Take_Home_10_3_2011

TARGET COURSE FOR IIT-JEE 2011

PHASE- ALL CHEMISTRY, MATHEMATICS & PHYSICS

TEST NO. 8 [TR-4(II)] (TAKE HOME) PAPER – II

Date : 10/3/2011Time : 3 : 00 Hrs. MAX MARKS: 243

Name : _________________________________________________________ Roll No. : __________________________

INSTRUCTIONS TO CANDIDATE

A. GENERAL : 1. Please read the instructions given for each question carefully and mark the correct answers against the question

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

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

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

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

answer. Section - III

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

C. FILLING THE OMR :

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

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

Section –I Section-II Section-III

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

A B C D E

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

A B C D E

the wrong method for filling the bubble are

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

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

P Q R S TA BCD

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

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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

SEA

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1

Space for rough work

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

CHEMISTRY

Section – I

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

Q.1 What is the correct sequence of osmotic pressure

of 0.01M aqueous solution of - (1) Al2(SO4)3 (2) K3PO4 (3) BaCl2 (4) Urea (A) π4 > π3 > π2 > π1 (B) π1 > π2 > π3 > π4 (C) π1 = π2 > π3 = π4 (D) π2 > π4 > π1 > π3

Q.2 The plot of AY

1 against AX

1 is linear with slope

and intercept respectively -

(A) 0

0

B

A

PP and 0

00

B

BA

PPP − (B) 0

0

B

A

PP and 0

00

B

AB

PPP −

(C) 0

0

A

B

PP and 0

00

A

BA

PPP − (D) 0

0

A

B

PP and 0

00

B

AB

PPP −

[k.M - I iz'u 1 ls 8 rd cgqfodYih iz'u gSaA izR;sd iz'u ds pkj

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

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

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

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

Q.1 fuEufyf[kr ds 0.01M tyh; foy;uksa ds ijklj.k

nkc dk lgh Øe D;k gS - (1) Al2(SO4)3 (2) K3PO4 (3) BaCl2 (4) Urea (A) π4 > π3 > π2 > π1 (B) π1 > π2 > π3 > π4 (C) π1 = π2 > π3 = π4 (D) π2 > π4 > π1 > π3

Q.2 AX

1ds lkisk

AY1

dk oØ Øe'k% fdl <ky o

vUr%[k.M ds lkFk js[kh; gS -

(A) 0

0

B

A

PP

rFkk 0

00

B

BA

PPP − (B) 0

0

B

A

PP

rFkk 0

00

B

AB

PPP −

(C) 0

0

A

B

PP

rFkk 0

00

A

BA

PPP − (D) 0

0

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B

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rFkk 0

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AB

PPP −



Q.3 ,d ;kSfxd PdCl4.6H2O blds 1 eksyy tyh; foy;u dk gkbMªsVsM ladqy gS, dk fgekad 269.28 KgSA ladqy dk 100% vk;uu ekurs gq, ladqy dk v.kqlw=k Kkr dhft,A

(ty dk Kf = 1.86 K kg mol–1) (A) [Pd(H2O)6]Cl4 (B) Pd(H2O)4Cl2]Cl2 .2H2O (C) Pd(H2O)3Cl3]Cl .3H2O (D) Pd(H2O)2Cl4] .4H2O Q.4 0.1M ,dykkjh; vEy dh pH, 2 ekih tkrh gSA fn;s

x;s rki T K ij bldk ijklj.k nkc gS - (A) 0.1 RT (B) 0.11 RT (C) 1.1 RT (D) 0.01 RT

Q.5 As2S3 ds LdUnu ds fy,] fefyeksy esa] izfrfyVj

oS/kqvi?kV; dh LdUnu kerk uhps nh xbZ gS: I. NaCl = 52 II. KCl = 51 III. BaCl2 = 0.69 IV. MgSO4 = 0.22 budh mRQqyu kerkvksa dk lgh Øe gS - (A) I > II > III > IV (B) I > II > III = IV (C) IV > III > II > I (D) IV = III > II > I Q.6 Lo.kZ ds 100 mL dksyk;Mh foy;u dk LdUnu

10% NaCl foy;u ds 1 mL feykus ls igys blesa inkFkZ ‘X’ ds 0.25g dks feykdj iq.kZr;k% jksdk tkrkgSA ‘X’ dh Lo.kZ la[;k gS &

(A) 0.25 (B) 25 (C) 250 (D) 2.5

Q.3 Compound PdCl4.6H2O is a hydrated complex of1 molal aqueous solution of it, has freezing point269.28 K. Assuming 100% ionization of complex,calculate the molecular formula of the complex -

(Kf for water = 1.86 K kg mol–1) (A) [Pd(H2O)6]Cl4 (B) Pd(H2O)4Cl2]Cl2 .2H2O (C) Pd(H2O)3Cl3]Cl .3H2O (D) Pd(H2O)2Cl4] .4H2O Q.4 pH of a 0.1M monobasic acid is measured to be 2.

Its osmotic pressure at a given temperature T K is - (A) 0.1 RT (B) 0.11 RT (C) 1.1 RT (D) 0.01 RT Q.5 The coagulation value in millimoles per litre of

electrolytes used for the coagulation of As2S3 are as below :

I. NaCl = 52 II. KCl = 51 III. BaCl2 = 0.69 IV. MgSO4 = 0.22 The correct order of their flocculating power is - (A) I > II > III > IV (B) I > II > III = IV (C) IV > III > II > I (D) IV = III > II > I Q.6 The coagulation of 100 mL of colloidal solution of

gold is completely prevented by addition of 0.25 gof a substance ''X'' to it before addition of 1 mL of10% NaCl solution. The gold number of ''X'' is -

(A) 0.25 (B) 25 (C) 250 (D) 2.5



Q.7 ,d /kkrq 'X' rFkk 'Y' nks DyksjkbM nsrh gSA X, NH4OH ds lkFk dkyk voksi rFkk Y, KI ds lkFk 'osr voksi nsrk gSA Y yky voksi nsrk gS tks KI ds vkf/kD; esa foys; gSA X o Y Øe'k% gS &

(A) HgCl2 rFkk Hg2Cl2 (B) Hg2Cl2 rFkk HgCl2

(C) HgCl2 rFkk ZnCl2 (D) ZnCl2 rFkk HgCl2

Q.8 tc SnCl2 ds vkf/kD; dks] HgCl2 foy;u esa feyk;k tkrk gSA 'osr voksi] /kwlj (;k dkyk) gks tkrk gSA fuEu esa ls fdlsds fuekZ.k ds dkj.k /kwlj jax gksrk gS -

(A) Hg2Cl2 (B) SnCl4 (C) Sn (D) Hg

iz'u 9 ls 13 rd cgqfodYih iz'u gaSA izR;sd iz'u ds pkj

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

vf/kd fodYi lgh gaSA OMR 'khV esa iz'u dh iz'u la[;k ds

lek vius mÙkj vafdr dhft,A izR;sd lgh mÙkj ds fy,

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

Q.9 vEyh; KMnO4 foy;u fdlds kjk fojaftr fd;k

tk ldrk gS (A) 'osr foVªksy (B) gjk foVªkWy (C) eksgj yo.k (D) uhyk foVªkWy

Q.7 A metal gives two chloride 'X' and 'Y'. X gives black precipitate with NH4OH and Y gives whiteppt. With KI, Y gives a red precipitate which issoluble in excess KI. X and Y respectively are -

(A) HgCl2 and Hg2Cl2 (B) Hg2Cl2 and HgCl2

(C) HgCl2 and ZnCl2 (D) ZnCl2 and HgCl2 Q.8 When excess of SnCl2 is added to a solution of

HgCl2, a white precipitate turning to grey (or black)is obtained. The grey colour is due to the formationof -

(A) Hg2Cl2 (B) SnCl4 (C) Sn (D) Hg Questions 9 to 13 are multiple choice questions. Eachquestion has four choices (A), (B), (C) and (D), out ofwhich MULTIPLE (ONE OR MORE) is correct. Markyour response in OMR sheet against the questionnumber of that question. + 3 marks will be given foreach correct answer and no negative marks. Q.9 Acidified KMnO4 solution can be decolourised by - (A) White vitriol (B) Green vitriol (C) Mohr's salt (D) Blue vitriol



