IIT Advance Paper 1 MockTest 2013

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) TO DOWNLOAD PAPER-II ¾ 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)

www.ecareerpoint.com

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2

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

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 BO 3 units and two tetrahedral BO 4 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


[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 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 ckWjsDl ds lanHkZ esa dkSulk lR; ugha gS ? Q.1 (A) ;g vEyksa ds vuqekiu ds fy, ,d mi;ksxh izkFkfed ekud gksrk gS (B) ,d eksy ckWjDs l] cQj ds :i es a iz;Dq r gks ldrk gS (C) ckWjsDl dk tyh; foy;u] cQj ds :i esa iz;qDr gks ldrk gS (D) ;g nks s 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 (C) FTTT

(B) TFFT (D) TFTF

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3

Q.3

When negatively charged colloid like As 2S3 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 (k p) 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 NO 2 ? (A) 1.33 atm (B) 0.1665 atm (C) 0.133 atm (D) None of these

Q.4

–

BH3 H 2 O 2 , HO    →      → (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 (k p), 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 –

BH3 H 2O 2 , HO    →      → (A)

Q.5

OR

OR

mRikn (A) gS -

Product (A) is OH

OH

(A)

(B) OR

OH

(A)

OR

(B) OR

OH

(C)

OH OR

OH OH

(D)

OR


(C) OR OH

OH OR

(D) OR OH



4

Q.6

Major product (P) in following reaction

Q.6

OH  → (P) CH3 – CO — (CH 2 — )4 CHO  

OH CH3 – CO — (CH 2 — )4 CHO    → (P) -

∆

(A)

O || CH3 CHO

(C)

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

(B)

(A)

(D)

O || CH3 CHO

(C)

CHO

(B)

(D) CHO

CH3

CH3 SO 2 Cl 2

NBS

hν

∆

KSH

 → (B)    → (C),     → (A)  

Q.7

∆

hν

mRikn (C) gS -

major product (C) is SH

(A)

NBS SO 2 Cl 2 KSH  → (B)    → (C),     → (A)  

Q.7

SH

SH

(A)

(B) Br

SH

(B) Br

Cl

Cl

Br Br

(C)

Br

(C)

(D)

Br

(D) SH

SH




5

OH Q.8

OH

OH

PCC(excess)

Q.8

(A)

OH OH 1 equivalent

(B)

NaBH4

(C) HO Product (D) will be 2

OH

(A)

CH3MgBr

CH3MgBr

+

H3O

H3O+

(C)

NaBH4 H2O

(A)

(B)

(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.


OH O

O

(D)

PCC(excess)

mRikn (D) gS -

(D)

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



6

Q.9

Choose the correct statement (s) (A) BeCO3 is kept in the atmosphere of CO 2 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

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 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

dkSulh vfHkfØ;k;s lEHko gS ? (A) MgCl2 + NaNO3 → (B) BaSO4 + HCl → (C) ZnSO4 + BaS → (D) BaCO3 + CH3COOH → ;gh dFkuks dk p;u dhft, W ] ukfHkd }kjk (A) ukfHkd ds fudV fLFkr ,d bysDVªku 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




7

Q.12

Which of the following cycloalkanes will show cis-trans isomerism CH3

Q.12

CH3

CH3

(A)

(B)

(B)

CH3 CH3

H3C

CH3

(D)

(C)

H3C Q.13

CH3

(A) H3C

CH3

(C)

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

CH3 CH3

(D) H3C

Which of the following combinations give Ph – CH3 ? (A) Ph – CH 2 – MgBr + NO2 –

Q.13

Ph – CH3 ? (A) Ph – CH 2 – MgBr + NO2 –

– OH →

– OH →

CaO – + (B) PhCH2COONa    →

CaO – + (B) PhCH2COONa    →

NaOH

NaOH

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

(C) CH3MgBr rFkk

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.

O || H 2O / H +    → C6H5 –C–OC2H5 

H 3PO 4   → (D) Ph – CH 2MgBr  

H 3PO 4 (D) Ph – CH2MgBr     →


dkSuls la;kstu fuEu ;kSfxd nsxsaA

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



8

x|ka'k

Passage # 1 (Ques. 14 & 15)

Q.14

Q.15

(i) P + C(carbon) + Cl 2 → Q + CO↑ (ii) Q + H 2O → R + HCl (iii) BN + H2O → R + NH3↑ (iv) Q + LiAlH 4 → S + LiCl + AlCl 3 (v) S + H2 → R + H2↑ (vi) S + NaH → T (P, Q, R, S and T do not represent their chemical symbols) 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 Compound T is used as a/an (A) oxidising agent (B) complexing agent (C) bleaching agent (D) reducing agent

