TARGET COURSE FOR IIT-JEE 2011 - Career Point

CAREER POINT

TARGET COURSE FOR IIT-JEE 2011

PHASE- ALL CHEMISTRY, MATHEMATICS & PHYSICS TEST NO. 14 [TM-7(II)] (TAKE HOME)

PAPER – II Date : 30/3/2011Time : 3 : 00 Hrs. MAX MARKS: 237

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. 5 marks will be awarded for each correct answer and

–2 mark for each wrong answer. 5 Passage based single correct type questions. 3 marks will be awarded for each correct answer and –1 mark for each wrong

answer. Section - II 6. Column matching type questions (4 × 5 type). 8 marks will be awarded for the complete correctly matched answer and No

Negative marking for wrong answer. However, 2 marks will be given for a correctly marked answer in any row. 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 in this section are to be given in the form of single integers i.e. 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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'6' should be filled as 0006

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0 0 0 00 1 2

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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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2

Space for rough work

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

CHEMISTRY

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. + 5 marks will be given for each correct answer and – 2 mark for each wrong answer.

Q.1 Identify the product C in the following reaction sequence

o–HOOC–C6H4–CH2–C6H5 → 2SOCl

A → 3AlCl.anhyd B → HCl/Hg–Zn C

(A) CH2

CH Cl

(B) C

CH2

O COCl

(C) CH2

CH2

(D)

[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, + 5 vad fn;s

tk;asxs rFkk izR;sd xyr mÙkj ds fy, 2 vad dkVk tk;sxkA

Q.1 fuEu vfHkfØ;k Øe esa mRikn C igpkfu,

o–HOOC–C6H4–CH2–C6H5 → 2SOCl

A → 3AlCl.anhyd B → HCl/Hg–Zn C

(A) CH2

CHCl

(B) C

CH2

OCOCl

(C) CH2

CH2

(D)



Q.2 vfHkfØ;k dk vafre mRikn ‘ ‘A’ gS

O

O∆

∆++

→,OH)ii(

,NaOHI)i(

3

2 CHI3 + A ;

(A) O

COOH (B)

O

CHO

(C) COOHCOOH (D)

O

Q.3 uhps nh xbZ vfHkfØ;k esa mRikn A o B gS O

O

O

LiAIH4

NaBH4

A

B

(A) OH

OHOH

(A rFkk B nksuks)

(B)

OO

O

(A rFkk B nksuks )

(C)

OH

OHOH

rFkk

O O

OH

(D) buesa ls dksbZ ugha

Q.2 The end product of the reaction

O

O ∆

∆++

→,OH)ii(

,NaOHI)i(

3

2 CHI3 + A ;

A is

(A) O

COOH (B)

O

CHO

(C) COOH COOH (D)

O

Q.3 The product A and B in the reaction given below are :

O O

O

LiAIH4

NaBH4

A

B

(A) OH

OH OH

(both A and B)

(B)

O O

O

(both A and B)

(C)

OH

OH OH

and

OO

OH

(D) None of these



Q.4 1 eksy ds ,d vkn'kZ fd foy;u (binary solution) ds fuekZ.k ds nkSjku Å"ekxfrdh; xq.k/keksZ esa ifjorZu ls lEcfU/kr fuEu esa dkSulk lgh izn'kZu gS -

(A)

∆ Gmix

T∆ Smix

∆ Hmix

eksy fHkUu

J mol–1 0

+

–

(B)

∆ Gmix

T∆ Smix

∆ Hmix

eksy fHkUu

J mol–1 0

+

–

(C)

∆ Gmix

T∆ Smix

∆ Hmix

eksy fHkUu

J mol–1 0

+

–

(D)

∆ Gmix

T∆ Smix

∆ Hmix

eksy fHkUu

J mol–1 0

+

–

Q.4 Which of the following represents correctly the changes in thermodynamic properties during the formation of 1 mole of an ideal binary solution.

(A)

∆ Gmix

T∆ Smix

∆ Hmix

mole fraction

J mol–1 0

+

–

(B)

∆ Gmix

T∆ Smix

∆ Hmix

mole fraction

J mol–1 0

+

–

(C)

∆ Gmix

T∆ Smix

∆ Hmix

mole fraction

J mol–1 0

+

–

(D)

∆ Gmix

T∆ Smix

∆ Hmix

mole fraction

J mol–1 0

+

–



Q.5 Study the following figure and choose the correct options. Assuming complete dissociation of electrolyte :

0.05M aq. Al2(SO4)3 solution

0.10M aq. MgCl2 solution

SPM

(A) There will be no net moment of any substance

across the membrane (B) MgCl2 will flow towards the Al2(SO4)3 solution (C) Al2(SO4)3 will flow towards the MgCl2 solution (D) The π(osmotic pressure) of 0.1 M MgCl2 is

higher than the π of 0.05 M Al2(SO4)3 Q.6 In the radioactive decay

ZXA → Z+1YA →energyhigh

4–A1–Z Z →

energylow

4–A1–Z Z the

sequence of the radiation emitted is : (A) α, β, γ (B) γ, α, β (C) β, γ, α (D) β, α, γ This section contains 2 paragraphs, each has 3 multiple choice questions. (Questions 7 to 12) Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. + 3 marks will be given for each correct answer and – 1 mark for each wrong answer.

Q.5 fo|qr vi?kV~; dk iw.kZ fo;kstu ekurs gq, fuEu fp=k

dk v/;;u djsa rFkk lgh fodYi pqus -

0.05M tyh; Al2(SO4)3 foy;u

0.10M tyh; MgCl2 foy;u

SPM

(A) f>Yyh ls fdlh Hkh inkFkZ dh dksbZ xfr ugha gksxh

(B) MgCl2 , Al2(SO4)3 foy;u dh vksj xfr djsxk

(C) Al2(SO4)3 , MgCl2 foy;u dh vksj xfr djsxk

(D) 0.1M MgCl2 dk π(ijklj.k nkc) 0.05 M

Al2(SO4)3 ds π ls vf/kd gksrk gS

Q.6 fuEu jsfM;ks/kehZ k; esa]

ZXA → Z+1YA →energyhigh

AZ Z 4–

1– → energylow

4–A1–Z Z

mRlftZr fofdj.kksa dk Øe gSA

(A) α, β, γ (B) γ, α, β (C) β, γ, α (D) β, α, γ

bl [k.M esa 2 vuqPNsn fn;s x;s gSa] izR;sd esa 3 cgqfodYih

iz'u gSaA (iz'u 7 ls 12) 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- 7 ls 9)

NBS ,fyfyd czksfeuhdj.k nsrk gS] tcfd ,YdksgkWyh KOH fogkbMªks gSykstuhdj.k nsrk gSA MkbZbZu] ,Ydhu ds lkFk MhYl-,YMj vfHkfØ;k kjk pfØ; ;kSfxd nsrk gSA ,Ydhu dk (CH3)2S ds lkFk vfHkfØ;k i'pkr vkstksuhvi?kVu ij ,fYMgkbM nsrk gSA

→ ∆,NBS A → KOHalcoholic B → = 22 CHCH C

S)CH()ii(

CCl/O)i(

23

43 → D

Q.7 mijksDr vfHkfØ;k Øe esa ;kSfxd A gS -

(A)

Br

(B)

Br

(C) Br

(D)

S

NH2

Q.8 vfHkfØ;k esa mRikn B gS - (A) ,Ydhu (B) ,Ydkbu (C) MkbZbZu (D) ,fyu Q.9 mRikn D gS

(A) 2CH3CH2CHO (B)

H5C2–C–C–CH2–CH3

O O

(C) (CH2)2CHOCHO

(D) (CH2)4 CHOCHO

Passage # 1 (Ques. 7 to 9) NBS brings allylic bromination while alc. KOH

brings dehydrohalogenation. Diene undergoes Diel’s-alder reaction with alkene to give a cyclic compound. Ozonolysis of alkene followed by reaction with (CH3)2S gives aldehydes.