Q.10 Which of the following is/are not true ? (A) Among halide ions, iodide ion is the most

powerful reducing agent (B) Fluorine is the only halogen which does not

show a variable oxidation state (C) HOCl is stronger acid than HOBr (D) HCl is a stronger acid than HBr Q.11 For the following oxoacids of phosphours H3PO2, H3PO3 and H3PO4 Which of the following statements are false ? (A) All of these are tribasic acids (B) All of these are reducing in nature (C) All of these have tetrahedral geometry of

phosphorus (D) The order of acidity is H3PO4 > H3PO3 > H3PO2

Q.12 Which of the following is/are correct ? (A) Hydrolysis of NCl3 gives NH3 and HOCl (B) Nitric oxide in solid state exhibit diamagnetism (C) NH3 is a weak reducing agent than PH3 (D) NH3 is less stable than PH3 Q.13 Which of the following are correct statements ? (A) Solid PCl5 exists as tetrahedral [PCl4]+ and

octahedral [PCl6]– ions (B) Solid PBr5 exists as [PBr4]+ Br– (C) Solid N2O5 exists as [NO2]+ [NO3]– (D) Oxides of phosphorus P2O3 and P2O5 exist as

monomers

Q.10 fuEu esa ls dkSuls lgh ugha gS ? (A) gsykbM vk;uksa esa] vk;ksMkbM vk;u lokZf/kd

izcy vipk;d gS (B) dsoy ¶yksjhu ,slk gsykstu gS tks ifjofrZr

la;kstdrk ugha n'kkZrk (C) HOCl, HOBr esa izcy vEy gS (D) HCl, HBr eas izcy vEy gS Q.11 QkLQksjl ds fuEu vkWDlks vEyksa ds fy, H3PO2, H3PO3 rFkk H3PO4 fuEu esa ls dkSulk dFku xyr gS ? (A) lHkh f=kkkjh; vEy gS (B) lHkh vipk;d izÑfr ds gS (C) lHkh esa QkWLQksjl dh prq"Qydh; T;kferh gS (D) vEyh;rk dk Øe H3PO4 > H3PO3 > H3PO2 Q.12 fuEu esa ls dkSulk lgh gS ? (A) NCl3 dk ty vi?kVu NH3 o HOCl nsrk gS (B) Bksl voLFkk esa ukbfVªd vkWlkbM izfrpqEcdRo n'kkZrk gS (C) NH3, PH3 ls nqcZy vipk;d gS (D) NH3, PH3 ls de LFkk;h gS

Q.13 fuEu esa ls dkSulk dFku lR; gS ? (A) Bksl PCl5 prq"Qydh; [PCl4]+ rFkk v"VQydh;

[PCl6]– vk;u dh rjg jgrk gS (B) Bksl PBr5, [PBr4]+ Br– ds :i esa jgrk gS (C) Bksl N2O5, [NO2]+ [NO3]– ds :i esa jgrk gS (D) QkWLQksjl ds vkWDlkbM P2O3 ,oa P2O5 ,dyd

ds :i esa jgrs gS



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

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

(iz'u 16 ls 18) gSaA izR;sd iz'u ds pkj fodYi (A), (B), (C) rFkk (D) gSa, ftuesa ls dsoy ,d fodYi lgh gSA OMR 'khV esa iz'u dh iz'u la[;k ds lek viuk mÙkj vafdr dhft;sA

izR;sd lgh mÙkj ds fy;s + 3 vad fn;s tk,saxs rFkk izR;sd

xyr mÙkj ds fy, 1 vad dkVk tk;sxkA

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

tc ok;q ds vkf/kD; esa ghfy;e xeZ gksdj lkekU; vkWDlkbM o lksfM;e ijkWDlkbM cukrk gSA tcfd ikWVsf'k;e o ifjokj ds vU; lnL; esa lqij vkWDlkbM cukus dh izofÙk gksrh gSA okLro esa NksVk /kuk;u] NksVs _.kk;u dks LFkkbZ djrk gSA rFkk cM+k /kuk;u] cMs+ _.kk;u dks LFkk;h djrk gSA vkWDlkbM o ijkWDlkbM] ty ds lkFk fØ;k djds lksfM;e gkbMªkWDlkbM cukrs gSA KO2, CO2 dks vo'kksf"kr djds vkWDlhtu nsrk gSA

Q.14 fuEu esa ls dkSulh kkjh;@kkjh; enk /kkrq] lqij vkWDlkbM ugha cukrh ?

(A) Li (B) Na (C) Mg (D) mijksDr lHkh Q.15 tc lksfM;e o iksVsf'k;e ok;q ds vkf/kD; esa tyrk

gS] izkIr mRikn gS - (A) Na2O, KO2 (B) Na2O2, KO2 (C) NaO2, K2O (D) NaO, K2O2

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

Passage # 1 (Ques. 14 & 15) When heated in excess of air lithium forms normal

oxide, sodium forms peroxide, while potassium andother members of the family have a tendency toform superoxides. Actually smaller cation canstabilise smaller anion and larger cation canstabilise anion. Oxides and peroxides reacts withwater forming sodium hydroxide. KO2 absorbs CO2

and produces oxygen. Q.14 Which alkali/alkaline earth metal do/does not form

superoxides ? (A) Li (B) Na (C) Mg (D) All of these Q.15 When sodium and potassium are burnt in excess of

air, the products obtained are - (A) Na2O, KO2 (B) Na2O2, KO2 (C) NaO2, K2O (D) NaO, K2O2



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

AlF3 futZy HF esa vfoys; gS ijUrq KF dh dqN ek=kk dh mifLFkfr esa ;g foys; gks tkrk gSA ifj.kkeh foy;u esa cksjksu MkbZDyksjkbM ds feykus ij AlF3 dk fdlh vU; inkFkZ 'X' ds lkFk iquZvoksi.k gks tkrk gSA

Q.16 KF dh mifLFkr esa fuEu esa fdlds fuekZ.k ds dkj.k

AlF3 foys; gS - (A) AlF4

– (B) AlF63–

(C) K[AlF3] (D) Al(OH)3 Q.17 BF3 dk vEyh; lkeF;Z, AlF3 dh rqyuk esa gksrk gS - (A) vf/kd (B) de (C) cjkcj (D) ugh ik;k tk ldrk Q.18 X gS - (A) B2F6 (B) BF4

+ (C) KBF4 (D) K2BF4

[k.M - III

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

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

[k.M esa izR;sd iz'u dk mÙkj 0 ls 9 rd bdkbZ ds ,d iw.kk±d

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

lgh mÙkj okys cqYyksa dks dkyk fd;k tkuk gSA mnkgj.k ds

fy, ;fn iz'u la[;k ¼ekusa½ X, Y, Z rFkk W ds mÙkj 6, 0, 9

Passage # 2 (Ques. 16 to 18) AlF3 is insoluble in anhydrous HF but it becomes

soluble in presence of little amount of KF. Additionof boron trifluoride to the resulting solution causesreprecipitation of AlF3 along with some other substance 'X'.

Q.16 AlF3 is soluble in presence of KF due to the

formation of - (A) AlF4

– (B) AlF63–

(C) K[AlF3] (D) Al(OH)3 Q.17 Acidic strength of BF3 as compared to AlF3 - (A) is more (B) is less (C) is equal (D) cannot be found Q.18 X is - (A) B2F6 (B) BF4

+ (C) KBF4 (D) K2BF4

Section - III This section contains 9 questions (Q.1 to 9). +3 markswill be given for each correct answer and no negativemarking. The answer to each of the questions is aSINGLE-DIGIT INTEGER, ranging from 0 to 9. Theappropriate bubbles below the respective questionnumbers 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 correctdarkening of bubbles will look like the following :

0 1 2

3

4

5 6

7

8

9

0 1 2

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X Y Z W

Q.1 The number of N—O—N bonds is N2O5, a white

crystalline solid are. Q.2 How many hydroxyl groups are present in

pyrophosphoric acid ? Q.3 The number of P = O bonds in P4O10 are. Q.4 The basicity of phosphorus acid (H3PO3) is. Q.5 Sulphurous acid reduces acidified potassium

dichromatie solution to green Cr3+ ions. How manymoles of sulpurous acid reacts with one mole ofK2Cr2O7.

rFkk 2 gkas, rks lgh fof/k ls dkys fd;s x;s cqYys ,sls fn[krs gSa

tks fuEufyf[kr gSA

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X Y Z W

Q.1 ,d 'osr fØLVy Bksl N2O5 esa N—O—N cU/kksa dh

la[;k gSA

Q.2 ik;jksQkWLQksfjd vEy esa mifLFkr gkbMªksfDly lewgksa

dh la[;k gS ?

