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CHEMISTRY

Section – I Questions 1 to 8 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. + 3 marks will be given for each correct answer and – 1 mark for each wrong answer.

Q.1 40 mL gaseous mixture of CO, CH4 and Ne was exploded with 10 mL of oxygen. On cooling the gases occupied 36.5 mL. After treatment with KOH the volume reduced by 9 mL and again on treatment with alkaline pyrogallol, the volume further reduced. Percentage of CH4 in the original mixture is -

(A) 22.5 (B) 77.5 (C) 7.5 (D) 15

[k.M - I iz'u 1 ls 8 rd cgqfodYih iz'u gSaA izR;sd iz'u ds pkj fodYi (A), (B), (C) rFkk (D) gSa, ftuesa ls dsoy ,d fodYi lghOMR 'khV esa iz'u dh iz'u la[;k ds le{k viuk mÙkj vafdr dhft;sA izR;sd lgh mÙkj ds fy, + 3 vad fn;s tk;asxs rFkk

1

Q.1 CO,CH4 Ne 40 mL xSlh; feJ.k dks 10mL vkWDlhtu ds lkFk j[kk tkrk gSA 'khryu ij xSl 36.5 mL vk;ru xzg.k djrh gSA KOH ds lkFk mipkfjr djus ds i'pkr 9 mL vk;ru de gks tkrk gS rFkk iqu% {kkjh; ik;jksxsyksy ds lkFk mipkfjr djus ij vk;ru esa vksj de gks tkrh gSA okLrfod feJ.k esa CH4 dk izfr'kr gS -

(A) 22.5 (B) 77.5 (C) 7.5 (D) 15



Q.2 For a 3s-orbital

Ψ(3s) = 39

1 2/3

0a1

(6 – 6σ + σ2)e–σ/2;

where σ = 0a3Z.r2

What is the maximum radial distance of node from nucleus ?

(A) Z

a)33( 0+ (B)

Za0

(C) Z

a)33(23 0+ (D)

Za2 0

Q.3 11 moles N2 and 12 moles of H2 mixture reacted in 20 litre vessel at 800 K. After equilibrium was reached, 6 mole of H2 was present. 3.58 litre of liquid water is injected in equilibrium mixture and resultant gaseous mixture suddenly cooled to 300 K. What is the final pressure of gaseous mixture ? Negative vapour pressure of liquid solution. Assume (i) all NH3 dissolved in water (ii) no change in volume of liquid (iii) no reaction of N2 and H2 at 300 K

N2 : 11 moles

H2 : 12 moles T = 800 K

V = 20 L

Initial condition

T = 300 K; P = ? NH3(aq) solution

N2 , H2

(A) 18.47 atm (B) 60 atm (C) 22.5 atm (D) 45 atm

Q.2 3s-

Ψ(3s) = 39

1 2/3

0a1

(6 – 6σ + σ2)e–σ/2;

σ = 0a3Z.r2

ukfHkd ls uksM+ dh vf/kdre f=kT;k nwjh D;k gksxh ?

(A) Z

a)33( 0+ (B) Za0

(C) Z

a)33(23 0+ (D)

Za2 0

Q.3 800 K ij 20 yhVj ik=k esa 11 eksy N2 rFkk 12 eksy H2 dk feJ.k fd;k djrk gSA lkE; izkIr gksus ds i'pkr H2 ds 6 eksy mifLFkr FksA lkE; feJ.k ls nzo ty ds 3.58 yhVj fudky fy;k tk;s rFkk ifj.kkeh xSlh; feJ.k dks vpkud 300 K rd B.Mk fd;k tk;s rks xSlh; feJ.k dk vafre nkc D;k gksxkA

nzo foy;u dk ok"i nkc] _.kkRed gSA ekuk (i) lHkh NH3, ty esa foys; gSA (ii) nzo ds vk;ru esa ifjorZu ugh gksrk gSA (iii) 300 K ij N2 rFkk H2 vfHkfØ;k ugh djrsA

N2 : 11 moles

H2 : 12 molesT = 800 K

V = 20 L

okLrfod fLFkfr

T = 300 K; P = ? NH3(aq) solution

N2 , H2

(A) 18.47 atm (B) 60 atm (C) 22.5 atm (D) 45 atm



Q.4 In the cyclic process shown in P–V diagram, the magnitude of the work done is -

V1

V2

V

P2 P1P

(A) π2

12

2P–P

(B) π

212

2V–V

(C) 4π (P2–P1)(V2–V1) (D) π(V2 – V1)2

Q.5

CN CN

CN CN

→ ∆⊕ /OH3 (A);

Product (A) of the reaction is -

(A)

O

O

O

(B)

O

O

O

(C) CO2H

CO2H (D)

O

O

O

Q.4 P–V fp=k esa n'kkZ;s x;s pØh; izØe esa fd;s x;s dk;Z dk ifjek.k gksxk -

V1

V2

V

P2 P1P

(A) π2

12

2P–P

(B) π

212

2V–V

(C) 4π (P2–P1)(V2–V1) (D) π(V2 – V1)2

Q.5

CNCN

CNCN

→ ∆⊕ /OH3 (A);

vfHkfØ;k dk mRikn (A) gS -

(A)

O

O

O

(B)

O

O

O

(C) CO2H

CO2H (D)

O

O

O



Q.6

CH3 Ph

H

O

NH2 →NaOBr

)Major()A( ;

Product (A) of the reaction is -

(A)

CH3 Ph

H

NH2 (B)

CH3 Ph

NH2

H

(C)

CH3 Ph

NH2

H (D) CH3

O

H

C–OΘ||

Ph

Q.7 Which of the following sets of reagents, used in the order shown, would successfully accomplish the conversion shown ?

CCH3 || O

CHCH2CH2CH3

| CH3

(A) CH3CH2CH2MgBr; H3O+; PCC; CH2Cl2 (B) CH3CH2CH2MgBr; H3O+; H2SO4 heat, PCC,

CH2Cl2

(C) (C6H5)3P– CHCH2CH3, B2H6; CH3CO2H+ –

(D) (C6H5)3P– CHCH2CH3; H2O + –

Q.6

CH3

Ph

H

O

NH2 →NaOBr

)Major()A( ;

vfHkfØ;k dk mRikn (A) gS -

(A)

CH3

Ph

H

NH2 (B)

CH3 Ph

NH2

H

(C)

CH3

Ph

NH2

H (D) CH3

O

H

C–OΘ||

Ph

Q.7 fn[kk;s x;s vfHkdeZdks ds Øe dk fuEu esa ls dkSulk ;qXe n'kkZ;s x;s :ikUrj.k dks lQyrk iwoZd lEiUu djok;sxk ?

CCH3

||O

CHCH2CH2CH3

| CH3

(A) CH3CH2CH2MgBr; H3O+; PCC; CH2Cl2 (B) CH3CH2CH2MgBr; H3O+; H2SO4 heat, PCC,

CH2Cl2

(C) (C6H5)3P– CHCH2CH3, B2H6; CH3CO2H+ –

(D) (C6H5)3P– CHCH2CH3; H2O+ –



Q.8

OH

OH

(Resorcinol)

→ +NaOHI2

)Iodoform(3CHI + (A);

Compound (A) is

(A)

CH2

O

CO2Na ONa Θ ⊕

Θ ⊕

(B)

O

H–O–C–(CH2)2–C–OH || ||

O

(C)

CH3–CH2–CH2–CH2–C–OH || O

(D)

CH3–(CH2)3–C–ONa || O

Θ ⊕

Questions 9 to 12 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct. Mark your response in OMR sheet against the question number of that question. + 4 marks will be given for each correct answer and no negative marks.