→ ∆,NBS A → KOHalcoholic B → = 22 CHCH C

S)CH()ii(

CCl/O)i(

23

43 → D

Q.7 Compound A in the above reaction sequence is

(A)

Br

(B)

Br

(C)

Br (D)

S

NH2

Q.8 The product B in the reaction is a/an (A) alkene (B) alkyne (C) diene (D) allene Q.9 Product D is

(A) 2CH3CH2CHO (B)

H5C2–C–C–CH2–CH3

O O

(C) (CH2)2 CHO CHO

(D) (CH2)4 CHOCHO



x|ka'k # 2 (iz- 10 ls 12)

ek¡l dks lajfkr j[kus gsrq lksfM;e lYQkbV (Na2SO3) dks feyk;k tkrk gSA Na2SO3 dh mifLFkfr dks ruq H2SO4 feykdj fu/kkZfjr fd;k tkrk gS] tc mRlftZr gksus okyh nqxZU/k pwus ds ikuh dks nwf/k;k dj nsrh gSA fueqZDr xSl dks lYQj MkbZvkWDlkbM ds #i esa fuf'pr fd;k tkrk gSA fueqZDr SO2 dks xSl dks ty esa ?kksyk tkrk gS rFkk bl SO2 dks –2

4SO esa vkWDlhdj.k gsrq vuqekiu izØe esa I2 foy;u dh vko'drk gksrh gSA

SO2 + 2H2O + I2 → 4H+ + –24SO + 2I–

vuqekiu dk ifj.kke tkuus ds fy, vafre foy;u esa csfj;e DyksjkbM dk vkf/kD; feyk;k tkrk gSA ifj.kkeh voksi dks ,df=kr dj Hkkj fy;k tkrk gSA

Q.10 SO2 rFkk CO2 nksuks pwus ds ikuh dks nwf/k;k dj nsrs gSA fuEu esa ls dkSuls vfHkdeZd dk mi;ksx bu nksuks xSlks dks foHksfnr djus esa fd;k tkrk gSA

(I) K2Cr2O7/H2SO4 (II) KMnO4/H+ (III) I2 foy;u

(A) I, II, III lgh (B) dsoy I, III lgh (C) dsoy II, III lgh (D) dsoy III lgh

Q.11 SO2 xSl dk mi;ksx ,d fojtad dkjd ds #i esa gksrk gSA bldh fojtau fØ;k gksrh gSA

(A) vLFkkbZ rFkk bldh vkWDlhdkjh izdfr ds dkj.k (B) vLFkkbZ rFkk bldh vipk;d fØ;k ds dkj.k (C) LFkkbZ rFkk bldh vkWDlhdkjh fØ;k ds dkj.k (D) LFkkbZ rFkk bldh vipk;d fØ;k ds dkj.k

Passage # 2 (Ques. 10 to 12) Sodium sulphite (Na2SO3) is added to meat as a

preservative. The presence of Na2SO3 can be detected by adding dil. H2SO4 when the pungent smelling gas evolved turns the lime water milky. The gas evolved was detected as sulphur dioxide. The SO2 evolved was dissolved in water and it requires I2 solution in order to oxidize SO2 to –2

4SO

in titration. SO2 + 2H2O + I2 → 4H+ + –2

4SO + 2I–

In order to check the results of titration, excess barium chloride is added to the final solution. The resulting precipitate is collected and weighed.

Q.10 SO2 and CO2 both turns lime water milky. Which of the following reagent can be used to distinguish these two gases ?

(I) K2Cr2O7/H2SO4 (II) KMnO4/H+ (III) I2 solution (A) I, II, III correct (B) I, III only correct (C) II, III only correct (D) III only correct Q.11 SO2 gas is used as a bleaching agent. Its bleaching

action is : - (A) temporary and due to its oxidizing nature (B) temporary and due to its reducing action (C) permanent and due to its oxidizing action (D) permanent and due to its reducing action



Q.12 Which of the following compounds is formed, when Na2SO3 is boiled with sulphur.

(A) Na2SO4 (B) Na2S2O5

(C) Na2S2O6 (D) Na2S2O3

Section - II

This section contains 2 questions (Questions 1, 2). Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S, T) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, A-T; B-Q, B-R; C-P, C-Q and D-S, D-T then the correctly bubbled 4 × 5 matrix should be as follows :

A B C D

P Q R S T

T S

P

P P Q R

R R

Q Q

S S T

T

P Q R S T

Mark your response in OMR sheet against the question number of that question in section-II. + 8 marks will be given for complete correct answer (i.e. +2 marks for each correct row) and NO NEGATIVE MARKING for wrong answer.

Q.12 tc Na2SO3 dks lYQj ds lkFk xeZ fd;k tkrk gS rks

fuEu esa ls dkSulk ;kSfxd curk gS (A) Na2SO4 (B) Na2S2O5

(C) Na2S2O6 (D) Na2S2O3

[k.M - II bl [k.M esa 2 iz'u (iz'u 1, 2) gSaA izR;sd iz'u esa nks LrEHkksa esa dFku fn;s x;s gSa] ftUgsa lqesfyr djuk gSA LrEHk-I (Column I ) esa fn;s x;s dFkuksa (A, B, C, D) dks LrEHk-II (Column II) esa fn;s x;s dFkuksa (P, Q, R, S,T) ls lqesy djuk gSA bu iz'uksa ds mÙkj uhps fn;s x;s mnkgj.k ds vuqlkj mfpr xksyksa dks dkyk djds n'kkZuk gSA ;fn lgh lqesy A-P, A-S, A-T; B-Q, B-R; C-P, C-Q rFkk D-S, D-T gS, rks lgh fof/k ls dkys fd;s x;s xksyksa dk 4 × 5 eSfVªDl uhps n'kkZ;s vuqlkj gksxk :

