Time : 3 : 00 Hrs. MAX MARKS: 246 Name : Roll No. : DateSpace for rough work CAREER POINT, CP Tower,...

TARGET – IIT JEE

CHEMISTRY, MATHEMATICS & PHYSICS

Time : 3 : 00 Hrs. MAX MARKS: 246

Name : _____________________________________ Roll No. : __________________________ Date : _____________

INSTRUCTIONS TO CANDIDATE

A. GENERAL :

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

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

B. MARKING SCHEME :

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

each wrong answer. 5. Multiple choice questions with multiple correct option. 4 marks will be awarded for each correct answer and No negative

marking for wrong answer. 6. Reason and Assertion type questions with only one correct answer in each. 3 marks will be awarded for each correct

answer and –1 mark for each wrong answer.

7. Passage based single correct type questions. 4 marks will be awarded for each correct answer and –1 mark for each wrong answer.

C. FILLING THE OMR :

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

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

Section –I Section-II Section-III

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

A B C D E

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

A B C D E

the wrong method for filling the bubble are

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

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

P Q R S TA BCD

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

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CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-3040000, Fax (0744) 3040050 email : [email protected]; Website : www.careerpointgroup.com 7

RS -11- I -6

SEA

L

Space for rough work

CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-3040000 Page # 2

Important Data (egRoiw.kZ vk¡dM+s)

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

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

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



CHEMISTRY

Section - I Questions 1 to 6 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 correct answerand – 1 mark for each wrong answer.

Q.1 The end product (A) of the reaction is

O

O (i) I2 + NaOH, ∆

(ii) H3O+, ∆ CHI3 + A ; A is

(A)

COOH

O (B)

CHO

O

(C) COOH

COOH (D)

O

Q.2 The end product in the reaction sequence is

HBr Benzoyl peroxide

Mg etherA B CO2

H3O+ C ; C is

(A)

COOH (B) COOH

(C) CH2COOH

(D) None of these

[k.M - I iz'u 1 ls 6 rd cgqfodYih iz'u gSaA izR;sd iz'u ds pkj

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

lgh gSA OMR 'khV esa iz'u dh iz'u la[;k ds lek viuk

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

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

Q.1 fuEu vfHkfØ;k dk vfUre mRikn A gS

O

O(i) I2 + NaOH, ∆

(ii) H3O+, ∆ CHI3 + A

(A)

COOH

O (B)

CHO

O

(C) COOH

COOH (D)

O

Q.2 fuEu vfHkfØ;k vuqØe esa vfUre mRikn C gS &

HBrBenzoylperoxide

Mg ether A B CO2

H3O+ C

(A)

COOH (B) COOH

(C) CH2COOH

(D) buesa ls dksbZ ugha



Q.3 Among the following acids which has the lowest pKa value ?

(A) CH3COOH (B) HCOOH (C) (CH3)2CH – COOH (D) CH3CH2COOH

Q.4 Compared to common colloidal sols micelles have : (A) higher colligative properties (B) lower colligative properties (C) same colligative properties (D) none of these

Q.5 The solubilities of MX, MX2, and M3X at 25ºC are 10–6 mol dm–3, 10–4 mol dm–3 and 10–3 mol dm–3

respectively. Ksp values of the salt at 25ºC are in the

order. (A) MX > MX2 > M3X (B) M3X > MX2 > MX (C) MX2 > M3X > MX (D) MX > M3X > MX2

Q.6 Electrolysis of water was carried out by passing 0.965 A current for 2000 seconds. Amount of O2

liberated is (in milli moles) (A) 2.5 (B) 5.0 (C) 10 (D) 20

Questions 7 to 10 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct. Mark your response in OMR sheet against the question number of that question. + 4 marks will be given for each correct answer and NO NEGATIVE marks for wrong answer.

Q.3 fuEu esa ls fdldk pKa eku U;wure gS ?

(A) CH3COOH (B) HCOOH

(C) (CH3)2CH – COOH (D) CH3CH2COOH

Q.4 lkekU; dksykbMh lksyksa dh rqyuk esa felsy ds & (A) v.kqla[;d xq.k/keZ mPp gksrs gS (B) v.kqla[;d xq.k/keZ de gksrs gSa (C) v.kqla[;d xq.k/keZ leku gksrs gS (D) buesa ls dksbZ ugha

Q.5 25ºC ij MX, MX2, rFkk M3X ds foys;rk xq.kuQy

Øe'k% 10–6 mol dm–3, 10–4 mol dm–3 rFkk 10–3 mol

dm–3 gSA 25ºC ij yo.k ds Ksp ekuksa dk Øe gS

(A) MX > MX2 > M3X (B) M3X > MX2 > MX (C) MX2 > M3X > MX (D) MX > M3X > MX2

Q.6 0.965 A /kkjk dks 2000 lsd.M rd izokfgr dj tydk fo|qqr vi?kVu fd;k x;kA eqDr gqbZ O2 dh ek=kk(feyh eksy esa) gS

(A) 2.5 (B) 5.0 (C) 10 (D) 20

iz'u 7 ls 10 rd cgqfodYih iz'u gaSA izR;sd iz'u ds pkj

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

fodYi lgh gaSA OMR 'khV esa iz'u dh iz'u la[;k ds lek

vius mÙkj vafdr dhft,A izR;sd lgh mÙkj ds fy, + 4 vad

fn;s tk;saxs rFkk xyr mÙkj ds fy, dksbZ _.kkRed vadu

ugha gSA



Q.7 When a mixture of calcium acetate and calciumformat is subjected to dry distillation, it yields.

(A) acetone (B) acetaldehyde (C) formaldehyde (D) acetic acid

Q.8 The reagent(s) used to distinguish but-1-yne from but-2-yne is/are

(A) Br2/CCl4 (B) KMnO4/OH– (C) Cu2Cl2 + NH4OH (D) [Ag(NH3)2]OH

Q.9 A solution of colourless salt H on boiling withexcess NaOH produces a non-flammable gas. The gas evolution stop after some time. Upon additionof Zn dust to the same solution, the gas evolutionrestarts. The colourless salt(s) H is (are)

(A) NH4NO3 (B) NH4NO2 (C) NH4Cl (D) (NH4)2SO4

Q.10 A gas described by van der waals equation (A) behave similar to an ideal gas in the limit of

large molar volumes (B) behaves similar to an ideal gas is in limit of

large pressures (C) is characterized by van der waals coefficients

that are dependent on the identity of the gas butare independent of the temperature.

(D) has the pressure that is lower than the pressureexerted by the same gas behaving ideally

Q.7 tc dsfY'k;e ,flVsV rFkk dsfY'k;e QkWesZV ds feJ.k dk 'kq"d vklou fd;k tkrk gSa] rks mRikn izkIr gksaxs &

(A) ,flVksu (B) ,flVsfYMgkbM (C) QkWesZfYMgkbM (D) ,flfVd vEy

Q.8 C;wV-1-vkbu rFkk C;wV-2-vkbu esa foHksn djus ds fy,

fdu vfHkdeZdksa dk mi;ksx fd;k tkrk gSa (A) Br2/CCl4 (B) KMnO4/OH– (C) Cu2Cl2 + NH4OH (D) [Ag(NH3)2]OH

Q.9 jaxghu yo.k H ds foy;u dks vf/kD; NaOH ds lkFk mckyus ij vToyu'khy xSl mRiUu gksrh gSAdqN le; i'pkr~ xSl mRiUu gksuk :d tkrk gSA bl foy;u esa Zn pw.kZ Mkyus ij xSl iqu% mRiUu gksrh gSA jaxghu yo.k H gSaA

(A) NH4NO3 (B) NH4NO2 (C) NH4Cl (D) (NH4)2SO4

Q.10 ok.Mjoky lehdj.k ds kjk xSl dks le>krs gS &

(A) cgqr vf/kd eksyj vk;ru dh lhek esa vkn'kZ xSl dh rjg O;ogkj djrh gSA

(B) cgqr vf/kd nkc dh lhek esa vkn'kZ xSl dh rjg O;ogkj djrh gSA

(C) ;g ok.Mjoky fu;rkadkssa kjk le>k;h tkrh gSa tks fd xSl dh fo'ks"krk ij fuHkZj djrs gS ysfdu rki ij fuHkZj ugha djrs gSa

(D) bldk nkc] vkn'kZ O;ogkj dh leku xSl kjk vkjksfir fd;k tkrk gS] ls de gksrk gS



This section contains 4 questions numbered 11 to 14,(Reason and Assertion type question). Each questioncontains Assertion and Reason. Each question has 4choices (A), (B), (C) and (D) out of which ONLY ONE iscorrect. Mark your response in OMR sheet against thequestion number of that question. + 3 marks will begiven for each correct answer and – 1 mark for eachwrong answer. The following questions given below consist of

an "Assertion" (A) and "Reason" (R) Typequestions. Use the following Key to choose theappropriate answer.

