TARGET – IIT JEE - Career Point

TARGET – IIT JEE

CHEMISTRY, MATHEMATICS & PHYSICS

Time : 3 : 00 Hrs. MAX MARKS: 243

Name : _____________________________________ Roll No. : __________________________ Date : _____________

INSTRUCTIONS TO CANDIDATE

A. GENERAL :

1. Please read the instructions given for each question carefully and mark the correct answers against the question numbers on the answer sheet in the respective subjects.

2. The answer sheet, a machine readable Optical Mark Recognition (OMR) is provided separately. 3. Do not break the seal of the question-paper booklet before being instructed to do so by the invigilators.

B. MARKING SCHEME :

Each subject in this paper consists of following types of questions:- Section - I 4. Multiple choice questions with only one correct answer. 3 marks will be awarded for each correct answer and –1 mark for

each wrong answer. 5. Multiple choice questions with multiple correct option. 3 marks will be awarded for each correct answer and No negative

marking. 6. Passage based single correct type questions. 3 marks will be awarded for each correct answer and –1 mark for each wrong

answer. Section - III

7. Numerical response (single digit integer answer) questions. 3 marks will be awarded for each correct answer and No negative marking for wrong answer. Answers to this Section are to be given in the form of single integer only (0 to 9)

C. FILLING THE OMR :

8. Fill your Name, Roll No., Batch, Course and Centre of Examination in the blocks of OMR sheet and darken circle properly. 9. Use only HB pencil or blue/black pen (avoid gel pen) for darking the bubbles. 10. While filling the bubbles please be careful about SECTIONS [i.e. Section-I (include single correct, reason type, multiple

correct answers), Section –II ( column matching type), Section-III (include integer answer type)]

Section –I Section-II Section-III

For example if only 'A' choice is correct then, the correct method for filling the bubbles is

A B C D E

For example if only 'A & C' choices are correct then, the correct method for filling the bublles is

A B C D E

the wrong method for filling the bubble are

The answer of the questions in wrong or any other manner will be treated as wrong.

For example if Correct match for (A) is P; for (B) is R, S; for (C) is Q; for (D) is P, Q, S then the correct method for filling the bubbles is

P Q R S TA BCD

Ensure that all columns are filled. Answers, having blank column will be treated as incorrect. Insert leading zeros (s)

012

3

4

56

7

8

9

012

3

4

56

7

8

9

012

3

4

56

7

8

9

012

3

4

56

7

8

9

'6' should be filled as 0006

012

3

4

56

7

8

9

0 1 2

3

4

5 6

7

8

9

0 1 2

3

4

5 6

7

8

9

0 1 2

3

4

5 6

7

8

9


0 0 0 00 1 2

3

4

5 6

7

8

9

0 1 2

3

4

5 6

7

8

9

012

3

4

56

7

8

9

012

3

4

56

7

8

9


CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-3040000, Fax (0744) 3040050 email : [email protected]; Website : www.careerpointgroup.com 1

RS -11- I - 4

SEA

L

CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-3040000 Page # 1

Space for rough work

Important Data (egRoiw.kZ vk¡dM+s)

Atomic Masses: H = 1, C = 12, K = 39, O = 16, Fe= 56, N= 14, I = 127, Ca = 40, Mg = 24, Al = 27, F = 19, Cl = 35.5, S = 32, (ijek.kq nzO;eku) Na = 23 Constants : R = 8.314 Jk–1mol–1, h = 6.63 × 10–34 Js , C = 3 × 108 m/s, e = 1.6 × 10–19 Cb,me = 9.1 × 10–31Kg,

(fu;rkad) : RH = 1.1 × 107 m–1, log 2 = 0.3010, log 3 = 0.4771, log(5.05) = 0.7032; ln2 = 0.693; ln 1.5 = 0.405; ln3 = 1.098

Space for Rough Work (jQ+ dk;Z gsrq LFkku)



CHEMISTRY

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 thatquestion. + 3 marks will be given for correct answer and– 1 mark for each wrong answer.

Q.1 A bubble of gas released at the bottom of a lakeincreases to eight times its volume when it reachesthe surface. Assuming that atmospheric pressure is equivalent to the pressure exerted by a column ofwater 10 m high, what is the depth of the lake ?

(A) 90 m (B) 10 m (C) 70 m (D) 80 m Q.2 100 ml of a buffer of 1 M NH3 and 1 M NH4

+ are placed in two voltaic cells separately. A current of1.5 A is passed through both cells for 20 minutes.If electrolysis of water only takes place

2H2O + O2 + 4e– → 4(OH–) (R.H.S.)

2H2O → 4H+ + O2 + 4e– (L.H.S.) then pH of the : (A) L.H.S. will increase (B) R.H.S. will increase (C) Both sides will increase (D) Both sides will decrease

[k.M - I iz'u 1 ls 8 rd cgqfodYih iz'u gSaA izR;sd iz'u ds pkjfodYi (A), (B), (C) rFkk (D) gSa, ftuesa ls dsoy ,d fodYi lgh gSA OMR 'khV esa iz'u dh iz'u la[;k ds lek 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 vad ?kVk;k tk;sxkA

Q.1 xSl ds ,d cqycqys dks >hy ds iSans ls NksM+k tkrk gS]rks lrg ij ig¡qpus ij ;g vk;ru dk vkB xquk c<+tkrk gSA ;g ekurs gq, fd nkc] ty ds ,d 10 ehVj Å¡ps LrEHk ds kjk vkjksfir nkc ds rqY; gS] rks >hydh xgjkbZ D;k gS ?

(A) 90 m (B) 10 m (C) 70 m (D) 80 m Q.2 1 M NH3 o 1 M NH4

+ ds 100 ml cQj dks iFkd&iFkd nks okWYVh; lsyksa esa Hkjk tkrk gSA 1.5 Adh /kkjk dks nksuksa lsyksa esa ls 20 feuV rd xqtkjk tkrk gSA ;fn dsoy ty dk fo|qr vi?kVu gks

2H2O + O2 + 4e– → 4(OH–) (R.H.S.) 2H2O → 4H+ + O2 + 4e– (L.H.S.) rks pH : (A) ck¡;h rjQ c<+sxh (B) nk¡;h rjQ c<+sxh (C) nksuksa rjQ c<+sxh (D) nksuksa rjQ ?kVsxh



Q.3 The rate constant for two parallel reactions were found to be 1.0 × 10–2 litre mol–1 s–1 and 3.0 × 10–2

litre mol–1 s–1 . If the corresponding energies of activation of the parallel reactions are 60.0 kJ mol–1

and 70.0 kJ mol–1 respectively, what is apparent overall energy of activation ?

(A) 130.0 kJ mol–1 (B) 67.5 kJ mol–1 (C) 100.0 kJ mol–1 (D) 65.0 kJ mol–1

Q.4 Which of the following statement is incorrect? (A) B(OH)3 partially reacts with water to form

H3O+ and [B(OH)4]–, and behaves like a weak acid

(B) B(OH)3 behaves like a strong monobasic acid in the presence of sugars, and this acid can be titrated against an NaOH solution using phenolphthalein as an indicator

(C) B(OH)3 does not donate a proton and does not form any salt with NaOH

(D) B(OH)3 reacts with NaOH, forming Na [B(OH)4]

Q.3 nks lekukUrj vfHkfØ;kvksa ds osx fu;rkad

1.0 × 10–2 litre mol–1 s–1 o 3.0 × 10–2 litre mol–1 s–1

Ikk;s x;sA ;fn lEcfU/kr lekukUrj vfHkfØ;kvksa dh

lfØ;.k ÅtkZ,sa Øe'k% 60.0 kJ mol–1 o 70.0 kJ mol–1

gks] rks lEiw.kZ lfØ;.k ÅtkZ D;k gS ? (A) 130.0 kJ mol–1 (B) 67.5 kJ mol–1 (C) 100.0 kJ mol–1 (D) 65.0 kJ mol–1

Q.4 fuEu esa ls dkSulk dFku xyr gS ? (A) B(OH)3 vkaf'kd :Ik ls ty ds lkFk fØ;k dj

H3O+ o [B(OH)4]– cukrk gS] rFkk ,d nqcZy

vEy dh rjg O;ogkj djrk gS (B) B(OH)3 'kdZjk dh mifLFkfr esa ,d izcy

,dkkjh; vEy dh rjg O;ogkj djrk gS] rFkk

bl vEy dk vuqekiu NaOH foy;u ds lkFk

fQuks¶Fksyhu dks lwpd ds :Ik esa iz;qDr djrs

gq, fd;k tk ldrk gS (C) B(OH)3 izksVksu ugha nsrk gS rFkk NaOH ds lkFk

dksbZ Hkh yo.k ugha cukrk gS (D) B(OH)3 ,NaOH ds lkFk fØ;k dj Na[B(OH)4]

cukrk gS



Q.5 Which is correct about the nucleophilicity of halideion ?