Q.15

Passage # 2 (Ques. 16 to 18) – O O O || | || HO – Ph– C–H + C–PH Ph– C | | O–H H


Q.14

x|ka'k slow

# 1 (iz- 14 ,oa 15)

(i) P + C(carbon) + Cl 2 → Q + CO↑ (ii) Q + H 2O → R + HCl (iii) BN + H 2O → R + NH3↑ (iv) Q + LiAlH 4 → S + LiCl + AlCl 3 (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 (B) I, II, IV (C) I, II, III (D) I, II, III, IV

;kSfxd T fdl :i esa iz;qDr gksrk gS (A) vkWDlhdkjd (B) ladqy dkjd (C) fojatd dkjd (D) vipk;d # 2 (iz- 16 ls 18)

O || Ph– C | H

–

HO –

O O || | Ph– C–H + C–PH | O–H

slow



9

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 || (A) H–C–H

(C) O Q.17 Q.18

Q.16

O || (A) H–C–H

(B) R 3CCHO (D) All of these

(C)

CHO

O

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

Q.17

CHO

(A)

Q.18

(A)

(C) 3

(B)

(C)

(D)

NO2 CHO

(D) CH3

Cl

(D) 4

CHO

OMe CHO

NO2 CHO


(B) 2

CHO

(B)

CH3

CHO

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

CHO

OMe CHO

(B) R 3CCHO (D) buesa ls dksbZ ugha

mi;qZDr vfHkfØ;k dh dksfV gS (A) 1

(D) 4

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

(C)

fuEu esa ls dkSuls fØ;kdkjd dsuhtkjks vfHkfØ;k lEiUu dj ldrs gS -

Cl



10

Section - III

[k.M -

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 : X Y Z W

Q.1

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 dks bZ _.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

Y

Z

W

0 1 2 3

0 1 2 3

0 1 2

4 5 6 7 8

4 5 6

0 1 2

0 1 2

0 1 2

0 1 2

0 1 2

3

3

3

3

4

4

4

4

5 6

5 6

5 6

5 6

7

7

7

7

8

8

8

8

9

9

9

9

3 4 5 6 7 8 9

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

Q.1

H 2S / H +

 → A (Light pink colour complex). HMNO4  Calculate CFSE value in light pink colour complex.

9

7 8 9

3 4 5 6 7 8 9

Pb 3O 4 / dil.HNO3 A(pedhyk xqykch ladqy)       →

∆

∆


III

+

2S / H  → A (pedhyk xqykch ladqy). HMNO4 H 

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



11

Q.2

Consider the following oxyanions :

Q.2

PO34 – , P2 O 64 – , SO 24 – , MnO – 4 , CrO 24 – , S2 O 52 – ,

PO34 – , P2 O 64 – , SO 24 – , MnO – 4 , CrO 24 – , S2 O 52 – ,

S2 O 72 –

S2 O 27 –

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/cm 3. The edge length of Li is 3.5 Å. Find out the number of Li atoms in a unit cell. (NA = 6.0 × 10 23 mol –1, M = 6.94 g mol –1) Two flask A and B of equal volumes maintained

Q.3

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 × 10 23 mol –1, M = 6.94 g mol –1) Q.4

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 N 2(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 ?

gksxk ?




12

Q.5

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

C(g) + D(g);

Q.5

12

K C = 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

Q.6

correctly matched ?

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

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

⊕

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

⊕

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

| CH3

| CH3

Conc.HI (c) CH3 –CH–O–CH3     → S N1

Conc.HI (c) CH3 –CH–O–CH3     → S N1

| CH3

| CH3

HI (d) CH3 –O–CH2CH3  → S N1

HI

(d) CH3 –O–CH2CH3  → S N1 (e)

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

K C = 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)

Conc.HI

–O–CH2 –CH = CH2     → S N1 ⊕

⊕

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

(g)

–O–CH2 –


⊕

H / H 2O     → S N1

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

(g)

–O–CH2 –

⊕

H / H 2O     → S N1



13

Q.7

How

many reactions the CH3 | CH3 –C–OH can be achieved | C2H5

synthesis

of Q.7

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

O || H ⊕ / H 2O (a) PhMgBr + CH3 –C–C2H5     → O || H⊕ / H 2O   → (b) C2H5MgBr + Ph–C–CH3   O || H ⊕ / H 2O (c) CH3MgBr + Ph–C– CH2CH3     → O || H ⊕ / H 2O   → (d) PhMgBr + CH3 –C–Cl  

