allen 0 1 3 1 4 0 7 4 clas amme & leade coue/sce/21-04-2015/pae-1 physics ka/01ct314074 5/40...

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 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 40 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 nks [kaM gSaAThe question paper consists of 3 parts (Physics, Chemistry and Mathematics). Each part consists of two 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 which ONEor MORE are correct.

(ii) Hkkx esa fl¼kUrksa] iz;ksxksa vkSj vk¡dM+ksa vkfn dks n'kkZus okys 3 vuqPNsn gSA rhuksa vuqPNsnksa ls lacfU/kr N% 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 3 paragraphs each describing theory, experiment, data etc. Six questions relate to three 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. [k.M–II o III es a ,d Hkh iz'u ugha gS / There is no questions in SECTION-II & III13. [kaM-IV es a 6 iz'u gSaA izR;sd iz'u dk mÙkj 0 ls 9 rd (nksuksa 'kkfey) ds chp dk ,dy vadh; iw.kk±d gSA

Section-IV contains 6 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)

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 1 C T 3 1 4 0 7 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

le; : 3 ?k.Vs egÙke vad : 222Time : 3 Hours Maximum Marks : 222

DATE : 21 - 04 - 2015ALLEN JEE (Advanced ) TEST

TARGET : JEE (Advanced) 2015 ENTHUSIAST & LEADER COURSE : SCORE

T M

isij – 1PAPER – 1

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) vuqPNsn izdkj 10 - 13Paragraph Type

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

Hkkx-2 jlk;u foKku I(i) ,d ;k vf/kd lgh fodYi izdkj 18 - 21Part-2 Chemistry One or More Options Correct Type



Hkkx-3 xf.kr I(i) ,d ;k vf/kd lgh fodYi izdkj 28 - 31Part-3 Mathematics One or More Options 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,

Kota/01CT3140742/40

Target : JEE (Advanced) 2015/21-04-2015/Paper-1

· 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

Enthusiast & Leader Course/Score/21-04-2015/Paper-1

PHYS

ICS

3/40Kota/01CT314074

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. Hard rods form right triangle with base angle a, which rotates at a constant angular velocity w about avertical axis AB (see figure). On the rod AC a spring with stiffness k = mw2 and unstretched length l iscoiled. One end of the spring is hinged to A and the other end of spring is attached to a sleeve of mass 'm'.

(A) For a

w =a l

1 cos

sin

g the mass 'm' be in equilibrium with the spring remaining undeformed.

(B) For option (A) the equilibrium will be stable.(C) For option (A) the equilibrium will be unstable.(D) For small oscillation, 'm' will perform SHM.

B C

A

a

w

m

fp= esa dBksj NM+ks ls ,d ledks.k f=Hkqt cuk;k x;k gS ftldk vk/kkj dks.k a gS rFkk ;g fp=kuqlkj Å/okZ/kj v{k AB ds

lkis{k fu;r dks.kh; osx w ds lkFk ?kw.kZu djrk gSA NM+ AC ij k = mw2 dBksjrk rFkk vfoLrkfjr yEckbZ l okyh ,d

fLizax yxk nh tkrh gSA bl fLizax ds ,d fljs dks A ij dhydhr dj nsrs gS rFkk nwljs fljs ls 'm' æO;eku dh ,d Lyho

tksM+ nh tkrh gSA

(A) a

w =a l

1 cos

sin

g ds fy;s æO;eku 'm' lkE;koLFkk esa gksxk rFkk fLizax fo:fir ugh gksxhA

(B) fodYi (A) ds fy;s lkE;koLFkk LFkk;h gksxhA

(C) fodYi (A) ds fy;s lkE;koLFkk vLFkk;h gksxhA

(D) vYi nksyu ds fy;s 'm' ljy vkorZ xfr djsxkA

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

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

4/40

PHYS

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Kota/01CT314074


2. Water jet is projected at an angle to the horizontal. At the point of projection, the area of the jet is S1 and

at the highest point, the area of the jet is S2. The initial velocity of projection is u.

(A) The angle of projection is 1 1

2

ScosS

- æ öç ÷è ø

(B) The range on the level ground is 2 2

1 12

2 2

2u S S1g S S

-

(C) The maximum height reached from the ground is 2 2

122

2u S1g S

æ ö-ç ÷

è ø(D) The rate of volume flow is S

2u

S2

S1

ty dh ,d /kkjk {kSfrt ls fdlh dks.k ij iz{ksfir dh tkrh gSA iz{ksi.k fcUnq ij /kkjk dk {ks=Qy S1 rFkk mPpre fcUnq ij

S2 gSA iz{ksi.k dk izkjfEHkd osx u gS%&

(A) iz{ksi.k dks.k dk eku 1 1

2

ScosS

- æ öç ÷è ø

gSA

(B) lery /kjkry ij ijkl 2 2

1 12

2 2

2u S S1g S S

- gksxhA

(C) /kjkry ls izkIr dh x;h vf/kdre Å¡pkbZ 2 2

122

2u S1g S

æ ö-ç ÷

è ø gksxhA

(D) vk;ru izokg dh nj S2u gSA


PHYS

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5/40Kota/01CT314074


3. Figure shows two cylinders connected with a common light piston as shown in figure. Cylinder wallsand piston are thermally insulated. Cross-sectional area of cylinders are A and 2A and they containdifferent gases H

2 and O

2 respectively. Pistons are open to atmosphere at one end as shown in figure.

Lycopodium powder is sprinkled on the lateral surface of the two cylinders. Pistons can be set intolongitudinal vibrations such that longitudinal standing wave can be set in the cylinders due to whichpowder at the base of cylinder gets collected at the certain positions. The length of gas columns is suchthat both the air columns resonate with the external driver frequency. Initially both the gasses are at sametemperature and H

2 chamber oscillates in first overtone and O

2 chamber in third overtone. Mark correct

statements (Take P0 = 105 N/m2, A = 10–4 m2) :-

(A) After standing wave is set, if the separation between consecutive heaps of powder collected in H2

chamber is 7 cm, then the separation in O2 chamber is

74

cm

(B) Now H2 chamber is supplied heat such that gas expands and piston shifts by 1 cm work done by

atmosphere as piston shifts is 0.2 J(C) Process in O

2 chamber is isothermal

(D) Process in O2 chamber is adiabatic

AH 2 O 2

2A\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

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

fp= esa ,d gYds fiLVu ls tqM+s nks csyu n'kkZ;s x;s gSA csyu dh nhokjs rFkk fiLVu Å"eh; dqpkyd gSA bu csyuksa dk

vuqizLFkdkV {ks=Qy A o 2A gS rFkk buesa Øe'k% H2 o O

2xSls Hkjh gq;h gSA ;s fiLVu fp=kuqlkj ,d fljs ls ok;qe.My

esa [kqyrs gSA nksuksa csyuksa dh ik'oZ lrg ij ykbdksikWfM;e pw.kZ fNM+dk tkrk gSA bu fiLVuksa dks vuqnS/; Z dEiUu djk;s tkrs

gS rkfd bu csyuksa esa vuqnS/;Z vizxkeh rjax mRiUu gks lds ftlds dkj.k csyu ds isnsa ij fo|eku pw.kZ dqN fo'ks"k fLFkfr;ksa

ij ,df=r gks tkrk gSA xSl LrEHk dh yEckbZ bl izdkj gS fd nksuksa ok;q LrEHk ckg~; pkyu vkofr ij vuqukn djrs gSA

izkjEHk esa nksuksa xSls leku rki ij gS rFkk H2 d{k izFke vf/kLojd esa rFkk O

2 d{k r`rh; vf/kLojd esa nksyu djrk gSA lgh

dFkuksa dks pqfu;s%& (P0 = 105 N/m2, A = 10–4 m2) :-

(A) vizxkeh rjax mRiUu gksus ds i'pkr~ ;fn H2 d{k esa ,df=r pw.kZ ds Øekxr <s +jksa ds e/; nwjh 7 cm gks rks O