(A) If both (A) and (R) are true, and (R) is thecorrect explanation of (A).

(B) If both (A) and (R) are true but (R) is not thecorrect explanation of (A).

(C) If (A) is true but (R) is false. (D) If (A) is false but (R) is true.

Q. 11 Assertion : The pKa value of acetic acid is lowerthan that of phenol.

Reason : Phenoxide is more resonance stabilizedthan carboxylate ion.

Q.12 Assertion : Upon heating an amide with Br2 and KOH, primary amine with one carbon atom less isformed.

Reason : The reaction occurs through carbene intermediate.

bl [k.M esa 11 ls 14 rd 4 iz'u gSa, (dFku rFkk dkj.k izdkj

ds iz'u)A izR;sd iz'u esa dFku rFkk dkj.k fn;s x;s gSaA

izR;sd iz'u ds 4 fodYi (A), (B), (C) rFkk (D) fn;s x;s gSa] ftuesa ls dsoy ,d 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;asxs rFkk izR;sd xyr mÙkj ds fy,

1 vad ?kVk;k tk;sxkA

uhps fn;s x;s fuEufyf[kr iz'u "dFku" (A) rFkk "dkj.k" (R) izdkj ds iz'u gSaA vr% mfpr mÙkj dk p;u djus ds fy;s fuEu rkfydk dk mi;ksx dhft;sA

(A) ;fn dFku rFkk dkj.k nksuksa lR; gSa rFkk dkj.k, dFku dk lgh Li"Vhdj.k gSA

(B) ;fn dFku rFkk dkj.k nksuksa lR; gSa rFkk dkj.k, dFku dk lgh Li"Vhdj.k ugha gSA

(C) ;fn dFku lR; gS ysfdu dkj.k vlR; gSA (D) ;fn dFku vlR; gS ysfdu dkj.k lR; gSA

Q. 11 dFku : ,flfVd vEy dk pKa eku fQukWy ls de gksrk gSA

dkj.k : fQukWDlkbM] dkcksZDlhysV vk;u dh viskk vf/kd vuquknh LFkk;h gksrk gSA

Q.12 dFku : ,ekbM dks Br2 rFkk KOH ds lkFk xeZ djus

ij ,d dkcZu de okyh izkFkfed ,ehu curh gSA

dkj.k : vfHkfØ;k dkchZu e/;orhZ kjk gksrh gSA



Q.13 dFku : nzo veksfu;k esa kkj /kkrqvksa dks ?kksyus ij

uhyk foy;u izkIr gksrk gS

dkj.k : nzo veksfu;k esa kkj /kkrq,sa [M(NH3)n]+ izdkj

dh foyk;dhÑr Lih'kht nsrh gSA (M = kkj /kkrq,sa)

Q.14 dFku : ,.Vh¶yksjkbV lajpuk esa] vkWDlkbM vk;u c.c.p. lajpuk (cUn ?kuh; fufcM+) rFkk Li⊕ vk;u100% prq"Qydh; fjfä;k¡ ?ksjrs gSaA

dkj.k : ,.Vh¶yksjkbV lajpuk esa fudVre ijek.kqvksa

ds e/; nwjh 43a gksrh gS] tgk¡ a ?ku dh Hkqtk

yEckbZ gSA bl [k.M esa 3 vuqPNsn fn;s x;s gSa] izR;sd esa 3 cgqfodYih

iz'u gSaA (iz'u 15 ls 23) izR;sd iz'u ds pkj fodYi (A), (B), (C) rFkk (D) gSa, ftuesa ls dsoy ,d fodYi lgh gSA OMR 'khV esa iz'u dh iz'u la[;k ds lek viuk mÙkj vafdr

dhft;sA izR;sd lgh mÙkj ds fy;s + 4 vad fn;s tk,saxs rFkk

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

vuqPNsn # 1 (iz- 15 ls 17) gkWQeku cksesekbM vfHkfØ;k esa NaOH + Br2 ds mi;ksx

kjk isrd ,ekbM ls ,d dkcZu de okyh ,ehu cuk;h tk ldrh gSA lkekU; vfHkfØ;k bl rjg gS :

RCONH2 + Br2 + 4NaOH → R–NH2 + 2NaBr + Na2CO3 +2H2O

Q.13 Assertion : Alkali metals dissolve in liquidammonia to give blue solutions.

Reason : Alkali metals in liquid ammonia givesolvated species of the type [M(NH3)n]+

(M = alkali metals)

Q.14 Assertion : In antifluorite structure, the oxide ionsoccupy c.c.p. (cubic close packing) and Li⊕ ions 100% tetrahedral voids.

Reason : The distance of the nearest neighbours in

antifluorite structure is 43a , where a is the edge

length of the cube.

This section contains 3 paragraphs, each has 3 multiplechoice questions. (Questions 15 to 23) Each question has4 choices (A), (B), (C) and (D) out of which ONLY ONEis correct. Mark your response in OMR sheet againstthe question number of that question. + 4 marks will begiven for each correct answer and – 1 mark for eachwrong answer. Passage # 1 (Ques. 15 to 17) Amines with one carbon less than the parent amide

can be prepared by Hofmann's bromamide reactionusing NaOH + Br2. The general reaction is :

RCONH2 + Br2 + 4NaOH → R–NH2 + 2NaBr + Na2CO3 +2H2O



Mechanism :

Br2+ 2NaOH → NaBr + NaOBr + H2O

R – C – NH2

O + OBr– →

−−OH

R – C – N – Br

O

H → −− OH,OH 2

R –– C –– NO

Nitrene

Alkyl shift R – N = C = O →

−OH2 R – NH2 + CO3–2

Isocyanate Amides can be converted into amines with the

same number of carbons using LiAlH4.

Q.15 When acetamide is treated with sodium hypobromite, the product formed is

(A) CH3CN (B) CH3OH (C) CH3NH2 (D) CH3COBr Q.16 Among the following which one does not act as an

intermediate in Hoffmann rearrangement ?

(A) RNCO (B) ••

NRCO

(C) RCO••

N Br (D) RNC

fØ;kfof/k :

Br2+ 2NaOH → NaBr + NaOBr + H2O

R – C – NH2

O+ OBr– →

−−OH

R – C – N – Br

O

H → −− OH,OH 2

R –– C –– NO

Nitrene

Alkylshift R – N = C = O →

−OH2 R – NH2 + CO3–2

Isocyanate ,ekbM dks LiAlH4 ds mi;ksx ls leku dkcZu okyh

,ehu esa ifjofrZr fd;k tk ldrk gSA

Q.15 tc ,flVkekbM dks lksfM;e gkbiksczksekbV ds lkFk

xeZ fd;k tkrk gS rks mRikn cusxk

(A) CH3CN (B) CH3OH (C) CH3NH2 (D) CH3COBr

Q.16 fuEu esa ls dkSulk gkWQeku iwufoZU;kl esa e/;orhZ ds

:i esa dk;Z ugha djrk gS ?

(A) RNCO (B) ••

NRCO

(C) RCO••

N Br (D) RNC



Q.17 Identify the product of the reaction C2H5 – C – C

CH3

H NH2

O

(R)-2-methylbutanamide

Br2, OH–

∆

(A) (R)-S-butylamine (B) (S)-S-butylamine (C) (R)-S-pentylamine (D) (S)-S-pentylamine Passage # 2 (Ques. 18 to 20)

Redox reactions play a pivotal role in chemistry and

biology. The values of standard redox potential (Eº) of

two half-cell reactions decide which way the reaction is

expected to proceed. A simple example is a Daniel cell

in which zinc goes into solution and copper gets

deposited. Given below are a set of half-cell reactions

(acidic medium) along with their Eº (V with respect to

normal hydrogen electrode) values. Using this data

obtain the correct explanations to questions 18 to 20.

