Allen Paper 1

i;k bu fun sZ'kks a dks /;ku 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 - 06Part-1 Physics Only One Option Correct Type

I(ii) vuqPNsn izdkj 07 - 09Paragraph Type

II eSfVDl&esy izdkj 10 - 13Matrix Match Type

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

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




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




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

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

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

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

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

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Boltzmann constant k = 1.38 1023 JK1

Coulomb's law constant pe9

0

1 = 9104

Universal gravitational constant G = 6.67259 1011 Nm2 kg2 Speed of light in vacuum c = 3 108 ms1 StefanBoltzmann constant s = 5.67 108 Wm2K4 Wien's displacement law constant b = 2.89 103 mK Permeability of vacuum 0 = 4p 107 NA2

Permittivity of vacuum 0 = 20

1cm

Planck constant h = 6.63 1034 Js

Kota/00CT214005

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

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PART-1 : PHYSICS

Hkkx-1 : HkkSfrd foKkuSECTIONI : (i) Only One option correct Type

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

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

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

2(B) 3 3

2(C) (D) Data insufficient

60

y

x

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

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

(A) 32

(B) 3 32

(C) (D) vkadM+s vi;kZIr gSA

BEWARE OF NEGATIVE MARKINGHAVE CONTROL HAVE PATIENCE HAVE CONFIDENCE 100% SUCCESS

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

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2. Consider a network shown in figure. Initially the switch S is open. The amount of charge flowthrough earth wire after closing the switch is -fp= esa iznf'kZr ifjiFk ij fopkj dhft;sA izkjEHk esa fLop S pkyw gSA fLop S dks can djus ds i'pkr~ HkwlaifdZrrkj ls izokfgr vkos'k dh ek=k gksxh&

6C 3C

3C 6C

V

S

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

CV (D) 3CV2

3. A parallel beam of light is incident on a thin prism of prism angle of 4p degrees. The refractive

index of the prism is 1.5. The focal length of the lens is 60 cm. The coordinates of converging pointof the beam is-

,d lekUrj izdk'k iaqt 4p fMxzh fizTe dks.k okys ,d irys fizTe ij vkifrr gSA fizTe dk viorZukad 1.5 gSA

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

q80cm

x0

y

(A) 260cm, cm3 (B)

160cm, cm

3 (C)

160cm, cm

3 - (D)

260cm, cm

3 -

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4. Two cylindrical straight and very long non magnetic conductors A and B, insulated from each other,carry a current I in the positive and the negative zdirection respectively. The direction of magnetic fieldat origin is :

nks csyukdkj lh/ks rFkk cgqr yEcs vpqEcdh; pkyd A rFkk B ,d nwljs ls foyfxr gSa rFkk muesa e'k% /kukRed rFkk.kkRed z- fn'kk esa I /kkjk izokfgr gksrh gSA ewy fcUnq ij pqEcdh; {ks= dh fn'kk gksxh%&

BAx

y

(A) i- (B) i+ (C) j (D) j5. Two boys of masses 50 kg and 60kg are moving along a vertical massless rope, the former climbing

up with an acceleration of 2ms2 while later coming down with constant velocity of 2ms1. Thetension in rope at fixed support will be(g= 10ms2).50kg rFkk 60kg nzO;eku ds nks yM+ds /okZ/kj nzO;ekughu jLlh ij xfr dj jgs gSA igyk yM+dk 2ms2 dsRoj.k ls ij dh vksj p

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6. A charge q moves with velocity 1v ai=r

$ (ms1) experiences a force 1F aqj(N)= -

r$

at a point in magneticfield region. If charge is moving with velocity 1

2v (ai bj)(ms )-= +r $ $ at the same point, it experience force2F q(ai bj)(N)= +

r$ $

. The force experienced by it at the same point if it is moving withvelocity 1 2(v v )

r r

-

1v ai=r

$ (ms1) ds osx ls xfr'khy ,d vkos'k q pqEcdh; {ks= esa fdlh fcUnq ij 1F aqj(N)= -

r$ cy vuqHko

djrk gSA ;fn vkos'k leku fcUnq ij 12v (ai bj)(ms )-= +r $ $ osx ls xfr djs rks ;g 2F q(ai bj)(N)= +r

$ $ cy

vuqHko djrk gSA ;fn ;g 1 2(v v )r r

osx ls xfr djs rks leku fcUnq ij blds }kjk vuqHko fd;k x;k cy gksxk&

(A) q(ai b j)+$ $ (B) bqj- $ (C) q(bi a j)+$ $ (D) None of these7. Two tuning forks A and B give 6 beats/second. A resound with a closed column of air 15 cm long

and B with an open column 30.5 cm long. The frequencies of A and B are respectively-nks Lofj= A rFkk B , 6 foLiUn/lSd.M nsrs gSaA A, 15 cm yEcs can ok;q LrEHk ds lkFk rFkk B, 30.5 cm yEcs [kqysLrEHk ds lkFk vuqukfnr gksrk gSA A rFkk B dh vko`fk;k e'k% gS%&(A) 550 Hz, 544 Hz (B) 500 Hz, 494 Hz (C) 400 Hz, 394 Hz (D) 366 Hz, 360 Hz

8. A screw gauge has a screw having 2 threads in 1 mm. The circular scale has 50 divisions. Find thediameter of wire, if the main scale shows 6th division and the vernier reads 46.,d Lwxst ds Lw esa 2 pwfM+;k (thread)1 mm nwjh ij gSA o`kh; iSekus esa 50 Hkkx gSaA ;fn eq[; iSekuk 6 Hkkx rFkkofuZ;j 46 Hkkx i

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(ii) Paragraph Type

(ii) vuqPNsn izdkjThis section contains 2 paragraphs each describing theory, experiment, data etc. Four questions relateto two paragraphs with two questions on each paragraph. Each question of a paragraph has only onecorrect answer among the four choices (A), (B), (C) and (D).bl [k.M esa flkUrksa] iz;ksxksa vkSj vkdM+ksa vkfn dks n'kkZus okys 2 vuqPNsn gSA nksuksa vuqPNsn ls lacaf/kr pkj iz'ugSa] ftuesa ls gj vuqPNsn ij nks iz'u gSaA vuqPNsn esa gj iz'u ds pkj fodYi (A), (B), (C) vkSj (D) gSa ftuesals dsoy ,d lgh gSA