2 d{k

esa ;g nwjh 74

cm izkIr gksrh gSA

(B) vc H2 d{k dks bl izdkj Å"ek nh tkrh gS fd xSl izlkfjr gksrh gS ,oa fiLVu 1 cm foLFkkfir gks tkrk gSA fiLVu ds

foLFkkfir gksus ij ok;qe.My }kjk fd;k x;k dk;Z 0.2 J gSA

(C) O2 d{k esa lerkih; izØe lEiUu gksrk gSA

(D) O2 d{k esa :¼ks"e izØe lEiUu gksrk gSA

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Kota/01CT314074

4. In which of the following figures, stream lines shown is/are correct ?

fuEu esa ls dkSuls fp=@fp=ksa esa /kkjk js[kk;sa lgh gS\

(A)

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

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

b u m p in r iv e r b ed

(B) \\\\\\\\\\\\\\\\\\\\\\\

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

b u m p in r iv e r b ed

(C) u p p e rlo w er

(D) u p p e rlo w er

5. In the circuit shown in the figure C1 = C, C

2 = 2C, C

3 = 3C, C

4 = 4C. Then select CORRECT

alternative(s):(A) Maximum potential difference is across C

1.

(B) Minimum potential difference is across combination of C3 and C

4

(C) Maximum potential energy is in C1.

(D) Minimum potential energy is in C2.

C4

C3

C2C1

iznf'kZr ifjiFk esa C1 = C, C

2 = 2C, C

3 = 3C, C

4 = 4C gSA lgh dFku@dFkuksa dks pqfu;s%&

(A) vf/kdre foHkokUrj C1 ds fljksa ij gSA

(B) U;wure foHkokUrj C3 rFkk C

4 ds la;kstu ij gSA

(C) vf/kdre fLFkfrt ÅtkZ C1 esa lafpr gSA

(D) U;wure fLFkfrt ÅtkZ C2 esa lafpr gSA



PHYS

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7/40Kota/01CT314074

6. A closed conducting loop, having resistance R, is being rotated about an axis perpendicular to the

magnetic field. Magnetic flux through the closed conducting loop is continuously changing according

to the graph shown in the adjacent figure. Then, which of the following statement(s) is/are correct?

(A) The electric current through the loop is minimum (zero) at t = 1 s, 3s and 5s.

(B) The electric current through the loop is minimum (zero) at t = 0 s, 2s and 6s.

(C) Total charge flown through any cross-section of a closed conducting loop between 0 and 6 s is zero

(D) Total work done in rotating the loop in the magnetic field is zero

time t(in sec)

Flux (in T-m ) f 2

10

0 1 2 3 4 5 6

-10

izfrjks/k R okyk ,d can pkyd ywi pqEcdh; {ks= ds yEcor~ v{k ds lkis{k ?kqek;k tkrk gSA bl can pkyd ywi ls fuxZr

pqEcdh; ¶yDl n'kkZ;s x;s vkjs[k ds vuqlkj yxkrkj ifjofrZr gks jgk gSA rc lgh dFku@dFkuksa dks pqfu;s%&

(A) t = 1 s, 3s rFkk 5s ij ywi ls fuxZr fo|qr /kkjk U;wure ('kwU;) gksxhA

(B) t = 0 s, 2s rFkk 6s ij ywi ls fuxZr fo|qr /kkjk U;wure ('kwU;) gksxhA

(C) 'kwU; ls 6 s ds e/; can pkyd ywi ds fdlh Hkh vuqizLFkdkV ls izokfgr gksus okyk dqy vkos'k 'kwU; gksxkA

(D) pqEcdh; {ks= esa ywi dks ?kqekus esa fd;k x;k dqy dk;Z 'kwU; gksxkA


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Kota/01CT314074

7. A particle of specific charge 'a' is projected from origin at t = 0 with a velocity ( )0ˆ ˆV V i k= +

r in a

magnetic field 0ˆB B k= -

r. Then :

(A) at 0

tB

p=

a, velocity of the particle is ( )0

ˆ ˆV i k- -

(B) at 0

t4 B

p=

a, speed of the particle is VV0

(C) at 0

2t

B

p=

a, magnitude of displacement of the particle is more than 0

0

2V

Ba

(D) at 0

2t

B

p=

a, distance travelled by the particle is less than 0

0

2 2 V

B

pa

.

fof'k"V vkos'k 'a' okys ,d d.k dks t = 0 ij ewy fcUnq ls osx ( )0ˆ ˆV V i k= +

r ds lkFk pqEcdh; {ks= 0

ˆB B k= -r

esa

iz{ksfir fd;k tkrk gSA rc %&

(A) 0

tB

p=

a ij d.k dk osx ( )0

ˆ ˆV i k- - gksxkA

(B) 0

t4 B

p=

a ij d.k dh pky V0 gksxhA

(C) 0

2t

B

p=

a ij d.k ds foLFkkiu dk ifjek.k 0

0

2V

Ba ls vf/kd gksxkA

(D) 0

2t

B

p=

a ij d.k }kjk r; nwjh 0

0

2 2 V

B

pa ls de gksxhA



PHYS

ICS

9/40Kota/01CT314074

8. Two small steel balls are suspended on weightless inextensible threads, as shown in figure. Mass of the

balls 'm' are equal, the lengths of the thread is l1 and l

2 (l

1 > l

2) respectively. The first ball m

1 is

deflected by an angle q and released. The collision is perfectly elastic. The air resistance can be neglected.

[The questions asked are in regard to first collision only].

(A) The speed of m1 before it collides with m

2 is ( )12 1 cosg - ql

(B) The magnitude of change in momentum of m1 due to collision with m

2 is ( )12 1 cosm g - ql

(C) The momentum of m2 just after collision is ( )12 1 cosm g - ql

(D) The angle a deviated by m2 after being hit by m

1 is q.

l1 l2

m1 m2

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

nks NksVh LVhy dh xsanksa dks fp=kuqlkj Hkkjghu vforkU; /kkxksa ij yVdk;k x;k gSA xsanksa dk æO;eku leku rFkk 'm' gS ,oa

èkkxksa dh yEckbZ Øe'k% l1 o l

2 (l

1 > l

2) gSA izFke xsan m

1 dks q dks.k ij fo{ksfir dj NksM+ fn;k tkrk gSA VDdj iw.kZr;k

izR;kLFk gSA ok;q izfrjks/k dks ux.; ekuk tkr ldrk gSA [;g iz'u dsoy izFke VDdj ds lanHkZ esa gh iwNk x;k gS].

(A) m2 ls Vdjkus ls iwoZ m

1 dh pky ( )12 1 cosg - ql gSA

(B) m2 ls VDdj ds dkj.k m

1 ds laosx esa ifjorZu dk ifjek.k ( )12 1 cosm g - ql gSA

(C) VDdj ds rqjUr i'pkr~ m2 dk laosx ( )12 1 cosm g - ql gSA

(D) m1 ls Vdjkus ds i'pkr~ m

2 }kjk fo{ksfir dks.k a dk eku q gSA


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Kota/01CT314074

(ii) Paragraph Type (ii) vuqPNsn izdkj

This section contains 3 paragraphs each describing theory, experiment, data etc. Six questions relateto three paragraphs with two questions on each paragraph. Each question of a paragraph has onlyone correct answer among the four choices (A), (B), (C) and (D).bl [k.M esa fl¼kUrksa] iz;ksxksa vkSj vk¡dM+ksa vkfn dks n'kkZus okys 3 vuqPNsn gSA rhuksa vuqPNsnksa ls lacaf/kr N% 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 10Even if surface tension is most important only for liquids, we shall here consider a simple regular "solid"surrounded by vacuum. The molecules are placed in a cubic grid (see the figure given below) with gridlength equal to the molecular separation length L

mol. Each molecule in the interior has six bonds to its

neighbors whereas a surface molecule has only five. If the total binding energy of a molecule in the bulk

is Î, a surface molecule will only be bound by 5

6Î. The missing binding energy corresponds to adding

an extra positive energy 1

6Î for each surface molecule.

