Time : 3 : 00 Hrs. MAX MARKS: 240 Name : Roll No. : Date · CAREER POINT, CP Tower, Road No.1,...

1

TARGET – IIT JEE

CHEMISTRY, MATHEMATICS & PHYSICS

Time : 3 : 00 Hrs. MAX MARKS: 240

Name : _____________________________________ Roll No. : __________________________ Date : _____________

INSTRUCTIONS TO CANDIDATE

A. GENERAL :

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

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

B. MARKING SCHEME :

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

each wrong answer. 5. Multiple choice questions with multiple correct option. 4 marks will be awarded for each correct answer and –1 mark for

each wrong answer. 6. Passage based single correct type questions. 4 marks will be awarded for each correct answer and –1 mark for each wrong answer.

Section - II 7. Column Matching type questions (4 × 5 type). 8 marks will be awarded for the complete correctly matched answer (i.e. +2

marks for each correctly matched row) and No Negative marking for wrong answer.

C. FILLING THE OMR :

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

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

Section –I Section-II Section-III

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

A B C D E

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

A B C D E

the wrong method for filling the bubble are

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

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

P Q R S TA BCD

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

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

SEA

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

Space for rough work

CHEMISTRY

Section – I

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

Q.1

At what temperature will helium atoms have the same root mean square speed as hydrogen molecules have at 300 K ?

(A) 300 K (B) 400 K (C) 500 K (D) 600 K

Q.2 The osmotic pressure of a solution of a synthetic polyisobutylene in benzene was determined at 25ºC. A sample containing 0.20 g of solute/100cm3 of solution developed a rise of 2.4 mm at osmotic equilibrium. The density of the solution was 0.88 g/cm3. The molecular weight of the polyisobutylene is -

(A) 2.4 × 105 (B) 3.6 × 104

(C) 4.8 × 103 (D) 6.4 × 102

[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 ?kVk;k tk;sxkA

Q.1

fdl rki ij ghfy;e ijek.kq dh oxZ ek/; ewy pky

300 K ij gkbMªkstu v.kqvksa ds leku gS ?

(A) 300 K (B) 400 K (C) 500 K (D) 600 K

Q.2 csUthu esa la'ysf"kr ikWyhvkblksC;wfVyhu ds ,d

foy;u ds ijklj.k nkc dks 25ºC ij fu/kkZfjr fd;k

x;kA 0.20 g foys;/100cm3 foy;u ;qDr uewus esa

ijklj.k lkE; ij 2.4 mm dh of) gksrh gSA foy;u

dk ?kuRo 0.88 g/cm3 FkkA ikWyhvkblksC;wfVyhu dk

v.kqHkkj gS

(A) 2.4 × 105 (B) 3.6 × 104

(C) 4.8 × 103 (D) 6.4 × 102



Q.3 A lead bullet weighing 18.0 gram and travelling at 500 m/s is embedded in a wooden block weighing

1.00 kg. If both the block and the bullet were

initially at 25ºC, what is the final temperature of the

block containing the bullet ? Assume no heat loss

to the surrounding. (Heat capacity of wood, 0.500

kcal/Kg.K : of lead, 0.030 kcal/Kg.K)

(A) 20.3ºC (B) 26.1ºC

(C) 39.2ºC (D) 75.4ºC

Q.4 Calculate the equilibrium constant at 25ºC for the

disproportionation of 3 mol aqueous HNO2 to yield

gaseous NO and aqueous −3NO . The standard

potential for the reduction of HNO2 to NO is 0.99 V ;

that for the reduction of −3NO to HNO2 is 0.94 V.

(A) 1 (B) 2 (C) 3 (D) 4

Q.5 In compound A(C30H60O) following tests are

observed negatively, A can be : Br2/H2O

–Ve

–Ve

–Ve Na metal

2, 4 DNPC30H60O (A)

Q.3 18.0 xzke Hkkj okyh ,d lhls dh xksyh gS rFkk

500 m/s ls pyrh gqbZ 1.00 kg Hkkj okys ydM+h ds

,d CykWd esa /k¡l tkrh gSA ;fn CykWd o xksyh izkjEHk

esa 25ºC ij Fks rks xksyh ;qDr CykWd dk vfUre rki

D;k gS\ ;g ekurs gq, fd ikfjik'oZ dks Å"ek dh

dksbZ gkfu ugha gksrh gSA (ydM+h dh Å"ek /kkfjrk]

0.500 kcal/Kg.K : lhls dh, 0.030 kcal/Kg.K)