Paragraph for Questions 9 and 10

iz'u 9 ,oa 10 ds fy;s vuqPNsnn identical rods each of mass m are welded at their ends to form a regular polygon and the corners arethen welded to a metal ring of radius R and mass M, such that the plane of polygon and plane of ring arein same plane and centres of polygon and ring coincide.,d le cgqHkqt cukus ds fy;s izR;sd m nzO;eku dh n le:i NM+ksa ds fljksa dks tksM+k tkrk gS rFkk R f=T;k rFkkM nzO;eku dh /kkfRod oy; dks blds fljksa ls bl izdkj tksM+k tkrk gS fd cgqHkqt rFkk oy; dk ry leku ry esa gks rFkkbuds dsUnz laikrh gksA

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

(A) 2

2 2 2sin

nnmR cos MR3 n

p p+ +

(B) 2

2 2 2tan

nnmR sin MR4 n

p p+ +

(C) 2

2 2 2sin

nnmR cos MR4 n

p p+ + (D)

2 2

2 2sin cos

n nnmR MR4 3

p p + +

10. If the rigid assembly of rods and hoop is allowed to roll down the incline of inclination q, the minimumvalue of the coefficient of static friction that will prevent slipping will be (Where I is moment of inertiaabout centre of mass) :-;fn NM+ksa rFkk fNnz ; qDr pdrh ds n`

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

w

B

E

Assuming mean free path l for the electron, they move in helical orbits as shown under the influenceof magnetic field. An electric field of constant magnitude, whose direction rotates with angular frequencyw, is also imposed on the solid in the plane shown.In the experiment w is fixed and the value of B is adjusted until the amount of input energy requiredto keep the amplitude of electric field constant reaches maximum.fdlh Bksl esa {ks= yxkus ij mlesa fo|eku vkos'k okgd bl izdkj O;ogkj djrs gSa tSls fd mudk izHkkoh O;eku m*gS tks fd fdlh Hkh izdkj ds {ks= dh vuqifLFkfr esa vkos'k okgdksa ds O;eku m ls fHkUu gksrk gSA fdlh Bksl esa pkyudh foLrr` O;k[;k djus ;k le>us ds fy, bl O;eku m* dh tkudkjh gksuk vR;Ur vko';d gSA izk;ksfxd rkSj ij;g lc ,d lkbDyksVkWu }kjk izkIr fd;k tk ldrk gS ftlesa pqEcdh; {ks= ds vykok vU; ckg~; fo|qr {ks= Hkh yxk;ktkrk gS tks fd vuqukn izkIr gksus rd ifjofrZr fd;k tkrk gSA bl voLFkk dks lkbDyksVkWu vuqukn dgk tkrk gSA

lkbDyksVkWu vuqukn vkosf'kr d.kksa dh ckg~; cyksa ds lkFk vU;ksU; f;k dks n'kkZrk gSA bu vkosf'kr d.kksa ij ,dpqEcdh; {ks= yx jgk gS] ftlds dkj.k ; s igys ls gh ,d o`kkdkj iFk ij xfr'khy gksrs gSaAvc ge fuEu fLFkfr ij fopkj djrs gSaA

w

B

E

ekuk pqEcdh; {ks= ds izHkko esa gsfydy d{kkvksa esa xfr'khy bysDVkWu ds fy, ek/; eq iFk l gS] fp= ns[ksaAfu;r ifjek.k okys ,d fo|qr {ks= ftldh fn'kk dks.kh; vko`fk w ds lkFk ifjofrZr gksrh gS] dks Hkh bl ry esa fp=kuqlkjbl Bksl ij vkjksfir fd;k tkrk gSA

bl iz;ksx essa w fu;r dj fn;k tkrk gS rFkk B dk eku rc rd O;ofLFkr fd;k tkrk gS tc rd fd fo|qr {ks= dsvk;ke dks fu;r cuk, j[kus ds fy, vko';d fuos'kh tkZ dh ek=k vius vf/kdre eku rd uk igqp tk,A

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

(A) 2qBw (B) qBw

(C) qB2w (D) Depends on electric field amplitude

bysDVkWu dk izHkkoh O;eku m* gS :-

(A) 2qBw (B) qBw

(C) qB2w (D) fo|qr {ks= vk;ke ij fuHkZj djrk gSA12. In actual situation l . So in order to make l practically very large so that resonance effect is prominent

we can conduct the experiment at :-(A) High temperature and low frequency (w) (B) High temperature and high frequency (w)(C) Low temperature and high frequency (w) (D) Low temperature and low frequency (w)okLrfod fLFkfr esa l gksrk gSA vr% iz;ksfxd rkSj ij l dk eku bruk vf/kd fd;k tk, fd vuquknh izHkko vis{kkd`rvf/kd izHkkoh gks rks blds fy, ;g iz;ksx djuk gksxk %&(A) mPp rkieku rFkk fuEu vko`fk (w) ij (B) mPp rkieku rFkk mPp vko`fk (w) ij(C) fuEu rkieku rFkk mPp vko`fk (w) ij (D) fuEu rkieku rFkk fuEu vko`fk (w) ij


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SECTIONII : Matrix-Match Type

[k.M - II : eSfVDl&esy izdkjThis Section contains 2 questions. Each question has four statements (A, B, C and D) given inColumn I and five statements (P, Q, R, S and T) in Column II. Any given statement in Column I can havecorrect matching with ONE or MORE statement(s) given in Column II. For example, if for a givenquestion, statement B matches with the statements given in Q and R, then for the particular question, againststatement B, darken the bubbles corresponding to Q and R in the ORS.bl [k.M esa 2 iz'u gSA izR;sd iz'u esa dkWye I esa 4 dFku (A, B, C vkSj D) vkSj dkWye II esa 5 dFku (P, Q, R, S vkSj T)gSaA dkWye I dk dksbZ Hkh dFku dkWye II ds ,d dFku ;k ,d ls vf/kd dFkuksa ls esy [kkrk gSA mnkgj.k ds fy,] fn, gq, iz'uesa ;fn dFku B dFkuks a Q vkSj R ls esy [kkrk gS] rks vksvkj,l (ORS) esa ml iz'u ds fy;s dFku B ds lkeus Q vkSj R lslEcfUkr cqycqyksa dks dkyk dhft;sA

1. Column-I gives some current distributions and a point P in the space around these current distributions.Column-II gives some expressions of magnetic field strength. Match column-I to corresponding fieldstrength at point P given in column-II

Column I Column II

(A) A conducting loop shaped as regular hexagon of side x, (P) 03 i32 xmp

carrying current i. P is the centroid of hexagon

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

mp

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

(C) Two coaxial cylinders of radii x and 2x, each carrying (R) 0i2xm

current i, but in opposite is hollow. P is a point at distance1.5x from the axis of the cylinders

(D) Magnetic field at the centre of an n-sided regular (S) 0i3 xmp

polygon, of circum circle of radius x, carrying currenti, n , P is centroid of the polygon.