(A) In DMSO order of nucleophilicity is F(–) > Cl(–) > Br(–) > I(–) while in water it is I(–) > Br(–) > Cl(–) > F(–) (B) In DMSO order of nucleophilicity is I(–) > Br(–) > Cl(–) > F(–) while in water it is F(–) > Cl(–) > Br(–) > I(–) (C) Order of nucleophilicity is same F(–) > Cl(–) > Br(–) > I(–) in both water and DMSO

(D) Order of nucleophilicity is same I(–) > Br(–) > Cl(–) > F(–) in both water and DMSO

Q.6 Consider the following four isomers

CH3

H C = C

CH3

H

(1)

CH3

H C = C

H

CH3

(2)

C2H2 BF3 (3) (4) Which is correct - (A) 2 and 4 possess centre of symmetry (B) 1, 2, 3 and 4 contain axis of symmetry (C) 2 and 3 possess alternating axis of symmetry (D) 4 contains centre of symmetry while 2 contains

plane of symmetry

Q.5 gSykbM vk;u dh ukfHkdLusgrk ds lUnHkZ esa dkSulk

lgh gS ? (A) DMSO esa ukfHkdLusgrk dk Øe

F(–) > Cl(–) > Br(–) > I(–) tcfd ty esa

I(–) > Br(–) > Cl(–) > F(–) gS

(B) DMSO esa ukfHkdLusgrk dk Øe

I(–) > Br(–) > Cl(–) > F(–) tcfd ty esa

F(–) > Cl(–) > Br(–) > I(–) gS

(C) ukfHkdLusgrk dk Øe F(–) > Cl(–) >

Br(–) > I(–) ty o DMSO nksuksa esa leku gS

(D) ukfHkdLusgrk dk Øe I(–) > Br(–) > Cl(–) > F(–)

ty o DMSO nksuksa esa leku gS

Q.6 fuEu pkj leko;oh;ksa ij fopkj dhft,

CH3

HC = C

CH3

H

(1)

CH3

H C = C

H

CH3

(2) C2H2 BF3 (3) (4) dkSulk lgh gS - (A) 2 o 4 eas dsUnzh; leferh mifLFkr gS (B) 1, 2, 3 o 4 esa vkh; leferh mifLFkr gS (C) 2 o 3 esa ,dkUrfjr vkh; leferh mifLFkr gS (D) 4 esa dsUnzh; leferh tcfd 2 esa ry leferh

mifLFkr gS



Q.7 )ii(

H)i( )( →

+ [X], [X] may be

(A) φ

(B)

φ

(C)

φ (D) None

Q.8 Cis-but-2-ene → )cold(KMnO4 (X)

(A) Me

H HO Me

H

OH

(B) OH

H Me OH

H

Me

(C) Me

H HO OH

H

Me

(D) Both (A) and (B) Questions 9 to 12 are multiple choice questions. Eachquestion has four choices (A), (B), (C) and (D), out ofwhich MULTIPLE (ONE OR MORE) is correct. Markyour response in OMR sheet against the questionnumber of that question. + 4 marks will be given forcorrect answer and – 1 mark for each wrong answer.

Q.7 )ii(

H)i( )( →

+ [X], [X] gks ldrk gS

(A) φ

(B)

φ

(C)

φ (D) dksbZ ugha

Q.8 leik-C;wV-2-bZu → )cold(KMnO4 (X)

(A) Me

HHO Me

H

OH

(B) OH

HMe OH

H

Me

(C) Me

HHO OH

H

Me

(D) (A) o (B) nksuksa

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 lgh gaSA OMR 'khV esa iz'u dh iz'u la[;k ds lek vius mÙkj vafdr dhft,A izR;sd lgh mÙkj ds fy, + 4 vad fn;s tk;saxs rFkk izR;sd xyr mÙkj ds fy, 1 vad ?kVk;k tk;sxkA



Q.9 fuEu esa ls dkSulk feJ.k ,d cQj dh rjg dk;Z dj

ldrk gS ?

(A) NaOH + CH3COONa (1 : 1 eksyj vuqikr)

(B) CH3COOH + NaOH (2 : 1 eksyj vuqikr)

(C) CH3COOH + NaOH (3 : 1 eksyj vuqikr)

(D) CH3COOH + NaOH (1 : 1 eksyj vuqikr)

Q.10 fLFkj vk;ru ij] tc ,d eksy xSl ds rki dks 298 ls 309 K rd c<+k;k tkrk gSaA xSl dks nh xbZ Å"ek 500 twy gSA rc dkSulk dFku lgh gS ?

(A) q = w = 500 J, ∆U = 0

(B) q = ∆U = 500 J w = 0

(C) q = w ≠ 500 J, ∆U = 0

(D) ∆U = 0, q = w = –500 J

Q.11 ,lhVksu (CH3COCH3) fdlesa eq[; mRikn gS :

(I) CH2=C=CH2 →⊕OH3

(II) CH3C≡CH → OH/HgSO/SOH 2442

(III) CH3C≡CH −

→OH/OH

THF.BH

22

3

(A) I (B) II

(C) III (D) buesa ls dksbZ ugha

Q.9 Which of the following mixtures can act as a

buffer ?

(A) NaOH + CH3COONa (1 : 1 molar ratio)

(B) CH3COOH + NaOH (2 : 1 molar ratio)

(C) CH3COOH + NaOH (3 : 1 molar ratio)

(D) CH3COOH + NaOH (1 : 1 molar ratio)

Q.10 When one mole of gas is heated at constant volumetemperature is raised from 298 to 309 K. Heat supplied to the gas is 500 joule. Then whichstatement is correct ?

(A) q = w = 500 J, ∆U = 0 (B) q = ∆U = 500 J w = 0 (C) q = w ≠ 500 J, ∆U = 0 (D) ∆U = 0, q = w = –500 J Q.11 Acetone (CH3COCH3) is the major product in :

(I) CH2=C=CH2 →⊕OH3

(II) CH3C≡CH → OH/HgSO/SOH 2442

(III) CH3C≡CH −

→OH/OH

THF.BH

22

3

(A) I (B) II

(C) III (D) None of these



Q.12 Which of the following ion will be aromatic innature ?

(A)

N |

H

⊕ (B)

Cl H H

+

(C) N

⊕

(D)

N +

H H

This section contains 2 paragraphs, each has 3 multiplechoice questions. (Questions 13 to 18) Each question has4 choices (A), (B), (C) and (D) out of which ONLY ONEis correct. Mark your response in OMR sheet against thequestion number of that question. + 4 marks will begiven for correct answer and – 1 mark for each wrong answer..

Passage # 1 (Ques. 13 to 15)

The percentage labeling of oleum is a uniqueprocess by means of which, the percentagecomposition of H2SO4, SO3 (free) and SO3

(combined) is calculated. Oleum is nothing but it is a mixture of H2SO4 and

SO3 i.e., H2S2O7, which is obtained by passing SO3

in solution of H2SO4. In order to didssolve free SO3

in oleum, dilution of oleum is done, in which oleumconverts into pure H2SO4. It is shown by thereaction as under :

Q.12 fuEu esa ls dkSuls vk;u dh ,sjksesfVd izdfr gksxh ?

(A)

N|

H

⊕ (B)

ClHH

+

(C) N

⊕

(D)

N +

H H

bl [k.M esa 2 vuqPNsn fn;s x;s gSa] izR;sd esa 3 cgqfodYih iz'u gSaA (iz'u 13 ls 18) izR;sd iz'u ds pkj fodYi (A), (B), (C) rFkk (D) gSa, ftuesa ls dsoy ,d fodYi lgh gSA OMR 'khV esa iz'u dh iz'u la[;k ds lek viuk mÙkj vafdr dhft;sA izR;sd lgh mÙkj ds fy;s + 4 vad fn;s tk,saxs rFkk izR;sd xyr mÙkj ds fy, 1 vad ?kVk;k tk;sxkA

vuqPNsn # 1 (iz'u la[;k 13 ls 15)

vkWfy;e dh izfr'krrk dk vadu fo'ks"k fof/k kjk fd;k tkrk gS ftlds QyLo:i H2SO4, SO3 (eqä) rFkk SO3

(la;qä) ds laxBu dks ifjdfyr fd;k tkrk gSA vkWfy;e okLro esa dqN ugha cfYd H2SO4 o SO3 dk

feJ.k gS vFkkZr~ H2S2O7 tks H2SO4 foy;u esa SO3 dks xqtkjus ls izkIr gksrk gSA vkWfy;e esa eqä SO3 dks ?kksyus ds lUnHkZ esa vkWfy;e dk ruqdj.k gksrk gSA ftlesa vkWfy;e 'kq) H2SO4 esa cny tkrk gSA bls fuEu vfHkfØ;k kjk n'kkZ;k x;k gS :



H2SO4 + SO3 + H2O → 2 H2SO4 (pure)

or, SO3 + H2O → H2SO4 (pure) When 100 g sample of oleum is diluted with

desired weight of H2O (in g), then the total mass ofpure H2SO4 obtained after dilution is known aspercentage labeling in oleum.

For example, if the oleum sample is labelled as"109% H2SO4" it means that 100 g of oleum ondilution with 9 g of H2O provides 109 g pureH2SO4, in which all free SO3 in 100 g of oleum isdissolved.

Q.13 For 109% labeled oleum if the number of moles of

H2SO4 and free SO3 be x and y respectively, then

what will be the value of yxyx

−+ ?

(A) 1 (B) 18 (C) 1/3 (D) 9.93

Q.14 In the above question, what is the percentage offree SO3 and H2SO4 in the oleum sample respectively?

(A) 60%, 40% (B) 30%, 70% (C) 85%, 15% (D) 40%, 60% Q.15 What volume of 1 M NaOH (in mL) will be

required to react completely with H2SO4 and SO3 in 109% labelled oleum ?

(A) 250 mL (B) 2224 mL (C) 750 mL (D) 1800 mL

H2SO4 + SO3 + H2O → 2 H2SO4 ('kq))

;k] SO3 + H2O → H2SO4 ('kq)) tc 100 g vkWfy;e ds uewus dks H2O (g esa), ds

,fPNd Hkkj dk ruq fd;k tkrk gS rc ruqrk ds i'pkr~ o 'kq) H2SO4 dk dqy nzO;eku izkIr gksrk gS ftls vkWfy;e esa izfr'krrk vadu dgk tkrk gSA

mnkgj.k ds fy,] ;fn vkWfy;e ds uewus dk vadu"109% H2SO4" gks rks bldk rkRi;Z gS fd 100 g vkWfy;e 9 g H2O ds lkFk ruqrk ij 109 g 'kq) H2SO4 nsrk gS ftlesa lHkh eqä SO3, 100 g vkWfy;e esa ?kqyh gksrh gSA

Q.13 109% vafdr vkWfy;e ds fy, ;fn H2SO4 o eqä SO3

ds eksyksa dh la[;k Øe'k% x o y gS] rks yxyx

−+ dk eku

D;k gksxk ? (A) 1 (B) 18 (C) 1/3 (D) 9.93

Q.14 mijksä iz'u esa] vkWfy;e uewus esa eqä SO3 rFkk H2SO4

dh izfr'krrk D;k gS ? (A) 60%, 40% (B) 30%, 70% (C) 85%, 15% (D) 40%, 60% Q.15 109% vafdr vkWfy;e esa H2SO4 o SO3 ds lkFk iw.kZ

vfHkfØ;k djus ds fy, 1 M NaOH (mL esa) dk vko';d vk;ru D;k gksxk ?

