ALLEN CLAS AMME · 2015. 12. 21. · ALL INDIA OPEN TEST/JEE (Adaced)/08-02-2015/PAPER-1 PHYSICS...

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

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

Hkkx-1 HkkSfrd foKku I(i) dsoy ,d lgh fodYi izdkj 03 - 08Part-1 Physics Only One Option Correct Type

I(ii) vuqPNsn izdkj 09 - 15Paragraph Type

IV iw.kk±d eku lgh izdkj (0 ls 9) 16 - 18Integer Value Correct Type (0 to 9)

Hkkx-2 jlk;u foKku I(i) dsoy ,d lgh fodYi izdkj 19 - 22Part-2 Chemistry Only One Option Correct Type



Hkkx-3 xf.kr I(i) dsoy ,d lgh fodYi izdkj 30 - 32Part-3 Mathematics Only One Option Correct Type



Kota/00CT2140032/40

ALL INDIA OPEN TEST/JEE (Advanced)/08-02-2015/PAPER-1

SOME USEFUL CONSTANTSAtomic No. H = 1, B = 5, C = 6, N = 7, O = 8, F = 9, Al = 13, P = 15, S = 16, Cl = 17,Br = 35, Xe = 54, Ce = 58,

Atomic masses : H = 1, Li = 7, B = 11, C = 12, N = 14, O = 16, F = 19, Na = 23, Mg = 24,Al = 27, P = 31, S = 32, Cl = 35.5, Ca=40, Fe = 56, Br = 80, I = 127,

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

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

· Coulomb's law constantpe

9

0

1 = 9×104

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

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

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

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

· Permittivity of vacuum Î0 = 20

1

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


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PART-1 : PHYSICS Hkkx-1 : HkkSfrd 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 gSaA izR;sd iz'u esa pkj fodYi (A), (B), (C) vkSj (D) gSa ftuesa ls dsoy ,dlgh gSA

1. A magnetic field (B), uniform between two magnets can be determined by measuring the inducedvoltage in the loop as it is pulled through the gap at uniform speed 20 m/sec. Size of magnet is2 cm × 1cm × 2cm and size of coil 4 cm × 6 cm as shown in figure. The correct variation of induced emfwith time is : (Assume at t = 0, the coil enters the magnetic field) :-nks pqEcdksa ds e/; le:i pqEcdh; {ks= (B) dks pqEcdks ds e/; LFkku esa fdlh ywi dks ,d leku pky 20 m/sec ls[khapus ij ywi esa izsfjr oksYVrk ds ekiu }kjk Kkr fd;k tk ldrk gSA pqEcd dk vkdkj 2 cm × 1cm × 2cm rFkkdq.Myh dk vkdkj 4 cm × 6 cm fp=kuqlkj gSA le; ds lkFk izsfjr fo|qr okgd cy dk lgh ifjorZu gksxk%&(ekuk fd t = 0 ij dq.Myh pqEcdh; {ks= esa izos'k djrh gS) :-

1cm 2cm

N

S

2cm

1cm

4cm

6cm

(A) 5002500 3000

E

0.40B

–0.40B

µsec (B) 5002500 3000

E

0.40B

–0.40B

t(µsec)0

(C) 5002500 3000

E

0.80B

–0.80B

µsec (D) 5002500 3000

E

0.80B

–0.80B

t(µsec)0

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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2. A solid sphere of radius r is floating at the interface of two immiscible liquids of densities r1 andr2 (r2 > r1), half of its volume lying in each. The height of the upper liquid column from the interface ofthe two liquids is h. The force exerted on the sphere by the upper liquid is (atmospheric pressure = p0 &acceleration due to gravity is g) :-f=T;k r okyk ,d Bksl xksyk ?kuRo r1 rFkk r2 (r2 > r1) okys nks vfeJ.kh; nzoksa ds vUrjki`"B ij bl izdkj ls rSj jgkgS fd izR;sd æo esa bldk vk/kk vk;ru jgrk gSA nksuksa æoksa ds vUrjki`"B ls Åijh æo LrEHk dh Å¡pkbZ h gSA Åijh æo}kjk xksys ij vkjksfir cy gksxk%& (ok;qe.Myh; nkc = p0 rFkk xq:Roh; Roj.k g gS)

Solidsphere

hr1

r2

r r2 > 1

(A) p0pr2 + (h – 2/3r)pr2r1g (B) (h – 2/3r)pr2r1g

(C) 2/3 pr3 r1g (D) p0 × pr2

3. A uniform disc of mass m = 12 kg slides down along smooth, frictionless hill, which ends in a horizontalplane without break. The disc is released from rest at a height of h = 1.25 m (it has no initial speed andit does not rotate), and lands on the top of a cart of mass M = 6 kg, which can move on a frictionlesssurface. The coefficient of kinetic friction between the cart and the disc is µ = 0.4. Find minimum lengthof the cart (in m) so that the disc begins to roll without slipping before loosing contact with the cart.æO;eku m = 12 kg okyh ,d le:i pdrh fpduh ?k"kZ.kjfgr igkM+h tks fd {kSfrt ry ij tkdj [kRe gksrh gS] dsvuqfn'k uhps dh vksj fQlyrh gSA pdrh dks h = 1.25 m dh Å¡pkbZ ij fojkekoLFkk ls NksM+k tkrk gS (bldh dksbZizkjfEHkd pky ugh gS rFkk ;g yq


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4. A screw gauge has some zero error but its value is unknown. We have two identical rods. When the firstrod is inserted in the screw, the state of the instrument is shown by diagram (I). When both the rods areinserted together in series then the state is shown by the diagram (II). What is the zero error of theinstrument? 1 msd = 100 csd = 1 mm :-,d LØwxst esa dqN 'kwU;kadh = qfV gS ijUrq bldk eku vKkr gSA gekjs ikl nks ,d tSlh NMs+ gSA tc igyh NM+ dks LØw esaizfo"V djkrs gS rks midj.k dh voLFkk dks fp= (I) }kjk n'kkZ;k x;k gSA tc nksuksa NM+ksa dks Js.khØe esa ,dlkFk LØw esaizfo"V djkrs gS rks midj.k dh voLFkk dks fp= (II) }kjk n'kkZ;k x;k gSA midj.k dh 'kwU;kadh =qfV D;k gksxh\1 msd = 100 csd = 1 mm :-

12

10

(A) –0.16 mm (B) +0.16 mm (C) +0.14 mm (D) –0.14 mm5. Block A of mass 2m, B and C each of mass m are in equilibrium and at rest shown in the figure. The

spring balance, pulley and strings are massless. Now, a force mg is applied on block C in downwarddirection. Find the new reading of spring balance. Assume all pulleys to be frictionless. (Given m = 2kg)

CykWd A dk æO;eku 2m gS rFkk CykWd B o C izR;sd dk æO;eku m gSA ; s lHkh lkE;koLFkk esa gS rFkk fp=kuqlkj

fojkekoLFkk esa gSA fLizax rqyk] f?kjuh rFkk jfLl;k¡ æO;ekughu gSA vc CykWd C ij uhps dh fn'kk esa cy mg yxk;k tkrkgSA fLizax rqyk dk u;k ikB~;kad Kkr dhft;sA lHkh f?kjfu;ksa dks ?k"kZ.kjfgr ekfu;sA (fn;k gS% m = 2kg)

2mA

m mCB

(A) 7 kg (B) 5 kg (C) 11 kg (D) 12 kg

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6. A uniform flexible chain of mass m and length R

