ALLEN 0 0 2 1 4 0 0 5 CLAS AMME · A andida writ i / e we i th ORS ee darkenin ropri ubbl wit hel f...

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

A. lkekU; / 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 40 i`"B gSa vkSj izR;sd fo"k; ds lHkh 22 iz'u vkSj muds mÙkj fodYi Bhd

ls i<+ s tk ldrs gSaA lHkh [kaMksa ds izkjEHk esa fn;s gq, funsZ'kksa dks /;ku ls i<+ sAOn breaking the seal of the booklet check that it contains 40 pages and all the 22 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 provide 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 dsoy ,d lgh gSaA

Contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONLYONE is correct.

(ii) Hkkx esa fl¼kUrksa] iz;ksxksa vkSj vk¡dM+ksa vkfn dks n'kkZus okys 2 vuqPNsn gSA nksuksa vuqPNsnksa ls lacfU/kr pkj iz'u gSaA ftuesa ls gjvuqPNsn ij nks iz'u gSaA fdlh Hkh vuqPNsn esa gj iz'u ds pkj fodYi (A), (B), (C) vkSj (D) gSa ftuesa ls dsoy ,d lgh gSaAContains 2 paragraphs each describing theory, experiment, data etc. Four questions relate to two paragraphswith two questions on each paragraph. Each question of a paragraph has ONLY ONE correct answer amongthe four choices (A), (B), (C) and (D)

12. [kaM–II esa 2 iz'u eSfVªDl lqesy izdkj ds gSA / SECTION – II contains 2 questions of Matrix Match Type.13. [kaM–III esa ,d Hkh iz'u ugha gSA / There is no questions in SECTION-III14. [kaM-IV es a 8 iz'u gSaA izR;sd iz'u dk mÙkj 0 ls 9 rd (nksuksa 'kkfey) ds chp dk ,dy vadh; iw.kk±d gSA

Section-IV contains 8 questions The answer to each question is a single digit integer, ranging from0 to 9 (both inclusive)

funs Z'k / INSTRUCTIONS

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

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

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

isij – 1PAPER – 1

Path to Success

ALLENCAREER INSTITUTEKOTA (RAJASTHAN)

T M

CLASSROOM CONTACT PROGRAMME(ACADEMIC SESSION 2014-2015)

PAPER CODE 0 0 C T 2 1 4 0 0 5

PATTERN : JEE (Advanced)TEST TYPE : MAJOR

ENTHUSIAST & LEADER COURSE

Date : 10 - 05 - 2015TARGET : JEE (Advanced) 2015

ALL INDIA OPEN TEST #02

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

Hkkx-1 HkkSfrd foKku I(i) dsoy ,d lgh fodYi izdkj 03 - 06Part-1 Physics Only One Option Correct Type

I(ii) vuqPNsn izdkj 07 - 09Paragraph Type

II eSfVªDl&esy izdkj 10 - 13Matrix Match Type

IV iw.kk±d eku lgh izdkj (0 ls 9) 14 - 17Integer Value Correct Type (0 to 9)

Hkkx-2 jlk;u foKku I(i) dsoy ,d lgh fodYi izdkj 18 - 20Part-2 Chemistry Only One Option Correct Type




Hkkx-3 xf.kr I(i) dsoy ,d lgh fodYi izdkj 29 - 31Part-3 Mathematics Only One Option Correct Type




SOME USEFUL CONSTANTSAtomic No. H = 1, B = 5, C = 6, N = 7, O = 8, F = 9, Al = 13, P = 15, S = 16, Cl = 17,

Br = 35, Xe = 54, Ce = 58,

Atomic masses : H = 1, Li = 7, B = 11, C = 12, N = 14, O = 16, F = 19, Na = 23, Mg = 24,

Al = 27, P = 31, S = 32, Cl = 35.5, Ca=40, Fe = 56, Br = 80, I = 127,

Xe = 131, Ba=137, Ce = 140,

2/40

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

· Coulomb's law constant pe9

0

1 = 9×104

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

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

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

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

· Permittivity of vacuum Î0 = 20

1

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

Kota/00CT214005

ALL INDIA OPEN TEST/JEE (Advanced)/10-05-2015/PAPER-1

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Kota/00CT214005

PART-1 : PHYSICS

Hkkx-1 : HkkSfrd foKkuSECTION–I : (i) Only One option correct Type

[k.M-I : (i) dsoy ,d lgh fodYi izdkjThis section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and

(D) out of which ONLY ONE is correct.

bl [k.M esa 8 cgqfodYi iz'u gSA izR;sd i'u esa pkj fodYi (A), (B), (C) vkSj (D) gSa ftuesa ls dsoy ,d

lgh gSA

1. A ray of light enters into a thick glass slab from air as shown in figure. The refractive index varies

as m= ( )2 3 y- . If width of slab is very large then maximum value of y-coordinate of the ray is -

(A) 3

2(B)

3 3

2(C) ¥ (D) Data insufficient

60°

y

x

fp= eas n'kkZ;svuqlkj ,d izdk'k fdj.k ok;q ls ,d eksVh dkap dh ifV~Vdk esa izos'k djrh gSA viorZukad

m = ( )2 3 y- ds vuqlkj ifjofZrZr gksrk gSA ;fn ifV~Vdk dh pkSM+kbZ cgqr vf/kd gS rks fdj.k ds y-funsZ'kkad dk

vf/kdre eku gS&

(A) 3

2(B)

3 3

2(C) ¥ (D) vkadM+s vi;kZIr 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

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PHYS

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Kota/00CT214005


2. Consider a network shown in figure. Initially the switch S is open. The amount of charge flow

through earth wire after closing the switch is -

fp= esa iznf'kZr ifjiFk ij fopkj dhft;sA izkjEHk esa fLop S pkyw gSA fLop S dks can djus ds i'pkr~ HkwlaifdZr

rkj ls izokfgr vkos'k dh ek=k gksxh&

6C 3C

3C 6C

V

S

(A) CV (B) 2CV (C) 12

CV (D) 3CV2

3. A parallel beam of light is incident on a thin prism of prism angle of 4

p degrees. The refractive

index of the prism is 1.5. The focal length of the lens is 60 cm. The coordinates of converging point

of the beam is-

,d lekUrj izdk'k iaqt 4p

fMxzh fizTe dks.k okys ,d irys fizTe ij vkifrr gSA fizTe dk viorZukad 1.5 gSA

ysal dh Qksdl nwjh 60 cm gSA iqat ds vfHklj.k fcUnq ds funsZ'kkad gksxs a&

q

80cm

x0

y

(A) 2

60cm, cm3

æ öç ÷è ø (B)

160cm, cm

3æ öç ÷è ø (C)

160cm, cm

3æ ö-ç ÷è ø (D)

260cm, cm

3æ ö-ç ÷è ø

PHYS

ICS

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Kota/00CT214005


4. Two cylindrical straight and very long non magnetic conductors A and B, insulated from each other,

carry a current I in the positive and the negative z–direction respectively. The direction of magnetic field

at origin is :–

nks csyukdkj lh/ks rFkk cgqr yEcs vpqEcdh; pkyd A rFkk B ,d nwljs ls foyfxr gSa rFkk muesa Øe'k% /kukRed rFkk

½.kkRed z- fn'kk esa I /kkjk izokfgr gksrh gSA ewy fcUnq ij pqEcdh; {ks= dh fn'kk gksxh%&

BAx

y

(A) i- (B) i+ (C) j (D) – j

5. Two boys of masses 50 kg and 60kg are moving along a vertical massless rope, the former climbing

up with an acceleration of 2ms–2 while later coming down with constant velocity of 2ms–1. The

tension in rope at fixed support will be

(g= 10ms–2).

50kg rFkk 60kg nzO;eku ds nks yM+ds Å/okZ/kj nzO;ekughu jLlh ij xfr dj jgs gSA igyk yM+dk 2ms–2 ds

Roj.k ls Åij dh vksj p<+rk gS tcfd ckn okyk yM+dk 2ms–1 ds fu;r osx ls uhps dh vksj mrjrk gSaA fLFkj

vkèkkj ij jLlh esa ruko gksxk&(g= 10ms–2).

