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INSTRUCTIONS

MARKING SCHEME Each subject in this paper consists of following types of questions:-

Section - I Multiple choice questions with only one correct answer. 3 marks will be awarded for each correct answer and –1 mark for each wrong answer. Multiple choice questions with multiple correct option. 3 marks will be awarded for each correct answer and No negative marking. Passage based single correct type questions. 3 marks will be awarded for each correct answer and –1 mark for each wrong answer.

Section - II

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)

TO DOWNLOAD PAPER-II To download Paper-II of JEE Advanced Pattern, please visit www.careerpoint.ac.in , and click on link – JEE Advance Mock Test Paper (Paper-II)



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 each correctanswer and – 1 mark for each wrong answer.

Q.1 Which is not true about borax ? (A) It is a useful primary standard for titrating

against acids (B) One mole of borax can be used as a buffer (C) Aqueous solution of borax can be used as

buffer (D) It is made up of two triangular BO3 units and

two tetrahedral BO4 units

Q.2 Give the correct order of initials T or F offollowing statements. Use T if statements is trueand F if it is false.

(i) In Gold schmidt thermite process aluminiumacts as a reducing agent.

(ii) Mg is extracted by electrolysis of aq.solution of MgCl2

(iii) Extraction of Pb is possible by carbon reductionmethod

(iv) Red Bauxite is purified by reduction method (A) TTTF (B) TFFT (C) FTTT (D) TFTF

[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 dkVk tk;sxkA

Q.1 ckWjsDl ds lanHkZ esa dkSulk lR; ugha gS ? (A) ;g vEyksa ds vuqekiu ds fy, ,d mi;ksxh

izkFkfed ekud gksrk gS (B) ,d eksy ckWjsDl] cQj ds :i esa iz;qDr gks ldrk gS (C) ckWjsDl dk tyh; foy;u] cQj ds :i esa iz;qDr

gks ldrk gS (D) ;g nkss f=kdks.kh; BO3 bdkbZ;ks rFkk nks prq"Qydh;

BO4 bdkbZ;ks dk cuk gksrk gS

Q.2 fuEu dFkuks ds izkjfEHkd T ;k F dk lgh Øe nhft,A ;fn dFku lR; gS rks T rFkk vlR; gS rks F dk mi;ksx dhft,A

(i) xksYM f'eM~V FkekZekbV izØe esa ,Y;wfefu;e vipk;d dh rjg dk;Z djrk gS

(ii) MgCl2 ds tyh; foy;u ds oS|qr vi?kV~u kjk Mg fu"df"kZr gksrk gS

(iii) dkcZu viPk;u fof/k kjk Pb dk fu"d"kZ.k lEHko gS (iv) vip;u fof/k kjk yky ckWDlkbV 'kqf)d`r gks

ldrk gS (A) TTTF (B) TFFT (C) FTTT (D) TFTF



Q.3 When negatively charged colloid like As2S3 sol is added to positively charged Fe(OH)3 sol in stoichiometric amounts ?

(A) Both the sols are precipitated simultaneously (B) They becomes positively charged colloid (C) They become negatively charged colloid (D) None of these

Q.4 At a certain temperature and 2 atm pressure equilibrium constant (kp) is 25 for the reaction SO2(g) + NO2(g) SO3(g) + NO(g)

Initially if we take 2 moles of each of the fourgases and 2 moles of inert gas, what would be the equilibrium partial pressure of NO2 ?

(A) 1.33 atm (B) 0.1665 atm (C) 0.133 atm (D) None of these

Q.5

OR

→ 3BH →–

22 HO,OH (A)

Product (A) is -

(A)

OR

OH

(B)

OR

OH

(C)

OR

OH

OH (D)

OR OH

Q.3 tc LVkWbfd;ksesfVªd ek=kkvksa esa _.kkosf'kr dkWyksbM tSls As2S3 lkWy dks /kukosf'kr dkWyksbM Fe(OH)3

lkWy esa feyk;k tkrk gS rks ? (A) nksuksa lkWy ,d lkFk voksfir gksrs gS (B) ;s /kukosf'kr dkWyksbM gks tkrs gS (C) ;s _.kkosf'kr dkWyksbM gks tkrs gS (D) buesa ls dksbZ ugha

Q.4 fuf'pr rki o 2 atm nkc ij fuEu vfHkfØ;k ds fy,lkE; fLFkjkad (kp), 25 gSA

SO2(g) + NO2(g) SO3(g) + NO(g) izkjEHk esa ;fn pkj xSlksa ds izR;sd ds 2 eksy rFkk

vfØ; xSl ds 2 eksy fy;s x;s rks NO2 dk lkE; vkaf'kd nkc D;k gksxk ?

(A) 1.33 atm (B) 0.1665 atm (C) 0.133 atm (D) buesa ls dksbZ ugha

Q.5

OR

→ 3BH →–

22 HO,OH (A)

mRikn (A) gS -

(A)

OR

OH

(B)

OR

OH

(C)

OR

OH

OH (D)

OR OH



Q.6 Major product (P) in following reaction

CH3 – CO —)CH(— 42 CHO∆ → HO (P) -

(A) O

|| CH3

(B)

(C) CHO

(D)

CHO

Q.7

CH3

ν →

h

ClSO 22 (A) ∆ →NBS (B) →KSH (C),

major product (C) is -

(A) SH

Br

(B) SH

Cl

(C)

Br

SH

(D)

Br

Q.6 fuEu vfHkfØ;k esa eq[; mRikn (P) gS s-

CH3 – CO —)CH(— 42 CHO∆ → HO (P) -

(A) O

||CH3

(B)

(C) CHO

(D)

CHO

Q.7

CH3

ν →

h

ClSO 22 (A) ∆ →NBS (B) →KSH (C),

mRikn (C) gS -

(A) SH

Br

(B) SH

Cl

(C)

Br

SH

(D)

Br



Q.8

OH

OH PCC(excess)

(A)

1 equivalent

OH OH

(B) CH3MgBr

H3O+

(C) NaBH4

H2O (D) Product (D) will be -

(A)

OH CH–CH3 | OH

(B) OH

OH

(C) O

OH

(D) OH

OH

Questions 9 to 13 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. + 3 marks will be given for each correct answer and no negative marks.

Q.8

OH

OHPCC(excess)

(A)

1 equivalent

OHOH

(B)

CH3MgBr

H3O+ (C) NaBH4

H2O (D)

mRikn (D) gS -

(A)

OHCH–CH3 |OH

(B) OH

OH

(C) O

OH

(D) OH

OH

iz'u 9 ls 13 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,

+ 3 vad fn;s tk;saxs rFkk dksbZ _.kkRed vadu ugha gSA



Q.9 lgh dFkuksa dk p;u dhft, - (A) BeCO3 dks ok;qe.Myh; CO2 esa j[kk tkrk gS

pwafd ;g U;wure rkih; LFkkbZ gSA (B) kkjh; foy;u esa Be, [Be(OH)4]2– ds :i esa

?kqyrk gS (C) BeF2, NaF ds lkFk ladqy vk;u cukrk gSA

ftlesa Be, /kuk;u ;qDr gksrk gSA (D) BeF2, NaF ds lkFk ladqy vk;u cukrk gS ftlesa

Be, _.kk;u ;qDr gksrk gS

Q.10 dkSulh vfHkfØ;k;s lEHko gS ? (A) MgCl2 + NaNO3 → (B) BaSO4 + HCl → (C) ZnSO4 + BaS → (D) BaCO3 + CH3COOH → Q.11 ;gh dFkuks dk p;u dhft, - (A) ukfHkd ds fudV fLFkr ,d bysDVªkWu] ukfHkd kjk

vkdf"kZr gksrk gS rFkk fLFkfrt ÅtkZ U;wu gksrh gS (B) cksj fl)kUr ds vuqlkj ,d bysDVªkWu yxkrkj

ÅtkZ fu"dkflr djrk gS ;fn ;g ,d dks'k esa fLFkr gks

(C) cksj ekWMy] cgqr bysDVªkWuh ijek.kqvksa ds LisDVªk dh O;k[;k ugh dj ldrk

(D) cksj ekWMy] ÅtkZ ds Dok.Vhdj.k ij vk/kkfjr izFke ijek.kq ekWMy Fkk

Q.9 Choose the correct statement (s) - (A) BeCO3 is kept in the atmosphere of CO2

since, it is least thermally stable (B) Be dissolves in an alkali solution forming

[Be(OH)4]2– (C) BeF2 forms complex ion with NaF in which

Be goes with cation (D) BeF2 forms complex ion with NaF in which

Be goes with anion Q.10 Which reaction is/are possible ? (A) MgCl2 + NaNO3 → (B) BaSO4 + HCl → (C) ZnSO4 + BaS → (D) BaCO3 + CH3COOH →

Q.11 Select the correct statement (s) - (A) An electron near the nucleus is attracted by

the nucleus and has a low potential energy (B) According to Bohr's theory, an electron

continuously radiate energy if it stayed inone orbit

(C) Bohr's model could not explain the spectra ofmultielectron atoms

(D) Bohr's model was the first atomic modelbased on quantization of energy



Q.12 Which of the following cycloalkanes will showcis-trans isomerism -

(A)

CH3

(B)

CH3

CH3

(C)

CH3

H3C

(D)

H3C CH3

Q.13 Which of the following combinations give Ph – CH3 ?