I2+ 2e– → 2I– Eº = 0.54

Cl2 + 2e– → 2Cl– Eº = 1.36

Mn3+ + e– → Mn2+ Eº = 1.50

Fe3+ + e– → Fe2+ Eº = 0.77

O2 + 4H+ + 4e– → 2H2O Eº = 1.23

Q.17 fuEu vfHkfØ;k ds mRikn dh igpku dhft, C2H5 – C – C

CH3

H NH2

O

(R)-2-esfFky C;wVsusekbM

Br2, OH–

∆

(A) (R)-S-C;wfVy,ehu (B) (S)-S-C;wfVy,ehu

(C) (R)-S-isfUVy,ehu (D) (S)-S-isfUVy,ehu vuqPNsn # 2 (iz- 18 ls 20)

jlk;u foKku rFkk tho foKku esa jsMkWDl vfHkfØ;k,sa

vk/kkjHkwr Hkwfedk fuHkkrh gSA nks v)Zlsy vfHkfØ;kvksa ds

ekud jsMkWDl foHko (Eº) ds eku ;g fuf'pr djrs gS fd

vfHkfØ;k fdl iFk ij vxzlj gSA ljy mnkgj.k gS Msfu;y

lSy ftlesa ftad foy;u esa fey tkrk gS rFkk dkWij tek

gksrk gSA uhps vZlsy vfHkfØ;kvksa ¼ vEyh; ek/;e esa½ ds lsV

buds Eº ¼ V lkekU; gkbMªkstu bysDVªkWM ds lanHkZ esa ½ eku

ds lkFk fn;s x;s gSA bu vkdM+ksa dk mi;ksx iz'u 18 ls 20 esas

dhft, I2+ 2e– → 2I– Eº = 0.54

Cl2 + 2e– → 2Cl– Eº = 1.36

Mn3+ + e– → Mn2+ Eº = 1.50 Fe3+ + e– → Fe2+ Eº = 0.77

O2 + 4H+ + 4e– → 2H2O Eº = 1.23



Q.18 Among the following, identify the correct

statement :

(A) Chloride ion is oxidised by O2

(B) Fe2+ is oxidized by iodine

(C) Iodide ion is oxidized by chlorine

(D) Mn2+ is oxidized by chlorine

Q.19 While Fe3+ is stable, Mn3+ is not stable in acid

solution because :

(A) O2 oxidises Mn2+ to Mn3+

(B) O2 oxidises both Mn2+ to Mn3+ and Fe2+ to Fe3+

(C) Fe3+ oxidises H2O to O2

(D) Mn3+ oxidises H2O to O2

Q.20 Sodium fusion extract, obtained from aniline, on

treatment with iron (II) sulphate and H2SO4 in

presence of air gives a Prussian blue precipitate.

The blue colour is due to the formation of :

(A) Fe4[Fe(CN)6]3 (B) Fe3[Fe(CN)6]2

(C) Fe4[Fe(CN)6]2 (D) Fe3[Fe(CN)6]3

Q.18 fuEu esa lgh dFku dh igpku dhft,A

(A) Dyksjhu vk;u] O2 kjk vkWDlhÑr gksrk gS

(B) Fe2+, vk;ksMhu kjk vkWDlhÑr gksrk gS

(C) vk;ksMkbM vk;u] Dyksjhu kjk vkWDlhÑr gksrk gS

(D) Mn2+, Dyksjhu kjk vkWDlhÑr gksrk gS

Q.19 vEyh; foy;u esa Fe3+ LFkk;h gS tcfd Mn3+ LFkk;h

ugha gksrk gS D;ksafd

(A) O2, Mn2+ dks Mn3+ esa vkWDlhÑr djrk gS

(B) O2, Mn2+ dks Mn3+ esa rFkk Fe2+ dks Fe3+ esa

vkWDlhÑr djrk gS

(C) Fe3+, H2O dks O2 esa vkWDlhÑr djrk gS

(D) Mn3+, H2O dks O2 esa vkWDlhÑr djrk gS

Q.20 ,fuyhu ls izkIr lksfM;e laxfyr fu"d"kZ dks ok;q dh

mifLFkfr esa vk;ju (II) lYQsV rFkk H2SO4 ds lkFk

xeZ djus ij izqf'k;u uhyk voksi izkIr gksrk gSA

uhyk jax fdlds cuus ds dkj.k vkrk gS :

(A) Fe4[Fe(CN)6]3 (B) Fe3[Fe(CN)6]2

(C) Fe4[Fe(CN)6]2 (D) Fe3[Fe(CN)6]3



Passage # 3 (Ques. 21 to 23)

NiCl2 HCNKCN → complex A

NiCl2 excessHCl → complex B

A & B complexes have the co-ordination number 4.

Q.21 The IUPAC name of complexes ‘A’ & ‘B’ are

respectively : (A) Potassium tetracyanonicklate (II) and Potassium tetrachloronicklate (II) (B) Potassium tetracyanonickel (II) and Potassium tetrachloronickel (II) (C) Potassium cyanonicklate (II) and Potassium chloronicklate (II) (D) Potassium cyanonickel (II) and Potassium chloronickel (II)

Q.22 The hybridization of both complexes are : (A) dsp2 (B) sp2 & dsp2 (C) dsp2 & sp3 (D) both sp3

Q.23 What are the magnetic nature of ‘A’ & ‘B’ ? (A) Both diamagnetic (B) ‘A’ is diamagnetic & ‘B’ is paramagnetic with

one unpaired electrons (C) ‘A’ is diamagnetic & ‘B’ is paramagnetic with

two unpaired electrons (D) Both are paramagnetic

vuqPNsn # 3 (iz- 21 ls 23)

NiCl2 HCNKCN → ladqy A

NiCl2 excessHCl → ladqy B

A o B ladqyksa dh milgla;kstu la[;k 4 gSA

Q.21 ladqy ‘A’ o ‘B’ dk IUPAC uke Øe'k% gS (A) iksVsf'k;e VsVªklk;uksfudysV (II) rFkk iksVsf'k;e VsVªkDyksjksfudysV (II) (B) iksVsf'k;e VsVªklk;uksfudy (II) rFkk iksVsf'k;e VsVªklk;uksfudy (II) (C) iksVsf'k;e lkbuksfudysV (II) rFkk iksVsf'k;e DyksjksfudysV (II) (D) iksVsf'k;e lkbuksfudy (II) rFkk iksVsf'k;e Dyksjksfudy (II)

Q.22 nksuksa ladqyksa dk ladj.k gSa (A) dsp2 (B) sp2 o dsp2 (C) dsp2 o sp3 (D) nksuksa dk sp3

Q.23 ‘A’ rFkk ‘B’ dh pqEcdh; izÑfr D;k gS ? (A) nksuksa izfrpqEcdh; (B) ‘A’ izfrpqEcdh; rFkk ‘B’ ,d v;qfXer bysDVªkWu

;qDr vuqpqEcdh; (C) ‘A’ izfrpqEcdh; rFkk ‘B’ nks v;qfXer bysDVªkWuksas

;qDr vuqpqEcdh; (D) nksuksa vuqpqEcdh; gSa



MATHEMATICS

Section - I Questions 1 to 6 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 correct answer and – 1 mark for each wrong answer.

Q. 1 The polynomial f(x) = x4 + ax3 + bx2 + cx + d hasreal coefficients and f (2i) = f(2 + i) = 0. The valueof (a + b + c + d) equals to-

(A) 1 (B) 4 (C) 9 (D) 10

Q. 2 If the sum ∑∞

= +++1k 2kkk)2k(1 =

cba +

where a, b, c ∈ N and lie in [1, 15], then a + b + cequals to-

(A) 6 (B) 8 (C) 10 (D) 11 Q. 3 Triangle ABC is isosceles with AB = AC and BC =

65 cm. P is a point on BC such that theperpendicular distances from P to AB and AC are24 cm and 36 cm, respectively. The area oftriangle ABC in sq cm is-

(A) 1254 (B) 1950

(C) 2535 (D) 5070

[k.M - I iz'u 1 ls 6 rd cgqfodYih iz'u gSaA izR;sd iz'u ds pkjfodYi (A), (B), (C) rFkk (D) gSa, ftuesa ls dsoy ,d fodYilgh gSA OMR 'khV esa iz'u dh iz'u la[;k ds lek viukmÙkj vafdr dhft;sA izR;sd lgh mÙkj ds fy, + 3 vad fn;stk;asxs rFkk izR;sd xyr mÙkj ds fy, 1 vad ?kVk;k tk;sxkA

Q. 1 cgqin f(x) = x4 + ax3 + bx2 + cx + d ds xq.kk¡dokLrfod gksa rFkk f (2i) = f(2 + i) = 0 rks(a + b + c + d) dk eku gS -

(A) 1 (B) 4 (C) 9 (D) 10

Q. 2 ;fn ∑∞

= +++1k 2kkk)2k(1 =

cba + tgk¡

a, b, c ∈ N ,oa varjky [1, 15] esa fLFkr gaS, rksa + b + c dk eku gS -

(A) 6 (B) 8 (C) 10 (D) 11 Q. 3 f=kHkqt ABC lefckgq gS] ftlesa AB = AC ,oa

BC = 65 cm gSA ;fn fcUnq P Hkqtk BC ij bl izdkj lsgS fd fcUnq P dh AB o AC ls yEcor~ nwfj;k¡ Øe'k% 24cm o 36 cm gks rks f=kHkqt ABC dk kS=kQy oxZ lseh esagksxk -

(A) 1254 (B) 1950

(C) 2535 (D) 5070



Q. 4 Three boxes are labelled A, B and C and each box contains four balls numbered 1, 2, 3 and 4. The balls in each box are well mixed. A child chooses one ball at random from each of the three boxes. If a, b and c are the numbers on the balls chosen from the boxes A, B and C resepctively, the child wins a toy helicopter when a = b + c. The odds in favour of the child to receive the toy helicopter are-

(A) 3 : 32 (B) 3 : 29 (C) 1 : 15 (D) 5 : 59

Q. 5 If N = 7p+4.5q.23 is a perfect cube, where 'p' and 'q' are positive integers, the smallest possible value of 'p + q' is-

(A) 5 (B) 2 (C) 8 (D) 6

Q. 6 The radius of the circle inscribed in a triangle with sides 12, 35 and 37, is-

(A) 4 (B) 5 (C) 6 (D) 7

Questions 7 to 10 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct. Mark your response in OMR sheet against the question number of that question. + 4 marks will be given for each correct answer and NO NEGATIVE marks for wrong answer.