L2

pæ ö£ç ÷è ø

rests on a fixed smooth cylindrical surface of

radius R such that one end A of the chain is at the top of the cylinder while the other end B is free. Thechain is held stationary by a horizontal thread PA as shown in the figure. Calculate the tension in the

thread. (where LR

a = ) :-

æO;eku m rFkk yEckbZ R

L2

pæ ö£ç ÷è ø

okyh ,d le:i yphyh tathj f=T;k R okyh fLFkj fpduh csyukdkj lrg ij

bl izdkj fojkekoLFkk esa j[kh gS fd tathj dk ,d fljk A csyu ds 'kh"kZ ij jgrk gS tcfd nwljk fljk B eqDr jgrk gSA

tathj dks fp=kuqlkj {kSfrt /kkxs PA }kjk jksddj j[kk tkrk gSA /kkxs esa ruko Kkr dhft;sA (;gk¡ LR

a = ) :-

BOR

AP

(A) ( )2sin

T mga

=a

(B) ( )a=a

2sinT mg

2(C) ( )

2sin2T mg

2

a

=a

(D) ( )2sinT mga=a



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7. A beam of light has a small wavelength spread dl about a central wavelength l. The beam travels invacuum until it enters a glass plate at an angle q relative to the normal to the plate, as shown in the figure.The index of refraction of the glass is given by n(l). The angular spread dq' of the refracted beam isgiven by :-,d izdk'k iqat dk dsUnzh; rajxnS/; Z l ds lkis{k de rjaxnS/; Z QSyko dl gSA iqat fuokZr esa rc rd xeu djrk gS tc rdfd ;g fp=kuqlkj IysV ds vfHkyEc ds lkis{k q dks.k ij dk¡p dh IysV esa izos'k ugh dj tkrkA dk¡p dk viorZukadn(l) }kjk fn;k tkrk gSA viofrZr iqat dk dks.kh; QSyko dq' gksxk%&

q'

dq'

Glass

Vacuum

q

(A) 1

'n

dq = dl (B) ( )dn

'd

ldq = dl

l(C)

1 d'

dnl

dq = dll

(D) ( )dntan '

'n d

lqdq = dl

l


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8. Two long, thin, solid cylinders are identical in size, but they are made of different substances with twodifferent thermal conductivities. The two cylinders are connected in series between a reservoir attemperature Thot and a reservoir at temperature Tcold. The temperature at the boundary between the twocylinders is Tb. One can conclude that :-(A) Tb is closer to Thot than it is to Tcold(B) Tb is closer to Tcold than it is to Thot(C) Tb is closer to the temperature of the reservoir that is connected through the cylinder of lower

thermal conductivity(D) Tb is closer to the temperature of the reservoir that is connected through the cylinder of higher

thermal conductivity

nks yEcs] irys] Bksl csyu ,dleku vkdkj esa gS ijUrq ; s nks fHkUu rkih; pkydrkvksa okys fHkUu inkFkksZ ls cus gSA nksuksa

csyuksa dks nks ik=ksa ds e/; Js.khØe eas tksM+k tkrk gS buesa ls ,d ik= Thot rki ij rFkk nwljk ik= Tcold rki ij gSA nksuksa

csyuksa ds e/; ifjlhek ij rki Tb gSA ge ;g fu"d"kZ fudky ldrs gS fd%&

(A) Tb, Tcold dh rqyuk esa Thot ds fudV gSA

(B) Tb, Thot dh rqyuk esa Tcold ds fudV gSA

(C) Tb ml ik= ds rki ds fudV gS tks fuEu rkih; pkydrk okys csyu ls tqM+k gS

(D) Tb ml ik= ds rki ds fudV gS tks mPp rkih; pkydrk okys csyu ls tqM+k gS



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(ii) Paragraph Type (i i ) vuqPNsn izdkj

This section contains 4 paragraphs each describing theory, experiment, data etc. Eight questionsrelate to four paragraphs with two questions on each paragraph. Each question of a paragraph hasonly one correct answer among the four choices (A), (B), (C) and (D).bl [k.M esa fl¼kUrks a] iz;ksxks a vkSj vk¡dM+ks a vkfn dks n'kkZus okys 4 vuqPNsn gSA pkjksa vuqPNsnksa ls lacaf/kr vkBiz'u gSa] ftuesa ls gj vuqPNsn ij nks iz'u gSA vuqPNsn esa gj iz'u ds pkj fodYi (A), (B), (C) vkSj (D) gSaftuesa ls dsoy ,d lgh gSA

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

There are two homogeneous, solid disks of radius R and mass m mounted by two parallel, horizontalaxes at the ends of a horizontal rod of negligible mass. The distance between these axes is d and thedisks can freely rotate around them. The rod itself, with the disks mounted on it, can also freely rotatearound a horizontal axis in it midpoint. (See the figures. All the three axes are perpendicular to the rod.)On the rim of each disk there is a small pin, and between them there is a spring of spring constant kwhich is initially compressed by Dl. (The spring is in contact with the pins until it extends to its unstretchedposition and then falls down.)f=T;k R rFkk æO;eku m okyh nks lekaxh] Bksl pdfr;ksa dks ux.; æO;eku okyh {kSfrt NM+ ds fljksa ij nks lekUrj {kSfrtv{kksa }kjk tksM+k x;k gSA bu v{kksa ds e/; nwjh d gS rFkk pdfr;k¡ buds lkis{k Lora= :i ls ?kw.kZu dj ldrh gSA og NM+ftl ij ; s pdfr;k¡ tqM+h gqbZ gS] Loa; Hkh blds e/; fcUnq ls tqM+h {kSfrt v{k ds lkis{k Lora= :i ls ?kw.kZu dj ldrhgSA (lHkh rhuksa v{ksa NM+ ds yEcor~ gS) izR;sd pdrh dh ifj/kh ij ,d NksVh fiu yxh gS rFkk buds e/; fLizax fu;rakdk okyh ,d fLizax tqM+h gqbZ gS tks izkjEHk esa Dl }kjk lEihfM+r gSA (fLizax fiuksa ds lkFk rc rd lEidZ esa jgrh gS tc rdfd ;g bldh vfoLrkfjr fLFkfr rd ugh QSy tkrh rFkk rc ;g uhps fxj tkrh gS)

d

D mm

R R

(a) (b)


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Kota/00CT214003

9. Find the angular velocity of the disks after we burn the thread that hold the spring in it compressed

position, provided that its initial position corresponds to figure (a) :-

tc ge ml /kkxs dks tyk nsrs gS tks fd fLizax dks bldh lEihfM+r voLFkk esa cuk;s j[krk gS rc pdfr;ksa ds dks.kh; osx

Kkr dhft;s] tcfd bldh izkjfEHkd fLFkfr fp= (a) ds laxr gSA

(A) 2 kR mDl

(B) k

2R mDl

(C) k

R mDl

(D) 2k

R 3mDl

10. Find the angular velocity of the disks after we burn the thread that hold the spring in it compressed

position, provided that its initial position corresponds to figure (b) :-

tc ge ml /kkxs dks tyk nsrs gS tks fd fLizax dks bldh lEihfM+r voLFkk esa cuk;s j[krk gS rc pdfr;ksa ds dks.kh; osx

Kkr dhft;s] tcfd bldh izkjfEHkd fLFkfr fp= (b) ds laxr gSA

(A) 2 2d 2k

R md 2R

D×

+

l

(B) 2 2d 2k

R 3md 2R

D×

+

l

(C) 2 22d k

R md R

D×

+

l

(D) 2 2d k

R md 2R

D×

+

l



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Paragraph for Questions 11 and 12

iz'u 11 ,oa 12 ds fy;s vuqPNsnFigure shows an arbitrary body through which electric charge is distributed in an arbitrary way. Thebody is divided in two region A and B. Using the result.The electric force that the charge in region A exerts on the charge in region B is equal to the force thatthe charge on the entire body exerts on the charge in region B or otherwise solve next two problems.

fp= esa ,d LoSfPNd vkd̀fr ds fi.M dks n'kkZ;k x;k gS ftles fo|qr vkos'k LoSfPNd rjhds ls forjhr gSA fi.M dks nks