(A) 20.3ºC (B) 26.1ºC (C) 39.2ºC (D) 75.4ºC

Q.4 25ºC ij 3 eksy tyh; HNO2 ds xSlh; NO o tyh; −3NO esa yfC/k ds fo"kekuqikru ds fy, lkE;

fu;rkad dh x.kuk dhft,A HNO2 ds NO esa

vip;u dk ekud foHko 0.99V gS] −3NO ds HNO2 esa

vip;u ds fy, 0.94 VgS

(A) 1 (B) 2 (C) 3 (D) 4 Q.5 ;kSfxd A(C30H60O) esa fuEu _.kkRed ijhk.k izsfkr

gq,] A gks ldrk gS : Br2/H2O

–Ve

–Ve

–VeNa /kkrq

2, 4 DNPC30H60O(A)



(A) an unsaturated ether (B) an epoxide

(C) a cyclic ketone (D) a cycloalkanol

Q.6 Consider the following reactions

C = C C2H5

HH

H5C2

C = CC2H5

HH5C2

H

H5C2 – C – CH2 – C2H5

O

2CH3 – CH2 – COOH

H5C2 – C ≡ C – C2H5

R1 R2

R3

R4

The correct set of reagents for these reactions is

R1 R2 R3 R4 (A) H2/Lindlar

catalyst Na/liq. NH3

(i) O3, (ii) H2O

H2O, H2SO4, HgSO4

(B) H2/Lindlar catalyst

Na/liq. NH3

H2O, H2SO4, HgSO4

(i) O3, (ii) H2O

(C) (i) O3, (ii) H2O

H2O, H2SO4, HgSO4

Na/liq. NH3

H2/Lindlar catalyst

(D) H2O, H2SO4, HgSO4

H2/Lindlar catalyst

(i) O3, (ii) H2O

Na/liq. NH3

(A) ,d vlarIr bZFkj (B) ,d biksDlkbM (C) ,d pØh; dhVksu (D) ,d lkbDyks,YdsukWy

Q.6 fuEu vfHkfØ;kvksa ij fopkj dhft,

C = CC2H5

HH

H5C2

C = CC2H5

HH5C2

H

H5C2 – C – CH2 – C2H5

O2CH3 – CH2 – COOH

H5C2 – C ≡ C – C2H5

R1 R2

R3

R4

bu vfHkfØ;kvksa ds fy, vfHkdeZdksa dk lgh leqPp; gS

R1 R2 R3 R4 (A) H2/fy.Mykj

mRiszjd Na/liq. NH3

(i) O3, (ii) H2O

H2O, H2SO4, HgSO4

(B) H2/ fy.Mykj mRiszjd

Na/liq. NH3

H2O, H2SO4, HgSO4

(i) O3, (ii) H2O

(C) (i) O3, (ii) H2O

H2O, H2SO4, HgSO4

Na/liq. NH3

H2/ fy.Mykj mRiszjd

(D) H2O, H2SO4, HgSO4

H2/ fy.Mykj mRiszjd

(i) O3, (ii) H2O

Na/liq. NH3



Q.7 Fastest rate of electrophilic addition will take placein -

(A) HO CH = CH2

(B) O2N CH = CH2

(C) CH3 CH = CH2

(D) CH = CH2

Q.8 Consider the following reaction

CH3 – CH – CH – CH3

D CH3 +

+ •Br → ‘X’+ HBr, X can

be -

(A) CH3 – CH – CH – CH2

D CH3

(B) CH3 – CH – C – CH3

D CH3

(C) CH3 – C – CH – CH3

D CH3

(D) CH3 – CH – CH – CH3

CH3

Q.7 fdl esa bysDVªkWuLusgh ;ksx dh nj rhoz gksxh -

(A) HO CH = CH2

(B) O2N CH = CH2

(C) CH3 CH = CH2

(D) CH = CH2

Q.8 fuEu vfHkfØ;k ij fopkj dhft,

CH3 – CH – CH – CH3

D CH3+

+ •Br → ‘X’+ HBr, X gksxk

(A) CH3 – CH – CH – CH2

D CH3

(B) CH3 – CH – C – CH3

D CH3

(C) CH3 – C – CH – CH3

D CH3

(D) CH3 – CH – CH – CH3

CH3



iz'u 9 ls 12 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 dslek vius mÙkj vafdr dhft,A izR;sd lgh mÙkj ds fy,+ 4 vad fn;s tk;saxs rFkk izR;sd xyr mÙkj ds fy, 1 vad ?kVk;k tk;sxkA

Q.9 fuEu esa ls dkSuls ewyd 0.30 M HCl foy;u esa blesa

H2S izokfgr djus ij voksfir gksaxs?