⊕

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

H⊕ / H 2O

(e) C2H5COOCH3 + CH 3MgCl     →

( excess)

(excess)

⊕

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

H⊕ / H 2O

(f) C2H5COCl + CH 3MgCl     →

(excess)

( excess)

⊕

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

H⊕ / H 2O

 → (g) CH3 –COCH3 + C2H5MgCl   

C2 H 5 MgCl CH 3MgCl    → (A)   (h) CH3 –CN     →(B) ⊕ ⊕

C H MgCl

CH 3MgCl 5   → (A)   (h) CH3 –CN 2     →(B) ⊕ ⊕ H 2O / H

Q.8

( major )

H 2O / H

H 2O / H

How many of the following compounds, yield yellow precipitate on reaction with I 2 and NaOH ?

Q.8

O

(a)

(d)

(e) CH3COOC2H5 (g) CH3CONH2

fuEu es a ls fdrus ;kSfxd] I2 rFkk NaOH ds lkFk vfHkfØ;k djus ij ihyk vo{ksi nsxs ? (a)

(b) O

O

Ph O | || Ph–C – C–C (c) 2H5 | Ph

H 2O / H

O

(b) O

( major )

O

Ph O | || Ph–C – C–C (c) 2H5 | Ph

(d)

(f) (CH3CO)2O

(e) CH3COOC2H5

(f) (CH3CO)2O

(h) CH3COCH2COCH3

(g) CH3CONH2

(h) CH3COCH2COCH3

OH


OH



14

Q.9

How many of the following reaction are correct

Q.9

methods for the preparation of propanoic acid ? HBr CO 2 Mg  →  → (a) CH3 –CH=CH2      → ⊕ Ether

H3 O

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

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

H3 O

O 3 , Zn (d) H3C–CH=C–CH3    →

| CH3


fuekZ.k ds fy, lgh lqesfyr gS ? HBr CO 2 Mg (a) CH3 –CH=CH2    →  →    → ⊕ Ether

H3 O

(1). KMnO 4 / OH BH3 .THF (b) H3C–C≡CH     →       → ⊕  Θ H 2O 2 , OH

CO2 Mg HBr (c) H2C=CH2    →  →    → ⊕ Ether

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

( 2). H

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

H3 O

O 3 , Zn (d) H3C–CH=C–CH3    →

H 2O

| CH3

H 2O



15

MATHEMATICS [k.M -

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

Q.2

Q.1

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.2

 

(A) 10 cm/min (C) 40 cm/min Q.3

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

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

depth of water is 70 cm is -  use π =

`

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 (C) 40 cm/min

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


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 (C) 7x + 24y = 0

22   7 

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

I

(B) 20 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



16

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

Q.4

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 Q.5

a 2 + a 2n +1 + 2 p 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) a2 b2c2 –2n Q.6

(B) 0 (D) (a2 + b2 + c2) –2n q Q.6

maximum of d(P, BC) is : (d(P, BC) represent distance between P and BC) (C) 2

(D) None of these

2

Q.7

2

∫1

4

(A) e – e (C) 2e4 – a

∫e

2

ln ( x )

dx is

Q.7

4

(B) e – a (D) 2e4 – e – a


(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 (C) 2

e4

If e x dx = a , then the value of

(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 )

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

(B) 1/2

(C) 28

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

(A) 1 (C) a2 b2c2 –2n

‘P’ is a point inside ∆ABC, such that

(A) 1

(B) 20

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

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

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

(B) 1/2 (D) buesa ls dksbZ ugha x2

;fn ∫ e dx = a gS] rks 1 4

(A) e – e (C) 2e4 – a

e4

∫e

ln ( x )

dx dk eku gS-

(B) e4 – a (D) 2e4 – e – a



17

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

 sin x  = x →0  x  (m ∈ I, and [.] denotes greatest integer function) (A) m if m ≤ 0 (B) m – 1 if m > 0 Limm

(C) m – 1 if m < 0

Q.10

Q.8

(A) 13 (C) 39


(B) 26 (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

(D) m if m > 0

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

;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-

Q.10

 sin x  = x →0  x  (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 Limm

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



18

Q.11

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

Q.11

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

greatest integer function, has fundamental period π for3 5 (A) a = (B) a = 2 4 2 4 (C) a = (D) a = 3 5 Q.12