(A) Ph – CH2 – MgBr + NO2 – – OH →

(B) PhCH2COONa – +

NaOH

CaO →

(C) CH3MgBr and

C6H5–C–OC2H5 || O

→+H/OH2

(D) Ph – CH2MgBr → 43POH

This section contains 2 paragraphs; passage- I has 2 multiple choice questions (No. 14 & 15) and passage-II has 3 multiple (No. 16 to 18). Each question has 4choices (A), (B), (C) and (D) out of which ONLY ONEis correct. Mark your response in OMR sheet againstthe question number of that question. + 3 marks willbe given for each correct answer and – 1 mark foreach wrong answer.

Q.12 fuEu esa ls dkSuls lkbDyks,Ydsu fll-Vªkal leko;ork

n'kkZ;saxkA -

(A)

CH3

(B)

CH3

CH3

(C)

CH3

H3C

(D)

H3C CH3

Q.13 dkSuls la;kstu fuEu ;kSfxd nsxsaA Ph – CH3 ?

(A) Ph – CH2 – MgBr + NO2 – – OH →

(B) PhCH2COONa – +

NaOH

CaO →

(C) CH3MgBr rFkk

C6H5–C–OC2H5

|| O

→+H/OH2

(D) Ph – CH2MgBr → 43POH

bl [k.M esa 2 x|ka'k fn;s x;s gSa] x|ka'k -I esa 2 cgqfodYih iz'u (iz'u 14 o 15) gSa rFkk x|ka'k-II esa 3 cgqfodYih iz'u (iz'u 16 ls 18) 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;s + 3 vad fn;s tk,saxs rFkk izR;sd xyr mÙkj ds fy, 1 vad dkVk tk;sxkA



Passage # 1 (Ques. 14 & 15) (i) P + C(carbon) + Cl2 → Q + CO↑ (ii) Q + H2O → R + HCl (iii) BN + H2O → R + NH3↑ (iv) Q + LiAlH4 → S + LiCl + AlCl3 (v) S + H2 → R + H2↑ (vi) S + NaH → T (P, Q, R, S and T do not represent their chemicalsymbols)

Q.14 Compound Q has - (I) zero dipole moment (II) a planar trigonal structure (III) an electron deficient compound (IV) a Lewis base Choose the correct code - (A) I, IV (B) I, II, IV (C) I, II, III (D) I, II, III, IV Q.15 Compound T is used as a/an - (A) oxidising agent (B) complexing agent (C) bleaching agent (D) reducing agent Passage # 2 (Ques. 16 to 18)

Ph– C HO– Ph– C–H + C–PH

|| |

|

O O–

O–H

slow|| O

| H

x|ka'k # 1 (iz- 14 ,oa 15) (i) P + C(carbon) + Cl2 → Q + CO↑ (ii) Q + H2O → R + HCl (iii) BN + H2O → R + NH3↑ (iv) Q + LiAlH4 → S + LiCl + AlCl3 (v) S + H2 → R + H2↑ (vi) S + NaH → T (P, Q, R, S T muds jklk;fud ladsrks dks ugha n'kkZrs)

Q.14 ;kSfxd Q j[krk gS - (I) 'kwU; f/kzqo vk/kw.kZ (II) ,d leryh; f=kdks.kh; lajpuk (III) ,d bysDVªkWu U;wu ;kSfxd (IV) ,d yqbZl kkj lgh dwV dk p;u dhft, - (A) I, IV (B) I, II, IV (C) I, II, III (D) I, II, III, IV Q.15 ;kSfxd T fdl :i esa iz;qDr gksrk gS - (A) vkWDlhdkjd (B) ladqy dkjd (C) fojatd dkjd (D) vipk;d

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

Ph– CHO–

Ph– C–H + C–PH || |

|

O O–

O–H

slow||O

|H



Ph– C–OH + Ph–CH2–O– || O

Ph– –

2CO + Ph – CH2 – OH

Q.16 Which of the following reactants can undergo Cannizaro's reaction.

(A)

H–C–H || O

(B) R3CCHO (C)

O CHO

(D) All of these

Q.17 Order of the above reaction is - (A) 1 (B) 2 (C) 3 (D) 4 Q.18 Which of the following is best hydride donor in

Cannizaro's reaction -

(A)

CHO

OMe

(B)

CHO

NO2

(C)

CHO

CH3

(D)

CHO

Cl

Ph– C–OH + Ph–CH2–O–||O

Ph– –

2CO + Ph – CH2 – OH Q.16 fuEu esa ls dkSuls fØ;kdkjd dsuhtkjks vfHkfØ;k

lEiUu dj ldrs gS -

(A)

H–C–H||O

(B) R3CCHO (C)

O CHO

(D) buesa ls dksbZ ugha

Q.17 mi;qZDr vfHkfØ;k dh dksfV gS - (A) 1 (B) 2 (C) 3 (D) 4

Q.18 dSuhtkjks vfHkfØ;k esa fuEu esa ls dkSulk mi;qDr gkbMªkbM nkrk gksrk gS -

(A)

CHO

OMe

(B)

CHO

NO2

(C)

CHO

CH3

(D)

CHO

Cl



Section - III This section contains 9 questions (Q.1 to 9).+3 marks will be given for each correct answer and nonegative marking. The answer to each of the questionsis a SINGLE-DIGIT INTEGER, ranging from 0 to 9.The appropriate bubbles below the respectivequestion numbers in the OMR have to be darkened.For example, if the correct answers to questionnumbers X, Y, Z and W (say) are 6, 0, 9 and 2, respectively, then the correct darkening of bubbleswill look like the following :

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 1 2

3

4

5 6

7

8

9

X Y Z W

Q.1 A(Light pink colour complex)

∆ → 343 HNO.dil/OPb

HMNO4 →+H/SH2 A (Light pink colour complex).

Calculate CFSE value in light pink colour complex.

[k.M - III bl [k.M esa 9 (iz-1 ls 9) iz'u gaSA izR;sd lgh mÙkj ds fy;s

+3 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 gSaA OMR esa iz'u la[;k ds laxr uhps fn;s x;s

cqYyksa esa ls lgh mÙkj okys cqYyksa dks dkyk fd;k tkuk gSA

mnkgj.k ds fy, ;fn iz'u la[;k ¼ekusa½ X, Y, Z rFkk W ds

mÙkj 6, 0, 9 rFkk 2 gkas, rks lgh fof/k ls dkys fd;s x;s cqYys

,sls fn[krs gSa tks fuEufyf[kr gSA

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

X Y Z W

Q.1 A(pedhyk xqykch ladqy) ∆

→ 343 HNO.dil/OPb

HMNO4 →+H/SH2 A (pedhyk xqykch ladqy).

pedhyk xqykch ladqy ds CFSE eku dh x.kuk dhft,A



Q.2 Consider the following oxyanions :

–34PO , –4

62OP , –24SO , –

4MnO , –24CrO , –2

52OS ,

–272OS

and find the value of R + Q – P

Where

P = Number of oxy anions having three equivalent

X – O bonds per central atom

Q = Number of oxy anions having two equivalent


R = Number of oxy anions having four equivalent


Q.3 Density of Li atom is 0.53 g/cm3. The edgelength of Li is 3.5 Å. Find out the number ofLi atoms in a unit cell.

(NA = 6.0 × 1023 mol–1, M = 6.94 g mol–1)

Q.4 Two flask A and B of equal volumes maintained

at temperature 300 K and 700 K contain equal

mass of He(g) and N2(g) respectively. What is

the ratio of total translational kinetic energy of

gas in flask A to that of flask B ?