Hkkxksa A rFkk B esa foHkkftr dj fn;k tkrk gSA

og fo|qr cy tks Hkkx A esa mifLFkr vkos'k Hkkx B esa mifLFkr vkos'k ij yxkrk gS] ml cy ds cjkcj gS tks lEiw.kZ fi.M

ij fo|eku vkos'k] Hkkx B esa mifLFkr vkos'k ij yxkrk gSA bl ifj.kke dk mi;ksx dj vki fuEu iz'u gy djsA

BA

11. Consider a uniformly charged spherical shell of radius R having total charge Q. Find electric force Fr

exerted on each hemisphere by the charge on the other hemisphere.

f=T;k R okys ,d le:i vkosf'kr xksykdkj dks'k ij fopkj dhft;s ftl ij dqy vkos'k Q gSA ,d v¼Zxksys ij

fo|eku vkos'k }kjk izR;sd v¼Zxksys ij yxk;s x;s fo|qr cy Fr

dk eku gksxk%&

(A) 2

20

Q4 RpÎ (B)

2

20

Q16 RpÎ (C)

2

20

Q8 RpÎ (D)

2

20

Q32 RpÎ

12. Consider a uniformly charged solid dielectric sphere of total charge Q and radius R. The force exertedby one hemisphere on other hemisphere is :

dqy vkos'k Q rFkk f=T;k R okys ,d le:i vkosf'kr Bksl ijkoS|qr xksys ij fopkj dhft;sA ,d v¼Zxksys }kjk nwljs

v¼Zxksys ij yxk;k x;k cy gksxk%&

(A) 2

20

3Q16 RpÎ (B)

2

20

3Q64 RpÎ (C)

2

20

Q9 RpÎ (D)

2

20

Q36 RpÎ


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A very careful inspection of the hydrogen spectrum emission lines shows that many of them are infact

not single lines but very closely spaced combination of two lines known as fine structure. Examine's the

hydrogen atom from the electron's frame of reference, in which the proton appears to travel around the

electron. In electron's reference frame the motion of proton is in a circular orbit that causes magnetic

f ield B at the electron. The interaction energy of the spin magnetic moment µs in a magnetic field is :

U = –µs·B

We define z-direction to be direction of magnetic field B, with a parallel se2m

m =h

; h

2é ù=ê úpë ûh

The effect of magnetic field is to split each level into two; (i) a higher state with orbital angular momentum

L and spin angular momentum S parallel and (ii) a lower state with L and S antiparallel. Thus energy

difference between the sates is DE = 2µsB

gkbMªkstu LiSDVªeh mRltZu js[kkvksa dk ,d cgqr lko/kkuhiwoZd fy;k x;k izs{k.k ;g n'kkZrk gS fd buesa ls vf/kdrj js[kk;sa

okLro esa dksbZ ,d js[kk uk gksdj nks cgqr utnhd fLFkr js[kkvksa dk la;kstu gksrh gS ftls eghu lajpuk (fine structure)

dgk tkrk gSA gkbMªkstu ijek.kq dk izs{k.k bysDVªkWu ds funsZ'k ra= ls fd;k tkrk gS] ftlesa izksVkWu bysDVªkWu ds pkjksa vksj xfr

djrk gqvk fn[kkbZ nsrk gSA bysDVªkWu ds funsZ'k ra= esa izksVkWu dh xfr ,d o`Ùkkdkj d{kk esa gksrh gS ftlds QyLo:i

bysDVªkWu ij pqEcdh; {ks= B mRiUu gks tkrk gSA pqEcdh; {ks= esa pØ.k pqEcdh; vk?kw.kZ µs dh vU;ksU; ÅtkZ fuEu izdkj

nh tkrh gS:

U = –µs·B

;gk¡ z-fn'kk lekUrj se2m

m =h

; h

2é ù=ê úpë ûh ds lkFk pqEcdh; {ks= B dh fn'kk ds :i esa ifjHkkf"kr dh tkrh gSA

pqEcdh; {ks= ds izHkko ds dkj.k izR;sd Lrj nks Hkkxksa esa foHkkftr gks tkrk gS ; (i) ,d mPprj voLFkk gksrh gS ftlesa d{kh;

dks.kh; laosx L rFkk pØ.k dks.kh; laosx S lekUrj gksrs gSa tcfd (ii) ,d fuEure voLFkk gksrh gS ftlesa L o S izfr

lekUrj gksrs gSaA bl izdkj voLFkkvksa ds e/; ÅtkZ vUrjky DE = 2µsB gksrk gSA



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13/40Kota/00CT214003

13. Apply Bohr's quantization rule of angular momentum and value of magnetic field of proton on electronand obtain the energy difference between states :-

cksgj dk dks.kh; laosx Dok.Ve fu;e rFkk bysDVªkWu ij izksVkWu ds pqEcdh; {ks= ds eku dk mi;ksx djrs gq, voLFkkvksa ds

e/; ÅtkZ vUrjky gksxk %&

(A) 8

04 3 4 5

0

me 1E

256 nm

D =p Î h (B)

403 3 3 4

0

me 1E

64 nm

D =p Î h

(C) 6

03 4 4 5

0

me 1E

128 nm

D =p Î h (D)

40

3 4 3 40

me 1E

256 nm

D =p Î h

14. When an external magnetic field B is applied to a hydrogen atom three photons can be emittedcorresponding to transition from n = 2 to n = 1. The energies can be 10.2 eV + µ

SB, 10.2 eV,

10.2eV – µSB where S

e2m

m =h

is Bohr's magneton. By finding hcE =l

find expression for Dl (change

in wavelength)

tc gkbMªkstu ijek.kq ij ,d ckg~; pqEcdh; {ks= B yxk;k tkrk gS rks ;gk¡ laØe.k n = 2 ls n = 1 ds laxr rhu QksVksu