Q.13

f(x) = sin (2( [a ]) x ) ( tgk¡ [.] egÙke iw.kkZad Qyu 3 ds fy, 2 2 (C) a = ds fy, 3 (A) a =

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

A unit vector which is equally inclined to the ˆ ˆ 2î + jˆ + 2k – 4 jˆ – 3k vectors î , and is 2 5 ˆ î + 5 jˆ – 5k ˆ – (î – 5 jˆ + 5k (A) (B) 51 51 î + 5 jˆ + 5k ˆ (C) (D) None of these 51

Q.12

ˆ 2î + jˆ + 2k og bdkbZ lfn'k tks lfn'kksa î , ,oa

If Q(x) = x 2 – 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)

Q.13

ˆ – 4 jˆ – 3k ds lkFk leku >qdk gqvk gS] gksxk 5 ˆ î + 5 jˆ – 5k ˆ – (î – 5 jˆ + 5k (A) (B) 51 51 î + 5 jˆ + 5k ˆ (C) (D) buesa ls dksbZ ugha 51 ;fn Q(x) = x2 – mx + 1; x ds eku (1, 2) ds fy,

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.


2

_.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



19

x|ka'k

Passage # 1 (Ques. 14 & 15)

# 1 (iz- 14 ,oa 15)

Two variable chords AB and BC of a circle

`

x2 + y2 = a2

x2 + y 2 = a 2 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 the circle at P and Q where P is closer to AB and

BC MN

P

Q

P; AB

O

`

∠OAB is (A) 30°

Q.15

BC

AB

N `

O is the centre of the circle Q.14

AB

(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°

Q.15

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

x|ka'k

 x – 1  + 1 log (16x2 – 8x + 1),   4 2  4 

be another curve.


A

(B) 60°

(C) 45°

(D) 15°

C

(A) 90°

Passage # 2 (Ques. 16 to 18)

C ≡ y = log1/4

∠OAB -

(B) 120°

(C) 60°

(D) 150°

# 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



20

Q.16

The range of the function g ( x) = log2 ( f ( x)) is (A) [1, ∞) (B) [2, ∞) (C) [0, ∞)

Q.17

Q.18

(D) None of these

The area bounded by the curve C, parabola 1 1 2 x = y + and the line x = is 4 4 (A) 1 (B) 3 2 1 (C) (D) 3 3 If the area bounded by the curve y = f ( x), curve C, ordinate x = 1/4 & the ordinate x = a is 1 4 – ln 4 + 1 / 4 – e1/4, then value of a is e

log4(16x2 – 8x + 1) g ( x) = log2 ( f ( x))

Q.16

(A) [1, ∞) (C) [0, ∞)

(B) [2, ∞) (D) 1 2 x = y + 4

C,

Q.17

1 4

x =

ifjc) {ks=kQy gksxk (A) 1 2 (C) 3 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 (C) 4

1 e

– e1/4,

1/ 4

(B) ln 4 (D) ln 12

Section - III [k.M -

This section contains 9 questions (Q.1 to 9). +3 marks will be given for each correct answer and

9( 1

no negative marking. The answer to each of the

9)

vad fn;s tk,saxs rFkk dks bZ _.kkRed vadu ugha gSA bl

questions is a SINGLE-DIGIT INTEGER, ranging

+3

from 0 to 9. The appropriate bubbles below the

[k.M esa izR;sd iz'u dk mÙkj

respective question numbers in the OMR have to be

iw.kk±d gSaA

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

III

OMR esa

0

ls

9

rd bdkbZ ds ,d

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

will look like the following :




21

ds

Q.1

X

Y

Z

W

mÙkj

0 1 2

0 1 2

0 1 2

0 1 2

,sls fn[krs gSa tks fuEufyf[kr gSA

3

3

3

3

4

4

4

4

5 6

5 6

5 6

5 6

7

7

7

7

8

8

8

8

9

9

9

9

6, 0, 9

rFkk

2 gkas,

If the chord of the parabola y 2 = 4ax, whose

rks lgh fof/k ls dkys fd;s x;s cqYys X

Y

Z W

0 1 2 3 4 5 6 7 8 9

0 1 2 3

0 1 2 3 4

4 5 6 7 8 9

5 6 7 8 9

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

λ

Q.2

. Find λ.