(A) 250 mL (B) 2224 mL (C) 750 mL (D) 1800 mL



Passage # 2 (Ques. 16 to 18) Ethyl acetoacetate and malonic ester are used for

the synthesis of various important organiccompound like ketone and substituted carboxylic acid. It contains active methylene (CH2) group. If contains acidic hydrogen and can be replaced as -

CH3COCH2COOC2H5 →B&& EtHCOCCOCH 2

)(

3

−

CH3COCHCO2Et | R

RX –X(–)

This alkyl substituted ethyl acetoacetate on hydrolysis produces acids

R|

EtCOCHCOCH 23)(H

NaOHdil+ →

R|

COOHCOCHCH3 −

Q.16 Malonic ester CH2(COOEt)2 can be converted in to CH3CH = CH – COOH by treating malonic ester with :

(A) EtONa / CH3CHO / ∆, H3O+ / ∆ (B) EtONa / CH3Cl, H3O+ / ∆

(C) EtONa / CH2Cl2, H3O+, ∆, alc KOH (D) All of these

vuqPNsn # 2 (iz'u la[;k 16 ls 18) ,fFky ,lhVks,lhVsV o esyksfud ,LVj dk mi;ksx

fofHkUu egRoiw.kZ dkcZfud ;kSfxdksa tSls dhVksu o

izfrLFkkfir dkckZsfDlfyd vEy ds la'ys"k.k esa fd;k

tkrk gSA buesa lfØ; esFkhfyu (CH2) lewg mifLFkr

gksrk gSA ;fn vEyh; gkbMªkstu mifLFkr gks] rks fuEu

izdkj ls foLFkkfir gksrk gS -

CH3COCH2COOC2H5 →B&& EtHCOCCOCH 2

)(

3

−

CH3COCHCO2Et |R

RX –X(–)

;g ,fYdy izfrLFkkfir ,fFky ,lhVks,lhVsV

tyvi?kVu ij vEy nsrk gS

R|

EtCOCHCOCH 23)(H

NaOHdil+ →

R|

COOHCOCHCH3 −

Q.16 esyksfud ,LVj CH2(COOEt)2 dk ifjorZu

CH3CH = CH – COOH esa esyksfud ,LVj dh fdlds

lkFk vfHkfØ;k ls gksrk gS : (A) EtONa / CH3CHO / ∆, H3O+ / ∆ (B) EtONa / CH3Cl, H3O+ / ∆ (C) EtONa / CH2Cl2, H3O+, ∆, alc KOH (D) mijksä lHkh



Q.17 CH3COCH2COOC2H5 RX

EtONa →

CH3CO

R|C HCOOC2H5 →

+ )(3OH X

OH

O

2

3→ CH3COCOOH + HCOOH

+ CH3COCH(CH2COOH)COOH RX would be -

(A) XCH=CH–

3

2

CH|

CHC =

(B) XCH2CH = CCH3CH = CH2

(C) XCH2–

2CH||

CHC − =CH–CH3

(D) XCH2–

3CH|

CHC = –CH=CH2

Q.18 CH3COCH2COOC2H5 can be converted in to 1, 3 di

ketone by treating it with -

(A) EtONa / CH3COCl, H3O+ / ∆

(B) EtONa / CH3COCH2COCl, H3O+ / ∆

(C) EtONa / CH3CH2COCl, H3O+ / ∆

(D) EtONa / CH3COCH2CH2COCl, H3O+ / ∆

Q.17 CH3COCH2COOC2H5 RX

EtONa →

CH3CO

R|C HCOOC2H5 →

+ )(3OH X

OH

O

2

3→ CH3COCOOH + HCOOH

+ CH3COCH(CH2COOH)COOH

RX gksxk -

(A) XCH=CH–

3

2

CH|

CHC =

(B) XCH2CH = CCH3CH = CH2

(C) XCH2–

2CH||

CHC − =CH–CH3

(D) XCH2–

3CH|

CHC = –CH=CH2

Q.18 CH3COCH2COOC2H5 dk ifjorZu 1, 3 MkbZdhVksu esa

bldh fdlds lkFk vfHkfØ;k ls gks ldrk gS -

(A) EtONa / CH3COCl, H3O+ / ∆

(B) EtONa / CH3COCH2COCl, H3O+ / ∆ (C) EtONa / CH3CH2COCl, H3O+ / ∆

(D) EtONa / CH3COCH2CH2COCl, H3O+ / ∆



Section - II This section contains 2 questions (Questions 1, 2). Each

question contains statements given in two columns which

have to be matched. Statements (A, B, C, D) in Column I

have to be matched with statements (P, Q, R, S, T) in

Column II. The answers to these questions have to be

appropriately bubbled as illustrated in the following

example. If the correct matches are A-P, A-S, A-T;

B-Q, B-R; C-P, C-Q and D-S, D-T then the correctly

bubbled 4 × 5 matrix should be as follows :

A B C D

P Q R S T

T S

P

P P Q R

RR

Q Q

S S T

T

P Q R S T

Mark your response in OMR sheet against the question

number of that question in section-II. + 8 marks will be

given for complete correct answer (i.e. +2 marks for each

correct row) and No Negative marks for wrong answer.

[k.M - II

bl [k.M esa 2 iz'u (iz'u 1, 2) gSaA izR;sd iz'u esa nks LrEHkksa

esa dFku fn;s x;s gSa] ftUgsa lqesfyr djuk gSA LrEHk-I

(Column I ) esa fn;s x;s dFkuksa (A, B, C, D) dks LrEHk-II

(Column II) esa fn;s x;s dFkuksa (P, Q, R, S,T) ls lqesy

djuk gSA bu iz'uksa ds mÙkj uhps fn;s x;s mnkgj.k ds

vuqlkj mfpr xksyksa dks dkyk djds n'kkZuk gSA ;fn lgh

lqesy A-P, A-S, A-T; B-Q, B-R; C-P, C-Q rFkk D-S, D-T gS,

rks lgh fof/k ls dkys fd;s x;s xksyksa dk 4 × 5 eSfVªDl uhps

n'kkZ;s vuqlkj gksxk :

ABCD

P Q R S T

T S

P

P P Q R

R R

Q Q

S S T

T

P Q R S T

vr% OMR 'khV esa iz'u dh iz'u la[;k ds lek viuk mÙkj

[k.M-II esa vafdr dhft;sA izR;sd iw.kZ lgh mÙkj ds fy;s + 8

vad fn;s tk;saxs (vFkkZr~ izR;sd lgh iafDr feyku ds fy, +2

vad fn, tk,asxs) rFkk xyr mÙkj ds fy;s dksbZ _.kkRed

vadu ugha gS (vFkkZr~ dksbZ vad ugha ?kVk;k tk;sxk)A



Q.1 Match the column Column-I Column-II (A) [Co(NH3)6]2+ (P) sp3 (B) [Cu(NH3)4]2+ (Q) dsp2 (C) [Mn Cl4]2– (R) d2sp3 (D) [PtCl4]2– (S) Diamagnetic (T) Paramagnetic Q.2 Column-I Column-II (Pair of compounds) (Reagent by which distinction can be

made)

(A)

OH and

OH

(P) Na (B) and Acetone

OH

(Q) [Ag(NH3)2](+)OH(–)

(C) and CH3–C–HOH

|| O

(R) NaBrO(Br2/NaOH)

(D) and

OH (S) PhNHNH2

(T) None

Q.1 LrEHk lqesfyr dhft,

LrEHk-I LrEHk-II (A) [Co(NH3)6]2+ (P) sp3 (B) [Cu(NH3)4]2+ (Q) dsp2 (C) [Mn Cl4]2– (R) d2sp3 (D) [PtCl4]2– (S) izfrpqEcdh; (T) vuqpqEcdh; Q.2 LrEHk-I LrEHk-II (;kSfxdksa dk ;qXe) (vfHkdeZd ftlds kjk foHksn dj ldrs gS)

(A)

OH o

OH

(P) Na (B) o ,lhVksu

OH

(Q) [Ag(NH3)2](+)OH(–)

(C) o CH3–C–HOH

|| O

(R) NaBrO(Br2/NaOH)

(D)

o

OH

(S) PhNHNH2

(T) dksbZ ugha



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 thatquestion. + 3 marks will be given for each correctanswer and – 1 mark for each wrong answer.

