;'khy ladqy cuk

ldrs gS

Pb+2, Ag+, Fe+2, Fe+3, Cu+2, Cd+2.

4. Phenyl triphenylmethyl ether on treatment with acid gives organic product (A). What is the molecular

weight of A.


Qsfuy VªkbZQsfuyesfFky bZFkj] vEy ds lkFk mipkfjr gksdj dkcZfud mRikn (A) nsrk gSA A dk vkf.od Hkkj D;k gS\

vius mÙkj ds vadks a dks (n'keyo LFkku dks NksM+dj) rc rd ;ksx dhft, tc rd vkidks bdkbZ vad izkIr

u gks tk,A


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SALL INDIA OPEN TEST/JEE (Advanced)/08-02-2015/PAPER-2

Kota/00CT214004

PART-3 : MATHEMATICS Hkkx-3 : xf.kr


This section contains 8 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 8 cgqfodYi iz'u gSaA izR;sd iz'u esa pkj fodYi (A), (B), (C) vkSj (D) gSa] ftuesa ls ,d ;kvf/kd lgh gSA

1. If ƒ : (0,1) ® R is defined by ( )n

n 1

2

2n 0

xƒ x

1 x+

¥

=

=-

å , then -

;fn ƒ : (0,1) ® R, ( )n

n 1

2

2n 0

xƒ x

1 x+

¥

=

=-

å }kjk ifjHkkf"kr gks] rc -

(A) ( )1 xƒ x

x 1- =

+(B) ( )

12

0

2 n2 1ƒ x dx

2-

=òl

(C) ( )1

0

ƒ x dx n2 1= -ò l (D) ( )1

0

1 1 1ƒ x dx .....

2 3 4= + + + ¥ò

2. For the equation 22014

2cos2x cos2x cos cos 4x 1x

æ öæ öp- = -ç ÷ç ÷

è øè ø; which of the following is/are correct ?

(A) The number of solutions of the equation is 3.(B) The sum of all the positive solutions is 1080p(C) The number of positive solutions is 4(D) The sum of all positive solutions is 1008p

lehdj.k 22014

2cos2x cos2x cos cos 4x 1x

æ öæ öp- = -ç ÷ç ÷

è øè ø ds fy, fuEu esa ls dkSulk@dkSuls dFku lgh gksxk@gksaxs ?

(A) lehdj.k ds gyksa dh la[;k 3 gksxhA

(B) lHkh /kukRed gyksa dk ;ksxQy 1080p gksxkA

(C) /kukRed gyksa dh la[;k 4 gksxhA

(D) lHkh /kukRed gyksa dk ;ksxQy 1008p gksxkA



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3. A fair die is rolled 'n' times. If P(n) denotes the probability that there are atleast two equal numbersamong the result obtained then-

,d fu"i{k ikls dks n ckj mNkyk tkrk gSA ;fn P(n), izkIr ifj.kkeksa esa de ls de nks leku la[;k vkus dh izkf;drk dks

n'kkZrk gks] rc -

(A) ( )P n 7 1³ = (B) ( ) 1P n 2

6= = (C) ( ) 13

P n 418

= = (D) ( ) 317P n 6

324= =

4. Let ƒ be a differentiable function satisfying the relation ƒ(xy) = xƒ(y) + yƒ(x) – 2xy (where x,y > 0) andƒ'(1) = 3, then

(A) ƒ(x) = xlnx + 3x – 2x

2(B) ƒ(x) = xlnx

(C) x = e–3 is the abscissa of the point of inflection of ƒ(x)

(D) The equation ƒ(x) = k has two solutions if x Î (–e–3,0)

ekuk ƒ ,d vodyuh; Qyu gS] tks lEcU/k ƒ(xy) = xƒ(y) + yƒ(x) – 2xy (tgk¡ x,y > 0) dks lUrq"V djrk gS rFkkƒ'(1) = 3 gS] rc