\\\\\\\\\\\\\\\\\50kg

60kg

(A) 1.2 kN (B) 1.1 kN (C) 0.60 kN (D) 1.6 kN

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PHYS

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Kota/00CT214005

6. A charge q moves with velocity 1v ai=r $ (ms–1) experiences a force

1F aqj(N)= -r

$ at a point in magnetic

field region. If charge is moving with velocity 12v (ai bj)(ms )-= +

r $ $ at the same point, it experience force

2F q(ai bj)(N)= +r

$ $ . The force experienced by it at the same point if it is moving with

velocity 1 2(v v )´r r

-

1v ai=r $ (ms–1) ds osx ls xfr'khy ,d vkos'k q pqEcdh; {ks= esa fdlh fcUnq ij

1F aqj(N)= -r

$ cy vuqHko

djrk gSA ;fn vkos'k leku fcUnq ij 12v (ai bj)(ms )-= +

r $ $ osx ls xfr djs rks ;g 2F q(ai bj)(N)= +

r$ $ cy

vuqHko djrk gSA ;fn ;g 1 2(v v )´r r osx ls xfr djs rks leku fcUnq ij blds }kjk vuqHko fd;k x;k cy gksxk&

(A) q(ai b j)+$ $ (B) bqj- $ (C) q(bi a j)+$ $ (D) None of these

7. Two tuning forks A and B give 6 beats/second. A resound with a closed column of air 15 cm long

and B with an open column 30.5 cm long. The frequencies of A and B are respectively-

nks Lofj= A rFkk B , 6 foLiUn/lSd.M nsrs gSaA A, 15 cm yEcs can ok;q LrEHk ds lkFk rFkk B, 30.5 cm yEcs [kqys

LrEHk ds lkFk vuqukfnr gksrk gSA A rFkk B dh vko`fÙk;k¡ Øe'k% gS%&

(A) 550 Hz, 544 Hz (B) 500 Hz, 494 Hz (C) 400 Hz, 394 Hz (D) 366 Hz, 360 Hz

8. A screw gauge has a screw having 2 threads in 1 mm. The circular scale has 50 divisions. Find the

diameter of wire, if the main scale shows 6th division and the vernier reads 46.

,d LØwxst ds LØw esa 2 pwfM+;k¡ (thread)1 mm nwjh ij gSA o`Ùkh; iSekus esa 50 Hkkx gSaA ;fn eq[; iSekuk 6 Hkkx rFkk

ofuZ;j 46 Hkkx i<+rk gS rks rkj dk O;kl Kkr djks\

(A) 6.46 mm (B) 3.46 min (C) 6.54 mm (D) 3.04 mm


PHYS

ICS

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Kota/00CT214005

(ii) Paragraph Type

(ii) vuqPNsn izdkjThis 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'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 vuqPNsnn identical rods each of mass m are welded at their ends to form a regular polygon and the corners arethen welded to a metal ring of radius R and mass M, such that the plane of polygon and plane of ring arein same plane and centres of polygon and ring coincide.,d le cgqHkqt cukus ds fy;s izR;sd m nzO;eku dh n le:i NM+ksa ds fljksa dks tksM+k tkrk gS rFkk R f=T;k rFkkM nzO;eku dh /kkfRod oy; dks blds fljksa ls bl izdkj tksM+k tkrk gS fd cgqHkqt rFkk oy; dk ry leku ry esa gks rFkkbuds dsUnz laikrh gksA

9. The moment of inertia of the system about an axis passing through the centre of mass of system andperpendicular to the plane of system will be :-fudk; ds ry ds yEcor~ rFkk fudk; ds nzO;eku dsUnz ls gksdj xqtjus okyh v{k ds lkis{k fudk; dk tM+Ro vk?kw.kZgksxk&

(A)

2

2 2 2sin

nnmR cos MR3 n

pé ùê úp

+ +ê úë û(B)

2

2 2 2tan

nnmR sin MR4 n

pé ùê úp

+ +ê úë û

(C)

2

2 2 2sin

nnmR cos MR4 n

pé ùê úp

+ +ê úë û(D)

2 2

2 2sin cos

n nnmR MR4 3

p pé ùê ú

+ +ê úë û

10. If the rigid assembly of rods and hoop is allowed to roll down the incline of inclination q, the minimum

value of the coefficient of static friction that will prevent slipping will be (Where I is moment of inertia

about centre of mass) :-

;fn NM+ksa rFkk fNnz ; qDr pdrh ds n`<+ la;kstu dks q dks.k ds urry ij uhps dh vksj yq<+dus fn;k tkrk gS rks fQlyu

dks jksdus ds fy;s LFkSfrd ?k"kZ.k xq.kkad dk U;wure eku gksxk (tgk¡ I nzO;eku dsUnz ds lkis{k tM+Ro vk?kq.kZ gSA)

(A) 2

I tan

2I (M nm)R

q+ + (B) 2

I sin

I (M nm)R

q+ + (C) 2

I cos

2I (M nm)R

q+ + (D) 2

I tan

I (M nm)R

q+ +


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Kota/00CT214005


iz'u 11 ,oa 12 ds fy;s vuqPNsnIn presence of the field the charge carriers in a solid behave as if they had an effective mass m* differentfrom the mass m of a charge carriers in absence of any field. The knowledge of value of m* is veryimportant for the understanding of the detailed theory of conduction in solid. The same can beexperimentally obtained by using a cyclotron in which apart from magnetic field some external electricfield is also applied which is varied till resonance is obtained. This is called cyclotron resonance.Cyclotron resonance describes the interaction of external forces with charged particles experiencing amagnetic field thus already moving on a circular path.Consider the case shown below :

w

B

E

Assuming mean free path l ® ¥ for the electron, they move in helical orbits as shown under the influenceof magnetic field. An electric field of constant magnitude, whose direction rotates with angular frequencyw, is also imposed on the solid in the plane shown.In the experiment w is fixed and the value of B is adjusted until the amount of input energy requiredto keep the amplitude of electric field constant reaches maximum.fdlh Bksl esa {ks= yxkus ij mlesa fo|eku vkos'k okgd bl izdkj O;ogkj djrs gSa tSls fd mudk izHkkoh æO;eku m*

gS tks fd fdlh Hkh izdkj ds {ks= dh vuqifLFkfr esa vkos'k okgdksa ds æO;eku m ls fHkUu gksrk gSA fdlh Bksl esa pkyudh foLrr O;k[;k djus ;k le>us ds fy, bl æO;eku m* dh tkudkjh gksuk vR;Ur vko';d gSA izk;ksfxd rkSj ij;g lc ,d lkbDyksVªkWu }kjk izkIr fd;k tk ldrk gS ftlesa pqEcdh; {ks= ds vykok vU; ckg~; fo|qr {ks= Hkh yxk;ktkrk gS tks fd vuqukn izkIr gksus rd ifjofrZr fd;k tkrk gSA bl voLFkk dks lkbDyksVªkWu vuqukn dgk tkrk gSAlkbDyksVªkWu vuqukn vkosf'kr d.kksa dh ckg~; cyksa ds lkFk vU;ksU; fØ;k dks n'kkZrk gSA bu vkosf'kr d.kksa ij ,dpqEcdh; {ks= yx jgk gS] ftlds dkj.k ; s igys ls gh ,d o`Ùkkdkj iFk ij xfr'khy gksrs gSaAvc ge fuEu fLFkfr ij fopkj djrs gSaA

w

B

E

ekuk pqEcdh; {ks= ds izHkko esa gsfydy d{kkvksa esa xfr'khy bysDVªkWu ds fy, ek/; eqä iFk l ® ¥ gS] fp= ns[ksaAfu;r ifjek.k okys ,d fo|qr {ks= ftldh fn'kk dks.kh; vkofÙk w ds lkFk ifjofrZr gksrh gS] dks Hkh bl ry esa fp=kuqlkjbl Bksl ij vkjksfir fd;k tkrk gSAbl iz;ksx essa w fu;r dj fn;k tkrk gS rFkk B dk eku rc rd O;ofLFkr fd;k tkrk gS tc rd fd fo|qr {ks= dsvk;ke dks fu;r cuk, j[kus ds fy, vko';d fuos'kh ÅtkZ dh ek=k vius vf/kdre eku rd uk igq¡p tk,A

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Kota/00CT214005

11. The effective mass m* of electron is :-

(A) 2qB

w(B)

qB

w

(C) qB

2w(D) Depends on electric field amplitude

bysDVªkWu dk izHkkoh æO;eku m* gS :-

(A) 2qB

w(B)

qB

w

(C) qB

2w(D) fo|qr {ks= vk;ke ij fuHkZj djrk gSA

12. In actual situation l ¹ ¥. So in order to make l practically very large so that resonance effect is prominentwe can conduct the experiment at :-

(A) High temperature and low frequency (w) (B) High temperature and high frequency (w)

(C) Low temperature and high frequency (w) (D) Low temperature and low frequency (w)

okLrfod fLFkfr esa l ¹ ¥ gksrk gSA vr% iz;ksfxd rkSj ij l dk eku bruk vf/kd fd;k tk, fd vuquknh izHkko vis{kkd`rvf/kd izHkkoh gks rks blds fy, ;g iz;ksx djuk gksxk %&

(A) mPp rkieku rFkk fuEu vkofÙk (w) ij (B) mPp rkieku rFkk mPp vkofÙk (w) ij

(C) fuEu rkieku rFkk mPp vko`fÙk (w) ij (D) fuEu rkieku rFkk fuEu vko`fÙk (w) ij


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SECTION–II : Matrix-Match Type

[k.M - II : eSfVªDl&esy izdkj

This Section contains 2 questions. Each question has four statements (A, B, C and D) given in

Column I and five statements (P, Q, R, S and T) in Column II. Any given statement in Column I can have

correct matching with ONE or MORE statement(s) given in Column II. For example, if for a given

question, statement B matches with the statements given in Q and R, then for the particular question, against

statement B, darken the bubbles corresponding to Q and R in the ORS.

bl [k.M esa 2 iz'u gSA izR;sd iz'u esa dkWye I esa 4 dFku (A, B, C vkSj D) vkSj dkWye II esa 5 dFku (P, Q, R, S vkSj T)

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

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

lEcfUèkr cqycqyksa dks dkyk dhft;sA

1. Column-I gives some current distributions and a point P in the space around these current distributions.

Column-II gives some expressions of magnetic field strength. Match column-I to corresponding field

strength at point P given in column-II

Column – I Column – II

(A) A conducting loop shaped as regular hexagon of side x, (P) 03 i

32 x

mp

carrying current i. P is the centroid of hexagon

(B) A cylinder of inner radius x and outer radius 3x, carrying (Q) 03 i

x

mp

current i. Point P is at a distance 2x from the axis of

the cylinder

(C) Two coaxial cylinders of radii x and 2x, each carrying (R) 0i

2x

m

current i, but in opposite is hollow. P is a point at distance

1.5x from the axis of the cylinders

(D) Magnetic field at the centre of an n-sided regular (S) 0i

3 x

mp

polygon, of circum circle of radius x, carrying current

i, n ® ¥, P is centroid of the polygon.