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

–1

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


34

λ

rks λ dk eku

Kkr dhft,A



22

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

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

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 one corner in common is

;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 x − 1 y + 1 z + 10 ;fn fcUnq P(1, 0, 0) dk js[kk = = 2 −3 8

Q.6

k , then k is equal to144

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

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

Q.8

2

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 2

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

kc 4

gS] rks k dk eku gksxk Q.9

x + y + 2x + 4y – 4 = 0 and pass through the point (2, 2) is

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

s


k gS, rks k dk eku gS 144

s

s S



23

PHYSICS [k.M -

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

A 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 2 R. The distance r < R. The electric potential at the centre of the shell will be –

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

r

dks'k ds dsUnz ij fo|qr foHko gksxk – Conductor

+q r O R

+q r O R

(B)

2R q

 1 − 1    4πε0  r 2 R  q  1 1  (C)  +  4πε0  r 2 R  (A)

4πε 0 r

(D) None of these


2 R r < R

R

Conductor

 1 1   −  4πε0  r 2 R  q  1 1  (C)  +  4πε0  r 2 R  q

O

q

Q.1

2R

(A)

I

q

(B)

q

4πε0 r

(D) buesa ls dksbZ ugha



24

Q.2

Q.3

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.2

The ratio between the energy stored in 5 µF capacitor to the 4 µF capacitor in the given circuit is –

Q.3

4Ω

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

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

5µF

5µF 2Ω

2Ω 4Ω

4µF

4Ω

4µF

10V

10V

(A) 1.2

(B) 1

(C) 1.25

(A) 1.2

(D) 3.6

→

Q.4

A uniform magnetic field B = B0 jˆ exists in

Q.4

gSA

projected towards negative x-axis with speed v

ls pky

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

2 Bq dm

(B)

Bqd m

(C)


Bd

2d m

(D)

Bqd

2m

(C) 1.25

,dleku pqEcdh; {ks=k

space. A particle of mass m and charged q is from a point ( d , 0, 0). The maximum value of v

(B) 1

m

nzO;eku rFkk v

q

→

B = B0 jˆ

(D) 3.6

Lisl esa fo|eku

vkos'k dk d.k fcUnq (d , 0, 0)

ls _.kkRed x-v{k dh vksj iz{ksfir gSA

dk vf/kdre eku ftlds fy, d.k y- z ry ls ugha Vdjk,xk] gS (A)

2 Bq dm

(B)

Bqd m

(C)

Bd

2d m

(D)

Bqd

2m



v

25

Q.5

An experiment measures quantities a, b, c and then X is calculated from X =

a1 / 2 b 2 c

3

Q.5

. If the

X =

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

X esa

in X can be -

Q.6

Q.7

(C) 1 %

c3

ls Kkr gksrk gSA ;fn

a, b

rFkk

c

es a

izfr'kr =kqfV gks ldrh gS -

(A) 12.5 % (B) 7 %

(D) 4 %

A projectile is given an initial velocity of iˆ + 2 jˆ .

a1 / 2 b 2

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

and ± 2%, respectively, then the percentage error (A) 12.5 % (B) 7 %

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

Q.6

(C) 1 %

(D) 4 %

,d iz{ksI; dk izkjfEHkd osx iˆ + 2 jˆ }kjk fn;k x;k

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

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

(A) y = 2 x – 5 x2

(B) y = x – 5 x2

(A) y = 2 x – 5 x2

(B) y = x – 5 x2

(C) 4 y = 2 x – 5 x2

(D) y = 2 x – 25 x2

(C) 4 y = 2 x – 5 x2

(D) y = 2 x – 25 x2

If the acceleration of wedge in the figure shown is a m/s2 towards left, then at this instant acceleration of the block (magnitude only) would be -

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 M

M

α

α

(A) 4a m/s2

(B) a 17 − 8 cos α m/s2

(C) 17 a m/s2

(D) 17 cos


α 2

× a m/s2

(A) 4a m/s2 (C) 17 a m/s2

(B) a 17 − 8 cos α m/s2 (D) 17 cos

2

× a m/s2



26

Q.8

A uniform chain AB of mass m and length l is placed with one end A at the highest point of a hemisphere of radius R. Referring to the top of the hemisphere as the datum level, the potential energy of the chain is – (given that l < A

π R 2

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 – π R (fn;k x;k gS l < ) m

Q.8

)

2

l

A

R B

(A)

(C) (D)

B

l   − sin  R   R

2

(B)

R

mR 2 g  l l

l   − sin  R   R

mR 2 g 

2l

l l   sin −   R R 

mR 2 g  l

l l   sin −   R R 

mR 2 g  l

l   − sin  R  l  R mR 2 g  l l  (B)  − sin  R  2l  R mR 2 g  l l  (C)  sin −  2l  R R  mR 2 g  l l  (D)  sin −  l  R R  ls 13 rd cgqfodYih iz'u gaSA izR;sd iz'u ds pkj

(A)

mR g  l

2l

l

Questions 9 to 13 are multiple choice questions. Each

iz'u

question has four choices (A), (B), (C) and (D), out of

fodYi (A), (B), (C) rFkk (D) gSa] ftuesa ls ,d ;k ,d ls

which MULTIPLE (ONE OR MORE) is correct.

vf/kd fodYi lgh gaSA

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.