(T) 03 i32 x

mp


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

LrEHk I LrEHk II

(A) Hkqtk x okys le"kV~Hkqt ds vkdkj okys pkyd ywi esa (P) 03 i32 xmp

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

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

mp

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

(C) f=T;k x rFkk 2x okys lek{kh; [kks[kys csyu ftlesa ijLij foijhr (R) 0i2xm

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

(D) f=T;k x okys ifjo`k okys n-Hkqtkvksa okys le cgqHkqt ds (S) 0i3 xmp

dsU ij pqEcdh; {ks= ftlesa i /kkjk izokfgr gksrh gSn gSA fcUnq P cgqHkqt dk dsUd gSA

(T) 03 i32 x

mp


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

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

ColumnI ColumnII

(A)

C

S

Density of field lines first increases then decreases

as we move radially outward

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

(Q)C

H.d 0rr

l

where Hr

represents electric or magnetic field

(R)C

H.d 0=rr

l

where Hr


(S)SH.dA 0=

rr

(B)C

S

Density of field lines continuously decreases as we move radially outward

where Hr


(T)SH.dA 0

rr

where Hr


(C)C

S

Axi

s

(D)

Axi

s

C

S

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

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

LrHkI LrEHkII

(A)C

S

f=T;h; ckgj dh rjQ tkus ij {ks= js[kkvksa dk ?kuRo igys c

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SECTION III : Integer Value Correct Type [k.M III : iw.kkd eku lgh izdkj

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

[k.M-IV : (iw.kkd eku lgh izdkj)This section contains 8 questions. The answer to each question is a single digit Integer, ranging from0 to 9 (both inclusive)bl [k.M esa 8 iz'u gSaA izR;sd iz'u dk mkj 0 ls 9 rd (nksuksa 'kkfey) ds chp dk ,dy vadh; iw.kk d gSA

1. A pendulum consist of a disk of mass M and radius R and a massless rod of length l. Find approximatetime period (in sec) of the system if the disk is mounted to the rod by a frictionless bearing so that it isperfectly free to spin? Assume small oscillation. (Take : l = 0.993m, g = 9.8 m/s2)iznf'kZr yksyd O;eku M rFkk f=T;k R okyh ,d pdrh rFkk yEckbZ l okyh ,d O;ekughu NM+ ls feydj cuk gSA;fn bl pdrh dks ?k"kZ.k jfgr fc;fjax dh lgk;rk ls NM + ij bl izdkj yxk fn;k tk;s fd ;g p.k djus ds fy;siw.kZr; Lora= gks rks bl fudk; dk yxHkx vkorZdky (lsd.M esa) Kkr dhft;sA ekuk vYi nksyu gksrs gSA(fn;k gS : l = 0.993m, g = 9.8 m/s2)

R

Ml

2. Two charges q and 2q are placed at (3,0) and (3,0) in xy plane. The locus of the point in the plane ofthe charges, where the field potential is zero is (x + a)2 + y2 = 4b2. Find the value of (a+b).nks vkos'kksa q rFkk 2q dks x-y ry esa (3,0) rFkk (3,0) ij j[kk x;k gSA vkos'kksa ds ry esa tgk {ks= foHko 'kwU; gS] fcUnqdk fcUnq iFk (x + a)2 + y2 = 4b2 gSA (a + b) dk eku Kkr dhft,A


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3. A plastic rod of length 1.0 m carries uniform positive charge +4.0 mC/m on half of its length and uniformnegative charge 4.0 m C/m on the remaining half of its length. Find magnitude of its net dipole momentin mCm.,d IykfLVd NM+ dh yEckbZ 1.0 m gSA bldh yEckbZ ds vk/ks Hkkx ij ,dleku /kukRed vkos'k +4.0 mC/m rFkkyEckbZ ds 'ks"k vk/ks Hkkx ij ,dleku .kkRed vkos'k -4.0 m C/m gSA bldk dqy f}/kq zo vk?kw.kZ dk ifjek.k mC-m esaKkr dhft,A

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

across the tube. Let lK be Ka line produced by tube and lmin be cut off wavelength. Calculate ( )1 K min2 l - lin the order of 108 m.dwyht uyh esa iz;qDr y{; inkFkZ dk ijek.kq ekad 41 gSA uyh ds fljksa ij 20kV foHkokUrj vkjksfir fd;k tkrk gSA

ekuk lK ] uyh }kjk mRiUu Ka js[kk rFkk vard rjaxnS/;Z lmin gSA ( )1 K min2 l - l dh x.kuk 10-8 m dksfV esa dhft,A


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5. A thermonuclear device consists of a torus of mean diameter 3 m with a tube of diameter 1 m, containingdeuterium gas at 102 mm mercury pressure and at room temperature (20C). A bank of capacitors of1200 mF is discharged through the tube at 50 kV. If only 20% of the electrical energy is transformed toplasma kinetic energy, the maximum temperature attained is equals to 1.18 10a K. Assuming that theenergy is equally shared between the deuterons and electrons in the plasma. Find the value of a .,d rki&ukfHkdh; ;qfDr ds Vksjl (torus) dk ek/; O;kl 3 m gSA blesa 1m O;kl dh ,d uyh yxh gqbZ gS ftlesa102 mm ikjn nkc rFkk 20C dejs ds rki ij M~;wVhfj;e xSl Hkjh gqbZ gSA bl uyh }kjk 50kV ij 1200 mF dsla/kkfj=ksa ds ,d lewg dks fujkosf'kr fd;k tkrk gSA ;fn fo|qr tkZ dk dsoy 20% Hkkx gh IykTek xfrt tkZ esaifjofrZr gksrk gS rks vf/kdre rkieku 1.18 10a K ds cjkcj izkIr gksrk gSA ekuk fd IykTek esa tkZ M~;wVsjksuks rFkkbysDVkWuksa ds e/; leku :i ls forfjr gksrh gSA a dk eku Kkr dhft,A

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

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

5W5W 5W

5W 5W5W5W5W

5W

5W

10V

I

5V 5W


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7. An electric dipole is placed along xaxis at the origin as shown in the figure. Electric field at the point Pis parallel to the yaxis. If xcoordinate of P is 2 m, what is its ycoordinate in meters.