mRlftZr gks ldrs gSaA ÅtkZvksa ds eku 10.2 eV + µSB, 10.2 eV, 10.2eV – µ

SB gks ldrs gSa tgk¡ S

e2m

m =h

cksgj

eSXusVkWu gSA hcE =l

Kkr djrs gq, Dl (rjaxnS/;Z esa ifjorZu) ds fy, O;atd Kkr dhft, %&

(A) l

Dl =p

2e Bm c

(B) l

Dl =p

22 e Bm c

(C) l

Dl =p

2e B2 m c

(D) l

Dl =p

2e B4 m c


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iz'u 15 ,oa 16 ds fy;s vuqPNsnTHE ATMOSPHERIC LAPSE RATEFor small volumes of gas, according to KINETIC THEORY OF GASES, all parts of the gas are at thesame temperature. But for huge volumes of gas like atmosphere, assumption of a uniform temperaturethroughout the gas is not valid. Different parts of the atmosphere are at different temperature. Apart fromthe surface of the earth, variations also occur in temperature at different heights in the atmosphere.The decrease in temperature with height called the ATMOSPHERIC LAPSE RATE is similar atvarious locations across the surface of the Earth. By analyzing the data collected at various locations, itis found that average global lapse rate is – 6.7 °C/km.The linear decrease with temperature only occurs in the lower part of the atmosphere called theTROPOSPHERE. This is the part of the atmosphere in which weather variation occurs and our planesfly. Above the troposphere is the stratosphere, with an imaginary boundary separating the two layers. Inthe stratosphere, temperature tends to the relatively constant.Absorption of sunlight at the Earth's surface warms the troposphere from below, so vertical convectioncurrents are continually mixing in the air. As a parcel of air rises, its pressure drops and it expands. Theparcel does work on its surrounding, so that its internal energy and, therefore, its temperature drops.Assume that the vertical mixing is so rapid as to be adiabatic and quantity TP(1-g)/g has a uniform valuethrough the layers of troposphere.ok;qe.Myh; LAPSE njxSl ds de vk;ru ds fy;s] xSlks ds xR;kRed fl¼kUr ds vuqlkj] xSl ds lHkh Hkkxksa ds rki leku gksrs gSA ijUrq xSlds vf/kd vk;ru ds fy;s tSls fd ok;qe.My ds fy;s xSl ds lHkh Hkkxksa esa leku rki dk vuqeku ykxw ugh gksrk gSAok;qe.My ds vyx&vyx Hkkxksa dk rki vyx&vyx gksxkA i`Foh dh lrg ds vykok] ok;qe.My dh vyx&vyxÅ¡pkbZ ij rki dk eku vyx&vyx gksrk gSA Å¡pkbZ ds lkFk rki esa deh ok;qe.Myh; lapse nj dgykrh gSA ;g nji`Foh dh lrg ij vyx&vyx LFkkuksa ij leku gksrh gSA vyx&vyx LFkkuksa ij lwpuk dks ,df=r dj ;g Kkr fd;ktkrk gS fd vkSlr fo'oO;kid ysIl (lapse) nj – 6.7 °C/km gSAdsoy ok;qe.My ds fupys Hkkx esa rki esa js[kh; deh gksrh gSA ;g Hkkx {kksHke.My dgykrk gSA ;g ok;qe.My dk ogHkkx gS ftlesa ekSle lEcU/kh ifjorZu gksrs gS rFkk ok;q;ku mM+rs gSA {kksHke.My ds Åij lerki e.My gksrk gSA ,ddkYifud lhek nks lrgksa dk foHkkftr djrh gSA lerki e.My esa rki yxHkx fu;r jgrk gSAi`Foh lrg ij lw;Z izdk'k ds vo'kks"k.k ls {kksHke.My dk uhps dk Hkkx xeZ gksrk gSA vr% Å/okZ/kj laogu /kkjk,sa yxkrkjgok esa fefJr gksrh jgrh gSA tSls&tSls gok ds iklZy (parcel) Åij mBrs gS] bldk nkc ?kVrk gS rFkk ;g QSyrh gSA iklZy(parcel) blds pkjksa vksj ds okrkoj.k ij dk;Z djrk gS rkfd bldh vkUrfjd ÅtkZ rFkk bldk rki ?kVrk gSA ekuk fdÅ/okZ/kj feJ.k bruk rhoz gksrk gS tSls fd :¼ks"e_ rFkk jkf'k TP(1-g)/g dk eku {kksHke.My dh ijrksa esa fu;r jgrk gSA



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15. Choose INCORRECT option :

(A) In above paragraph heat conduction is not dominating mechanism of heat transfer.

(B) At equator temperature lapse rate is maximum as compared to any other latitude.

(C) Bulk modulus for the process involved in the above paragraph is P.

(D) According to above paragraph weather variation is due to temperature variation in vertical direction.

vlR; dFku dk p;u dhft;sA

(A) bl x|ka'k esa Å"ek LFkkukUrj.k ds fy, Å"ek pkyu rjhdk izHkkoh ugha gSA

(B) Hkwe/; js[kk ij rki lapse nj vU; v{kka'k dh rqyuk esa vf/kdre gksrh gSA

(C) bl x|ka'k esa of.kZr izØe ds fy;s vk;ru xq.kkad dk eku P gSA

(D) bl x|ka'k ds vuqlkj ekSle ifjorZu Å/okZ/kj fn'kk esa rki ifjorZu ds dkj.k gksrk gSA

16. If behavior of the mixing of parcels of air is approximately assumed to be adiabatic then lapse rate can be

expressed as :

;fn gok ds iklZy dk feJ.k yxHkx :¼ks"e eku fy;k tk, rks lapse nj O;Dr dh tk ldrh gSA

(A) ( )P dP

1T dy

- g - (B) T 1 dPP dy

æ ö- g- ç ÷gè ø

(C) P 1 dPT dy

æ ög -- ç ÷gè ø

(D) ( )T dP

1P dy

- g -

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 4 questions. The answer to each question is a single digit Integer, ranging from0 to 9 (both inclusive)bl [k.M esa 4 iz'u gSaA izR;sd iz'u dk mÙkj 0 ls 9 rd (nksuksa 'kkfey) ds chp dk ,dy vadh; iw.kk ±d gSA

1. In a physics lab, a small cube slides down a frictionless incline as shown in figure, it strikes elasticallyand horizontally with another cube that is only one-half of its mass. If the incline is 20 cm high and thetable is 90 cm off the floor, big and small cubes strikes the ground at a distance x & y meter respectively

from the table . Then yx

equal to.

HkkSfrdh dh iz;ksx'kkyk esa ,d NksVk ?ku fp=kuqlkj ,d ?k"kZ.kjfgr urry ij uhps dh vksj fQlyrk gSA ;g blds vk/ks

æO;eku okys ,d vU; ?ku ls izR;kLFk rFkk {kSfrt :i ls Vdjkrk gSA ;fn urry 20 cm Å¡pk gks rFkk Vscy Q'kZ ls

90 cm Åij gks rks cM+k o NksVk nksuksa ?ku Vscy ls Øe'k% x rFkk y ehVj nwjh ij /kjkry ls Vdjkrs gSaA yx

dk eku gksxkA

90cm////////////////////////////////////////

20cm

m

m



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2. The amount of power delivered by the ac source in the capacitor & resistance R1

of the circuit given

below is 3X 10

8´ watt, then x is :

uhps fn,s x, ifjiFk esa ac L=ksr }kjk la/kkfj= rFkk izfrjks/k R1

esa nh xbZ 'kfä dh ek=k ;fn 3X

108

´ okWV gks rks x gksxk :

~

X =12 c W R =51 W

R =62 WX =8L W

E = 130 Vrms

3. Water at temperature 40°C flows from a tap T into a heated container C. The container has a heatingelement (a resistor R) which is generating heat at the rate of P, that may be varied. The rate of water in

flow from tap is m = 1000 / min7

l . The heat generated is sufficient so that the water in the container is

boiling and getting converted into steam at a steady rate. What is the minimum power P (in MW) thatmust be generated as heat in the steady state in resistor R so that the amount of liquid water in thecontainer neither increases nor decreases with time ? (Neglect other losses of heat, such as conductionfrom the container to the air and heat capacity of container)

For water, specific heat c = 4.2 kJ kg–1 K–1, latent heat of vaporization Lvap = 2.268 MJ kg–1, density

r = 1000 kg m–3.iznf'kZr fp= esa 40°C rkieku okyk ty ,d uy T ls fdlh xeZ fd, gq, ik= C esa izokfgr gksrk gSA bl ik= esa ,dÅ"eu ;qfä (,d izfrjks/kd R) yxh gqbZ gS tks P nj ij Å"ek mRiUu dj jgh gS rFkk bls ifjofrZr Hkh fd;k tk ldrk gSA

uy ls ty cgus dh nj m = 1000 / min

7l gSA ;gk¡ mRiUu Å"ek bruh i;kZIr gS fd ik= esa Hkjk ty mcy dj LFkk;h

nj ij Hkki esa ifjofrZr gks tkrk gSA izfrjks/k R esa LFkk;h voLFkk esa Å"ek ds :i esa fdruh U;wure 'kfä P (MW esa)mRiUu gksuk pkfg, rkfd ik= esa æo ty dh ek=k le; ds lkFk uk rks c

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4. At a smooth horizontal table are two identical cubes of mass m, connected by a spring of rigidity k. Thelength of spring in the unstretched state is l

0. The right cube is linked to the load mass m at the end. At

some time the system is released and the system moves without initial velocity. Find the maximumdistance (in cm) between blocks during the motion of the system.[l