ABCD

P Q R S T

T S

P

P P Q R

R R

Q Q

S S T

T

P Q R S T

vr% OMR 'khV esa iz'u dh iz'u la[;k ds lek viuk mÙkj [k.M-II esa vafdr dhft;sA izR;sd iw.kZ lgh mÙkj ds fy;s +8 vad fn;s tk;saxs ¼vFkkZr~ izR;sd lgh iafDr feyku ds fy, +2 vad fn, tk,saxs½ o xyr mÙkj ds fy, dksbZ _.kkRed vadu ugha gSA



Q.1

Column -I Column-II

(A) [Pt(NH3)2Cl2] (P) paramagnetic

(B) [CuCl4]2– (Q) homoleptic complex

(C) [Fe(CN)6]4– (R) square planar

(D) [FeF6]3– (S) heteroleptic

(T) octahedral

Q.2

Column –I Column-II (A) Co-ordination no. of

Ca2+ and F– in fluorite structure

(P) 8, 4

(B) C.No. of Zn2+ and S–2

in zinc blend structure (Q) 8, 8

(C) C.No. of Cs+ and Cl– in CsCl (bcc type) structure

(R) 4, 8

(D) C.No. of Li+ and O2–

in antifluorite structure (S) 4, 4 (T) 6, 6

Q.1

LrEHk-I LrEHk -II

(A) [Pt(NH3)2Cl2] (P) vuqpqEch;

(B) [CuCl4]2– (Q) gkseksysifVd ladqy

(C) [Fe(CN)6]4– (R) oxkZdkj leryh;

(D) [FeF6]3–

(S) fgVsjksysifVd (T) v"VQydh;

Q.2 LrEHkksa dk feyku dhft,:

LrEHk-I LrEHk -II (A) ¶yksjkbV lajpuk esa Ca2+ rFkk

F– dh milgla;kstu la[;k (P) 8, 4

(B) ftad Cys.M lajpuk esa Zn2+ rFkk S–2 milg;kstu la[;k

(Q) 8, 8

(C) CsCl ( bcc izdkj) lajpuk esa Cs+ rFkk Cl– dh milgal;kstu la[;k

(R) 4, 8

(D) izfr¶yksjkbV lajpuk esa Li+ rFkk O2– dh milgla;kstu la[;k

(S) 4, 4 (T) 6, 6



Section - III

This section contains 5 questions (Q.1 to 5).+3 marks will be given for each correct answer and nonegative 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 has 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 The number of possible isomers of an octahedralcomplex, [Co(C2O4)2(NH3)2]– are.

Q.2 Titanium shows magnetic moment of 1.73 B.M. inits compound. What is the oxidation number of Tiin the compound ?

[k.M - III bl [k.M esa 5 (iz-1 ls 5) iz'u gaSA izR;sd lgh mÙkj ds fy;s

+3 vad fn;s tk,saxs 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 ,d v"VQydh; ladqy [Co(C2O4)2(NH3)2]– dh lEHko

leko;fo;ksa dh l[;ka gksrh gSA

Q.2 VkbVsfu;e blds ;kSfxd esa 1.73 B.M. pqEcdh; vk?kw.kZ

n'kkZrk gSA ;kSfxd esa Ti dh vkWDlhdj.k l[;k¡ D;k

gksxhA



Q.3 In the disproportionation reaction of NaOH withone molecule of P4, number of molecules of NaOHreacting are …… .

Q.4 When a 20 mL of 0.08 M weak base BOH is titrated

with 20 mL 0.08 M HCl, the pH of the solution at theend point is 5. What will be pOH if 10 mL of 0.04 MNaOH is added to the resulting solution ? [Given : log2 = 0.30 and log 3 = 0.48] Give the answer in nearestwhole number.

Q.5 When sulphur (in the form of S8) is heated at

temperature T, at equilibrium, the pressure of S8

falls by 30% from 1.0 atm, because S8(g) is partially converted into S2(g). Find the value of Kp

for this reaction. Give the answer in nearest wholenumber.

Q.3 NaOH dh] P4 ds ,d v.kq ds lkFk fo"kekuqikrh

vfHkfØ;k esa NaOH v.kqvks dh la[;k gksrh gSA Q.4 tc 0.08 M nqcZy kkj BOH ds 20 mL dks] 0.08 M

HCl ds 20 mL ds lkFk vuqekfir fd;k tkrk gS] rks

vfUre fcUnw ij foy;u dk pH, 5 gSA tc ifj.kkeh

foy;u esa 0.04 M NaOH dk 10 mL feyk;k tk,] rks

pOH D;k gksxh? [fn;k x;k : log 2 = 0.30 rFkk log 3 =

0.48] viuk mÙkj fudVre iw.kZ la[;k esa nhft,A

Q.5 tc lYQj (S8 dh voLFkk esa) dks lkE; ij rki T ij

xeZ fd;k tkrk gS] rks S8 dk nkc 1.0 atm ls 30%

de gks tkrk gS] D;ksfd vkaf'kd #i ls S8(g), S2(g) esa

ifjofrZr gks tkrk gSA bl vfHkfØ;k ds fy, Kp dk

eku crkb,A viuk mÙkj fudVre iw.kZ la[;k esa

nhft,A



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. + 5 marks will be given for each correct answerand – 2 mark for each wrong answer. Q.1 If A = [aij]n×n where aij= i100 + j100, then

∞→nlim

101

n

1iii

n

a∑= equals -

(A) 501 (B)

1011 (C)

1012 (D)

1013

Q.2 Let f(x) be a continuous function on [0, 4]

satisfying f(x) . f(4 – x) = 1. The value of the

definite integral ∫ +

4

0

dx)x(f1

1 equals -

(A) 0 (B) 1 (C) 2 (D) 4 Q.3 Let a and b be positive numbers not equal to 1 and

23π < θ < 2π. If atanθ > btanθ > 1, then which one of

the following is always true ? (A) a > b > 1 (B) a < b < 1 (C) b < a < 1 (D) b > a > 1

[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, + 5 vad fn;s tk;asxs rFkk izR;sd xyr mÙkj ds fy, 2 vad dkVk tk;sxkA

Q.1 ;fn A = [aij]n×n tgk¡ aij= i100 + j100 gS] rc

∞→nlim

101

n

1iii

n

a∑= cjkcj gS -

(A) 501 (B)

1011 (C)

1012 (D)

1013

Q.2 ekuk f(x), [0, 4] esa larr~ Qyu gS tks f(4 – x) = 1 dks larq"V djrk gS] rks fuf'pr lekdyu

∫ +

4

0

dx)x(f1

1 dk eku cjkcj gS -

(A) 0 (B) 1 (C) 2 (D) 4

Q.3 ekuk /kukRed la[;k,¡ a rFkk b, 1 rFkk 2

3π < θ < 2π ds

cjkcj ugha gSA ;fn atanθ > btanθ > 1, rc fuEu esa ls dkSulk lnSo lR; gS –

(A) a > b > 1 (B) a < b < 1 (C) b < a < 1 (D) b > a > 1



Q.4 ;fn lfEeJ la[;k z lehdj.k |z – (1 + i) |2 = 2 dks

larq"V djrh gS rFkk w = z2 gSA rc lfEeJ lery esa

'w' kjk vuqlfjr fcUnqiFk gS - (A) x – y – 1 = 0 (B) x + y – 1= 0 (C) x – y + 1 = 0 (D) x + y + 1 = 0

