INSTRUCTIONS ™ MARKING SCHEME ¾ Each subject in this paper consists of following types of questions:Section - I ¾ Multiple choice questions with only one correct answer. 3 marks will be awarded for each correct answer and –1 mark for each wrong answer. ¾ Multiple choice questions with multiple correct option. 3 marks will be awarded for each correct answer and No negative marking. ¾ Passage based single correct type questions. 3 marks will be awarded for each correct answer and –1 mark for each wrong answer. Section - II ¾ 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)

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CHEMISTRY Section – I

[k.M - I

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

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 le{k viuk mÙkj vafdr dhft;sA izR;sd lgh mÙkj ds fy, + 3 vad fn;s tk;asxs rFkk izR;sd xyr mÙkj ds fy, 1 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.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 of following statements. Use T if statements is true and F if it is false. (i) In Gold schmidt thermite process aluminium acts as a reducing agent. (ii) Mg is extracted by electrolysis of aq. solution of MgCl2 (iii) Extraction of Pb is possible by carbon reduction method (iv) Red Bauxite is purified by reduction method (A) TTTF (B) TFFT (C) FTTT (D) TFTF

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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 (C) FTTT

(B) TFFT (D) TFTF

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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.3

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 four gases 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.4

–

2 O 2 , HO BH  3 → H  → (A)

Q.5

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 vo{ksfir gksrs gS (B) ;s /kukosf'kr dkWyksbM gks tkrs gS (C) ;s _.kkosf'kr dkWyksbM gks tkrs gS (D) buesa ls dksbZ ugha fuf'pr rki o 2 atm nkc ij fuEu vfHkfØ;k ds fy, lkE; fLFkjkad (kp), 25 gSA SO2(g) + NO2(g)

(A) 1.33 atm (C) 0.133 atm

(B) 0.1665 atm (D) buesa ls dksbZ ugha –

2 O 2 , HO BH  3 → H  → (A)

Q.5

OR

OR Product (A) is -

mRikn (A) gS -

OH

OH

(A)

(B) OR

OH

(A)

OR

(B) OR

OH

(C)

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 ?

OH OR

OH OH

(D)

OR

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OH

(C) OR OH

OR

(D)

OR OH

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Q.6

Major product (P) in following reaction

Q.6

OH

)4 CHO  → (P) CH3 – CO —(CH 2—

OH

)4 CHO  → (P) CH3 – CO —(CH 2—

∆

(A)

O || CH3 CHO

(C)

∆

(B)

(A)

(D)

O || CH3 CHO

(C)

CHO

(B)

(D)

CHO

CH3

CH3 SO 2 Cl 2

  → (A)

Q.7

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

hν

NBS → ∆

KSH

(B)  → (C),

hν

mRikn (C) gS -

major product (C) is SH

SH

(B)

(A)

NBS ∆

Cl 2  → (C), SO 2 → (A) → (B) KSH

Q.7

Br

SH

(A)

SH

(B) Br

Cl

Cl

Br Br (C)

Br (D)

Br (D)

(C)

SH

SH

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OH Q.8

OH

OH

PCC(excess)

Q.8 (A)

OH OH 1 equivalent

(B)

NaBH4

(C) H2O Product (D) will be -

OH (A)

CH3MgBr

CH3MgBr H3O

NaBH4 H2O

(A)

(B)

(D)

mRikn (D) gS -

(D)

OH

OH | CH–CH3

(A)

OH | CH–CH3

OH

(B)

(B)

OH (C)

OH OH OH

(D)

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.

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OH O

O

(D)

PCC(excess)

(C)

H3O+

+

OH

(C)

OH

OH OH 1 equivalent

OH 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 le{k vius mÙkj vafdr dhft,A izR;sd lgh mÙkj ds fy, + 3 vad fn;s tk;saxs rFkk dksbZ _.kkRed vadu ugha gSA

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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.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

Which reaction is/are possible ? (A) MgCl2 + NaNO3 → (B) BaSO4 + HCl → (C) ZnSO4 + BaS → (D) BaCO3 + CH3COOH →

Q.10

dkSulh vfHkfØ;k;s lEHko gS ?