(A) ƒ(x) = xlnx + 3x – 2x

2(B) ƒ(x) = xlnx

(C) x = e–3, ƒ(x) ds ufr ifjorZu fcUnq dk Hkqt gSA

(D) lehdj.k ƒ(x) = k ds nks gy gksaxs ;fn x Î (–e–3,0) gksA

5. Let ƒ : (0, ¥) ® R be a function defined as ( )2

0

sin xƒ dx

x

p

l =+ lò , then

ekuk ƒ : (0, ¥) ® R ,d Qyu gS] tks ( )2

0

sin xƒ dx

x

p

l =+ lò }kjk ifjHkkf"kr gS] rc

(A) ƒ(l) > 0 (B) ( )lim ƒ 0l®¥

l = (C) ( )2lim ƒ 0l®¥

l l = (D) ( )2lim ƒ 2l®¥

l l = p


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6. If a and b are the roots of the equation ax2 + bx + c = 0 and ab = 3, where a,b,c are in arithmeticprogression, then (a2 + b2) is equal to-;fn a rFkk b lehdj.k ax2 + bx + c = 0 ds ewy rFkk ab = 3 gS] tgk¡ a,b,c lekUrj Js.kh esa gks] rks (a2 + b2) dk ekugksxk -(A) 2 (B) –2 (C) a3 + b3 (D) a + b

7. Let ( ) ( )2

2

x 2 3 x 2ƒ x

k x x 2

é - + ³= ê

ê + <ë, then which of the following statements is/are true ?

(A) If ƒ(x) is continuous at x = 2, then k = –1(B) There exists at least one value of k for which ƒ(x) is derivable " x Î R(C) If ƒ(x) has a local minima at x = 2, then k > –1

(D) ( )3

2

10ƒ x dx

3=ò

ekuk ( ) ( )2

2

x 2 3 x 2ƒ x

k x x 2

é - + ³= ê

ê + <ë gks] rks fuEu esa ls dkSulk@dkSuls dFku lgh gksxk@gksaxs\

(A) ;fn x = 2 ij ƒ(x) larr~ gks] rks k = –1 gksxkA(B) k dk de ls de ,d eku fo|eku gksxk ftlds fy;s ƒ(x) lHkh x Î R ds fy;s vodyuh; gksA(C) ;fn x = 2 ij ƒ(x) dk ,d LFkkuh; fufEu"B gks] rks k > –1 gksxkA

(D) ( )3

2

10ƒ x dx

3=ò

8. If a variable chord of the hyperbola x2 – y2 = 9 touches the parabola y2 = 12x, then locus of middlepoints of these chords is expressed as x3 + l1xy2 + l2y2 = 0 (l1,l2 Î I), then-

(A) l1 + l2 = 2 (B) 2 21 2 10l + l =

(C) Number of divisors of ( )2 21 2l + l is 4 (D) Number of divisors of ( )2 2

1 2l + l is 6

;fn vfrijoy; x2 – y2 = 9 dh ,d pj thok ijoy; y2 = 12x dks Li'kZ djrh gS] rc bl thok ds e/; fcUnq dsfcUnqiFk dks x3 + l1xy2 + l2y2 = 0 (l1,l2 Î I) ds :i esa O;Dr fd;k tkrk gS] rc -

(A) l1 + l2 = 2 (B) 2 21 2 10l + l =

(C) ( )2 21 2l + l ds Hkktdksa dh la[;k 4 gksxhA (D) ( )2 2

1 2l + l ds Hkktdksa dh la[;k 6 gksxhA



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This Section contains 4 multiple choice questions. Each question has matching lists. The codes for the

lists. have choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

bl [k.M esa 4 cgqfodYi iz'u gSaA izR; sd iz'u esa lqesyu lwpha gSA lwfp;ksa ds fy, dksM ds fodYi (A), (B), (C) vkSj (D)

gSa ftuesa ls dsoy ,d lgh gSA9. Let P be the plane parallel to y-axis and containing the points (1,0,1) and (3,2,–1). Also A º (4,0,0)

and B º (6,0,–2) are two points and P º (x0,y

0,z

0) is a variable point on the plane P = 0.