(A) Cu2+ (B) Sb3+ (C) Cd2+ (D) As3+

Q.10 veksfu;e dkcsZesV ds fo;kstu dks fuEu lehdj.k

kjk iznf'kZr dj ldrs gS, NH4COONH2(s) 2NH3(g) + CO2(g)

vxz vfHkfØ;k ds fy, ∆Hº _.kkRed gSA lkE; nka;s

ls cka;h vksj f'k¶V gksxk ;fn

(A) nkc esa deh gks

(B) rki esa of) gks

(C) veksfu;k dh lkUnzrk esa of) gks (D) CO2 dh lkUnzrk esa of) gks

Questions 9 to 12 are multiple choice questions. Eachquestion has four choices (A), (B), (C) and (D), out ofwhich MULTIPLE (ONE OR MORE) is correct. Markyour response in OMR sheet against the questionnumber of that question. + 4 marks will be given for each correct answer and – 1 mark for each wrong answer.

Q.9 Which of the following radicals in a 0.30 M HClsolution will be precipitated by passing H2S through it ?

(A) Cu2+ (B) Sb3+ (C) Cd2+ (D) As3+

Q.10 The dissociation of ammonium carbamate may be

represented by the equation,

NH4COONH2(s) 2NH3(g) + CO2(g)

∆Hº for the forward reaction is negative. The

equilibrium will shift from right to left, if there is -

(A) a decrease in pressure

(B) an increase in temperature

(C) an increase in the concentration of ammonia (D) an increase in the concentration of CO2



Q.11 In which of the following compounds first is moreacidic than second :

(A)

COOH

CH3

,

COOH

CH3

(B)

OH

NO2

,

OH NO2

(C)

OH

, H2CO3

(D)

COOH

OH

,

COOH

OH

Q.12 In which of the following pairs the first one is thestronger base than second -

(A) CH3COOΘ, HCOOΘ (B) HOΘ, Θ

2NH

(C) CH2 = ΘΘ

≡− :: CCH,CH (D) CH3NH2, CH3OH

Q.11 fuEu ;kSfxdksa esa ls fdlesa izFke] frh; ls vf/kd

vEyh; gS :

(A)

COOH

CH3

,

COOH

CH3

(B)

OH

NO2

,

OHNO2

(C)

OH

, H2CO3

(D)

COOH

OH

,

COOH

OH

Q.12 fuEu ;qXeksa esa ls fdlesa izFke] frh; ls izcy kkj gS-

(A) CH3COOΘ, HCOOΘ

(B) HOΘ, Θ2NH

(C) CH2 = ΘΘ

≡− :: CCH,CH

(D) CH3NH2, CH3OH



This section contains 2 paragraphs; each has 3 multiple choice questions. (Questions 13 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. + 4 marks will be given for each correct answer and – 1 mark for each wrong answer.

Passage # 1 (Ques. 13 to 15) For a general reaction given, below the value of

solubility product can be given as : AxBy xAy+ + yBx– a 0 0 a – s xs ys Ksp

= (xs)x. (ys)y or

Ksp = xxyy(s)x + y Solubility product gives us not only an idea about

the solubility of an electrolyte in a solvent but also helps in explaining concept of precipitation and calculation of H+ ion, OH–, ion. It is also useful in qualitative analysis for the identification and separation of basic radicals.

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

iz'u gSaA (iz'u 11 ls 16) izR;sd iz'u ds pkj fodYi (A), (B), (C) rFkk (D) gSa, ftuesa ls ,d ;k ,d ls vf/kd fodYi lgh

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

vafdr dhft;sA izR;sd lgh mÙkj ds fy;s + 5 vad fn;s

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

x|ka'k # 1 (iz. 13 ls 15)

uhps nh xbZ ,d lkekU; vfHkfØ;k ds fy, foys;rk

xq.kuQy ds eku dks fuEu izdkj fn;k tk ldrk gS : AxBy xAy+ + yBx– a 0 0 a – s xs ys Ksp

= (xs)x. (ys)y or

Ksp = xxyy(s)x + y foys;rk xq.kuQy gesa u dsoy ,d foyk;d esa

fo|qrvi?kV~; dh foys;rk ds ckjs esa tkudkjh nsrk

gS cfYd voksi.k dh vo/kkj.kk dk le>kus o H+

vk;u, OH– vk;u dh x.kuk djus esa lgk;d gSA ;g

kkjh; ewydksa ds iFkDdj.k rFkk igpku ds fy,

xq.kkRed fo'ys"k.k esa Hkh mi;ksxh gSA.



Q.13 K2CrO4 is slowly added to a solution containing 0.20 M AgNO3 and 0.20 M Ba(NO3)2. Describe what happens, if the Ksp

for Ag2CrO4 is 1.1 × 10–12 and the Ksp of BaCrO4 is 1.2 × 10–10.