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

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

y

P(x,y)

p x

8. In the given circuit for what value of n, power generated in resistance R will be minimum.fn;s x;s ifjiFk esa n ds fdl eku ds fy, izfrjks/k R esa mRiUu 'kfDr dk eku U;wure gksxk\

(n)W

(n)W (n+2)W(n+3)W

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

(R)W

(n/5)WE

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Kota/00CT214005

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

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

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

1. Calculate the percentage hydrolysis in 0.01M aqueous solution of NaOCN(Kb for OCN = 1010)NaOCN ds 0.01M tyh; foy;u esa ty vi?kVu dk izfr'kr crkb;s

(OCN ds fy, Kb = 1010 gS)(A) 0.1 % (B) 0.01 % (C) 0.0001 % (D) 0.001 %

2. Identify the correct statement-(A) In an isolated system DU > 0 and DS > 0 for an irrversible process(B) In an isolated system DU = 0 and DS = 0 for an irrversible process(C) In an isolated system DU > 0 and DS = 0 for an irrversible process(D) In an isolated system DU = 0 and DS > 0 for an irrversible processlgh dFku dks igpkfu,sa -

(A) ,d foyfxr ra= esa DU > 0 rFkk vuqRe.kh; ize ds fy, DS > 0(B) ,d foyfxr ra= esas DU = 0 rFkk vuqRe.kh; ize ds fy, DS = 0(C) ,d foyfxr ra= esa DU > 0 rFkk vuqRe.kh; ize ds fy, DS = 0(D) ,d foyfxr ra= esa DU = 0 rFkk vuqRe.kh; ize ds fy, DS > 0

3. The resistance of 0.1 M solution of oxalic acid is 200 ohm and cell constant is 2.0 cm1, the molarconductance (in S cm2 mole1) of 0.1 M oxalic acid is :-vkWDlsfyd vEy ds 0.1 M foy;u dk izfrjks/k 200 vkse rFkk lsy fu;rkad 2.0 cm1 gS rks 0.1 M vkWDlsfyd vEy

dk eksyj pkydRo (S cm2 mol1 esa) gS :-(A) 100 (B) 10 (C) 1 (D) 0.1


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4. Which of the following is not possible structure for any oxide of nitrogenfuEu esa ls dkSulh lajpuk ukbVkstu ds fdlh Hkh vkWDlkbM ds fy;s lEHko ugha gS

(I) :

N O N+ :

+ O

::

:O

::O

:::O

:: (II) :N O N: : +

O

O....

..

..O....

(III) O = NN+O....O ..

..

....

..

..(IV) N = O N++

2:

:

:

:

(A) II, IV (B) III, IV (C) I only (dsoy) (D) II, III and (rFkk) IV5. Select reaction in which anion of main ionic product must have s-bond :

(A) Li + O2(Excess) (B) Na(Excess) + O2 (C) Ca + F2(Excess) (D) NaOH(cold) + Cl2 og vfHkf;k pqfu, ftlds eq[; vk;fud mRikn ds .kk;u esa s-ca/k mifLFkr gks :(A) Li + O2(vkf/kD;) (B) Na(vkf/kD;) + O2 (C) Ca + F2(vkf/kD;) (D) NaOH(BaMk) + Cl2

6. CH OH2

CHOH

CH OH2When above compound is treated with excess HI product formed is :-(A) 3-iodoprop-1-ene (B) 1, 2, di-iodopropane(C) Propene (D) 2-iodopropaneCH OH2

CHOH

CH OH2

tc mijksDr ;kSfxd dks HI vkf/kD; ds lkFk mipkfjr djrs gS rks cuus okyk mRikn gS&(A) 3-vk;Mksizksi-1-bZu (B) 1, 2, Mkb-vk;Mksizksisu(C) izksihu (D) 2-vk;Mksizksisu

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7.O

H O3

+ NH NH2 2 (i) HBrXOH / D (ii) OEt / DY Z (Major product) / eq[; mRikn

"Z" is (gS)

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

8. CH2CO2NaCH2CO2Na

Electrolysis (A) Br2(CCl4)

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

C is -

(A) CH2CH2

(B) CH3CH3

(C) CHCH

(D) CH2CHCH3

CH2CO2NaCH2CO2Na

(A) Br2(CCl4)

(B) alc.1.KOH (C)

NaNH22. H+

fo|qr vi?kVu

lksfM;e lDlhusV

ds ckn

C gS-

(A) CH2CH2

(B) CH3CH3

(C) CHCH

(D) CH2CHCH3

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(ii) Paragraph Type (ii) vuqPNsn izdkj

This section contains 2 paragraphs each describing theory, experiment, data etc. Four questions relateto two paragraphs with two questions on each paragraph. Each question of a paragraph has only onecorrect answer among the four choices (A), (B), (C) and (D).bl [k.M esa flkUrksa] iz;ksxksa vkSj vkdM+ksa vkfn dks n'kkZus okys 2 vuqPNsn gSA nksuksa vuqPNsn ls lacaf/kr pkj iz'ugSa] ftuesa ls gj vuqPNsn ij nks iz'u gSaA vuqPNsn esa gj iz'u ds pkj fodYi (A), (B), (C) vkSj (D) gSa ftuesals dsoy ,d lgh gSA