Q.5 f(x, y) = x2 – 4x + y2 + 6y dk U;wure eku tcfd x ,oa y izfrca/k 0 ≤ x ≤ 1 ,oa 0 ≤ y ≤ 1 ls lEcaf/kr gks] gksxk -

(A) –1 (B) –2 (C) –3 (D) –5

Q.6 lfn'k →a = i + j + 2 k rFkk →

b = i + 2 j + k ds

leryh; rFkk lfn'k →c = i + j + k ds yEcor~ lfn'k

ftldk ifjeki 2 gS] gksxk - (A) – j + k (B) i – k (C) i – j (D) j + k

bl [k.M esa 2 vuqPNsn fn;s x;s gSa] izR;sd esa 3 cgqfodYih iz'u gSaA (iz'u 7 ls 12) 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- 7 ls 9) ,d dkk esa n fo|kFkhZ gS rFkk ifjkk esa n esa ls Bhd k

fo|kFkhZvksa ds mÙkhZ.k gksus dh izkf;drk lh/kh leuqikrh k2 (0 ≤ k ≤ n) ds gSA rc

Q.4 If z is a complex number satisfying the equation |z

– (1 + i) |2 = 2 and w = z2 , then the locus traced by

'w' in the complex plane is - (A) x – y – 1 = 0 (B) x + y – 1= 0 (C) x – y + 1 = 0 (D) x + y + 1 = 0 Q.5 The minimum value of f(x, y) = x2 – 4x + y2 + 6y

when x and y are subjected to the restriction0 ≤ x ≤ 1 and 0 ≤ y ≤ 1, is -

(A) –1 (B) –2 (C) –3 (D) –5 Q.6 A vector of magnitude 2 coplanar with the

vectors →a = i + j + 2 k and →

b = i + 2 j + k

and perpendicular to the vector →c = i + j + k is -

(A) – j + k (B) i – k (C) i – j (D) j + k This section contains 2 paragraphs, each has 3 multiplechoice questions. (Questions 7 to 12) Each question has 4 choices (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.

Passage # 1 (Ques. 7 to 9) There are n students in a class and the probability

that exactly k out of n pass the examination isdirectly proportional to k2 (0 ≤ k ≤ n). Then.



Q.7 ;fn n = 5 gS rFkk Bhd nks fo|kFkhZ;ksa ds ifjkk mÙkh.kZ

djus dh izkf;drk nm gS (m rFkk n ijLij vHkkT;

gS) rc m + n gS - (A) 59 (B) 61 (C) 57 (D) buesa ls dksbZ ugha

Q.8 ;knfPNd :i ls pqus x, ,d fo|kFkhZ ds kjk ijhkk esa mÙkh.kZ gksus dh izkf;drk gS -

(A) 2n1n

++ (B)

)1n2(2)1n(3

++

(C) )2n(2)1n(3

++ (D) buesa ls dksbZ ugha

Q.9 ;fn pquk x;k fo|kFkhZ ijhkk esa mÙkh.kZ ik;k tkrk gS] rc ifjkk esa mÙkh.kZ gksus okyk og vdsyk fo|kFkhZ gksus dh izkf;drk gS -

(A) 2)1n(n

1+

(B) 2)1n(n

2+

(C) 2))1n(n(

4+

(D) none of these

x|ka'k # 2 (iz- 10 ls 12)

js[kk L : 2

1x − = 1y =

21z

−+ ,oa fcUnq A(1, 1, 1) ij

fopkj dhft,A ekuk P, A ls L ij Mkys x;s yEc dk ikn gS ,oa Q, A dk js[kk L ij izfrfcEc gS, "O" ewy fcUnq gSA

Q.7 If n = 5 and the probability that exactly two

students pass the examination is nm (m and n are

relatively prime) then m + n is - (A) 59 (B) 61 (C) 57 (D) none of these Q.8 The probability that a student selected at random

has passed the examination is -

(A) 2n1n

++ (B)

)1n2(2)1n(3

++

(C) )2n(2)1n(3

++ (D) none of these

Q.9 If the selected student has been found to pass theexamination then the probability that he is the onlystudent to have passed the examination is -

(A) 2)1n(n

1+

(B) 2)1n(n

2+

(C) 2))1n(n(

4+

(D) none of these

Passage # 2 (Ques. 10 to 12)

Consider the line L : 2

1x − = 1y =

21z

−+ and a

point A(1, 1, 1). Let P be the foot of theperpendicular from A on L and Q be the image ofthe point A in the line L, "O" being the origin.



Q.10 The distance of the origin from the plane passing through the point A and containing the line L, is -

(A) 31 (B)

31 (C)

32 (D)

21

Q.11 The distance of the point A from the line L, is -

(A) 1 (B) 2 (C) 3 (D) 34

Q.12 The distance of the origin from the point Q, is -

(A) 3 (B) 6

17 (C) 3

17 (D) 3

1

Section - II

This section contains 2 questions (Questions 1, 2). Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S, T) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, A-T; B-Q, B-R; C-P, C-Q and D-S, D-T then the correctly bubbled 4 × 5 matrix should be as follows :

A B C D

P Q R S T

T S

P

P P Q R

R R

Q Q

S S T

T

P Q R S T

Q.10 js[kk L j[kus okys rFkk fcUnq A ls xqtjus okys lery dh ewy fcUnq ls nwjh gksxh -

(A) 31 (B)

31 (C)

32 (D)

21

Q.11 js[kk L ls fcUnq A dh nwjh gS -

(A) 1 (B) 2 (C) 3 (D) 34

Q.12 fcUnq Q ls ewy fcUnq dh nwjh gS -

(A) 3 (B) 6

17 (C) 3

17 (D) 3

1

[k.M - II bl [k.M esa 2 iz'u (iz'u 1, 2) gSaA izR;sd iz'u esa nks LrEHkksa esa dFku fn;s x;s gSa] ftUgsa lqesfyr djuk gSA LrEHk-I (Column I ) esa fn;s x;s dFkuksa (A, B, C, D) dks LrEHk-II (Column II) esa fn;s x;s dFkuksa (P, Q, R, S,T) ls lqesy djuk gSA bu iz'uksa ds mÙkj uhps fn;s x;s mnkgj.k ds vuqlkj mfpr xksyksa dks dkyk djds n'kkZuk gSA ;fn lgh lqesy A-P, A-S, A-T; B-Q, B-R; C-P, C-Q rFkk D-S, D-T gS, rks lgh fof/k ls dkys fd;s x;s xksyksa dk 4 × 5 eSfVªDl uhps n'kkZ;s vuqlkj gksxk :