The binding energy may be estimated as Î » hm where h is the latent heat of evaporation and

3molmol

A

Mm L

N= = r is the mass of a single molecule. Dividing the molecular surface energy

1

6Î with the

molecular area scale 2molL , we arrive the following estimate of the surface energy density..

2 2

11 166 6 mol

mol mol

hmh L

L L

Îa » » » r

The appearance of the molecular separation scale is quite understandable because rLmol

represents theeffective surface mass density of layer of thickness L

mol. One might expect that the smallness of the

molecular scale would make the surface energy density insignificant in practice, but the small thicknessis offset by the fairly large values of r and h in normal liquids, for example water.Consider 3 liquids(P) h = 105 J/kg, r = 104 kg/m3, L

mol = 10–7 m

(Q) h = 4 × 105 J/kg, r = 103 kg/m3, Lmol

= 3 × 10–7 m

(R) h = 8 × 104 kg, r = 4000 kg/m3, Lmol

= 1.5 × 10–6 m

9. The correct order of surface tension is :-

(A) aP > a

Q > a

R(B) a

Q > a

P > a

R(C) a

P > a

R > a

Q(D) a

R > a

Q > a

P

10. Consider completely wetting liquid. The height of liquid rising in capillary of sufficient length will notdepend on :-(A) Length of capillary tube (B) Latent heat of evaporation

(C) Thickness of layer (Lmol

) (D) Density of liquid.


PHYS

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11/40Kota/01CT314074

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

;|fi i`"B ruko dsoy æoksa ds fy;s gh vfr egRoiw.kZ gksrk gS] ijUrq ge ;gk¡ ,d lkekU; "Bksl" ij fopkj djsxsa tks fuokZr

}kjk f?kjk gqvk gSA ;gk¡ uhps fn, x;s fp= esa v.kq ,d ?kuh; fxzM esa j[ks gq;s gS rFkk fxzM yEckbZ] vkf.od nwjh Lmol

ds rqY;

gSA blds vUnj izR;sd v.kq vius iMkSfl;ksa ds lkFk N% ca/k cukrk gS] tcfd ,d i`"Bh; v.kq dsoy ik¡p ca/k gh j[krk gSA

;fn vk;ru esa ,d v.kq dh dqy ca/ku ÅtkZ Î gks rks ,d i`"Bh; v.kq dsoy 5

6Î ls gh ca/kk gksxkA ;gk¡ ij foyqIr ;g

caèku ÅtkZ izR;sd i"Bh; v.kq ds fy;s ,d vfrfjDr /kukRed ÅtkZ 1

6Î ds tqM+us ds laxr gksrh gSA

ca/ku ÅtkZ dk ifjdyu Î » hm }kjk fd;k tk ldrk gS] tgk¡ h ok"iu dh xqIr Å"ek rFkk 3molmol

A

Mm L

N= = r ; ,d

vq.k dk æO;eku gSA vkf.od i`"Bh; ÅtkZ 1

6Î dks vkf.od {ks=Qy iSekus 2

molL ls foHkkftr djus ij i`"Bh; ÅtkZ

?kuRo fuEu izdkj fy[kk tk ldrk gS %

2 2

11 166 6 mol

mol mol

hmh L

L L

Îa » » » r

;gk¡ vkf.od nwjh iSekus dh mifLFkfr vklkuh ls le>h tk ldrh gS] D;ksafd rLmol

, Lmol

eksVkbZ okyh ijr ds izHkkoh

i`"Bh; æO;eku ?kuRo dks n'kkZrh gSA ;gk¡ ;g lkspk tk ldrk gS fd vkf.od iSekus ds vR;Yi gksus ij O;ogkfjd :i ls

i`"Bh; ÅtkZ ?kuRo Hkh lkFkZd :i esa izkIr ugha gksxkA ijUrq ;gk¡ æoksa ds r o h ds mPp eku vYi eksVkbZ ds mijksä izHkko

dks ux.; dj nsrs gSa] tSls fd tyA vc fuEu rhu æoksa ij fopkj dhft;s %

(P) h = 105 J/kg, r = 104 kg/m3, Lmol

= 10–7 m

(Q) h = 4 × 105 J/kg, r = 103 kg/m3, Lmol

= 3 × 10–7 m

(R) h = 8 × 104 kg, r = 4000 kg/m3, Lmol

= 1.5 × 10–6 m

9. i`"B ruko dk lgh Øe gS %&(A) a

P > a

Q > a

R(B) a

Q > a

P > a

R(C) a

P > a

R > a

Q(D) a

R > a

Q > a

P

10. ,d iw.kZr;k Hkhxs æo ij fopkj dhft;sA fdlh i;kZIr yEckbZ dh ds'kuyh esa p<+ s gq;s æo dh Å¡pkbZ fdl ij fuHkZj ugh

djsxh\

(A) ds'kuyh dh Å¡pkbZ (B) ok"iu dh xqIr Å"ek

(C) ijr dh eksVkbZ (Lmol

) (D) æo dk ?kuRo


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

Energy is often releases or absorbed in a nuclear reaction. A reaction releases energy when kineticenergy of product nuclei after reaction is greater than the kinetic energy of interacting nuclei beforereaction, the increase results from the transformation of rest mass into kinetic energy. The Q value ofnuclear reaction (amount of energy released)Q Value = SE

kf – SE

ki

SEkf = final kinetic energy

SEki = initial kinetic energy

Q Value = SBi – SB

f

SBi = initial binding energy

SBf = final binding energy

For an exoergic reaction Q > 0, such reaction can proceed even the reacting particles are at rest. Forendoergic reaction Q < 0, such reaction requires threshold kinetic energy. The nuclear reaction followscertain conservation of(i) charge (ii) linear and angular momentum (iii) mass and energy.ukfHkdh; vfHkfØ;k esa ÅtkZ lkekU;r;k vo'kksf"kr ;k mRlftZr gksrh gSA ;fn vfHkfØ;k ds ckn mRikn ukfHkdksa dh xfrtÅtkZ dk eku] vfHkfØ;k ls iwoZ fØ;kdkjd ukfHkdksa dh xfrt ÅtkZ ls vf/kd gks rks Å"ek mRlftZr gksrh gS rFkk ;gvfrfjDr ÅtkZ fojke æO;eku ds xfrt ÅtkZ esa :ikUrj.k ds QyLo:i izkIr gksrh gSA ukfHkdh; vfHkfØ;k dk Q eku(mRlftZr ÅtkZ dh ek=k) fuEu izdkj fy[kk tkrk gSAQ eku = SE

kf – SE

ki

SEkf = vafre xfrt ÅtkZ]

SEki = izkjfEHkd xfrt ÅtkZ

Q eku = SBi – SB

f

SBi = izkjfEHkd ca/ku ÅtkZ

SBf = vafre ca/ku ÅtkZ

,d Å"ek{ksih vfHkfØ;k ds fy;s Q > 0 gksrk gS rFkk ;g vfHkfØ;k rc Hkh laEiUu gksrh gS tc fØ;kdkjd d.k fojke esagksA Å"ek'kks"kh vfHkfØ;k ds fy;s Q < 0 gksrk gS rFkk ;g vfHkfØ;k lEiUu gksus ds fy;s nSgyh xfrt ÅtkZ dh vko';drkgksrh gSA ukfHkdh; vfHkfØ;k esa fuEu dk laj{k.k gksrk gS(i) vkos'k (ii) js[kh; rFkk dks.kh; laosx (iii) æO;eku rFkk ÅtkZ