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 in one orbit (C) Bohr's model could not explain the spectra of multielectron atoms (D) Bohr's model was the first atomic model based on quantization of energy

Q.11

Q.11

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(A) MgCl2 + NaNO3 → (B) BaSO4 + HCl → (C) ZnSO4 + BaS → (D) BaCO3 + CH3COOH →

;gh dFkuks dk p;u dhft, (A) ukfHkd ds fudV fLFkr ,d bysDVªku W ] 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

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Q.12

Which of the following cycloalkanes will show cis-trans isomerism CH3 CH3 (B) (A)

H3C

CH3

Q.12

CH3 (B)

H3C

CH3

CH3 CH3

(D)

(C)

H3C Q.13

CH3

(A)

CH3 CH3

(D)

(C)

fuEu esa ls dkSuls lkbDyks,Ydsu fll-Vªkl a leko;ork n'kkZ;saxkA -

H3C

Which of the following combinations give Ph – CH3 ? (A) Ph – CH2 – MgBr + NO2 –

– OH →

Q.13

dkSuls la;kstu fuEu ;kSfxd nsxsaA Ph – CH3 ? (A) Ph – CH2 – MgBr + NO2 –

– OH →

– + CaO (B) PhCH2COONa  →

– +  (B) PhCH2COONa →

O || H O / H+ (C) CH3MgBr and C6H5–C–OC2H5 2 →

O || H O / H+ (C) CH3MgBr rFkk C6H5–C–OC2H5 2 →

CaO

NaOH

NaOH

This section contains 2 paragraphs; passage- I has 2 multiple choice questions (No. 14 & 15) and passageII 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.

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H PO

(D) Ph – CH2MgBr 3  4 →

H PO

(D) Ph – CH2MgBr 3  4 →

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 le{k 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

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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 chemical symbols) 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) O– O O || | || HO– Ph– C–H + C–PH Ph– C | | O–H H

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

Q.14

(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) ;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 (C) I, II, III 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) slow

(B) I, II, IV (D) I, II, III, IV

O || Ph– C | H

HO–

O– O || | Ph– C–H + C–PH | O–H

slow

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O || Ph– C–OH + Ph–CH2–O–

O || Ph– C–OH + Ph–CH2–O–

Ph– CO 2– + Ph – CH2 – OH

Ph– CO 2– + Ph – CH2 – OH Q.16

Which of the following reactants can undergo Cannizaro's reaction. O || (B) R3CCHO (A) H–C–H (C) (D) All of these

O Q.17 Q.18

fuEu esa ls dkSuls fØ;kdkjd dsuhtkjks vfHkfØ;k lEiUu dj ldrs gS O || (A) H–C–H (C)

CHO

O

Order of the above reaction is (A) 1 (B) 2 (C) 3

Q.17

(A)

CH3

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(B) 2

CHO

(D) 4

CHO (B)

OMe CHO (C)

Cl

(C) 3

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

NO2 CHO (D)

(D) buesa ls dksbZ ugha

CHO

(A)

(B)

OMe CHO

Q.18

(B) R3CCHO

mi;qZDr vfHkfØ;k dh dksfV gS (A) 1

(D) 4

Which of the following is best hydride donor in Cannizaro's reaction CHO CHO

(C)

Q.16

NO2 CHO (D)

CH3

Cl

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Section - III

[k.M - 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 the respective question numbers in the OMR have to be darkened. For example, if the correct answers to question numbers X, Y, Z and W (say) are 6, 0, 9 and 2, respectively, then the correct darkening of bubbles will look like the following :

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

X 0 1 2 3 4 5 6 7 8 9

Q.1

Y 0 1 2 3 4 5 6 7 8 9

Z 0 1 2 3 4 5 6 7 8 9

W

X

0 1 2 3 4 5 6 7 8 9

0 1 2 3 4 5 6 7 8 9

4 / dil.HNO 3 A(Light pink colour complex) Pb 3O  →

Q.1

H S/ H+

Calculate CFSE value in light pink colour complex.