ABCD

P Q R S T

T S

P

P P Q R

R R

Q Q

S S T

T

P Q R S T



vr% OMR 'khV esa iz'u dh iz'u la[;k ds lek viuk mÙkj [k.M-II esa vafdr dhft;sA izR;sd iw.kZ lgh mÙkj ds fy;s +8 vad fn;s tk;saxs ¼vFkkZr~ izR;sd lgh iafDr feyku ds fy, +2 vad fn, tk,saxs½ o xyr mÙkj ds fy, dksbZ _.kkRed vadu ugha gSA

Q.1 ,d ijyo; y2 = lx (l > 0) ij fopkj dhft, ftldk 'kh"kZ ewy fcUnq 'O' ij gSA

LrEHk–I LrEHk-II

(A) ;fn fn, x, ijoy; ij fLFkr fcUnq P (P) 4l

dks ewy fcUnq ls feyk;k tkrk gS rFkk P ls xqtjus okyh OP ds yEcor~ js[kk vk dks Q ij feyrh gS rc PQ dk vk ij izksi cjkcj gS

(B) ;fn ijoy; ds fdlh LoSPN fcUnq P ij (Q) 2l

[khapk x;k vfHkyEc tks vk dks G ij feyrk gS rks PG ds e/; fcUnq dk fcUnqiFk ,d ijoy; gS ftldk ukfHkyEc gS (C) ,d pj ijoy; dk ukfHkyEc l gS tks fn, (R) l x, ijoy; dks Li'kZ djrk gS] nksuksa oØksa ds vk lekUrj gS rks pj ijoy; ds 'kh"kZ dk fcUnq iFk ,d vU; ijoy; gS ftldk ukfHkyEc gS (D) fn, x, ijoy; ij pj fcUnq P dh (S) 2l ukfHk; f=kT;k ds e/; fcUnq dk fcUnqiFk

,d ijoy; gS ftldk ukfHkyEc gS (T) 3l

Mark your response in OMR sheet against the questionnumber of that question in section-II. + 8 marks will begiven for complete correct answer (i.e. +2 marks for eachcorrect row) and NO NEGATIVE MARKING for wronganswer.

Q.1 Consider a parabola y2 = lx (l > 0) with its vertexat the origin 'O'.

Column –I Column-II

(A) If origin is joined to any point P on the (P) 4l

given parabola and a line perpendicular to OP through P meet the axis at Q, then projection of PQ on the axis is equal to

(B) A normal is drawn at any arbitrary (Q) 2l

point P on the parabola which meets the axis at G. Locus of the middle point of PG is a parabola whose latus rectum is (C) A variable parabola of latus rectum l, (R) l touches the given parabola, the axes of the two curves being parallel. The locus of the vertex of the variable parabola is another parabola whose latus rectum is (D) The locus of the mid point of the focal (S) 2l radii of variable point P moving on the

given parabola, is a parabola whose (T) 3l

latus rectum is



Q.2

LrEHk–I LrEHk-II

(A) ekuk f(x) = 3 x + tan–1x ds kjk (P) ,dSdh

f : R → R ifjHkkf"kr gS, rc f(x) gS

(B) ekuk f(x) = sin3π (sgn (x2 + 3x + 5)) (Q) vUrZksih

ds kjk f : (–∞, ∞) → –1, 0, 1ifjHkkf"kr

gS] rc f(x) gS (tgk¡ sgnx flXue Qyu

dks iznf'kZr djrk gS)

(C) ekuk f(x) = 2xe ds kjk (R) fo"ke

f : [–2, 2] → (0, e2] ifjHkkf"kr gS]

rc f(x) gS

(D) ekuk f(x) = 2xx45 −+ ds kjk (S) vO;qRØe.kh;

f : (–1, 5) → [0, 3] ifjHkkf"kr gS] (T) vukorhZ

rc f(x) gS

Q.2

Column –I Column-II

(A) Let f : R → R be defined as f(x) (P) one-one

= 3 x + tan–1x, then f(x) is

(B) Let f : (–∞, ∞) → –1, 0, 1be defined (Q) into

as f(x) = sin3π (sgn (x2 + 3x + 5)) then

f(x) is (where sgnx denotes signum

function of x.)

(C) Let f : [–2, 2] → (0, e2] be defined (R) odd

as f(x) = 2xe , then f(x) is

(D) Let f : (–1, 5) → [0, 3] be (S) non

defined as f(x) = 2xx45 −+ , then invertible

f(x) is (T) aperiodic



Section - III

This section contains 5 questions (Q.1 to 5).

+3 marks will be given for each correct answer and no

negative marking. The answer to each of the questions

is a SINGLE-DIGIT INTEGER, ranging from 0 to 9.

The appropriate bubbles below the respective question

numbers in the OMR has to be darkened. For example,

if the correct answers to question numbers X, Y, Z and

W (say) are 6, 0, 9 and 2, respectively, then the correct

darkening of bubbles will look like the following :

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

[k.M - III

bl [k.M esa 5 (iz-1 ls 5) iz'u gaSA izR;sd lgh mÙkj ds

fy;s +3 vad fn;s tk,saxs 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 If the value of the definite integral

∫ +−

−1

02

1

1xxxsin dx =

n

2π (where n ∈ N), then find

the value of 12

"n" .

Q.2 Let C be a curve passing through M(2, 2) such

that the slope of the tangent at any point to the

curve is reciprocal of the ordinate of the point. If

the area bounded by curve C and line x = 2 is

expressed as a rational qp (where p and q are in

their lowest form), then find

+ q

4p .

Q.3 If in a ∆ABC 11

as − = 12

bs − = 13

cs − and

λ tan2 2A = 455 then find the value of

231λ .

Q.1 ;fn fuf'pr lekdyu

∫ +−

−1

02

1

1xxxsin dx =

n

2π (tgk¡ n ∈ N) gS] rc 12

"n" dk

eku Kkr dhft,A

Q.2 ekuk oØ C fcUnq M(2, 2) ls bl izdkj xqtjrk gS

fd oØ ds fdlh fcUnq ij Li'kZ js[kk dk <ky fcUnq

dh dksVh dk O;qRØe gS ;fn oØ C rFkk js[kk x = 2

kjk ifjc) ks=kQy ifjes; qp (tgk¡ p rFkk q

U;wure :i esa gS) ds :i esa iznf'kZr fd;k tkrk gS

rks

+ q

4p Kkr djksA

Q.3 ;fn ,d ∆ABC esa 11

as − = 12

bs − = 13

cs − rFkk

λ tan2 2A = 455 gS] rc

231λ dk eku Kkr dhft,A



Q.4 Let ai , i = 1, 2, 3 ..., n denotes the integers in the

domain of function f(x) =

−−21x25x4log 2/1

where ai < ai+1 ∀ i ∈ N. If the line

L : 4

ax2 1− = 2

1

aay + =

5

3

aaz − meets the xy, yz

and zx planes at A, B and C respectively, and if

volume of the tetrahedron OABD is V, where 'O',

is the origin and D is the image of C in the x-axis,

then find the value of 7V9 .