11. The binding energy per nucleon of 73 Li and 4

2 He nuclei are 5.6 MeV and 7.06 MeV, then in the

reaction ( )7 43 2p Li 2 He+ ® , the energy of p must be

73 Li rFkk 4

2 He ukfHkdks ds fy;s izfr U;wfDy;kWu ca/ku ÅtkZ dk eku Øe'k% 5.6 MeV rFkk 7.06 MeV gksrk gSA rc

vfHkfØ;k ( )7 43 2p Li 2 He+ ® esa p dh ÅtkZ fuf'pr :i ls gksxh%&

(A) 28.24 MeV (B)17.28 MeV (C) 1.46 Me (D) 39.2 MeV

12. A nucleus with mass no. 220 initially at rest emits an a particle. If the Q value of reaction is 5.5 MeV,

then kinetic energy of a particle is :

izkjEHk esa fojkekoLFkk esa fLFkr 220 æO;eku la[;k okyk ,d ukfHkd ,d a d.k mRlftZr djrk gSA ;fn vfHkfØ;k dk Q

eku 5.5 MeV gks rks a d.k dh xfrt ÅtkZ gS%&(A) 4.4 MeV (B) 5.4 MeV (C) 5.6 MeV (D) 6.5 MeV


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Paragraph for Questions 13 and 14

iz'u 13 ,oa 14 ds fy;s vuqPNsnA massless square loop of side a is kept in xz plane as shown. Magnetic field in space is non uniform

given by 0B y ˆB ka

= . The loop is rotated about x-axis with constant angular velocity w.

Hkqtk a okyk ,d æO;ekughu oxkZdkj ywi fp=kuqlkj xz ry esa j[kk gSA ;gk¡ ,d vle:i pqEcdh; {ks= 0B y ˆB ka

=

fo|eku gSA ywi dks x-v{k ds lkis{k fu;r dks.kh; osx w ls ?kqek;k tkrk gSA

x

y

z

B

a

13. e.m.f. induced in the loop as function of time is equal to :-

ywi esa izsfjr fo|qr okgd cy le; ds Qyu ds :i esa gksxk%&

(A) 2

0 sin 22

B atw w (B) ( )

20 1 sin 22

B at

w+ w (C) 2B

0a2w cos wt (D) B

0a2w cos2wt

14. Torque required to rotate the loop with constant angular velocity (as a function of time). Take resistanceof loop = R.

ywi dks fu;r dks.kh; osx ls ?kqekus ds fy;s vko';d cyk?kw.kZ le; ds Qyu ds :i eas gksxk (ywi dk izfrjks/k = R)

(A) ( )2 4

20 1 sin 24

w+ w

B at

R(B)

2 420 sin 2

4

ww

B at

R

(C) 2 4

204cos

ww

B at

R(D)

2 440 cos

ww

B at

R

SECTION –II / [k.M – II & SECTION –III / [k.M – IIIMatrix-Match Type / eSfVªDl&esy izdkj Integer Value Correct Type / iw.kk±d eku lgh izdkj

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


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SECTION-IV : (Integer Value Correct Type)

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

0 to 9 (both inclusive)

bl [k.M esa 6 iz'u gSaA izR;sd iz'u dk mÙkj 0 ls 9 rd (nksuksa 'kkfey) ds chp dk ,dy vadh; iw.kk ±d gSA1. As shown in figure, there is a thick spherical shell with the walls coated with 'lamp black'. A point

source which generates thermal energy at a constant rate 'P' is placed at the centre S of the shell. Derivean expression for the temperature T at point X in steady state, where SX = 1.5 R. Your expression wouldbe as follows :

1/ 4

2

P PT16 R 3x KR

æ ö æ ö= +ç ÷ ç ÷s p pè ø è ø.

Here K is the coefficient of thermal conductivity of material of shell and s is Stefen's constant. Find thevalue of x.

fp= esa ,d eksVk xksyh; dks'k n'kkZ;k x;k gS ftldh nhokjs dkyh iqrh gqbZ gSA bl dks'k ds dsUæ 'S' ij ,d fcUnq L=ksr j[kk

tkrk gS tks fu;r nj P ij rkih; ÅtkZ mRiUu djrk gSA LFkk;h voLFkk esa fcUnq X ij rkieku T fuEu izdkj fy[kk tkrk gSA1/ 4

2

P PT16 R 3x KR

æ ö æ ö= +ç ÷ ç ÷s p pè ø è ø.

tgk¡ SX = 1.5 R, K dks'k ds inkFkZ dk rkih; pkydrk xq.kkad gS o s LVhQu fu;rkad gSA x dk eku Kkr dhft;sA

X

2R

S

R



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2. An object O is located 40 cm from the first of two thin converging lenses (each of focal length 20 cm),as shown in figure below. If speed of the right lens is 7 cm/s rightward, speed of final image is7x/4 cm/s at the instant shown, find the value of x.

fp= esa ,d fcEc O dks izR;sd 20 cm Qksdl nwjh okys nks irys vfHklkjh yaslksa esa ls ,d ls 40 cm dh nwjh ij j[kk x;kgSA ;fn nka;s ysal dh pky 7 cm/s nka;h vksj gks rFkk vafre izfrfcEc dh pky iznf'kZr {k.k ij 7x/4 cm/s gks rks x Kkrdhft;sA

O

20cm

v=7cm/s

f =20cm2f =20cm1

40cm

3. A long wire bend into the shape of a right angle is held stationary on a horizontal frictionless plane. Avery long rod of mass 1 kg initially starts with velocity v

0 = 4 m/s from the apex A of the bend wire. The

resistance per unit length of the wire and the rod is ( ) 22 1 10-- ´ W /m. The whole arrangement is put in

a region of uniform magnetic field of 0.05 T directed normally into the horizontal plane. Find the distancetravelled by the rod before it comes to rest.

,d yEcs rkj dks ledks.k ij eksM+dj bls ,d {kSfrt ?k"kZ.kjfgr ry ij fLFkj jksddj j[kk x;k gSA bl eqM+s gq, rkj ds 'kh"kZ

A ls 1 kg æO;eku dh ,d cgqr yEch NM+ izkjEHk esa v0 = 4 m/s osx ls xfr djuk izkjEHk djrh gSA rkj rFkk NM+ dh izfr

bdkbZ yEckbZ dk izfrjks/k ( ) 22 1 10-- ´ W /m gSA bl lEiw.kZ O;oLFkk dks 0.05 T ds le:i pqEcdh; {ks= esa j[kk tkrk

gS] ftldh fn'kk {kSfrt ry esa yEcor~ uhps dh vksj gSA fojkekoLFkk esa vkus ls iwoZ NM + }kjk r; nwjh Kkr dhft;sA

A

´ ´

´ ´

´ ´

´ ´


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4. Consider the E versus x graph. What is the minimum velocity that should be given to a pointcharge –Q of mass M at x = 3L so that it can reach the origin? Take Q = 1 mC, mass of the chargeas 1g, E

0 = 5 N/C and L = 5m.

fuEu E-x vkjs[k ij fopkj dhft;sA x = 3L ij fLFkr M æO;eku ds fcUnq vkos'k –Q dks fdruk U;wure osx fn;k tk;s

rkfd ;g ewyfcUnq rd igq¡p tkos\ Q = 1 mC, vkos'k dk æO;eku 1g, E0 = 5 N/C rFkk L = 5m ysaA

–E0

3L2LL

E0

x

5. A uniform solid sphere of mass m = 400 gm and radius R = 2cm is released from rest from a point A ofa rough slide AB. Initially, the centre O of the sphere is at the horizontal level of A. At the lower end B,the slide passes to smooth horizontal plane. A spring is attached to a wall on the horizontal plane.Find the maximum compression (in cm) of the spring in the process of motion of thesphere. (Take g = 10 m/s2),d m = 400 gm æO;eku rFkk R = 2cm f=T;k ds le:i Bksl xksys dks [kqjnjh <yku AB ds fcUnq A ls fojkekoLFkk

ls NksM+k tkrk gSA izkjEHk essa xksys dk dsUæ O, A dh {kSfrt lh/k esa gSA fupys fljs B ij ;g <yku ,d fpdus {kSfrt ry

ls feyrh gSA bl {kSfrt ry ij fLFkr nhokj ls ,d fLizax tksM + nh tkrh gSA xksys dh xfr ds nkSjku fLizax esa mRiUu vf/kdre

lEihM+u (cm esa) Kkr dhft;sA (g = 10 m/s2)

0.3m

Sufficientrough

B

AO

R = 2cmm = 400 gm

Smoothlight strip

K = 1000 N/m

Smooth Horizontalfloor

2cm



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6. Ten moles of a gas (molar heat capacity for constant volume process is CV) is enclosed in rigid hollow

sphere of inner radius 'a' and outer radius '5a' and it's temperature is 3T0 at t = 0. Heat is conducted out

to the environment (temperature T0) through the sphere material of conductivity K and negligible heat

capacity. At t = t1 (second) the temperature of the gas is found to be 2T

0 then find the value of t

1.