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

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

9. Given the data below :0combHD (CH4, g) = 900 kJ/mole ; 0combHD (C, graphite) = 400 kJ/mole0combHD (H2, g) = 300 kJ/mole ; 0atomizationHD (C, graphite) = 700 kJ/mole

and 0fH (H,g)D = 220 kJ/moleWhat should be the bond enthalpy of CH bond in kJ/mole.(A) 400 (B) 420 (C) 310 (D) 345uhpsa dqN vkdM+s fn;s x, gS :

0HD ngu (CH4, g) = 900 kJ/mole ; 0HD ngu (C, xzsQkbV) = 400 kJ/mole0HD ngu (H2, g) = 300 kJ/mole ; 0HD ijekf.o;dj.k (C, xzsQkbV) = 700 kJ/mole

rFkk 0fH (H,g)D = 220 kJ/moleCH cU/k dh cU/k ,saUFkSYih kJ/mole esa D;k gksuh pkfg,A(A) 400 (B) 420 (C) 310 (D) 345

10. Identify the incorrect statement -(A) 0combHD (C, graphite) is equal to 0fHD (CO2, g)(B) 0 0Cl Cl atom 2H H [Cl (g)]-D = D(C) The standard enthalpy of combustion is always negative.(D) The standard enthalpy of formation is always negativexyr dFku dk p;u dhft, -(A) 0combHD (C, xzsQkbV) , 0fHD (CO2, g) ds cjkcj gksrh gS(B) 0 0Cl Cl atom 2H H [Cl (g)]-D = D(C) ngu dh ekud , sUFkSYih lnSo .kkRed gksrh gS(D) fuekZ.k dh ekud , sUFkSYih lnSo .kkRed gksrh gS

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

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

Solution of S1 + S2 in dil HCl gives white ppt with BaCl2 which is not soluble in dil HNO3 ?

Mixture of S1 and S2 not soluble in water but soluble in dil HCl with the evolution of a colourlessand odourless gas(G) which turns Ba(OH)2 milky.

Solution of S1 + S2 in dil HCl does not give ppt with H2S but gives white ppt(p) with NaOH solutionwhich is soluble in exess NaOH(aq).

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

S1 + S2 dk ruq HCl esa foy;u BaCl2 ds lkFk 'osr vo{ksi nsrk gS tks ruq HNO3 esa foys; ugha gSa ?

S1 rFkk S2 dk feJ.k ty esa foys;'khy ugha gS ysfdu ruq HCl esa foys; gksdj ,d jaxghu rFkk xU/kghu xSl(G)nsrk gS tks Ba(OH)2 dks nwf/k;k dj nsrh gSA

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

nsrk gS tks NaOH(aq) ds vkf/kD; esa foys; gSA

11. Salt S1 and S2 are respectively :

yo.k S1 rFkk S2 e'k% gSa :

(A) PbSO4, PbCO3 (B) PbCO3, ZnSO4 (C) ZnS, ZnSO4 (D) ZnSO4 + ZnCO312. Solution of S1 + S2 in dil. HCl gives white ppt with :

(A) K4[Fe(CN)6] (B) Excess NH4OH (C) H2S(g) (D) All of the aboveS1 + S2 dk ruq HCl esa foy;u] fuEu esa ls fdlds lkFk 'osr vo{ksi nsrk gS :

(A) K4[Fe(CN)6] (B) NH4OH vkf/kD; (C) H2S(g) (D) mijksDr lHkhSpace for Rough Work / dPps dk;Z ds fy, LFkku

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SECTIONII : Matrix-Match Type[k.M - II : eSfVDl&esy izdkj

This Section contains 2 questions. Each question has four statements (A, B, C and D) given inColumn I and five statements (P, Q, R, S and T) in Column II. Any given statement in Column I can havecorrect matching with ONE or MORE statement(s) given in Column II. For example, if for a givenquestion, statement B matches with the statements given in Q and R, then for the particular question, againststatement B, darken the bubbles corresponding to Q and R in the ORS.bl [k.M esa 2 iz'u gSA izR;sd iz'u esa dkWye I esa 4 dFku (A, B, C vkSj D) vkSj dkWye II esa 5 dFku (P, Q, R, S vkSj T)gSaA dkWye I dk dksbZ Hkh dFku dkWye II ds ,d dFku ;k ,d ls vf/kd dFkuksa ls esy [kkrk gSA mnkgj.k ds fy,] fn, gq, iz'uesa ;fn dFku B dFkuks a Q vkSj R ls esy [kkrk gS] rks vksvkj,l (ORS) esa ml iz'u ds fy;s dFku B ds lkeus Q vkSj R lslEcfUkr cqycqyksa dks dkyk dhft;sA

1. Column-I Column-II(Conversion) (Process / Furnace involved in given conversion)

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

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

(C) Concentrated Galena PbO (R) Carbon reduction

(D) Red Bauxite Al(Pure) (S) S2 (sulphide) is oxidised(T) Blast Furnace

LrEHk-I LrEHk-II

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

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

(B) 'kq ghesVkbV Fe(dPpk) (Q) Lo&vip;u

(C) lkfUr xSysuk PbO (R) dkcZu vip;u

(D) yky ckWDlkbV Al('kq) (S) S2 (lYQkbM) vkWDlhd`r gksrk gS(T) okR;k HkV~Vh


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

(A)OH

HPCl3 (P) Optically inactive mixture

(B)OH

HHI (Q) Optically active product

(C)OH

HPDC (R) Product will give precipitate when treated with

alcoholic AgNO3

(D)OH

HNa

Me I(S) Product formation involve substitution reaction.