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Z 0 1 2 3 4 5 6 7 8 9

W 0 1 2 3 4 5 6 7 8 9

4 / dil.HNO 3 A(pedhyk xqykch ladqy) Pb 3O  →

∆

∆

2 → A (Light pink colour complex). HMNO4 

Y 0 1 2 3 4 5 6 7 8 9

H S/ H+

2 → A (pedhyk xqykch ladqy). HMNO4 

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

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Q.2

Consider the following oxyanions :

Q.2

PO 34– , P2 O 64 – , SO 24 – , MnO 4– , CrO 24 – , S2 O 52 – ,

PO 34– , P2 O 64 – , SO 24 – , MnO 4– , CrO 24 – , S2 O 52 – ,

S2 O 72 –

S2 O 72 –

rFkk R + Q – P dk eku Kkr dhft,A

and find the value of R + Q – P

tgk¡

Where

P = izfr dsfUnz; ijek.kq] rhu leku

P = Number of oxy anions having three equivalent

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

X – O bonds per central atom

Q = izfr dsfUnz; ijek.kq] nks leku

Q = Number of oxy anions having two equivalent

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

X – O bonds per central atom

R = izfr dsfUnz; ijek.kq] pkj leku

R = Number of oxy anions having four equivalent

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

X – O bonds per central atom Q.3

Q.4

fuEu vkWDlh_.kk;uks ij fopkj dhft,A

Density of Li atom is 0.53 g/cm3. The edge length of Li is 3.5 Å. Find out the number of Li atoms in a unit cell. (NA = 6.0 × 1023 mol–1, M = 6.94 g mol–1)

Q.3

Two flask A and B of equal volumes maintained

Q.4

Li ijek.kq dk ?kuRo 0.53 g/cm3 gSA Li dh dksj yEckbZ 3.5 Å gSA ,d bdkbZ dksf"Vdk esa Li

ijek.kqvksa dh la[;k Kkr dhft,A (NA = 6.0 × 1023 mol–1, M = 6.94 g mol–1)

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

at temperature 300 K and 700 K contain equal

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

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

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

the ratio of total translational kinetic energy of

esa dqy LFkkukUrfjr xfrt ÅtkZ dk vuqikr D;k

gas in flask A to that of flask B ?

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gksxk ?

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Q.5

C(g) + D(g);

For the reaction 2A(g) + B(g)

Q.5

12

KC = 10 . If the initial moles of A, B, C and D are

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

What is the equilibrium concentration of A in

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

term of n × 10–12 ? How many of the following reactions are correctly matched ?

Q.6

fuEu esa ls fdruh vfHkfØ;k,s lgh lqesfyr gS ? CH3 | HI S N1 (a) CH3–C–O–CH3 → | CH3

CH3 | HI (a) CH3–C–O–CH3 → S N1 | CH3

⊕

H 2O → S N1 (b) CH3–CH–O–CH3 H/ | CH3

⊕

H 2O → S N1 (b) CH3–CH–O–CH3 H/ | CH3

Conc.HI

(c) CH3–CH–O–CH3  → S N1 | CH3

Conc.HI

(c) CH3–CH–O–CH3  → S N1 | CH3

HI (d) CH3–O–CH2CH3 → S N1

HI

(d) CH3–O–CH2CH3 → S N1 (e)

C(g) + D(g); ds fy,

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

2, 1, 7 and 3 moles respectively in a one litre vessel.

Q.6

vfHkfØ;k 2A(g) + B(g)

Conc.HI

–O–CH2–CH = CH2  → S N1

(e)

.HI –O–CH2–CH = CH2 Conc  → S N1 ⊕

⊕

H 2O → S N1 (f) CH2 = CH – O – CH = CH2 H/

(g)

–O–CH2–

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⊕

H 2O H/ → S N1

H 2O → S N1 (f) CH2 = CH – O – CH = CH2 H/

(g)

–O–CH2–

⊕

H 2O H/ → S N1

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Q.7

How

many reactions the synthesis of CH3 | CH3–C–OH can be achieved | C 2 H5 O || ⊕ H 2O (a) PhMgBr + CH3–C–C2H5 H/ → O || ⊕ H 2O → (b) C2H5MgBr + Ph–C–CH3 H/ O || ⊕ H 2O → (c) CH3MgBr + Ph–C– CH2CH3 H/ O || ⊕ H 2O → (d) PhMgBr + CH3–C–Cl H/

Q.7

CH3 | fuEu esa ls fdruh vfHkfØ;kvksa }kjk CH3–C–OH | C 2 H5 dk fuekZ.k gks ldrk gS O || ⊕ H 2O → (a) PhMgBr + CH3–C–C2H5 H/ O || ⊕ H 2O → (b) C2H5MgBr + Ph–C–CH3 H/ O || ⊕ H 2O → (c) CH3MgBr + Ph–C– CH2CH3 H/ O || ⊕ H 2O → (d) PhMgBr + CH3–C–Cl H/ ⊕