(A) The Ag2CrO4 precipitates first out of solution and then BaCrO4 precipitates

(B) The BaCrO4 precipitates first out of solution

(C) Both Ag2CrO4 and BaCrO4 precipitate simultaneously out of solution

(D) Neither Ag2CrO4 nor BaCrO4 precipitates out of solution

Q.14 What is the molar solubility of Cu(OH)2 in

1.0 M NH3, if the deep blue complex ion, [Cu(NH3)4]2+ forms ? The Ksp for Cu(OH)2 is 1.6 × 10–19 and the Kf for [Cu(NH3)4]2+ is 1.1 × 1013.

(A) 7.1 × 10–4 M (B) 7.6 × 10–3 M (C) 6.7 × 10–3 M (D) 5.6 × 10–3 M

Q.13 K2CrO4 dks /khjs&/khjs 0.20 M AgNO3 o 0.20 M

Ba(NO3)2 ;qDr foy;u esa feyk;k tkrk gSA ;fn

Ag2CrO4 ds fy, Ksp, 1.1 × 10–12 o BaCrO4 ds

fy, Ksp, 1.2 × 10–10 gks] rks D;k gksxk

(A) Ag2CrO4 foy;u ls igys voksfir ckn esa

BaCrO4 voksfir gksrk gS

(B) BaCrO4 foy;u ls igys voksfir gksrk gS

(C) Ag2CrO4 o BaCrO4 nksuksa foy;u ls

lkFk&lkFk voksfir gksrs gS (D) u rks Ag2CrO4 uk gh BaCrO4 foy;u ls

voksfir gksrs gS

Q.14 Cu(OH)2 dh eksyj foy;srk 1.0 M NH3 esa D;k gS

;fn xgjk uhyk ladqy vk;u [Cu(NH3)4]2+ curk

gS\ Cu(OH)2 dk Ksp = 1.6 × 10–19 rFkk

[Cu(NH3)4]2+ dk Kf = 1.1 × 1013 gS

(A) 7.1 × 10–4 M (B) 7.6 × 10–3 M (C) 6.7 × 10–3 M (D) 5.6 × 10–3 M



Q.15 The solubility product of AgI (mol.wt. = 235) in water is 4.9 × 10–11 at a given temperature. The solubility of AgI in 0.001 M KI solution (in gm/litre)is -

(A) 1.15 × 10–5 (B) 4.90 × 10–8 (C) 4.55 × 10–6 (D) 7.90 × 10–8 Passage # 2 (Ques. 16 to 18) The aldol condensation also offer a convenient

way to synthesize molecules with five and six membered ring. This can be done by an intramolecular aldol condensation using a dialdehyde, a keto aldehyde or a diketone as the substrate. The major product is formed by the attack of the enolate from the ketone side of the molecule that adds to the aldehyde group. The reason the aldehyde group undergoes addition preferentially may arise from the greater reactivity of aldehyde towards nucleophilic addition generally.

In reaction of this type five membered rings form far more readily than seven membered rings and six membered rings are more favourable than four or eight membered rings when possible.

Q.15 fn;s x;s rki ij ty esa AgI (v.kqHkkj = 235) dk

foys;rk xq.kuQy] 4.9 × 10–11 gSA AgI dh

0.001 M KI foy;u esa foys;rk (gm/litre esa) gS

(A) 1.15 × 10–5 (B) 4.90 × 10–8 (C) 4.55 × 10–6 (D) 7.90 × 10–8 x|ka'k # 2 (iz. 16 ls 18)

,YMkWy la?kuu ik¡p rFkk N% lnL;h oy; okys

v.kqvksa ds la'ys"k.k ds fy, mi;ksxh gSA budksa

Mkb,fYMgkbM] dhVks ,sfYMgkbM ;k ,d

MkbdhVksu dks fØ;k/kj ds :i esa iz;qDr djrs

gq, vUr% vkf.od ,YMkWy la?kuu kjk izkIr

fd;k tk ldrk gSA v.kq ds dhVksu rjQ ls

buksysV ds vkØe.k kjk tks ,fYMgkbM lewg ij

tqM+rk gS] ls eq[; mRikn curk gSA ,sfYMgkbM

lewg ij ;ksx dk eq[; dkj.k lkekU;r;k

ukfHkdLusgh ;ksx ds izfr ,sfYMgkbM dh vf/kd

fØ;k'khyrk gSA bl izdkj dh vfHkfØ;k esa ik¡p

lnL;h oy; lkr lnL;h oy; dh rqyuk esa

vf/kd 'kh?kzrk ls curh gSA rFkk N% lnL;h oy;

pkj ;k vkB lnL;h oy; ds viskk lEHko gks]

rks vf/kd vuqdwy gksrh gSA



Q.16 1-Ethylcyclopent-1-ene on reductive

ozonolysis followed by aq. NaOH/∆ gives -

(A) CH3

O (B)

O

(C)

OCH3

(D)