9

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



27

Q.9

A point object is placed at 30 cm from a convex 3   glass lens  µ g =  of focal length 20 cm. The 2  

3  yasl   µ g =  ls 30 cm ij fLFkr gSA fcEc dk 2  

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

A

C 12V

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

gS fp=k es a n'kkZ, ifjiFk es a A

4V

4Ω

2Ω

i1 1/2A

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

F

E

B

vafre izfrfcEc vuUr ij fufeZr gksxk ;fn -

Q.10

In the circuit shown in figure -

20 cm Qksdl nwjh ds vory dk¡p

Q.9

D

F 12V

E

i1 i2

B

G

1/2A

4V

4Ω

2Ω

6Ω

D

6Ω

i2

G

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

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


C



28

Q.11

A thermally insulated chamber of volume 2 V 0 is

Q.11

2V 0 vk;ru dk ,d Å"ekjks/kh psEcj S {ks =kQy ds

divided by a frictionless piston of area S into two

?k"kZ.k jfgr fiLVu }kjk nks cjkcj Hkkxks a

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

foHkkftr fd;k x;k gSA Hkkx A nkc P 0 rFkk rki

T 0 ij

pressure P 0 and temperature T 0 and in part B is

vkn'kZ xSl j[krk gS rFkk Hkkx B es a fuokZr gSA

k cy

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 -

ls fp=kkuqlkj tksM h+ xbZ gSA izkjEHk es a fLiz ax vfolkfjr gSA psEcj A es a xSl izlkfjr gksus ds fy, Lora =k gSA ekuk lkE;koLFkk es a fLiz ax x0 }kjk lEihM+; gSA rc – B

A

A

(A) final pressure of the gas is

kx0

(A) xSl dk vafre nkc

S

1 2 kx0 2

kx0 S

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

1 (C) change in internal energy of the gas is kx02 2 (D) temperature of the gas is decreased


o B es a

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

B

(B) work done by the gas is

A

gS 1 2 kx gS 2 0

(C) xSl dh vkUrfjd ÅtkZ ea s ifjorZu

1 2 kx gS 2 0

(D) xSl dk rkieku ?kVrk gS



29

Q.12

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

Q.12

B

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) Weight of the rod is lowered by

l

B′

2

(A) NM+ dk Hkkj

1 (B) Loss of gravitational potential energy is mgl 2 (C) Angular velocity is 3 g / l (D) Rotational kinetic energy is Q.13

2

}kjk de gks tkrk gS

(B) xq:Rokd"kZ.k fLFkfrt ÅtkZ dh gkfu

6

(D) ?kw.kZu xfrt ÅtkZ Q.13

ml 2 ω2

X

A2

h ρ

gS

A1

h


6

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 -

A1

h

1 mgl gS 2

(C) dks.kh; osx 3 g / l gS

ml 2 ω2

The vessel shown in the figure has two sections of areas of cross section A1 and A2. A liquid of density ρ fills both the sections, up to a height h in each. Neglect atmospheric pressure -

h

l

X

A2 ρ



30

(A) The pressure at the base of the vessel is 2 hρ g (B) The force exerted by the liquid on the base of the vessel is 2 hρ gA2 (C) The weight of the liquid is < 2 hρ gA (D) The walls of the vessel at the level X exert a downward force hρ g ( A2 – A1) on the liquid

2hρ g

(A) (B) 2hρ gA2 (C) (D) X

< 2hρ gA hρ g ( A2 – A1)

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.

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

Passage # 1 (Q.14 to 15)

x|ka'k

Figure shows a schematic diagram for a device known as a potentiometer. The potentiometer is useful for measuring an unknown electromotive force (emf) ε x 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 r x, r 0 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.