Q. 4 rhu lanwd ftu ij A, B o C vafdr gS rFkk izR;sd lanwd esa pkj xsansa ftu ij 1, 2, 3 o 4 vad n'kkZ;s x;s gSa] j[kh xbZ gSA izR;sd lanwd dh xsnsa Hkyh Hkk¡fr lqesfyr gSA ,d cPpk bu rhuksa lanwdksa esa ls izR;sd ls ;knPN;k ,d xsan dk p;u djrk gS] ;fn a, b o cØe'k% lanwd A, B o C ls p;u dh xbZ xsanksa ij vk;s vad gksaA cPpk ,d gSyhdkWIVj f[kykSuk thrrk gS ;fn a = b + c rks cPps ds gSyhdkWIVj f[kykSuk thrus ds ik esa la;ksxkuqikr gS -

(A) 3 : 32 (B) 3 : 29 (C) 1 : 15 (D) 5 : 59

Q. 5 ;fn N = 7p+4.5q.23 ,d iw.kZ ?ku gS] tgk¡ 'p' o 'q' /kukRed iw.kkZd gS] rks 'p + q' dk U;wure laHkkfor eku gksxk-

(A) 5 (B) 2 (C) 8 (D) 6

Q. 6 ,d f=kHkqt ftldh Hkqtk,a 12, 35 rFkk 37 gS] ds vUnj cuk, x, oÙk dh f=kT;k gS -

(A) 4 (B) 5 (C) 6 (D) 7

iz'u 7 ls 10 rd cgqfodYih iz'u gaSA izR;sd iz'u ds pkj fodYi

(A), (B), (C) rFkk (D) gSa] ftuesa ls ,d ;k vf/kd fodYi lgh gaSA

OMR 'khV esa iz'u dh iz'u la[;k ds lek vius mÙkj vafdr

dhft,A izR;sd lgh mÙkj ds fy, + 4 vad fn;s tk;saxs rFkk xyr

mÙkj ds fy, dksbZ _.kkRed vadu ugha gSA



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

f(x) = ∫ +−x

1

2 dx)2x3x(x , 1 ≤ x ≤ 4. Then the

(A) maximum value of f(x) is 4

53

(B) maximum value of f(x) is 463

(C) minimum value of f(x) is 21

−

(D) minimum value of f(x) is 41

−

Q. 8 ∫π

0

dx)x(sinfx is equal to-

(A) ∫π

π

0

dx)x(sinf2

(B) ∫π

π2/

0

dx)x(sinf

(C) ∫π

π2/

0

dx)x(sinf2 (D) None of these

Q. 9 Let the algebraic sum of perpendicular distancesfrom the points A(2, 0), B(0, 2), C(1, 1) to avariable line be zero, then all such lines

(A) are concurrent (B) passes through fixed point (C) touches a given circle (D) passes through orthocenter of triangle ABC

Q. 7 ekukfd f(x) ,d Qyu gS tks bl izdkj ifjHkkf"kr gS

fd f(x) = ∫ +−x

1

2 dx)2x3x(x , 1 ≤ x ≤ 4 rks

(A) f(x) dk vf/kdre eku 4

53 gS

(B) f(x) dk vf/kdre eku 463 gS

(C) f(x) dk U;wure eku 21

− gS

(D) f(x) dk U;wure eku 41

− gS

Q. 8 ∫π

0

dx)x(sinfx cjkcj gS -

(A) ∫π

π

0

dx)x(sinf2

(B) ∫π

π2/

0

dx)x(sinf

(C) ∫π

π2/

0

dx)x(sinf2 (D) buesa ls dksbZ ugh

Q. 9 ekuk fcUnqvksa A(2, 0), B(0, 2), C(1, 1) ls fdlh pj

js[kk ij Mkys x, yEcksa dk chth; ;ksx 'kwU; gS rc

bl izdkj dh js[kk,a (A) laxkeh gS (B) vpj fcUnq ls xqtjsxh (C) ,d fn, x, oÙk dks Li'kZ djsaxh (D) f=kHkqt ABC ds yEc dsUnz ls xqtjsaxh



Q. 10 Value of

π−

−

451tan2tan 1 is-

(A) 177

− (B) 173

−

(C) 3

17− (D) less than

31

−

This section contains 4 questions numbered 11 to 14,(Reason and Assertion type question). Each questioncontains Assertion and Reason. Each question has 4choices (A), (B), (C) and (D) out of which ONLY ONE iscorrect. Mark your response in OMR sheet against thequestion number of that question. + 3 marks will begiven for each correct answer and – 1 mark for eachwrong answer.

The following questions given below consist ofan "Assertion" (A) and "Reason" (R) Typequestions. Use the following Key to choose theappropriate answer.



(C) If (A) is true but (R) is false.

(D) If (A) is false but (R) is true.]

Q. 10

π−

−

451tan2tan 1 dk eku gS-

(A) 177

− (B) 173

−

(C) 3

17− (D)

31

− ls de

bl [k.M esa 11 ls 14 rd 4 iz'u gSa, (dFku rFkk dkj.k izdkj

ds iz'u)A izR;sd iz'u esa dFku rFkk dkj.k fn;s x;s gSaA

izR;sd iz'u ds 4 fodYi (A), (B), (C) rFkk (D) fn;s x;s gSa]

ftuesa ls dsoy ,d 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;asxs rFkk izR;sd xyr mÙkj ds fy,

1 vad ?kVk;k tk;sxkA

uhps fn;s x;s fuEufyf[kr iz'u "dFku" (A) rFkk "dkj.k" (R) izdkj ds iz'u gSaA vr% mfpr mÙkj dk p;u djus ds fy;s fuEu rkfydk dk mi;ksx dhft;sA

(A) ;fn (A) rFkk (R) nksuksa lR; gSa rFkk (R), (A) dk lgh Li"Vhdj.k gSA

(B) ;fn (A) rFkk (R) nksuksa lR; gSa rFkk (R), (A) dk lgh Li"Vhdj.k ugha gSA

(C) ;fn (A) lR; gS ysfdu (R) vlR; gSA

(D) ;fn (A) vlR; gS ysfdu (R) lR; gSA



Q. 11 dFku (A) : ekuk ABCD ,d pØh; prqHkqZt gS tks bdkbZ f=kT;k ds oÙk esa cuk;k x;k gS ;fn AB.CD+ BC.DA ≥ 4 rc ABCD ,d oxZ gS

dkj.k (R) : ,d pØh; prqHkqZt oxZ gksrk gS ;fn blds fod.kZ oÙk ds O;kl gS

Q. 12 dFku (A): ;fn izFke 30 izkdr la[;kvksa ds leqPp; esa ls nks fo"ke la[;k,sa pquh tk, rks mudk ;ksx le gksus

dh izkf;drk 297 gS

dkj.k (R): ekuk A rFkk B nks ?kVuk,a gS rc B ds ?kVus

dh izkf;drk tcfd A ?kfVr gks pqdh gks )A(P

)BA(P ∩ gS

Q. 13 dFku (A): a × b = a × c rc b = λ a + c

dkj.k (R): ;fn a × b = 0 ⇒ a o b lajs[kh; gSA

;fn a rFkk b v'kwU; lfn'k gSaA

Q. 14 dFku (A) : ∫ + dx)1x(e)xe(F xx = lnx – x + c

fn;k gS fd ∫ dx)x(F = lnx

dkj.k (R) : ∫ ′+ dx)]x(f)x(f[ex = exf(x) + c

Q. 11 Assertion (A) : Suppose ABCD is a cyclicquadrilateral inscribed in a circle of radius one unitwith AB.CD+ BC.DA ≥ 4, then ABCD is a square.