H

O

Q.17 Which of the following compound on reaction

with O3/Zn, H2O followed by aq. NaOH/∆ will

form

O

(A)

(B)

(C) (D)

Q.16 1-,fFkylkbDyksisUV-1-bZu vksthuhvi?kVu rFkk

blds ckn tyh; NaOH/∆ ij nsrh gS-

(A) CH3

O (B)

O

(C)

OCH3

(D)

H

O

Q.17 fuEu esa ls dkSulk ;kSfxd O3/Zn, H2O ds lkFk

vfHkfØ;k rFkk blds ckn tyh; NaOH/∆ ls

O

cuk,sxk

(A)

(B)

(C) (D)



Q.18 The true statement about the major product of

CH3 – C – CH2 – CH2 – CH2 – CH2 – C – H

O O

in reaction with aq. NaOH followed by heating is -

(A) It gives yellow ppt with I2/ΘOH

(B) It gives silver mirror with Tollen's reagent (C) It shows stereoisomerism (D) It does not give yellow ppt with 2, 4 DNP

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 :

Q.18

CH3 – C – CH2 – CH2 – CH2 – CH2 – C – H

O O

dh tyh; NaOH ds lkFk vfHkfØ;k rFkk blds

ckn xeZ djus esa eq[; mRikn ds lUnHkZ esa lgh

dFku gS

(A) ;g I2/ΘOH ds lkFk ihyk voksi nsrk gS

(B) ;g VkWysu vfHkdeZd ds lkFk jtr niZ.k nsrk gS

(C) ;g f=kfoeleko;ork n'kkZrk gS (D) 2, 4 DNP ds lkFk ihyk voksi ugha nsrk gS

[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 :



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 marks for wrong answer.

Q.1 Match the column for H2O2 Column -I Column -II

(A) ‘10 volume’ (P) 5.358 N

(B) ‘20 volume’ (Q) 3.036%(w/v)

(C) ‘30 volume’ (R)3.4 g H2O2/100 mL

solution

(D) ‘11.2 volumes’ (S) 1.785 M

(T) 9.09% (w/v)

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 H2O2 ds fy, LrEHk feyku dhft,

LrEHk -I LrEHk -II

(A) ‘10 vk;ru’ (P) 5.358 N

(B) ‘20 vk;ru’ (Q) 3.036%(w/v)

(C) ‘30 vk;ru’ (R)3.4 g H2O2/100 mL

foy;u

(D) ‘11.2 vk;ru’ (S) 1.785 M

(T) 9.09% (w/v)



Q.2 Match the column Column -I Column -II (Compounds) (Type of reaction given by the compound)

(A)

CH=O

(P) Aldol condensation

(B)

C – CH3 O

(Q) Cannizzaro reaction

(C) CH3CHO (R) Fehling test (D) CH3 – CH2OH (S) Chloroform reaction

(T) Reaction with grignard reagent

Q.2 LrEHk feyku dhft,

LrEHk -I LrEHk -II

(;kSfxd) (;kSfxd kjk nh tkus

okyh vfHkfØ;k dk izdkj)

(A)

CH=O

(P) ,YMkWy la?kuu

(B)

C – CH3 O

(Q) dsfutkjksa vfHkfØ;k

(C) CH3CHO (R) Qsgfyax ijhk.k

(D) CH3 – CH2OH (S) DyksjksQkWeZ vfHkfØ;k

(T) fxzXukMZ vfHkdeZd

ds lkFk vfHkfØ;k



MATHEMATICS

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 answerand – 1 mark for each wrong answer.

Q.1 If (3 + x2008 + x2009)2010 = a0 + a1x + a2x2 +….+ anxn,

then the value of a0 –21 a1 –

21 a2 + a3 –

21 a4 –

21 a5 + a6 –….. is

(A) 32010 (B) 1

(C) 22010 (D) None of these

Q.2

++∑= 2

2tan 241

1–

mmmn

m

is equal to

(A) tan–1

+++

22

2

nnnn (B) tan–1

+ 2––

2

2

nnnn

(C) tan–1

+++nn

nn2

2 2 (D) None of these

[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 ?kVk;k tk;sxkA

Q.1 ;fn (3 + x2008 + x2009)2010 = a0 + a1x + a2x2 +….+ anxn gS,

rc a0 –21 a1 –

21 a2 + a3 –

21 a4 –

21 a5 + a6 –….. dk

eku gS -

(A) 32010 (B) 1

(C) 22010 (D) buesa ls dksbZ ugha

Q.2

++∑= 2

2tan 241

1–

mmmn

m

cjkcj gS -

(A) tan–1

+++

22

2

nnnn (B) tan–1

+ 2––

2

2

nnnn

(C) tan–1

+++nn

nn2

2 2 (D) buesa ls dksbZ ugha



Q.3 If the radius of the circle (x – 1)2 + (y – 2)2 = 1 and (x – 7)2 + (y – 10)2 = 4 are increasing uniformly w.r.t. time as 0.3 and 0.4 unit/sec, then they will touch each other at t equal to -