0 = 1 cm, m = 3 kg, k = 1000 N/m]

,d fpduh {kSfrt est ij æO;eku m okys nks ,d tSls ?kuksa dks k fu;rkad okyh ,d fLizax dh lgk;rk ls tksM+k tkrk gSA

vrfur voLFkk esa fLizax dh yEckbZ l0 gSA nka;s okys ?ku dks fp=kuqlkj ,d yksM æO;eku m ls blds fljs ls tksM + fn;k

tkrk gSA fdlh le; fudk; dks fojkekoLFkk ls NksM+s tkus ij ;g fudk; fcuk fdlh izkjfEHkd osx ds xfr djrk gSA

fudk; dh xfr ds nkSjku CykWdkas ds e/; vf/kdre nwjh (cm esa) Kkr dhft;sA [l0=1 cm, m = 3 kg, k = 1000 N/m]

m

mm


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PART-2 : CHEMISTRY

Hkkx-2 : jlk;u foKkuSECTION–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 gSaA izR;sd iz'u esa pkj fodYi (A), (B), (C) vkSj (D) gSa ftuesa ls dsoy ,d

lgh gSA

1. Which among the following aqueous solution have highest boiling point -(A) 1 molal KCl solution (B) 2 molal glucose

(C) 1 molal Co(NO3)2 solution (D) 1 molal NaCl solution

fuEu tyh; foy;uksa esa ls fdldk DoFkukad lokZf/kd gksxk -

(A) 1 eksyy KCl foy;u (B) 2 eksyy Xywdkst

(C) 1 eksyy Co(NO3)2 foy;u (D) 1 eksyy NaCl foy;u

2. Which of the following is a correct statement -(A) High temperature is favourable for physical adsorption

(B) Flocculation value is measure of protective power

(C) Fe(OH)3 aqueous solution can be used for cogulation of As2S3 aqueous solution

(D) Aqueous gold solution is an example of positively charged sol

fuEu esa ls dkSulk dFku lgh gS&

(A) HkkSfrd vf/k'kks"k.k ds fy;s mPp rki mi;qDr gksrk gSA

(B) m.kZu eku] j{k.k {kerk dk ekiu gksrk gSA

(C) As2S3 tyh; foy;u ds LdUnu ds fy, Fe(OH)3 tyh; foy;u dk mi;ksx fd;k tk ldrk gSA

(D) tyh; xksYM foy;u] /kukRed vkosf'kr lkWy dk ,d mnkgj.k gSA


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YALL INDIA OPEN TEST/JEE (Advanced)/08-02-2015/PAPER-1

Kota/00CT214003

3. A process must be spontaneous (feasible) if(A) Entropy of system increases (B) Energy of system decreases

(C) Gibbs free energy decreases (D) Entropy of universe increases

,d izØe Lor% (laHko) gksuk pkfg, ;fn

(A) ra= dh ,.VªkWih c


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5. Which of the following ligand is not p-acceptor ligand-

fuEu esa ls dkSulk fyxs.M] p-xzkgh fyxs.M ugha gS -

(A) p-C3H

5– (B) p-C

5H

5– (C) CH—CH2 2

CH..

(D) PF3

6. Which of the following metal oxide is more stable if the free energies of formation of Cr2O3 and Al2O3per mole of oxygen consumed are –540 kJ and –827 kJ respectively :

(A) Cr2O3 (B) Al2O3

(C) Both have equal stability (D) None of these

fuEu esa ls dkSulk /kkrq vkWDlkbM vf/kd LFkk;h gksxk ;fn Cr2O3 rFkk Al2O3 ds fuekZ.k dh] izfr eksy vkWDlhtu [kpZ

ij eqä ÅtkZ,sa Øe'k% –540 kJ rFkk –827 kJ gSa

(A) Cr2O3 (B) Al2O3

(C) nksuksa dk LFkkf;Ro leku gS (D) buesa ls dksbZ ugha

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7. OO2N

Cl

H N–OH2 P + QNa

Cyclicproduct

H /+

RH O3

+

S + T

CO2NaHCO3

(pØh;mRikn)

Product S is :

mRikn S gS&

(A)

O N2 NH2

Cl(B) CH3 – NH2 (C)

H N2 NO2

Cl(D) CH3 – CH2 – NH2

8. Major product obtained during reactionfuEu vfHkfØ;k ds nkSjku izkIr eq[; mRikn gS&

O

O

O(i) DI-BALH(–78ºC) (excess)(ii) H O2

(A)

OH

O

O (B)

O

OH

O (C)

OH

OH

O(D)

OH

OH

OH



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(ii) Paragraph Type (ii) vuqPNsn izdkj



Order of a reaction depends on many factors like temperature. Concentration , reaction condition etc eg.

1NS

4 9 2 4 9(2º)

C H Cl H O C H OH+ ¾¾¾® + HCl

2NS

4 9 4 9(2º)

C H Cl NaOH C H OH+ ¾¾¾® + NaCl

,d vfHkfØ;k dh dksfV dbZ dkjdks tSls rki ] lkUnzrk] vfHkfØ;k fLFkfr vkfn ij fuHkZj djrh gS :

1NS

4 9 2 4 9(2º)

C H Cl H O C H OH+ ¾¾¾® + HCl

2NS

4 9 4 9(2º)

C H Cl NaOH C H OH+ ¾¾¾® + NaCl

9. Find time taken to complete 50% reaction according to SN1 mechanism if time taken to complete 75%

reaction is 40 min -

;fn vfHkfØ;k dks 75% iw.kZ gksus esa 40 feuV yxrs gS] rks SN1 fØ;kfo/kh ds vuqlkj vfHkfØ;k dks 50% iw.kZ gksus

esa yxk le; gS -(A) 40 min (B) 20 min (C) 10 min (D) 80 min

10. Starting with [C4H9Cl] = 1M and [NaOH] = 2M what is the rate in (M sec–1) according to SN

2 mechanismwhen [C4H9Cl] changes to 0.25 M ,

Given : k = 4 × 10–4 M–1sec–1

[C4H9Cl] = 1M rFkk [NaOH] = 2M ds lkFk izkjEHk gksus ij SN2 fØ;kfo/kh ds vuqlkj nj (M sec–1) esa D;k gksxh

tc [C4H9Cl] dh la[;k 0.25 M gks tkrh gSA

fn;k gS : k = 4 × 10–4 M–1sec–1

(A) 1.25 × 10–4 (B) 10–4 (C) 5 × 10–4 (D) 2.5 × 10–4


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When testing for a specific ion with a reagent, interferences may occur owing to the presence of otherions in the solution, which also react with reagent. In some cases it is possible to prevent this interferencewith the addition of reagents, so called masking agents, which form stable complexes with interferingions. There is no need for the physical seperation of the ion involved and therefore the time for the testcan be cut considerable. Masking may also be achieved by dissolving precipitates or by the selectivedissolution of a precipitate from a mixture :tc fdlh vfHkdeZd }kjk ,d fof'k"V vk;u dk ijh{k.k fd;k tkrk gS rks foy;u esa mifLFkr vU; vk;u tks HkhvfHkdeZd ds lkFk fØ;k djrs gSa ] dh mifLFkfr ds dkj.k ck/kk (interference) mRiUu gks ldrh gSA dqN ifjfLFkr;ksaesa , sls vfHkdeZd dks feykdj bl ck/kk dks jksdk tkuk lEHko gS ftUgsa ekWfLdax vfHkdeZd dgk tkrk gS rFkk tksvojks/kdkjh vk;uksa ds lkFk LFkk;h ladqy cuk ysrs gSA bl izdkj] lfEefyr vk;u ds HkkSfrd i`FkDdj.k dh vko';drkugha iM+rh gS rFkk ijh{k.k esa yxus okys le; dks Hkh dkQh de fd;k tk ldrk gSA vo{ksiksa ds feJ.k ls ] vo{ksidks p;ukRed :i ls foys; djds Hkh i`Fkd fd;k tk ldrk gS :