Q.1 Vectors b,arr and c

r with magnitude 2, 3 & 4 respectively

are coplanar. A unit vector dr

is perpendicular to all of

them. If 3k

3j

6i)dc()ba( +−=×××

rrrr and the angle

between ar and b

r is 30º then |k.c||j.c||i.c|

rrr++ is

equal to

(A) 35 (B)

95 (C)

125 (D)

185

Q.2 Every line of the family x (a + 2λ) + y(1 + 3λ) = (1 + 2λ)

intersects the line 3x – 2y = 3 at the same point,

where λ is a parameter and 'a' is constant, then thevalue of a is

(A) 0 (B) 4

(C) 2 (D) None of these

[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 gSA OMR 'khV esa iz'u dh iz'u la[;k ds lek 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 vad ?kVk;k tk;sxkA

f

Q.1 lfn'k b,arr ,oa c

r ftuds ifjek.k Øe'k% 2, 3 ,oa 4 gS]

leryh; gSA ,d bdkbZ lfn'k dr

bu lHkh lfn'kksa ds

yEcor~ gSA ;fn 3k

3j

6i)dc()ba( +−=×××

rrrr rFkk ar

,oa br

ds e/; dks.k 30º gS] rc |k.c||j.c||i.c|rrr

++ dk

eku gS -

(A) 35 (B)

95 (C)

125 (D)

185

Q.2 js[kk lewg x (a + 2λ) + y(1 + 3λ) = (1 + 2λ) dh izR;sd js[kk] js[kk 3x – 2y = 3 dks leku fcUnq ij izfrPNsnu djrh gS tgk¡ λ ,d izkpy rFkk 'a' vpjkad gS] rc a dk eku gS -

(A) 0 (B) 4

(C) 2 (D) buesa ls dksbZ ugha



Q.3 )x4sin1(n

x9cosx6cosxcosx4cosLim 20x +−

→ l is equal to

(A) 425 (B)

825

(C) 225 (D)

425−

Q.4 Circumradius of a ∆ABC is 2, O is the circumcentre, H is the orthocentre then

641 (AH2

+ BC2) (BH2 + AC2) (CH2 + AB2) is

equal to (A) 64 (B) 16

(C) 641 (D) 1

Q.5 The sum of the infinitely decreasing geometric

progression is equal to the greatest value of the

function f(x) = 3x3 – x – 76 on interval [0, 3]; the

first term of the progression is equal to the square

of the common ratio. The common ratio of the

G.P. is-

(A) 12 − (B) 13 −

(C) 12 + (D) 13

1+

Q.3 )x4sin1(n

x9cosx6cosxcosx4cosLim 20x +−

→ l dk eku gS -

(A) 425 (B)

825

(C) 225 (D)

425−

Q.4 ,d ∆ABC dh ifjf=kT;k 2 gS, O ifjdsUnz gS] H yEcdsUnz

gS] rc 641 (AH2

+ BC2) (BH2 + AC2) (CH2 + AB2) dk

eku gS - (A) 64 (B) 16

(C) 641 (D) 1

Q.5 ,d vuUr Ðkleku xq.kksÙkj Js<+h dk ;ksx] vUrjky

[0, 3] esa Qyu f(x) = 3x3 – x – 76 ds egÙke eku ds

cjkcj gSA Js<+h dk izFke in lkoZvuqikr ds oxZ ds

cjkcj gSA xq- Js- dk lkoZ vuqikr gS -

(A) 12 − (B) 13 −

(C) 12 + (D) 13

1+



Q.6 The number of values of k for which, the equationf(x) = x3 – 12x + k = 0 has two real distinct roots inthe interval (2, 3) is-

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

Q.7

++++∞→ n2

1.....n3

1n2

1n

1limn

is equal to

(A) 0 (B) 1 (C) 2 (D) 4

Q.8 Consider f(x) = log441 (sin2 x – 20 cos x + 1), then- (A) f(x) is many one & bounded function (B) f(x) is many one & unbounded function (C) f(x) is one-one & bounded function (D) f(x) is one-one & unbounded function Questions 9 to 12 are multiple choice questions. Eachquestion has four choices (A), (B), (C) and (D), out ofwhich MULTIPLE (ONE OR MORE) is correct. Markyour response in OMR sheet against the questionnumber of that question. + 4 marks will be given for eachcorrect answer and – 1 mark for each wrong answer.

Q.9 If f(x) and g(x) are two increasing and differentiablefunction such that f(x), g(x) > 0 ∀ x ∈ R, then-

(A) (f(x))g(x) is always increasing (B) If f(x) > 1, then (f(x))g(x) is increasing (C) If (f(x))g(x)

is decreasing then f(x) < 1 (D) If (f(x))g(x) is decreasing then f(x) > 1

Q.6 k ds ekuksa dh la[;k ftuds fy, lehdj.k f(x) = x3 – 12x + k = 0 ds nks fofHkUu okLrfod ewy

vUrjky (2, 3) esa fLFkr gks] gksxh - (A) 0 (B) 1

(C) 2 (D) 3

Q.7

++++∞→ n2

1.....n3

1n2

1n

1limn

dk eku gS -

(A) 0 (B) 1 (C) 2 (D) 4

Q.8 ekukfd f(x) = log441 (sin2 x – 20 cos x + 1), rc - (A) f(x) cgq,sdh ,oa ifjc) Qyu gS (B) f(x) cgq,sdh ,oa vifjc) Qyu gS (C) f(x) ,dSdh ,oa ifjc) Qyu gS (D) f(x) ,dSdh ,oa vifjc) Qyu gS

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 lgh gaSA OMR 'khV esa iz'u dh iz'u la[;k ds lek vius mÙkj vafdr dhft,A izR;sd lgh mÙkj ds fy,+ 4 vad fn;s tk;saxs rFkk izR;sd xyr mÙkj ds fy, 1 vad ?kVk;k tk;sxkA

Q.9 ;fn f(x) ,oa g(x) nks o/kZeku rFkk vodyuh; Qyu bl izdkj gSa] fd f(x), g(x) > 0 ∀ x ∈ R, rc -

(A) (f(x))g(x) ges'kk o/kZeku gS (B) ;fn f(x) > 1, rc (f(x))g(x) o/kZeku gS (C) ;fn (f(x))g(x)

Ðkleku gS] rc f(x) < 1 (D) ;fn (f(x))g(x) Ðkleku gS] rc f(x) > 1



Q. 10 If f(x) = ln )1x9x3( 2 ++ +

++

+−

1m2x

1313 2

x

x

is an odd function ∀ x ∈ ]8,8[– , (where [·]denotes greatest integer function) then possibleinteger value(s) of m is (are)-

(A) 4 (B) 5 (C) 6 (D) 9

Q.11 The general solution of the differential equationex dx = x2e1 − dy is-

(A) y = k – cos–1 (ex) (B) y = sin–1(ex) + c (C) y = sin–1 ce1 x2 +−

(D) ce1cosy x21 +−= −

Q.12 Consider the equation k|1x|

12 =−

− , now which

of the following is true- (A) The given equation has 4 solution if k ∈ (0, 2) (B) There is no value of k for which given equation

has exactly two solutions (C) There is no value of k for which given equation

has exactly one solution (D) None of these

Q. 10 ;fn f(x) = ln )1x9x3( 2 ++ +

++

+−

1m2x

1313 2

x

x,

∀ x ∈ ]8,8[– , ,d fo"ke Qyu gS (tgk¡ [·] egÙke

iw.kk±d Qyu dks O;Dr djrk gS) rc m ds lEHko iw.kk±d eku gSa -

(A) 4 (B) 5 (C) 6 (D) 9

Q.11 vody lehdj.k ex dx = x2e1 − dy dk O;kid

gy gS - (A) y = k – cos–1 (ex) (B) y = sin–1(ex) + c (C) y = sin–1 ce1 x2 +−

(D) ce1cosy x21 +−= −

Q.12 ekukfd lehdj.k k|1x|

12 =−

− , fuEu esa ls dkSulk

dFku lR; gS - (A) nh xbZ lehdj.k ds 4 gy gksaxs ;fn k ∈ (0, 2) (B) ;gk¡ k dk dksbZ eku ugha gS ftlds fy, nh xbZ

lehdj.k ds Bhd nks gy gks (C) ;gk¡ k dk dksbZ eku ugha gS ftlds fy, nh xbZ

lehdj.k dk Bhd ,d gy gks (D) buesa ls dksbZ ugha




x|ka'k # 1 (iz- 13 ls 15)

,d fijkfeM dk vk/kkj vk;rkdkj gS] blds vk/kkj ds rhu 'kh"kZ A(2, 2, –1), B(3, 1, 2) ,oa C(1, 1, 1) gS (Øe esa gks Hkh ldrs gSa vkSj ugha Hkh), blds f'k[kj dk 'kh"kZ

−−

310,

326,4P gS rFkk vk/kkj dk pkSFkk 'kh"kZ D gSA

Q.13 D ds funsZ'kkad gS -

(A) (4, 0, 2) (B) (4, 2, 0)

(C) (2, 0, 4) (D) (0, 2, 4)

Q.14 P ls fijkfeM ds vk/kkj ij Mkys x, yEc ds ikn ds

funsZ'kkad gS -

(A)

32,

34,2 (B)

34,

32,2

(C)

32,2,

34 (D)

2,

34,

32

This section contains 2 paragraphs; each has 3 multiplechoice questions. (Questions 13 to 18) Each question has4 choices (A), (B), (C) and (D) out of which ONLY ONEis correct. Mark your response in OMR sheet against thequestion number of that question. + 4 marks will begiven for each correct answer and – 1 mark for eachwrong answer.