Q.3 P4O10 esa P = O ca/kksa dh la[;k gSA Q.4 QkWLQksjl vEy (H3PO3) dh kkjh;rk gSA

Q.5 lY¶;qjl vEy] vEyh;Ñr iksVsf'k;e MkbZØksesV

foy;u dks gjs Cr3+ vk;u esa cny nsrk gSA lY¶;qjl

vEy ds fdrus eksy] K2Cr2O7 ds ,d eksy ls fØ;k

djrs gSA



Q.6 Sulphur is paramagnetic in the vapour state due topresence of ………. unpaired electrons in theantibonding molecular orbitals.

Q.7 The number of Se—Se bonds in SeO3 tetramer are. Q.8 The number of S—S bonds in dithionic acid are. Q.9 Iron sulphide is heated in air to form A, an oxide of

sulphur. A is dissolved in water to give an acid.The basicity of this acid is.

Q.6 izfrcfU/kr vkf.od dkd esa………. v;qfXer bysDVªkWuksa dh mifLFkfr ds dkj.k lYQj] ok"i voLFkk esa vuqpqEcdRo gksrk gSA

Q.7 SeO3 esa prq"yd esa Se—Se ca/kksa dh la[;k gSA Q.8 MkbZFkk;ksfud vEy esa S—S ca/kksa dh la[;k gSA Q.9 vk;ju lYQkbM] ok;q esa] xeZ gksdj lYQj dk

vkWDlkbM (A) cukrk gSA A ty esa ?kqydj ,d vEy

nsrk gSA vEy dh kkjh;rk gSA



MATHEMATICS

Section – I

Questions 1 to 8 are multiple choice questions. Eachquestion has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. Mark your response inOMR 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 If k ∈ N and Ik = ∫π

π−

k

k

dxxx2

2

][sin|sin| , (where [.]

denotes the greatest integer function) then ∑=

100

1kkI

equals to (A) –10100 (B) –40400 (C) 20200 (D) none of these

Q.2 If Sn = ∑= ++

+n

r rrrr

1234 2

12 then S20 is equal to

(A) 221220 (B)

441420 (C)

221439 (D)

441440

Q.3 The coefficient of x5 in the expansion of (x2 – x – 2)5

is (A) –83 (B) –82 (C) –86 (D) –81

[k.M - I iz'u 1 ls 8 rd cgqfodYih iz'u gSaA izR;sd iz'u ds pkj





Q.1 ;fn k ∈ N rFkk Ik = ∫π

π−

k

k

dxxx2

2

][sin|sin| , (tgk¡ [.]

egÙke iw.kkZad Qyu dks iznf'kZr djrk gS) rc ∑=

100

1kkI

cjkcj gS (A) –10100 (B) –40400 (C) 20200 (D) buesa ls dksbZ ugha

Q.2 ;fn Sn = ∑= ++

+n

r rrrr

1234 2

12 gS] rc S20 cjkcj gS

(A) 221220 (B)

441420 (C)

221439 (D)

441440

Q.3 (x2 – x – 2)5 ds izlkj esa x5 dk xq.kkad gksxk

(A) –83 (B) –82 (C) –86 (D) –81



Q.4 vfrijoy; xy = 4 ij fLFkr fcUnq P ij vfrijoy; ds vfHkyEc dk lehdj.k tks js[kk 2x – y = 5 ds lekUrj gS] gS

(A) 2x – y + 32 = 0 (B) 2x – y – 32 = 0

(C) 2x – y – 23 = 0 (D) 2 x + y – 3 = 0

Q.5 ;fn nh?kZoÙk 2

2

2

2

by

ax

+ = 1 dh ,d Li'kZjs[kk ftldh

<+ky 2 gS] oÙk x2 + y2 + 4x + 1 = 0 dk vfHkyEc gS] rcab dk vf/kdre eku gS

(A) 4 (B) 2 (C) 1 (D) buesa ls dksbZ ugha

Q.6 ijoy; y2 = 4ax dh fdlh ukHkh; thok dks O;kl

ekudj [khapk x;k oÙk ijoy; dks nks fcUnqvksa ‘t’ rFkk

‘t' ’ ij dkVrk gS (ukHkh; thok ds fljksa ds vykok) rc (A) tt′ = –1 (B) tt′ = 2 (C) tt′ = 3 (D) buesa ls dksbZ ugha

Q.7 ekuk f(x) = [x] + x2; R → R rc y = f –1(x), y = 0 ls

dksfV;ksa x = 21 rFkk x = 5 ds e/; ifjc) ks=k dk

ks=kQy gS (tgk¡ [.] rFkk . Øe'k% egÙke iw.kkZad Qyu

rFkk fHkUukRed Hkkx Qyu dks iznf'kZr djrs gS)

Q.4 The equation of normal to the rectangularhyperbola xy = 4 at the point P on the hyperbolawhich is parallel to the line 2x – y = 5 is

(A) 2x – y + 32 = 0 (B) 2x – y – 32 = 0

(C) 2x – y – 23 = 0 (D) 2 x + y – 3 = 0

Q.5 If a tangent of slope 2 of the ellipse 2

2

2

2

by

ax

+ = 1 is

normal to the circle x2 + y2 + 4x + 1 = 0, then themaximum value of ab is

(A) 4 (B) 2 (C) 1 (D) none of these

Q.6 A circle drawn on any focal chord of the parabola y2 = 4ax as diameter cuts the parabola at two points ‘t’ and ‘t' ’ (other than the extremity of focal chord), then

(A) tt′ = –1 (B) tt′ = 2 (C) tt′ = 3 (D) none of these

Q.7 Let f(x) = [x] + x2; R → R then area of figurebounded by y = f –1(x), y = 0 between the ordinates

x = 21 and x = 5 is (where [.] and . represent the

greatest integer and fractional part functionsrespectively)



(A) )1240(23

1+ oxZ bdkbZ

(B) 3

40 oxZ bdkbZ

(C) 23

40 oxZ bdkbZ

(D) )1240(23

1− oxZ bdkbZ

Q.8 i – 2 – 3i + 4 .... 100 inks rd] dk ;ksxQy gS tgk¡

i = 1− (A) 50(1 – i) (B) 25 i (C) 25(1 + i) (D) 100(1 – i)

iz'u 9 ls 13 rd cgqfodYih iz'u gaSA izR;sd iz'u ds pkj fodYi (A), (B), (C) rFkk (D) gSa] ftuesa ls ,d ;k ,d ls vf/kd fodYi lgh gaSA OMR 'khV esa iz'u dh iz'u la[;k ds lek vius mÙkj vafdr dhft,A izR;sd lgh mÙkj ds fy,+ 3 vad fn;s tk;saxs rFkk dksbZ _.kkRed vadu ugha gSA

Q.9 ;fn ∫x

dttf0

)( = x2 + 2x – ∫x

dttft0

)( , x ∈ (0, ∞),

rc ;g gS (A) vkorhZ (B) vkorhZ ijarq eq[; vkorZukad fo|eku ugha (C) vkorhZ iajrq eq[; vkorZukad fo|eku (D) dqN ugha dg ldrs

(A) )1240(23

1+ sq units

(B) 3

40 sq. units

(C) 23

40 sq. units

(D) )1240(23

1− sq units

Q.8 The sum of i – 2 – 3i + 4 .... up to 100 terms, where

i = 1− , is (A) 50(1 – i) (B) 25 i (C) 25(1 + i) (D) 100(1 – i) Questions 9 to 13 are multiple choice questions. Eachquestion has four choices (A), (B), (C) and (D), out ofwhich MULTIPLE (ONE OR MORE) is correct. Markyour response in OMR sheet against the questionnumber of that question. + 3 marks will be given foreach correct answer and no negative marks.