(T) 03 i

32 x

mp


PHYS

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Kota/00CT214005

LrEHk-I esa dqN /kkjk forj.k rFkk bu /kkjk forj.kksa ds pkjksa vksj lef"V esa ,d fcUnq P fn;k x;k gSA LrEHk–II esa pqEcdh;{ks= lkeF;Z ds dqN O;atd fn;s x;s gSA LrEHk–I dk LrEHk–II esa fcUnq P ij laxr {ks= lkeF;Z ds lkFk feyku dhft,A

LrEHk– I LrEHk– II

(A) Hkqtk x okys le"kV~Hkqt ds vkdkj okys pkyd ywi esa (P) 03 i

32 x

mp

/kkjk i izokfgr gksrh gSA "kVHkqt dk dsUæd P gSA

(B) vkarfjd f=T;k x rFkk ckg~; f=T;k 3x okys csyu esa i /kkjk (Q) 03 i

x

mp

izokfgr gksrh gSA fcUnq P csyu dh v{k ls 2x nwjh ij gSA

(C) f=T;k x rFkk 2x okys lek{kh; [kks[kys csyu ftlesa ijLij foijhr (R) 0i

2x

m

fn'kk esa /kkjk i izokfgr gksrh gSA fcUnq P csyuksa dh v{k ls1.5 x nwjh ij gSA

(D) f=T;k x okys ifjoÙk okys n-Hkqtkvksa okys le cgqHkqt ds (S) 0i

3 x

mp

dsUæ ij pqEcdh; {ks= ftlesa i /kkjk izokfgr gksrh gSn ® ¥ gSA fcUnq P cgqHkqt dk dsUæd gSA

(T) 03 i

32 x

mp


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2. Column-I shows some curves, and their properties in part of space are given in column-II.Here C – represents a closed loop

S – represents a closed surfaceMatch the Column-I with Column-II.

Column–I Column–II

(A)

C

S

Density of field lines first increases then decreases

as we move radially outward

(P) Such curves can represent both electric andmagnetic field lines

(Q)C

H.d 0¹òrrlÑ

where Hr

represents electric or magnetic field

(R)C

H.d 0=òrrlÑ

where Hr


(S)SH.dA 0=ò

rr

Ñ

(B)C

S

Density of field lines continuously decreases as we move radially outward

where Hr


(T)SH.dA 0¹ò

rr

Ñ

where Hr


(C)C

S

Axi

s

(D)

Axi

s

C

S

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LrEHk-I esa dqN oØ fn[kk;s x;s gS rFkk lef"V ds fdlh Hkkx esa muds xq.kksa dks LrEHk-II esa n'kkZ;k x;k gSA;gk¡ C – can ywi dks iznf'kZr djrk gSA

S – can i`"B dks iznf'kZr djrk gSALrEHk-I dk LrEHk-II ds lkFk feyku dhft,

LrHk–I LrEHk–II

(A)C

S

f=T;h; ckgj dh rjQ tkus ij {ks= js[kkvksa dk ?kuRo igys c<+rk gS rFkk fQj ?kVrk gSA

(P) bl izdkj ds oØ fo|qr rFkk pqEcdh; nksuksa rjg dh{ks= js[kkvksa dks iznf'kZr dj ldrs gSA

(Q)C

H.d 0¹òrrlÑ

;gk¡ Hr

fo|qr vFkok pqEcdh; {ks= dks iznf'kZr djrk gSA

(R)C

H.d 0=òrrlÑ

;gk¡ Hr


(S)SH.dA 0=ò

rr

Ñ

;gk¡ Hr


(T)SH.dA 0¹ò

rr

Ñ

(B)

C

S

f=T;h; ckgj dh rjQ tkus ij {ks= js[kkvksa dk ?kuRo yxkrkj ?kVrk tkrk gSA

;gk¡ Hr


(C) C

S

Axi

s

(D)

Axi

s

C

S

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

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

[k.M-IV : (iw.kk±d eku lgh izdkj)

This section contains 8 questions. The answer to each question is a single digit Integer, ranging from

0 to 9 (both inclusive)

bl [k.M esa 8 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 pendulum consist of a disk of mass M and radius R and a massless rod of length l. Find approximate

time period (in sec) of the system if the disk is mounted to the rod by a frictionless bearing so that it is

perfectly free to spin? Assume small oscillation. (Take : l = 0.993m, g = 9.8 m/s2)

iznf'kZr yksyd æO;eku M rFkk f=T;k R okyh ,d pdrh rFkk yEckbZ l okyh ,d æO;ekughu NM+ ls feydj cuk gSA

;fn bl pdrh dks ?k"kZ.k jfgr fc;fjax dh lgk;rk ls NM + ij bl izdkj yxk fn;k tk;s fd ;g pØ.k djus ds fy;s

iw.kZr; Lora= gks rks bl fudk; dk yxHkx vkorZdky (lsd.M esa) Kkr dhft;sA ekuk vYi nksyu gksrs gSA

(fn;k gS : l = 0.993m, g = 9.8 m/s2)

R

Ml

2. Two charges q and –2q are placed at (–3,0) and (3,0) in x–y plane. The locus of the point in the plane of

the charges, where the field potential is zero is (x + a)2 + y2 = 4b2. Find the value of (a+b).

nks vkos'kksa q rFkk –2q dks x-y ry esa (–3,0) rFkk (3,0) ij j[kk x;k gSA vkos'kksa ds ry esa tgk¡ {ks= foHko 'kwU; gS] fcUnq

dk fcUnq iFk (x + a)2 + y2 = 4b2 gSA (a + b) dk eku Kkr dhft,A


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3. A plastic rod of length 1.0 m carries uniform positive charge +4.0 mC/m on half of its length and uniform

negative charge –4.0 m C/m on the remaining half of its length. Find magnitude of it’s net dipole moment

in mC–m.

,d IykfLVd NM+ dh yEckbZ 1.0 m gSA bldh yEckbZ ds vk/ks Hkkx ij ,dleku /kukRed vkos'k +4.0 mC/m rFkk

yEckbZ ds 'ks"k vk/ks Hkkx ij ,dleku ½.kkRed vkos'k -4.0 m C/m gSA bldk dqy f}/kq zo vk?kw.kZ dk ifjek.k mC-m esa

Kkr dhft,A

4. In a Coolidge tube the atomic number of target material is 41. A potential difference of 20 kV is applied

across the tube. Let lK be Ka line produced by tube and l

min be cut off wavelength. Calculate ( )1

K min2 l - l

in the order of 10–8 m.

dwyht uyh esa iz;qDr y{; inkFkZ dk ijek.kq Øekad 41 gSA uyh ds fljksa ij 20kV foHkokUrj vkjksfir fd;k tkrk gSA

ekuk lK ] uyh }kjk mRiUu Ka js[kk rFkk vard rjaxnS/;Z l

min gSA ( )1

K min2 l - l dh x.kuk 10-8 m dksfV esa dhft,A


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5. A thermonuclear device consists of a torus of mean diameter 3 m with a tube of diameter 1 m, containing

deuterium gas at 10–2 mm mercury pressure and at room temperature (20°C). A bank of capacitors of

1200 mF is discharged through the tube at 50 kV. If only 20% of the electrical energy is transformed to

plasma kinetic energy, the maximum temperature attained is equals to 1.18 × 10a K. Assuming that the

energy is equally shared between the deuterons and electrons in the plasma. Find the value of a .