11. For the test of Cadmium ion with H2S, "_______" ion can be masked with excess KCN solution :H2S ds lkFk dsMfe;e vk;u ds ijh{k.k esa "_______" vk;u dks KCN foy;u dks vkf/kD; esa feykdj ekWLdfd;k tk ldrk gS :(A) Al+3 (B) Zn+2 (C) Cu+2 (D) Ba+2

12. Select the correct statement :(A) When testing for Pb2+ in the presence of Ag+, we may produce a mixture of AgCl and PbCl2

precipitates. If ammonia is added silver chloride dissolves in the form of diammine Silver ion whileremaining ppt. does not change its colour

(B) When H2S gas is passed into sodium nitroprusside then purple colour is obtained(C) When CrO2Cl2 is passed into NaOH solution then colour of solution becomes green(D) Brown ring complex has d2sp3 hybridisation with 3.87 B.M. magnetic moment.lgh dFku pqfu, :(A) Ag+ dh mifLFkfr esa Pb+2 dk ijh{k.k djus ij AgCl rFkk PbCl2 ds vo{ksiksa dk feJ.k cu ldrk gSA ;fn

blesa veksfu;k feyk;h tkrh gS rks flYoj DyksjkbM] Mkb,sEehu flYoj vk;u ds :i esa foys; gks tkrk gS tcfd'ks"k vo{ksi dk jax ugha cnyrk gS

(B) tc lksfM;e ukbVªksiz qlkbM esa H2S xSl dks izokfgr fd;k tkrk gS rks cSaxuh jax izkIr gksrk gS

(C) tc CrO2Cl2 dks NaOH foy;u esa ls izokfgr fd;k tkrk gS rks foy;u dk jax gjk gks tkrk gS

(D) Hkwjh oy; ladqy esa d2sp3 ladj.k mifLFkr gksrk gS o pqEcdh; vk?kw.kZ 3.87 B.M. gksrk gS



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iz'u 13 ,oa 14 ds fy;s vuqPNsn

'X'(3º Alcohol) /(3º ),YdksgkWy

SOCl2 P

CH –C–OK3

CH3

CH3 Q O3Zn / H O2R + S

Conc.KOH

Major /

HCOO K + T– +

CHI + U3Ca(OH)2

I2

dry distillation / 'kq"d vklou

O

eq[;

13. Unknown alcohol 'X' is :

vKkr ,sYdksgkWy 'X' gS&

(A)OH

(B)

OH

(C) OH

(D) None

14. R & S can not be distinguished by

(A) Iodoform test (B) Tollen's reagent (C) Fehling solution (D) 2,4 DNP test

R rFkk S fuEu esa ls fdlds }kjk foHksfnr ugha fd;s tk ldrs gSa&

(A) vk;ksMksQkeZ ijh{k.k (B) VkWysUl vfHkdeZd (C) Qsgfyax foy;u (D) 2,4 DNP ijh{k.k


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N-bromo succinamide is used for bromination of allylic & benzylic Position. Reaction is proceeded byfree radical mechanism.In given reactionEthylbenzene + NBS ¾®,sfyfyd rFkk csfUtfyd fLFkfr ds czksehfudj.k ds fy;s N-czkseks lfDlusekbM dk iz;ksx fd;k tkrk gSA vfHkfØ;k eqDrewyd fØ;kfof/k }kjk gksrh gSA

nh xbZ vfHkfØ;k esa

,sfFkycsfUtu + NBS ¾®15. Incorrect statement for above reaction

(A) One of the product can show positive iodoform test(B) One of the reactant can participate in hofmann bromide degradation(C) One of the product can participate in hofmann bromide degradation(D) One of the reactant can show stereoisomerism

mijksDr vfHkfØ;k ds fy;s xyr dFku gS&

(A) mRikn esa ls ,d /kukRed vk;ksMksQkeZ ijh{k.k iznf'kZr dj ldrk gS

(B) fØ;kdkjd esa ls ,d gkWQeku czksekbM ds fuEuhdj.k esa Hkkx ys ldrk gS

(C) mRikn esa ls ,d gkWQeku czksekbM ds fuEuhdj.k esa Hkkx ys ldrk gS

(D) fØ;kdkjd esa ls ,d f=foe~ leko;ork iznf'kZr dj ldrk gS16. Reaction is example of

(A) Free radical addition reaction (B) Free radical substitution reaction(C) Electrophilic addition reaction (D) None of these

vfHkfØ;k fuEu esa ls fdldk mnkgj.k gS&

(A) eqDr ewyd ;ksxkRed vfHkfØ;k (B) eqDr ewyd izfrLFkkiu vfHkfØ;k

(C) bysDVªkWuLusgh ;ksxkRed vfHkfØ;k (D) buesa ls dksbZ ugha





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SECTION-IV : (Integer Value Correct Type)

[k.M-IV : (iw.kk±d eku lgh izdkj)

This section contains 4 questions. The answer to each question is a single digit Integer, ranging from0 to 9 (both inclusive)

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

1. Calculate the magnitude of work done by 2mole ideal gas in kJ, when expansion is taking place from2 litre to 22 litre against constant external pressure of 1 atm.

[Given :1 litre-atm = 100 J]

nks eksy vkn'kZ xSl dk fu;r ckg~; nkc 1atm ds fo:¼ 2 litre ls 22 litre rd izlkj gksrk gS] rks xSl }kjk fd;s

x;s dk;Z ds ifjek.k dh x.kuk kJ esa dhft,A

[fn;k x;k gS :1 litre-atm = 100 J]


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2. Quinhydrone is a sparingly soluble one-to-one addition compound formed from hydroquinone andquinone. When solid quinhydrone is dissolved in an aqueous medium, equal concentrations ofhydroquinone and quinone result. If a platinum wire is dipped into the solution, the potential of theelectrode is governed by the reversible reaction:

O

O

(Q)

+ 2H+ + 2e l

(H Q)2

OH

OH

Where Q denotes quinone and H2Q denotes hydroquinone. The E° value of this half cell reaction is+0.46 V with respect to the saturated calomel reference electrode (SCE). Because the potential of thequinone-hydroquinone half reaction depends on the concentration of hydrogen ion, it is possible to usethis system to measure pH. A sample solution of unknown pH was saturated with quinhydrone and aplatinum electrode was dipped into it. If the reduction potential of the such electrode was found to be+0.22 V with respect to SCE, what was the pH of the sample solution?

[Given : 2.303 RT = 0.06

F]

gkbMªksDohukWu rFkk DohukWu ls fufeZr Dohu gkbMªkWu ,d vYi foys; ;ksxkRed ;kSfxd gksrk gS] tc Bksl DohugkbMªkWu dks,d tyh; ek/;e esa ?kksyk tkrk gSA rc leku lkUnzrk ds gkbMªksDohukWu rFkk DohukWu izkIr gksrs gSa ;fn ,d IysVhue ds rkjdks foy;u esa Mqck;k tkrk gS rc bysDVªkWMksa dk foHko mRØe.kh; vfHkfØ;k }kjk fu;af=r fd;k tkrk gSA

O

O

(Q)

+ 2H+ + 2e l

(H Q)2

OH

OH

tgk¡ Q-DohukWu dks rFkk H2Q gkbMªksDohukWu dks crkrk gSA bl v/kZlsy vfHkfØ;k dk ekud dSyksesy funsZ'kh bysDVªkWM(SCE)ds lkis{k E° eku +0.46 V gSA D;ksafd DohukWu & gkbMªksDohukWu v/kZ lSYk vfHkfØ;k dk foHko gkbMªkstu vk;u dh lkUnzrkij fuHkZj djrk gS] pH dks ekius ds fy, bl ra= dk mi;ksx fd;k tkuk lEHko gSA DohugkbMªksu ls lar`Ir vKkr pH dk,d uewuk foy;u vkSj blesa Mqck;k x;k ,d IysVhue bysDVªkWM fy;k x;k gSA ;fn bl bysDVªkWM dk foHko SCE dslkis{k +0.22 V ik;k x;k gks] rks bl uewus foy;u dh pH D;k gksxh\

[fn;k gS %& 2.303 RT = 0.06

F]



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3. Find the number of ions which are soluble in excess of NaOH as well as excess of NH4OHAg+, Pb+2, Al+3, Sn+2, Zn+2, Cd+2.

fuEu esa ls ,sls vk;uksa dh la[;k crkbZ;sa tks NaOH rFkk NH4OH nksuksa ds vkf/kD; esa foys; gS

Ag+, Pb+2, Al+3, Sn+2, Zn+2, Cd+2.