Q.5 Let S denote sum of the series

32

3 + 3.2

44

+ 3.2

56

+ 5.2

67

+ .... ∞

compute the value of S–1.

Q.4 ekuk ai , i = 1, 2, 3 ..., n Qyu

f(x) =

−−21x25x4log 2/1 ds izkUr esa iw.kkZadksa dks

iznf'kZr djrk gS tgk¡ ai < ai+1 ∀ i ∈ N ;fn js[kk

L : 4

ax2 1− = 2

1

aay + =

5

3

aaz − leryksa xy, yz

rFkk zx dks Øe'k% A, B rFkk C ij feyrh gS rFkk

;fn prq"Qyd OABD dk vk;ru V gS tgk¡ 'O'

ewy fcUnq rFkk D fcUnq C dk x-vk esa izfrfcEc gS

rks 7V9 dk eku Kkr djksA

Q.5 ekuk Js.kh

32

3 + 3.2

44

+ 3.2

56

+ 5.2

67

+ .... ∞

dk ;ksx S ls iznf'kZr djrs gS] rc S–1 dk eku

gksxkA



PHYSICS

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. + 5 marks will be given for each correct answerand – 2 mark for each wrong answer.

Q.1 As shown the time taken by projectile to reachfrom A to B is t. Then the distance AB is equal to-

60º30º

u • B

A

(A) 3

ut (B) 2ut3 (C) ut3 (D) 2ut

Q.2 As shown the acceleration of M with respect toground is 2 m/s2 , then (use θ = 37º)

m

θ M

(A) Acceleration of m w.r.t. M is 5 m/s2

(B) Acceleration of m w.r.t. ground is 5 m/s2

(C) Acceleration of m w.r.t. M is 2 m/s2

(D) Acceleration of m w.r.t. ground is 10 m/s2

[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, + 5 vad fn;s tk;asxs rFkk izR;sd xyr mÙkj ds fy, 2 vad dkVk tk;sxkA

Q.1 n'kkZ, vuqlkj izksI; kjk A ls B rd igq¡pus esa fy;k x;k le; t gSA rc nwjh AB cjkcj gS -

60º30º

u• B

A

(A) 3

ut (B) 2ut3 (C) ut3 (D) 2ut

Q.2 n'kkZ, vuqlkj /kjkry ds lkisk M dk Roj.k 2 m/s2

gS, rc (θ = 37º mi;ksx djsa)

m

θ M

(A) M ds lkisk m dk Roj.k 5 m/s2 gS

(B) /kjkry ds lkisk m dk Roj.k 5 m/s2 gS

(C) M ds lkisk m dk Roj.k 2 m/s2 gS

(D) /kjkry ds lkisk m dk Roj.k 10 m/s2 gS



Q.3 4l yEckbZ rFkk 4m nzO;eku dh ,d iryh NM+ n'kkZ, vuqlkj eksM+h tkrh gSA rc O ls xqtjus okyh rFkk dkxt ds ry ds yEcor~ vk ds lkisk NM+ dk tM+Ro vk?kw.kZ gS -

90º 90º

l l l l

O

(A) 3

m 2l (B) 3m10 2l

(C) 12

m 2l (D) 24

m 2l

Q.4 ifjiFk ds fy, A rFkk C ds chp rqY; izfrjks/k gS - r r

r r

rrr

r

C

D

B

A

r

(A) r1112 (B) r

1113 (C) r

1114 (D) r

1115

Q.5 R izfrjks/k ds ,d fLFkj ywi ls lEcfU/kr pqEcdh; ¶yDl T vkorZdky ds nkSjku le; ds lkFk fuEukuqlkj ifjofrZr gksrk gS &

φ = at(T – t) rc T le; ds nkSjku ywi esa mRiUu m"ek dh ek=kk gS

(ekuk dq.Myh dk izsjdRo ux.; gS)

(A) R3

aT3 (B)

R3Ta 22

(C)RTa 32

(D) R3Ta 32

Q.3 A thin rod of length 4l and mass 4m is bent asshown. Then the moment of inertia of rod about theaxis passing through O and perpendicular to theplane of paper is -

90º 90º

l l l l

O

(A) 3

m 2l (B) 3m10 2l

(C) 12

m 2l (D) 24

m 2l

Q.4 For the circuit equivalent resistance between A andC is -

r r

r r

rrr

r

C

D

B

A

r

(A) r

1112 (B) r

1113 (C) r

1114 (D) r

1115

Q.5 Magnetic flux linked with a stationary loop ofresistance R varies with time during time period Tas follows

φ = at(T – t) then the amount of heat generated in the loop

during time T is (assume inductance of coil isnegligible)

(A) R3

aT3 (B)

R3Ta 22

(C)RTa 32

(D) R3Ta 32



Q.6 A ek/;e esa xfreku ,d izdk'k fdj.k A rFkk B nks ek/;eksa dh lery vUrZlrg ij vkifrr gksrh gS rFkk B ek/;e esa viofrZr gksrh gSA vkiru dks.k i gS rFkk viorZu dks.k r gSA sin i rFkk sin r ds chp vkjs[k n'kkZ;k x;k gS] rc -

53º

sin i

sin r (A) B ek/;e esa izdk'k dh pky] A ek/;e dh rqyuk

esa th

43

gS

(B) ifjlhek ij iw.kZ vkUrfjd ijkorZu gksrk gS

(C) µA > µB (D) mijksDr lHkh bl [k.M esa 2 vuqPNsn fn;s x;s gSa] izR;sd esa 3 cgqfodYih iz'u gSaA (iz'u 7 ls 12) 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- 7 ls 9)

,d vkn'kZ fijek.kqd xSl izlkfjr gksrh gS rkfd xSl

dks LFkkukUrfjr m"ek dh ek=kk bldh vkUrfjd ÅtkZ

esa deh ds cjkcj gksrh gSA

Q.6 A ray of light travelling in medium A, is incidenton plane interface of two mediums A and B andgets refracted into medium B. The angle ofincidence is i and angle of refraction is r. The graph between sin i and sin r is shown, then -

53º

sin i

sin r

(A) speed of light in medium B is th

43

of that in

medium A (B) total internal reflection take place at boundary (C) µA > µB (D) All of above This section contains 2 paragraphs, each has 3 multiplechoice questions. (Questions 7 to 12) 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 be given for each correct answer and – 1 mark for eachwrong answer. Passage # 1 (Ques. 7 to 9) An ideal diatomic gas is expanded so that the

amount of heat transferred to the gas is equal to thedecrease in its internal energy.