Passage # 1 (Ques. 13 to 15) The base of a pyramid is rectangular, three of its

vertices of the base are A(2, 2, –1), B(3, 1, 2) and C(1, 1, 1) (may or may not be in order). Its vertex at

the top is

−−

310,

326,4P and fourth vertex of the

base is D. Q.13 Coordinates of D are

(A) (4, 0, 2) (B) (4, 2, 0)

(C) (2, 0, 4) (D) (0, 2, 4)

Q.14 Coordinates of foot of the normal drawn from P on

the base of the pyramid are-

(A)

32,

34,2 (B)

34,

32,2

(C)

32,2,

34 (D)

2,

34,

32



Q.15 Volume of the pyramid is (in cubic units) (A) 20 (B) 10 (C) 40 (D) 30 Passage # 2 (Ques. 16 to 18) If set of values of 'p' for which the straight line

2px – 2y – p = 0 intersects the curve y = sin–1(2x – 1)at three distinct points is [α, β]. Let C1 & C2 be two curves as:

C1 : 1)1]([

yx2

2

2

2=

−β+

α, C2: y = [β] +

α−x4

Where [·] denotes greatest integer function

Q.16 ∫β+α

α

−2

dx)xcos|xsin(| is equal to-

(A) 0 (B) 2 (C) 4 (D) 8 Q.17 The probability that a point is chosen which lie

inside the curve C1 but at a distance greater than2 units from the origin is-

(A) 21 (B)

31 (C)

41 (D)

32

Q.18 If a tangent is drawn to the curve C2 which is

passing through the point ([β], 3α + 1) and meetsthe lines x = α and y = [β] at P and Q, then midpoint of PQ is-

(A) (α, [β]) (B) (2α – 1, 2[β] + 1) (C) ([β], α) (D) (2α + 1, 3[β] – 1)

Q.15 fijkfeM dk vk;ru (?ku bdkbZ esa) gS (A) 20 (B) 10 (C) 40 (D) 30

x|ka'k # 2 (iz- 16 ls 18)

;fn 'p' ds ekuksa dk leqPp; ftuds fy, ljy js[kk 2px – 2y – p = 0 oØ y = sin–1(2x – 1) dks rhu fofHkUu fcUnqvksa ij izfrPNsnu djrh gS] [α, β] gSA ekuk C1 ,oaC2 nks oØ bl izdkj gS] fd

C1 : 1)1]([

yx2

2

2

2=

−β+

α, C2: y = [β] +

α−x4

tgk¡ [·]egÙke iw.kk±d Qyu dks O;Dr djrk gSA

Q.16 ∫β+α

α

−2

dx)xcos|xsin(| dk eku gS -

(A) 0 (B) 2 (C) 4 (D) 8

Q.17 ,d fcUnq dks pquus dh izkf;drk tks oØ C1 ds vUnj fLFkr gks ijUrq ewyfcUnq ls ftldh nwjh 2 bdkbZ ls vf/kd gks] gksxh -

(A) 21 (B)

31 (C)

41 (D)

32

Q.18 ;fn oØ C2 dh ,d Li'kZjs[kk [khaph tkrh gS] tks fcUnq ([β], 3α + 1) ls xqtjrh gS rFkk js[kkvksa x = α ,oa y = [β]dks fcUnq P ,oa Q ij feyrh gS] rc PQ dk e/; fcUnq gS -

(A) (α, [β]) (B) (2α – 1, 2[β] + 1) (C) ([β], α) (D) (2α + 1, 3[β] – 1)



Section - II This section contains 2 questions (Questions 1, 2).

Each question contains statements given in two

columns which have to be matched. Statements (A, B,

C, D) in Column I have to be matched with statements

(P, Q, R, S, T) in Column II. The answers to these

questions have to be appropriately bubbled as

illustrated in the following example. If the correct

matches are A-P, A-S, A-T; B-Q, B-R; C-P, C-Q and

D-S, D-T then the correctly bubbled 4 × 5 matrix

should be as follows :

A B C D

P Q R S T

T S

P

P P Q R

R R

Q Q

S S T

T

P Q R S T

Mark your response in OMR sheet against the

question number of that question in section-II. + 8

marks will be given for complete correct answer (i.e.

+2 marks for each correct row) and No Negative

marks for wrong answer.

[k.M - II

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

LrEHkksa esa dFku fn;s x;s gSa] ftUgsa lqesfyr djuk gSA

LrEHk-I (Column I ) esa fn;s x;s dFkuksa (A, B, C, D) dks

LrEHk-II (Column II) esa fn;s x;s dFkuksa (P, Q, R, S,T) ls

lqesy djuk gSA bu iz'uksa ds mÙkj uhps fn;s x;s mnkgj.k

ds vuqlkj mfpr xksyksa dks dkyk djds n'kkZuk gSA ;fn

lgh lqesy A-P, A-S, A-T; B-Q, B-R; C-P, C-Q rFkk

D-S, D-T gS, rks lgh fof/k ls dkys fd;s x;s xksyksa dk

4 × 5 eSfVªDl uhps n'kkZ;s vuqlkj gksxk

ABCD

P Q R S T

T S

P

P P Q R

R R

Q Q

S S T

T

P Q R S T




vad fn, tk,asxs) rFkk xyr mÙkj ds fy;s dksbZ _.kkRed vadu

ugha gS (vFkkZr~ dksbZ vad ugha ?kVk;k tk;sxk)A



Q.1 Column-I Column-II

(A) Let x be the set of all 15 digit (P) 1

integers. An integer is chosen

at random for x. The probability

that it is a palindrome is n10

1

then the value of n is

(B) If the coefficient of xn in the expansion (Q) 4

of 3

3

)x1()x1(

−+ is 102, then the value

of n equals to

(C) The least distance of a point on (R) 5

the ellipse x2 + 2y2 = 6 from the

line x + y = 7 is 2

n , then n is

(D) f : R → R and satisfies f(2) = – 1, (S) 7

f '(2) = 4. If 7)x("f)x3(3

2

=−∫

then f(3) has the value equal to (T) 10

Q.1 LrEHk-I LrEHk-II

(A) ekuk x lHkh 15 vadh; iw.kk±dksa (P) 1

dk leqPp; gSA x esa ls ;knPN;k

,d iw.kk±d pquk tkrk gSA blds

palindrome (og iw.kk±d tks mYVk

rFkk lh/kk ,d leku i<+k tk,)

gksus dh izkf;drk n10

1 gS] rc n

dk eku gS

(B) ;fn 3

3

)x1()x1(

−+ ds izlkj esa xn dk (Q) 4

xq.kkad 102 gS] rc n dk eku gS

(C) nh?kZoÙk x2 + 2y2 = 6 ij fLFkr (R) 5

,d fcUnq dh js[kk x + y = 7 ls

U;wure nwjh 2

n gS] rc n cjkcj gS

(D) f : R → R rFkk f(2) = – 1, f '(2) = 4 (S) 7

dks lUrq"V djrk gSA ;fn

7)x("f)x3(3

2

=−∫ rc f(3)

dk eku gS (T) 10



Q.2 Column-I Column-II

(A) A line l intersect the graph of (P) 0

f(x) = 2x4 + 7x3 + 3x – 5 at

four points (xr, yr) where

r = 1, 2, 3, 4 then ∑=

−4

1rrx2 is

(B) f(x) = 10 + 4x ln9 – 33–x – 3x–1 (Q) 7

is greatest at x = a,

then a is

(C) If y = ax3 + bx2 + cx + d has (R) 5

horizontal tangent at (–2, 6)

and (2, 0) then 'd' is

(D) If ∫ +

2/1

02

x2

)x21(ex8 dx = a then [a] is (S) 3

equal to where [·] is greatest integer

function

(T) 9

Q.2 LrEHk-I LrEHk-II

(A) ,d js[kk l Qyu (P) 0

f(x) = 2x4 + 7x3 + 3x – 5 ds xzkQ

dks pkj fcUnqvksa (xr, yr) tgk¡

r = 1, 2, 3, 4 ij izfrPNsn

djrh gS] rc ∑=

−4

1rrx2 dk eku gS

(B) f(x) = 10 + 4x ln9 – 33–x – 3x–1 (Q) 7 dk egÙke eku x = a ij gS] rc

a cjkcj gS

(C) ;fn y = ax3 + bx2 + cx + d (R) 5

fcUnq (–2, 6) ,oa (2, 0) ij

Li'kZ js[kk kSfrt gS] rc

'd' dk eku gS

(D) ;fn ∫ +

2/1

02

x2

)x21(ex8 dx = a rc [a] (S) 3

cjkcj gS tgk¡ [·] egÙke iw.kk±d

Qyu gS (T) 9



PHYSICS

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 thatquestion. + 3 marks will be given for each correctanswer and – 1 mark for each wrong answer.

Q.1 An ideal gas obey the relation of P1/2V2 = constant.

The coefficient of volume expansion of the gas is :

(where T is the absolute temperature)

(A) 1/3T (B) –1/3T (C) 1/2T (D) –1/2T

Q.2 Two lights beams of different colours pass through

the prism. Inside the prism, their paths are shown

in the figure. Which of the following statements is

wrong :

100º

40º40º

40º 40ºYellow light

beam

blue light beam

Air Air

[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 gSA OMR 'khV esa iz'u dh iz'u la[;k ds lek 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 vad ?kVk;k tk;sxkA

Q.1 ,d vkn'kZ xSl lEcU/k P1/2V2 = fu;rkad dk ikyu

djrh gSA xSl dk vk;ru izlkj xq.kkad gksxk (tgk¡ T

ije rki gS) (A) 1/3T (B) –1/3T (C) 1/2T (D) –1/2T

Q.2 fHkUu-fHkUu jaxksa dh nks izdk'k iqat fizTe ls xqtjrh gSA

fizTe ds vUnj] muds iFk fp=k esa n'kkZ;s x;s gSA fuEUk

esa ls dkSulk dFku xyr gS

100º

40º40º

40º 40ºYellow light

beam

blue light beam

Air Air



(A) The yellow light beam suffers minimum

deviation

(B) The blue light beam suffers deviation larger

than yellow

(C) Both the beams have same magnitude of

minimum deviation

(D) Both the beams suffer deviation in clockwise

sense

Q.3 If the power dissipated in the resister configuration

I, II, III is P1, P2, P3 respectively, then correct order

of them is

(I)

5V2Ω2Ω4Ω

3Ω

(II)

1Ω

1Ω1Ω

1Ω 1Ω1Ω

1Ω1Ω

5V

(A) ihyh izdk'k iqat U;wure fopyu iznf'kZr djrh gS

(B) uhyh izdk'k iaqt dk fopyu ihys ls vf/kd gS

(C) nksuksa iaqt U;wure fopyu dk leku ifjek.k

j[krh gS

(D) nksuksa iqat nfk.kkorZ fn'kk eas fopyu iznf'kZr

djrh gS

Q.3 ;fn izfrjks/kksa ds vfHkfoU;kl I, II, III esa O;f;r 'kfDr

Øe'k% P1, P2, P3 gS] rc mudk lgh Øe gS

(I)