# 1 (Q.14 ls 15)

fp=k foHkoekih ds :i es a Kkr ;qfä ds fy, fp=k-izk:i iznf'kZr djrk gSA foHkoekih vKkr fo|qr okgd cy (fo-ok-cy) ε x dk Kkr fo-ok-cy ε0 ds :i es a fu/kkZj.k ds fy, mi;ksxh gSA bu nksu aks ds la;kstu es a rhljk fo-okcy ε /kkjk i ds mRiknu es a mi;qä gksrk gSA rhuks a fo-okcy Øe'k% r x, r 0 rFkk r vkUrfjd izfrjks/k j[krs gSA foHkoekih dk dsUnz h; ifjiFk rRo ,d yEck iz frjks/k gS tks dqy izfrjks/k R j[krk gS] D fcUnq ij fQly ldrk gSA



31

The resistivity ρ of this variable resistor is constant. Therefore, the resistance between points A and D is given by α R, where α is the distance from A to D divided by the distance from A to E . A very sensitive ammeter called a galvanometer is connected between points C and D in order to measure the current through that section of the circuit. B εx or ε0

ix or i0

r x or r 0 αR

A

C Galvanometer (1 – α) R E D

i G

ε

r

B εx or ε0

ix or i0

r x or r 0 αR

A

C xsYosuksehVj

(1 – α) R E D

i F

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 so that the galvanometer reads ix = 0. Using the loop theorem for the current loop ABCD, it can be shown that the current i satisfies the equation i = εx/(αx R). In the second step, the unknown emf εx is replaced by the known ε0, and the sliding contact at point D is again adjusted so that the galvanometer reads zero. Applying the loop theorem again leads to the result that the current i is also given i = ε0/(αx R). Equating the two expression for i we get for the unknown emf εx = ε0αx/α0.


bl ifjorhZ izfrjks/k dh izfrjks/kdrk ρ fu;r gSA vr% fcUnqv aks A rFkk D ds e/; izfrjks/k α R }kjk fn;k x;k gS ] tgk¡ α, A ls D rd nw jh }kjk A ls E rd foHkkftr nwj h gSA ,d cgqr laons h vehVj tks xsYoksuksehVj dgykrk gS bls ifjiFk ds bl Hkkx ls xqtjus okyh /kkjk ds fu/kkZj.k dh dksfV es a C o D fcUnqv aks ds e/; tksMk+ tkrk gSA

G

ε

r

F

foHkoekih dh fØ;k nks pj.kks a es a pkfyr gksrh gSA izFke vKkr fo-ok-cy εx ifjiFk es a B o C fcUnqv aks ds e/; tksMk+ tkrk gS rFkk D fcUnq ij rkj dks bl izdkj O;ofLFkr djrs gS fd xsYosuksehVj ix = 0 ikB~;kad nsrk gSA /kkjk yw i ABCD ds fy, ywi ize;s dk mi;ksx djus ij ;g iznf'kZr gksrk gS fd /kkjk i, i = εx/(αx R) lehdj.k dks lar"q V djrh gSA f}rh; pj.k es a] vKkr fo-ok-cy ε0 }kjk cny fn;k tkrk gS rFkk rkj dks D ij bl izdkj O;ofLFkr djrs gS fd xsYosuksehVj 'kwU ; ikB~;kad nsrk gSA iqu% yw i ize;s yxkus ij ;g ifj.kke gksrk gS fd /kkjk i = ε0/(αx R) }kjk nh xbZ gSA i ds fy, nksu aks O;atdks dh x.kuk ls ge ikrs gS fd vKkr fo-ok-cy εx = ε0αx/α0 gSA



32

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 + αx R) (B) ε/ R (C) ε/(r + R) (D) ε/(r + α0 R)

Q.14

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

Q.15

(A) 2 V (C) 6 V

x|ka'k

Passage # 2 (Ques. 16 to 18)

Find the time taken by wave to reach from source end to fixed end 25 10 L L (A) × (B) 12 F / µ F / µ 4 L L (C) (D) F / µ F / µ


(B) 3 V (D) 12 V

# 2 (Q.16 ls 18)

L 4 L

Four pieces of string each of length L are joined end to end to make a long string of length 4 L. The linear mass density of the strings are µ, 4 µ, 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 (ignore any reflection and absorptions). String has been stretched under a tension F .

Q.16

/kkjk i leku ekuh tkrh gS tc ε0 blls tksMk+ tkrk gS tc εx blls tqMk+ gqvk gSA ;fn xsYosuksehVj nksu aks fLFkfr;ks a es a 'kwU ; ikB~;kad nsrk gS] fuEu es a ls dkSulk i ds fy, lgh O;atd Hkh gS ? (A) ε/(r + αx R) (B) ε/ R (C) ε/(r + R) (D) ε/(r + α0 R) fn;k x;k mnkgj.k tgk¡ ε0 = 6V gSA ;fn AD yEckbZ 0.2 m ds cjkcj gS tc εx tqMk+ gqvk gS rFkk 0.4 m ds cjkcj gS tc ε0 tqMk+ gqvk gS εx dk eku D;k gS ?