Q.2 fuEu vkWDlh_.kk;uks ij fopkj dhft,A

–3

4PO , –462OP , –2

4SO , –4MnO , –2

4CrO , –252OS ,

–272OS

rFkk R + Q – P dk eku Kkr dhft,A

tgk¡

P = izfr dsfUnz; ijek.kq] rhu leku

X – O ca/kksa ;qDr vkWDlh _.kk;uks dh la[;k

Q = izfr dsfUnz; ijek.kq] nks leku

X – O ca/kks ;qDr vkWDlh _.kk;uksa dh la[;k

R = izfr dsfUnz; ijek.kq] pkj leku

X – O ca/kks ;qDr vkWDlh _.kk;uksa dh la[;k

Q.3 Li ijek.kq dk ?kuRo 0.53 g/cm3 gSA Li dh dksjyEckbZ 3.5 Å gSA ,d bdkbZ dksf"Vdk esa Liijek.kqvksa dh la[;k Kkr dhft,A

(NA = 6.0 × 1023 mol–1, M = 6.94 g mol–1)

Q.4 leku vk;ru ds nks ¶ykLd A rFkk B, rki 300 K

rFkk 700 K ij fLFkr gS ftlesa Øe'k% He(g) rFkk

N2(g) ds leku nzO;eku gSA ¶ykLd A ls ¶ykLd B

esa dqy LFkkukUrfjr xfrt ÅtkZ dk vuqikr D;k

gksxk ?



Q.5 For the reaction 2A(g) + B(g) C(g) + D(g);KC = 1012. If the initial moles of A, B, C and D are2, 1, 7 and 3 moles respectively in a one litre vessel.What is the equilibrium concentration of A interm of n × 10–12 ?

Q.6 How many of the following reactions arecorrectly matched ?

(a)

CH3–C–O–CH3

CH3

|

|

CH3

1NHI S→

(b) CH3–CH–O–CH3

CH3 |

12

NOH/H S →

⊕

(c) CH3–CH–O–CH3

CH3 |

1NHI.Conc S →

(d) CH3–O–CH2CH3 1NHI S→

(e)

–O–CH2–CH = CH2 1NHI.Conc S →

(f) CH2 = CH – O – CH = CH2 12

NOH/H S →

⊕

(g)

–O–CH2– 12

NOH/H S →

⊕

Q.5 vfHkfØ;k 2A(g) + B(g) C(g) + D(g); ds fy,

KC = 1012 gSA ;fn ,d yhVj ik=k esa A, B, C rFkk D

ds izkjfEHkd eksy Øe'k% 2, 1, 7 rFkk 3 eksy gS] rks A

dh lkE; lkUnzrk n × 10–12 ds inksa esa D;k gksxh ?

Q.6 fuEu esa ls fdruh vfHkfØ;k,s lgh lqesfyr gS ?

(a)

CH3–C–O–CH3

CH3

|

|

CH3

1NHI S→

(b) CH3–CH–O–CH3

CH3

|1

2N

OH/H S →⊕

(c) CH3–CH–O–CH3

CH3

|1N

HI.Conc S →

(d) CH3–O–CH2CH3 1NHI S→

(e)

–O–CH2–CH = CH2 1NHI.Conc S →

(f) CH2 = CH – O – CH = CH2 12

NOH/H S →

⊕

(g)

–O–CH2– 12

NOH/H S →

⊕



Q.7 How many reactions the synthesis of

CH3–C–OH

C2H5

|

|

CH3

can be achieved -

(a) PhMgBr +

CH3–C–C2H5 || O

→⊕ OH/H 2

(b) C2H5MgBr +

Ph–C–CH3 || O

→⊕ OH/H 2

(c) CH3MgBr +

Ph–C– CH2CH3 || O

→⊕ OH/H 2

(d) PhMgBr +

CH3–C–Cl || O

→⊕ OH/H 2

(e) C2H5COOCH3 +)excess(

3MgClCH →⊕ OH/H 2

(f) C2H5COCl +)excess(


(g) CH3–COCH3 + C2H5MgCl →⊕ OH/H 2

(h) CH3–CN ⊕ →

H/OH

MgClHC

2

52 )B()A(H/OH

MgClCH

)major( 2

3⊕

→

Q.8 How many of the following compounds, yieldyellow precipitate on reaction with I2 and NaOH ?

(a) O

(b)

O

O

(c)

Ph–C – C–C2H5 | || Ph O

| Ph

(d) OH

(e) CH3COOC2H5 (f) (CH3CO)2O (g) CH3CONH2 (h) CH3COCH2COCH3

Q.7 fuEu esa ls fdruh vfHkfØ;kvksa kjk

CH3–C–OH

C2H5

|

|

CH3

dk fuekZ.k gks ldrk gS -

(a) PhMgBr +

CH3–C–C2H5 || O

→⊕ OH/H 2

(b) C2H5MgBr +

Ph–C–CH3 || O

→⊕ OH/H 2

(c) CH3MgBr +

Ph–C– CH2CH3 || O

→⊕ OH/H 2

(d) PhMgBr +

CH3–C–Cl || O

→⊕ OH/H 2

(e) C2H5COOCH3 +)excess(


(f) C2H5COCl +)excess(


(g) CH3–COCH3 + C2H5MgCl →⊕ OH/H 2

(h) CH3–CN ⊕ →

H/OH

MgClHC

2

52 )B()A(H/OH

MgClCH

)major( 2

3⊕

→

Q.8 fuEu esa ls fdrus ;kSfxd] I2 rFkk NaOH ds lkFk

vfHkfØ;k djus ij ihyk voksi nsxs ?

(a) O

(b)

O

O

(c)

Ph–C – C–C2H5 | ||Ph O

|Ph

(d) OH

(e) CH3COOC2H5 (f) (CH3CO)2O (g) CH3CONH2 (h) CH3COCH2COCH3



Q.9 How many of the following reaction are correct

methods for the preparation of propanoic acid ?

(a) CH3–CH=CH2 →HBr

Ether

Mg→OH

CO

3

2⊕ →

(b) H3C–C≡CH OH,OH

THF.BH

22

3Θ → ⊕ →

H).2(HO/KMnO).1( 4

(c) H2C=CH2 →HBr

Ether

Mg→OH

CO

3

2⊕ →

(d) H3C–CH=C–CH3 |

CH3 OH

Zn,O

2

3 →

Q.9 fuEu esa ls fdruh vfHkfØ;k,s] izksisuksbd vEy ds

fuekZ.k ds fy, lgh lqesfyr gS ?

(a) CH3–CH=CH2 →HBr

Ether

Mg→OH

CO

3

2⊕ →

(b) H3C–C≡CH OH,OH

THF.BH

22

3Θ → ⊕ →

H).2(HO/KMnO).1( 4

(c) H2C=CH2 →HBr

Ether

Mg→OH

CO

3

2⊕ →

(d) H3C–CH=C–CH3 |

CH3 OH

Zn,O

2

3 →



MATHEMATICS

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 marks for each wrong answer.

Q.1 Equation of chord of the circle x2 + y2 – 3x – 4y – 4 = 0 which passes through

the origin such that origin divides it in the ratio 4 : 1, is -

(A) x = 0 (B) 24x + 7y = 0 (C) 7x + 24y = 0 (D) 7x – 24y = 0

Q.2 Water is poured into an inverted conical vessel whose radius of the base is 2 m and height is 4m at the rate of 77 litre/minute. The rate at which the water level is rising at the instant when the

depth of water is 70 cm is -

=π

722use

(A) 10 cm/min (B) 20 cm/min (C) 40 cm/min (D) none of these Q.3 Number of ways in which 6 different toys can be

distributed among two brothers in ratio 1 : 2, is - (A) 30 (B) 60 (C) 20 (D) 40

[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 dkVk tk;sxkA

Q.1 o`Ùk x2 + y2 – 3x – 4y – 4 = 0 dh ml thok dk

lehdj.k tks ewyfcUnq ls gksdj bl izdkj xqtjrh gS

fd ewy fcUnq bls 4 : 1 ds vuqikr esa foHkkftr djrk

gS] gksxk- (A) x = 0 (B) 24x + 7y = 0 (C) 7x + 24y = 0 (D) 7x – 24y = 0 Q.2 ,d mYVs 'kaDokdkj crZu ftlds vk/kkj dh f=kT;k

2m gS rFkk Å¡pkbZ 4m gS] esa 77 yhVj/feuV dh nj

ls ty Mkyk tkrk gS] rks og nj ftl ij ty dh

lrg Åij mBrh gS] tcfd ty dh xgjkbZ 70 cm gS

gksxh (π = 22/7) (A) 10 cm/min (B) 20 cm/min (C) 40 cm/min (D) buesa ls dksbZ ugha Q.3 6 fofHkUu f[kykSuksa dks nks HkkbZ;ksa ds e/; 1 : 2 ds

vuqikr esa ck¡Vus ds rjhdks dh la[;k gS - (A) 30 (B) 60 (C) 20 (D) 40



Q.4 In a polygon no three diagonals are concurrent.If the total number of points of intersection ofdiagonals interior to the polygon be 70, then thenumber of diagonals of the polygon, is

(A) 8 (B) 20 (C) 28 (D) 12 Q.5 If p, q, r are in A.P., then the determinant

rcq22bp2a

q2q2p2pcq32bp2aa

21n2n2

1nn

22n21n22

−++++++

+++++

+

+

++

=

(A) 1 (B) 0 (C) a2b2c2 –2n (D) (a2 + b2 + c2) –2n q Q.6 A(3, 4), B(0, 0) & C(3, 0) are vertices of ∆ABC.