Q.9 If ∫x

dttf0

)( = x2 + 2x – ∫x

dttft0

)( , x ∈ (0, ∞).

Then it is (A) periodic (B) periodic but fundamental period does not exist (C) periodic but fundamental period exist (D) nothing can be said



Q.10 If the straight line 3x + 4y = 24 intersects the axesat A and B and the straight line 4x + 3y = 24 at Cand D, then points A, B, C, D lies on

(A) circle (B) parabola (C) ellipse (D) hyperbola

Q.11 Let Sn = ∑=

++n

r nrnrn

13

22 and

Tn = ∑−

=

++1

03

22n

r nrnrn for n = 1, 2, 3 ... then

(A) Tn < 6

11 (B) Tn > 6

11

(C) Sn < 6

11 (D) Sn > 6

11

Q.12 Let A(k) be the area bounded by the curves y = x2 – 3

and y = kx + 2, then

(A) The range of A(k) is

∞,

3510

(B) The range of A(k) is

∞,

3520

(C) If function k → A(k) is defined for k ∈ [–2, ∞), then A(k) is many-one function

(D) The value of k for which area is minimum is 1.

Q.10 ;fn ljy js[kk 3x + 4y = 24 vkksa dks A rFkk B ij rFkk ljy js[kk 4x + 3y = 24 vkksa dks C rFkk D ij izfrPNsn djrh gS] rc fcUnq A, B, C, D fuEu ij fLFkr gS

(A) oÙk (B) ijoy; (C) nh?kZoÙk (D) vfrijoy;

Q.11 ekuk n = 1, 2, 3 ... ds fy;s Sn = ∑=

++n

r nrnrn

13

22

rFkk Tn = ∑−

=

++1

03

22n

r nrnrn gS] rc

(A) Tn < 6

11 (B) Tn > 6

11

(C) Sn < 6

11 (D) Sn > 6

11

Q.12 ekuk A(k) oØ y = x2 – 3 rFkk y = kx + 2 ls ifjc)

ks=kQy gS] rc

(A) A(k) dk izkUr

∞,

3510

gS

(B) A(k) dk izkUr

∞,

3520

gS

(C) ;fn Qyu k → A(k), k ∈ [–2, ∞) ds fy,ifjHkkf"kr gS] rc A(k) cgq,sdh Qyu gS

(D) k dk eku ftlds fy, ks=kQy U;wure gS] 1 gS



Q.13 Qyu y = f (x) dk xzkQ tks fcUnq (0, 1) ls xqtjrk gS

rFkk vodyu lehdj.k dxdy + y cos x = cos x dks

lUrq"V djrk gS] bl izdkj gS fd (A) ;g vpj Qyu gS (B) ;g vkorhZ Qyu gS (C) ;g uk rks le uk gh fo"ke Qyu gS (D) ;g lHkh x ds fy, larr~ ,oa vodyuh; Qyu gS

bl [k.M esa 2 x|ka'k fn;s x;s gSa] x|ka'k -I esa 2 cgqfodYih iz'u (iz'u 14 o 15) gSa rFkk x|ka'k-II esa 3 cgqfodYih iz'u (iz'u 16 ls 18) gSaA izR;sd iz'u ds pkj fodYi (A), (B), (C) rFkk (D) gSa, ftuesa ls dsoy ,d fodYi lgh gSA OMR 'khV esa iz'u dh iz'u la[;k ds lek viuk mÙkj vafdr dhft;sA izR;sd lgh mÙkj ds fy;s + 3 vad fn;s tk,saxs rFkk izR;sd xyr mÙkj ds fy, 1 vad dkVk tk;sxkA

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

ekuk f(x) ,d vodyuh; QYku bl izdkj gS fd

f ′(x) = f(x) + ∫2

0

)( dxxf , ;fn f(0) = 3

4 2e− , rc

f ′(x) = f (x) + ∫2

0

)( dxxf

ekuk ∫2

0

)( dxxf = k ⇒ f ′(x) – f (x) = k

Q.13 The graph of the function y = f (x) passing throughthe point (0, 1) and satisfying the differential

equation dxdy + y cos x = cos x is such that

(A) It is a constant function (B) It is periodic (C) It is neither an even nor an odd function (D) It is continuous and differentiable for all x

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

Passage # 1 (Ques. 14 & 15) Let f(x) is a differentiable function such that

f ′(x) = f(x) + ∫2

0

)( dxxf , if f(0) = 3

4 2e− , then

f ′(x) = f (x) + ∫2

0

)( dxxf

Let ∫2

0

)( dxxf = k ⇒ f ′(x) – f (x) = k



⇒ e–x f (x) = k ∫ − dxe x ⇒ f (x) = cex – k

pwafd ∫2

0

)( dxxf = k ⇒ k = 3

)1( 2 −ec rFkk

f(0) = 3

4 2e− ⇒ c = 1, k = 3

12 −e

Q.14 f (x) cjkcj gS

(A) ex –

−3

12e (B) ex +

−3

12e

(C) ex –

+3

12e (D) buesa ls dksbZ ugha

Q.15 x + f (x) = 0 ds gyksa dh la[;k gS (A) 0 (B) 1 (C) 2 (D) buesa ls dksbZ ugha x|ka'k # 2 (iz- 16 ls 18)

;fn ax2 + by2 + 2hxy + 2gx + 2fy + c = 0 ,d nh?kZoÙk iznf'kZr djrk gS rc h2 < ab ,oa abc + 2fgh – af 2 –bg2 – ch2 ≠ 0 gSA ;fn mijksDr lehdj.k dks lUrq"V djus okys izR;sd fcUnq (x1, y1) ds fy, (2h – x1, 2k – y1)Hkh bls lUrq"V djrk gS] rc (h, k) bldk dsUnz gSA v)Z nh?kZvk ,oa v)Z y?kq vk dh yEckbZ oØ ij fLFkr fdlh fcUnq dh blds dsUnz ls nwjh ds vf/kdre ,oa U;wure eku gS

⇒ e–x f (x) = k ∫ − dxe x ⇒ f (x) = cex – k

Since ∫2

0

)( dxxf = k ⇒ k = 3

)1( 2 −ec &

f(0) = 3

4 2e− ⇒ c = 1, k = 3

12 −e

Q.14 f (x) is equal to

(A) ex –

−3

12e (B) ex +

−3

12e

(C) ex –

+3

12e (D) None of these

Q.15 The number of solution of x + f (x) = 0 is (A) 0 (B) 1 (C) 2 (D) None of these

Passage # 2 (Ques. 16 to 18) If ax2 + by2 + 2hxy + 2gx + 2fy + c = 0 represents an

ellipse, then h2 < ab and abc + 2fgh – af 2 – bg2

– ch2 ≠ 0. If for every point (x1, y1) satisfying aboveequation (2h – x1, 2k – y1) also satisfy it, then (h, k) is centre of it. The length of semi major axis andminor axis is the maximum and minimum value ofthe distance of a point lying on the curve from itscentre.