H 2O (e) C2H5COOCH3 + CH 3 MgCl H/ →

H ⊕ / H 2O

(e) C2H5COOCH3 + CH 3 MgCl  →

( excess)

( excess)

⊕

H 2O → (f) C2H5COCl + CH 3 MgCl H/

H ⊕ / H 2O

(f) C2H5COCl + CH 3 MgCl  →

( excess)

( excess)

⊕

H 2O → (g) CH3–COCH3 + C2H5MgCl H/

H ⊕ / H 2O

(g) CH3–COCH3 + C2H5MgCl  → C H MgCl

C H MgCl

H 2O / H

Q.8

( major )

(b)

(a) O

(d)

(e) CH3COOC2H5 (g) CH3CONH2

Q.8

( major )

H 2O / H

fuEu esa ls fdrus ;kSfxd] I2 rFkk NaOH ds lkFk vfHkfØ;k djus ij ihyk vo{ksi nsxs ? O

(a)

(b) O

O

Ph O | || (c) Ph–C – C–C2H5 | Ph

H 2O / H

H 2O / H

How many of the following compounds, yield yellow precipitate on reaction with I2 and NaOH ? O

CH MgCl

2 5 3   → (A)     →(B) (h) CH3–CN ⊕ ⊕

CH MgCl

2 5 3   → (A)     →(B) (h) CH3–CN ⊕ ⊕

O

Ph O | || (c) Ph–C – C–C2H5 | Ph

(d)

(f) (CH3CO)2O

(e) CH3COOC2H5

(f) (CH3CO)2O

(h) CH3COCH2COCH3

(g) CH3CONH2

(h) CH3COCH2COCH3

OH

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OH

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Q.9

How many of the following reaction are correct methods for the preparation of propanoic acid ? HBr

Mg

CO 2

Ether

H3 O

(a) CH3–CH=CH2  → →   → ⊕ .THF (b) H3C–C≡CH BH 3 → Θ H 2 O 2 , OH

HBr

(1). KMnO 4 / OH  ⊕  → ( 2). H CO 2

Ether

H3 O

(c) H2C=CH2  → →   → ⊕ (d) H3C–CH=C–CH3  → | H 2O CH3

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fuEu esa ls fdruh vfHkfØ;k,s] izksisuksbd vEy ds fuekZ.k ds fy, lgh lqesfyr gS ? 2  → Mg   → (a) CH3–CH=CH2 HBr → CO ⊕

Ether

H3 O

(1). KMnO / OH ( 2). H

.THF 4 (b) H3C–C≡CH BH 3 →  ⊕  → Θ H 2 O 2 , OH

Mg

O 3 , Zn

Q.9

2 (c) H2C=CH2 HBr  → Mg → CO   → ⊕

Ether

H3 O

O , Zn

(d) H3C–CH=C–CH3 3 → | H 2O CH3

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MATHEMATICS Section – I

[k.M - 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.

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 le{k viuk mÙkj vafdr dhft;sA izR;sd lgh mÙkj ds fy, + 3 vad fn;s tk;asxs rFkk izR;sd xyr mÙkj ds fy, 1 vad dkVk tk;sxkA

Q.1

Q.2

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 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

Q.1

(A) x = 0 (C) 7x + 24y = 0 Q.2

22   depth of water is 70 cm is -  use π =  7   (A) 10 cm/min (C) 40 cm/min Q.3

(B) 20 cm/min (D) none of these

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

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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-

Q.3

(B) 24x + 7y = 0 (D) 7x – 24y = 0

,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

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

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Q.4

Q.5

In a polygon no three diagonals are concurrent. If the total number of points of intersection of diagonals interior to the polygon be 70, then the number of diagonals of the polygon, is (A) 8 (B) 20 (C) 28 (D) 12

Q.4

If p, q, r are in A.P., then the determinant

Q.5

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

a 2 + a 2 n +1 + 2p b 2 + 2n + 2 + 3q c 2 + p 2n + p 2n +1 + q 2q = 2 n 2 n +1 2 a +2 +p b + 2 + 2q c − r