5V2Ω2Ω4Ω

3Ω

(II)

1Ω

1Ω1Ω

1Ω 1Ω1Ω

1Ω1Ω

5V



(III)

1Ω

2Ω 3Ω

2Ω

1Ω

5V

(A) P1 > P2 > P3 (B) P3 > P2 > P1 (C) P1 > P2 = P3 (D) P2 > P3 > P1

Q.4 Three point charges are located on circumference ofcircle of radius 2.5 m as shown in figure. Theelectrostatic interaction potential energy of charge 2q is-

(0, 0) O x

(metre)

y (metre)

2q

q

)2,5(

q

(A) 2kq1516 (B) 2kq

3037

(C) 2kq3043 (D) 2

4528 kq

(III)

1Ω

2Ω 3Ω

2Ω

1Ω

5V

(A) P1 > P2 > P3 (B) P3 > P2 > P1 (C) P1 > P2 = P3 (D) P2 > P3 > P1

Q.4 fp=kkuqlkj 2.5 m f=kT;k ds oÙk dh ifjf/k ij nks fcUnq vkos'k j[ks x;s gSA vkos'k 2q dh fLFkjoS/kqfrdh var%fØ;k fLFkfrt ÅtkZ gksxh

(0, 0) Ox

(metre)

y (metre)

2q

q

)2,5(

q

(A) 2kq1516 (B) 2kq

3037

(C) 2kq3043 (D) 2

4528 kq



Q.5 The open glass tube has uniform cross-section anda valve separating the two identical ends. The valveis initially closed end 1 has planer film of soapbubble and end 2 has a soap bubble having radiusof curvature ‘r’. Immediately after opening thevalve.

valve

Planer soap film

2r

1

(A) Air from end 1 flows towards end 2. The

volume of soap bubble at end 2 increases (B) Air from end 2 flows towards end 1. The

volume of soap bubble at end 2 increases (C) Air from end 2 flow towards end 1. The

volume of soap bubble at end 2 decreases (D) no change occurs

Q.6 A small block describes the vertical circular motioninside the smooth spherical cavity of radius R. Thespeed at the lowest position A is just sufficient tomake it reach the highest point B. The angle φ at which the speed of the block is three fifth of thelargest speed satisfies –

Q.5 dk¡p dh [kqyh uyh ,dleku vuqizLFk dkV j[krh gS

rFkk nks le:i fljksa ds ,d okYo iFkd djrk gSA

izkjEHk eas okYo cUn gS rc fljk 1 lkcqu ds cqycqys

dh lery fQYe gS rFkk fljk 2, r oØrk f=kT;k dk

lkcwu dk cqycqyk gSA okYo dks [kksyus ds rqjUr ckn

valve

Planer soap film

2r

1

(A) ok;q fljs 1 ls fljs 2 dh vksj izokfgr gksrh gSA

fljs 2 ij lkcwu ds cqycqys dk vk;ru c<+rk gS

(B) ok;q fljs 2 ls fljs 1 dh vksj izokfgr gksrh gSA

fljs 2 ij lkcwu ds cqycqys dk vk;ru c<+rk gS (C) ok;q fljs 2 ls fljs 1 dh vksj izokfgr gksrh gSA

fljs 2 ij lkcwu ds cqycqys dk vk;ru ?kVrk gSA

(D) dksbZ ifjorZu izkIr ugha gksrk gS

Q.6 ,d NksVk CykWd R f=kT;k dh ?k"kZ.kghu xksykdkj xqfgdk

ds vUnj Å/okZ/kj oÙkh; xfr djrk gSA fuEure fLFkr

A ij pky mls mPpre fcUnq B rd igq¡pus ds fy;s

Bhd i;kZIr gSA dks.k φ ftl ij CykWd dh pky

vf/kdre pky dh 3/5 gS] rks φ lUrq"V djrk gS



R C

B

A

φ

(A) 34π

<φ<π (B)

43

3π

<φ<π

(C) 6

54

3 π<φ<

π (D) π<φ<π6

5

Q.7 ,d lekUrj IysV la/kkfj=k dk IysV ks=kQy A rFkk iFkdj.k d gSA izkjEHk esa mls K = 3 ijkoS|qrkad ds inkFkZ ls iw.kZr;k Hkjk x;k gSA Hkjs gq;s inkFkZ dh pkSM+kbZ x fp=kkuqlkj fu;r pky v ls ?kVrh gSA le;

vUrjky 0 < t < vl ds fy;s] ifjiFk dk le;

fu;rkad fuEu ds vuqlkj ifjofrZr gksrk gS

d

l

xv

R

(A)

−

εl

vt23d

AR 0 (B)

+

εl

vt21d

AR 0

(C)

+

εl

vt23d

AR 0 (D)

−

εl

vt21d

AR 0

R C

B

A

φ

(A) 34π

<φ<π (B)

43

3π

<φ<π

(C) 6

54

3 π<φ<

π (D) π<φ<π6

5

Q.7 A parallel plate capacitor has plate area A andseparation ‘d’. Initially it is completely filled withmaterial of dielectric constant K = 3. The width xof filled material decreases at a constant speed v as

shown in figure. For time interval 0 < t < vl , time

constant of circuit varies as

d

l

x v

R

(A)

−

εl

vt23d

AR 0 (B)

+

εl

vt21d

AR 0

(C)

+

εl

vt23d

AR 0 (D)

−

εl

vt21d

AR 0



Q.8 A transparent cube of 0.21 m edge contains a small

air bubble. Its apparent distance when viewed

through one face of cube is 0.1 m and viewed from

opposite face is 0.04 m. The actual distance of the

bubble from the second face of the cube is-

(A) 0.06 m (B) 0.17 m

(C) 0.05 m (D) 0.04 m

Questions 9 to 12 are multiple choice questions. Eachquestion has four choices (A), (B), (C) and (D), out ofwhich MULTIPLE (ONE OR MORE) is correct. Markyour response in OMR sheet against the questionnumber of that question. + 4 marks will be given for eachcorrect answer and – 1 mark for each wrong answer.

Q.9 The coefficient of friction between the block andplank is µ and its value is such that block becomesstationary with respect to plank before it reachesthe other end. Then-

M m v0

(A) the work done by friction on the block is

negative (B) the work done by friction on the plank is

positive (C) the net work done by friction is negative (D) net work done by the friction is zero

Q.8 0.21 ehVj Hkqtk dk ,d ikjn'khZ ?ku ,d NksVk ok;q

dk cqycwyk j[krk gSA bldh vkHkklh nwjh tc bls

?ku ds ,d Qyd ls ns[kk tk,] 0.1 m gS rFkk

foijhr Qyd ls ns[kus ij 0.04 m gSA ?ku ds frh;

Qyd ls cqycwys dh okLrfod nwjh gS - (A) 0.06 m (B) 0.17 m

(C) 0.05 m (D) 0.04 m

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 lgh gaSA OMR 'khV esa iz'u dh iz'u la[;k ds lek vius mÙkj vafdr dhft,A izR;sd lgh mÙkj ds fy,+ 4 vad fn;s tk;saxs rFkk izR;sd xyr mÙkj ds fy, 1 vad ?kVk;k tk;sxkA

Q.9 CykWd rFkk r[rsa ds e/; ?k"kZ.k xq.kkad µ gS rFkk

mldk eku bl izdkj gS fd CykWd r[rsa ds nwljs fljs

rd igq¡pus rd r[rs ds lkisk fLFkj gks tkrk gSA rc

M m v0

(A) CykWd ij ?k"kZ.k kjk fd;k x;k dk;Z _.kkRed gS

(B) r[rs ij ?k"kZ.k kjk fd;k x;k dk;Z /kukRed gS

(C) ?k"kZ.k kjk fd;k x;k usV dk;Z _.kkRed gS

(D) ?k"kZ.k kjk fd;k x;k usV dk;Z 'kwU; gS



Q.10 A charged particle of unit mass and unit charge

moves with velocity of )j6i8(v ^^ +=→ m/s in a

magnetic field of →B = Tk2 ^ . Choose the correct

alternative(s). (A) The path of the particle may be x2 + y2 – 4x – 21 = 0 (B) The path of the particle may be x2 + y2 = 25 (C) The path of the particle may be y2 + z2 = 25 (D) The time period of the particle will be 3.14s

Q.11 An object of length 1 cm is placed on a principle axis of biconvex lens of radius 5 cm. Distance between the lens and object is 20 cm. Space between the lens and object is filled with medium of two different refractive index 2 and 1 as shown in the figure. Refractive index is 1 on the left of the object and on the right side of the lens. Boundary of both medium is mid-way between the object and lens as shown in figure.