Q.7 The molar specific heat of the gas in this process is

given by C whose value is -

(A) – 2R5 (B) –

2R3

(C) 2 R (D) 2R5

Q.8 The process can be represented by the equation

TVn = constant, where the value of n is -

(A) n = 57 (B)

51n =

(C) n = 23 (D) n =

53

Q.9 If in the above process the initial temperature of the

gas be T0 and the final volume be 32 times the

initial volume, the work done (in joules) by the gas

during the process will be -

(A) RT0 (B) 2

RT5 0

(C) 2RT0 (D) 2

RT0

Q.7 bl izØe esa xSl dh eksyj fof'k"V m"ek C kjk nh

tkrh gS ftldk eku gS -

(A) – 2R5 (B) –

2R3

(C) 2 R (D) 2R5

Q.8 izØe dks lehdj.k TVn = fu;rkad kjk iznf'kZr fd;k

tk ldrk gS] tgk¡ n dk eku gS -

(A) n = 57 (B)

51n =

(C) n = 23 (D) n =

53

Q.9 ;fn mijksDr izØe esa] xSl dk izkjfEHkd rkieku T0 gS

rFkk vfUre vk;ru izkjfEHkd vk;ru dk 32 xquk gS]

izØe ds nkSjku xSl kjk lEiUu dk;Z (twy esa)

gksxk -

(A) RT0 (B) 2

RT5 0

(C) 2RT0 (D) 2

RT0



x|ka'k # 2 (iz- 10 ls 12)

,d T;k rjax x-vk ds vuqfn'k [khaph gqbZ ,d Mksjh

esa _.kkRed x-fn'kk esa lapfjr gksrh gSA x = 2 cm ij

Mksjh dk ,d d.k bldh ek/; fLFkfr ij ik;k tkrk

gS rFkk ;g t = 1 sec ij /kukRed y-fn'kk esa xfr dj

jgk gSA rjax dk vk;ke] rjax-yEckbZ rFkk rjax dh

dks.kh; vkofÙk Øe'k% 0.1 m, 4π m rFkk 4π rad/s gSA

Q.10 rjax dh lehdj.k gS -

(A) y = 0.1 sin (4π (t – 1) + 8 (x – 2)) (B) y = 0.1 sin ((t – 1) – (x – 2)) (C) y = 0.1 sin (4π (t – 1) – 8 (x – 2))

(D) buesa ls dksbZ ugha

Q.11 x = 2 m rFkk t = 1sec ij d.k dh pky gS -

(A) 0.2 π m/s (B) 0.6 π m/s (C) 0.4 π m/s (D) 0

Q.12 x = 2m rFkk t = 1.125s ls LFkkukUrfjr rkRkf.kd

'kfDr gS -

(A) 10 J/s (B) 3

4π J/s

(C) 3

2π J/s (D) 0

Passage # 2 (Ques. 10 to 12) A sinusoidal wave is propagating in negative

x-direction in a string stretched along x-axis. Aparticle of string at x = 2 cm is found at its meanposition and it is moving in positive y-direction at t= 1s. The amplitude of the wave, the wave-lengthand the angular frequency of the wave are 0.1 m,

4π m and 4π rad/s, respectively.

Q.10 The equation of the wave is - (A) y = 0.1 sin (4π (t – 1) + 8 (x – 2)) (B) y = 0.1 sin ((t – 1) – (x – 2)) (C) y = 0.1 sin (4π (t – 1) – 8 (x – 2)) (D) None of these

Q.11 The speed of particle at x = 2 m and t = 1sec is - (A) 0.2 π m/s (B) 0.6 π m/s (C) 0.4 π m/s (D) 0

Q.12 The instantaneous power transfer through x = 2mand t = 1.125s is -

(A) 10 J/s (B) 3

4π J/s

(C) 3

2π J/s (D) 0



[k.M - II

bl [k.M esa 2 iz'u (iz'u 1, 2) gSaA izR;sd iz'u esa nks LrEHkksa esa

dFku fn;s x;s gSa] ftUgsa lqesfyr djuk gSA LrEHk-I (Column

I ) esa fn;s x;s dFkuksa (A, B, C, D) dks LrEHk-II (Column II) esa

fn;s x;s dFkuksa (P, Q, R, S,T) ls lqesy djuk gSA bu iz'uksa

ds mÙkj uhps fn;s x;s mnkgj.k ds vuqlkj mfpr xksyksa dks

dkyk djds n'kkZuk gSA ;fn lgh lqesy A-P, A-S, A-T; B-Q,

B-R; C-P, C-Q rFkk D-S, D-T gS, rks lgh fof/k ls dkys fd;s

x;s xksyksa dk 4 × 5 eSfVªDl uhps n'kkZ;s vuqlkj gksxk :

ABCD

P Q R S T

T S

P

P P Q R

R R

Q Q

S S T

T

P Q R S T

vr% OMR 'khV esa iz'u dh iz'u la[;k ds lek viuk mÙkj

[k.M-II esa vafdr dhft;sA izR;sd iw.kZ lgh mÙkj ds fy;s +8 vad

fn;s tk;saxs ¼vFkkZr~ izR;sd lgh iafDr feyku ds fy, +2 vad

fn, tk,saxs½ o xyr mÙkj ds fy, dksbZ _.kkRed vadu ugha gSA

Section - II

This section contains 2 questions (Questions 1, 2).

Each question contains statements given in two

columns which have to be matched. Statements (A, B,

C, D) in Column I have to be matched with

statements (P, Q, R, S, T) in Column II. The answers

to these questions have to be appropriately bubbled

as illustrated in the following example. If the correct

matches are A-P, A-S, A-T; B-Q, B-R; C-P, C-Q and

D-S, D-T then the correctly bubbled 4 × 5 matrix

should be as follows :

A B C D

P Q R S T

T S

P

P P Q R

R R

Q Q

S S T

T

P Q R S T

Mark your response in OMR sheet against the

question number of that question in section-II. + 8

marks will be given for complete correct answer (i.e.

+2 marks for each correct row) and NO NEGATIVE

MARKING for wrong answer.



Q.1 For the photoelectric experiment, match the

following column-I with column-II if frequency and intensity of incident photon are f and I while work function of metal is φ.

Column -I Column-II (A) If frequency is increased (P) Stopping potential keeping intensity and increases work function constant (B) If I is increased keeping (Q) Saturation f and φ are constant photocurrent increases (C) If φ is decreased keeping (R) Maximum kinetic f and I are constant energy of the photoelectrons increases (D) If the distance between (S) Stopping potential anode and cathode remains the same increases (T) Maximum kinetic energy of the photoelectrons decreases

Q.1 izdk'k fo|qr iz;ksx ds fy,] fuEufyf[kr LrEHk-I dk

LrEHk-II ds lkFk feyku dhft,A ;fn vkifrr

QksVksu dh vkofÙk rFkk rhozrk] f rFkk I gS tcfd

/kkrq dk dk;ZQyu φ gSA

LrEHk-I LrEHk-II

(A) ;fn rhozrk rFkk dk;ZQyu (P) fujks/kh foHko c<+rk gS

dks fu;r j[krs gq, vkofÙk

c<+k;h tkrh gSA

(B) ;fn f rFkk φ dks fu;r j[krs (Q) larIr izdk'k /kkjk

gq, I dks c<+k;k tkrk gS c<+rh gS

(C) ;fn f rFkk I dks fu;r j[krs (R) QksVksbysDVªkWuksa dh

gq, φ dks ?kVk;k tkrk gS vf/kdre xfrt ÅtkZ

c<+rh gS

(D) ;fn ,uksM rFkk dSFkksM+ ds (S) fujks/kh foHko leku

chp nwjh c<+rh gS jgrk gS

(T) QksVksbysDVªkWuksa dh

vf/kdre xfrt ÅtkZ

?kVrh gS



Q.2 For the following circuit match the following

columns at steady state.