Match List-I with List-II and select the correct answer using the code given below the list.List-I List-II

(P) If the equation of the plane P = 0 is (1) 16x + ay + bz = c, then |a + b + c| is

(Q) If (PA + PB) is minimum, then |4x0 + y

0 + 2z

0| is (2) 12

(R) If |PA – PB| )0, NéÎ ë , then N is (3) 3

(S) If the reflection of the line AB in the plane P = 0 (4) 8

is x 2 y z

1 0 1- - a + b

= =-

, then (a4 + b4) is

ekuk P, y-v{k ds lekUrj og lery gS tks fcUnq (1,0,1) rFkk (3,2,–1) dks j[krk gSA lkFk gh A º (4,0,0)rFkk B º (6,0,–2) nks fcUnq gSa rFkk P º (x

0,y

0,z

0), lery P = 0 ij ,d pj fcUnq gSA

lwph-I dks lwph-II ls lqesfyr dhft, rFkk lwfp;ksa ds uhps fn, u, dksM dk iz;ksx djds lgh mÙkj pqfu;s %lwph-I lwph-II

(P) ;fn lery P = 0 dk lehdj.k x + ay + bz = c gks] (1) 16rks |a + b + c| dk eku gksxk

(Q) ;fn (PA + PB) U;wure gks] rks |4x0 + y

0 + 2z

0| dk eku gksxk (2) 12

(R) ;fn |PA – PB| )0, NéÎ ë gks] rks N dk eku gksxk (3) 3

(S) ;fn lery P = 0 ij js[kk AB dk izfrfcEc (4) 8

x 2 y z1 0 1- - a + b

= =-

gks] rks (a4 + b4) dk eku gksxk

Codes :P Q R S

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


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10. Let z = a + ib (where a,bÎR and i 1= - ) such that |2z + 3i| = |z2|.

Match List-I with List-II and select the correct answer using the code given below the list.List-I List-II

(P) |z|max

is equal to (1) 2(Q) |z|

min is equal to (2) 1

(R) If |z| is maximum Þ z = a + ib, then (a3 + b3) is equal to (3) 3

(where a,b Î R and i 1= - ) (4) 27

(S) If |z| is minimum Þ z = x + iy, then (x2 + 2y2) is equal to

(where x,y Î R and i 1= - )

ekuk z = a + ib (tgk¡ a,bÎR rFkk i 1= - ) bl izdkj gS fd |2z + 3i| = |z2|

lwph-I dks lwph-II ls lqesfyr dhft, rFkk lwfp;ksa ds uhps fn, u, dksM dk iz;ksx djds lgh mÙkj pqfu;s %

lwph-I lwph-II

(P) |z|max

dk eku gksxk (1) 2

(Q) |z|min

dk eku gksxk (2) 1

(R) ;fn |z| vf/kdre gS Þ z = a + ib gks] rks (a3 + b3) dk eku gksxk (3) 3

(tgk¡ a,b Î R rFkk i 1= - ) (4) 27

(S) ;fn |z| U;wure gS Þ z = x + iy gks] rks (x2 + 2y2) dk eku gksxk

(tgk¡ x,y Î R rFkk i 1= - )

Codes :P Q R S

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



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11. Consider ( )1

0

ƒ x t t x dt= -òMatch List-I with List-II and select the correct answer using the code given below the list.

List-I List-II

(P) ( )x R

min ƒ xÎ

is equal to (1) 0

(Q)[ ]

( )x 0,1max ƒ xÎ

is equal to (2)2 2

6-

(R) ( )x 1

min ƒ x³

is equal to (3)16

(S) Number of points where ƒ(x) is non-derivable is (4)13

ekuk ( )1

0

ƒ x t t x dt= -ò gSA

lwph-I dks lwph-II ls lqesfyr dhft, rFkk lwfp;ksa ds uhps fn, u, dksM dk iz;ksx djds lgh mÙkj pqfu;s %lwph-I lwph-II

(P) ( )x R

min ƒ xÎ

dk eku gksxk (1) 0

(Q)[ ]

( )x 0,1max ƒ xÎ

dk eku gksxk (2)2 2

6-

(R) ( )x 1

min ƒ x³

dk eku gksxk (3)16

(S) fcUnqvksa dh la[;k] tgk¡ ƒ(x) vodyuh; ugha gS] gksxh (4)13

Codes :P Q R S

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


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12. Consider the word "BREAKAGE". The number of ways in which the letters of this word can bearranged such thatMatch List-I with List-II and select the correct answer using the code given below the list.