(A) 45 sec (B) 90 sec (C) 11 sec (D) 135 sec Q.4 The angle of intersection of the normal at the point

23,

25– of the curves x2 – y2 = 8 and

9x2 + 25y2 = 225 is

(A) 0 (B) 2π (C)

3π (D)

4π

Q.5 The lengths of the perpendicular from the points

(m2, 2m), (mm′, m + m′) and (m′2, 2m′) to the line x + y + 1 = 0 form

(A) an A.P. (B) a G.P. (C) a H.P. (D) none of these

Q.6 Let f(x) = g(x) x/1x/1

x/1x/1

eeee

−

−

+− and x ≠ 0 where g is a

continuous function. Then )x(flim0x→

exists if

(A) g(x) is any polynomial

(B) g(x) = x + 4

(C) g(x) = x2

(D) g(x) = 2 + 3x + 4x2

Q.3 ;fn oÙk (x – 1)2 + (y – 2)2 = 1 rFkk (x – 7)2 + (y – 10)2

= 4 dh f=kT;k le; 0.3 rFkk 0.4 [email protected] ds lkisk ,d leku :i ls c<+ jgh gS] rc og fuEu le; t ij ,d nwljs dks Li'kZ djsaxh -

(A) 45 sec (B) 90 sec (C) 11 sec (D) 135 sec Q.4 oØksa x2 – y2 = 8 rFkk 9x2 + 25y2 = 225 ds fcUnq

23,

25– ij [khaps x;s vfHkyEcksa dk izfrpNsnu

dks.k gksxk -

(A) 0 (B) 2π (C)

3π (D)

4π

Q.5 fcUnqvksa (m2, 2m), (mm′, m + m′) rFkk (m′2, 2m′) ls

js[kk x + y + 1 = 0 ij Mkys x;s yEcksa dh yEckbZ;k¡ gSa

(A) A.P. esa (B) G.P. esa

(C) H.P. esa (D) buesa ls dksbZ ugha

Q.6 ekuk fd f(x) = g(x) x/1x/1

x/1x/1

eeee

−

−

+−

,oa x ≠ 0 tgk¡ g

lrr Qyu gS] rks )x(flim0x→

fo|eku gS] ;fn

(A) g(x) dksbZ cgqin gS

(B) g(x) = x + 4 (C) g(x) = x2 (D) g(x) = 2 + 3x + 4x2



Q.7 The solutions of the equation z4 + 4z3i – 6z2 – 4zi – i = 0 are the vertices of a

convex polygon in the complex plane. The area ofthe polygon is :

(A) 23/4 (B) 23/2 (C) 25/4 (D) 2 Q.8 Number of 4 digit positive integer if the product of

their digits is divisible by 3 is: (A) 2700 (B) 6628 (C) 7704 (D) 5464

Questions 9 to 12 are multiple choice questions. Eachquestion has four choices (A), (B), (C) and (D), out ofwhich MULTIPLE (ONE OR MORE) is correct. Markyour response in OMR sheet against the questionnumber of that question. + 4 marks will be given for eachcorrect answer and – 1 mark for each wrong answer.

Q.9 Let the parabolas y = x (c – x) and y = x2 + ax + btouch each other at the point (1, 0), then -

(A) a + b + c = 0 (B) a + b = 2 (C) b – c = 1 (D) a + c = – 2

Q.10 If f(x) = ∫ −x

dtt0

|1| where 0 ≤ x ≤ 2, then

(A) range of f(x) is [0, 1] (B) f (x) is differentiable at x = 1 (C) f (x) = cos–1 x has two real roots (D) f ' (1/2) = 1/2

Q.7 lehdj.k z4 + 4z3i – 6z2 – 4zi – i = 0 ds gy lfEeJ

lery esa ,d mÙky cgqHkqt ds 'kh"kZ gSaA cgqHkqt dk

ks=kQy gS : (A) 23/4 (B) 23/2 (C) 25/4 (D) 2

Q.8 4 vadh; /kukRed iw.kk±dksa dh la[;k ;fn muds vadksa

dk xq.kuQy 3 ls foHkkftr gks] gS: (A) 2700 (B) 6628 (C) 7704 (D) 5464

iz'u 9 ls 12 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 dslek vius mÙkj vafdr dhft,A izR;sd lgh mÙkj ds fy,+ 4 vad fn;s tk;saxs rFkk izR;sd xyr mÙkj ds fy, 1 vad ?kVk;k tk;sxkA