Q.9 A sample of H2O2 solution labeled as "28

volume" has density of 26.5 g/L. Mark the correct

option (s) representing concentration of same

solution in other units -

(A) 22OHM = 2.5 (B) %

vw = 17

(C) Mole fraction of H2O2 = 0.2 (D)

22OHm = 13.88

Q.8

OH

OH

(fjlksflZukWy)

→ +NaOHI2

)Iodoform(3CHI + (A);

;kSfxd (A) gS -

(A)

CH2

O

CO2Na ONaΘ ⊕

Θ ⊕

(B)

O

H–O–C–(CH2)2–C–OH || ||

O

(C)

CH3–CH2–CH2–CH2–C–OH || O

(D)

CH3–(CH2)3–C–ONa || O

Θ ⊕

iz'u 9 ls 12 rd cgqfodYih iz'u gaSA izR;sd iz'u ds pkj fodYi (A), (B), (C) rFkk (D) gSa] ftuesa ls ,d ;k ,d ls vf/kd fodYi OMR 'khV esa iz'u dh iz'u la[;k ds le{k vius mÙkj vafdr dhft,A izR;sd lgh mÙkj ds fy, + 4 vad fn;s tk;saxs rFkk dksbZ _.kkRed vadu ugha

Q.9 H2O2 foy;u ds ,d uewus ij "28 vk;ru" vafdr gS

ftldk ?kuRo 26.5 g/L gSA vU; bdkb;ksa esa leku

foy;u dh iznf'kZr lkUnzrkvksa ds lgh fodYiks dk

p;u dhft,A -

(A) 22OHM = 2.5 (B) %

vw = 17

(C) H2O2 ds eksy fHkUu = 0.2 (D)

22OHm = 13.88



Q.10 Following represent the Maxwell distribution curve for an ideal gas at two temperature T1 and T2. Which of the following option (s) are true ?

2mpsU II

1mpsU I T1

T2 A2

dU2dU1

A1

u

dUdN

N1

×

A1= Area of small rectangle I A2 = Area of small rectangle II

(A) Total area under the two curves is independent of moles of gas

(B) If dU1 = 1mpsfU and dU2 = 2mpsfU then A1 = A2

(C) T1 > T2 and hence higher the temperature, sharper the curve

(D) The fraction of molecules having speed = Umps decreases as temperature increases

Q.11 Which of the following anions are more stable than benzyl anion ?

(A)

CH3

CH2 Θ

(B)

NO2

CH2 Θ

(C)

OH

CH2 Θ

(D)

CN

CH2 Θ

Q.10 nks rki T1 rFkk T2 ij ,d vkn'kZ xSl ds fy, eSDlosy forj.k oØ uhps iznf'kZr gSA fuEu esa ls dkSuls fodYi lR; gS ?

2mpsUII

1mpsU I T1

T2 A2

dU2dU1

A1

u

dUdN

N1

×

A1= NksVs vk;r I dk {ks=kQy A2 = NksVs vk;r II dk {ks=kQy

(A) nks oØks esa dqy {ks=kQy] xSl ds eksyks ls Lora=k gksrk gS

(B) ;fn dU1 = 1mpsfU rFkk dU2 = 2mpsfU rc A1 = A2

(C) T1 > T2 rFkk rki mPp gksus ij oØ uqdhyk gksrk gS (D) pky = Umps ;qDr v.kqvksa ds fHkUu] rki c<+us ij

?kVrs gS

Q.11 fuEu esa ls dkSuls _.kk;u] csUtkby _.kk;u dh vis{kk vf/kd LFkkbZ gS ?

(A)

CH3

CH2Θ

(B)

NO2

CH2Θ

(C)

OH

CH2Θ

(D)

CN

CH2Θ



Q.12 Which of the following are more reactive than

diphenyl in electrophilic aromatic substitution

reaction ?

(A)

CH3

O

N |

H

(B) CH3

O

O

(C)

CH3

O

O (D)

OR

Section - II This section contains 2 questions. Each question has four statements (A, B, C and D) given in Column-I and five statements (P, Q, R, S and T) in Column-II. Any given statement in Column–I can have correct matching with One or More statement(s) given in Column II. For example, if for a given question, statement B matches with the statements given in Q and R, then for the particular question, against statement B, darken the bubbles corresponding to Q and R in the OMR. +8 marks will be given for each correct answer (i.e. +2 marks for each correct row) and no negative marking for each wrong answer.

Q.12 bysDVªkWuLusgh ,sjksesfVd izfrLFkkiu vfHkfØ;k esa fuEu

esa ls dkSulk MkbZQsfuy vf/kd fØ;k'khy gS ?

(A)

CH3

O

N|

H

(B) CH3

O

O

(C)

CH3

O

O (D)

OR

[k.M – II

bl [k.M esa 2 iz'u ( 1 ,oa 2) gSaA izR;sd iz'u esa pkj dFku (A, B, C, D) -I ik¡p dFku (P, Q, R, S,T)

-II esa fn;s x;s gSaA LrEHk-I esa fn, x, dFkuksa dks LrEHk-II esa fn, x, dFkuksa esa ls ,d ;k ,d ls vf/kd dFkuksa ls lqesy djuk gSA mnkgj.k ds fy, ;fn fn, x, iz'u esa dFku B; Q rFkk R esa fn, x, dFkuksa ls lqesy djuk gS] rks fof'k"V iz'u ds fy, dFku B ds le{k OMR esa Q rFkk R ds laxr fn, x, xksys dks dkyk djuk gSA izR;sd lgh mÙkj ds fy;s +8 vad fn;s tk,saxs (vFkkZr~ izR;sd lgh iafDr feyku ds fy, +2 vad fn, tk,asxs) _.kkRed vadu ugha



Q.1 Match the column : Column -I Column-II (A)C(s, graphite) + O2(g) → CO2(g)

(P) °∆ combustionH

(B)C(s, graphite) → C(g) (Q) °∆ formationH

(C)CO(g)+21 O2(g) →

CO2(g)

(R) °∆ natomizatioH

(D)CH4(g) → C(g) + 4H(g)

(S) °∆ ationlimsubH

(T) °∆ tionneutralisaH

Q.1 LrEHk lqesfyr dhft,A LrEHk-I LrEHk-II (A)C(s, xszQkbV) + O2(g) → CO2(g)

(P) °∆ combustionH

(B)C(s, xszQkbV) → C(g) (Q) °∆ formationH

(C)CO(g)+21 O2(g) →

CO2(g)

(R) °∆ natomizatioH

(D)CH4(g) → C(g) + 4H(g)

(S) °∆ ationlimsubH

(T) °∆ tionneutralisaH



Q.2 Match the column :

Column –I (Reaction)

Column-II (Type of reaction)

(A)