(A) 1 (C) a2b2c2 –2n Q.6

Q.6

distance between P and BC) (C) 2

(D) None of these

2

Q.7

If

∫e

x

2

dx = a , then the value of

(A) e – e (C) 2e4 – a

ln ( x ) dx is

e

1

4

∫

4

(B) e – a (D) 2e4 – e – a

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c2 − r

(B) 0 (D) (a2 + b2 + c2) –2n q

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

(A) 1

(B) 1/2

(C) 2

(D) buesa ls dksbZ ugha

2

e4

(D) 12

;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)

maximum of d(P, BC) is : (d(P, BC) represent (B) 1/2

b 2 + 2n +1 + 2q

(A) 1 (C) a2b2c2 –2n

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

(A) 1

(C) 28

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

a 2 + 2n + p

‘P’ is a point inside ∆ABC, such that

If

(B) 20

a 2 + a 2 n +1 + 2p b 2 + 2n + 2 + 3q c2 + p 2n + p 2n +1 + q 2q =

(B) 0 (D) (a2 + b2 + c2) –2n q

A(3, 4), B(0, 0) & C(3, 0) are vertices of ∆ABC.

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

Q.7

2

;fn ∫ e x dx = a gS] rks 1

(A) e4 – e (C) 2e4 – a

e4

∫

ln ( x ) dx dk eku gS-

e

(B) e4 – a (D) 2e4 – e – a

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Q.8

If circumradius and inradius of triangle be 10 and 3 respectively then value of a 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. 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.9

Q.10

 sin x  Limm = x →0  x 

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 eku gS(A) 13

(B) 26

(C) 39

(D) buesa ls dksbZ ugha

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 le{k 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

 sin x  Limm = x →0  x 

(m ∈ I, and [.] denotes greatest integer function)

(m ∈ I

(A) m if m ≤ 0 (C) m – 1 if m < 0

djrk gS) (A) m ;fn m ≤ 0 (C) m – 1 ;fn m < 0

2

(B) m – 1 if m > 0 (D) m if m > 0

2

The equation, 3x + 4y – 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

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Q.10

[.] egÙke iw.kkZad Qyu dks iznf'kZr (B) m – 1 ;fn m > 0 (D) m ;fn m > 0

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

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Q.11

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

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

π for-

(A) a =

3 2 2 (C) a = 3

3 ds fy, 2 2 (C) a = ds fy, 3

5 4 4 (D) a = 5

(B) a =

A unit vector which is equally inclined to the 2ˆi + ˆj + 2kˆ – 4ˆj – 3kˆ vectors ˆi , and is 2 5

(A) (C) Q.13

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

greatest integer function, has fundamental period

(A) a =

Q.12

Q.11

– (ˆi – 5ˆj + 5kˆ 51 ˆi + 5ˆj + 5kˆ 51

(B)

Q.12

(D) None of these

(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 passageII 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.

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2ˆi + ˆj + 2kˆ og bdkbZ lfn'k tks lfn'kksa ˆi , ,oa 2

– 4ˆj – 3kˆ ds lkFk leku >qdk gqvk gS] gksxk 5 ˆi + 5ˆj – 5kˆ – (ˆi – 5ˆj + 5kˆ (A) (B) 51 51 ˆi + 5ˆj + 5kˆ (C) (D) buesa ls dksbZ ugha 51

ˆi + 5ˆj – 5kˆ 51

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

5 ds fy, 4 4 (D) a = ds fy, 5 (B) a =

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 le{k 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

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Passage # 1 (Ques. 14 & 15) Two variable chords AB and BC of a circle 2

2

x|ka'k # 1 (iz- 14 ,oa 15) ` x2 + y2 = a2

x + y = a are such that AB = BC = a, and M

AB = BC = a

and N are the mid points of AB and BC

M

respectively such that line joining MN intersect

BC MN

P

Q O `

P; AB

O is the centre of the circle ∠OAB is (A) 30° Q.15

BC

AB

N

`

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

Q.14

AB

2

(B) 60°

(C) 45°

(D) 15°

Q.14

(A) 30°

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 the point (0, 2). Let

1 1  C ≡ y = log1/4  x –  + log4 (16x2 – 8x + 1), 2 4  be another curve.