,d rki&ukfHkdh; ;qfDr ds Vksjl (torus) dk ek/; O;kl 3 m gSA blesa 1m O;kl dh ,d uyh yxh gqbZ gS ftlesa

10–2 mm ikjn nkc rFkk 20°C dejs ds rki ij M~;wVhfj;e xSl Hkjh gqbZ gSA bl uyh }kjk 50kV ij 1200 mF ds

la/kkfj=ksa ds ,d lewg dks fujkosf'kr fd;k tkrk gSA ;fn fo|qr ÅtkZ dk dsoy 20% Hkkx gh IykTek xfrt ÅtkZ esa

ifjofrZr gksrk gS rks vf/kdre rkieku 1.18 × 10a K ds cjkcj izkIr gksrk gSA ekuk fd IykTek esa ÅtkZ M~;wVsjksuks rFkk

bysDVªkWuksa ds e/; leku :i ls forfjr gksrh gSA a dk eku Kkr dhft,A

6. In the given circuit find the value of I in amperes.

fn;s x;s ifjiFk esa /kkjk I dk eku (,fEi;j esa) Kkr dhft,A5V 5W

5W5W

5W

5W 5W

5W5W5W

5W

5W

10V

I

5V 5W


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7. An electric dipole is placed along x–axis at the origin as shown in the figure. Electric field at the point P

is parallel to the y–axis. If x–coordinate of P is 2 m, what is its y–coordinate in meters.

,d fo|qr f}/kqo fp=kuqlkj x-v{k ds vuqfn'k ewy fcUnq ij fLFkr gSA fcUnq P ij fo|qr {ks=] y- v{k ds lekUrj gSA ;fn

P dk x- funsZ'kkad 2 m gks rks y-funsZ'kkad dk eku ehVj esa Kkr dhft,A

y

P(x,y)

p x

8. In the given circuit for what value of n, power generated in resistance R will be minimum.

fn;s x;s ifjiFk esa n ds fdl eku ds fy, izfrjks/k R esa mRiUu 'kfDr dk eku U;wure gksxk\

(n)W

(n)W (n+2)W

(n+3)W

(n+1)W (n+4)W

(R)W

(n/5)W

E

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

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

This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and(D) out of which ONLY ONE is correct.bl [k.M esa 8 cgqfodYi iz'u gSA izR;sd i'u esa pkj fodYi (A), (B), (C) vkSj (D) gSa ftuesa ls dsoy ,dlgh gSA

1. Calculate the percentage hydrolysis in 0.01M aqueous solution of NaOCN

(Kb for OCN– = 10–10)

NaOCN ds 0.01M tyh; foy;u esa ty vi?kVu dk izfr'kr crkb;s

(OCN– ds fy, Kb

= 10–10 gS)

(A) 0.1 % (B) 0.01 % (C) 0.0001 % (D) 0.001 %

2. Identify the correct statement-

(A) In an isolated system DU > 0 and DS > 0 for an irrversible process

(B) In an isolated system DU = 0 and DS = 0 for an irrversible process

(C) In an isolated system DU > 0 and DS = 0 for an irrversible process

(D) In an isolated system DU = 0 and DS > 0 for an irrversible process

lgh dFku dks igpkfu,sa -

(A) ,d foyfxr ra= esa DU > 0 rFkk vuqRØe.kh; izØe ds fy, DS > 0

(B) ,d foyfxr ra= esas DU = 0 rFkk vuqRØe.kh; izØe ds fy, DS = 0

(C) ,d foyfxr ra= esa DU > 0 rFkk vuqRØe.kh; izØe ds fy, DS = 0

(D) ,d foyfxr ra= esa DU = 0 rFkk vuqRØe.kh; izØe ds fy, DS > 0

3. The resistance of 0.1 M solution of oxalic acid is 200 ohm and cell constant is 2.0 cm–1, the molar

conductance (in S cm2 mole–1) of 0.1 M oxalic acid is :-

vkWDlsfyd vEy ds 0.1 M foy;u dk izfrjks/k 200 vkse rFkk lsy fu;rkad 2.0 cm–1 gS rks 0.1 M vkWDlsfyd vEy

dk eksyj pkydRo (S cm2 mol–1 esa) gS :-

(A) 100 (B) 10 (C) 1 (D) 0.1


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4. Which of the following is not possible structure for any oxide of nitrogen

fuEu esa ls dkSulh lajpuk ukbVªkstu ds fdlh Hkh vkWDlkbM ds fy;s lEHko ugha gS

(I) :

N O N——+ :

+ O

– ::

–:O

::O

:::O

::

(II)

:

N O N——: : +

O

O....

..

..–O....

(III) O = N—N+ O

..

..–O ......

......

(IV) N = O N—++ –2: :

:

:

(A) II, IV (B) III, IV (C) I only (dsoy) (D) II, III and (rFkk) IV

5. Select reaction in which anion of main ionic product must have s-bond :

(A) Li + O2(Excess) ® (B) Na(Excess) + O

2 ®

(C) Ca + F2(Excess) ® (D) NaOH(cold) + Cl

2 ®

og vfHkfØ;k pqfu, ftlds eq[; vk;fud mRikn ds ½.kk;u esa s-ca/k mifLFkr gks :

(A) Li + O2(vkf/kD;) ® (B) Na(vkf/kD;) + O

2 ®

(C) Ca + F2(vkf/kD;) ® (D) NaOH(BaMk) + Cl

2 ®

6. CH –OH2

CH–OH

CH –OH2

When above compound is treated with excess HI product formed is :-

(A) 3-iodoprop-1-ene (B) 1, 2, di-iodopropane

(C) Propene (D) 2-iodopropane

CH –OH2

CH–OH

CH –OH2

tc mijksDr ;kSfxd dks HI vkf/kD; ds lkFk mipkfjr djrs gS rks cuus okyk mRikn gS&

(A) 3-vk;Mksizksi-1-bZu (B) 1, 2, Mkb-vk;Mksizksisu

(C) izksihu (D) 2-vk;Mksizksisu

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7.O

H O3

+ NH –NH2 2 (i) HBrX

OH /— D (ii) OEt /— DY Z (Major product) / eq[; mRikn

"Z" is (gS)

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

8. CH2CO2NaCH2CO2Na

Electrolysis (A) Br2(CCl4)

(B) alc.1.KOH (C)followed by NaNH22. H+Sodium succinate

C is -

(A) CH2

CH2

(B) CH3

CH3

(C) CHCH

(D) CH2

CH–CH3

CH2CO2NaCH2CO2Na

(A) Br2(CCl4)

(B) alc.1.KOH (C)

NaNH22. H+

fo|qr vi?kVu

lksfM;e lDlhusV

ds ckn

C gS-

(A) CH2

CH2

(B) CH3

CH3

(C) CHCH

(D) CH2

CH–CH3

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(ii) Paragraph Type (ii) 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

Enthalpy of combustion of a given compounds is defined as following. It is the change in enthalpy when1 mole of this compound combines with required amount of O2 to give product in their stable forms.The enthalpy of combustion is usually measured by placing a known mass of the compound in a closedsteel cointainer ( known as bomb-calorimeter)fn;s x, ;kSfxdks ds ngu dh ,sUFkSYih;k¡ fuEu izdkj ls ifjHkkf"kr dh xbZ gS ,sUFkSYih esa ifjoRkZu gksrk gS tc bl ;kSfxddk 1 eksy blds LFkk;h :i esa mRikn izkIr djus ds fy, O2 dh vko';d ek=k ds lkFk ;ksx djrk gSA ngu dh,sUFkSYih dks lkekU;r% LVhy ds ,d cUn ik= esa ;kSfxd ds Kkr nzO;eku dks j[kus ls ekik tkrk gSA (ftls ce dSyksjhehVj ds :i esa tkuk tkrk gSA)

9. Given the data below :0combHD (CH4, g) = –900 kJ/mole ; 0

combHD (C, graphite) = – 400 kJ/mole

0combHD (H2, g) = – 300 kJ/mole ; 0

atomizationHD (C, graphite) = 700 kJ/mole

and 0fH (H,g)D = 220 kJ/mole

What should be the bond enthalpy of C–H bond in kJ/mole.

(A) 400 (B) 420 (C) 310 (D) 345

uhpsa dqN vk¡dM+s fn;s x, gS :0HD ngu (CH4, g) = –900 kJ/mole ; 0HD ngu (C, xzsQkbV) = – 400 kJ/mole

0HD ngu (H2, g) = – 300 kJ/mole ; 0HD ijekf.o;dj.k (C, xzsQkbV) = 700 kJ/mole

rFkk 0fH (H,g)D = 220 kJ/mole

C–H cU/k dh cU/k ,saUFkSYih kJ/mole esa D;k gksuh pkfg,A(A) 400 (B) 420 (C) 310 (D) 345

10. Identify the incorrect statement -

(A) 0combHD (C, graphite) is equal to 0

fHD (CO2, g)

(B) 0 0Cl Cl atom 2H H [Cl (g)]-D = D

(C) The standard enthalpy of combustion is always negative.(D) The standard enthalpy of formation is always negativexyr dFku dk p;u dhft, -

(A) 0combHD (C, xzsQkbV) , 0

fHD (CO2, g) ds cjkcj gksrh gS

(B) 0 0Cl Cl atom 2H H [Cl (g)]-D = D

(C) ngu dh ekud , sUFkSYih lnSo ½.kkRed gksrh gS(D) fuekZ.k dh ekud , sUFkSYih lnSo ½.kkRed gksrh gS

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

Mixture of two salts S1 and S2 of same metal gives following observation

• Solution of S1 + S2 in dil HCl gives white ppt with BaCl2 which is not soluble in dil HNO

3 ?