OR

C–O–CH2–CH3 || O

∆

(P) Pyrolysis of xanthate

(B)

OR

N–Me |

Me

| OΘ

⊕

→∆

(Q) Pyrolysis of ester

(C)

CH3–CH2–O–C–S–Me || O

→

∆

(R) Cope reaction

(D) Cl

EtONaEtOH →

(S)E2 (elimination

bimolecular )

(T) Hofmann

amminolysis

Q.2 LrEHk lqesfyr dhft,A

LrEHk–I (vfHkfØ;k)

LrEHk II (vfHkfØ;k dk

izdkj)

(A)

OR

C–O–CH2–CH3 ||O

∆

(P) tSUFksV dk rki vi?kVu

(B)

OR

N–Me|

Me

|OΘ

⊕

→∆

(Q) ,LVj dk rki vi?kVu

(C)

CH3–CH2–O–C–S–Me || O

→

∆

(R) dkWih vfHkfØ;k

(D) Cl

EtONaEtOH →

(S) E2 (f}v.kqd

foyksiu )

(T) gkWQeku

veksuhvi?kV~u



Section – III This section contains 6 questions (Q.1 to 6). +4 marks will be awarded for each correct answer and no negativemarking for wrong answer. The answer to each questionis a single-digit integer, ranging from 0 to 9. The bubblecorresponding to the correct answer is to be darkened inthe OMR. Q.1 Consider the following reaction Na3PO4 + (NH4)2 MoO4 + HNO3(dil.) → 'X' (canary yellow ppt) Then calculate total number of atoms of 15th

group elements which are sp3 hybridized incompound 'X'.

Q.2

Water P + Q

P + R↑ + S

Na2O2

0°C

Water25°C

Calculate sum of bond order between samebonded atoms in Q and R compounds.

Q.3 The standard reduction potential of normal

calomel electrode and reduction potential ofsaturated calomel electrodes are 0.27 & 0.33 voltrespectively. What is the concentration of Cl– in saturated solution of KCl in multiple of 10 ?

[k.M - III

bl [k.M esa 6 (iz-1 ls 6) iz'u gaSA izR;sd lgh mÙkj ds fy;s +4 vad fn;s tk,saxs rFkk dksbZ _.kkRed vadu ugha gSA bl [k.M esa izR;sd iz'u dk mÙkj 0 ls 9 rd bdkbZ ds ,d iw.kk±d

OMR esa iz'u la[;k ds laxr uhps fn;s x;s xksyksaa esa ls lgh mÙkj okys xksys dks dkyk fd;k tkuk gSA

Q.1 fuEu vfHkfØ;k ij fopkj dhft,A Na3PO4 + (NH4)2 MoO4 + HNO3(dil.) → 'X' (canary yellow ppt) 15th lewg rRoksa ds mu ijek.kqvksa dh dqy la[;k dh

x.kuk dhft, tks ;kSfxd 'X' esa sp3 ladfjr gksrs gS

Q.2

Water P + Q

P + R↑ + S

Na2O2

0°C

Water25°C

Q rFkk R ;kSfxdks esa leku caf/kr ijek.kqvksa ds e/; ca/k Øe ds ;ksx dh x.kuk dhft,A

Q.3 lekU; dsyksey bysDVªkWM ds ekud vip;u foHko

rFkk larIr dsyksey bysDVªkWM ds vip;u foHko Øe'k% 0.27 rFkk 0.33 volt gSA KCl ds larIr foy;u esa Cl– dh lkUnzrk] 10 ds xq.kt esa D;k gksxh ?



Q.4 How many of the following sequence of

reactions can give

Br

COOH

(a) ClCH

AlCl3

3 →

HO/KMnOH

4Θ

⊕→ 2

3

BrFeBr →

(b) ClCH

AlCl3

3 → 2

3

BrFeBr →

HO/KMnOH

4Θ

⊕→

(c)

CH3

NO2

2BrFe→ → C–Pd,H2

∆ →

,POH).2(

HNO).1(

23

2

HO

KMnO4Θ

→

(d) Ph MgBr OH/H

CO

2

2⊕

→3

2

AlBr

Br→

Q.5 How many moles of acetone from in the reductive

ozonolysis of one mole of hexa–2, 4–diene.

Q.4 fuEu esa ls fdruh vfHkfØ;kvksa dk vuqØe ;kSfxd

Br

COOH

ns ldrk gS

(a) ClCH

AlCl3

3 →

HO/KMnOH

4Θ

⊕→ 2

3

BrFeBr →

(b) ClCH

AlCl3

3 → 2

3

BrFeBr →

HO/KMnOH

4Θ

⊕→

(c)

CH3

NO2

2BrFe→ → C–Pd,H2

∆ →

,POH).2(

HNO).1(

23

2

HO

KMnO4Θ

→

(d) Ph MgBr OH/H

CO

2

2⊕

→3

2

AlBr

Br→

Q.5 gSDlk–2, 4–MkbZbu ds ,d eksy ds vipk;h vkstksuh

vi?kVu esa ,flVksu ds fdrus eksy curs gS :



Q.6 How many of the following compound are more

basic than NH2 ?

(a) CH3–CH2– 2HN••

(b)CH3–C–NH2

|| NH

(c)

C–NH2 || O

(d) CH2 = CH– 2HN••

Q.6 fuEu esa ls fdrus ;kSfxd, NH2

dh vis{kk

vf/kd {kkjh; gS ?

(a) CH3–CH2– 2HN••

(b)CH3–C–NH2||NH

(c)

C–NH2 ||O

(d) CH2 = CH– 2HN••



MATHEMATICS

Section - I Questions 1 to 8 are multiple choice questions. Eachquestion has four choices (A), (B), (C) and (D), out ofwhich ONLY ONE is correct. Mark your response inOMR sheet against the question number of that question.+ 3 marks will be given for each correct answer and – 1 mark for each wrong answer.

Q.1 Let ABC be a triangle. Let A be the point (1, 2),y = x is the perpendicular bisector of AB andx – 2y + 1 = 0 is the angle bisector of ∠C. If equation of BC is given by ax + by – 5 = 0, thenthe value of a + b is-

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

Q.2 Maximum value of x2lnx1 is -

(A) 2e (B) e (C) e1 (D)

e21

Q.3 The coefficient of x4 in the expansion(1 + 5x + 9x2 +…….) (1 + x2)11 is -

(A) 11C2 + 4 11C1 + 3 (B) 11C2 + 3 11C1 + 4 (C) 3 11C2 + 4 11C1 + 3 (D) 171

[k.M - I iz'u 1 ls 8 rd cgqfodYih iz'u gSaA izR;sd iz'u ds pkj

fodYi (A), (B), (C) rFkk (D) gSa, ftuesa ls dsoy ,d fodYi

lgh OMR 'khV esa iz'u dh iz'u la[;k ds le{k viuk

mÙkj vafdr dhft;sA izR;sd lgh mÙkj ds fy, + 3 vad fn;s

tk;asxs rFkk izR;sd xyr mÙkj ds fy, 1 ?kVk;k tk;sxkA

Q.1 ekuk ABC ,d f=kHkqt gS rFkk A fcUnq (1, 2) gS y = x; AB dk yEc v/kZd gS rFkk x – 2y + 1 = 0 ; ∠C dk

dks.kk)Zd gSA ;fn BC dk lehdj.k ax + by – 5 = 0 gS] rks a + b dk eku gS-

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

Q.2 x2lnx1 dk vf/kdre eku g -

(A) 2e (B) e (C) e1 (D)

e21

Q.3 (1 + 5x + 9x2 +…….) (1 + x2)11 ds izlkj esa x4 dk

xq.kkad gS - (A) 11C2 + 4 11C1 + 3 (B) 11C2 + 3 11C1 + 4 (C) 3 11C2 + 4 11C1 + 3 (D) 171