Q.9 ekuk ijoy; y = x (c – x) rFkk y = x2 + ax + b ,d nwljs dks (1, 0) ij Li'kZ djrs gS, rc -

(A) a + b + c = 0 (B) a + b = 2 (C) b – c = 1 (D) a + c = – 2

Q.10 ;fn f(x) = ∫ −x

dtt0

|1| tgk¡ 0 ≤ x ≤ 2, rc -

(A) f (x) dk ifjlj [0, 1] gS

(B) x = 1 ij f (x) vodyuh; gS

(C) f (x) = cos–1 x ds nks okLrfod ewy gS

(D) f ' (1/2) = 1/2



Q.11 If non-zero vectors →a and

→b are equally inclined to

vector →c , then

→c is -

(A) →

→→

→→

→→

→

++

+b

|b||a|

|b|a|b|2|a|

|a|

(B) ++

→

→→

→

aba

b

||||

|| →

→→

→

+b

ba

a

||||

||

(C) ++

→

→→

→

aba

a

||2||

|| →

→→

→

+b

ba

b

||2||

||

(D) ++

→

→→

→

aba

b

||||2

||+

+

→

→→

→

aba

a

||||2

||

Q.12 Equation of a line passing through (1, – 1) and perpendicular to a line given by x2 – 5xy + 4y2 = 0 is

(A) 4x + y = 3 (B) x – 4y = 5 (C) x + y = 0 (D) x – y = 2

This section contains 2 paragraphs; each has 3 multiplechoice questions. (Questions 13 to 18) Each question has4 choices (A), (B), (C) and (D) out of which ONLY ONEis correct. Mark your response in OMR sheet against thequestion number of that question. + 4 marks will begiven for each correct answer and – 1 mark for eachwrong answer.

Q.11 ;fn v'kwU; lfn'k →a rFkk

→b lfn'k

→c ds lkFk

leku :i ls >qds gq, gS, rc →c gS -

(A) →

→→

→→

→→

→

++

+b

|b||a|

|b|a|b|2|a|

|a|

(B) ++

→

→→

→

aba

b

||||

|| →

→→

→

+b

ba

a

||||

||

(C) ++

→

→→

→

aba

a

||2||

|| →

→→

→

+b

ba

b

||2||

||

(D) ++

→

→→

→

aba

b

||||2

||+

+

→

→→

→

aba

a

||||2

||

Q.12 ,d js[kk tks (1,–1) ls xqtjrh gS rFkk x2 – 5xy + 4y2 = 0

dh ,d js[kk ds yEcor~ gS] gksxh

(A) 4x + y = 3 (B) x – 4y = 5 (C) x + y = 0 (D) x – y = 2

bl [k.M esa 2 vuqPNsn fn;s x;s gSa] izR;sd esa 3 cgqfodYih iz'u gSaA (iz'u 13 ls 18) izR;sd iz'u ds pkj fodYi (A), (B), (C) rFkk (D) gSa, ftuesa ls dsoy ,d fodYi lgh gSA OMR 'khV esa iz'u dh iz'u la[;k ds lek viuk mÙkj vafdr dhft;sA izR;sd lgh mÙkj ds fy;s + 4 vad fn;s tk,saxs rFkk izR;sd xyr mÙkj ds fy, 1 vad ?kVk;k tk;sxkA



x|ka'k # 1 (iz- 13 ls 15) ,d 'kkWfiad ekWy eas ,d Ldhe py jgh gS : 'SURF'

fMVtsZUV ds izR;sd iSdsV esa ,d dwiu gS ftl ij 'kCn 'SURF' ds vkj Nis gS] ;fn ,d O;fDr 'SURF' fMVtsZUV ds de ls de pkj iSdsV [kjhnrk gS rFkk lHkh vkjks ls 'SURF' 'kCn curk gS] rks og fMVtsZUV dk ,d iSdsV Ýh ikrk gSA

Q.13 ;fn ,d O;fDr ,d lkFk ,sls vkB iSdsV [kjhnrk gS] rc mlds ikl dwiuks ds fofHkUu lap;ks dh la[;k gksxh -

(A) 48 (B) 84 (C) 11C3 (D) 12C4

Q.14 ;fn ,d O;fDr ,sls vkB iSdsV [kjhnrk gS] rc mls Bhd ,d iSdsV Ýh feyus dh izkf;drk gS -

(A) 7/33 (B) 102/495 (C) 13/55 (D) buesa ls dksbZ ugha

Q.15 ;fn ,d O;fDr ,sls vkB iSdsV [kjhnrk gS] rc mls nks iSdsV Ýh feyus dh izkf;drk gS -