Q.16 For the ellipse 2x2 – 2xy + 4y2 – (3 + 2 ) = 0, the inclination of major axis of it with x-axis is

(A) 12π (B)

8π

(C) 8

3π (D) 8

5π

Q.17 The equation of tangent to

2x2 – 2xy + 4y2 – (3 + 2 ) = 0 such that sum ofperpendicular dropped from foci is 2 units, is

(A) y cos 4

3π – x sin4

3π = 1

(B) y sin8

3π – x cos8

3π = 1

(C) x cos 8π – y sin

8π = 1

(D) y cos8

5π + x sin8

5π = 1

Q.18 The product of perpendiculars from the foci to anytangent to above given ellipse is

(A) 4 (B) 2

(C) 1 (D) 21

Q.16 nh?kZoÙk 2x2 – 2xy + 4y2 – (3 + 2 ) = 0 ds fy,] blds nh?kZvk dk x-vk ls >qdko gS

(A) 12π (B)

8π

(C) 8

3π (D) 8

5π

Q.17 2x2 – 2xy + 4y2 – (3 + 2 ) = 0 dh Li'kZ js[kk dk

lehdj.k ftl ij ukfHk;ksa ls Mkys x, yEcks dk

;ksxQy 2 bdkbZ gS] gS

(A) y cos 4

3π – x sin4

3π = 1

(B) y sin8

3π – x cos8

3π = 1

(C) x cos 8π – y sin

8π = 1

(D) y cos8

5π + x sin8

5π = 1

Q.18 mijksDr iz'u esa fn;s x, nh?kZoÙk dh fdlh Li'kZjs[kk

ij ukfHk;ksa ls Mkys x, yEcksa dk xq.kuQy gS (A) 4 (B) 2

(C) 1 (D) 21



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 isa SINGLE-DIGIT INTEGER, ranging from 0 to 9. The appropriate bubbles below the respective questionnumbers in the OMR have to be darkened. For example,if the correct answers to question numbers X, Y, Z andW (say) are 6, 0, 9 and 2, respectively, then the correctdarkening of bubbles will look like the following :

0 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

0 1 2

3

4

5 6

7

8

9

X Y Z W

Q.1 If the tangent at the point P(2, 4) to the parabola

y2 = 8x meets the parabola y2 = 8x + 5 at Q and R, then the sum of the abscissa and ordinate of themidpoint of QR is ....

[k.M - III

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

012

3

4

56

7

8

9

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

X Y Z W

Q.1 ;fn ijoy; y2 = 8x ij fLFkr fcUnq P(2, 4) ij [khaph xbZ Li'kZjs[kk ijoy; y2 = 8x + 5 dks fcUnq Q ,oa R ij feyrh gS] rc QR ds e/; fcUnq ds Hkqt rFkk dksfV dk ;ksxQy gS ....



Q.2 The number of integral values of a for which a unique circles passes through the points of intersection of the rectangular hyperbola x2 – y2 = a2

and the parabola y = x2 is ........

Q.3 If f (x) = ∫−+ 1)1(

12xx

dx and f (1) = 0, then the

value of 5 (f (3/2))2 is .....

Q.4 The value of

∫∫∞→

dxedxex xx x

x 0

22

0

22 /lim is ..

Q.5 If the area bounded by the curves y = x2 and

y = )1(

22x+

is k sq. units, then the value of [k] is ....

Q.6 A pair of curves such that (a) the tangents drawn at points with equal

abscissas intersect on the y-axis, (b) the normal drawn at points with equal abscissas

intersect on the x-axis, (c) one curve passes through (1, 1) and other passes

through (2, 3) Then the number of points of intersection of the

curves is ....

Q.2 a ds iw.kkZad ekuksa dh la[;k ftlds fy, vfrijoy;

x2 – y2 = a2 rFkk ijoy; y = x2 ds izfrPNsnu fcUnqvksa ls

,d vfrh; oÙk xqtjrk gS] gS …….

Q.3 ;fn f (x) = ∫−+ 1)1(

12xx

dx rFkk f (1) = 0 gS] rc

5 (f (3/2))2 dk eku gksxk .....

Q.4

∫∫∞→

dxedxex xx x

x 0

22

0

22 /lim dk eku gksxk ..

Q.5 ;fn oØksa y = x2 rFkk y = )1(

22x+

ls ifjc) ks=kQy

k oxZ bdkbZ gS] rc [k] dk eku gS .....

Q.6 ,d oØ ;qXe bl izdkj gS fd (a) leku Hkqt okys fcUnqvksa ij [khaph xbZ Li'kZjs[kk,

y-vk ij izfrPNsnu djrh gS (b) leku Hkqt okys fcUnqvksa ij [khaps x, vfHkyEc

x-vk ij izfrPNsnu djrs gS (c) ,d oØ] fcUnq (1, 1) ls xqtjrk gS rFkk nwljk oØ]

fcUnq (2, 3) ls xqtjrk gS rc oØksa ds izfrPNsnu fcUnqvksa dh la[;k gS ....



Q.7 If y = y(x) and

++

dxdy

yx

1sin2 = – cos x, y(0) = 1,

then 9 y

π

2 equals .....

Q.8 The value of 20C0 – 2

120 C +

32

20 C –4

320 C + ... is

λ/21 then λ is ... Q.9 The sequence a1, a2, a3, .... satisfies a1 = 19,

a9 = 99 and for all n ≥ 3, an is the arithmetic meanof the first (n – 1) terms, then the value of [a2/20] is (where [.] represents greatest integer function) ...

Q.7 ;fn y = y(x) rFkk

++

dxdy

yx

1sin2 = – cos x, y(0) = 1,

rc 9 y

π

2 cjkcj gS .....

Q.8 20C0 – 2

120 C +

32

20 C –4

320 C + ... dk eku λ/21 gS]

rc λ dk eku gS ...

Q.9 vuqØe a1, a2, a3, .... ; a1 = 19, a9 = 99 dks lUrq"V djrk gS rFkk lHkh n ≥ 3 ds fy, an izFke (n – 1) inks dk lekUrj ek/; gS] rc [a2/20] dk eku gS (tgk¡ [.]egÙke iw.kkZad Qyu dks iznf'kZr djrk gS) ...



PHYSICS

Section – I

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

Q.1 In the circuit shown in the figure switch S is closed attime t = 0. Select the wrong statement –

C

2C R

2R

(A) Rate of increase of charge is not same in boththe capacitors

(B) Ratio of charge stored in capacitors C and 2C at any time t would be 1 : 2

(C) Time constants of both the capacitors are equal

(D) Steady state charge in capacitors C and 2C arein the ratio of 2 : 1

[k.M - I

iz'u 1 ls 8 rd cgqfodYih iz'u gSaA izR;sd iz'u ds pkj





Q.1 fp=k esa n'kkZ, x, ifjiFk esa] le; t = 0 ij fLop S can

fd;k tkrk gSA xyr dFku pqfu;s –

C

2C R

2R

(A) vkos'k esa of) dh nj nksuksa la/kkfj=kksa esa leku

ugha gksrh gS

(B) fdlh le; t esa la/kkfj=k C o 2C esa laxzfgr vkos'k

dk vuqikr 1 : 2 gksxk

(C) nksuksa la/kkfj=kksa ds le; fu;rkad leku gS

(D) la/kkfj=k C o 2C essa LFkk;h vkos'k esa vuqikr 2 : 1

gSA



Q.2 Consider a toroid of circular cross-section of radius b, major radius R much greater than minor radius b, (see diagram) find the total energy stored in magnetic field of toroid [B is magnetic field which is uniform across crossection] –

(A) 0

222

2RbB

µπ (B)

0

222

4RbB

µπ

(C) 0

222

8RbB

µπ (D)

0

222 RbBµπ

Q.3 Two long concentric cylindrical conductors of radii a & b (b < a) are maintained at a potential difference V& carry equal & opposite currents I. An electron with a particular velocity "U" parallel to the axis will travel undeviated in the evacuated region between the conductors. Then U =

(A)

µ

π

abnI

V4

0 l

(B)

banI

V2

0 lµ

π

(C)

µ

π

abnI

V2

0 l

(D)

µ

π

banI

V8

0 l

Q.2 b f=kT;k okys oÙkh; vuqizLFk dkV dh ,d VksjkWbM ijfopkj dhft,] nh?kZ f=kT;k R, y?kq f=kT;k b ls cgqrvf/kd gSA (fp=kkuqlkj) VksjkbM ds pqEcdh; ks=k esalaxzfgr dqy ÅtkZ Kkr dhft, [B pqEcdh; ks=k gS tks fljksa ds e/; ,dleku gS] –

(A) 0

222

2RbB

µπ (B)

0

222

4RbB

µπ

(C) 0

222

8RbB

µπ (D)