Q.4 ekuk P ,oa Q ijoy; y2 = 4a(x – b) ij fLFkr fcUnq(4, – 4) ,oa (9, 6) gSA ekuk R; P ,oa Q ds e/; ijoy; ds pki ij fLFkr fcUnq gS] rks ∆PRQ dk {ks=kQy vf/kdre gksxk tc-

(A) ∠PRQ = 90° (B) fcUnq R (4, 4) gS

(C) fcUnq R

1,

41 gS (D) buesa ls dksbZ ugha

Q.5 ;fn y = e–x cos x rFkk y4 + ky = 0 tgk¡

y4 = 4

4

dxyd gS] rks k =

(A) 4 (B) – 4 (C) 2 (D) –2 Q.6 ;fn m rFkk n (n > m) /kukRed iw.kkZad gSa] rks varjky

[0, 2π] esa lehdj.k n|sin x| = m |cos x| dh la[;k gS- (A) m (B) n (C) mn (D) buesa ls dksbZ ugha Q.7 ekuk Qyu f(x) tks fuEu izdkj ifjHkkf"kr gS

f(x) = 1x

1x0;x2;xsin 21

≥<<

+λ−

; f(x), x = 1 ij

LFkkuh; fufEu"B j[krk gS ;fn x dk eku fuEu varjky esa fLFkr gS

(A) [sin 1, 1] (B) (– sin 1, 1) (C) (sin 1, 1) (D) [0, sin 1]

Q.8 2

2

2

dxdy

dxydy

=⋅ dk O;kid gy gS-

(A) y = c1x + c2 (B) y = xc1

2ec

(C) y = c1x + c2ex (D) y = xcxc 21 ee +

Q.4 Let P and Q be points (4, – 4) and (9, 6) of theparabola y2 = 4a(x – b). Let R be a point on thearc of the parabola between P and Q. Then thearea of ∆PRQ is largest when -

(A) ∠PRQ = 90° (B) the point R is (4, 4)

(C) The point R is

1,

41 (D) None of these

Q.5 If y = e–x cos x and y4 + ky = 0, where

y4 = 4

4

dxyd , then k =

(A) 4 (B) – 4 (C) 2 (D) –2

Q.6 If m and n (n > m) are positive integers, thennumber of the equation n |sin x| = m |cos x| in[0, 2π] is

(A) m (B) n (C) mn (D) none of these

Q.7 Let a function f(x) be defined as

f(x) = 1x

1x0;x2;xsin 21

≥<<

+λ−

; f(x) can have

local minimum at x = 1 if the value of x lies inthe interval

(A) [sin 1, 1] (B) (– sin 1, 1) (C) (sin 1, 1) (D) [0, sin 1]

Q.8 The general solution of 2

2

2

dxdy

dxydy

=⋅ is

(A) y = c1x + c2 (B) y = xc1

2ec

(C)y = c1x + c2ex (D) y = xcxc 21 ee +



iz'u 9 ls 12 rd cgqfodYih iz'u gaSA izR;sd iz'u ds pkj fodYi (A), (B), (C) rFkk (D) gSa] ftuesa ls ,d ;k ,d ls vf/kd OMR 'khV esa iz'u dh iz'u la[;k dsle{k vius mÙkj vafdr dhft,A izR;sd lgh mÙkj ds fy,+4 vad fn;s tk;saxs rFkk izR;sd xyr mÙkj ds fy, dksbZ _.kkRed vadu ugha

Q.9 →→b,a →c rhu bdkbZ lfn'k gSa tks ,d nwljs ls

dks.k α ij leku >qds gS] rks →a ,oa→b o →c ds lery

ds e/; dks.k gS -

(A) θ = cos–1

αα

2cos

cos (B) θ = sin–1

αα

2cos

cos

(C) θ = cos–1

α

α

sin2

sin(D) θ = sin–1

α

α

sin2

sin

.Q.10 ;fn f(2–x) = f(2 + x) rFkk f(4 – x) = f(4 + x) gS ,oa

f(x) Qyu gS ftlesa ∫2

0

)x(f dx = 5 gS] rks ∫50

0

)x(f dx

dk eku gS -

(A) 125 (B) ∫46

4–

dx)x(f

(C) ∫51

1

dx)x(f (D) ∫52

2

dx)x(f

Questions 9 to 12 are multiple choice questions. Eachquestion has four choices (A), (B), (C) and (D), out ofwhich MULTIPLE (ONE OR MORE) may be correct.Mark your response in OMR sheet against the questionnumber of that question. +4 marks will be given for eachcorrect answer and no negative marks for wrong answer.

Q.9 If →→b,a and →c are three unit vectors equally

inclined to each other at an angle α, then the

angle between →a and plane of →b and →c is -

(A) θ = cos–1

αα

2cos

cos (B) θ = sin–1

αα

2cos

cos

(C) θ = cos–1

α

α

sin2

sin(D) θ = sin–1

α

α

sin2

sin

Q.10 If f(2–x) = f(2 + x) and f(4 – x) = f(4 + x) and

f(x) is function of which ∫2

0

)x(f dx = 5, then

∫50

0

)x(f dx is equal to -

(A) 125 (B) ∫46

4–

dx)x(f

(C) ∫51

1

dx)x(f (D) ∫52

2

dx)x(f



Q.11 Let ur be the number of ways in which r personscan form a line x be the number of ways in whichA1 is at third rank and y be the number of ways inwhich A1 is not at third rank then

(A) x + y = ur (B) x = y (C) x + y = r! (D) y ≥ x if n ≥ 3

Q.12 ln x + x1 is not less than

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

Section - II

This section contains 2 questions. Each question has four statements (A, B, C and D) given in Column-I and five statements (P, Q, R, S and T) in Column-II. Any givenstatement in Column–I can have correct matching withOne or More statement(s) given in Column II. Forexample, if for a given question, statement B matcheswith the statements given in Q and R, then for theparticular question, against statement B, darken thebubbles corresponding to Q and R in the OMR. +8 markswill be given for each correct answer (i.e. +2 marks foreach correct row) and no negative marking for each wrong answer.