If ‘P’ is a point inside ∆ABC, such that

d(P, BC) ≤ min. d(P,AB), d(P,AC). Then

maximum of d(P, BC) is : (d(P, BC) represent

distance between P and BC)

(A) 1 (B) 1/2

(C) 2 (D) None of these

Q.7 If ∫ =2

1

x adxe2

, then the value of dx)x(n4e

e∫ l is

(A) e4 – e (B) e4 – a (C) 2e4 – a (D) 2e4 – e – a

Q.4 ,d cgqHkqt esa dksbZ rhu fod.kZ laxkeh ugha gSA ;fn

cgqHkqt ds vUrLFk fod.kksZ ds izfrPNsn fcUnqvksa dh

dqy la[;k 70 gS] rks cgqHkqt ds fod.kksZ dh la[;k gS -

(A) 8 (B) 20 (C) 28 (D) 12

Q.5 ;fn p, q, r l- Js- esa gS] rks lkjf.kd

rcq22bp2a

q2q2p2pcq32bp2aa

21n2n2

1nn

22n21n22

−++++++

+++++

+

+

++

=

(A) 1 (B) 0 (C) a2b2c2 –2n (D) (a2 + b2 + c2) –2n q

Q.6 A(3, 4), B(0, 0) ,oa C(3, 0) f=kHkqt ABC ds 'kh"kZ gSA

;fn ∆ABC ds vUnj fcUnq P bl izdkj gS fd

d(P, BC) ≤ min. d(P,AB), d(P,AC) rc

d(P, BC) dk vf/kdre eku gS : (d(P, BC) ; P ,oa

BC ds e/; nwjh dks iznf'kZr djrk gS) (A) 1 (B) 1/2 (C) 2 (D) buesa ls dksbZ ugha

Q.7 ;fn ∫ =2

1

x adxe2

gS] rks dx)x(n4e

e∫ l dk eku gS-

(A) e4 – e (B) e4 – a (C) 2e4 – a (D) 2e4 – e – a



Q.8 If circumradius and inradius of triangle be10 and 3 respectively then value ofa cot A + b cot B + c cot C is equal to

(A) 13 (B) 26 (C) 39 (D) None of these Questions 9 to 13 are multiple choice questions. Eachquestion has four choices (A), (B), (C) and (D), out ofwhich MULTIPLE (ONE OR MORE) is correct.Mark your response in OMR sheet against thequestion number of that question. + 3 marks will be given for each correct answer and no negative marks.

Q.9

→ xxsinmLim

0x =

(m ∈ I, and [.] denotes greatest integer function) (A) m if m ≤ 0 (B) m – 1 if m > 0 (C) m – 1 if m < 0 (D) m if m > 0

Q.10 The equation, 3x2 + 4y2 – 18x + 16y + 43 = c (A) cannot represent real pair of straight lines for

any value of c (B) represent empty set, if c < 0 (C) represent a point, if c = 0 (D) None of these

Q.8 ;fn f=kHkqt dh ifjf=kT;k rFkk vUr% f=kT;k Øe'k%10 ,oa 3 gSa] rks a cot A + b cot B + c cot C dk ekugS-

(A) 13 (B) 26 (C) 39 (D) buesa ls dksbZ ugha



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, + 3 vad fn;s tk;saxs rFkk dksbZ _.kkRed vadu ugha gSA

Q.9

→ xxsinmLim

0x =

(m ∈ I [.] egÙke iw.kkZad Qyu dks iznf'kZr

djrk gS)

(A) m ;fn m ≤ 0 (B) m – 1 ;fn m > 0

(C) m – 1 ;fn m < 0 (D) m ;fn m > 0

Q.10 lehdj.k 3x2 + 4y2 – 18x + 16y + 43 = c

(A) c ds fdlh Hkh eku ds fy, okLrfod ljy js[kk;qXe dks iznf'kZr ugha djrk gS

(B) fjDr leqPp; dks iznf'kZr djrk gS ;fn c < 0 (C) fcUnq dks iznf'kZr djrk gS ;fn c = 0

(D) buesa ls dksbZ ugha



Q.11 f(x) = sin )x])a[(2( , where [.] denote the

greatest integer function, has fundamental period π for-

(A) a = 23 (B) a =

45

(C) a = 32 (D) a =

54

Q.12 A unit vector which is equally inclined to the

vectors i , 2

k2ji2 ++ and 5

k3–j4– is -

(A) 51

k5j5–i(– + (B) 51

k5–j5i +

(C) 51

k5j5i ++ (D) None of these

Q.13 If Q(x) = x2 – mx + 1 is negative for value of x is (1, 2), then m lies in the interval

(A) (–3/2, 1/2) (B) (5/2, ∞) (C) (1/2, 5/2) (D) (– ∞, –3/2)

This section contains 2 paragraphs; passage- I has 2 multiple choice questions (No. 14 & 15) and passage-II has 3 multiple (No. 16 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. + 3 marks will be given for each correct answer and – 1 marks for each wrong answer.

Q.11 f(x) = sin )x])a[(2( ( tgk¡ [.] egÙke iw.kkZad Qyu

dks iznf'kZr djrk gS) ewy vkorZukad π j[krk gS-

(A) a = 23 ds fy, (B) a =

45 ds fy,

(C) a = 32 ds fy, (D) a =

54 ds fy,

Q.12 og bdkbZ lfn'k tks lfn'kksa i , 2

k2ji2 ++ ,oa

5k3–j4– ds lkFk leku >qdk gqvk gS] gksxk -

(A) 51

k5j5–i(– + (B) 51

k5–j5i +

(C) 51

k5j5i ++ (D) buesa ls dksbZ ugha

Q.13 ;fn Q(x) = x2 – mx + 1; x ds eku (1, 2) ds fy,

_.kkRed gS] rks m fuEu varjky esa fLFkr gksxk

(A) (–3/2, 1/2) (B) (5/2, ∞) (C) (1/2, 5/2) (D) (– ∞, –3/2)

bl [k.M esa 2 x|ka'k fn;s x;s gSa] x|ka'k -I esa 2 cgqfodYih iz'u (iz'u 14 o 15) gSa rFkk x|ka'k-II esa 3 cgqfodYih iz'u (iz'u 16 ls 18) 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;s + 3 vad fn;s tk,saxs rFkk izR;sd xyr mÙkj ds fy, 1 vad dkVk tk;sxkA



Passage # 1 (Ques. 14 & 15) Two variable chords AB and BC of a circle

x2 + y2 = a2 are such that AB = BC = a, and M

and N are the mid points of AB and BC

respectively such that line joining MN intersect

the circle at P and Q where P is closer to AB and

O is the centre of the circle

Q.14 ∠OAB is -

(A) 30° (B) 60° (C) 45° (D) 15°

Q.15 Angle between tangents at A and C is -

(A) 90° (B) 120° (C) 60° (D) 150°

Passage # 2 (Ques. 16 to 18) Consider the differential equation ex (ydx – dy) = e–x (ydx + dy). Let y = f(x) be a particular solution to this

differential equation which passes through thepoint (0, 2). Let

C ≡ y = log1/4

41–x +

21 log4 (16x2 – 8x + 1),

be another curve.

x|ka'k # 1 (iz- 14 ,oa 15)