• Mixture of S1 and S2 not soluble in water but soluble in dil HCl with the evolution of a colourless

and odourless gas(G) which turns Ba(OH)2 milky.

• Solution of S1 + S2 in dil HCl does not give ppt with H2S but gives white ppt(p) with NaOH solution

which is soluble in exess NaOH(aq)

.

leku /kkrq ds nks yo.kksa S1 rFkk S2 ds feJ.k ls fuEu izs{k.k izkIr gq;s

• S1 + S2 dk ruq HCl esa foy;u BaCl2 ds lkFk 'osr vo{ksi nsrk gS tks ruq HNO

3 esa foys; ugha gSa ?

• S1 rFkk S2 dk feJ.k ty esa foys;'khy ugha gS ysfdu ruq HCl esa foys; gksdj ,d jaxghu rFkk xU/kghu xSl(G)

nsrk gS tks Ba(OH)2 dks nwf/k;k dj nsrh gSA

• S1 + S2 dk ruq HCl esa foy;u H2S ds lkFk dksbZ vo{ksi ugha nsrk ysfdu NaOH foy;u ds lkFk 'osr vo{ksi

nsrk gS tks NaOH(aq)

ds vkf/kD; esa foys; gSA

11. Salt S1 and S2 are respectively :

yo.k S1 rFkk S2 Øe'k% gSa :

(A) PbSO4, PbCO

3(B) PbCO

3, ZnSO

4(C) ZnS, ZnSO

4(D) ZnSO

4 + ZnCO

3

12. Solution of S1 + S2 in dil. HCl gives white ppt with :

(A) K4[Fe(CN)

6] (B) Excess NH

4OH (C) H

2S

(g)(D) All of the above

S1 + S2 dk ruq HCl esa foy;u] fuEu esa ls fdlds lkFk 'osr vo{ksi nsrk gS :

(A) K4[Fe(CN)

6] (B) NH

4OH vkf/kD; (C) H

2S

(g)(D) mijksDr lHkh


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

This Section contains 2 questions. Each question has four statements (A, B, C and D) given inColumn I and five statements (P, Q, R, S and T) in Column II. Any given statement in Column I can havecorrect matching with ONE or MORE statement(s) given in Column II. For example, if for a givenquestion, statement B matches with the statements given in Q and R, then for the particular question, againststatement B, darken the bubbles corresponding to Q and R in the ORS.


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

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

lEcfUèkr cqycqyksa dks dkyk dhft;sA1. Column-I Column-II

(Conversion) (Process / Furnace involved in given conversion)

(A) Copper matte ¾¾® Cu (Blister) (P) Electrolytic reduction

(B) Pure Heamatite ¾¾® Fe(Pig) (Q) Self reduction

(C) Concentrated Galena ¾¾® PbO (R) Carbon reduction

(D) Red Bauxite ¾¾® Al(Pure) (S) S2– (sulphide) is oxidised

(T) Blast Furnace

LrEHk-I LrEHk-II

(:ikUrj.k) (fn;s x;s :ikUrj.k es a lfEefyr izØe @ HkV~Vh)

(A) dkWij eSV ¾¾® Cu (QQksysnkj) (P) fo|qrvi?kVuh; vip;u

(B) 'kq¼ ghesVkbV ¾¾® Fe(dPpk) (Q) Lo&vip;u

(C) lkfUær xSysuk ¾¾® PbO (R) dkcZu vip;u

(D) yky ckWDlkbV ¾¾® Al('kq¼) (S) S2– (lYQkbM) vkWDlhd`r gksrk gS

(T) okR;k HkV~Vh


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2. Column - I Column - II(Reactions) (About major organic product)

(A)

OHH

PCl3 (P) Optically inactive mixture

(B)

OHH

HI (Q) Optically active product

(C)

OHH

PDC (R) Product will give precipitate when treated with

alcoholic AgNO3

(D)

OHH

NaMe – I

(S) Product formation involve substitution reaction.

(T) Product will react with CH3MgCl to produce

hydrocarbon

LrEHk lqesfyr dhft,&LrEHk- I LrEHk- II(vfHkfØ;k,sa) (eq[; dkcZfud mRikn ds lnaHkZ esa)

(A)

OHH

PCl3 (P) izdkf'kd vfØ; feJ.k

(B)

OHH

HI (Q) izdkf'kd lfØ; mRikn

(C)

OHH

PDC (R) mRikn vo{ksi nsxk tc ,YdksgkWfy; AgNO3 ds lkFk mipkfjr

djrs gSA

(D)

OHH

NaMe – I

(S) mRikn fuekZ.k esa izfrLFkkiu vfHkfØ;k lfEefyr gksrh gSA

(T) mRikn dh CH3MgCl ds lkFk vfHkfØ;k djkus ij gkbMªksdkcZu

dk fuekZ.k gksxkA


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


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SECTION-IV : (Integer Value Correct Type)

[k.M-IV : (iw.kk±d eku lgh izdkj)

This section contains 8 questions. The answer to each question is a single digit Integer, ranging from


bl [k.M esa 8 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. When a metal is exposed with light of wavelength l, the maximum kinetic energy of electron

produced was found to be 2 eV. When the same metal was exposed with light of wavelength l/2,

the maximum kinetic energy of elelctron produced was 6 eV. What is the value of work function of

metal in eV.

tc ,d /kkrq dks l rjaxnS/; Z ds izdk'k esa j[kk tkrk gS] rc mRikfnr bySDVªkWu dh vf/kdre xfrt ÅtkZ 2 eV ik;h x;hA

tc blh /kkrq dks l/2 rjaxnS/; Z okys izdk'k esa j[kk x;k rks mRikfnr bySDVªkWu dh vf/kdre xfrt ÅtkZ 6 eV FkhA /kkrq

ds dk;ZQyu dk eku eV esa D;k gksxkA

2. Calculate the change in pressure (in atm) when 2 mol of NO and 16 gm O2 in a 6.25 litre originally at

27º C react to produce the maximum quantity of NO2 possible according to the equation,

2 NO (g) + O2 (g) ® 2NO

2 (g)

(Take R = 1

12 litre atm /mol K)

nkc esa ifjorZu dh x.kuk (atm esa ) dhft, tc 27º C ij 6.25 yhVj okLrfod ek=k esa 2 eksy NO rFkk 16 gm O2

vfHkfØ;k djds NO2 dh vf/kdre ek=k fuEu lehdj.k ds vuqlkj mRikfnr djrs gS ,

2 NO (g) + O2 (g) ® 2NO

2 (g)

(fn;k gS R = 1

12 litre atm /mol K)


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3. Element X can crystllizes in fcc lattice as well as in bcc lattice. If density of bcc lattice is 3 3

2g/cm3 then

what is density of fcc lattice in gm/cm3.

rRo X, fcc tkyd ds lkFk&lkFk bcc tkyd esa Hkh fØLVyhd`r gks ldrk gSA ;fn bcc tkyd dk ?kuRo 3 3

2g/cm3

gS rks fcc tkyd dk ?kuRo] gm/cm3 esa D;k gS \

4. Out of NiS, CoS, MnS, ZnS

P = Number of black precipitates.

Q = Number of sulphides which is / are soluble in dil. HCl

R = Number of sulphides which is / are soluble in dil. CH3COOH

S = Number of sulphides which is / are soluble in hot / conc. HNO3.