Q.11 ekuk ur mu rjhdksa dh la[;k gS ftuesa r O;fDr ,d

js[kk fufeZr djrs gS rFkk x mu rjhdksa dh la[;k gS

ftuesa A1 rhljh jSad ij gS rFkk y mu rjhdks dh

la[;k gS ftuesa A1 rhljh jSad ij ugh gS] rc (A) x + y = ur (B) x = y (C) x + y = r! (D) y ≥ x ;fn n ≥ 3

Q.12 ln x + x1 fuEu ls NksVk ugha gksxk

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

[k.M – II

bl [k.M esa 2 iz'u (iz'u 1 ,oa 2) gSaA izR;sd iz'u esa pkj dFku

(A, B, C, D) LrEHk-I esa rFkk ik¡p dFku (P, Q, R, S,T) -II

-I -II ,d ;k ,d ls vf/kd dFkuksa ls lqesy djuk gSA

mnkgj.k ds fy, ;fn fn, x, iz'u esa dFku B; Q rFkk R esa fn, x, dFkuksa ls lqesy djuk gS] rks fof'k"V iz'u ds fy, dFku Bds le{k OMR esa Q rFkk R ds laxr fn, x, xksys dks dkyk

djuk gSA izR;sd lgh mÙkj ds fy;s +8 vad fn;s tk,saxs (vFkkZr~

izR;sd lgh iafDr feyku ds fy, +2 vad fn, tk,asxs)

_.kkRed vadu ugha



Q.1 Match the following. Column-I Column-II (A) If x is real, then the greatest (P) 3 and least value of the

expression 6x3x2

2x2 ++

+ are

(B) If a + b = 1; a > 0, b > 0, (Q) 1/3 Then the minimum value of

+

+

b11

a11 is

(C) The maximum value (R) 5 Attained by y = 10–|x–10|,

–1 ≤ x ≤ 3, is

(D) If P(t2, 2t), t ∈ [0, 2] is an (S) –1/13 arbitrary point on parabola y2 = 4x. Q is foot of perpendicular from focus S on the tangent at, (T) –3 then maximum area of triangle PQS is

Q.1 LrEHk feyku dhft, LrEHk-I LrEHk-II

(A) x (P) 3

6x3x2

2x2 ++

+

U;wure eku gS

(B) ;fn a + b = 1; a > 0, b > 0, (Q) 1/3

gS] rks

+

+

b11

a11 dk

U;wure eku gS

(C) y = 10–|x–10| , –1 ≤ x ≤ 3 (R) 5 }kjk xzg.k fd;k x;k vf/kdre

eku gksxk

(D) ;fn P(t2, 2t), t ∈ [0, 2] ijoy; (S) –1/13 y2 = 4x ij LoSPN fcUnq gS rFkk bl ij [khaph xbZ Li'kZ js[kk ij ukfHk S ls Mkys x;s yEc dk ikn Q gS] rks f=kHkqt PQS dk (T) –3 vf/kdre {ks=kQy gS



Q.2 Match the following. Column-I Column-II (A) If set of all possible values of k (P) 3

for which every solution of the

inequation

x2 – (3k – 1) x + 2k2 – 3k – 2 ≥ 0

is also a solution of the inequation

x2 – 1 ≥ 0 is [l , m], then l + m is

equal to

(B) If a, b, c and d are four positive (Q) 1

real numbers such that abcd = 1

and minimum value of

(1 + a)(1 + b)(1 + c)(1 + d)

is 16λ, then λ is equal to

(C) If solution set of the inequality (R) 2

5x+2 > x/1

251

is (l, ∞), then l is

equal to

(D) Let f(x) = x3 + 3x + 1. If g(x) (S) 0

is the inverse function of f(x) and

g′(5) = 6λ, then λ is equal to (T) 4

Q.2 -I –II

(A) ;fn k ds lHkh lEHko ekuksa dk (P) 3

leqPp; ftlds fy, vlfedk

x2 – (3k – 1) x + 2k2 – 3k – 2 ≥ 0

dk izR;sd gy vlfedk x2 – 1 ≥ 0

dk Hkh gy gS [l , m] gS] rks l + m

dk eku gS

(B) ;fn pkj /kukRed okLRfod la[;k,¡ (Q) 1

a, b, c ,oa d bl izdkj gS fd abcd = 1

rFkk (1 + a)(1 + b)(1 + c)(1 + d)

dk U;wure eku 16λ gS] rks λ dk eku gS

(C) ;fn vlfedk 5x+2 > x/1

251

dk (R) 2

gy leqPp; (l, ∞) gS] rks l dk eku gS

(D) ekuk f(x) = x3 + 3x + 1 ;fn g(x); (S) 0

f(x) dk izfrykse Qyu gS rFkk

g′(5) = 6λ gS] rks λ dk eku gS (T) 4



Section - III This section contains 6 questions (Q.1 to 6). The answerto each of the questions is a Single-digit integer, ranging from 0 to 9. The bubble corresponding to the correctanswer is to be darkened in the OMR. +4 marks will begiven for each correct answer and no negative markingfor each wrong answer.

Q.1 If f(x) =

≤<+π

≤≤++π

2x1,xtanxcos2

1x0,baxax

]xtan[

1–

32

2

is

differentiable in [0, 2], then a = 1k

1 and b = 4π –

2k26 .

Find 90

kk 22

21 + {where [.] denotes greatest

integer function}.

Q.2 Let the curve y = f(x) passes through (4, –2) satisfy the differential equation,

y (x +y3) dx = x(y3–x) dy and

y = g(x) = ∫ ∫+xsin

8/1

xcos

8/1

1–1–

2 2

dttcosdttsin ,

0≤ x ≤ π/2.Then if the area of the region bounded by curves. y = f(x), y = g(x) and x = 0 is

81

λ

π

163 , Find λ.

[k.M – III bl [k.M esa 6 (iz-1 ls 6) iz'u gaSA bl [k.M esa izR;sd iz'u dkmÙkj 0 ls 9 rd bdkbZ ds ,d iw.kk±d OMR la[;k ds laxr uhps fn;s x;s xksyksaa esa ls lgh mÙkj okys xksysdks dkyk fd;k tkuk gSA izR;sd lgh mÙkj ds fy;s +4 vad fn;s tk,saxs rFkk izR;sd xyr mÙkj ds fy, dksbZ _.kkRed vadu ugha

Q.1 f(x) =

≤<+π

≤≤++π

2x1,xtanxcos2

1x0,baxax

]xtan[

1–

32

2

[0, 2] esa vodyuh; gS] rks a = 1k

1 ,oa

b = 4π –

2k26 gS] rks

90kk 2

22

1 + dk eku dhft, {tgk¡

[.] egÙke iw.kkZad Qyu dks iznf'kZr djrk gS}

Q.2 ekuk oØ y =f(x) fcUnq (4, –2) ls gksdj xqtjrk gS rFkk

vody lehdj.k y (x +y3) dx = x(y3–x) dy dks larq"V djrk gS rFkk

y = g(x) = ∫ ∫+xsin

8/1

xcos

8/1

1–1–

2 2

dttcosdttsin ,

0≤ x ≤ π/2 gS] rc ;fn oØksa y = f(x), y = g(x) rFkk

x = 0 }kjk ifjc} {ks=k dk {ks=kQy 81 λ

π

163

gS] rks

λ Kkr dhft,A



Q.3 If |z – 1| + |z + 3| ≤ 8, then the range of values of|z – 4| is [x1, x2], then find the value of x1.x2.

Q.4 Let α be the angle in radians between 4y

36x 22

+ =1

and the circle x2 + y2 = 12 at their points of

intersection. If α = tan–1

32k , then find the value

of 4

k 2

Q.5 The length of x intercept of common tangent tothe curves y2 = 8x and xy = – 1 is equal to.