` x2 + y2 = a2 AB BC

AB = BC = a AB BC

M N MN

` P Q

P; AB O `

Q.14 ∠OAB -

(A) 30° (B) 60° (C) 45° (D) 15°

Q.15 A C - (A) 90° (B) 120° (C) 60° (D) 150°

x|ka'k # 2 (iz- 16 ls 18) ex (ydx – dy) = e–x (ydx + dy)

y = f(x)

(0, 2)

C ≡ y = log1/4

41–x +

21



Q.16 The range of the function g(x) = log2 (f (x)) is (A) [1, ∞) (B) [2, ∞) (C) [0, ∞) (D) None of these

Q.17 The area bounded by the curve C, parabola

x = y2 + 41 and the line x =

41 is

(A) 1 (B) 3

(C) 32 (D)

31

Q.18 If the area bounded by the curve y = f (x), curve C, ordinate x = 1/4 & the ordinate x = a is

4 – ln 4 + 4/1e

1 – e1/4, then value of a is

(A) ln 6 (B) ln 4

(C) 4 (D) ln 12

Section - III

This section contains 9 questions (Q.1 to 9). +3 marks will be given for each correct answer and no negative marking. The answer to each of the questions is a SINGLE-DIGIT INTEGER, ranging from 0 to 9. The appropriate bubbles below therespective question numbers in the OMR have to bedarkened. For example, if the correct answers toquestion numbers X, Y, Z and W (say) are 6, 0, 9 and2, respectively, then the correct darkening of bubbleswill look like the following :

log4(16x2 – 8x + 1)

Q.16 g(x) = log2 (f (x))

(A) [1, ∞) (B) [2, ∞) (C) [0, ∞) (D)

Q.17 C, x = y2 + 41 x =

41

ifjc) ks=kQy gksxk (A) 1 (B) 3

(C) 32 (D)

31

Q.18 ;fn oØ y = f (x), oØ C, dksfV x = 1/4 ,oa dksfV

x = a kjk ifjc) ks=kQy 4 – ln 4 + 4/1e

1 – e1/4,

gS] rks a dk eku gksxk

(A) ln 6 (B) ln 4 (C) 4 (D) ln 12

[k.M - III

9 ( 1 9)

+3 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 gSaA OMR esa iz'u la[;k ds laxr uhps fn;s x;s

cqYyksa esa ls lgh mÙkj okys cqYyksa dks dkyk fd;k tkuk gSA

mnkgj.k ds fy, ;fn iz'u la[;k ¼ekusa½ X, Y, Z rFkk W ds



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 1 2

3

4

5 6

7

8

9

X Y Z W

Q.1 If the chord of the parabola y2 = 4ax, whose

equation is y – x 2 + 2a4 = 0, is a normal

the curve, and that its length is λa3 . Find the

value of λ

Q.2 Solve the following equation for x (where [x]

and x denotes integral and fractional part of x)

|x – 1| = 2[x] – 3x. If the solution of equation

is x1 and x2 where x1 + x2 = λ34 . Find λ.

Q.3 If α is real number for which f(x) = ln cos–1x is defined, then find the possible value of sum ofthe squares of integral values of α.

mÙkj 6, 0, 9 rFkk 2 gkas, rks lgh fof/k ls dkys fd;s x;s cqYys

,sls fn[krs gSa tks fuEufyf[kr gSA

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

X Y Z W

Q.1 y2 = 4ax

y – x 2 + 2a4 = 0

λa3 λ dk eku Kkr dhft,

Q.2 fuEu lehdj.k dks x ds fy, gy dhft, (tgk¡ [x] rFkk x; x ds iw.kkZad ,oa fHkUukRed Hkkx dks iznf'kZr

djrs gS) |x – 1| = 2[x] – 3x ;fn lehdj.k dk

gy x1 ,oa x2 gS tgk¡ x1 + x2 = λ34 rks λ dk eku

Kkr dhft,A



Q.4 If the reflection of the point P(1, 0, 0) in the line

810z

31y

21x +

=−+

=− is (α, β, γ). Find – (α + β + γ)

Q.5 Chords of the hyperbola x2 – y2 = a2 touch the parabola y2 = 4ax. If the locus of their middlepoints is the curve ym(x – a) = xn. Find m + n.

Q.6 Two squares are chosen at random from small

squares (1×1) drawn on chess board and the

probability that two squares chosen have exactly

one corner in common is 144

k , then k is equal to-

Q.7 The number of solutions of the equation

4sin2x+10cosec2x + 9tan2x – 31 = 0 in x ∈ [0, 2π] is-

Q.8 log (a + c), log(a + b), log(b + c) are in A.P. anda, c, b are in H.P., Where a, ,b, c > 0.

If a + b = 4kc , then the value of k is

Q.9 The difference between the radius of the smallest

and largest circles which have centers on the

circumference of the circle

x2 + y2 + 2x + 4y – 4 = 0 and pass through the point

(2, 2) is

Q.3 ;fn α okLrfod la[;k gS ftlds fy, f(x) = ln cos–1xifjHkkf"kr gS] rks α ds iw.kkZadh; ekuksa ds oxksZa ds ;ksxdk lEHko eku Kkr dhft;sA

Q.4 ;fn fcUnq P(1, 0, 0) dk js[kk 810z

31y

21x +

=−+

=−

esa izfrfcEc (α, β, γ) gS] rks – (α + β + γ) Kkr dhft,

Q.5 vfrijoy; x2 – y2 = a2 dh thok;sa ijoy; y2 = 4ax dks Li'kZ djrh gSA ;fn muds e/; fcUnqvksa dk fcUnqiFk ym(x – a) = xn gS] rks m + n Kkr dhft,A

Q.6 ,d 'karjat cksMZ ij cuk;s x;s NksVs oxksZa (1×1) ls

;kn`PN;k nks oxksZa dk p;u fd;k tkrk gS rFkk

p;fur nks oxksZ dk Bhd ,d fdukjk (corner)

mHk;fu"B gksus dh izkf;drk 144

k gS, rks k dk eku gS

Q.7 x ∈ [0, 2π] esa lehdj.k

4sin2x+10cosec2x + 9tan2x – 31 = 0 ds gyksa dh la[;k gksxh-

Q.8 log (a + c), log(a + b), log(b + c) l- Js- esa gS rFkk

a, c, b g- Js- esa gS] tgk¡ a, ,b, c > 0 ;fn a + b = 4kc

gS] rks k dk eku gksxk

Q.9 lcls NksVs rFkk lcls cM+s o`Ùk dk dsUnz o`Ùk

x2 + y2 + 2x + 4y – 4 = 0 dh ifjf/k ij fLFkr gS rFkk

fcUn (2 2) ls gksdj xtjrs gS dh f=kT;k dk vUrj



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 in OMR sheet against the question number of thatquestion. + 3 marks will be given for each correctanswer and – 1 mark for each wrong answer.

Q.1 A point charge q is placed at a distance r from the centre O of a uncharged spherical shell ofinner radius R and outer radius 2R. The distancer < R. The electric potential at the centre of theshell will be –

Conductor

+q r OR

2R

(A) 04πε

q

−

Rr 211 (B)

rq

04πε

(C) 04πε

q

+

Rr 211 (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 dkVk tk;sxkA

Q.1 q OR 2R

r r < R dks'k ds dsUnz ij fo|qr foHko gksxk –

Conductor

+q r OR

2R

(A) 04πε

q

−

Rr 211 (B)

rq

04πε

(C) 04πε

q

+

Rr 211 (D) buesa ls dksbZ ugha



Q.2 A capacitor of capacitance 10 µF is charged to a potential 50 V with a battery. The battery is now disconnected and an additional charge 200 µC is given to the positive plate of the capacitor. The potential difference across the capacitor will be -