(T) Product will react with CH3MgCl to producehydrocarbon

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

(A)OH

HPCl3 (P) izdkf'kd vf; feJ.k

(B)OH

HHI (Q) izdkf'kd lf; mRikn

(C)OH

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

djrs gSA

(D)OH

HNa

Me I(S) mRikn fuekZ.k esa izfrLFkkiu vfHkf;k lfEefyr gksrh gSA

(T) mRikn dh CH3MgCl ds lkFk vfHkf;k djkus ij gkbMksdkcZudk fuekZ.k gksxkA


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


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SECTION-IV : (Integer Value Correct Type)

[k.M-IV : (iw.kkd eku lgh izdkj)This section contains 8 questions. The answer to each question is a single digit Integer, ranging from0 to 9 (both inclusive)bl [k.M esa 8 iz'u gSaA izR;sd iz'u dk mkj 0 ls 9 rd (nksuksa 'kkfey) ds chp dk ,dy vadh ; iw.kk d gSA

1. When a metal is exposed with light of wavelength l, the maximum kinetic energy of electronproduced was found to be 2 eV. When the same metal was exposed with light of wavelength l/2,the maximum kinetic energy of elelctron produced was 6 eV. What is the value of work function ofmetal in eV.tc ,d /kkrq dks l rjaxnS/; Z ds izdk'k esa j[kk tkrk gS] rc mRikfnr bySDVkWu dh vf/kdre xfrt tkZ 2 eV ik;h x;hAtc blh /kkrq dks l/2 rjaxnS/; Z okys izdk'k esa j[kk x;k rks mRikfnr bySDVkWu dh vf/kdre xfrt tkZ 6 eV FkhA /kkrqds dk;ZQyu dk eku eV esa D;k gksxkA

2. Calculate the change in pressure (in atm) when 2 mol of NO and 16 gm O2 in a 6.25 litre originally at27 C react to produce the maximum quantity of NO2 possible according to the equation,2 NO (g) + O2 (g) 2NO2 (g)(Take R = 1

12 litre atm /mol K)

nkc esa ifjorZu dh x.kuk (atm esa ) dhft, tc 27 C ij 6.25 yhVj okLrfod ek=k esa 2 eksy NO rFkk 16 gm O2vfHkf;k djds NO2 dh vf/kdre ek=k fuEu lehdj.k ds vuqlkj mRikfnr djrs gS ,2 NO (g) + O2 (g) 2NO2 (g)(fn;k gS R = 1

12 litre atm /mol K)


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3. Element X can crystllizes in fcc lattice as well as in bcc lattice. If density of bcc lattice is 3 32

g/cm3 then

what is density of fcc lattice in gm/cm3.

rRo X, fcc tkyd ds lkFk&lkFk bcc tkyd esa Hkh fLVyhd`r gks ldrk gSA ;fn bcc tkyd dk ?kuRo 3 32

g/cm3

gS rks fcc tkyd dk ?kuRo] gm/cm3 esa D;k gS \ 4. Out of NiS, CoS, MnS, ZnS

P = Number of black precipitates.Q = Number of sulphides which is / are soluble in dil. HClR = Number of sulphides which is / are soluble in dil. CH3COOHS = Number of sulphides which is / are soluble in hot / conc. HNO3.(P + Q + R + S = Total number is your answer)NiS, CoS, MnS, ZnS esa ls

P = dkys vo{ksiksa dh la[;kQ = ,sls lYQkbMksa dh la[;k tks ruq HCl esa foys; gS @ gSaR = ,sls lYQkbMksa dh la[;k tks ruq CH3COOH esa foys; gS @ gSaS = ,sls lYQkbMksa dh la[;k tks xeZ @ lkU HNO3 esa foys; gS @ gSa(P + Q + R + S = dqy la[;k gh vkidk mkj gS)


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5. "Number of geometrical isomers are NOT changed in [Pt(Cl) (Br) (PMe3)2]" When one PMe3 is replaced by NH3 When Clfl is replaced by Brfl

When Brfl is replaced by CN When both PMe3 are replaced by one en(ethylene diamine) When one Cl and one PMe3 are replaced by one en(ethylene diamine) When all the Megroups are replaced by EtgroupsFind number of condition(s) given above which make the underlined statement CORRECT :"[Pt(Cl) (Br) (PMe3)2] esa T;kfefr leko;fo;ksa dh la[;k ugha cnyrh gS" tc ,d PMe3 dks NH3 }kjk foLFkkfir fd;k tkrk gS tc Clfl dks Brfl }kjk foLFkkfir fd;k tkrk gS

tc Brfl dks CN }kjk foLFkkfir fd;k tkrk gS

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

mijksDr esa ls , slh fLFkfr;ksa dh la[;k crkbZ;sa tks js[kk afdr dFku ds fy, lgh gSa :6. PhCO2H + NaHCO3 A (gas) xSl

PhOH + NaNH2 B (gas) xSlCH3OH + NaH C (gas) xSlCH3SH + EtLi D (gas) xSlPhSO3H + Na E (gas) xSlIf the molecular weight of A, B, C, D ,E is x, y, z, w, v then find the value of (y + z + w + v) x;fn A, B, C, D rFkk E ds v.kqHkkj e'k% x, y, z, w, v gS rks (y + z + w + v) x dk eku Kkr dhft,A


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7. For the given reaction sequence :

fn;s x;s vfHkf;k e esa

COOEtCOOEt

NaOEtD

P H O3

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

Number of monohalogenated products is/are -

eksuksgsykstuhd`r mRiknksa dh la[;k gSA8. How many these reactions will give 1 alcohol as major product :

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

(1) O 2H / H O+ (2) O

+ CH3MgBr 2H O

(3) HCHO + CH3MgBr 2H O (4) CH3CHO + CH3MgBr 2H O

(5) CH CCH CCH +CH MgBr3 2 3 3||O

||O

2H O (6) ||O

Cl + CH3MgBr 2H O Excess (vkf/kD;)

(7) O + CH3MgBr 2H O (8) CH3CH2Br NaOH(Aq.)

(9) CH3MgBr + CO2 2H O


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PART-3 : MATHEMATICS

Hkkx-3 : xf.krSECTIONI : (i) Only One option correct Type

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

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

x 1 05! 4! 3! 2!

+ + + + + = have -

(A) 5 distinct real roots (B) exactly 3 distinct real roots(C) exactly one real root (D) exactly one positive real root

lehdj.k 5 4 3 2x x x x

x 1 05! 4! 3! 2!