List-I List-II(P) The two A's are not together is (1) 720(Q) The two E's are together but no two A's are together is (2) 1800(R) Neither two A's nor two E's are together is (3) 5760(S) No two vowels are together is (4) 7560

ekuk ,d 'kCn "BREAKAGE" gSA bl 'kCn ds v{kjksa dks O;ofLFkr djus ds rjhdks a dh la [;k] ftuesalwph-I dks lwph-II ls lqesfyr dhft, rFkk lwfp;ksa ds uhps fn, u, dksM dk iz;ksx djds lgh mÙkj pqfu;s %

lwph-I lwph-II(P) nks A lkFk&lkFk uk gksa (1) 720(Q) nks E lkFk&lkFk gksa ijUrq nks A lkFk&lkFk uk gkas (2) 1800(R) uk rks nks A uk gh nks E lkFk&lkFk gks (3) 5760(S) nks Loj lkFk&lkFk uk gksa (4) 7560

Codes :P Q R S

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





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SECTION–III : (Integer Value Correct Type)

[kaM-III : (iw.kk±d eku lgh izdkj)This section contains 4 questions. The answer to each question is a three digit Integer, ranging from000 to 999.


1. Consider the curves C1 : |z – 2| = 2 + Re(z) and C2 : |z| = 3 (where z = x + iy, x,yÎR and i 1= - ).

They intersect at P and Q in the first and fourth quadrants respectively. Tangents to C1 at P and Q intersect

the x-axis at R and tangents to C2 at P and Q intersect the x-axis at S. If area of DPRS is 2l sq.units,

then (l2) is

ekuk oØ C1 : |z – 2| = 2 + Re(z) rFkk C2 : |z| = 3 (tgk¡ z = x + iy, x,yÎR rFkk i 1= - ) gSA ;g izFke rFkk

prqFkZ prqFkk¥±'k esa Øe'k% P rFkk Q ij izfrPNsn djrs gSA C1 ij P rFkk Q ij Li'kZjs[kk;sa x-v{k ij R ij feyrs

gS rFkk C2 ij P rFkk Q ij Li'kZjs[kk;sa x-v{k ij S ij feyrs gSaA ;fn f=Hkqt PRS dk {ks=Qy 2l oxZ bdkbZ

gks] rks (l2) dk eku gksxk

2. Let w and z be complex numbers such that |w| = 1 and |z| = 10. Let w z

argz-æ öq = ç ÷

è ø. The maximum

possible value of tan2q can be expressed as pq (where p and q are relative prime), then (p +2q) is

ekuk w rFkk z lfEeJ la[;k;sa bl izdkj gS fd |w| = 1 rFkk |z| = 10 gSA ekuk w z

argz-æ öq = ç ÷

è ø gSA tan2q ds

vf/kdre lEHko eku dks pq (tgk¡ p rFkk q ijLij vHkkT; la[;k;sa gSa ) ds :i esa O;Dr fd;k tk ldrk gks ] rks

(p +2q) dk eku gksxk


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3. The equation x4 + ax3 + bx2 + ax + 1 = 0 has at least one real root and if the minimum value of

E(a,b) = a2 + b2 can be expressed as pq (where p & q are relatively prime), then (p2 + q2) is

lehdj.k x4 + ax3 + bx2 + ax + 1 = 0 dk de ls de ,d okLrfod ewy rFkk ;fn E(a,b) = a2 + b2 ds U;wure

eku dks pq (tgk¡ p rFkk q ijLij vHkkT; la[;k;sa gSa) ds :i esa O;Dr fd;k tk ldrk gks] rks (p2 + q2) dk eku

gksxk

4. A bag contains (2n + 1) coins. It is known that 'n' of these coins have tail on both the sides, whereas

the remaining (n + 1) coins are fair. A coin is randomly drawn from the bag and tossed. If the probability

that the toss results in a tail is 3142

, then the value of 'n' is

,d FkSys esa (2n + 1) flDds gSaA Kkr gS fd buesa ls 'n' flDdksa ds nksuksa vksj iV gS rFkk 'ks"k (n + 1) fu"i{kikrh flDds

gSaA ,d flDds dk ;kn`PN;k p;u dj mNkyk tkrk gSA ;fn ifj.kke esa iV vkus dh izkf;drk 3142

gks] rks 'n' dk

eku gksxk



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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. The tangent at a point R on the hyperbola 2 2