[All quantities are expressed in SI units and take Cv × (ln2) = 2pKa]

,d xSl] ftlds fy;s fu;r vk;ru izØe dh eksyj Å"ek /kkfjrk CV

gS] ds 10 eksy dks vkarfjd f=T;k 'a' rFkk ckg~;

f=T;k '5a' okys ,d n`<+ [kks[kys xksys ds vUnj j[kk x;k gS rFkk t = 0 ij bldk rkieku 3T0 gSA bl xksys ds inkFkZ dh

Å"eh; pkydrk K rFkk Å"ek /kkfjrk ux.; gSA bl xksys ls Å"ek T0 rkieku okys ifjos'k esa izokfgr gksrh gSA t = t

1 sec

ij xSl dk rkieku 2T0 gks rks t

1 dk eku Kkr dhft;sA leLr jkf'k;k¡ SI bdkbZ esa yh xbZ gS rFkk C

v × (ln2) = 2pKa ysaA

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PART-2 : CHEMISTRY

Hkkx-2 : jlk;u foKkuSECTION–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 ;k

vf/kd lgh gSA

1. Two solutions S1 and S

2 containing 0.1M NaCl(aq.) and 0.08M BaCl

2(aq.) are separated by

semipermeable membrane. Which among the following statement(s) is/are correct-

(A) S1 and S

2 are isotonic (B) S

1 is hypertonic and S

2 is hypotonic

(C) S1 is hypotonic and S

2 is hypertonic (D) Osmosis will take place from S

1 to S

2

S1 S2

0.1M NaCl 0.08M BaCl2

SPM

nks foy;uksa S1 rFkk S

2 esa 0.1M NaCl(aq.) rFkk 0.08M BaCl

2(aq.) gS] dks ,d v/kZikjxE; f>Yyh }kjk i`Fkd

fd;k x;k gSaA fuEu dFkuksa esa ls dkSuls lgh gS@gSa-

(A) S1 rFkk S

2 leijkljh (isotonic) gS (B) S

1 mPp ijkljh rFkk S

2 U;quijkljh gS

(C) S1 U;quijkljh rFkk S

2 mPpijkljh gS (D) S

1 ls S

2 esa ijklj.k gksxk



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2. Rate law expression of a reaction isR = k[A]2/3[B]1

Select correct statement (s) for above expression

(A) Order of reaction = 53

(B) Above reaction must be a complex reaction(C) Units of rate constant is atm–2/3sec–1

(D) Units of rate constant will be sec–1 if reactant A is taken in large excess

,d vfHkfØ;k dk nj fu;e dk O;atd gSR = k[A]2/3[B]1

mijksDr O;atd ds fy, lgh dFku dk p;u dhft;sA

(A) vfHkfØ;k dh dksVh = 53

(B) mijksDr vfHkfØ;k fuf'pr gh ladqy vfHkfØ;k gksuh pkgh;s

(C) nj fu;rkad dh bdkbZ atm–2/3sec–1 gS

(D) ;fn vfHkdkjd A dks vkf/kD; esa fy;k x;k gks rks nj fu;rkad dh bdkbZ sec–1 gksxhA3. Select the correct statement(s)

(A) In high pressure zone for real gas, slope is b

RTæ öç ÷è ø

for 'Z' vs 'P' graph

(B) Rate of effusion of gas is in order He > Ne > O2 > SO

2 under similar conditions

(C) Charle's law states that for a given amount of gas, volume is directly proportional to its absolute temperature at constant pressure.

(D) In NaCl crystal, a 3

2æ öç ÷è ø

= rNa+ + r

Cl–

lgh dFku dk p;u dhft;s

(A) okLrfod xSl ds fy;s mPp nkc {ks= esa] 'Z' vs 'P' oØ ds fy, <ky b

RTæ öç ÷è ø

gS

(B) xSl ds fulj.k dh nj leku ifjfLFkfr;ksa esa He > Ne > O2 > SO

2 ds Øe esa gksrh gS

(C) pkYlZ fu;e dgrk gS fd xSl dh ,d nh xbZ ek=k ds fy;s vk;ru fu;r nkc ij blds ije rki ds lh/ks lekuqikrh

gksrk gS

(D) NaCl fØLVy esa, a 3

2æ öç ÷è ø

= rNa+ + r

Cl–

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4. Which of the following statements are INCORRECT :

(A) Atoms which have same number of neutrons they are called isosters of each other

(B) Molecular ions which have same number of atoms and same number of electrons they are called

as isotones

(C) Atoms which have same isotopic axis, they are called isodiaphers

(D) Atoms of different elements which has same mass and atomic number, they are called as isobars

fuEu esa ls dkSuls dFku xyr gS :

(A) ijek.kq tks leku la[;k esa U;wVªk Wu j[krs gS osa ,d nwljs ds lelajpukRed (isosters) dgykrs gS

(B) vkf.od vk;u ftuesa ijek.kqvksa dh leku la[;k rFkk bysDVªkWuksa dh leku la[;k mifLFkr gksrh gS osa leU;wVªkWfud

(isotones) dgykrs gS

(C) ijek.kq tks leku leLFkkfud v{k j[krs gS osa vkWblksMkbQlZ (isodiaphers) dgykrs gS

(D) vyx&vyx rRoksa ds ,sls ijek.kq ftudk æO;eku rFkk ijek.kq Øekad leku gksrs gS osa leHkkfjd dgykrs gS

5. P4O

6 + H

2O ® A D¾¾® Final products

P4O

6 + H

2O ® A D¾¾® vafre mRikn

(A) H3PO

4(B) H

3PO

3(C) PH

3(D) P

4

6. Fe4[Fe(CN)

6]

3 is / are known as :

(A) Ferro-ferri cyanide (B) Ferri-Ferro cyanide

(C) Prussian blue (D) Iron(II) Hexacyanidoferrate(III)

Fe4[Fe(CN)

6]

3 dks tkuk tkrk gS @ gSa :

(A) Qsjks&Qsjh lk;ukbM (B) Qsjh&Qsjks lk;ukbM

(C) izqf'k;u uhyk (D) vk;ju(II) gSDlklk;ukbMksQsjsV(III)


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7.