Reason (R) : A cyclic quadrilateral is a square ifits diagonals are the diameters of the circle.

Q. 12 Assertion (A): If two odd numbers are selectedfrom the set of first 30 natural numbers, then

probability that their sum is even is 297 .

Reason (R): Let A and B are two events, thenprobability of occurrence of event B when event A

has already happened is )A(P

)BA(P ∩ .

Q. 13 Assertion (A): a × b = a × c then b = λ a + c

Reason (R): If a × b = 0 ⇒ a is collinear to b .

If a & b are non-zero vector.

Q. 14 Assertion (A) : ∫ + dx)1x(e)xe(F xx = lnx – x + c

given that ∫ dx)x(F = lnx

Reason (R) : ∫ ′+ dx)]x(f)x(f[ex = exf(x) + c



This section contains 3 paragraphs, each has 3 multiplechoice questions. (Questions 15 to 23) Each question has4 choices (A), (B), (C) and (D) out of which ONLY ONEis correct. Mark your response in OMR sheet against thequestion number of that question. + 4 marks will begiven for each correct answer and – 1 mark for eachwrong answer.

Passage # 1 (Q. 15 to 17) The equations of the sides AB, BC, CA

of a triangle ABC are 2x + y = 0, x + py = q,

x – y = 3 respectively. The point P is (2, 3).

Q.15 If P is the centroid, then p + q =

(A) 47 (B) 50

(C) 65 (D) 74

Q.16 If P is the orthocentre, then p + q =

(A) 47 (B) 50

(C) 65 (D) 74

Q.17 If P is the circumcentre, then p + q =

(A) 47 (B) 50

(C) 65 (D) 74

bl [k.M esa 3 vuqPNsn fn;s x;s gSa] izR;sd esa 3 cgqfodYih iz'u gSaA (iz'u 15 ls 23) izR;sd iz'u ds pkj fodYi (A), (B), (C) rFkk (D) gSa, ftuesa ls dsoy ,d fodYi lgh gSA OMR 'khV esa iz'u dh iz'u la[;k ds lek viuk mÙkj vafdr dhft;sA izR;sd lgh mÙkj ds fy;s + 4 vad fn;s tk,saxs rFkk izR;sd xyr mÙkj ds fy, 1 vad ?kVk;k tk;sxkA

x|ka'k # 1 (Q. 15 ls 17) ;fn ,d f=kHkqt ABC dh Hkqtkvksa AB, BC, CA

ds lehdj.k Øe'k% 2x + y = 0, x + py = q, x – y = 3 gksa rFkk fcUnq P ds funsZ'kkad (2, 3) gSA rc

Q.15 ;fn P dsUnzd gks rks p + q =

(A) 47 (B) 50

(C) 65 (D) 74

Q.16 ;fn P yEcdsUnz gks rks p + q =

(A) 47 (B) 50

(C) 65 (D) 74

Q.17 ;fn P ifjdsUnz gks, rks p + q =

(A) 47 (B) 50

(C) 65 (D) 74



Passage # 2 (Q.18 to 20) An acute angled triangle ABC is inscribed in a

circle of radius R. The altitudes from the vertices

A, B, C meet the circle at D, E, F respectively. Let

s', r' and r'1 be the semiperimeter, inradius and

exradius opposite to D of triangle DEF. Then'

Q.18 R's =

(A) Σ sin 2A (B) Σ sinA (C) Σ cosA (D) Σ cos2A

Q.19 R'r =

(A) 1 – Σ cos2A (B) –1 – Σ cos2A

(C) 1 + Σ sinA (D) –1 + Σ cos2A

Q.20 R'r 1 =

(A) 1 + cos2A – cos2B – cos2C

(B) – cos2A + cos2B + cos2C

(C) 1 + sin2A – sin2B – sin 2C

(D) 2sinA – sinB – sinC

x|ka'k # 2 (Q. 18 ls 20) ,d U;wu dks.k f=kHkqt ABC, R f=kT;k ds oÙk esa

cuk;k x;k gSA 'kh"kksZ A, B, C ls 'kh"kZ yEc oÙk ij

Øe'k% D, E, F ij feyrs gSaA ekuk s', r' rFkk r'1 Øe'k%

v/kZifjeki] vUr%f=kT;k rFkk f=kHkqt DEF ds 'kh"kZ D

ds lEeq[k cká f=kT;k gS rc

Q.18 R's =

(A) Σ sin 2A (B) Σ sinA (C) Σ cosA (D) Σ cos2A

Q.19 R'r =

(A) 1 – Σ cos2A (B) –1 – Σ cos2A (C) 1 + Σ sinA (D) –1 + Σ cos2A

Q.20 R'r 1 =

(A) 1 + cos2A – cos2B – cos2C (B) – cos2A + cos2B + cos2C (C) 1 + sin2A – sin2B – sin 2C (D) 2sinA – sinB – sinC



Passage # 3 (Q. 21 to 23) Two vectors a

r and br

having unit modulus and angle between them is θ.

Now φ(θ) = ∫×

−

2

2

)ba(

)b.a(

2 dx)x(f

rr

rr

and f satisfies the

condition f(y) + f(x) = xy

yx + for all x, y ∈ R –

0 and h(θ) = – φ(θ) + 21

21 |b.a||ba|

rrrr× , where

b2b1

rr= .

Q.21 Fundamental period of φ(θ) is-

(A) 2π (B) π (C) 2π (D) 6π

Q.22 The volume of the parallelopiped formed by a

r , br

and barr

× , where angle between ar and b

r is

taken for which h(θ) is minimum is-

(A) 1 (B) 21

(C) 41 (D)

81

Q.23 If ar and b

r are non collinear vectors, then

number of solution of the equation )('3)(4 θφ+θφ = 0 in [0, 2π] are-

(A) 2 (B) 4 (C) 6 (D) 8

x|ka'k # 3 (Q. 21 ls 23)

nks bdkbZ ifjek.k ds lfn'k ar rFkk b

r gS rFkk

muds e/; dks.k θ gS

vc φ(θ) = ∫×

−

2

2

)ba(

)b.a(

2 dx)x(f

rr

rr

rFkk lHkh x, y ∈ R –0

ds fy,] f izfrcU/k f(y) + f(x) = xy

yx + dks lUrq"V

djrk gS rFkk h(θ) = – φ(θ) + 21

21 |b.a||ba|

rrrr×

tgk¡ b2b1

rr= .

Q.21 φ(θ) dk ewyHkwr vkorZukad gS-

(A) 2π (B) π (C) 2π (D) 6π

Q.22 ar , b

r rFkk ba

rr× kjk cuk, x, lekUrj "kV~Qyd

dk vk;ru] tgk¡ ar rFkk b

r ds e/; dks.k og gS

ftlds fy, h(θ) U;wure gS] gksxk -

(A) 1 (B) 21

(C) 41 (D)

81

Q.23 ;fn ar rFkk b

rvlajs[kh; lfn'k gS rc lehdj.k

)('3)(4 θφ+θφ = 0 ds [0, 2π] esa gyksa dh la[;k gS-

(A) 2 (B) 4 (C) 6 (D) 8



PHYSICS

Section - I

Questions 1 to 6 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 conveyer belt of length l is moving with velocity

v. A block of mass m is pushed against the motion

of conveyer belt with velocity v0 from end B with

respect to conveyer belt. Co-efficient of friction

between block and belt is µ. The value of v0 so that

the amount of heat liberated as a result of

retardation of the block by conveyer belt is

maximum is –

v0 v AB

l (A) lµg (B) lµg2

(C) lµg2 (D) lµg3

[k.M - I iz'u 1 ls 6 rd cgqfodYih iz'u gSaA izR;sd iz'u ds pkj

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

lgh gSA OMR 'khV esa iz'u dh iz'u la[;k ds lek viuk

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

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

Q.1 l yEckbZ dh ,d dUos;j csYV v osx ls xfr dj jgh

gSA m nzO;eku ds ,d CykWd dks dUos;j csYV dh xfr

ds fo:) B fljs ls dUos;j csYV ds lkisk v0 osx ls

/kDdk fn;k tkrk gSA CykWd rFkk csYV ds e/; ?k"kZ.k

xq.kkad µ gSA v0 dk og eku ftlls fd dUos;j csYV

kjk CykWd ds eanu ds ifj.kkeLo:i eqDr gqbZ Å"ek

dh ek=kk vf/kdre gks] gksxk &

v0 v AB

l

(A) lµg (B) lµg2

(C) lµg2 (D) lµg3



Q.2 A body of mass m1 moving at a constant speed

undergoes an elastic head on collision with a body

of mass m2 initially at rest. The ratio of the kinetic

energy of mass m1 after the collision to that before

the collision is -

(A) 2

21

21

mmm–m

+

(B) 2

21

21

m–mmm

+

(C) 2

21

1

mmm2

+

(D) 2

21

2

mmm2

+

Q.3 In the figure shown, the heavy cylinder (radius R)

resting on a smooth surface separates two liquids of

densities 2ρ and 3ρ. The height h for the translatory

equilibrium of cylinder must be –

R3ρ 2ρ h R

(A) 2R3 (B)