0

222 RbBµπ

Q.3 a o b (b < a) f=kT;k okyh nks yEcs ladsUnzh; csyukdkj pkyd V foHkokUrj o leku vkos'k ,oa foijhr /kkjk I j[ks gq, gSA ,d bysDVªkWu pkydksa ds e/; fuokZfrr ks=k esa fcuk fopfyr gq, vk ds lekUrj fuf'pr osx "U" ls xfr djrk gSA rc U =

(A)

µ

π

abnI

V4

0 l

(B)

banI

V2

0 lµ

π

(C)

µ

π

abnI

V2

0 l

(D)

µ

π

banI

V8

0 l



Q.4 A D.C. supply of 120 V is connected to a largeresistance X. A voltmeter of resistance 10 kΩplaced in series in circuit reads 4V. This is anunusual use of voltmeter for measuring very highresistance. The value of X is -

120V

V

10kΩ X

(A) 390 kΩ (B) 290 kΩ (C) 190 kΩ (D) 300 kΩ

Q.5 A circuit element is placed in a closed box. A time t = 0, a constant current generator supplying acurrent of I amp is connected across the box.Potential difference across the box varies accordingto graph as shown in the figure. The element in thebox is -

2

8

3time t (sec)

Volts (V)

(A) a resistance of 2Ω (B) a battery of emf 6V (C) an inductance of 2H (D) a capacitance

Q.4 120 V dh D.C. lIykbZ ls ,d vf/kd eku dk izfrjks/kX la;ksftr gSA ifjiFk esa Js.khØe esa fLFkr 10 kΩizfrjks/k dk ,d oksYVehVj 4V iBu djrk gSA cgqr mPp izfrjks/k ekius ds fy, ;g oksYVehVj dk vlkekU; mi;ksx gSA X dk eku gS -

120V

V

10kΩ X

(A) 390 kΩ (B) 290 kΩ (C) 190 kΩ (D) 300 kΩ

Q.5 ifjiFk rRo ,d can ckWDl esa fLFkr gSA le; t = 0 ij] I ,sfEi;j dh /kkjk lIykbZ dj jgs ,d fu;r /kkjk tujsVj ckWDl ds fljska ij la;ksftr gSaA ckWDl ds fljs ij foHkokUrj fp=k esa n'kkZ, vuqlkj vkjs[k ds vuqlkj ifjorhZ gSaA ckWDl esa rRo gS -

2

8

3time t (sec)

Volts (V)

(A) 2Ω dk izfrjks/k (B) 6V fo-ok-c- dh ,d cSVjh (C) 2H dk izsjdRo (D) ,d la/kkfj=k



Q.6 A uniform current carrying ring of mass m andradius R is connected by a massless string asshown. A uniform magnetic field B0 exist in theregion to keep the ring in horizontal position, thenthe current in the ring is (l = length of the string) -

B0

(A) 0RB

mgπ

(B) 0RB

mg

(C)0RB3

mgπ

(D) 0

2BRmg

π

Q.7 An inductor is placed in series with resistor. An emf E is applied to the combination. The rate atwhich power delivered by the battery is P1. The rate at which power dissipated in the resistor in P2. Rate at which energy stored in inductor in P3. Then which of the following statements is correct -

(A) P1 = P3 – P2 (B) P1 = P3 (C) P1 = P2 + P3 (D) P1 < P2 + P3

Q.6 m nzO;eku o R f=kT;k okyh ,d ,dleku /kkjkokgh

oy; fp=k esa n'kkZ, vuqlkj ,d nzO;ekughu Mksjh

kjk la;ksftr gSA kSfrt fLFkfr esa oy; dks cuk,

j[kus ds fy, ,dleku pqEcdh; ks=k B0 mifLFkr gS] rks oy; esa /kkjk gS (l = Mksjh dh yEckbZ) -

B0

(A) 0RB

mgπ

(B) 0RB

mg

(C)0RB3

mgπ

(D) 0

2BRmg

π

Q.7 ,d izsjd Js.khØe esa ,d izfrjks/k ls la;ksftr gSa ,d

fo-ok-cy E la;kstu dks vkjksfir fd;k tkrk gSA og

nj ftl ij cSVjh kjk 'kfDr iznku dh tkrh gS] P1

gSA og ij ftl ij izfrjks/k esa 'kfDr O;f;r gS] P2 gSA

og nj ftl ij izsj.k esa ÅtkZ laxzfgr gS] P3 gSA rc

fuEu esa ls dkSulk dFku lgh gS - (A) P1 = P3 – P2 (B) P1 = P3 (C) P1 = P2 + P3 (D) P1 < P2 + P3



Q.8 λ js[kh; vkos'k ?kuRo okys R f=kT;k dh ,d oy;

2R f=kT;k ds ,d Bksl dkYifud xksys dh vkSj xfr

djrh gSa rkfd oy; dk dsUnz xksys ds dsUnz ls gksdj

xqtjsA oy; dk vk oy; o xksys ds dsUnzksa dks

tksM+us okyh js[kk ds yEcor~ gksrk gSA bl izfØ;k esa

xksys ls izokfgr vf/kdre ¶yDl gS

R/2 R

(A) 0ε

λR (B) 02ε

λR

(C) 04ε

λπR (D) 03ε

λπR

iz'u 9 ls 13 rd cgqfodYih iz'u gaSA izR;sd iz'u ds pkj

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

vf/kd fodYi lgh gaSA OMR 'khV esa iz'u dh iz'u la[;k ds

lek vius mÙkj vafdr dhft,A izR;sd lgh mÙkj ds fy,

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

Q.8 A ring of radius R having a linear charge density

λ moves towards a solid imaginary sphere of radius

2R , so that the centre of ring passes through the

centre of sphere. The axis of the ring is

perpendicular to the line joining the centres of the

ring and the sphere. The maximum flux through the

sphere in this process is

R/2 R

(A) 0ε

λR (B) 02ε

λR

(C) 04ε

λπR (D) 03ε

λπR

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



Q.9 In the figure initial status of capacitors and theirconnection is shown. Which of the following iscorrect about this circuit ?

15V+ –

2µFA B

10V+ –

3µFC D

A B

C D (A) Final charge on each capacitor is zero (B) Final total electrical energy of the capacitor

will be non zero (C) total charge flown from A to D is 30µC (D) total charge flown from A to D is – 30µC Q.10 R = 10Ω & E = 13 V and voltmeter & Ammeter are

ideal then -

V

A

3Ω

E

R

c6V

a

b

8V

(A) Reading of Ammeter is 2.4 A (B) Reading of Ammeter is 8.4 A (C) Reading of voltmeter is 8.4 V (D) Reading of voltmeter is 27 V

Q.9 fuEu fp=kksa esa la/kkfj=kksa dh izkjfEHkd voLFkk,sa ,oa

buds la;kstu n'kkZ, x, gSa fuEu esa ls dkSulk ifjiFk

ds ckjs esa lgh gS ? 15V

+ –

2µFA B

10V+ –

3µFC D

A B

C D (A) izR;sd la/kkfj=k ij vfUre vkos'k 'kwU; gS

(B) la/kkfj=k dk vfUre dqy fo|qrh; ÅtkZ v'kwU; gksxh

(C) A ls D dks izokfgr dqy vkos'k 30µC gS

(D) A ls D dks izokfgr dqy vkos'k – 30µC gSA

Q.10 R = 10Ω o E = 13 V rFkk oksYVehVj o vehVj vkn'kZ gS] rc -

V

A

3Ω

E

R

c6V

a

b

8V

(A) vehVj dk ikB;kad 2.4 A gSA (B) vehVj dk ikB;kad 8.4 A gS (C) oksYVehVj dk ikB~;kad 8.4 V gSA (D) oksYVehVj dk ikB~;kad 27 V gSA



Q.11 fp=k esa n'kkZ, x,s lSy dk vkarfjd izfrjks/k ux.; gSA dqath K dks can djus ds i'pkr vehVj ikB~;kad