Q.6 Family of lines x(a + b) + y = 1 where a and b are

the roots of the equation x3 – 3x2 + x + 1 = 0 and [a + b] = 1 (where [.] denotes the greatest integerfunction), such that it intercept a triangle of areaA with co-ordinate axes. Find the maximumvalue of 2A.

Q.3 ;fn |z – 1| + |z + 3| ≤ 8 gS] rks |z – 4| ds ekuksa dk

ifjlj [x1, x2] gS] rks x1.x2 dk eku Kkr dhft,A

Q.4 ekuk 4y

36x 22

+ =1 rFkk oÙk x2 + y2 = 12 ds e/; buds

izfrPNsn fcUnq ij dks.k α (jsfM;u esa) gSA ;fn

α = tan–1

32k gS] rks

4k 2

dk eku Kkr dhft,A

Q.5 oØksa y2 = 8x rFkk xy = – 1 dh mHk;fu"B Li'kZ js[kk

ds x vUr% [k.M dh yEckbZ Kkr dhft,A

Q.6 js[kkvksa dk fudk; x(a + b) + y = 1, tgk¡ a ,oa b

lehdj.k x3 – 3x2 + x + 1 = 0 ds ewy gS rFkk

[a + b] = 1 gS (tgk¡ [.] egÙke iw.kkZad Qyu dks

iznf'kZr djrk gS) bl izdkj gS fd bldk funsZ'kh v{kksa

ds lkFk vUr% [k.M {ks=kQy A dk f=kHkqt gS] rks 2A

dk vf/kdre eku Kkr dhft,A



PHYSICS

Section - I Questions 1 to 8 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. + 3 marks will be given for each correct answer and – 1 mark for each wrong answer.

Q.1 A satellite of mass ms revolving in a circular orbit of radius rs round the earth of mass M has a total energy E. Then its angular momentum will be -

(A) (2 Emsrs2)1/2 (B) (2 Emsrs

2) (C) (2 Emsrs)1/2 (D) (2 Emsrs)

[k.M - I 1 ls 8 rd cgqfodYih iz'u gSaA izR;sd iz'u ds pkj fodYi

(A), (B), (C) rFkk (D) gSa, ftuesa ls dsoy ,d fodYi lghOMR

+ 3 vad

1 vad ?kVk;k tk;sxkA

Q.1 ms nzO;eku dk ,d mixzg dqy ÅtkZ E rFkk nzO;eku

M dh iFoh ds pkjksa vksj rs f=kT;k dh oÙkkdkj d{kk

esa pôj yxk jgk gSA rc bldk dks.kh; laosx gksxk - (A) (2 Emsrs

2)1/2 (B) (2 Emsrs2)

(C) (2 Emsrs)1/2 (D) (2 Emsrs)



Q.2 ,d f}ijek.oh; vkn'kZ xSl fp=kkuqlkj P-V fp=k ds

vuqlkj Å"ekxfrd ifjorZu ds vUrxZr gSA xSl dks

nh xbZ dqy Å"ek yxHkx gS - (ln 2 = 0.7 mi;ksx djs) P2P0

P0

V0

A

B

C

V 2V0

Isothermal

(A) 2.5 P0V0 (B) 1.4 P0V0

(C) 3.9 P0V0 (D) 1.1 P0V0

Q.3 leku inkFkZ dh cuh 12 le:i NMsaa ?ku ds :i esa

O;ofLFkr dh xbZ gSA P rFkk R dk rki 90ºC rFkk

30ºC ij O;ofLFkr gSA rc V fcUnq dk rki] tc ;g

LFkkbZ voLFkk rd igq¡prk gS -

90ºC

30ºCS R

V

a QUT

P

W

(A) 65ºC (B) 60ºC (C) 20ºC (D) 50ºC

Q.2 A diatomic ideal gas undergoes a thermodynamic change according to the P-V diagram shown in the figure. The total heat given to the gas is nearly (use ln 2 = 0.7) -

P 2P0

P0

V0

A

B

C

V 2V0

Isothermal

(A) 2.5 P0V0 (B) 1.4 P0V0

(C) 3.9 P0V0 (D) 1.1 P0V0

Q.3 12 identical rods made of same material are arranged in the form of a cube. The temperature of P and R are maintioned at 90ºC and 30ºC respectively. Then the temperature of point V, when steady state is reached –

90ºC

30ºCS R

V

a QUT

P

W

(A) 65ºC (B) 60ºC (C) 20ºC (D) 50ºC



Q.4 1.5 viorZukad ds irys voryksÙky ysal dh mÙky

lrg n'kkZ, vuqlkj ikWfy'k dh xbZ gSA ,d NksVh

oLrq ysal ds ck¡;h 30 cm ij blds eq[; v{k ij ok;q

esa j[kk gSA vafre izfrfcEc dh nwjh gS - r1 = 60 cm

r2 = 20 cm

(A) 20 cm (B) 30 cm (C) 10 cm (D) 15 cm

Q.5 ,d ifjiFk esa mRikfnr Å"ek Q = I2Rt }kjk nh xbZ gS, tgk¡ I /kkjk] R izfrjks/k rFkk t le; gSA ;fn I, R o t esa izsf{kr izfr'kr =kqfV;k¡ Øe'k% 2%, 1% rFkk 1% gS, rc Å"ek esa izsf{kr vf/kdre =kqfV gksxh -

(A) 2% (B) 3% (C) 4% (D) 6%

Q. 6 ,d fi.M ,d ehukj ds Åijh fljs ij A ls Å/okZ/kj

Åij dh vksj QSadk x;k gSA ;g t1 le; esa /kjkry ij igq¡prk gSA ;fn bls blh pky ls A ls Å/okZ/kj uhps dh vksj QSadk tk,] rks ;g t2 le; esa /kjkry ij igq¡prk gSA mls A ls eqä :i ls fxjus fn;k tkrk gS rc /kjkry rd igq¡pus esa blds }kjk fy;k x;k le; -

(A) t = 2

21 tt + (B) t = 2

21 tt −

(C) t = 21tt (D) t =2

1

tt

Q.4 Convex surface of thin concavo-convex lens of refractive index 1.5 is silvered as shown. A small object is kept in air at 30 cm left of the lens on its principal axis. The distance of the final image is -

r1 = 60 cm

r2 = 20 cm

(A) 20 cm (B) 30 cm (C) 10 cm (D) 15 cm

Q.5 The heat generated in a circuit is given by Q = I2Rt, where I is current, R is resistance and t is time. If the percentage errors in measuring I, R and t are 2%, 1% and 1%, respectively, then the maximum error in measuring heat will be -

(A) 2% (B) 3% (C) 4% (D) 6%

Q. 6 A body is thrown vertically upwards from A, the

top of a tower. It reaches the ground in time t1. If it is thrown vertically downwards from A with the same speed, it reaches the ground in time t2. It is allowed to fall freely from A, then the time it takes to reach the ground is given by -