23R

(C) 2R (D) 43R

Q.2 m1 nzO;eku dk ,d fi.M tks fu;r pky ls xfr'khy

gS ] fojkekoLFkk esa mifLFkr m2 nzO;eku ds nwljs fi.M

ls lEeq[k izR;kLFk VDdj djrk gSA VDdj ds i'pkr~

rFkk VDdj ls iwoZ m1 nzO;eku dh xfrt ÅtkZvksa dk

vuqikr gksxk &

(A) 2

21

21

mmm–m

+

(B) 2

21

21

m–mmm

+

(C) 2

21

1

mmm2

+

(D) 2

21

2

mmm2

+

Q.3 n'kkZ, x, fp=k esa ,d fpduh lrg ij fojke dj jgk

,d flys.Mj (f=kT;k R) 2ρ rFkk 3ρ ?kuRo ds nks nzoksa

dks iFkd djrk gSA flys.Mj dh LFkkukUrjh;

lkE;oLFkk ds fy, Å¡pkbZ h gksuh pkfg, &

R3ρ 2ρh R

(A) 2R3 (B)

23R

(C) 2R (D) 43R



Q.4 In the diagram shown, the charge + Q is fixed.Another charge +2q, is projected from a distance Rfrom the fixed charge. Minimum separationbetween the two charges, if the velocity becomes

31 times of the projected velocity, at this moment

is (Assume gravity to be absent)

+Q R M, +2q

V30º

(A) R23 (B) R3 (C) R

21 (D) 4 R

Q.5 A cube made of wires of equal length is connectedto a battery as shown in the figure. The magneticfield at the centre of the cube is -

L I

I

+ –

(A) L2Iµ12 0

π (B)

L2Iµ6 0

π

(C) L

Iµ6 0

π (D) zero

Q.4 n'kkZ, x, fp=k esa] vkos'k + Q fLFkj gSA bl fLFkj

vkos'k ls R nwjh ij ls ,d nwljs +2q vkos'k dks

izksfir fd;k tkrk gSA ml k.k ij nksuksa vkos'kksa ds

e/; U;wure nwjh] ;fn osx izksfir osx dk 3

1 xquk

gks tk,] gksxh ¼ekuk xq:Ro vuqifLFkr gS½ &

+Q R M, +2q

V 30º

(A) R23 (B) R3 (C) R

21 (D) 4 R

Q.5 fp=k esa n'kkZ, vuqlkj leku yEckbZ ds rkjksa ds cus

,d ?ku dks ,d cSVjh ls tksM+k tkrk gSA ?ku ds

dsUnz ij pqEcdh; ks=k gksxk &

LI

I

+ –

(A) L2Iµ12 0

π (B)

L2Iµ6 0

π

(C) L

Iµ6 0

π (D) 'kwU;



Q.6 At a moment (t = 0), when the charge on capacitorC1 and C2 is zero, the switch is closed. If I0 be the current through inductor at t = 0, for t > 0 (initiallyC2 is uncharged)

C2

C1

S

L

(A) maximum current through inductor equals I0/2 (B) maximum current through inductor equals

21

01

CCIC

+

(C) maximum charge on C1 = 21

201

CCLCIC

+

(D) maximum charge on C1 = 21

01 CCLIC+

Questions 7 to 10 are multiple choice questions. Eachquestion has four choices (A), (B), (C) and (D), out ofwhich MULTIPLE (ONE OR MORE) is correct. Markyour response in OMR sheet against the questionnumber of that question. + 4 marks will be given foreach correct answer and NO NEGATIVE marks forwrong answer.

Q.6 fdlh k.k (t = 0) ij, tc la?kkfj=k C1 o C2 ij vkos'k

'kwU; gks] rc fLop dks pkyw fd;k tkrk gSA ;fn t = 0

ij izsjd ls izokfgr /kkjk I0 gks] rks t > 0 ds fy,

(izkjEHk esa C2 vukosf'kr gS)

C2

C1

S

L

(A) izsjd ls izokfgr vf/kdre /kkjk I0/2 ds rqY; gksxh

(B) izsjd ls izokfgr vf/kdre /kkjk 21

01

CCIC

+ ds rqY;

gksxh

(C) C1 ij vf/kdre vkos'k = 21

201

CCLCIC

+

(D) C1 ij vf/kdre vkos'k = 21

01 CCLIC+

iz'u 7 ls 10 rd cgqfodYih iz'u gaSA izR;sd iz'u ds pkj

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

fodYi lgh gaSA OMR 'khV esa iz'u dh iz'u la[;k ds lek

vius mÙkj vafdr dhft,A izR;sd lgh mÙkj ds fy, + 4 vad

fn;s tk;saxs rFkk xyr mÙkj ds fy, dksbZ _.kkRed vadu

ugha gSA



Q.7 400 nm rjaxnS/;Z ds ,do.khZ rjaxksa ds fofdj.kksa dks Øe'k% 3.4 eV, 4.8 eV rFkk 5.9 eV dk;ZQyuksa dh /kkrqvksa Zn, Fe rFkk Ni dh lrgksa ij vkifrr fd;k tkrk gSA (hc = 12400 eV-Å yas) :

(A) fdlh /kkrq dh lrg ls QksVksbysDVªkWuksa ls lEc) vf/kdre KE, 0.3 eV gksrh gS

(B) Ni dh lrg ls dksbZ QksVksbysDVªkWu mRlftZr ugha gksaxs

(C) ;fn L=kksr ds fofdj.k dh rjaxnS/;Z dks nqxquk dj fn;k tk, rks QksVksbysDVªkWuksa dh KE Hkh nqxquh gks tkrh gS

(D) rhuksa /kkrqvksa dh lrg ls QksVks bysDVªkWu mRlftZr gksaxs ;fn vkifrr fofdj.kksa dh rjaxnS/;Z 200 nm ls de gks

Q.8 vory niZ.k dh ck;ha rFkk nk;ha lrg dh oØrk

f=kT;k Øe'k% 10 cm rFkk 15 cm gSA niZ.k dh oØrk

f=kT;k 15 cm gks rks &

ok;q ty (n = 4/3)

dk¡p (n = 3/2)

(A) la;kstu dh rqY; Qksdl nwjh – 18 cm gS

(B) la;kstu dh rqY; Qksdl nwjh +36 cm gS

(C) ra=k vory niZ.k dh rjg O;ogkj djrk gS

(D) ra=k mÙky niZ.k dh rjg O;ogkj djrk gS

Q.7 Radiations of monochromatic waves of wavelength400 nm are made incident on the surface of metalsZn, Fe and Ni of work functions 3.4 eV, 4.8 eV and5.9 eV respectively (take hc = 12400 eV-Å) :

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

(B) no photoelectrons are emitted from the surfaceof Ni

(C) if the wavelength of source of radiation isdoubled then KE of photoelectrons is alsodoubled

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

Q.8 The radius of curvature of the left and right surface

of the concave lens are 10 cm and 15 cmrespectively. The radius of curvature of the mirroris 15 cm –

Air Water (n = 4/3)

Glass (n = 3/2)

(A) equivalent focal length of the combination is

– 18 cm (B) equivalent focal length of the combination is

+36 cm (C) the system behaves like a concave mirror (D) the system behaves like a convex mirror



Q.9 n'kkZ, x, fp=k esa rkjksa P1Q1 rFkk P2Q2 dks iVfj;ksa ij 5 m/s dh leku pky ls f[kldk;k tkrk gSA bl ks=k

esa 1 T dk pqEcdh; ks=k fo|eku gSA 9 Ω ds izfrjks/k esa

fo|qr /kkjk gksxh &

× × × × × × ×

× × × × × × ×

× × × × × × ×

× × × × × × ×

4cm

P2 P1

9Ω2Ω2Ω

Q2 Q1 (A) 'kwU; ;fn nksuksa rkj ck;ha vksj fQlyrs gSa (B) 'kwU; ;fn nksuksa rkj foijhr fn'kk esa fQlyrs gSa (C) 20 mA ;fn nksuksa rkj ck;ha vksj xfr djrs gSa (D) 20 mA ;fn nksuksa rkj foijhr fn'kk esa xfr djrs gSa