(A) 50 V (B) 80 V (C) 100 V (D) 60 V Q.3 The ratio between the energy stored in 5 µF

capacitor to the 4 µF capacitor in the given circuit is –

4Ω

4µF

10V

4Ω

5µF

2Ω

(A) 1.2 (B) 1 (C) 1.25 (D) 3.6

Q.4 A uniform magnetic field →B = jB ˆ

0 exists in

space. A particle of mass m and charged q is

projected towards negative x-axis with speed v

from a point (d, 0, 0). The maximum value of v

for which the particle does not hit the y-z plane is

(A) dmBq2 (B)

mBqd (C)

mdBd

2 (D)

mBqd2

Q.2 10 µF /kkfjrk dk la/kkfj=k ,d cSVjh ls 50 V foHko ls vkosf'kr gSA vc cSVjh dks gVk ysrs gS rFkk

vfrfjDr vkos'k 200 µC la/kkfj=k dh /kukRed IysV

dks fn;k x;k gSA la/kkfj=k ij foHkokUrj gksxk - (A) 50 V (B) 80 V (C) 100 V (D) 60 V Q.3 fn;s x;s ifjiFk esa 5 µF la/kkfj=k ls 4 µF la/kkfj=k esa

lafpr ÅtkZvksa ds e/; vuqikr gS –

4Ω

4µF

10V

4Ω

5µF

2Ω

(A) 1.2 (B) 1 (C) 1.25 (D) 3.6

Q.4 ,dleku pqEcdh; ks=k →B = jB ˆ

0 Lisl esa fo|eku

gSA m nzO;eku rFkk q vkos'k dk d.k fcUnq (d, 0, 0)

ls pky v ls _.kkRed x-vk dh vksj izksfir gSA v

dk vf/kdre eku ftlds fy, d.k y-z ry ls ugha

Vdjk,xk] gS -

(A) dmBq2 (B)

mBqd (C)

mdBd

2 (D)

mBqd2



Q.5 An experiment measures quantities a, b, c and

then X is calculated from X = 3

22/1

cba . If the

percentage errors in a, b and c are ± 1%, ± 3%

and ± 2%, respectively, then the percentage error

in X can be -

(A) 12.5 % (B) 7 % (C) 1 % (D) 4 %

Q.6 A projectile is given an initial velocity of ji ˆ2ˆ + .

The cartesian equation of its path is - (g = 10 m/s2)

(A) y = 2x – 5x2 (B) y = x – 5x2

(C) 4y = 2x – 5x2 (D) y = 2x – 25x2

Q.7 If the acceleration of wedge in the figure shownis a m/s2 towards left, then at this instantacceleration of the block (magnitude only) wouldbe -

m M

α

(A) 4a m/s2 (B) a α− cos817 m/s2

(C) a17 m/s2 (D) 17 cos 2α × a m/s2

Q.5 ,d iz;ksx esa a, b, c jkf'k;k¡ fu/kkZfjr gksrh gS rFkk rc

X = 3

22/1

cba ls Kkr gksrk gSA ;fn a, b rFkk c esa

izfr'kr =kqfV;k¡ Øe'k% ± 1%, ± 3% rFkk ± 2% gS] rc

X esa izfr'kr =kqfV gks ldrh gS -

(A) 12.5 % (B) 7 % (C) 1 % (D) 4 %

Q.6 ,d izksI; dk izkjfEHkd osx ji ˆ2ˆ + kjk fn;k x;k

gSA blds iFk ds dkrhZ; funsZ'kkad gS - (g = 10 m/s2)

(A) y = 2x – 5x2 (B) y = x – 5x2

(C) 4y = 2x – 5x2 (D) y = 2x – 25x2

Q.7 n'kkZ, fp=k esa] ;fn xqVds dk ck¡;h vksj Roj.k a m/s2

gS] rc blh k.k ij CykWd dk Roj.k (dsoy ifjek.k) gksxk –

m M

α

(A) 4a m/s2 (B) a α− cos817 m/s2

(C) a17 m/s2 (D) 17 cos 2α × a m/s2



Q.8 A uniform chain AB of mass m and length l is placed with one end A at the highest point of ahemisphere of radius R. Referring to the top ofthe hemisphere as the datum level, the potential

energy of the chain is – (given that l <2Rπ )

l A

RB

(A)l

gmR2

−

Rl

Rl sin

(B) l

gmR2

2

−

Rl

Rl sin

(C) l

gmR2

2

−

Rl

Rlsin

(D) l

gmR 2

−

Rl

Rlsin

Questions 9 to 13 are multiple choice questions. Eachquestion has four choices (A), (B), (C) and (D), out ofwhich MULTIPLE (ONE OR MORE) is correct.Mark your response in OMR sheet against thequestion number of that question. + 3 marks will begiven for each correct answer and no negative marks.

Q.8 m nzO;eku rFkk l yEckbZ dh ,dleku pSu AB dk

,d fljk R f=kT;k ds v)Zxksys ds mPpre fcUnq ij

A fljs ij fLFkr gSA v)Zxksys ds mPp fljs dks

vk/kkj Lrj ds :i esa ysus ij pSu dh fLFkfrt ÅtkZ

gSA –

(fn;k x;k gS l <2Rπ )

l A

RB

(A)l

gmR2

−

Rl

Rl sin

(B) l

gmR2

2

−

Rl

Rl sin

(C) l

gmR2

2

−

Rl

Rlsin

(D) l

gmR 2

−

Rl

Rlsin



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, + 3 vad fn;s tk;saxs rFkk dksbZ _.kkRed vadu ugha gSA



Q.9 A point object is placed at 30 cm from a convex

glass lens

=µ

23

g of focal length 20 cm. The

final image of object will be formed at infinity if -

(A) another concave lens of focal length 60 cm is placed in contact with the previous lens

(B) another convex lens of focal length 60 cm is placed at a distance of 30 cm from the first lens

(C) the whole system is immersed in a liquid of refractive index 4/3

(D) the whole system is immersed in a liquid of refractive index 9/8

Q.10 In the circuit shown in figure -

2Ω

12V

4Ω 6Ω

4VE

A C F

B D

G i1

i2 1/2A

(A) E = 6.6 V (B) i1 = 1.1 A (C) i2 = 0.5 A (D) E = 4.4 A

Q.9 20 cm Qksdl nwjh ds vory dk¡p

yasl

=µ

23

g ls 30 cm ij fLFkr gSA fcEc dk

vafre izfrfcEc vuUr ij fufeZr gksxk ;fn -

(A) 60 cm Qksdl nwjh dk nwljk vory ysal izFke

ysal ls lEidZ esa fLFkr gS

(B) 60 cm Qksdl nwjh dk nwljk mÙky ysal izFke

ysal ls 30 cm dh nwjh ij gS

(C) lEiw.kZ fudk; 4/3 viorZukad ds nzo esa Mqck gqvk gS

(D) lEiw.kZ fudk; 9/8 viorZukad ds nzo esa Mqck gqvk gS

Q.10 fp=k esa n'kkZ, ifjiFk esa -

2Ω

12V

4Ω 6Ω

4VE

A C F

BD

Gi1

i2 1/2A

(A) E = 6.6 V (B) i1 = 1.1 A (C) i2 = 0.5 A (D) E = 4.4 A



Q.11 A thermally insulated chamber of volume 2V0 is

divided by a frictionless piston of area S into two

equal parts A and B. Part A has an ideal gas at

pressure P0 and temperature T0 and in part B is

vacuum. A massless spring of force constant k is

connected with piston and the wall of the container

as shown. Initially spring is unstretched. Gas in

chamber A is allowed to expand. Let in equilibrium

spring is compressed by x0. Then -

A

B

(A) final pressure of the gas is S

kx0

(B) work done by the gas is 202

1 kx

(C) change in internal energy of the gas is 202

1 kx

(D) temperature of the gas is decreased

Q.11 2V0 vk;ru dk ,d Å"ekjks/kh psEcj S ks=kQy ds

?k"kZ.k jfgr fiLVu kjk nks cjkcj Hkkxksa A o B esa

foHkkftr fd;k x;k gSA Hkkx A nkc P0 rFkk rki T0 ij

vkn'kZ xSl j[krk gS rFkk Hkkx B esa fuokZr gSA k cy

fu;rkad dh ,d fLizax fiLVu ls rFkk ik=k dh nhokj

ls fp=kkuqlkj tksM+h xbZ gSA izkjEHk esa fLizax vfolkfjr

gSA psEcj A esa xSl izlkfjr gksus ds fy, Lora=k gSA

ekuk lkE;koLFkk esa fLizax x0 kjk lEihM+; gSA rc –

A

B

(A) xSl dk vafre nkc S

kx0 gS

(B) xSl kjk fd;k x;k dk;Z 202

1 kx gS

(C) xSl dh vkUrfjd ÅtkZ eas ifjorZu 202

1 kx gS

(D) xSl dk rkieku ?kVrk gS



Q.12 Consider the rotation of a rod of mass m and length l from position AB to AB′. Which of thefollowing statements are correct ?