(P + Q + R + S = Total number is your answer)

NiS, CoS, MnS, ZnS esa ls

P = dkys vo{ksiksa dh la[;k

Q = ,sls lYQkbMksa dh la[;k tks ruq HCl esa foys; gS @ gSa

R = ,sls lYQkbMksa dh la[;k tks ruq CH3COOH esa foys; gS @ gSa

S = ,sls lYQkbMksa dh la[;k tks xeZ @ lkUæ HNO3 esa foys; gS @ gSa

(P + Q + R + S = dqy la[;k gh vkidk mÙkj gS)


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5. "Number of geometrical isomers are NOT changed in [Pt(Cl) (Br) (PMe3)

2]"

• When one PMe3 is replaced by NH

3

• When Cl¯ is replaced by Br¯

• When Br¯ is replaced by CN

• When both PMe3 are replaced by one en(ethylene diamine)

• When one Cl and one PMe3 are replaced by one en(ethylene diamine)

• When all the Me–groups are replaced by Et–groups

Find number of condition(s) given above which make the underlined statement CORRECT :

"[Pt(Cl) (Br) (PMe3)

2] esa T;kfefr leko;fo;ksa dh la[;k ugha cnyrh gS"

• tc ,d PMe3 dks NH

3 }kjk foLFkkfir fd;k tkrk gS

• tc Cl¯ dks Br¯ }kjk foLFkkfir fd;k tkrk gS

• tc Br¯ dks CN }kjk foLFkkfir fd;k tkrk gS

• tc nksuksa PMe3 dks ,d en(,sfFkyhu Mkb,sehu) }kjk foLFkkfir fd;k tkrk gS

• tc ,d Cl rFkk ,d PMe3 dks ,d en(,sfFkyhu Mkb,sehu) }kjk foLFkkfir fd;k tkrk gS

• tc lHkh Me–lewgksa dks Et–lewgksa }kjk foLFkkfir fd;k tkrk gS

mijksDr esa ls , slh fLFkfr;ksa dh la[;k crkbZ;sa tks js[kk afdr dFku ds fy, lgh gSa :

6. PhCO2H + NaHCO3 ® A (gas) xSl

PhOH + NaNH2 ® B (gas) xSl

CH3OH + NaH ® C (gas) xSl

CH3SH + EtLi ® D (gas) xSl

PhSO3H + Na ® E (gas) xSl

If the molecular weight of A, B, C, D ,E is x, y, z, w, v then find the value of (y + z + w + v) – x

;fn A, B, C, D rFkk E ds v.kqHkkj Øe'k% x, y, z, w, v gS rks (y + z + w + v) – x dk eku Kkr dhft,A


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7. For the given reaction sequence :

fn;s x;s vfHkfØ;k Øe esa

COOEtCOOEt

NaOEtD

Å – P

H O3Å

Q D

(R) Zn-Hg/HCl (S) NBS/hv ?

Number of monohalogenated products is/are -

eksuksgsykstuhd`r mRiknksa dh la[;k gSA

8. How many these reactions will give 1º alcohol as major product :

fuEu esa ls fdruh vfHkfØ;k,sa eq[; mRikn ds :i esa 1º ,YdksgkWy nsxhA

(1) O 2H / H O+

¾¾¾¾® (2)

O

+ CH3MgBr ¾® 2H O¾¾¾®

(3) HCHO + CH3MgBr ¾® 2H O¾¾¾® (4) CH3CHO + CH3MgBr ¾® 2H O¾¾¾®

(5) CH –C–CH –C–CH +CH MgBr3 2 3 3

||O

||O

¾® 2H O¾¾¾® (6) ||O

Cl + CH3MgBr ¾® 2H O¾¾¾®

Excess (vkf/kD;)

(7) O

+ CH3MgBr ¾® 2H O¾¾¾® (8) CH3CH2Br NaOH(Aq.)¾¾¾¾®

(9) CH3MgBr + CO2 ¾® 2H O¾¾¾®


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PART-3 : MATHEMATICS

Hkkx-3 : xf.krSECTION–I : (i) Only One option correct Type

[k.M-I : (i) dsoy ,d lgh fodYi izdkjThis section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and

(D) out of which ONLY ONE is correct.

bl [k.M esa 8 cgqfodYi iz'u gSA izR;sd i'u esa pkj fodYi (A), (B), (C) vkSj (D) gSa ftuesa ls dsoy ,d

lgh gSA

1. The equation 5 4 3 2x x x x

x 1 05! 4! 3! 2!

+ + + + + = have -

(A) 5 distinct real roots (B) exactly 3 distinct real roots

(C) exactly one real root (D) exactly one positive real root

lehdj.k 5 4 3 2x x x x

x 1 05! 4! 3! 2!

+ + + + + = ds -

(A) 5 fHkUu okLrfod ewy gksaxsA (B) Bhd 3 fHkUu okLrfod ewy gksaxsA

(C) Bhd ,d okLrfod ewy gksxkA (D) Bhd ,d /kukRed okLrfod ewy gksxkA

2. If 2

x1999

p= , then the value of cosx cos2x cos3x...........cos999x is equal to-

;fn 2

x1999

p= gks] rks cosx cos2x cos3x...........cos999x dk eku gksxk -

(A) 1 (B) 999

1

2(C) 1998

1

2(D) 999

1

2-


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3. If A =z 2z : 3, z Cz 2

ì - ü= Îí ý

+î þ and z1, z2, z3, z4 Î A are 4 complex numbers representing points

P, Q, R, S respectively on the complex plane such that z1 – z2 = z4 – z3, then maximum value of area ofquadrilateral PQRS is -

;fn A =z 2z : 3, z Cz 2

ì - ü= Îí ý

+î þ rFkk z1, z2, z3, z4 Î A, pkj lfEeJ la[;k;sa gSa] tks lfEeJ lery ij fLFkr fcUnqvksa

Øe'k% P, Q, R, S dks bl izdkj iznf'kZr djrh gS fd z1 – z2 = z4 – z3 gks] rks prqHkqZt PQRS ds {ks=Qy dk vf/kdreeku gksxk -

(A) 94 (B)

92 (C) 9 (D) 16

4. Let 2 2

2 2

x y1

a b+ = is an ellipse and ( )a 2 3 b= + ; (where a,b > 0). If P is a point on the ellipse, then the

least value of acute angle between the tangent of the ellipse at P and OP is (where O is origin)-

ekuk 2 2

2 2

x y1

a b+ = ,d nh?kZoÙk gS rFkk ( )a 2 3 b= + ; (tgk¡ a,b > 0) gSA ;fn P nh?kZoÙk ij ,d fcUnq gks] rks nh?kZoÙk

ds fcUnq P ij [khaph xbZ Li'kZjs[kk rFkk js[kk OP ds e/; U;wudks.k dk U;wure eku gksxk (tgk¡ O ewyfcUnq gS) -

(A) 12

p(B)

6

p(C)

4

p(D)

5

12

p

5. If all roots of equation x5 + ax4 + 90x3 + bx2 + cx – 243 = 0 are positive real numbers, then-

;fn lehdj.k x5 + ax4 + 90x3 + bx2 + cx – 243 = 0 ds lHkh ewy /kukRed okLrfod la[;k;sa gS] rks -

(A) a < b (B) a + b > 0 (C) b + c > 0 (D) c < 0

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6. Let ( )/ 2

13 1

1

0

I cosx dxp

+= ò and ( )

/ 213 1

2

0

I cosx dxp

-= ò , then value of 2

1

I

I

é ùê úë û

is equal to (where [.] greatest

integer function)(A) 0 (B) 1 (C) 2 (D) greater than 2

ekuk ( )/ 2

13 1

1

0

I cosx dxp

+= ò rFkk ( )

/ 213 1

2

0

I cosx dxp

-= ò gks] rks 2

1

I

I

é ùê úë û

dk eku gksxk (tgk¡ [.] egÙke iw.kk±d Qyu

dks n'kkZrk gS)

(A) 0 (B) 1 (C) 2 (D) 2 ls vf/kd

7. Let ƒ(x,y) = max(x2 + 2y, y2 + 4x) where x,y Î R, then least value of ƒ(x,y) is-

(A) –4 (B) –2 (C) 5

2- (D) Not defined

ekuk ƒ(x,y) = vf/kdre(x2 + 2y, y2 + 4x) tgk¡ x,y Î R gks] rks ƒ(x,y) dk U;wure eku gksxk -

(A) –4 (B) –2 (C) 5

2- (D) ifjHkkf"kr ugha

8. Raghav and his father are standing together on a circular track of radius 100 meters. When father givesa signal Raghav starts to run around the track at a speed of 10m/s. If the displacement of Raghav from

his father is increasing at the rate of l m/s when Raghav has run 1

4 of the way around the track then l

is equal to-

jk?ko rFkk mlds firk ,d 100 ehVj f=T;k ds o`Ùkh; iFk ij lkFk&lkFk [kM+s gq, gSaA firk ds b'kkjk djus ij jk?ko iFk ds

pkjksa rjQ 10m/s dh pky ls nkSM+uk 'kq: djrk gSA tc jk?ko o`Ùkkdkj iFk dk 1

4 Hkkx nkSM+ pqdk gks] ml le; ;fn jk?ko

dk mlds firk ls foLFkkiu l m/s dh nj ls c<+rk gS] rks l dk eku gksxk -

(A) 1

10(B) 5 (C) 10 2 (D) 5 2

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(ii) Paragraph Type (ii) 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


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

A slip of paper is given to A. who marks it with either a plus sign or a minus sign, the probability

of his writing a plus sign is known to be 1

3. He then passes the slip to B, who may either leave it

alone or change the sign before passing it on to C. Next C passes the slip to D after perhaps changing

the sign. Finally D passes it to an honest judge after perhaps, changing the sign. It is known that B,

C and D each change the sign with probability 2/3.

dkxt dh ,d iphZ A dks nh tkrh gSA tksfd ml ij ;k rks /kukRed fp ;k ½.kkRed fp vafdr djrk gSA

iphZ ij /kukRed fp ds fy[kus dh Kkr izkf;drk 1

3 gSA vc bl iphZ dks og B dks nsrk gSA tks bl iphZ dks

C dks nsus ls igys ;k rks ;Fkkor j[krk gS ;k fp ifjofrZr djrk gSA vc C bl iphZ dk fp ifjofrZr@vifjofrZrdjds D dks nsrk gSA vUr esa D fp ifjofrZr@vifjofrZr djds bl iphZ dks ,d bZekunkj U;k;/kh'k dks nsrk gSA;g Kkr gS fd B, C rFkk D izR;sd ds fp ifjofrZr djus dh izkf;drk 2/3 gSA