Q.10 fp=k esa n'kkZ;s x, ifjiFk esa & S1 S2

3Ω 10V 10V 2Ω1Ω

20V i

(A) i = 2.5 tc S1 dks pkyw rFkk S2 dks can fd;k tkrk gS

(B) i = A3

20 tc S1 dks can rFkk S2 dks pkyw fd;k tkrk gS

(C) i = A35 tc S1 rFkk S2 nksuksa dks can fd;k tkrk gS

Q.9 In the diagram shown, the wires P1Q1 and P2Q2 are made to slide on the rails with same speed of 5 m/s.In this region a magnetic field of 1 T exists. Theelectric current in 9 Ω resistor is –

× × × × × × ×

× × × × × × ×

× × × × × × ×

× × × × × × ×

4cm

P2 P1

9Ω2Ω2Ω

Q2 Q1 (A) zero if both wires slide towards left (B) zero if both wires slide in opposite direction (C) 20 mA if both wires move towards left. (D) 20 mA if both wires move in opposite direction

Q.10 In the circuit shown in figure : S1 S2

3Ω 10V 10V 2Ω 1Ω

20V i

(A) i = 2.5 A when S1 is closed and S2 is open

(B) i = A3

20 when S1 is open and S2 is closed

(C) i = A35 when S1 and S2 both are open

(D) i = 20A when both S1 and S2 are closed



(D) i = 20A tc S1 rFkk S2 nksuksa dks pkyw fd;k tkrk gS

bl [k.M esa 11 ls 14 rd 4 iz'u gSa, (dFku rFkk dkj.k izdkj ds iz'u)A izR;sd iz'u esa dFku rFkk dkj.k fn;s x;s gSaA izR;sd iz'u ds 4 fodYi (A), (B), (C) rFkk (D) fn;s x;s gSa] ftuesa ls dsoy ,d 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;asxs rFkk izR;sd xyr mÙkj ds fy, 1 vad ?kVk;k tk;sxkA uhps fn;s x;s fuEufyf[kr iz'u "dFku" (A) rFkk

"dkj.k" (R) izdkj ds iz'u gSaA vr% mfpr mÙkj dk p;u djus ds fy;s fuEu rkfydk dk mi;ksx dhft;sA

(A) ;fn (A) rFkk (R) nksuksa lR; gSa rFkk (R), (A) dk lgh Li"Vhdj.k gSA

(B) ;fn (A) rFkk (R) nksuksa lR; gSa rFkk (R), (A) dk lgh Li"Vhdj.k ugha gSA

(C) ;fn (A) lR; gS ysfdu (R) vlR; gSA (D) ;fn (A) vlR; gS ysfdu (R) lR; gSA Q. 11 dFku (A) : Q vkos'k okys ,d pkyd esa nks xksyh;

fNnz cuk, tkrs gaSA nks fcUnq vkos'kksa q1 rFkk q2 dks fNnzksa ds dsUnzksa ij j[kk tkrk gS rFkk nwljs vkos'k q dks pkyd ds ckgj j[kk tkrk gSA pkyd dh cká lrg ij vkos'k ds dkj.k pkyd ds vanj fdlh fcUnq ij fo|qr ks=k 'kwU; gksrk gSA

dkj.k (R) : mi;qZDr dFku esa LFkSfrd fLFkfr esa pkyd

This section contains 4 questions numbered 11 to 14,(Reason and Assertion type question). Each questioncontains Assertion and Reason. 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 questionnumber of that question. + 3 marks will be given for eachcorrect answer and – 1 mark for each wrong answer.

The following questions given below consist ofan "Assertion" (A) and "Reason" (R) Typequestions. Use the following Key to choose theappropriate answer.



(C) If (A) is true but (R) is false. (D) If (A) is false but (R) is true.

Q. 11 Assertion (A) : Two spherical cavities are madeinside a conductor having charge Q. Two pointcharges q1 and q2 are kept at the centres of cavityand another charge q is kept outside the conductor.Electric field due to charge on outer surface ofconductor is zero at a point inside the conductor.

Reason (R) : In the above assertion electricpotential is constant inside the conductor in staticcondition.



ds vanj fo|qr foHko fu;r jgrk gSA

Q. 12 dFku (A) : ,d kSfrt uyh esa ,d vkn'kZ nzO; dh xfr dh fLFkfr esa] tgk¡ vuqizLFk dkV dk ks=kQy U;wure gksrk gS] nkc vf/kdre gksrk gSA

dkj.k (R) : fofHkUu vkn'kZ nzoksa esa fofHkUu xgjkbZ ij fLFkr fcUnqvksa ij tyLFkSfrd nkc leku fd;k tk ldrk gSA

Q.13 dFku (A) : ,d izksI; blds mPpre fcUnq ij

foLQksfVr gks tkrk gSA nzO;eku dsUnz ml fcUnq ls

vkxs fdlh fcUnq ij fxjsxk tgk¡ izksI; vfoLQksV dh

fLFkfr esa fxjrkA

dkj.k (R) : izksI; dk Hkkj cká cy gSA

Q.14

R

L

S

ε

dFku (A) : fp=k esa fLop pkyw djus ds rqjar ckn

izsjd ds fljksa ds e/; foHko iru vf/kdre gksrk gSA

dkj.k (R) : fLop pkyw djus ds rqjar ckn /kkjk esa

Q. 12 Assertion (A) : In case of motion of an ideal fluid in a horizontal tube, where the area of cross-section is minimum, pressure is maximum.

Reason (R) : Hydrostatic pressure in different idealliquids at points of different depth can be same.

Q.13 Assertion (A) : A projectile gets exploded at itshighest point. The centre of mass will fall at a pointwhich is farther than the point where the projectilewould have fallen in unexploded condition.

Reason (R) : The weight of the projectile is theexternal force.

Q.14

R

L

S

ε

Assertion (A) : In the figure, just after closing the

switch the potential drop across inductor ismaximum.

Reason (R) : The rate of change of current justafter closing the switch is maximum.



ifjorZu dh nj vf/kdre gksrh gSA

bl [k.M esa 3 vuqPNsn fn;s x;s gSa] izR;sd esa 3 cgqfodYih

iz'u gSaA (iz'u 15 ls 23) izR;sd iz'u ds pkj fodYi (A), (B), (C) rFkk (D) gSa, ftuesa ls dsoy ,d fodYi lgh gSA OMR 'khV esa iz'u dh iz'u la[;k ds lek viuk mÙkj vafdr

dhft;sA izR;sd lgh mÙkj ds fy;s + 4 vad fn;s tk,saxs rFkk


x|ka'k # 1 (iz- 15 ls 17) 1kg rFkk 2kg nzO;eku ds nks CykWdksa dks k = 600 N/m

fLizax fu;rkad dh ,d vkn'kZ fLizax kjk tksM+k tkrk

gSA fp=k esa n'kkZ, vuqlkj ;s ,d fpdus kSfrt eSnku

ij fojkekoLFkk esa j[ks gSaA r = 50 cm rFkk m = 1 kg

nzO;eku dk ,d n<+ xksyk dkxt ds ry ds yEcor~

O;kl ds ifjr% ?kw.kZu djrs gq, h Å¡pkbZ ls eqDr :i

ls fxj jgk gS rFkk ;g 1 kg ds CykWd ds lkFk VDdj

djrk gSA izR;koLFkku xq.kkad rFkk xksys o CykWd ds

e/; ?k"kZ.k xq.kkad Øe'k% (1/2) rFkk (1/3) gSA(g = 10 m/s2, m = 1 kg)

m

ω

2m k

hm

Q.15 VDdj ds rqjar ckn 1 kg ds CykWd dk osx gksxk &

This section contains 3 paragraphs, each has 3 multiplechoice questions. (Questions 15 to 23) Each question has4 choices (A), (B), (C) and (D) out of which ONLY ONEis correct. Mark your response in OMR sheet against thequestion number of that question. + 4 marks will begiven for each correct answer and – 1 mark for each wrong answer.

Passage # 1 (Ques. 15 to 17) Two blocks of masses 1 kg and 2 kg are connected

by an ideal spring having spring constantk = 600 N/m. They are at rest on a smoothhorizontal ground as shown in the figure. A rigidsphere, rotating with initial angular speed ω = 30 rad/sec, about a diameter perpendicular to the planeof the paper having radius r = 50 cm and massm = 1 kg, is falling freely from a height h = 5m andcollides with the 1 kg block.