A B

B′ (A) Weight of the rod is lowered by

2l

(B) Loss of gravitational potential energy is 21 mgl

(C) Angular velocity is lg /3

(D) Rotational kinetic energy is 6

22ωml

Q.13 The vessel shown in the figure has two sections

of areas of cross section A1 and A2. A liquid ofdensity ρ fills both the sections, up to a height hin each. Neglect atmospheric pressure -

h A1

XA2h

ρ

Q.12 ekuk m nzO;eku rFkk l yEckbZ dh NM+ AB ls

AB′ fLFkfr rd ?kw.kZu djrh gSA fuEu esa ls dkSulk

dFku lgh gS ? A B

B′

(A) NM+ dk Hkkj 2l kjk de gks tkrk gS

(B) xq:Rokd"kZ.k fLFkfrt ÅtkZ dh gkfu 21 mgl gS

(C) dks.kh; osx lg /3 gS

(D) ?kw.kZu xfrt ÅtkZ 6

22ωml gS

Q.13 fp=k esa n'kkZ;k x;k ik=k A1 rFkk A2 vuqizLFk dkV

ks=kQy ds nks Hkkx j[krk gSA ρ ?kuRo dk nzo nksuksa

Hkkxksa esa h Å¡pkbZ rd Hkjk gS] ok;qe.Myh; nkc

ux.; gS -

h A1

XA2h

ρ



(A) The pressure at the base of the vessel is 2hρg (B) The force exerted by the liquid on the base

of the vessel is 2hρgA2 (C) The weight of the liquid is < 2hρgA (D) The walls of the vessel at the level X exert a

downward force hρg(A2 – A1) on the liquid This section contains 2 paragraphs; passage- I has 2 multiple choice questions (No. 14 & 15) and passage-II has 3 multiple (No. 16 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. + 3 marks will be given for each correct answer and – 1 mark for each wrong answer.

Passage # 1 (Q.14 to 15) Figure shows a schematic diagram for a device

known as a potentiometer. The potentiometer is useful for measuring an unknown electromotive force (emf) εx in terms of known emf ε0. In addition to these two, a third emf ε is used to produce the current i. The three emfs have internal resistances rx, r0 and r, respectively.

The central circuit element of the potentiometer is a long resistor, having total resistance R, with a sliding contact at point D.

(A) 2hρg (B)

2hρgA2 (C) < 2hρgA (D) X hρg(A2 – A1)

bl [k.M esa 2 x|ka'k fn;s x;s gSa] x|ka'k -I esa 2 cgqfodYih

iz'u (iz'u 14 o 15) gSa rFkk x|ka'k-II esa 3 cgqfodYih iz'u

(iz'u 16 ls 18) 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;s + 3 vad fn;s tk,saxs rFkk

izR;sd xyr mÙkj ds fy, 1 vad dkVk tk;sxkA

x|ka'k # 1 (Q.14 ls 15) fp=k foHkoekih ds :i esa Kkr ;qfä ds fy, fp=k-izk:i

iznf'kZr djrk gSA foHkoekih vKkr fo|qr okgd cy

(fo-ok-cy) εx dk Kkr fo-ok-cy ε0 ds :i esa fu/kkZj.k ds

fy, mi;ksxh gSA bu nksuksa ds la;kstu esa rhljk fo-ok-

cy ε /kkjk i ds mRiknu esa mi;qä gksrk gSA rhuksa fo-ok-

cy Øe'k% rx, r0 rFkk r vkUrfjd izfrjks/k j[krs gSA

foHkoekih dk dsUnzh; ifjiFk rRo ,d yEck izfrjks/k gS

tks dqy izfrjks/k R j[krk gS] D fcUnq ij fQly ldrk

gSA



The resistivity ρ of this variable resistor isconstant. Therefore, the resistance betweenpoints A and D is given by αR, where α is the distance from A to D divided by the distance from A to E. A very sensitive ammeter called agalvanometer is connected between points C and D in order to measure the current through thatsection of the circuit.

B C

ix or i0

A

i

G ε r F

E(1 – α) R

Galvanometerεx or ε0rx or r0

αR

D

The operation of the potentiometer is conducted

in two steps. First, the unknown emf εx is connected in the circuit between points B and C, and the sliding contact at point D is adjusted sothat the galvanometer reads ix = 0. Using theloop theorem for the current loop ABCD, it can be shown that the current i satisfies the equationi = εx/(αxR). In the second step, the unknownemf εx is replaced by the known ε0, and the sliding contact at point D is again adjusted sothat the galvanometer reads zero.

Applying the loop theorem again leads to theresult that the current i is also given i = ε0/(αxR). Equating the two expression for i we get for theunknown emf εx = ε0αx/α0.

bl ifjorhZ izfrjks/k dh izfrjks/kdrk ρ fu;r gSA vr%

fcUnqvksa A rFkk D ds e/; izfrjks/k αR kjk fn;k x;k gS]

tgk¡ α, A ls D rd nwjh kjk A ls E rd foHkkftr

nwjh gSA ,d cgqr laosnh vehVj tks xsYoksuksehVj

dgykrk gS bls ifjiFk ds bl Hkkx ls xqtjus okyh

/kkjk ds fu/kkZj.k dh dksfV esa C o D fcUnqvksa ds e/;

tksM+k tkrk gSA

B C

ix or i0

A

i

G ε r F

E(1 – α) R

xsYosuksehVjεx or ε0rx or r0

αR

D

foHkoekih dh fØ;k nks pj.kksa esa pkfyr gksrh gSA izFke

vKkr fo-ok-cy εx ifjiFk esa B o C fcUnqvksa ds e/; tksM+k tkrk gS rFkk D fcUnq ij rkj dks bl izdkj O;ofLFkr djrs gS fd xsYosuksehVj ix = 0 ikB~;kad nsrk gSA /kkjk ywi ABCD ds fy, ywi izes; dk mi;ksx djus ij ;g iznf'kZr gksrk gS fd /kkjk i, i = εx/(αxR) lehdj.k dks larq"V djrh gSA frh; pj.k esa] vKkr fo-ok-cy ε0 kjk cny fn;k tkrk gS rFkk rkj dks Dij bl izdkj O;ofLFkr djrs gS fd xsYosuksehVj 'kwU; ikB~;kad nsrk gSA

iqu% ywi izes; yxkus ij ;g ifj.kke gksrk gS fd /kkjk i = ε0/(αxR) kjk nh xbZ gSA i ds fy, nksuksa O;atdks dh x.kuk ls ge ikrs gS fd vKkr fo-ok-cy εx = ε0αx/α0 gSA



Q.14 The current i is assumed to be same when ε0 is connected as it is when εx is connected. If the galvanometer reads zero in both cases, which of the following is also a correct expression for i ?

(A) ε/(r + αxR) (B) ε/R (C) ε/(r + R) (D) ε/(r + α0R) Q.15 Consider the example where ε0 = 6V. If the

length AD is equal to 0.2 m when εx is connected and equal to 0.4 m when ε0 is connected. What is the value of εx ?

(A) 2 V (B) 3 V (C) 6 V (D) 12 V Passage # 2 (Ques. 16 to 18) Four pieces of string each of length L are joined

end to end to make a long string of length 4L. The linear mass density of the strings are µ, 4µ, 9µ and 16µ, respectively.

One end of the combined string is tied to a fixed support and a transverse wave has been generated at the other end having frequency f(ignore any reflection and absorptions). String has been stretched under a tension F.

Q.16 Find the time taken by wave to reach from source end to fixed end -

(A) 1225 ×

µ/FL (B)

µ/10F

L

(C) µ/

4F

L (D) µ/F

L

Q.14 /kkjk i leku ekuh tkrh gS tc ε0 blls tksM+k tkrk gS tc εx blls tqM+k gqvk gSA ;fn xsYosuksehVj nksuksa fLFkfr;ksa esa 'kwU; ikB~;kad nsrk gS] fuEu esa ls dkSulk ids fy, lgh O;atd Hkh gS ?

(A) ε/(r + αxR) (B) ε/R (C) ε/(r + R) (D) ε/(r + α0R)

Q.15 fn;k x;k mnkgj.k tgk¡ ε0 = 6V gSA ;fn AD yEckbZ 0.2 m ds cjkcj gS tc εx tqM+k gqvk gS rFkk 0.4 m ds cjkcj gS tc ε0 tqM+k gqvk gS εx dk eku D;k gS ?