4. Total number of dibromo product obtained in following reaction.

fuEu vfHkfØ;k esa izkIr MkbZ czkseks mRikn dh dqy la[;k gS&

Me

Br

CH

H

CH2

HBrCCl4


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

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 gSaA izR;sd iz'u esa pkj fodYi (A), (B), (C) vkSj (D) gSa ftuesa ls dsoy ,dlgh gSA

1.x

x

e sin x 4cos x 3dx

e 5sin x 3cos x 6+ + +

+ + +ò is equal to-x

x

e sin x 4cos x 3dx

e 5sin x 3cos x 6+ + +

+ + +ò dk eku gksxk -

(A) ( )( )x1 x n e 5sin x 3cos x 6 C2

+ + + + +l (B) ( )( )x1 x n e 5sin x 3cos x 6 C2

- + + + +l

(C) ( )x1x n e 5sin x 3cos x 6 C2

+ + + + +l (D) ( )x1x n e 5sin x 3cos x 6 C2

- + + + +l

(where C is constant of integration)

(tgk¡ C lekdyu vpj gS)

2. A function 'ƒ' is defined for all real numbers and satisfies ƒ(2 + x) = ƒ(2 – x) and ƒ(7 + x) = ƒ(7 – x) forall real x. If x = 0 is a root of ƒ(x) = 0, then the least number of roots of ƒ(x) = 0 for x Î [–1000,1000]is-

,d Qyu 'ƒ' lHkh okLrfod la[;kvksa ds fy, ifjHkkf"kr gS rFkk lHkh okLrfod x ds fy, ƒ(2 + x) = ƒ(2 – x) rFkk

ƒ(7 + x) = ƒ(7 – x) dks lUrq"V djrk gSA ;fn x = 0, ƒ(x) = 0 dk ,d ewy gks] rks x Î [–1000,1000] ds fy,

ƒ(x) = 0 ds ewyksa dh U;wure la[;k gksxh -

(A) 400 (B) 301 (C) 401 (D) 501



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3. The value of ( )4 2

n 4n 2

n 3n 10n 10

2 n 4

¥

=

+ + ++

å is-

( )4 2

n 4n 2

n 3n 10n 10

2 n 4

¥

=

+ + ++

å dk eku gksxk -

(A) 1 (B) 1110

(C) 32

(D) 2

4. The area of region in xy-plane consisting of all points (a,b) such that quadratic equation

ax2 + 2(a + b – 7)x + 2b = 0 has fewer than 2 real solutions is-

xy-lery esa lHkh fcUnqvksa (a,b) dks j[kus okys {ks= dk {ks=Qy rkfd f}?kkr lehdj.k ax2 + 2(a + b – 7)x + 2b = 0 ds

nks ls de okLrfod gy gks] gksxk -

(A) 4p (B) 9p (C) 25p (D) 49p

5. The value of 2

2

xdx

1 sin x 1 sin x

p

-p + + +ò is-

2

2

xdx

1 sin x 1 sin x

p

-p + + +ò dk eku gksxk -

(A) p2 (B) 2

3p

(C) 3

3p

(D) 0


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6. Let ( )

2tan sec

3 3A

2015cot cos 2014

2

p pé ùê úê ú=

pæ öê úpç ÷ê úè øë û

and P be a 2 × 2 matrix such that PPT = I (where I is identity

matrix of order 2). If Q = PAPT and R= [rij]2×2 = PTQ8P, then-

ekuk ( )

2tan sec

3 3A

2015cot cos 2014

2

p pé ùê úê ú=

pæ öê úpç ÷ê úè øë û

rFkk P, ,d 2 × 2 vkO;wg bl izdkj gS fd PPT = I gS] (tgk¡ I dksfV 2 dk

rRled vkO;wg gS)A ;fn Q = PAPT rFkk R= [rij]2×2 = PTQ8P gks] rc -

(A) r11 = 81 (B) 11r 27 3= (C) 11r 4 3= (D) 11r 3= -

7. Given two independent events, if the probability that exactly one of them occurs is2649

and the probability

that none of them occurs is 1549

, then the probability of least probable of the two events is-

nks LorU= ?kVuk;sa nh xbZ gS] ;fn muesa ls Bhd ,d ds ?kfVr gksus dh izkf;drk 2649

rFkk muesa ls fdlh Hkh ?kVuk ds ?kfVr

ugha gksus dh izkf;drk 1549

gks] rks nksuksa ?kVukvksa esa ls U;wure izkf;drk okyh ?kVuk dh izkf;drk gksxh -

(A) 27

(B) 37

(C) 47

(D) 57

8. If the function ƒ(x) satisfies ƒ'(x) + ƒ(x) cotx – 2 cosx = 0 and ƒ 12pæ ö =ç ÷

è ø, then ƒ

3pæ ö

ç ÷è ø

is equal to-

;fn Qyu ƒ(x), ƒ'(x) + ƒ(x) cotx – 2 cosx = 0 dks lUrq"V djrk gS rFkk ƒ 12pæ ö =ç ÷

è ø gks] rks ƒ

3pæ ö

ç ÷è ø

dk eku gksxk -

(A) 0 (B) 12

(C) 3

2(D) 1


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(ii) Paragraph Type (ii) vuqPNsn izdkj



Let ( ) ( )x

0

g x ƒ t dt= ò where ƒ(x) is a continuous function. Given ƒ(x) is an odd function "xÎR and is

periodic with period 2. Also , ƒ(x) is integrable on every interval.

ekuk ( ) ( )x

0

g x ƒ t dt= ò tgk¡ ƒ(x) ,d larr~ Qyu gSA fn;k x;k gS fd ƒ(x) lHkh xÎR ds fy, fo"ke Qyu gS rFkk

vkorZdky 2 ds lkFk vkorhZ gSA lkFk gh ƒ(x), izR;sd vUrjky ij lekdyuh; gSA

9. Which of the following is correct ?

(A) g(x) is an odd and periodic function (B) g(x) is an even and periodic function

(C) g(x) is an even and aperiodic function (D) g(x) is neither odd nor even function

fuEu esa ls dkSulk lgh gksxk -

(A) g(x) ,d fo"ke rFkk vkorhZ Qyu gSA (B) g(x) ,d le rFkk vkorhZ Qyu gSA

(C) g(x) ,d le rFkk vukorhZ Qyu gSA (D) g(x) uk rks fo"ke uk gh le Qyu gSA

10. g(4n) is equal to-

(A) 0 " n Î I (B) 0 only for n = 0

(C) odd integer " n Î I (D) cannot be determined for all values of n

g(4n) dk eku gksxk -

(A) 0 " n Î I (B) dsoy n = 0 ds fy, 'kwU;

(C) lHkh n Î I ds fy, fo"ke iw.kk±d (D) n ds lHkh ekuksa ds fy;s Kkr ugha fd;k tk ldrk