µ = 2 µ = 1 1.5 µ = 1

10 cm 10 cm

1 cm

(A) The image will be formed at distance of

7.5 cm from the optical centre (B) The image will be formed at distance of

10 cm from the optical centre (C) The size of the image is 0.5 cm (D) The size of the image is 0.4 cm

Q.10 bdkbZ nzO;eku rFkk bdkbZ vkos'k dk ,d vkosf'kr

d.k osx )j6i8(v ^^ +=→ m/s ls

→B = Tk2 ^ ds pqEcdh;

ks=k esa xfr djrk gSA lgh fodYi@fodYiksa dk p;udhft,A

(A) d.k dk iFk fuEu gks ldrk gS x2 + y2 – 4x – 21 = 0 (B) d.k dk iFk x2 + y2 = 25 gks ldrk gS (C) d.k dk iFk y2 + z2 = 25 gks ldrk gS (D) d.k dk vkorZdky 3.14s gksxk

Q.11 1 cm yEckbZ dh ,d oLrq dks 5 cm f=kT;k ds fmÙky ysUl dh eq[; vk ij j[kk tkrk gSA ysUl rFkk oLrq ds e/; nwjh 20 cm gSA ysUl rFkk oLrq ds e/; dsLFkku dks nks fHkUu-fHkUu viorZukadksa ds ek/;e 1 o 2 ls fp=kkuqlkj Hkjk x;k gSA oLrq ds ck;as ij rFkk ysUl dh nk;ha rjQ viorZukad 1 gSA nksuksa ek/;e dh lhek fp=kkuqlkj oLrq o ysUl ds ek/; esa gSA

µ = 2 µ = 1 1.5 µ = 1

10 cm 10 cm

1 cm

(A) izfrfcEc izdkf'kd dsUnz ls 7.5 cm dh nwjh ij

cusxk (B) izfrfcEc izdkf'kd dsUnz ls 10 cm dh nwjh ij cusxk (C) izfrfcEc dk vkdkj 0.5 cm gS (D) izfrfcEc dk vkdkj 0.4 cm gS



Q.12 Radiations of monochromatic waves of

wavelength 400 nm are made incident on the surface of metals Zn, Fe and Ni of work functions 3.4 eV, 4.8 eV and 5.9 eV respectively (take hc = 12400 eV-Å) :

(A) maximum KE associated with photoelectrons from the surface of any metal is 0.3 eV

(B) no photoelectrons are emitted from the surface of Ni

(C) if the wavelength of source of radiation is doubled then KE of photoelectrons is also doubled

(D) photoelectrons will be emitted from the surface of all the three metals if the wavelength of incident radiations is less than 200 nm

This section contains 2 paragraphs; each has 3 multiple choice questions. (Questions 13 to 18) Each question has 4 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. + 4 marks will be given for each correct answer and – 1 mark for each wrong answer.

Q.12 400 nm rjaxnS/;Z dh ,do.khZ rjaxksa dh fofdj.ksa

Zn, Fe rFkk Ni /kkrqvksa dh lrg ij vkifrr gksrh

gS rFkk ftudk dk;ZQyu Øe'k% 3.4 eV, 4.8 eV o

5.9 eV gS (hc = 12400 eV-Å ysa) :

(A) fdlh Hkh /kkrq dh lrg ls QksVksbysDVªkWuksa ls

lEcfU/kr vf/kdre KE, 0.3 eV gS

(B) Ni dh lrg ls QksVksbysDVªkWu mRlftZr ugha

gksrs gS

(C) ;fn fofdj.k ds L=kksr dh rjaxnS/;Z nqxquh dj

nh tk;s rc QksVksbysDVªkWuksa dh KE Hkh nqxquh

gks tkrh gS

(D) lHkh rhu /kkrqvksa dh lrg ls QksVksbysDVªkWu

mRlftZr gksxsa ;fn vkifrr fofdj.kksa dh

rjaxnS/;Z 200 nm ls de gksrh gS





A hydrogen atom in third excited state makes a

transition to first excited state and emit photon.

This emitted photon is absorbed by He+ ion

which was already in seventh excited state.

After absorption of photon He+ ion jumps from

seventh excited state to an excited state having

quantum number nf.

Q.13 The quantum number nf of the state finally

populated in He+ ion

(A) 8 (B) 10

(C) 16 (D) 20

Q.14 Now He+ ion jumps to lower state by emitting

single photon ‘P’ of visible light. The energy of

this photon is as possible as close to energy of

the absorbed photon by He+ ion (as mentioned

in paragraph). Then wavelength of this photon

‘P’ is nearly.

(A) 6760Å (B) 5000Å

(C) 4480 Å (D) 3500Å

x|ka'k # 1 (iz- 13 ls 15)

rhljh mÙksftr voLFkk esa fLFkr ,d gkbMªkstu ijek.kq igyh mÙksfrt voLFkk esa laØe.k dj QksVkWu mRlftZr djrk gSA ;g mRlftZr QksVkWu He+ vk;u kjk vo'kksf"kr gksrk gS tks igys ls lkroha mÙksftr voLFkk FkkA QksVkWu ds vo'kks"k.k ds ckn He+ vk;u lkroha mÙksftr voLFkk ls nƒ Dok.Ve la[;k okyh ,d mÙksftr voLFkk esa Nykax yxkrk gSA

Q.13 He+ vk;u esa vfUre voLFkk esa mifLFkr Dok.Ve

la[;k nƒ gS (A) 8 (B) 10 (C) 16 (D) 20

Q.14 vc He+ vk;u n'; izdk'k ds ,dy QksVkWu ‘P’

ds mRltZu kjk fupys Lrj esa Nykax yxkrk gSA

bl QksVkWu dh ÅtkZ He+ vk;u (x|ka'k esa O;k[;k

ds vuqlkj) kjk vo'kksf"kr QksVkWu dh ÅtkZ ds

yxHkx cjkcj gksrh gSA rc bl QksVkWu ‘P’ dh

rjaxnS/;Z yxHkx gS

(A) 6760Å (B) 5000Å (C) 4480 Å (D) 3500Å



Q.15 The ratio of the kinetic energy of the n = 4

electron for H atom to that He+ ion is P1 and the

ratio of the total energy of the n = 6 electron for

the H atom to that of He+ ion is P2. Then ratio

P1/P2 is

(A) 1 (B) 9/4

(C) 4/9 (D) 4


A uniform solid sphere of mass M and radius R

is attached to four identical light springs of

spring constant k which are fixed to walls as

shown in figure. The springs are attached

symmetrically to the axle of the sphere as

shown in figure. The axle is mass less and all

the springs and the axle are in horizontal plane.

The unstretched length of each spring is L. The

sphere is initially at its equilibrium position

with its centre of mass (CM) at a distance L

from the each wall. The sphere rolls without

slipping with initial velocity ^00 ivv =

→ and

angular velocity ω0. The coefficient of friction

is µ.

Q.15 H ijek.kq rFkk He+ vk;u ds fy;s n = 4 bysDVªkWu

dh xfrt ÅtkZ dk vuqikr P1 gS rFkk H ijek.kq

o He+ vk;u ds fy;s n = 6 bysDVªkWu dh dqy

ÅtkZ dk vuqikr P2 gSA rc vuqikr P1/P2 gS

(A) 1 (B) 9/4 (C) 4/9 (D) 4

x|ka'k # 2 (iz- 16 ls 18)

M nzO;eku rFkk R f=kT;k ds ,d ,dleku Bksl

xksys dks k cy fu;rkad dh pkj le:i gYdh

fLizaxksa tks fp=kkuqlkj nhokjksa ls dlh gqbZ gS ls

tksM+k x;k gSA fLiazxsa xksys dh vk (axle) ls

fp=kkuqlkj lefer :i ls tksM+h xbZ gSA vk

nzO;ekughu rFkk lHkh fLizaxsa rFkk vk kSfrt ry

esa gSA izR;sd fLiazx dh vfoLrkfjr yEckbZ L gSA

xksyk izkjEHk esa izR;sd nhokj ls L nwjh ij mlds

nzO;eku dsUnz (CM) ls viuh lkE;koLFkk fLFkfr

ij gSA xksyk izkjfEHkd osx ^00 ivv =

→ rFkk dks.kh;

osx ω0 ls fcuk fQlys yq<+drk gSA ?k"kZ.k xq.kkad µ

gSA



k k

2dv0 k

x

2dk

y

Q.16 The net external force acting on the sphere

when its centre of mass is at displacement x with respect to its equilibrium position is

(A) – 10 kx (B) –20/7 kx (C) –2/5 kx (D) –7/5 kx Q.17 The centre of mass of the sphere undergoes

simple harmonic motion time period T equal to

(A) k20

M72π (B) k10

M2π

(C) k2M52π (D)

k7M52π

Q.18 The maximum value of angular velocity ω0 about centre of mass for which sphere will roll without slipping is

(A) k5M7

Rgµ (B)

k20M7

Rgµ

(C) k5M49

Rgµ (D)

k16M35

Rgµ

kk

2dv0 k

x

2dk

y

Q.16 xksys ij dk;Zjr usV ckg~; cy tc mldh

lkE;koLFkk fLFkfr ds lkisk mldk nzO;eku dsUnz x foLFkkiu ij gS-

(A) – 10 kx (B) –20/7 kx (C) –2/5 kx (D) –7/5 kx

Q.17 xksys dk nzO;eku dsUnz T vkorZdky ls ljy

vkorZ xfr djrk gS] T cjkcj gS

(A) k20

M72π (B) k10

M2π

(C) k2M52π (D)

k7M52π

Q.18 nzO;eku dsUnz ds lkisk dks.kh; osx ω0 dk

vf/kdre eku ftlds fy,s xksyk fcuk fQlys

yq<dsxk] gS

(A) k5M7

Rgµ (B)

k20M7

Rgµ

(C) k5M49

Rgµ (D)

k16M35

Rgµ



Section - II This section contains 2 questions (Questions 1, 2).

Each question contains statements given in two

columns which have to be matched. Statements (A, B,

C, D) in Column I have to be matched with

statements (P, Q, R, S, T) in Column II. The answers

to these questions have to be appropriately bubbled

as illustrated in the following example. If the correct

matches are A-P, A-S, A-T; B-Q, B-R; C-P, C-Q and

D-S, D-T then the correctly bubbled 4 × 5 matrix

should be as follows :

A B C D

P Q R S T

T S

P

P P Q R

R R

Q Q

S S T

T

P Q R S T

Mark your response in OMR sheet against the

question number of that question in section-II. + 8

marks will be given for complete correct answer (i.e.

+2 marks for each correct row) and No Negative

marks for wrong answer.