A

C

B D

R = 3Ω X = 2Ω

Y = 4ΩS = 7Ω

3µF6V

Column -I Column-II

(A) Potential difference (P) zero

(in volts) across A and D

(B) Potential difference (Q) 59

(in volts) across capacitor

(C) Value of Y(in Ω) for (R) 51

which no energy is stored

in capacitor

(D) Steady state current (S) 14

(in amp) in the branch

containing capacitor (T) 3

14

Q.2 fuEufyf[kr ifjiFk ds fy, LFkk;h voLFkk ij

fuEufyf[kr LrEHkksa dk feyku dhft,

A

C

B D

R = 3Ω X = 2Ω

Y = 4ΩS = 7Ω

3µF6V

LrEHk-I LrEHk-II

(A) A rFkk D ds fljksa ij (P) 'kwU;

foHkokUrj (oksYV esa) gS

(B) la/kkfj=k ds fljksa ij foHkokUrj (Q) 59

(oksYV esa) gS

(C) Y(Ω esa) dk eku ftlds fy, (R) 51

la/kkfj=k esa dksbZ ÅtkZ lafpr

ugha gksrh (D) la/kkfj=k dks j[kus okyh (S) 14

'kk[kk esa LFkk;h voLFkk

/kkjk (,Eih;j esa) gS (T) 3

14



Section - III

This section contains 5 questions (Q.1 to 5).

+3 marks will be given for each correct answer and

no negative marking. The answer to each of the

questions is a SINGLE-DIGIT INTEGER, ranging

from 0 to 9. The appropriate bubbles below the

respective question numbers in the OMR has to be

darkened. For example, if the correct answers to

question numbers X, Y, Z and W (say) are 6, 0, 9 and

2, respectively, then the correct darkening of bubbles

will look like the following :

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

[k.M - III

bl [k.M esa 5 (iz-1 ls 5) iz'u gaSA izR;sd lgh mÙkj ds

fy;s +3 vad fn;s tk,saxs 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 An ideal gas follows the process T = T0 + αV2,

where T0 and α are positive constants and V is

volume of one mole of gas. If this process the

minimum pressure attained by gas is βR 0Tα

then find the value of β.

Q.2 A travelling wave is given by

)4t12xt12x3(8.0y 22 +++

= where x and y are m

and t is in sec then find the wave velocity in

m/s.

Q.3 Light traveling in air falls at an incidence angle

of 2º at one refracting surface of prism of

refractive index 1.5 and refracting angle 4º. The

medium on the other side of prism is water of

refractive index 34 . Find the deviation produced

by prism in degrees.

Q.1 ,d vkn'kZ xSl izØe T = T0 + αV2 dk vuqlj.k

djrk gS] tgk¡ T0 rFkk α /kukRed fu;rkad gS rFkk

V xSl ds ,d eksy dk vk;ru gSA ;fn ;g izØe

xSl kjk izkIr U;wure nkc βR 0Tα gS rc

β dk eku Kkr dhft,A

Q.2 ,d izxkeh rjax )4t12xt12x3(

8.0y 22 +++= kjk

nh tkrh gS tgk¡ x rFkk y ehVj esa gSa rFkk t

lsd.M esa gS rc rjax osx m/s esa Kkr dhft,A

Q.3 ok;q esa xfreku izdk'k 1.5 viorZukad rFkk 4º

viorZu dks.k ds fizTe dh ,d viorZd lrg ij

2º ds vkiru dks.k ij fxjrk gSA fizTe ds nwljh

vksj ij ek/;e ty gS ftldk viorZukad 34 gSA

fizTe kjk mRiUu fopyu fMxzh esa Kkr dhft,A



Q.4 A hydrogen atom (mass = 1.66 × 10–27 kg and

ionization potential 13.6 eV) moving with a

velocity 6.24 × 104 m/s makes a completely

inelastic head on collision with another

stationary hydrogen atom. Both atoms are in

ground state before collision, up to what state

either one atom may be excited ?

Q.5 Three point charges of q = 0.1 C each are placed

at the corners of an equilateral triangle of side

L = 1m. If this system is supplied energy at the

constant rate of P = 10 kW then how much time

in hours will be required to move one of the

charge on to the mid-point of the line joining

the other two fixed charges.

Q.4 6.24 × 104 m/s osx ls xfreku ,d gkbMªkstu

ijek.kq (nzO;eku = 1.66 × 10–27 kg rFkk vk;uu

foHko 13.6 eV) nwljs fLFkj gkbMªkstu ijek.kq ls

,d iw.kZr% izR;kLFk lEeq[k VDdj djrk gSA

VDdj ls igys nksuksa ijek.kq vk| voLFkk esa gSA

izR;sd ijek.kq fdl voLFkk rd mRrsftr gks

ldrk gS ?

Q.5 q = 0.1 C ds rhu fcUnq vkos'kksa dks L = 1m Hkqtk ds ,d leckgw f=kHkqt ds dksuksa ij j[kk x;k gSA ;fn bl fudk; dks P = 10 kW dh fu;r nj ij ÅtkZ lIykbZ dh tkrh gS rks fdlh ,d vkos'k dks vU; nks fLFkj vkos'kksa dks tksM+us okyh js[kk ds e/; fcUnq rd xfr djkus esa fdruk le; (?k.Vs esa) yxsxkA



Date : 30/3/2011Time : 3 : 00 Hrs. MAX MARKS: 237

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, 5 vad fn;s tk;saxs o izR;sd xyr mÙkj ds fy, 2 vad ?kVk;k tk,xkA

5. x|ka'k ij vk/kkfjr cgqfodYih izdkj ds iz'u ftuesa ls dsoy ,d fodYi lgh gSA izR;sd lgh mÙkj ds fy, 3 vad fn, tk,saxsa rFkk

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

[k.M – II

6. LrEHkksa dks lqesfyr djus okys iz'u (4 × 5 izdkj ds) gSaA iw.kZ :Ik ls lgh lqesfyr mÙkj ds fy, 8 vad fn;s tk;saxs rFkk xyr mÙkj ds fy, dksbZ _.kkRed vadu ugha gSA fdUrq] fdlh iafDr esa lgh :Ik ls fpfUgr mÙkj ds fy, 2 vad fn;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 ,d iw.kk±d esa nhft,A tSls (0 ls 9)

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. xksyksa 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 xksyksa 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


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

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