(A) t = 2

21 tt + (B) t = 2

21 tt −

(C) t = 21tt (D) t =2

1

tt



Q.7 In an arrangement shown in the figure, the acceleration of block A and B are given –

m

Bα=30º m

A

(A) g/3, g/6 (B) g/6, g/3 (C) g/2, g/2 (D) 0, 0

Q.8 A uniform rope of linear mass density λ and length l is coiled on a smooth horizontal surface. One end is pulled up with constant velocity v. Then the average power applied by the external agent in pulling the entire rope just off the ground is -

v

(A) 21

λlv2 + 2

2 glλ (B) λlgv

(C) 21

λv3 + 2lvgλ (D) λlgv +

21

λv3

Q.7 fp=k esa n'kkZbZ O;oLFkk esa A rFkk B ds Roj.k fn;s x;s

gS –

m

Bα=30º m

A

(A) g/3, g/6 (B) g/6, g/3 (C) g/2, g/2 (D) 0, 0

Q.8 λ js[kh; nzO;eku ?kuRo rFkk l yEckbZ dh ,dleku

jLlh fpdus {kSfrt ry ij dq.Mfyr gSA ,d fljs dks

fu;r osx v ls [khapk tkrk gSA rc cká dkjd }kjk

lewph jLlh dks tehu ls nwj [khapus esa yxk;h x;h

vkSlr 'kfä gS - v

(A) 21

λlv2 + 2

2 glλ (B) λlgv

(C) 21

λv3 + 2lvgλ (D) λlgv +

21

λv3



Questions 9 to 12 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) may be correct. Mark your response in OMR sheet against the question number of that question. +4 marks will be given for each correct answer and no negative marks for wrong answer. Q.9 There are three wire MO, NO and PQ, wires MO

and NO are fixed and perpendicular to each other. Wire PQ moves with a constant velocity v as shown in the figure and resistance per unit length of each wire is λ and magnetic field exists perpendicular and inside the paper then –

Q

B

NO

P

M

(A) current in loop is anticlockwise

(B) magnitude of current in the loop is)12( +λ

Bv

(C) current in the loop is independent of time (D) magnitude of current decreases as time

increases

iz'u 9 ls 12 rd cgqfodYih iz'u gaSA izR;sd iz'u ds pkj fodYi (A), (B), (C) rFkk (D) gSa] ftuesa ls ,d ;k ,d ls vf/kd OMR 'khV esa iz'u dh iz'u la[;k ds le{k vius mÙkj vafdr dhft,A izR;sd lgh mÙkj ds fy, +4 vad fn;s tk;saxs rFkk izR;sd xyr mÙkj ds fy, dksbZ _.kkRed vadu ugha Q.9 MO, NO PQ MO NO

ds yEcor~ gSA rkj PQ fp=k esa n'kkZ, vuqlkj fu;r osx v ls xfr djrk gS rFkk izR;sd rkj dh izfr bdkbZ yEckbZ izfrjks/k λ gS rFkk pqEcdh; {ks=k yEcor~ rFkk dkxt ds ry ds vUnj dh vksj fo|eku gS] rc -

Q

B

NO

P

M

(A) ywi esa /kkjk okekorZ gS

(B) ywi esa /kkjk dk eku )12( +λ

BvgS

(C) ywi esa /kkjk le; ls Lora=k gS (D) le; c<+us ds lkFk /kkjk dk eku ?kVrk gS



Q.10 In the circuit shown in the figure,if both the bulbs B1 and B2 are identical -

~

C = 500µF B1

B2 L = 10mH

320 V, 50 Hz (A) Their brightness will be the same (B) B2 will be brighter than B1 (C) As frequency of supply voltage is increased

brightness of B1 will increase and that of B2 will decreases

(D) Only B2 will glow because the capacitor has infinite impedance

Q.11 When a body is moving up with constant velocity -

(A) work done by lifting force is positive (B) work done by lifting force is negative (C) work done by force of gravity is positive (D) work done by force of gravity is negative

Q.10 fp=k esa n'kkZ, ifjiFk esa] ;fn B1 rFkk B2 nksuksa cYc leku gS -

~

C = 500µF B1

B2 L = 10mH

320 V, 50 Hz (A) budh nhIrrk leku gksxh (B) B2, B1 ls vf/kd nhIr gksxk (C) lIykbZ oksYVrk dh vkofr c<+us ij B1 dh

nhIrrk c<+rh gS rFkk B2 dh ?kVrh gS (D) dsoy B2 nhIr gksxk D;ksafd la/kkfj=k vuUr

izfrck/kk j[krk gS Q.11 tc ,d fi.M fu;r osx ls Åij xfr dj jgk gS -

(A) Åij mBkus okys cy }kjk fd;k x;k dk;Z /kukRed gS

(B) Åij mBkus okys cy }kjk fd;k x;k dk;Z _.kkRed gS

(C) xq:Roh; cy }kjk fd;k x;k dk;Z /kukRed gS (D) xq:Roh; cy }kjk fd;k x;k dk;Z _.kkRed gS



Q.12 Two inclined frictionless tracks of different inclinations meet at A from where two blocks P and Q of different masses are allowed to slide down from rest at the same time, one on each track, as shown in the figure. Then –

B θ1 Cθ2

AP Q

(A) both blocks will reach the bottom at the same time (B) block Q will reach the bottom earlier than

block P (C) both blocks will reach the bottom with the

same speed (D) block Q will reach the bottom with a higher

speed than block P

Section - II This section contains 2 questions. Each question has four statements (A, B, C and D) given in Column-I and five statements (P, Q, R, S and T) in Column-II. Any given statement in Column–I can have correct matching with One or More statement(s) given in Column II. For example, if for a given question, statement B matches with the statements given in Q and R, then for the particular question, against statement B, darken the bubbles corresponding to Q and R in the OMR. +8 marks will be given for each correct answer and no negative marking for each wrong answer.

Q.12 fHkUu vkur dh nks ?k"kZ.kjfgr vkur iFk A fcUnq ij feyrh gS tgk¡ ls fHkUu nzO;eku ds nks CykWd P rFkk Q leku le; ij fojke ls fp=kkuqlkj izR;sd iFk ij uhps fQlyus ds fy, Lora=k gS] rc -

B θ1 Cθ2

AP Q

(A) nksuksa CykWd ry ij leku le; ij igq¡psxsa

(B) CykWd Q, CykWd P ls igys ry ij igq¡psxk

(C) nksuksa CykWd leku pky ls ry ij igq¡psxsa

(D) CykWd Q, CykWd P ls mPp pky ls ry ij igq¡psxk

[k.M – II 2 iz'u ( 1 2) pkj

dFku (A, B, C, D) -I ik¡p dFku (P, Q, R, S,T) -II -I -II

,d ;k ,d ls vf/kd

B; Q R

B OMR Q R +8 vad

_.kkRed vadu ugha



Q.1 A square loop of uniform conducting wire is as shown in figure. A current-I (in amp.) enters the loop from one end and leaves the loop from opposite end as shown in figure. The length of one side of square loop is l metre. The wire has uniform cross section area and uniform linear mass density. In four situations of column-I, the loop is subjected to four different unifrom and constant magnetic field. Under the conditions of column-I, match the column-I with corresponding results of column-II. (B0 in column-I is a positive non-zero constant).

y

x

I/2

I/2I/2

I/2

I

I Column-I Column-II (A) B

r = iB0 in tesla (P) magnitude of net force

on loop is 2 lIB0 newton

(B) Br

= jB0 in tesla (Q) magnitude of net force

on loop is zero

(C) Br

= )ji(B0 + in (R) magnitude of net torque

tesla on loop about its centre

is zero

(D) Br

= kB0 in tesla (S) magnitude of net force

on loop is B0I l newton

(T) net force perpendicular to the plane of loop

Q.1 I ( )

l

I -I -I -II

(B0 -I )

y

x

I/2

I/2I/2

I/2

I

I LrEHk-I LrEHk-II

(A) Br

= iB0 (P)

2 lIB0

(B) Br

= jB0 (Q)

(C) Br

= )ji(B0 + (R) blds dsUnz ds ifjr% ywi ij

dqy cyk?kw.kZ dk eku 'kwU; gS

(D) Br

= kB0 Vslyk esa (S) ywi ij dqy cy dk eku

B0I l U;wVu gS

(T) ywi ds ry ds yEcor~ dqy

cy



Q.2 Assume sufficient friction to prevent slipping.