0.25 ,sfEi;j ls 125 ,sfEi;j rd ifjorZr gks tkrk gSA rc

R1

A

K

R = 10Ω

E

ideal ammeter (A) R1 = 10 Ω (B) R1 = 15 Ω (C) lSy ls yh xbZ 'kfDr esa of) gksrh gS (D) R ls izokfgr /kkjk 40% rd de gks tkrh gS Q.12 5 cm o 10 cm dh nks oÙkkdkj dq.Mfy;k¡ 2A dh

leku /kkjk izokfgr j[krh gSaA dq.Mfy;k¡ Øe'k% 50 o100 Qsjs j[krh gS rFkk bl izdkj j[kh gqbZ gS dh muds ry o lkFk gh muds dsUnz Hkh lEikrh gSA dq.Myh ds mHk;fu"B dsUnz ij pqEcdh; ks=k dk ifjek.k gS -

(A) 8π × 10–4 T gS] ;fn dq.Myh;ksa esa /kkjk leku fn'kk esa izokfgr gSa

(B) 4π × 10–4 T gS] ;fn dq.Myh;ksa esa /kkjk foijhrfn'kk esa izokfgr gS

(C) 'kwU; gS] ;fn dq.Mfy;ksa esa /kkjk foijhr fn'kk esa izokfgr gS

(D) 8π × 10–4 T gS] ;fn dq.Myh;ksa esa /kkjk foijhr fn'kk esa izokfgr gS

Q.11 The internal resistance of the cell shown in thefigure is negligible. On closing the key K, theammeter reading changes from 0.25 amp to

125 amp, then –

R1

A

K

R = 10Ω

E

ideal ammeter (A) R1 = 10 Ω (B) R1 = 15 Ω (C) power drawn from the cell increases (D) the current through R decreases by 40%

Q.12 Two circular coils of radii 5 cm and 10 cm carryequal currents of 2A. The coils have 50 and 100turns respectively and are placed in such a way thattheir planes as well as their centre coincide.Magnitude of magnetic field at the common centreof the coil is -

(A) 8π × 10–4 T if current in the coil are in same sense (B) 4π × 10–4 T if current in the coil are in opposite

sense (C) zero if currents in the coils are in opposite sense (D) 8π × 10–4 T if current in the coil are in opposite

sense



Q.13 vk;rkdkj vuqizLFk dkV (a = 2.00 cm kjk b = 3.00 cm)

rFkk vkarfjd f=kT;k R = 4.00 cm okyh ,d VksjkWbM

500 Qsjksa okys ,d rkj tks Imax = 50.0 A rFkk f =π

ω2

=

60.0 Hz vkofÙk okyh I = Imax sin ωt /kkjk ls ;qDr gS] dks

j[ks gq, gSA 20 Qsjkas okyh ,d dq.Myh fp=k esa n'kkZ,

vuqlkj VksjkbWM ls lEcfU/kr gS &

(A)

time

Indu

ced

emf

a b

R

N′= 20

N = 500Toroid

(B)

time

Indu

ced

emf

(C) izsfjr fo- ok- cy dk vf/kdre eku 0.422 V gS

(D) izsfjr fo- ok- cy dk vf/kdre eku 0.122 V gS

Q.13 A toroid having a rectangular cross-section

(a = 2.00 cm by b = 3.00 cm) and inner radius

R = 4.00 cm consists of 500 turns of wire that

carries a current I = Imax sin ωt with Imax = 50.0 A and

a frequency f =π

ω2

= 60.0 Hz. A coil that consists

of 20 turns of wire links with the toroid as shown

in figure –

(A)

time

Indu

ced

emf

a b

R

N′= 20

N = 500Toroid

(B)

time

Indu

ced

emf

(C) Maximum value of induced emf is 0.422 V

(D) Maximum value of induced emf is 0.122 V



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

Passage # 1 (Ques. 14 & 15) A particle of mass m and charge q is accelerated by

a potential difference V volt and made to enter a magnetic field region at an angle θ with the field.At the same moment, another particle of same massand charge is projected in the directions of the fieldfrom the same point. Magnetic field of induction is B.

Q.14 What would be the speed of second particle so that

both particle meet again and again after a regular

interval of time which should be minimum ?

(A) mqv cosθ (B)

mqv2 cosθ

(C) 2mqv sinθ (D) 2

mq2 cosθ

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

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

(iz'u 16 ls 18) gSaA izR;sd iz'u ds pkj fodYi (A), (B), (C)

rFkk (D) gSa, ftuesa ls dsoy ,d fodYi lgh gSA OMR 'khV

esa iz'u dh iz'u la[;k ds lek viuk mÙkj vafdr dhft;sA

izR;sd lgh mÙkj ds fy;s + 3 vad fn;s tk,saxs rFkk izR;sd

xyr mÙkj ds fy, 1 vad dkVk tk;sxkA

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

m nzO;eku o q vkos'k okyk ,d d.k V oksYV foHkokUrj

ls Rofjr gS rFkk kS=k ds lkFk θ dks.k cukrs gq, ,d

pqEcdh; kS=k esa izos'k djrk gSaaA mlh k.k ij] mlh

nzO;eku o vkos'k okyk nwljk d.k leku fcUnq ls pqEcdh;

ks=k dh fn'kk esa izksfir gSA pqEcdh; ks=k B gSA

Q.14 frh; d.k dh pky D;k gksxh] rkfd nksuksa d.k og le;

tks U;wure gks ds fu;fer le;kUrjky ds lkFk ,d&nwljs

ls ckj&ckj feyrs jgsa &

(A) mqv cosθ (B)

mqv2 cosθ

(C) 2mqv sinθ (D) 2

mq2 cosθ



Q.15 Find the time interval after which they meet -

(A) qB

mπ2 (B) qBm

2π

(C) qBmπ (D)

qBm

23π

Passage # 2 (Ques. 16 to 18) In the circuit shown in figure V1 & V2 are two

voltmeters of resistances 3000Ω and 2000Ωrespectively. In addition R1 = 2000Ω, R2 = 3000Ωand E = 200V.

Q.16 The reading of voltmeter V2 when S is open -

E

V1 V2

E

R1 R2 S

(A) 120 V (B) 80 V (C) 100 V (D) 140 V Q.17 The reading of voltmeter V1 when S is closed - (A) 120 V (B) 80 V (C) 100 V (D) 60 V

Q.15 og le;kUrjky ftlds i'pkr~ os feyrs gS] Kkr dfj,s &

(A) qB

mπ2 (B) qBm

2π

(C) qBmπ (D)

qBm

23π

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

fp=k esa n'kkZ, x, ifjiFk esa] 3000Ω o 2000Ω izfrjks/k ds

nks oksYVehVj V1 o V2 la;ksftr gSA R1 = 2000Ω, R2 =

3000Ω o E = 200V ifjiFk esa la;ksftr gSaA

Q.16 oksYVehVj V2 dk ikB~;kad tc S [kksyk tkrk gS &

E

V1 V2

E

R1 R2S

(A) 120 V (B) 80 V (C) 100 V (D) 140 V Q.17 oksYVehVj V1 dk ikB~;kad tc S can fd;k tkrk gS& (A) 120 V (B) 80 V (C) 100 V (D) 60 V



Q.18 Current through S when it is closed is -

(A) 501 A (B)

201 A

(C) 301 A (D)

601 A

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 isa SINGLE-DIGIT INTEGER, ranging from 0 to 9. The appropriate bubbles below the respective questionnumbers in the OMR have to be darkened. For example,if the correct answers to question numbers X, Y, Z andW (say) are 6, 0, 9 and 2, respectively, then the correctdarkening of bubbles will look like the following :

0 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

0 1 2

3

4

5 6

7

8

9

X Y Z W

Q.18 S ls izokfgr /kkjk tc bls can fd;k tkrk gS &

(A) 501 A (B)

201 A

(C) 301 A (D)

601 A

[k.M - III

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

012

3

4

56

7

8

9

0 1 2

3

4

5 6

7

8

9

0 1 2

3

4

5 6

7

8

9

0 1 2

3

4

5 6

7

8

9

X Y Z W



Q.1 In figure C1 = 2µF, C2 = 6µF & C3 = 3.5 µF. If break down voltages of the individual capacitorsare V1 = 100 V, V2 = 50V & V3 = 400 V.Maximum voltage can be placed across points a & b

is 3

100 × x then value of x.