9. The probability that judge see a plus sign on the slip is -

iphZ ij U;k;/kh'k ds }kjk /kukRed fp ns[kus dh izkf;drk gksxh -

(A) 1

81(B)

17

81(C)

25

81(D)

41

8110. If the judge see a plus sign on the slip, then the probability that A originally wrote a plus sign is -

;fn U;k;/kh'k iphZ ij /kukRed fp ns[krk gS] rks A ds }kjk okLro esa iphZ ij /kukRed fp fy[kus dh izkf;drk gksxh -

(A) 1

41(B)

12

41(C)

13

41(D)

14

41


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iz'u 11 ,oa 12 ds fy;s vuqPNsn

Let S1 : 12x2 – 7xy – 12y2 + 25 = 0 is a hyperbola, L : y = 2x is a line and S2 is reflection of S1

in L.

ekuk S1 : 12x2 – 7xy – 12y2 + 25 = 0 ,d vfrijoy ;, L : y = 2x ,d js[kk rFkk oØ S1 dk js[kk L esa

izfrfcEc oØ S2 gSA

11. Equation of S2 is xy = c, then c is -

S2 dk lehdj.k xy = c gks] rks c gksxk -

(A) –2 (B) –1 (C) 1 (D) 2

12. Equation of transverse axis of S1 is -

S1 ds fr;Zd v{k dk lehdj.k gksxk -

(A) y = 7x (B) x + 7y = 0 (C) 3x = 4y (D) 7x + y = 0


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

This Section contains 2 questions. Each question has four statements (A, B, C and D) given inColumn I and five statements (P, Q, R, S and T) in Column II. Any given statement in Column I can havecorrect matching with ONE or MORE statement(s) given in Column II. For example, if for a givenquestion, statement B matches with the statements given in Q and R, then for the particular question, againststatement B, darken the bubbles corresponding to Q and R in the ORS.


gSaA dkWye I dk dksbZ Hkh dFku dkWye II ds ,d dFku ;k ,d ls vf/kd dFkuksa ls esy [kkrk gSA mnkgj.k ds fy,] fn, gq, iz'uesa ;fn dFku B dFkuks a Q vkSj R ls esy [kkrk gS] rks vksvkj,l (ORS) esa ml iz'u ds fy;s dFku B ds lkeus Q vkSj R lslEcfUèkr cqycqyksa dks dkyk dhft;sA

1. Column-I Column-II

(A) If ( )[ ]

[ ]( )

max sin t, t , x x 0,2ƒ x

min cos t, t ,x x ( ,2 ]

ì æ öpé ùÎ - Î pï ç ÷ê ú= ë ûè øíï - Î p Î p pî

(P) ( )( )( ) ( )( )10

0

x 1 x 2 x 3 ... x 9 dx- - - -ò

then number of points in [0,2p], where ƒ(x) is (Q) 1not differentiable is

(B) If ( )1

5 4 4

0

n 3x x 4x 5dx

25+ + =ò , then n is (R) 13

(C) The maximum value of16sin3q – 4sin2q – 16sinq + 8, q Î R is (S) 9

(D) The number of integral values of c for whichequation sin–1x – x = c has a solution is (T) greater than number of

points in (0,9) where|sin px| is not derivable

LrEHk-I LrEHk-II

(A) ;fn ( )[ ]

[ ]( )

ì æ öpé ùÎ - Î pï ç ÷ê ú= ë ûè øíï - Î p Î p pî

vf/kdre

U;wure

sin t, t , x x 0,2ƒ x

cos t, t , x x ( ,2 ]

(P) ( )( )( ) ( )( )10

0

x 1 x 2 x 3 ... x 9 dx- - - -ò

gks] rks vUrjky [0,2p] esa mu fcUnqvksa dh la [;k] (Q) 1

tgk¡ ƒ(x) vodyuh ; ugha gS] gksxh

(B) ;fn ( )1

5 4 4

0

n 3x x 4x 5dx

25+ + =ò gks] rks n gksxk (R) 13

(C) 16sin3q – 4sin2q – 16sinq + 8, q Î R dk

vf/kdre eku gksxk (S) 9

(D) c ds iw.kk±d ekuks a dh la[;k ftlds fy,

lehdj.k sin–1x – x = c dk gy gS] gksxh (T) vUrjky (0, 9) esa fcUnqvksa dh la[;kls vf/kd] tgk¡ |sin px| vodyuh;

ugha gS

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2. Column-I Column-II

(A) In a DABC, 22tan A7

= and the altitude AD from (P) 3

A divides BC into segments of length 3 and 17.Then AD – 8 is equal to

(B) If ha, hb, hc are length of altitude and r is inradius (Q) 9

of DABC, then least value of ahr

å is equal to

(C) ABCD is a tetrahedron of volume 27. E, F & G are points (R) number of solutionson AB, BC and CA respectively dividing them in 2 : 1 of 2sin–1(sinx) = x(internally), then the volume of tetrahedron DEFG isequal to (S) 18

(D) If a, b, crr r

are three vectors such that | a | | b | 1= =rr

and c a b= ´rr r

,

then maximum value of (a 2b 3c).[(2a 3b c) (3a b 2c)]+ + + + ´ + +r r rr r r r r r

(T) number of pointsis equal to of discontinuity of

ƒ(x) = [x + [x + [x]]]in (1,20)(where [.] representgreatest intergerfunction)

LrEHk-I LrEHk-II

(A) f=Hkqt ABC esa, 22tan A7

= rFkk 'kh"kZ A ls [khapk x;k (P) 3

'kh"kZyEc AD Hkqtk BC dks yEckbZ 3 rFkk 17 ds [k.M esafoHkkftr djrk gSA rc AD – 8 cjkcj gksxk

(B) ;fn f=Hkqt ABC esa ha, hb, hc 'kh"kZyEc dh yEckbZ;k¡ rFkk (Q) 9

r vUr%f=T;k gks ] rks ahr

å dk U;wure eku gksxk

(C) prq"Qyd ABCD dk vk;ru 27 gSA E, F ,oa G Øe'k% (R) 2sin–1(sinx) = x

AB, BC rFkk CA ij fcUnq bl izdkj gS fd ; s budks vuqikr 2 : 1 ds gyksa dh la[;k(vUr% foHkkftr) esa foHkkftr djrh gS] rks prq"Qyd DEFG dkvk;ru gksxk (S) 18

(D) ;fn a, b, crr r

rhu lfn'k bl izdkj gS fd | a | | b | 1= =rr

rFkk

c a b= ´rr r

gks] rks (a 2b 3c).[(2a 3b c) (3a b 2c)]+ + + + ´ + +r r rr r r r r r

(T) vUrjky (1,20) esa

dk vf/kdre eku gksxk ƒ(x) = [x + [x + [x]]]

ds vlarr~ fcUnq(tgk¡ [.] egÙke iw.kk±d Qyudks n'kkZrk gS)


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

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SECTION-IV : (Integer Value Correct Type)

[k.M-IV : (iw.kk±d eku lgh izdkj)This section contains 8 questions. The answer to each question is a single digit Integer, ranging from


bl [k.M esa 8 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. Functions ƒ, g, h are differentiable on some open interval around 0 and satisfy the equations and initial

conditions

( )2 1ƒ ' 2ƒ gh , ƒ 0 1

gh= + =

( )2 4g' ƒg h , g 0 1

ƒh= + =

( )2 1h ' 3ƒgh , h 0 1

ƒg= + = , then

( ) ( ) ( )ƒ x g x h x tan nx4

pæ ö= +ç ÷è ø

, then n is equal to

Qyu ƒ, g, h, 'kwU; dks ysrs gq, fdlh foo`Ùk vUrjky esa vodyuh ; gS rFkk lehdj.kksa o izkjfEHkd izfrcU/kks a

( )2 1ƒ ' 2ƒ gh , ƒ 0 1

gh= + =

( )2 4g' ƒg h , g 0 1

ƒh= + =

( )2 1h ' 3ƒgh , h 0 1

ƒg= + = dks lUrq"V djrs gSaA rc ;fn

( ) ( ) ( )ƒ x g x h x tan nx4

pæ ö= +ç ÷è ø

gS] rks n dk eku gksxk

2. The number of values of a for which the curves y = ax2 + ax + 1

24 and x = ay2 + ay +

1

24 have

common tangent is

a ds ekuksa dh la[;k ftlds fy, oØ y = ax2 + ax +1

24 rFkk x = ay2 + ay +

1

24 dh mHk;fu"B Li'kZjs[kk

gS] gksxh


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3. A function ƒ is differentiable for all real numbers and satisfies ƒ(3 + x) = ƒ(3 – x) and ƒ(8 + x) = ƒ(8 – x)

for all real x. If ƒ(0) = ƒ'(0) = 0, then least number of roots of equation ƒ'(x) = 0 in (–11,11) is