2 2

x y1

a b- = (a,b Î R+) passes through the point (0,–b) and

the normal at the point R passes through the point ( )2 2a,0 . If 'e' denotes the eccentricity of the hyperbola,

then 'e2' is

vfrijoy; 2 2

2 2

x y1

a b- = (a,b Î R+) ds fcUnq R ij Li'kZjs[kk] fcUnq (0,–b) ls xqtjrh gS rFkk fcUnq R ij vfHkyEc

fcUnq ( )2 2a,0 ls xqtjrk gSA ;fn 'e' vfrijoy dh mRdsUærk gks] rks 'e2' dk eku gksxk

2. A point P is randomly selected from the region described by 3 z 5£ £ (where 'z' is a complex number).

If the probability that argument of P lies in the interval ,4 4p pé ù-ê úë û

is 1N

, then N is

3 z 5£ £ (tgk¡ 'z' ,d lfEeJ la[;k gS) }kjk ifjHkkf"kr {ks= essa fcUnq P ;kn`PN;k p;fur fd;k tkrk gSA ;fn izkf;drk

fd fcUnq P dk dks.kkad] vUrjky ,4 4p pé ù-ê úë û

esa fLFkr gks] 1N

gS] rc N dk eku gksxk


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Kota/00CT214004


3. If a normal chord of the parabola 2y2 = x has slope -(2 3) , then the area of triangle formed by normal

chord of the parabola and the tangents drawn at the extremities of the normal chord is

;fn ijoy; 2y2 = x dh vfHkyEc thok dh izo.krk -(2 3) gks] rc ijoy; dh vfHkyEc rFkk vfHkyEc thok

ds fljksa ij [khaph xbZ Lif'kZ;ksa }kjk fufeZr f=Hkqt dk {ks=Qy gksxk

4. If ƒ(x) is a twice differentiable function in x Î [0,p] such that ƒ(0) = ƒ(p) = 0 and

( ) ( )( )/ 2

0

ƒ 2x ƒ" 2xp

+ò sinx cosx dx = kp, then k is

;fn ƒ(x), x Î [0,p] esa ,d nks ckj vodyuh; Qyu bl izdkj gS fd ƒ(0) = ƒ(p) = 0

rFkk ( ) ( )( )/ 2

0

ƒ 2x ƒ" 2xp

+ò sinx cosx dx = kp gks] rks k dk eku gksxk


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D. vadu ;kstuk / Marking scheme :15. [kaM-I (i) ds gj iz'u esa dsoy lgh mÙkj okys cqycqys (BUBBLE) dks dkyk djus ij 3 vad vkSj dksbZ Hkh cqycqyk dkyk ugha djus ij

'kwU; (0) vad iznku fd;k tk;sxk bl [akM ds iz'uksa esa xyr mÙkj nsus ij dksbZ ½.kkRed vad ugha fn;s tk;saxsaAFor each question in Section-I (i), you will be awarded 3 marks if you darken the bubble corresponding to the correct answerand zero mark if no bubbles are darkened No negative marks will be awarded for incorrect answers in this section.