C3

b

C1

C2

a

Q.2 Figure shows a square loop 20 cm on each side inthe x-y plane with its centre at the origin. The loop carries a current of 7A. Above it at y = 0,z = 12cm an infinitely long wire parallel to the yaxis carrying a current of 10 A. The net force onthe loop is ------×10–4N.

x

C

D

A

B

y

z

d

Q.1 fuEu fp=k esa] C1 = 2µF, C2 = 6µF o C3 = 3.5 µF gSA ;fn vyx&vyx la/kkfj=kks dh Hkatd oksYVrk V1 = 100 V, V2 = 50V o V3 = 400 V gSA fcUnqvksa a o b ds fljksa

ij vf/kdre oksYVrk 3

100 × x gS rks x Kkr djks &

C3

b

C1

C2

a

Q.2 fuEu fp=k ,d oxkZdkj ywi tks ewy fcUnq ij blds dsUnz ds lkFk x-y ry esa izR;sd Hkqtk 20 cm j[krk gS] dks n'kkZrk gSA ywi 7A /kkjk j[krk gSa blds Åij y = 0 ijz = 12cm dk ,d vuUr yEckbZ dk rkj y vk ds lekUrj gS] ftlesa 10 A dh /kkjk izokfgr gSA ywi ij dqy cy ------×10–4N gS &

x

C

D

A

B

y

z

d



Q.3 A capacitor of capacity 2µF is charged to apotential difference of 12V. It is then connectedacross an inductor of inductance 0.6 mH. Thecurrent in the circuit at a time when the potentialdifference across the capacitor is 6.0 volt is ….. × 10–1 Amp.

Q.4 A series LCR circuit containing a resistance of

120Ω has angular resonance frequency 4 × 105

rad/s. At resonance the voltage across resistance

and inductance are 60V & 40V respectively. At

what frequency the current in the circuit lags the

voltage by 45°. Given answer in…… × 105 rad/sec.

Q.5

V ~

AC source

C

L

Current in the inductance is 0.8 A while in thecapacitance is 0.6 A. The current drawn from thesource is ------×10–1 Amp.

Q.3 2µF /kkfjrk okyk ,d la/kkfj=k 12V foHkokUrj ls vkosf'kr

gSA bls fQj 0.6 mH. izsjdRo okys ,d izsjd ds fljs ij

la;ksftr fd;k tkrk gSA ml le; ifjiFk esa /kkjk tc

la/kkfj=k ds fljksa ij foHkokUrj 6.0 oksYV gS] ….. × 10–1

Amp. gSA

Q.4 120Ω izfrjks/k ls ;qDr ,d Js.khØe LCR ifjiFk

4 × 105 rad/s dks.kh; vuquknh vkofÙk j[krk gSA vuqukn

ij izfrjks/k o izsjdRo ds fljks ij oksYVrk Øe'k% 60V o

40V gSA fdruh vkofÙk ij ifjiFk esa /kkjk] oksYVst ls 45°

dyk ls ihNs gks tkrh gSA mÙkj nhft,…… × 105

rad/sec. Q.5

V ~

AC source

C

L

izsjdRo esa /kkjk 0.8 A gS tcdh la/kkfj=k esa 0.6 A gSA

L=kksr ls yh xbZ /kkjk ------×10–1 Amp gSA



Q.6 A charged dust particle of radius 5 × 10–7 m is

located in a horizontal electric field having an

intensity of 6.28 × 105 V/m. The surrounding

medium is air with coefficient of viscosity

η = 1.6 × 10–5 N-s/m2. If the particle moves with a

uniform horizontal speed 0.02 m/s, The number of

electrons on it is a × 10 then a is -

Q.7 Two identical capacitors are connected as shown

and having initial charge Q0. Separation between

plates of each capacitor is d0. Suddenly the left

plate of upper capacitor and right plate of lower

capacitor start moving with speed v towards left

while other plate of each capacitor remains fixed.

(given 0

0

2dVQ = 10 amp)

The value of current in the circuit is x × 10 amp.

then determine x..

Q.6 5 × 10–7 m f=kT;k dk ,d vkosf'kr /kqyd.k

6.28 × 105 V/m rhozrk okys ,d kSfrt fo|qr ks=k esa

fLFkr gSA η = 1.6 × 10–5 N-s/m2 ';kurk xq.kkad ds lkFk

ifjos'k ek/;e ok;q gSA ;fn d.k 0.02 m/s ,dleku

kSfrt pky ls xfr djrk gS] rks bl ij bysDVªkWuksa dh

la[;k a × 10 gksxh rks a D;k gksxkA

Q.7 nks le:i la/kkfj=k n'kkZ, vuqlkj la;ksftr gS rFkk

izkjfEHkd vkos'k Q0 j[krs gSA izR;sd la/kkfj=k dh

IysVksa ds e/; iFkDdj.k d0 gSA vpkud Åijh

la/kkfj=k dh cka;h IysV rFkk fupys la/kkfj=k dh nk;ah

IysV ck;h vksj v pky ls xfr djuk izkjEHk dj nsrh

gSA tcfd izR;sd la/kkfj=k dh vU; IysV fQDl jgrh

gSA (fn;k gS 0

0

2dVQ = 10 amp)

ifjiFk esa /kkjk dk eku x × 10 amp gSA rks x Kkr

djksA



+ – + – + –

+ – + – + –

V ← Q0

Q0 ←V

Q.8 Consider the potentiometer circuit shown.

Potentiometer wire AB is 100 cm long and has

electrical resistance of 1Ω between its ends. With

key opened the balancing length AJ equals 40 cm.

Now, the key is closed. New balance length

is……× 10 cm.

A' 2Ω

2Ω

Key

A ideal ammeter

B

30V 2Ωduring circuit

J

+ – + – + –

+ – + – + –

V ←Q0

Q0 ← V

Q.8 n'kkZ, x, foHkoekih ifjiFk ij fopkj dhft,A

foHkoekih rkj AB 100 cm yEck ,oa blds fljks ds

e/; fo/kqr izfrjks/k 1Ω gSA dqath [kksyus ij larqyu

yEckbZ AJ, 40 cm izkIr gksrh gSA vc dqath can dj

nh tkrh gSA ubZ laarqyu yEckbZ gS ……× 10 cm.

A'2Ω

2Ω

Key

Aideal ammeter

B

30V 2Ωduring circuit

J



Q.9 The plane of a square loop of wire with edge length

a = 0.2 m is in perpendicular to the earth’s

magnetic field at a point, where B = 15.0 µT, as

shown in figure. The total resistance of the loop

and the wires connecting it to a sensitive ammeter

is 0.500Ω. If the loop is suddenly collapsed by

horizontal forces as shown what is the total charge

(Cb) passing through the ammeter ?

A

a a

a a

F F

Q.9 rkj ds ,d oxkZdkj ywi dk ry (fdukjks dh yEckbZ

a = 0.2 m ds lkFk) ,d fcUnq ij iFoh ds pqEcdh;

kS=k ds yEcor~ gS tgk¡ B = 15.0 µT gS] fp=k esa

n'kkZ;s vuqlkj gSA ywi rFkk lqxzkgh vehVj ls

la;ksftr rkjks dk dqy izfrjks/k 0.500Ω gSA ;fn ywi

vpkud n'kkZ, vuqlkj kSfrt cyksa kjk fldqM+

(collapse) gks tkrk gS] rks vehVj ls izokfgr dqy

vkos'k (Cb) D;k gksxk \

A

a a

a a

FF



Date : 10/3/2011Time : 3 : 00 Hrs. MAX MARKS: 243

Name : _________________________________________________________ Roll No. : __________________________

INSTRUCTIONS TO CANDIDATE

A. lkekU; :

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

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

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

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

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

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

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

C. OMR dh iwfrZ :

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

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

Section –I Section-II Section-III

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

A B C D E

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

A B C D E

the wrong method for filling the bubble are

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

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

P Q R S TA BCD

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

012

3

4

56

7

8

9

012

3

4

56

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

8

9


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Documents

correct method

wrong answer

correct match

multiple correct answers

single correct type

answer sheet

integer answer type

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