[k.M - II

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

LrEHkksa esa dFku fn;s x;s gSa] ftUgsa lqesfyr djuk gSA

LrEHk-I (Column I ) esa fn;s x;s dFkuksa (A, B, C, D) dks

LrEHk-II (Column II) esa fn;s x;s dFkuksa (P, Q, R, S,T) ls

lqesy djuk gSA bu iz'uksa ds mÙkj uhps fn;s x;s

mnkgj.k ds vuqlkj mfpr xksyksa dks dkyk djds n'kkZuk

gSA ;fn lgh lqesy A-P, A-S, A-T; B-Q, B-R; C-P, C-

Q rFkk D-S, D-T gS, rks lgh fof/k ls dkys fd;s x;s

xksyksa dk 4 × 5 eSfVªDl uhps n'kkZ;s vuqlkj gksxk

ABCD

P Q R S T

T S

P

P P Q R

R R

Q Q

S S T

T

P Q R S T




vad fn, tk,asxs) rFkk xyr mÙkj ds fy;s dksbZ _.kkRed vadu

ugha gS (vFkkZr~ dksbZ vad ugha ?kVk;k tk;sxk)A



Q.1 An real object ‘S’ is placed on the optic axis an optical instrument as shown in column-I. The distance between the object and the instrument can be varied. Some statements are given in column-II about image. f is the magnitude of focal length of the instrument. Match the optical instruments given in column-I with the appropriate properties of images given in column-II.

Column-I Column-II

(A) S

(P) If image distance

r from pole lies in range 0 < r < f, then image must be virtual

(B)S

(Q) If image distance

r from pole lies 0 < r < f, then image must be magnified

Q.1 ,d okLrfod oLrq ‘S’ dks LrEHk-I esa n'kkZ;s

izdkf'k;h midj.k dh izdkf'kd vk ij j[kk x;k

gSA oLrq rFkk midj.k ds e/; dh nwjh ifjofrZr

gks ldrh gSA LrEHk-II eas izfrfcEc ds lkisk dqN

dFku fn;s x;s gSA f midj.k dh Qksdl nwjh dk

ifjek.k gSA LrEHk-I esa fn;s x;s izdkf'k; midj.kksa

dks LrEHk-II esa fn;s x;s izfrfcEcksa ds mi;qDr

xq.kksa ds lkFk lqesfyr dhft,A LrEHk-I LrEHk-II

(A) S

(P) ;fn /kzqo ls

izfrfcEc dh nwjh r ijkl 0 < r < f eas gS, rks izfrfcEc vo'; vkHkklh gksxk

(B)S

(Q) ;fn /kzqo ls

izfrfcEc dh nwjh r ijkl 0 < r < f esa

gS, rks izfrfcEc vo'; vko/khZr gksxk



(C) S

(R) If object distance

r from pole is in

range f < r < 2f,

then image must

be real

(D) S

(S) If object distance r

from the pole is in

the range 2f < r < ∞,

then image must

be diminished

(T) If image distance

from the pole is in

the range of

2f < r < ∞, the image

must be real.

(C) S

(R) ;fn /kzqo ls oLrq

dh nwjh r ijkl

f < r < 2f esa

gS, rks izfrfcEc

vo'; okLrfod gksxk

(D) S

(S) ;fn /kqzo ls oLrq

dh nwjh r ijkl

2f < r < ∞ eas gksrh

gS, rks izfrfcEc

vo'; NksVk gksxk

(T);fn /kzqo ls izfrfcEc

dh nwjh ijkl

2f < r < ∞ esa gksrh

gS] rks izfrfcEc

vo'; okLrfod

gksxk



Q.2 In column I, some thermodynamic processes are given for an ideal gas. Match the processes given in column I with corresponding thermodynamic change given in column II.

Column- I Column-II (A) An insulated container (P) The pressure of has two chambers the gas decreases separated by a light piston. The chamber I contains an ideal gas and the chamber II has vacuum. Piston is frictionless. The piston is released from shown position

piston

Ideal gas vacuum

(B) An ideal monoatomic (Q) The internal gas expands to thrice energy of gas its original volume decreases such that its pressure

P ∝ 3V1 where V is

the volume of gas

Q.2 LrEHk-I esa] ,d vkn'kZ xSl ds fy;s dqN Å"ekxfrdh izØe fn;s x;s gSA LrEHk-I esa fn;s x;s izØeksa dks LrEHk-II esa fn;s x;s lEcfU/kr Å"ekxfrdh ifjorZuksa ls feyku dhft,A

LrEHk- I LrEHk-II (A) ,d dqpkyd ik=k esa (P) xSl dk nkc ?kVrk gS

nks psEcj gS tks gYds

fiLVu kjk iFkd gS

psEcj- I esa vkn'kZ xSl

rFkk psEcj-II esa fuokZr

gSA fiLVu ?k"kZ.kghu gSA

fiLVu dks n'kkZ;h fLFkfr

ls NksM+k x;k gS

piston

Ideal gas vacuum

(B) ,d vkn'kZ ,dyijek.kq (Q) xSl dh vkUrfjd

xSl dks mlds izkjfEHkd ÅtkZ ?kVrh gS

vk;ru ds rhu xqus

rd bl izdkj izlkfjr

fd;k tkrk gS fd mldk nkc

P ∝ 3V1

gS tgk¡ V xSl

dk vk;ru gS



(C) An ideal monoatomic (R) The gas loses heat

gas expands to twice

its original volume

such that its temperature

T ∝ 3/2V

1 , where V

is the volume of gas

(D) An ideal monoatomic (S) The gas neither

gas expands such that loses heat nor

its temperature T and gains heat

V follows the behaviour

shown in the graph (T) Work done is

and T ∝ V2 positive

V1 V2 V

T

(C) ,d vkn'kZ ,dy (R) xSl Å"ek [kksrh gS ijek.kq xSl dks mlds izkjfEHkd vk;ru ds nqxqus rd bl izdkj izlkfjr fd;k tkrk gS

mldk rki T ∝ 3/2V1

gS, tgk¡ V xSl dk vk;ru gS (D) ,d vkn'kZ ,dy (S) xSl u rks Å"ek ijek.kqd xSl bl [kksrh gS u gh xzg.k izdkj izlkfjr gksrh gS djrh gS fd mldk rki T o V xzkQ esa n'kkZ;k O;ogkj djrh gS rFkk T ∝ V2 (T) fd;k x;k dk;Z /kukRed gS

V1 V2 V

T



Space for Rough Work (jQ+ dk;Z gsrq LFkku)



Time : 3 : 00 Hrs. MAX MARKS: 243

INSTRUCTIONS TO CANDIDATE

A. lkekU; :

1. Ñi;k izR;sd iz'u ds fy, fn, x, funsZ'kksa dks lko/kkuhiwoZd if<+;s rFkk lEcfU/kr fo"k;kas esa mÙkj&iqfLrdk ij iz'u la[;k ds lek lgh mÙkj fpfUgr dhft,A

2. mRrj ds fy,] OMR vyx ls nh tk jgh gSA 3. ifjohkdksa kjk funsZ'k fn;s tkus ls iwoZ iz'u&i=k iqfLrdk dh lhy dks ugha [kksysaA

B. vadu i)fr: bl iz'ui=k esa izR;sd fo"k; esa fuEu izdkj ds iz'u gSa:- [k.M – I 4. cgqfodYih izdkj ds iz'u ftuesa ls dsoy ,d fodYi lgh gSA izR;sd lgh mÙkj ds fy, 3 vad fn;s tk;saxs o izR;sd xyr mÙkj ds

fy, 1 vad ?kVk;k tk,xkA 5. cgqfodYih izdkj ds iz'u ftuesa ls ,d ;k ,d ls vf/kd fodYi lgh gSaA izR;sd lgh mÙkj ds fy, 3 vad fn, tk;saxs rFkk dksbZ

_.kkRed vadu ugha gSA 6. x|ka'k ij vk/kkfjr cgqfodYih izdkj ds iz'u ftuesa ls dsoy ,d gh fodYi lgh gSaA izR;sd lgh mÙkj ds fy, 3 vad fn, tk;saxs

rFkk izR;sd xyr mÙkj ds fy, 1 vad ?kVk;k tk;sxkA [k.M – III 7. x.kukRed izdkj ds iz'u gSaA izR;sd lgh mÙkj ds fy, 3 vad fn;s tk;saxs rFkk xyr mÙkj ds fy, dksbZ _.kkRed vadu ugha gSA bl

[k.M esa mÙkj bdkbZ iw.kk±d esa nhft, (tSls 0 ls 9)A

C. OMR dh iwfrZ :

8. OMR 'khV ds CykWdksa esa viuk uke] vuqØek¡d] cSp] dkslZ rFkk ijhkk dk dsUnz Hkjsa rFkk xksyksa dks mi;qDr :i ls dkyk djsaA 9. xksyks dks dkyk djus ds fy, dsoy HB isfUly ;k uhys/dkys isu (tsy isu iz;ksx u djsa) dk iz;ksx djsaA 10. di;k xksyks dks Hkjrs le; [k.Mks dks lko/kkuh iwoZd ns[k ysa [vFkkZr [k.M I (,dy p;ukRed iz'u] dFku izdkj ds iz'u]

cgqp;ukRed iz'u), [k.M –II (LrEHk lqesyu izdkj ds iz'u), [k.M-III (iw.kkZd mÙkj izdkj ds iz'u½]

Section –I Section-II Section-III

For example if only 'A' choice is correct then, the correct method for filling the bubbles is

A B C D E

For example if only 'A & C' choices are correct then, the correct method for filling the bublles is

A B C D E

the wrong method for filling the bubble are

The answer of the questions in wrong or any other manner will be treated as wrong.

For example if Correct match for (A) is P; for (B) is R, S; for (C) is Q; for (D) is P, Q, S then the correct method for filling the bubbles is

P Q R S TA BCD

Ensure that all columns are filled. Answers, having blank column will be treated as incorrect. Insert leading zeros (s)

012

3

4

56

7

8

9

012

3

4

56

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012

3

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56

7

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012

3

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56

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012

3

4

56

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0 1 2

3

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5 6

7

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9

0 1 2

3

4

5 6

7

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0 1 2

3

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5 6

7

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9


0 0 0 00 1 2

3

4

5 6

7

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0 1 2

3

4

5 6

7

8

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012

3

4

56

7

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9

012

3

4

56

7

8

9


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