OH

(i) CO , OH2–

(ii) H+ A BBr , HOH2

The intermediate which is formed between A & B is/are :

,sls e/;orhZ tks A rFkk B ds e/; curs gS@gSa&

(A)

OBrC–O–

O(B)

O–

Br

Br

(C) C–O–

OOH

(D)

OHBr

Br

8. Correct statement is/are :

(A) DIBAL-H & H2-Pd/BaSO

4 both give aldehyde with acid chloride

(B) MnO2 oxidises all allylic & benzylic alcohols

(C) Pyruvic acid reduces H5IO

6

(D) Ozonolysis of ethyne is slower than ozonolysis of ethene

lgh dFku dk p;u dhft;s&

(A) DIBAL-H rFkk H2-Pd/BaSO

4 nksuksa vEyh; DyksjkbM ds lkFk ,sfYMgkbM nsrs gSa

(B) MnO2 , lHkh ,sfyyhd rFkk csfUtyhd , sYdksgkWyksa dks vkWDlhdr djrk gS

(C) ik;:fod vEy] H5IO

6 dks vipf;r djrk gS

(D) ,sFkkbu dk vkstksuh vi?kVu ,sfFku dh rqyuk esa /khek gksrk gS


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

Consider the following solution :Beaker-A Beaker-B Beaker-C Beaker-D

HCl NaOH CH COOHK =10

3

a–5

NH OH4

K =10b–5

If MAB

represent solution obtained by mixing content of beaker A and B completely.

fuEu foy;uks ij fopkj dhft;s :fcdj-A fcdj-B fcdj-C fcdj-D

HCl NaOH CH COOHK =10

3

a–5

NH OH4

K =10b–5

;fn MAB

] fcdj A rFkk B ds vo;oksa dks iw.kZ fefJr djus ls izkIr foy;u dks iznf'kZr djrk gS] rks9. Correct order of pH of mixed solution obtained on mixing above solution each having (same) equal

equivalent of solute.

mijksDr foy;u ftuesa izR;sd esa foys; ds leku rqY;kad mifLFkr gS] dks feykus ls izkIr fefJr foy;u dh pH

dk lgh Øe gS(A) M

AB > M

AC > M

BC > M

CD(B) M

AC > M

CD > M

BC > M

AB

(C) MBC

> MAB

> MCD

> MAC

(D) MBC

> MAB

= MCD

> MAC

10. Buffer solution can be obtained by mixing content of -(A) Beaker A and C in equivalent amount(B) Beaker A and D in equivalent amount(C) Beaker A and D where equivalent of solute in beaker A is less than that of D .(D) Beaker A and D where equivalent of solute in beaker D is less than that of A.fuEu dkSuls vo;oks dks feykus ls cQj foy;u izkIr gks ldrk gS -(A) rqY;kad ek=k esa fcdj A rFkk C ds vo;o(B) rqY;kad ek=k esa fcdj A rFkk D ds vo;o(C) fcdj A rFkk D tgk¡ fcdj A esa foys; dk rqY;kad D ls de gksrk gS(D) fcdj A rFkk D tgk¡ fcdj D esa foys; dk rqY;kad A ls de gksrk gS



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Molecular orbital theory is based on linear combination of atomic orbitals (LCAO). According to LCAO

when respective atomic orbitals of the atoms interact, they undergoes constructive and destructive

interference giving two types of molecular orbital i.e. bonding and antibonding molecular orbitals

respectively.

v.kq d{kd fl¼kar ] ijekf.o; d{kdksa ds js[kh; la;ksx (LCAO) ij vk/kkfjr gSA LCAO ds vuqlkj tc ijek.kqvksa

ds lacfU/kr ijekf.o; d{kd] vU;ksU; fØ;k djrs gSa rks os laiks"kh rFkk fouk'kh O;frdj.k }kjk nks izdkj ds vkf.od

d{kd Øe'k% ca/kh rFkk foijhr ca/kh vkf.od d{kd cukrs gSaA

11. Among the following the INCORRECT statement is :

(A) During N2

+ formation, one electron is removed from the bonding molecular orbital of N

2

(B) During O2

+ formation, one electron is removed from antibonding molecular orbitals of O

2

(C) During O2

– formation one electron is added to the bonding molecular orbital of O

2

(D) During CN– formation one electron is added to the bonding molecular orbital of CN

fuEu esa ls xyr dFku gS&

(A) N2

+ ds fuekZ.k ds nkSjku] N

2 ds ca/kh vkf.od d{kd ls ,d bysDVWªku gVk;k tkrk gS

(B) O2

+ ds fuekZ.k ds nkSjku] O

2 ds foijhr ca/kh vkf.od d{kd ls ,d bysDVWªku gVk;k tkrk gS

(C) O2

– ds fuekZ.k ds nkSjku] O

2 ds ca/kh vkf.od d{kd esa ,d bysDVªWku tksM+k tkrk gS

(D) CN– ds fuekZ.k ds nkSjku] CN ds ca/kh vkf.od d{kd esa ,d bysDVªku tksM+k tkrk gS

12. If Hund's rule does NOT hold good, then which of the following pairs is diamagnetic

;fn gqUM ds fu;e dk ikyu ugha gksrk gS rks fuEu esa ls dkSu lk ; qXe izfrpqEcdh; gS&

(A) B2, O

2

+(B) C

2, O

2(C) O

2

–, N

2(D) N

2, NO


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For given reaction sequence molecular formula for compound 'U' is C7H

6O

2 & P gives negative

Fehling test.

fn;s x;s vfHkfØ;k Øe ds fy;s ;kSfxd 'U' dk v.kqlw= C7H

6O

2 gS rFkk P ½.kkRed Qsgfyax ijh{k.k nsrk gSA

P + Q NaOH RO3 S + PZn, H O2

W

HgSO4H SO2 4

Red V NaOHHot Fe tube CaO, U

KMnO4 T

CrO Cl2 2

13. Compound which is not a hydrocarbon

;kSfxd tks gkbMªksdkcZu ugha gS&

(A) W (B) R (C) T (D) V

14. Compound S is :

;kSfxd S gS&

(A) CH3 – CH = O (B) Ph – CH = O (C)

CH = O

CH = O(D) CH – CH – CH2

O O

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

Matrix-Match Type / eSfVªDl&esy izdkj Integer Value Correct Type / iw.kk±d eku lgh izdkj




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[k.M-IV : (iw.kk±d eku lgh izdkj)

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


bl [k.M esa 6 iz'u gSaA izR;sd iz'u dk mÙkj 0 ls 9 rd (nksuksa 'kkfey) ds chp dk ,dy vadh; iw.kk ±d gSA

1. A radioactive sample undergoes a and b decay simultaneously. If half lifes of a and b decay are 4 years

and 12 years respectively. After how many years 6.25% of reactant is left.

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

,d jsfM;kslfØ; uequs esa a rFkk b {k; ,d lkFk gksrk gSA ;fn a rFkk b {k; dh v/kZ vk;q Øe'k% 4 o"kZ rFkk

12 o"kZ gS rks fdrus o"kksZ ds i'pkr~ 6.25% vfHkdkjd 'ks"k cpsxkA

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

2. Ratio of p

C

KK

for a reaction is 625 at 300K. The difference in heat of reaction at consant pressure and

constant volume (in calorie) is

(Take : R = 1

12 atm-L/mole-K, R = 2 cal./mole-K)

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

300K ij ,d vfHkfØ;k ds fy, p

C

KK

dk vuqikr 625 gS fu;r nkc rFkk fu;r vk;ru ij vfHkfØ;k dh Å"ek

esa vUrj (dSyksjh esa) gS

(fn;k gS : R = 1

12 atm-L/mole-K, R = 2 cal./mole-K)

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 sum of oxidation state of Ag in AgO(s)

AgO(s) esa Ag dh vkWDlhdj.k voLFkk dk ;ksx Kkr dhft,A

4. Find the number of molecule in which central atom gives 'ic' oxy acid on hydrolysis :

(i) XeF2 + H

2O at extreme

temperature¾¾¾¾® (ii) SF4 + H

2O ¾¾®

(iii) NCl3 + H

2O ¾¾® (iv) SO

2Cl

2 + H

2O ¾¾®

(v) P4S

10 + H

2O ¾¾® (iv) N

2O

4 + H

2O ¾¾®

fuEu esa ls , sls v.kqvksa dh la[;k Kkr dhft, ftuesa dsUæh; ijek.kq] tyvi?kVu ij 'bZd' vkWDlh vEy nsrk gS