(A) 2 V (B) 3 V (C) 6 V (D) 12 V

x|ka'k # 2 (Q.16 ls 18) L

4L µ, 4µ, 9µ 16µ

rFkk vkofr f dh vuqizLFk rjax Mksjh ds nwljs fljs esa mRiUu gksrh gS (ijkorZu o vo'kks"k.k dks ux.; ysa)A Mksjh F ruko kjk [khaph xbZ gSA

Q.16 fQDl fljs ls L=kksr fljs rd igq¡pus esa rjax kjk

fy;k x;k le; Kkr djks -

(A) 1225 ×

µ/FL (B)

µ/10F

L

(C) µ/

4F

L (D) µ/F

L



Q.17 Find the ratio of wavelengths of the waves fourstrings, starting from right hand side -

(A) 12 : 6 : 4 : 3

(B) 4 : 3 : 2 : 1 (C) 3 : 4 : 6 : 12 (D) 1 : 2 : 3 : 4 Q.18 Find the rate at which energy is transferred is

maximum for the string having mass density - (A) µ (B) 4µ (C) 9µ (D) 16µ

Section - III This section contains 9 questions (Q.1 to 9). +3 marks will be given for each correct answer and no negative marking. The answer to each of the questionsis a SINGLE-DIGIT INTEGER, ranging from 0 to 9.The appropriate bubbles below the respective question numbers in the OMR have to be darkened.For example, if the correct answers to questionnumbers X, Y, Z and W (say) are 6, 0, 9 and 2,respectively, then the correct darkening of bubbleswill look like the following :

Q.17 nk¡;s gkFk dh vksj ls izkjEHk dj] pkjksa Mksfj;ksa dh rjaxksa dh rjaxnS/;ksZ dk vuqikr Kkr djks -

(A) 12 : 6 : 4 : 3 (B) 4 : 3 : 2 : 1 (C) 3 : 4 : 6 : 12 (D) 1 : 2 : 3 : 4 Q.18 nj Kkr djks ftl ij fuEu nzO;eku ?kuRo okyh

Mksjh ds fy, LFkkukUrfjr ÅtkZ vf/kdre gksxh - (A) µ (B) 4µ (C) 9µ (D) 16µ

[k.M - III

bl [k.M esa 9 (iz-1 ls 9) iz'u gaSA izR;sd lgh mÙkj ds fy;s +3 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 gSaA OMR esa iz'u la[;k ds laxr uhps fn;s x;s cqYyksa esa ls lgh mÙkj okys cqYyksa dks dkyk fd;k tkuk gSA mnkgj.k ds fy, ;fn iz'u la[;k ¼ekusa½ X, Y, Z rFkk W ds mÙkj 6, 0, 9 rFkk 2 gkas, rks lgh fof/k ls dkys fd;s x;s cqYys ,sls fn[krs gSa tks fuEufyf[kr gSA



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 1 2

3

4

5 6

7

8

9

X Y Z W

Q.1 A parallel plate capacitor is maintained at acertain potential difference. When a 3 mm slab isintroduced between the plates, in order tomaintain the same potential difference, thedistance between the plates is increased by 2.4mm. Find the dielectric constant of the slab.

Q.2 Find the self inductance (in H) of a coil in which

an e.m.f. of 10 V is induced when the current inthe circuit changes uniformly from 1 A to 0.5 Ain 0.2 sec.

Q.3 A circuit is shown in figure.

3 µF 2 µF

5 µF

4 µF + –

6V

A B

Find the charge (in µC) on the condenser having

a capacity of 5 µF.

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

X Y Z W

Q.1 ,d lekUrj IysV la/kkfj=k fuf'pr foHkokUrj ij O;ofLFkr gSA tc 3 mm dh ,d Lysc IysVks ds e/; leku foHkokUrj ij Øe esa O;ofLFkr djds feykbZ tkrh gS IysVks ds e/; 2.4 mm kjk of) gksrh gSA Lysc dk ijkoS|qrkad Kkr djksA

Q.2 dq.Myh dk LoizsjdRo (H esa) ftlesa 10 V dk fo-ok-cy izsfjr gS tc ifjiFk esa 0.2 lsd.M esa /kkjk 1 A ls 0.5 A rd ,dleku :i ls ifjofrZr gksrh gSA

Q.3 ,d ifjiFk fp=k esa n'kkZ;k x;k gSA

3 µF2 µF

5 µF

4 µF+ –

6V

A B

5 µF dh /kkfjrk j[kus okys la/kkfj=k ij vkos'k (µC esa)

Kkr djksA



Q.4 A circular loop of radius R is bent along a

diameter and given a shape as shown in the

figure. One of the semicircles (KNM) lies in the

x-z plane and the other one (KLM) in the y-z

plane with their centres at the origin. Current I is

flowing through each of the semi circles as

shown in figure.

M L

K I

N

L L y

x

z

A particle of charge q is released at the origin

with a velocity →v = iv ˆ

0 . Find the magnitude of

instantaneous force→F on the particle if µ0qv0I = 8R.

Assume that space is gravity free.

Q.4 ,d R f=kT;k dk oÙkkdkj ywi O;kl ds vuqfn'k eksM+k

tkrk gSA nh xbZ vkdfr fp=k esa iznf'kZr gSA ,d

v)ZoÙk (KNM) x-z ry esa j[kk gS rFkk nwljk

(KLM) y-z ry esa j[kk gS buds dsUnz ewy fcUnw ij

gSA fp=kkuqlkj izR;sd v)ZoÙkks esa izokfgr /kkjk I gSA

M L

K I

N

LL y

x

z

q vkos'k dk d.k ewy fcUnq ls →v = iv ˆ

0 osx ls NksM+k

tkrk gSA d.k ij rkRkf.kd cy →F dk ifjek.k Kkr

djks ;fn µ0qv0I = 8R gSA ekfu, fd Lisl xq:Ro

Lora=k gSA



Q.5 If the intensity of sound is doubled, by how many decibels does the sound level increase ? (in dB)

Q.6 The electric potential between a proton and an

electron is given by, V = V0 ln

0rr

where V0 and r0 are constants and r is the radius of the electron orbit around the proton. Assuming Bohr's model to be applicable, it is found that r is proportional to nx, where n is the principal quantum number. Find the value of x.

Q.7 Heat at the rate of 200 W is produced in an

X-ray tube operating at 20 kV. Find the current (in × 10–2 A) in the circuit. Assume that only a small fraction of the kinetic energy of electrons is converted into X-rays.

Q.8 The half-life of a radioactive nuclide is 20 hours.

The fraction (1/x) of original activity will remain after 40 hours. What is the value of x ?

Q.9 Assume that the mass of a nucleus is

approximately given by M = Amp where A is the

mass number. Estimate the density of matter

(in × 1017 kg/m3) inside a nucleus.

Q.5 ;fn /ofu dh rhozrk nqxquh dh tk;s rks fdrus Mslhcy kjk /ofu Lrj c<+sxk ? (dB esa)

Q.6 ,d izksVksu rFkk ,d bysDVªkWu ds e/; fo|qr foHko

V = V0 ln

0rr kjk fn;k tkrk gS] tgk¡ V0 rFkk r0

fu;rkad gS rFkk r izksVkWu ds pkjksa vksj xfreku bysDVªkWu dh dkk dh f=kT;k gSA ekukfd cksgj ekWMy ekU; gS] ;g ik;k tkrk gS fd r, nx ds lekuqikrh gS] tgk¡ n eq[; Dok.Ve la[;k gSA x dk eku Kkr djksA

Q.7 20 kV ij pkfyr X-fdj.k ufydk esa 200 W dh nj ij Å"ek mRiUu gksrh gSA ifjiFk esa /kkjk (10–2 A esa) Kkr djksA ekukfd bysDVªkWu dh ÅtkZ dk dqN Hkkx gh X-fdj.k esa ifjofrZr gksrk gSA

Q.8 ,d jsfM;kslfØ; ukfHkd dh v)Zvk;q 20 ?k.Vs gSA 40 ?k.Vs i'pkr~ okLrfod lfØ;rk dk (1/x) Hkkx 'ks"k jgrk gSA x dk eku D;k gS ?

Q.9 ekfu, fd ukfHkd dk nzO;eku yxHkx M = Amp kjk

fn;k x;k gS] tgk¡ A nzO;eku la[;k gSA ukfHkd esa

inkFkZ ds ?kuRo (1017 kg/m3 esa) dk vkdyu dhft,A

IIT Advance Paper-1 Mock Test - JEE Main, JEE … To crack IIT‐JEE/AIEEE/BITSAT a unique Adaptive...

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Transcript of IIT Advance Paper-1 Mock Test - JEE Main, JEE … To crack IIT‐JEE/AIEEE/BITSAT a unique Adaptive...