16. [kaM-I (ii) ds gj iz'u esa dsoy lgh mÙkjksa (mÙkj) okys lHkh cqycqyksa (cqycqys) dks dkyk djus ij 3 vad vkSj dksbZ Hkh cqycqyk dkykugha djus ij 'kwU; (0) vad iznku fd;k tk;sxkA vU; lHkh fLFkfr;ksa esa ½.kkRed ,d (–1) vad iznku fd;k tk;sxkAFor each question in Section-I (ii), you will be awarded 3 marks if you darken all the bubble(s) corresponding to only thecorrect answer(s) and zero mark if no bubbles are darkened. In all other cases minus one (–1) mark will be awarded

17. [kaM-III ds gj iz'u esa dsoy lgh mÙkjksa (mÙkj) okys lHkh cqycqyksa (cqycqys) dks dkyk djus ij 3 vad vkSj dksbZ Hkh cqycqyk dkykugha djus ij 'kwU; (0) vad iznku fd;k tk;sxkA vU; lHkh fLFkfr;ksa esa ½.kkRed ,d (–1) vad iznku fd;k tk;sxkAFor each question in Section-III, you will be awarded 3 marks if you darken all the bubble(s) corresponding to only thecorrect answer(s) and zero mark if no bubbles are darkened. In all other cases minus one (–1) mark will be awarded

18. [kaM-IV ds gj iz'u esa dsoy lgh mÙkjksa (mÙkj) okys lHkh cqycqyksa (cqycqys) dks dkyk djus ij 3 vad vkSj dksbZ Hkh cqycqyk dkykugha djus ij 'kwU; (0) vad iznku fd;k tk;sxkA vU; lHkh fLFkfr;ksa esa ½.kkRed ,d (–1) vad iznku fd;k tk;sxkAFor each question in Section-IV, you will be awarded 3 marks if you darken all the bubble(s) corresponding to only thecorrect answer(s) and zero mark if no bubbles are darkened. In all other cases minus one (–1) mark will be awarded

17. g = 10 m/s2 iz;qDr djsa] tc rd fd vU; dksbZ eku ugha fn;k x;k gksATake g = 10 m/s2 unless otherwise stated.

Name of the Candidate / ijh{kkFkhZ dk uke

I have read all the instructions and shall abide by them.eSusa lHkh vuqns'kksa dks i<+ fy;k gS vkSj eSa mudk vo'; ikyu d:¡xk@d:¡xhA

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

Form Number / QkWeZ la[;k

I have verified all the information filled in by the Candidate.ijh{kkFkhZ }kjk Hkjh xbZ tkudkjh dks eSus a tk¡p fy;k gSA

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


Corporate Office : ALLEN CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan)-324005

+91-744-2436001 [email protected]

Appropriate way of darkening the bubble for your answer to be evaluatedvkids mÙkj ds ewY;kadu ds fy, cqycqys dk s dkyk djus dk mi;qDr rjhdk

a

a

a

a

a

a

The one and the only acceptable,d vkSj dsoy ,d Lohdk;Z

Part darkeningvkaf'kd dkyk djuk

Darkening the rimfje dkyk djuk

Cancelling after darkeningdkyk djus ds ckn jn~n djuk

Erasing after darkeningdkyk djus ds ckn feVkuk

Answer will not be evaluated -no marks, no negative marks

mÙkj dk ewY;kadu ugha gksxk&dksbZ vad ugha] dksbZ ½.kkRed vad ugha

Figure-1 : Correct way of bubbling for valid answer and a few examplex of invalid answers fp=&1 % oS/k mÙkj ds fy, cqycqyk Hkjus dk lgh rjhdk vkSj voS/k mÙkjks a ds dqN mnkgj.kAAny other form of partial marking such as ticking or crossing the bubble will be invalidvkaf'kd vadu ds vU; rjhds tSls cqycqys dks fVd djuk ;k ØkWl djuk xyr gksxkA

10

234

6789

4 2 0 0 0 20

2

0

2 2 23 3 3 3 3

0

4 4 4 45 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9

1 1 1 1 1

20

456789

1

54

5

3

1

3

Figure-2 : Correct Way of Bubbling your Form Number on the ORS. (Example Form Numebr : 14200022)fp=&2 % vks-vkj-,l (ORS) ij vkids QkWeZ uEcj ds ccy dks Hkjus dk lgh rjhdkA (mnkgj.k QkWeZ uEcj : 14200022)

Kota/00CT214004Your Target is to secure Good Rank in JEE 201544/44

allen clas amme · 2015-12-21 · physics ka/00ct214004 5/44 space for rough work / dpps dk;z ds...

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