+ + + + + = ds -

(A) 5 fHkUu okLrfod ewy gksaxsA (B) Bhd 3 fHkUu okLrfod ewy gksaxsA(C) Bhd ,d okLrfod ewy gksxkA (D) Bhd ,d /kukRed okLrfod ewy gksxkA

2. If 2x1999

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

;fn 2

x1999

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

(A) 1 (B) 9991

2 (C) 19981

2 (D) 9991

2-


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

- = + and z1, z2, z3, z4 A are 4 complex numbers representing points

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

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

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

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

(A) 94 (B) 92 (C) 9 (D) 16

4. Let 2 2

2 2x y 1a b

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

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

ekuk

2 2

2 2x y 1a b

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

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

(A) 12p (B) 6

p (C) 4p (D) 512

p

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

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

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

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6. Let ( )/ 2

13 11

0

I cosx dxp

+= and ( )/ 2

13 12

0

I cosx dxp

-= , then value of 21

II

is equal to (where [.] greatest

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

ekuk ( )/ 2

13 11

0

I cosx dxp

+= rFkk ( )/ 2

13 12

0

I cosx dxp

-= gks] rks 21

II

dk eku gksxk (tgk [.] egke iw.kkd Qyu

dks n'kkZrk gS)(A) 0 (B) 1 (C) 2 (D) 2 ls vf/kd

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

(A) 4 (B) 2 (C) 52- (D) Not defined

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

(A) 4 (B) 2 (C) 52- (D) ifjHkkf"kr ugha

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

his father is increasing at the rate of l m/s when Raghav has run 14 of the way around the track then l

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

pkjksa rjQ 10m/s dh pky ls nkSM+uk 'kq: djrk gSA tc jk?ko o`kkdkj iFk dk 14 Hkkx nkSM+ pqdk gks] ml le; ;fn jk?ko

dk mlds firk ls foLFkkiu l m/s dh nj ls c

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(ii) Paragraph Type (ii) vuqPNsn izdkj

This section contains 2 paragraphs each describing theory, experiment, data etc. Four questions relateto two paragraphs with two questions on each paragraph. Each question of a paragraph has only onecorrect answer among the four choices (A), (B), (C) and (D).bl [k.M esa flkUrksa] iz;ksxksa vkSj vkdM+ksa vkfn dks n'kkZus okys 2 vuqPNsn gSA nksuksa vuqPNsn ls lacaf/kr pkj iz'ugSa] ftuesa ls gj vuqPNsn ij nks iz'u gSaA vuqPNsn esa gj iz'u ds pkj fodYi (A), (B), (C) vkSj (D) gSa ftuesals dsoy ,d lgh gSA


iz'u 9 ,oa 10 ds fy;s vuqPNsnA slip of paper is given to A. who marks it with either a plus sign or a minus sign, the probability

of his writing a plus sign is known to be 13

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

alone or change the sign before passing it on to C. Next C passes the slip to D after perhaps changingthe sign. Finally D passes it to an honest judge after perhaps, changing the sign. It is known that B,C and D each change the sign with probability 2/3.

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

iphZ ij /kukRed fp ds fy[kus dh Kkr izkf;drk 13

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

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

9. The probability that judge see a plus sign on the slip is -iphZ ij U;k;/kh'k ds }kjk /kukRed fp ns[kus dh izkf;drk gksxh -

(A) 181 (B) 1781 (C)

2581 (D)

4181

10. If the judge see a plus sign on the slip, then the probability that A originally wrote a plus sign is -;fn U;k;/kh'k iphZ ij /kukRed fp ns[krk gS] rks A ds }kjk okLro esa iphZ ij /kukRed fp fy[kus dh izkf;drk gksxh -

(A) 141 (B) 1241 (C)

1341 (D)

1441


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

Let S1 : 12x2 7xy 12y2 + 25 = 0 is a hyperbola, L : y = 2x is a line and S2 is reflection of S1in L.

ekuk S1 : 12x2 7xy 12y2 + 25 = 0 ,d vfrijoy;, L : y = 2x ,d js[kk rFkk o S1 dk js[kk L esaizfrfcEc o S2 gSA

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

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

(A) 2 (B) 1 (C) 1 (D) 212. Equation of transverse axis of S1 is -

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

(A) y = 7x (B) x + 7y = 0 (C) 3x = 4y (D) 7x + y = 0Space for Rough Work / dPps dk;Z ds fy, LFkku

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SECTIONII : Matrix-Match Type[k.M - II : eSfVDl&esy izdkj

This Section contains 2 questions. Each question has four statements (A, B, C and D) given inColumn I and five statements (P, Q, R, S and T) in Column II. Any given statement in Column I can havecorrect matching with ONE or MORE statement(s) given in Column II. For example, if for a givenquestion, statement B matches with the statements given in Q and R, then for the particular question, againststatement B, darken the bubbles corresponding to Q and R in the ORS.bl [k.M esa 2 iz'u gSA izR;sd iz'u esa dkWye I esa 4 dFku (A, B, C vkSj D) vkSj dkWye II esa 5 dFku (P, Q, R, S vkSj T)gSaA dkWye I dk dksbZ Hkh dFku dkWye II ds ,d dFku ;k ,d ls vf/kd dFkuksa ls esy [kkrk gSA mnkgj.k ds fy,] fn, gq, iz'uesa ;fn dFku B dFkuks a Q vkSj R ls esy [kkrk gS] rks vksvkj,l (ORS) esa ml iz'u ds fy;s dFku B ds lkeus Q vkSj R lslEcfUkr cqycqyksa dks dkyk dhft;sA

1. Column-I Column-II

(A) If ( ) [ ][ ]( )

max sin t, t , x x 0,2

x

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

p - p = - p p p

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

0

x 1 x 2 x 3 ... x 9 dx- - - -

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

(B) If ( )1

5 4 4

0

n 3x x 4x 5dx

25+ + = , then n is (R) 13

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

(D) The number of integral values of c for whichequation sin1x x = c has a solution is (T) greater than number of

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

LrEHk-I LrEHk-II

(A) ;fn ( ) [ ][ ]( )

p - p = - p p p

vf/kdre

U;wure

sin t, t , x x 0,2

x

cos t, t , x x ( ,2 ](P) ( )( )( ) ( )( )

10

0

x 1 x 2 x 3 ... x 9 dx- - - -

gks] rks vUrjky [0,2p] esa mu fcUnqvksa dh la[;k] (Q) 1tgk (x) vodyuh ; ugha gS] gksxh