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∠OAB

Q.15

A

(B) 60°

(C) 45°

(D) 15°

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) 1 1  C ≡ y = log1/4  x –  + 4 2 

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Q.16

Q.17

The range of the function g(x) = log2 (f (x)) is (A) [1, ∞)

(B) [2, ∞)

(C) [0, ∞)

(D) None of these

Q.16

1 1 and the line x = is 4 4

(A) 1 (C)

Q.17

(B) 3

2 3

(D)

1 e

1/ 4

(B) [2, ∞) (D) 1 x = y2 + 4

C, (A) 1 2 (C) 3

1 3

– e1/4, then value of a is

1 4

Q.18

(B) 3 1 (D) 3

;fn oØ y = f (x), oØ C, dksfV x = 1/4 ,oa dksfV x = a }kjk ifjc) {ks=kQy 4 – ln 4 +

(A) ln 6

(B) ln 4

gS] rks a dk eku gksxk

(C) 4

(D) ln 12

(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 the respective question numbers in the OMR have to be darkened. For example, if the correct answers to question numbers X, Y, Z and W (say) are 6, 0, 9 and 2, respectively, then the correct darkening of bubbles will look like the following :

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x =

ifjc) {ks=kQy gksxk

If the area bounded by the curve y = f (x), curve C, ordinate x = 1/4 & the ordinate x = a is 4 – ln 4 +

g(x) = log2 (f (x))

(A) [1, ∞) (C) [0, ∞)

The area bounded by the curve C, parabola x = y2 +

Q.18

log4(16x2 – 8x + 1)

1 e1 / 4

– e1/4,

[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

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X 0 1 2 3 4 5 6 7 8 9

Q.1

Y 0 1 2 3 4 5 6 7 8 9

Z 0 1 2 3 4 5 6 7 8 9

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

W 0 1 2 3 4 5 6 7 8 9

,sls fn[krs gSa tks fuEufyf[kr gSA X 0 1 2 3 4 5 6 7 8 9

If the chord of the parabola y2 = 4ax, whose

Y 0 1 2 3 4 5 6 7 8 9

Z 0 1 2 3 4 5 6 7 8 9

W 0 1 2 3 4 5 6 7 8 9

equation is y – x 2 + 4a 2 = 0, is a normal the curve, and that its length is

3a λ . Find the

y – x 2 + 4a 2 = 0

value of λ Q.2

y2 = 4ax

Q.1

3a λ

Solve the following equation for x (where [x]

λ dk eku Kkr dhft,

and {x} denotes integral and fractional part of x) |x – 1| = 2[x] – 3{x}. If the solution of equation is x1 and x2 where x1 + x2 =

34 . Find λ. λ

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] – 3{x} ;fn lehdj.k dk gy x1 ,oa x2 gS tgk¡ x1 + x2 =

Q.3

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

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34 rks λ dk eku λ

Kkr dhft,A

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Q.4

If the reflection of the point P(1, 0, 0) in the line x − 1 y + 1 z + 10 is (α, β, γ). Find – (α + β + γ) = = 2 8 −3

Q.3

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

Q.5

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

Q.4

;fn fcUnq P(1, 0, 0) dk js[kk

Two squares are chosen at random from small

Q.5

Q.6

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

squares (1×1) drawn on chess board and the probability that two squares chosen have exactly

x − 1 y + 1 z + 10 = = 2 8 −3

Q.6

k , then k is equal toone corner in common is 144

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 ,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)

Q.7

Q.8

Q.9

The number of solutions of the equation 4sin2x+10cosec2x + 9tan2x – 31 = 0 in x ∈ [0, 2π] islog (a + c), log(a + b), log(b + c) are in A.P. and a, c, b are in H.P., Where a, ,b, c > 0. kc , then the value of k is If a + b = 4

mHk;fu"B gksus dh izkf;drk Q.7

Q.8

(2, 2) is

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 =

and largest circles which have centers on the x2 + y2 + 2x + 4y – 4 = 0 and pass through the point

x ∈ [0, 2π] esa lehdj.k 4sin2x+10cosec2x + 9tan2x – 31 = 0 ds gyksa dh la[;k gksxh-

The difference between the radius of the smallest circumference of the circle

k gS, rks k dk eku gS 144

kc 4

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

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PHYSICS Section – I

[k.M - I

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

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 le{k viuk mÙkj vafdr dhft;sA izR;sd lgh mÙkj ds fy, + 3 vad fn;s tk;asxs rFkk izR;sd xyr mÙkj ds fy, 1 vad dkVk tk;sxkA

Q.1

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

r

dks'k ds dsUnz ij fo|qr foHko gksxk –

Conductor

Conductor

+q r O R

+q r O R

q 1 1   −  4πε0  r 2 R  q 1 1  (C)   + 4πε0  r 2 R 

(B)

2R q 4πε 0 r

(D) None of these

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2R r