µ, 4µ, 9µ

16µ

rFkk vko`fr 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) (C)

25 L × 12 F / µ 4 L F / µ

(B) (D)

10 L F / µ L F / µ



33

Q.17

Q.18

Find the ratio of wavelengths of the waves four strings, 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 Find the rate at which energy is transferred is maximum for the string having mass density (A) µ (B) 4µ (C) 9µ (D) 16µ

Q.17

(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 -

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 :


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

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



34

X

Y

Z

W

X

Y

Z

W

0 1 2

0 1 2

0 1 2

0 1 2

3

3

3

4

4

4

4

5 6

5 6

5 6

5 6

0 1 2 3 4 5 6

0 1 2

3

0 1 2 3

7

7

7

7

0 1 2 3 4 5 6 7

8

8

8

8

9

9

9

9

4 5 6 7 8 9

7 8 9

3 4 5 6 7 8 9

8 9

Q.1

A parallel plate capacitor is maintained at a certain potential difference. When a 3 mm slab is introduced between the plates, in order to maintain the same potential difference, the distance between the plates is increased by 2.4 mm. Find the dielectric constant of the slab.

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 es a O;ofLFkr djds feykbZ tkrh gS IysVks ds e/; 2.4 mm }kjk o`f) gksrh gSA Lysc dk ijkoS|rq kad Kkr djksA

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 in the circuit changes uniformly from 1 A to 0.5 A in 0.2 sec.

Q.2

dq.Myh dk Loiz sjdRo (H es a) ftles a 10 V dk fo-ok-cy iz sfjr gS tc ifjiFk es a 0.2 lsd.M es a /kkjk 1 A ls 0.5 A rd ,dleku :i ls ifjofrZr gksrh gSA

Q.3

A circuit is shown in figure.

Q.3

,d ifjiFk fp=k es a n'kkZ;k x;k gSA

2 µF

2 µF

A B

B

3 µF

3 µF

5 µF

5 µF 4 µF

4 µF

+ – 6V

+ – 6V

Find the charge (in µC) on the condenser having a capacity of 5 µF.


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

Kkr djksA



35

Q.4

A circular loop of radius R is bent along a

Q.4

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

,d R f=kT;k dk o`Ùkkdkj yw i O;kl ds vuqfn'k eksMk+ tkrk gSA nh xbZ vkd`fr fp=k es a iznf'kZr gSA ,d v)ZoÙ` k (KNM) x- z ry es a j[kk gS rFkk nw ljk

plane with their centres at the origin. Current I is

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

flowing through each of the semi circles as

gSA fp=kkuqlkj izR;sd v)ZoÙ` kks es a izokfgr /kkjk I gSA

shown in figure.

L L

y

L

M

L

K

L

y

x

N z

I

K

A particle of charge q is released at the origin →

with a velocity v = v0iˆ . Find the magnitude of →

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

q

x

N I

vkos'k dk d.k ew y fcUnq ls

tkrk gSA d.k ij rkR{kf.kd cy

z →

v = v0iˆ →

F

osx ls NksMk+

dk ifjek.k Kkr

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

Assume that space is gravity free.


M

L



36

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

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  r  V = V0 ln   }kjk fn;k tkrk gS] tgk¡ V 0 rFkk r 0  r 0  fu;rkad gS rFkk r izksVkWu ds pkjksa vksj xfreku bysDVªkWu dh d{kk dh f=kT;k gSA ekukfd cksgj ekWMy ekU; gS] ;g ik;k tkrk gS fd r , n x ds lekuqikrh gS] tgk¡ n eq[; Dok.Ve la[;k gSA x dk eku Kkr djksA

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.7

20 kV ij pkfyr X -fdj.k ufydk esa 200 W dh

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.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

Assume that the mass of a nucleus is

Q.9

ekfu, fd ukfHkd dk nzO;eku yxHkx M = Am p }kjk

 r  electron is given by, V = V0 ln    r 0  where V 0 and r 0 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 n x, where n is the principal quantum number. Find the value of x. Q.7

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

approximately given by M = Am p where A is the

fn;k x;k gS] tgk¡

mass number. Estimate the density of matter

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

(in × 1017 kg/m3) inside a nucleus.


A

nzO;eku la[;k gSA ukfHkd esa



37

IIT Advance Paper 1 MockTest 2013

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