Column-I Column-II (A)

F

Ring

(P) Body accelerates forward

(B)

F

Disc

(Q) Rotation about the centre of mass is clockwise

(C)

F

Solid cylinder

R/2

(R) Friction force acts backwards

(D)

F

Solid cylinder

(S) No friction acts

(T) None

Q.2 ekukfd fQlyus ls jksdus ds fy, i;kZIr ?k"kZ.k gSA

LrEHk-I LrEHk-II

(A)

F

Ring

(P)

(B)

F

Disc

(Q)

(C)

F

Solid cylinder

R/2

(R) ?k"kZ.k cy ihNs dh vksj

dk;Zjr gS

(D)

F

Solid cylinder

(S) dksbZ ?k"kZ.k dk;Zjr ugha gS

(T) dksbZ ugha



Section - III This section contains 6 questions (Q.1 to 6). The answerto each of the questions is a Single-digit integer, ranging from 0 to 9. The bubble corresponding to the correctanswer is to be darkened in the OMR. +4 marks will begiven for each correct answer and no negative markingfor each wrong answer. Q.1 In Young's experiment, the source is red light of

wavelength 7 × 10–7 m. When at thin glass plateof refractive index 1.5 at this wavelength is putin the path of one of the interfering beams, thecentral bright fringe shifts by 10–3 m to theposition previously occupied by the 5th brightfringe. Find the thickness (in µm) of the plate.

Q.2 Four point charges +8µC, –1µC, –1µC and +8µC

are fixed at the points –2

27 m, –23 m, +

23 m

and 2

27 m respectively on the y-axis, A particle

of mass 6 × 10–4 kg and charge +0.1 µC moves along the x-direction. Its speed at x = +∞ is v0. Find the least value of v0 (in m/s) for which theparticle will cross the origin. Assume that space

is gravity free. Given 04

1πε

= 9 × 109 Nm2/C2.

– III bl [k.M esa 6 ( 1 6) gaSA bl [k.M esa izR;sd iz'u dk mÙkj 0 ls 9 rd esa gSaA OMR esa iz'u la[;k ds laxr uhps fn;s x;s xksyksaa esa ls lgh mÙkj okys xksys dks dkyk fd;k tkuk gSA izR;sd lgh mÙkj ds fy;s +4 vad fn;s tk,saxs rFkk izR;sd xyr mÙkj ds fy, dksbZ

gSA

Q.1 ;ax iz;ksx esa] 7 × 10–7 m rjaxnS/;Z ds yky izdk'k dk

L=kksr gSA bl rjaxnS/;Z ij tc 1.5 viorZukad dh

iryh dk¡p dh IysV dks O;frdj.k iq¡t ds ,d iFk esa

j[krs gS] rks dsUnzh; nhIr fÝat 5oha nhIr fÝat }kjk

izkIr dh xbZ fiNyh fLFkfr ls 10–3 m }kjk cny

tkrh gSA IysV dh eksVkbZ (µm esa) Kkr djksA

Q.2 pkj fcUnq vkos'k +8µC, –1µC, –1µC rFkk +8µC,

y-v{k ij Øe'k% –2

27 m, –23 m, +

23 m rFkk

227 m fcUnqvksa ij fQDl gS] 6 × 10–4 kg nzO;eku

rFkk +0.1 µC vkos'k dk ,d d.k x-fn'kk ds vuqfn'k xfr djrk gSA x = +∞ ij bldh pky v0 gSA v0 (m/s esa)dk U;wure eku Kkr djks ftlds fy, d.k ewy fcUnq dks ikj djsxkA ekuk fd Lisl xq:Ro Lora=k gSA fn;k

x;k gS 04

1πε

= 9 × 109 Nm2/C2



Q.3 0.05 kg steam at 373 K and 0.45 kg of ice at253 K are mixed in an insulated vessel. Find theequilibrium temperature in degree centigrade ofthe mixture. Given, Lfusion = 80 cal/g = 336 J/g,Lvaporization = 540 cal/g = 226g J/g, Sice = 2100 J/kgK = 0.5 cal/gK and Swater = 4200 J/kg K = 1 cal/gK.

Q.4 A loop and a disc roll without slipping with the

same linear velocity v. The mass of the loop andthe disc is same. If the total kinetic energy of theloop is 8 J, find the kinetic energy of the disc (in J).

Q.5 A plank with a bar on it executes a S.H.M. of

amplitude A = 10 cm. What should be thecoefficient of friction which prevents the slidingof the bar for time period 1.0 sec (in 10–1) ?

Q.6 An element of atomic number 9 emits Kα X-ray

of wavelength λ. Find the atomic number of theelements which emits Kα X-ray of wavelength4λ.

Q.3 373 K ij 0.05 kg Hkki rFkk 253 K ij cQZ ds 0.45 kg

,d fo|qrjks/kh ik=k esa fefJr fd;s x;s gSA feJ.k dk

rqY; rki fMxzh ls.VhxszM esa Kkr djksA fn;k x;k gS,

Lxyu = 80 cal/g = 336 J/g, Lok"ihdj.k = 540 cal/g

= 226g J/g, ScQZ = 2100 J/kg K = 0.5 cal/gK rFkk

Sty = 4200 J/kg K = 1 cal/gK

Q.4 ,d ywi rFkk ,d pdrh leku js[kh; osx v ls fcuk

fQlys ?kwe jgh gSA ywi rFkk pdrh dk nzO;eku

leku gSA ;fn ywi dh dqy xfrt ÅtkZ 8 J gS]

pdrh dh xfrt ÅtkZ Kkr djks (J esa)A

Q.5 ,d NM+ ;qDr r[rs ij A = 10 cm vk;ke dh ljy

vkorZ xfr djrk gSA ?k"kZ.k xq.kkad D;k gksxk ftlls

1.0 lsd.M le; vUrjky ds fy, NM+ dks fQlyus

ls jksdk tk lds (10–1 esa) ?

Q.6 ijek.kq Øekad 9 dk ,d rRo λ rjaxnS/;Z dh Kα

X-fdj.ksa mRlftZr djrk gSA ml rRo dk ijek.kq

Øekad Kkr djks tks 4λ rjaxnS/;Z dh KαX-fdj.k

mRlftZr djrk gSA

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IIT Advance Paper-2 - Career Point