(B) ;fn ( )1

5 4 4

0

n 3x x 4x 5dx

25+ + = gks] rks n gksxk (R) 13

(C) 16sin3q 4sin2q 16sinq + 8, q R dkvf/kdre eku gksxk (S) 9

(D) c ds iw.kkd ekuks a dh la[;k ftlds fy,lehdj.k sin1x x = c dk gy gS] gksxh (T) vUrjky (0, 9) esa fcUnqvksa dh la[;k

ls vf/kd] tgk |sin px| vodyuh;ugha gS

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2. Column-I Column-II

(A) In a DABC, 22tan A7

= and the altitude AD from (P) 3

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

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

of DABC, then least value of ahr

is equal to

(C) ABCD is a tetrahedron of volume 27. E, F & G are points (R) number of solutionson AB, BC and CA respectively dividing them in 2 : 1 of 2sin1(sinx) = x(internally), then the volume of tetrahedron DEFG isequal to (S) 18

(D) If a, b, cr

r r

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

r

and c a b= r

r r

,

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

r r r r r r (T) number of pointsis equal to of discontinuity of

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

LrEHk-I LrEHk-II

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

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

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

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

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

dk U;wure eku gksxk

(C) prq"Qyd ABCD dk vk;ru 27 gSA E, F ,oa G e'k% (R) 2sin1(sinx) = xAB, BC rFkk CA ij fcUnq bl izdkj gS fd ; s budks vuqikr 2 : 1 ds gyksa dh la[;k(vUr% foHkkftr) esa foHkkftr djrh gS] rks prq"Qyd DEFG dkvk;ru gksxk (S) 18

(D) ;fn a, b, cr

r r

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

r

rFkk

c a b= r

r r

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

r r r r r r (T) vUrjky (1,20) esadk vf/kdre eku gksxk (x) = [x + [x + [x]]]

ds vlarr~ fcUnq

(tgk [.] egke iw.kkd Qyudks n'kkZrk gS)


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

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SECTION-IV : (Integer Value Correct Type)

[k.M-IV : (iw.kkd eku lgh izdkj)This section contains 8 questions. The answer to each question is a single digit Integer, ranging from0 to 9 (both inclusive)bl [k.M esa 8 iz'u gSaA izR;sd iz'u dk mkj 0 ls 9 rd (nksuksa 'kkfey) ds chp dk ,dy vadh ; iw.kk d gSA

1. Functions , g, h are differentiable on some open interval around 0 and satisfy the equations and initialconditions

( )2 1 ' 2 gh , 0 1gh

= + =

( )2 4g' g h , g 0 1h

= + =

( )2 1h ' 3gh , h 0 1g

= + = , then

( ) ( ) ( ) x g x h x tan nx4p = + , then n is equal to

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

( )2 1 ' 2 gh , 0 1gh

= + =

( )2 4g' g h , g 0 1h

= + =

( )2 1h ' 3gh , h 0 1g

= + = dks lUrq"V djrs gSaA rc ;fn

( ) ( ) ( ) x g x h x tan nx4p = + gS] rks n dk eku gksxk

2. The number of values of a for which the curves y = ax2 + ax + 124 and x = ay2 + ay + 124 have

common tangent is

a ds ekuksa dh la[;k ftlds fy, o y = ax2 + ax + 124

rFkk x = ay2 + ay + 124

dh mHk;fu"B Li'kZjs[kk

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3. A function is differentiable for all real numbers and satisfies (3 + x) = (3 x) and (8 + x) = (8 x)for all real x. If (0) = '(0) = 0, then least number of roots of equation '(x) = 0 in (11,11) isQyu lHkh okLrfod la[;kvksa ds fy, vodyuh; rFkk (3 + x) = (3 x) ,oa (8 + x) = (8 x) dks lHkhokLrfod x ds fy, lUrq"V djrk gSA ;fn (0) = '(0) = 0 gks] rks vUrjky (11,11) esa lehdj.k '(x) = 0 dsewyksa dh U;wure la[;k gksxh

4. Let (x,y) be a variable point on the curve x2 + 4y2 4x + 8y + 4 = 0. Then ( )( )2 2

2 2

max x y 4x 2y 5min x y 4x 2y 5

+ + + ++ + + +

is equal to

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

+ + + +

2 2

2 2

x y 4x 2y 5x y 4x 2y 5

vf/kdre

U;wure dk eku gksxk

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

40 and 40z

have real

and imaginary parts lies in the interval [0, 1] is A, then least integer greater than or equal to A100 is

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

40 rFkk 40z

ds okLrfod rFkk dkYifud

Hkkx vUrjky [0, 1] esa gS] A gks] rks A100 ls cM+k ;k cjkcj U;wure iw.kkd gksxk


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

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

vlfedkvksa x2 + y2 < p

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

7. Let a, b, c are three distinct natural number less than 10 such that the system of linear equations in x, y, z3x + 4y + 5z = a4x + 5y + 6z = b5x + 6y + 7z = c

posses more than one solution, then the maximum value b can take isekuk 10 ls NksVh rhu fHkUu izkd`r la[;k;sa a, b, c bl izdkj gSa fd x, y, z esa jS[kh; lehdj.kksa ds fudk;

3x + 4y + 5z = a4x + 5y + 6z = b5x + 6y + 7z = c

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

of BD and CE, then the ratio of area of the quadrilateral BCDE and the area of the triangle AMN isequal toekuk f=Hkqt ABC esa Hkqtkvksa AC rFkk AB ij fcUnq D rFkk E gSA ;fn M rFkk N Hkqtkvkas BD rFkk CE ds e/; fcUnq gks]rks prqHkqZt BCDE ds {ks=Qy rFkk f=Hkqt AMN ds {ks=Qy dk vuqikr gksxk


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Proves

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

JEE NurtureEnthusiast

LeaderJEE Nurture

EnthusiastLeader

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

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

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

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

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

PRE-MEDICAL NurtureEnthusiast

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Pre-Nurture & For Class VI to XCareer Foundation

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

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-05-15 NA NA NA NA NA NA NA

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

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

(Ex-ALLEN / XII Passed before 2015)

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

D. vadu ;kstuk / Marking scheme :15. [kaM-I (i) ds gj iz'u esa dsoy lgh mkjksa (mkj) 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 .kkRed ,d (1) vad iznku fd;k tk;sxkAFor each question in Section-I (i), you will be awarded 3 marks if you darken all the bubble(s) corresponding to only thecorrect answer(s) and zero mark if no bubbles are darkened. In all other cases minus one (1) mark will be awarded

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

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

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

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

Name of the Candidate / ijh{kkFkhZ dk uke

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

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