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Let 10 10log x log yx, x , y and ( ) ( )10log xyxy are four consecutive terms of a geometric progression (x,y > 0).

ekuk 10 10log x log yx, x , y rFkk ( ) ( )10log xyxy fdlh xq.kksÙkj Js.kh ds pkj Øekxr in gS (x,y > 0)A

11. Number of ordered pair (x,y) is-

(A) 2 (B) 3 (C) 4 (D) more than four

Øfer ;qXe (x,y) dh la[;k gksxh -

(A) 2 (B) 3 (C) 4 (D) pkj ls vf/kd

12. If (x1,y1) and (x2,y2) are two ordered pair which satisfies the condition in comprehension

(where x1,x2,y1,y2 Ï N), then 1 2

1 2

x xy y

is-

;fn (x1,y1) rFkk (x2,y2) nks Øfer ;qXe gS] tks vuqPNsn esa fn, x, izfrcU/k dks lUrq"V djrs gSa, rc 1 2

1 2

x xy y dk eku gksxk

(tgk¡ x1,x2,y1,y2 Ï N gS) -

(A) 10 (B) 20 (C) 500 (D) 1000



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Consider the curves 318

C : y x27

= and C2 : y = (x + a)2.

ekuk oØ 318

C : y x27

= rFkk C2 : y = (x + a)2 gSA

13. The range of 'a' for which there exists two common tangents to the curves C1 and C2 other thanx-axis is -

a dk ifjlj ftlds fy, oØksa C1 rFkk C2 dh x-v{k ds vfrfjDr nks mHk;fu"B Li'kZjs[kk;sa fo|eku gks] gksxk -

(A) ( )0,¥ (B) { }1

, 02

æ ö- ¥ -ç ÷è ø

(C) { }3

, 02

æ ö- ¥ -ç ÷è ø

(D) [ )1,¥

14. If a = 4 ,then the area bounded by the two tangent lines mentioned in the previous question and the givenparabola C2 is-

;fn a = 4 gks] rks mijksDr iz'u esa nh xbZ nks Li'kZ js[kkvksa rFkk fn, x, ijoy; C2 }kjk ifjc¼ {ks=Qy gksxk -

(A) 100 (B) 144 (C) 180 (D) 200


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Let ƒ be a differentiable function satisfying

( ) ( )( ) ( )( )ƒ x x 1

1 1

0 0 0

ƒ t dt cos t ƒ t dt ƒ t dt- -- - =ò ò ò " x Î 0, 2pæ ö

ç ÷è ø

and ƒ(0) = 1.

ekuk ƒ ,d vodyuh; Qyu gS] tks

( ) ( )( ) ( )( )ƒ x x 1

1 1

0 0 0

ƒ t dt cos t ƒ t dt ƒ t dt- -- - =ò ò ò " x Î 0, 2pæ ö

ç ÷è ø

dks lUrq"V djrk gS rFkk ƒ(0) = 1 gSA

15. The number of solution(s) of the equation 2ƒ(2x) – (sinx)ƒ(x) = 0 in x Î 0,4pæ ö

ç ÷è ø

is -

x Î 0,4pæ ö

ç ÷è ø

esa] lehdj.k 2ƒ(2x) – (sinx)ƒ(x) = 0 ds gyksa dh la[;k gksxh -

(A) 0 (B) 1 (C) 2 (D) 3

16. The value of ( )/ 2

0

ƒ x dxp

ò lies in the interval -

( )/ 2

0

ƒ x dxp

ò dk eku fuEu vUrjky esa fLFkr gksxk -

(A) 2

,1æ öç ÷pè ø(B) 1, 2

pæ öç ÷è ø

(C) ,22pæ ö

ç ÷è ø

(D) 2

0,æ öç ÷pè ø





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

This section contains 4 questions. The answer to each question is a single digit Integer, ranging from0 to 9 (both inclusive)bl [k.M esa 4 iz'u gSaA izR;sd iz'u dk mÙkj 0 ls 9 rd (nksuksa 'kkfey) ds chp dk ,dy vadh ; iw.kk ±d gSA

1. Let ABCD be tetrahedron with AB = 41, AC = 7, AD = 18, BC = 36, BD = 27 and D27

1813

C

7A

41

B

CD = 13 as shown in figure. Let d be the distance between the midpoints of edges

AB and CD, then d2

é ùê úë û

is (where [.] denotes greatest integer function)

ekuk fp=kuqlkj ABCD prq"Qyd gS] ftlesa AB = 41, AC = 7, AD = 18, BC = 36, BD = 27 D27

1813

C

7A

41

BrFkk CD = 13 gSA ekuk d, dksjksa AB rFkk CD ds e/; fcUnqvksa ds e/; nwjh gks] rks

d2

é ùê úë û

dk

eku gksxk (tgk¡ [.] egÙke iw.kk±d Qyu gks iznf'kZr djrk gS)

2. Let (x,y) be a pair of real number satisfying 2 2

y56x 33y

x y+ = -

+ and

2 2

x33x 56y

x y- =

+. If

p| x | | y |

q+ = (where p and q are relatively prime), then (6p – q) is

ekuk (x,y) okLrfod la[;kvksa dk ,d ; qXe gS] tks 2 2

y56x 33y

x y+ = -

+ rFkk

2 2

x33x 56y

x y- =

+ dks lUrq"V

djrk gSA ;fn p

| x | | y |q

+ = (tgk¡ p rFkk q ijLij vHkkT; la[;k;sa gSa), rc (6p – q) dk eku gksxk


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3. If , 0,2pæ öa bÎç ÷

è ø satisfy the following simultaneous equation

2 sin2b = 3sin2a and tanb = 3 tana,then the value of (8cos2a) is

;fn , 0,2pæ öa bÎç ÷

è ø fuEu ; qxir lehdj.k dks lUrq"V djrs gS

2 sin2b = 3sin2a rFkk tanb = 3 tana,rc (8cos2a) dk eku gksxk

4. Consider ( ) ( )( )( )( )

( )( )( )

( )( )

2 2 2x a x b x c

ƒ xa b a c b a b c c a c b

+ + += + +

- - - - - - (where a,b,c are distinct real number).

If 'p' denotes the number of natural numbers in the range of ƒ(x), then unit digit of (p + 8)2015 is

ekuk ( ) ( )( )( )( )

( )( )( )

( )( )

2 2 2x a x b x c

ƒ xa b a c b a b c c a c b

+ + += + +

- - - - - - (tgk¡ a,b,c fofHkUu okLrfod la[;k;s a gSa)A

;fn 'p', ƒ(x) ds ifjlj esa izkd`r la[;kvksa dh la[;k dks n'kkZrk gks] rks (p + 8)2015 dk bdkbZ vad gksxk



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D. vadu ;kstuk / Marking scheme :14. [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

15. [kaM-I (ii) ds gj iz'u esa dsoy lgh mÙkj okys cqycqys (BUBBLE) dks dkyk djus ij 3 vad vkSj dksbZ Hkh cqycqyk dkyk ugha djus ij'kwU; (0) vad iznku fd;k tk;sxk bl [akM ds iz'uksa esa xyr mÙkj nsus ij dksbZ ½.kkRed vad ugha fn;s tk;saxsaAFor each question in Section-I (ii), you will be awarded 3 marks if you darken the bubble corresponding to the correct answerand zero mark if no bubbles are darkened No negative marks will be awarded for incorrect answers in this section.

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

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

Name of the Candidate / ijh{kkFkhZ dk uke

I have read all the instructions and shall abide by them.eSusa lHkh vuqns'kksa dks i

ALLEN CLAS AMME · 2015. 12. 21. · ALL INDIA OPEN TEST/JEE (Adaced)/08-02-2015/PAPER-1 PHYSICS...

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