The co-efficient of restitution and the co-efficient of friction between the sphere and the block are(1/2) and (1/3) respectively. (g = 10 m/s2, m = 1 kg)

m

ω

2m k

h m

Q.15 The velocity of 1 kg block just after collision is - (A) 5 m/s (B) 10 m/s (C) 15 m/s (D) 20 m/s



(A) 5 m/s (B) 10 m/s (C) 15 m/s (D) 20 m/s Q.16 fLizax dk vf/kdre foLRkkj gksxk &

(A) 21 m (B) m

41

(C) m61 (D) m

81

Q.17 VDdj ds rqjar ckn xksys dk dks.kh; osx gksxk & (A) 25 rad/s (B) 10 rad/s (C) 20 rad/s (D) 5 rad/s x|ka'k # 2 (iz- 18 ls 20)

m nzO;eku rFkk S0 ks=kQy dk ,d vpkyd fiLVu

fp=kkuqlkj vpkyd] can flys.Mj dks foHkkftr djrk

gSA m nzO;eku ds fiLVu dks k cy fu;rkad dh ,d

fLizax kjk flys.Mj dh 'kh"kZ nhokj ls tksM+k tkrk

gSA 'kh"kZ Hkkx fuokZfrr gS rFkk ryh okys Hkkx esa lkE;

fLFkfr esa ,d vkn'kZ xSl P0 nkc ij gSA :)ks"e

fu;rkad γ rFkk lkE;koLFkk esa izR;sd Hkkx dh yEckbZ

l gSA (?k"kZ.k dks ux.; ekus)

Q.16 The maximum elongation of the spring is -

(A) 21 m (B) m

41

(C) m61 (D) m

81

Q.17 Angular velocity of sphere just after impact is (A) 25 rad/s (B) 10 rad/s (C) 20 rad/s (D) 5 rad/s

Passage # 2 (Ques. 18 to 20) A non conducting piston of mass m and area S0

divides a non-conducting, closed cylinder as shown in figure. Piston having mass m is connected with top wall of cylinder by a spring of force constant k. Top part is evacuated and bottom part contains an ideal gas at pressure P0

in equilibrium position. Adiabatic constant γ

and in equilibrium length of each part l. (neglect friction)

P0 m

S0



P0

m S0

Q.18 lkE; fLFkfr ij fLizax esa lEihM+u Kkr dhft,

(ekuk S0P0 > mg )

(A) 'kwU; (B) k

mg–SP2 00

(C) k

mg–SP 00 (D) k2

mg–SP 00

Q.19 ;fn fiLVu lkE;koLFkk ls gYdk lk foLFkkfir gksrk gS

rc y?kq nksyuksa ds fy, dks.kh; vkofÙk Kkr dhft,-

(A) l

l

m2SPk 00γ+ (B)

l

l

mSPk2 00γ+

(C) mk (D)

l

l

mSPk 00γ+

Q.20 ;fn fLizax dks vla;ksftr dj fn;k tk, rFkk flys.Mj

ds 'kh"kZ Hkkx dks gVk fn;k tk, rks y?kq nksyuksa ds

fy, dks.kh; vkofÙk Kkr dhft,A (ekuk lkE; fLFkfr

ij xSl dk nkc P1 rFkk xSl LrEHk dh yEckbZ l1 gS)

Q.18 Find compression in the spring at equilibrium

position (Assuming S0P0 > mg )

(A) zero (B) k

mg–SP2 00

(C) k

mg–SP 00 (D) k2

mg–SP 00

Q.19 If piston is displaced slightly from equilibrium then

find angular frequency for small oscillation -

(A) l

l

m2SPk 00γ+ (B)

l

l

mSPk2 00γ+

(C) mk (D)

l

l

mSPk 00γ+

Q.20 If spring is disconnected and top part of cylinder is

removed then find the angular frequency for small

oscillation. (Assuming pressure of gas at

equilibrium position is P1 and length of gas column

is l1)

(A) 1

01

mSPl

γ (B) 1

01

mSP2

l

γ

(C) 1

01

m4SPl

γ (D) 1

01

m2SPl

γ



(A) 1

01

mSPl

γ (B) 1

01

mSP2

l

γ

(C) 1

01

m4SPl

γ (D) 1

01

m2SPl

γ

x|ka'k # 3 (iz- 21 ls 23) ,d oLrq dks ,d leksÙky ySal ls 3 ehVj dh nwjh ij

j[kk tkrk gSA fp=k esa n'kkZ;s vuqlkj bldk izfrfcEc ySal ls 2 ehVj dh nwjh ij curk gSA (ySal dk viorZukad 1.5 gS)

3 m

oLrq izfrfcEc

2 m

0.33 m

oLrq

(i)

3m (ii)

oLrq

3m

oLrq

(iii)

Q.21 izfrfcEc dh fLFkfr Kkr dhft, ;fn ySal ds ,d

Passage # 3 (Ques. 21 to 23) An object is placed at a distance 3 meter from an

equi-convex lens. Its image is formed at 2 m fromthe lens as shown in the figure. (Refractive index oflens is 1.5)

3 m

Object image

2 m

0.33 m

Object

(i)

3m

Object

(ii)

3m

Object

(iii)

Q.21 Find the position of image if one face of the lens issilvered [Figure (i)] -

(A) 12 m from the lens towards right (B) 12 from the lens towards left

(C) 6 m from the lens towards right (D) at infinity



Qyd dks flYoj ls ysfir dj fn;k tk, [fp=k (i)] -

(A) ySal ls 12 m ij nk;ha vksj

(B) ySal ls 12m ij ck;ha vksj

(C) ySal ls 6 m ij nk;ha vksj

(D) vuUr ij

Q.22 izfrfcEc dh fLFkfr Kkr dhft, ;fn ySal dks nks lefer leryksÙky ySal esa dkVk tk, rFkk leryksÙky ySal ds ,d Hkkx dks gVk fn;k tk,

[fp=k (ii)] - (A) ySal ls 12 m ij nk;ha vksj (B) ySal ls 12 m ij ck;ha vksj (C) ySal ls 6 m ij nk;ha vksj (D) ySal ls 6 m ij ck;ha vksj Q.23 izfrfcEc dh fLFkfr Kkr dhft, ;fn ySal dks nks

lefer leryksÙky ySal esa dkVk tk, rFkk

leryksÙky ySal ds ,d Hkkx dks gVk fn;k tk, rFkk

lery lrg dks flYoj ls ysfir dj fn;k tk, [fp=k(iii)]

(A) ySal ls 12 m ij nk;ha vksj

(B) ySal ls 12 m ij ck;ha vksj

(C) ySal ls 2 m ij ck;h vksj

(D) vuUr ij

Q.22 Find the position of image if lens is cut into two

symmetrical plano convex lens and one of the plano

convex lens is removed [figure (ii)] -

(A) 12 m from the lens towards right

(B) 12 m from the lens towards left

(C) 6 m from the lens towards right

(D) 6 m from the lens towards left

Q.23 Find the position of image if lens is cut into two

symmetrical plano-convex lens and one of the

plano-convex lens is removed and the plane surface

is silvered [figure (iii)]

(A) 12 m from the lens towards right

(B) 12 m from the lens towards left

(C) 2 m from the lens towards left

(D) at infinity



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



Time : 3 : 00 Hrs. MAX MARKS: 246

INSTRUCTIONS TO CANDIDATE

A. lkekU; :

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

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

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

fy, 1 vad ?kVk;k tk,xkA 5. cgqfodYih izdkj ds iz'u ftuesa ls ,d ;k ,d ls vf/kd fodYi lgh gSaA izR;sd lgh mÙkj ds fy, 4 vad fn, tk;saxs rFkk xyr

mÙkj ds fy, dksbZ _.kkRed vadu ugha gSA 6. dFku rFkk dkj.k izdkj ds iz'u] ftuesa ls izR;sd esa dsoy ,d mÙkj lgh gSA izR;sd lgh mÙkj ds fy, 3 vad fn;s tk;saxs rFkk

izR;sd xyr mÙkj ds fy, 1 vad ?kVk;k tk,xkA 7. x|ka'k ij vk/kkfjr cgqfodYih izdkj ds iz'u ftuesa ls dsoy ,d fodYi lgh gSA izR;sd lgh mÙkj ds fy, 4 vad fn, tk;saaxs rFkk


C. OMR dh iwfrZ :

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

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

Section –I Section-II Section-III

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

A B C D E

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

A B C D E

the wrong method for filling the bubble are

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

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

P Q R S TA BCD

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

012

3

4

56

7

8

9

012

3

4

56

7

8

9

012

3

4

56

7

8

9

012

3

4

56

7

8

9


012

3

4

56

7

8

9

0 1 2

3

4

5 6

7

8

9

0 1 2

3

4

5 6

7

8

9

0 1 2

3

4

5 6

7

8

9


0 0 0 00 1 2

3

4

5 6

7

8

9

0 1 2

3

4

5 6

7

8

9

012

3

4

56

7

8

9

012

3

4

56

7

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