(i) XeF2 + H

2O ¾¾¾¾®mPp rki ij (ii) SF

4 + H

2O ¾¾®

(iii) NCl3 + H

2O ¾¾® (iv) SO

2Cl

2 + H

2O ¾¾®

(v) P4S

10 + H

2O ¾¾® (iv) N

2O

4 + H

2O ¾¾®



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5. How many structures of 1,4-disubstituted benzene is/are possible with molecular formula C8H

8O

3

in which one –COOH group must be present :

v.kqlw= C8H

8O

3 ds 1,4-f}izfrLFkkfir csUthu dh fdruh lajpuk,sa laHko gS ftuesa ,d –COOH lewg mifLFkr gksuk

pkfg;sA

6. How many amide from molecular formula C4H

9NO can show the hydrogen bonding :

v.kqlw= C4H

9NO ls fdrus ,sekbM gkbMªkstu cU/ku iznf'kZr dj ldrs gSaA


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PART-3 : MATHEMATICS

Hkkx-3 : xf.krSECTION–I : (i) One or more options correct Type

[k.M-I : (i) ,d ;k vf/kd lgh fodYi izdkjThis 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 ;k

vf/kd lgh gSA

1. z is a complex number satisfying |z – 4 – 3i| = 1, then-(A) maximum value of |z| is 6(B) minimum value of |z| is 4

(C) maximum value of principal argument of z is 1 6 6 4tan

4 24 3- é ù+

ê ú-ë û

(D) minimum value of principal argument of z is 1 6 6 4tan

4 24 3- é ù-

ê ú-ë û

lfEeJ la[;k z, |z – 4 – 3i| = 1 dks lUrq"V djrh gS] rks -

(A) |z| dk vf/kdre eku 6 gksxkA

(B) |z| dk U;wure eku 4 gksxkA

(C) z ds eq[; dks.kkad dk vf/kdre eku 1 6 6 4tan

4 24 3- é ù+

ê ú-ë û

gksxkA

(D) z ds eq[; dks.kkad dk U;wure eku 1 6 6 4tan

4 24 3- é ù-

ê ú-ë û

gksxkA

2. If xr

and yr

are non zero vectors such that x y x 2y+ = -r r r r

, then-

(A) 2

2x.y y=r r r(B)

2x.y y=r r r

(C) The value of 2

1x.y

y 2+

+

r r

r can be 1

2(D) The value of 2

1x.y

y 2+

+

r r

r can be 1

;fn xr

rFkk yr

v'kwU; lfn'k bl izdkj gS fd x y x 2y+ = -r r r r

gks] rks -

(A) 2

2x.y y=r r r(B)

2x.y y=r r r

(C) 2

1x.y

y 2+

+

r r

r dk eku 1

2 gks ldrk gSA (D) 2

1x.y

y 2+

+

r r

r dk eku 1 gks ldrk gSA



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3. Consider the word 'INSTITUTE'

(A) Number of ways of selection of 4 letters is 41

(B) Total number of words beginning with vowel is 8!

23!

´

(C) Total number of words which are formed by 4 letters are 626

(D) Total number of words beginning with vowel is 8!

3

ekuk 'kCn INSTITUTE(A) 4 v{kjksa ds p;u ds rjhdksa dh la[;k 41 gksxhA

(B) Loj ls izkjEHk gksus okys 'kCnksa dh dqy la[;k 8!2

3!´ gksxhA

(C) 4 v{kjksa }kjk fufeZr 'kCnksa dh dqy la[;k 626 gksxhA

(D) Loj ls izkjEHk gksus okys 'kCnksa dh dqy la[;k 8!

3 gksxhA

4. If the sum of n terms of the series 1 2 3

5 7 9.......

1.2.3 2.3.3 3.4.3+ + + is Sn, then-

;fn Js.kh 1 2 3

5 7 9.......

1.2.3 2.3.3 3.4.3+ + + ds n inksa dk ;ksxQy Sn gks] rks -

(A) 3

107S

108= (B) n n

1S 1

3 .n= - (C) n

nlim S 1

®¥= (D) n

nlim S 3

®¥=

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5. Let ƒ(x) = sinx + cosx g(x) = sinx – cosx

( )( )

/ 2

0

g xA dx

ƒ x

p

= ò and B is area bounded by curve y = ƒ(x), y = g(x) and x-axis where x Î [0,p]

ekuk ƒ(x) = sinx + cosx g(x) = sinx – cosx

( )( )

/ 2

0

g xA dx

ƒ x

p

= ò rFkk B, oØ y = ƒ(x), y = g(x) rFkk x-v{k] tgk¡ x Î [0,p] gS] }kjk ifjc¼ {ks=Qy gks] rks

(A) A = 0 (B) A 2 n 2= - l (C) B 2 2= - (D) B 2 2 2= -

6. If ƒ(x) is continuous function having range [–2,2] " x Î R and ( )( ) ( )

( ) ( )

ƒ x |ƒ x |

ƒ xƒ x

e eg x

e e

-=

+, then-

(A) g(x) is many one function

(B) Range of g(x) is 4

4

1 e,0

1 e

é ù-ê ú+ë û

(C) [g(x)] can take only 2 elements ([.] denotes greatest integer function)

(D) g(x) + p = 0 has no solution

;fn ƒ(x) larr~ Qyu ftldk ifjlj [–2,2] " x Î R rFkk ( )( ) ( )

( ) ( )

ƒ x |ƒ x |

ƒ xƒ x

e eg x

e e

-=

+ gks] rks -

(A) g(x) cgq,dSdh Qyu gksxk

(B) g(x) dk ifjlj 4

4

1 e,0

1 e

é ù-ê ú+ë û

gksxkA

(C) [g(x)] dsoy 2 eku ys ldrk gS (tgk¡ [.] egÙke iw.kk±d Qyu dks n'kkZrk gS)

(D) g(x) + p = 0 dk dksbZ gy ugha gksxkA


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7. Maximum distance of any point on the ellipse x2 + 2y2 + 2xy – 1 = 0 from the centre is-

nh?kZoÙk x2 + 2y2 + 2xy – 1 = 0 ij fLFkr fdlh fcUnq dh blds dsUæ ls vf/kdre nwjh gksxh -

(A) 3 3+ (B) 2 2+ (C) 2

3 5-(D)

5 1

2

+

8. If |z – 2015| = 2015 and arg(z) = q where z ¹ 0, then-

(A) Range of q is same as sin–1x (B) Range of q is same as tan–1x

(C) 4030

1z

æ ö-ç ÷è ø

is equal to itanq (D) 4030

1z

æ ö-ç ÷è ø

is equal to icotq

;fn |z – 2015| = 2015 rFkk arg(z) = q tgk¡ z ¹ 0 gks] rks -

(A) q dk ifjlj sin–1x ds leku gksxkA (B) q dk ifjlj tan–1x ds leku gksxkA

(C) 4030

1z

æ ö-ç ÷è ø

= itanq (D) 4030

1z

æ ö-ç ÷è ø

= icotq


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L1,L2 are two tangents to the parabola y2 = 16x which makes an angle of 1 1tan

3- æ ö

ç ÷è ø

to the tangent (T)

to the same parabola at the end point of latus rectum with positive ordinate, then

L1,L2 ijoy; y2 = 16x ij nks Li'kZjs[kk;sa gS] tks ml ijoy; ds /kukRed dksfV okys ukfHkyEc ds vfUre fljs

ij [khaph xbZ Li'kZ js[kk (T) ds lkFk 1 1tan

3- æ ö

ç ÷è ø

dk dks.k cukrh gS] rks

9. Area of triangle formed by the lines L1,L2 & T is-

js[kkvksa L1,L2 rFkk T }kjk fufeZr f=Hkqt dk {ks=Qy gksxk

(A) 3 (B) 6 (C) 12 (D) 24

10. Chord of contact joining point of contacts of tangents L1 & L2 always passes through the point -

L1 rFkk L2 dh Li'kZjs[kkvksa ds Li'kZfcUnqvksa dks feykus okyh Li'kZ thok;sa lnSo fuEu fcUnq ls xqtjsxh -