Qyu ƒ lHkh okLrfod la[;kvksa ds fy, vodyuh; rFkk ƒ(3 + x) = ƒ(3 – x) ,oa ƒ(8 + x) = ƒ(8 – x) dks lHkh

okLrfod x ds fy, lUrq"V djrk gSA ;fn ƒ(0) = ƒ'(0) = 0 gks] rks vUrjky (–11,11) esa lehdj.k ƒ'(x) = 0 ds

ewyksa dh U;wure la[;k gksxh

4. Let (x,y) be a variable point on the curve x2 + 4y2 – 4x + 8y + 4 = 0. Then ( )( )

2 2

2 2

max x y 4x 2y 5

min x y 4x 2y 5

+ + + +

+ + + +is equal to

ekuk (x,y), oØ x2 + 4y2 – 4x + 8y + 4 = 0 ij pj fcUnq gS] rc ( )

( )+ + + +

+ + + +

2 2

2 2

x y 4x 2y 5

x y 4x 2y 5

vf/kdre

U;wure dk eku gksxk

5. If area of the region in the complex plane that consist of all points z such that both z

40 and

40

z have real

and imaginary parts lies in the interval [0, 1] is A, then least integer greater than or equal to A

100 is

;fn lfEeJ lery esa {ks= dk {ks=Qy] ftlesa lHkh fcUnq z bl izdkj gS fd z

40 rFkk

40

z ds okLrfod rFkk dkYifud

Hkkx vUrjky [0, 1] esa gS] A gks] rks A

100ls cM+k ;k cjkcj U;wure iw.kk±d gksxk


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6. The area of region containing the points (x,y) such that x2 + y2 < p

12 and sin(x + y) < 0 is equal to

vlfedkvksa x2 + y2 < p

12 rFkk sin(x + y) < 0 dks lUrq"V djus okys fcUnqvksa (x,y) ls cuus okys {ks= dk {ks=Qy gksxk

7. Let a, b, c are three distinct natural number less than 10 such that the system of linear equations in x, y, z

3x + 4y + 5z = a

4x + 5y + 6z = b

5x + 6y + 7z = c

posses more than one solution, then the maximum value b can take is

ekuk 10 ls NksVh rhu fHkUu izkd`r la[;k;sa a, b, c bl izdkj gSa fd x, y, z esa jS[kh; lehdj.kksa ds fudk;

3x + 4y + 5z = a

4x + 5y + 6z = b

5x + 6y + 7z = c

dk ,d ls vfèkd gy gks] rks b dk vf/kdre eku gksxk8. Let ABC be a triangle with D and E on the respective sides AC and AB. If M and N are the mid points

of BD and CE, then the ratio of area of the quadrilateral BCDE and the area of the triangle AMN isequal to

ekuk f=Hkqt ABC esa Hkqtkvksa AC rFkk AB ij fcUnq D rFkk E gSA ;fn M rFkk N Hkqtkvkas BD rFkk CE ds e/; fcUnq gks]rks prqHkqZt BCDE ds {ks=Qy rFkk f=Hkqt AMN ds {ks=Qy dk vuqikr gksxk


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Kota/00CT214005

Admission Announcement ( Session 2015-16 ) KOTA Center

JEE (Main), Board, Olympiads, KVPY & NTSEScholarship Upto 90% based on performance in ASAT, TALLENTEX, Apply Online ( 500/-) through our website or visit our office to obtain Application Form by paying 600/-` `

For details of ASAT dates kindly visit our website.

Corporate Office: "SANKALP" CP-6, Indra Vihar, KOTA (Rajasthan)-324005 ,Tel. : +91-744-5156100, 2436001 | E-Mail : [email protected] : www.allen.ac.in Mobile | Website : m.allen.ac.in

For getting a call back send sms INFO to 9829988255

Experience the Leadership of Trust of + Students Students currently studying in ALLEN Classroom Programmes

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227 Total SelectionsBIOLOGY : | CHEMISTRY : | MATHS :

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Kota Ahmedabad Bengaluru Bhilwara Chandigarh Indore Jaipur Mumbai Rajkot Ranchi Surat Vadodarau u u u u u u u u u u

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JEE (Adv.) / IIT-JEE 2014

All from Classroom

Total Selections

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AIPMT 2014Total Selections

10211Classroom : | DLP : 5 3

AMEY GUPTAAIR 8 - JEE (Adv)

GOVIND LAHOTIAIR 3 - JEE (Adv)

CHITRAANG MURDIAAIR 1 - JEE (Adv)

KUSHAL BABELAIR 4 - JEE (Main)

JEE (Main) 2014JEE (Advanced) 2014

SHREYA KESHARIAIR 9 - AIIMS

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AIIMS 2014AIPMT 2014

AIR-1 - Girl Category

TEJASWIN JHAAIR 1 - AIPMT

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From Classroom

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Proves

Stream Course Name (Eligibility) Admission Classes FromMode Phase-I Phase-II

JEE NurtureEnthusiast

LeaderJEE Nurture

EnthusiastLeader

(Main+Advanced) (X to XI Moving) ASAT 05-04-15 19-04-15 03 (XI to XII Moving) Direct 01-04-15 04-05-15

(XII Passed/Appeared) ASAT 15-04-15 06-05-15 01 (Main) (X to XI Moving) Direct 22-04-15 13-05-15 22

(XI to XII Moving) Direct 06-04-15 04-05-15 27

(XII Passed/Appeared) Direct 15-04-15 11-05-15 27

Phase-III Phase-IV Phase-V Phase-VI Phase-VII Phase-VIII Phase-IX Phase-X

PRE-MEDICAL NurtureEnthusiast

LeaderAchiever

Pre-Nurture & For Class VI to XCareer Foundation

-05-15 20-05-15 03-06-15 17-06-15 01-07-15 NA NA NANA NA NA NA NA NA NA NA

-06-15 17-06-15 29-06-15 06-07-15 20-07-15 NA NA NA-06-15 NA NA NA NA NA NA NA

-05-15 NA NA NA NA NA NA NA

-05-15 08-06-15 18-06-15 01-07-15 13-07-15 22-07-15 03-08-15 12-08-15

(X to XI Moving) Direct 20-04-15 17-05-15 21-06-15 NA NA NA NA NA NA NA (XI to XII Moving) Direct 06-04-15 13-05-15 NA NA NA NA NA NA NA NA

(XII Passed/Appeared) Direct 13-04-15 18-05-15 07-06-15 28-06-15 12-07-15 29-07-15 10-08-15 NA NA NADirect 10-06-15 30-06-15 16-07-15 30-07-15 11-08-15 NA NA NA NA NA

(Ex-ALLEN / XII Passed before 2015)

ASAT 08-04-15 08-07-15 08-10-15 NA NA NA NA NA NA NA

D. vadu ;kstuk / Marking scheme :15. [kaM-I (i) ds gj iz'u esa dsoy lgh mÙkjksa (mÙkj) okys lHkh cqycqyksa (cqycqys) dks dkyk djus ij 3 vad vkSj dksbZ Hkh cqycqyk dkyk 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;sxkAFor each question in Section-I (i), 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

16. [kaM-I (ii) esa gj iz'u esa dsoy lgh mÙkj okys cqycqys (BUBBLE) dks dkyk djus ij 3 vad vkSj dksbZ Hkh cqycqyk dkyk ugha djusij '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 (ii), you will be awarded 3 marks if you darken the bubble corresponding to the correctanswer and zero mark if no bubbles are darkened No negative marks will be awarded for incorrect answers inthis section.

17. [k.M–II ds iz'u ds fy,, flQZ mfpr mÙkj okys cqËs@cqËksa dks dkyk fd;k gqvk gS rks izR;sd iafä ds fy, vkidks 2 vad fn;s tk;s axsAvr% bl [k.M ds izR;sd iz'u ds vf/kdre 8 vad gSA bl [k.M esa xyr mÙkj @ mÙkjksa ds fy, dksbZ ½.kkRed vad ugha fn;s tk;s axsAFor question in Section – II, you will be awarded 2 marks for each row in which you have darkened the bubble(s)corresponding to the correct answer. Thus, each question in this section carries a maximum of 8 marks. Thereis no negative marking for incorrect answer(s) for this section.

18. [kaM-IV esa gj iz'u esa dsoy lgh mÙkj okys cqycqys (BUBBLE) dks dkyk djus ij 3 vad vkSj dksbZ Hkh cqycqyk dkyk ugha djusij '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-IV, you will be awarded 3 marks if you darken the bubble corresponding to the correctanswer and zero mark if no bubbles are darkened No negative marks will be awarded for incorrect answers inthis section.

19. 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;ZPart darkeningvkaf'kd dkyk djukDarkening the rimfje dkyk djukCancelling after darkeningdkyk djus ds ckn jn~n djukErasing 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

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234

6789

4 2 0 0 0 20

2

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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

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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)

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