(A) (4,0) (B) (2,0) (C) (–2,0) (D) (–4,0)



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iz'u 11 ,oa 12 ds fy;s vuqPNsnMr. X travels from Delhi to Muscat with change of airplanes in Chennai, Dubai and Sharjah respectively.He has one piece of luggage. At each stop the luggage is transferred from one airplane to another . At

the airport in Delhi there is a probability of 1

20 that luggage is not placed in right plane. This probability

is 1

10 at airport in Chennai,

1

5 at airport in Dubai and

1

10 at airport in Sharjah

fe- X fnYyh ls eLdV dh ;k=k Øe'k% psUubZ] nqcbZ rFkk 'kkjtkg ls ifjofrZr djds iwjh djrk gSA muds ikl ,dlkeku gSA izR;sd Bgjko ij lkeku dks ,d gokbZtgkt ls nwljs gokbZtgkt esa igq¡pk;k tkrk gSA fnYyh ,;jiksVZ ij

lkeku ds lgh LFkku ij ugha igq¡pus dh izkf;drk 1

20 gSA ;g izkf;drk psUubZ ,;jiksVZ ij

1

10, nqcbZ ,;jiksVZ ij

1

5 rFkk 'kkjtkg ,;jiksVZ ij

1

10 gSA

11. The probability that Mr. X luggage does not reach Sharjah with him -

fe- X ds lkFk mudk lkeku 'kkjtkg ugha igq¡pus dh izkf;drk] gksxh -

(A) 0.236 (B) 0.316 (C) 0.406 (D) 0.604

12. If Mr. X luggage does not reach muscat with him, then the probability that it was lost at airport of

Dubai is -

;fn fe- X dk lkeku muds lkFk eLdV ugha igq¡prk gS] rks muds lkeku ds nqcbZ ij NwVus dh izkf;drk gksxh -

(A) 855

1922(B)

900

1922(C)

950

1922(D)

1000

1922


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Consider a tetrahedron ABCD such that area of DABC = k1 sq. units, area DBCD = k2 sq. units, area

DACD = k3 sq. units and area DABD = k4 sq. units The volume of tetrahedron is 1

6 cubic units.

If each of the faces ACB, ACD, BCD of tetrahedron subtends a right angle at 'C' then

ekuk prq"Qy ABCD bl izdkj gS fd f=Hkqt ABC dk {ks=Qy k1 oxZ bdkbZ] f=Hkqt BCD dk {ks=Qy k2 oxZ

bdkbZ] f=Hkqt ACD dk {ks=Qy k3 oxZ bdkbZ rFkk f=Hkqt ABD dk {ks=Qy k4 oxZ bdkbZ gSA prq"Qyd dk

vk;ru 1

6 ?ku bdkbZ gSA

;fn prq"Qd ds izR;sd Qyd ACB, ACD, BCD, 'C' ij ledks.k vUrfjr djrs gSa] rks13. Geometric mean of k1,k2,k3 is -

k1,k2,k3 dk xq.kksÙkj ek/; gksxk -

(A) 1

4(B)

1

3(C)

1

2(D)

1

8

14. Which of the following expression is true-

fuEu esa ls dkSulk O;atd lR; gksxk -

(A) 3

2 2i 4

i 1

k k=

=å (B) 1 2 34

1 2 3

k k kk

k k k

æ ö= ç ÷+ +è ø

(C) 3

2 2i 4

i 1

k k=

=å (D) 1 2 34

1 2 3

k k kk

k k k

æ ö= ç ÷+ +è ø

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

Matrix-Match Type / eSfVªDl&esy izdkj Integer Value Correct Type / iw.kk±d eku lgh izdkj




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


bl [k.M esa 6 iz'u gSaA izR;sd iz'u dk mÙkj 0 ls 9 rd (nksuksa 'kkfey) ds chp dk ,dy vadh; iw.kk ±d gSA

1. There are n balls of different colours having same weights, if sum of weights of every combination of

balls taken three at a time, is 315 gm & sum of weights of every combination of balls taken five balls at

a time, is 1575

2gm, then possible value of n is

fofHkUu jax dh n xsanksa dk Hkkj leku gaSA ;fn rhu xsanksa ds lHkh lEHko lap;ksa ds Hkkjksa dk ;ksx 315 xzke gS rFkk ik¡p xsanksa ds

lHkh lEHko lap;ksa ds Hkkjksa dk ;ksx 1575

2 xzke gS] rks n dk lEHko eku gksxk

2. If 3a1 + 4b1 – 10 = 0 and 4a2 – 3b2 + 20 = 0, then area of triangle formed by (–2,4) (a1,b1) (a2,b2)

when 2 2i ia + b is minimum (for i = 1,2) is

;fn 3a1 + 4b1 – 10 = 0 rFkk 4a2 – 3b2 + 20 = 0 gks] rks (–2,4) (a1,b1) (a2,b2) }kjk fufeZr f=Hkqt dk {ks=Qy]

tc 2 2i ia + b U;wure gS (i = 1,2 ds fy,) gksxk


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3. An equilateral triangle is inscribed inside the circle of radius 12 units. If maximum area of circle which

touches one of the sides of triangle and given circle is A, then value of A

25é ùê úpë û

is

(where [.] denotes greatest integer function)

,d leckgq f=Hkqt 12 bdkbZ f=T;k ds o`Ùk ds vUrxZr gSA ;fn f=Hkqt dh fdlh ,d Hkqtk rFkk fn, x, o`Ùk dks Li'kZ djus

okys o`Ùk dk vf/kdre {ks=Qy A gks] rks A

25é ùê úpë û

dk eku gksxk

(tgk¡ [.] egÙke iw.kk±d Qyu dks n'kkZrk gS)

4. If / 2

nn

0

I x sin xdxp

= ò , then value of 3

14

p(I4 + 12I2) is

;fn / 2

nn

0

I x sin xdxp

= ò gks] rks 3

14

p(I4 + 12I2) dk eku gksxk



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5.

naa

n

8e n 1lim sin

e n 1 n®¥

æ öæ ö æ ö æ ö+ç ÷ç ÷ ç ÷ ç ÷ç ÷+è ø è øè ø è øwhere a Î Q (rational number) is equal to

naa

n

8e n 1lim sin

e n 1 n®¥

æ öæ ö æ ö æ ö+ç ÷ç ÷ ç ÷ ç ÷ç ÷+è ø è øè ø è ø, tgk¡ a Î Q (ifjes; la[;k) gksxk

6. If sum of n positive numbers is 13 then possible value of n such that their product is maximum, is

;fn n /kukRed la[;kvksa dk ;ksxQy 13 gks] rks n dk lEHko eku rkfd budk xq.kuQy vf/kdre gS] gksxk


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D. vadu ;kstuk / Marking scheme :14. [kaM-I (i) ds gj iz'u esa dsoy lgh mÙkj okys cqycqys (BUBBLE) dks dkyk djus ij 4 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 4 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 in thissection.

15. [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 toonly the correct answer(s) and zero mark if no bubbles are darkened. In all other cases minus one (–1) mark willbe awarded

16. [kaM-IV esa gj iz'u esa dsoy lgh mÙkj okys cqycqys (BUBBLE) dks dkyk djus ij 4 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-IV, you will be awarded 4 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. 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 gS A

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/01CT314074Your Target is to secure Good Rank in JEE 201540/40

allen 0 1 3 1 4 0 7 4 clas amme & leade coue/sce/21-04-2015/pae-1 physics ka/01ct314074 5/40...

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