(A) 1/7 (B)1/5 (C) 1/42 (D) 1/165

x|ka'k # 2 (iz- 16 ls 18) A(x1, y1) B(x2, y2), C(x3, y3) ,d f=kHkqt ABC. ds

'kh"kZ gSa lx + my + n = 0 js[kk L dk lehdj.k gSa Q.16 ;fn L f=kHkqt ABC dh Hkqtkvksa BC, CA rFkk AB dks

Øe'k% P, Q, R fcUnqvksa ij dkVrh gksa] rks

RBAR

QACQ

PCBP

×× dk eku gS

(A) –1 (B) – 2/1 (C) 2/1 (D) 1

Passage # 1 (Ques. 13 to 15) A shopping mall is running a scheme : Each packet

of detergent 'SURF 'contains a coupon which bearsletter of the word 'SURF', if a person buys at leastfour packets of detergent 'SURF' and produced allthe letters of the word 'SURF' then he gets one freepacket of detergent

Q.13 If a person buys eight such packet at a time, thennumber of different combinations of coupon he has

(A) 48 (B) 84 (C) 11C3 (D) 12C4 Q.14 If person buys eight such packets, then the

probability that he gets exactly one free packet is – (A) 7/33 (B) 102/495 (C) 13/55 (D) none of these Q.15 If a person buys eight such packets, then the

probability that he gets two free packets is - (A) 1/7 (B)1/5 (C) 1/42 (D) 1/165

Passage # 2 (Ques. 16 to 18) A(x1, y1) B(x2, y2), C(x3, y3) are the vertices of a

triangle ABC. lx + my + n = 0 is an equation of theline L.

Q.16 If L intersects the sides BC, CA and AB of thetriangle ABC at P, Q, R respectively then

RBAR

QACQ

PCBP

×× is equal to

(A) –1 (B) – 2/1 (C) 2/1 (D) 1



Q.17 If the centroid of the triangle ABC is at the originand algebraic sum of the lengths of theperpendiculars from the vertices of the triangleABC on the line L is equal to 1 then sum of the reciprocal of squares of the intercepts made by L onthe coordinate axes is equal to

(A) 0 (B) 4 (C) 9 (D) 16 Q.18 If P divides BC in the ratio 2 : 1 and Q divides CA

in the ratio 1 : 3 then R divides AB in the ratio (P,

Q, R are the point as in Q. 16)

(A) 3 : 2 (B) 2 : 3

(C) 4 : 3 (D) None of these

Section - II This section contains 2 questions (Questions 1, 2). Eachquestion contains statements given in two columns whichhave to be matched. Statements (A, B, C, D) in Column Ihave to be matched with statements (P, Q, R, S, T) inColumn II. The answers to these questions have to beappropriately bubbled as illustrated in the followingexample. 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 :

Q.17 ;fn f=kHkqt ABC dk dsUnzd ewyfcUnq gks rFkk f=kHkqt

ABC ds 'kh"kZ ls js[kk L ij Mkys x;s yEcksa dh

yEckbZ;ksa dk cht xf.krh; ;ksx 1 ds rqY; gks] rks js[kk

L kjk funsZ'kh vkksa ls dkVs x;s vUr%[k.Mksa ds oxksZ ds

O;qRØe dk ;ksx gS (A) 0 (B) 4 (C) 9 (D) 16 Q.18 ;fn P, BC dks 2 : 1 ds vuqikr esa ,oa Q; CA dks 1 : 3

ds vuqikr esa foHkkftr djrk gks rks R ; AB dks fuEu vuqikr esa foHkkftr djrk gS

(P, Q, R ogh fcUnq gS tks iz'u 16 ds gSa) (A) 3 : 2 (B) 2 : 3

(C) 4 : 3 (D) buesa ls dksbZ ugha

[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 lghlqesy 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



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 marks for wrong answer.

Q.1 If S ≡ (x – 2)2 + (y + 1)2 = 1 then match column I to

column II Column I Column II

(A) tangent to S (P) 3x + 4y – 2 = 0

(B) Diameter of S (Q) x = 0

(C) Line perpendicular to tangent (R) y = 0

obtained in ‘A’

(D) A chord of S, which is not (S) x – y – 2 = 0

diameter (T) 2x – 5 = 0

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 (vFkkZr~ izR;sd lgh iafDr feyku ds fy, +2

vad fn, tk,asxs) rFkk xyr mÙkj ds fy;s dksbZ _.kkRed

vadu ugha gS (vFkkZr~ dksbZ vad ugha ?kVk;k tk;sxk)A

Q.1 ;fn S ≡ (x – 2)2 + (y + 1)2 = 1 gS] rks LraHk I o LraHk II dk

feyku djks

Column I Column II

(A) S dh Li'kZ js[kk gksxh (P) 3x + 4y – 2 = 0

(B) S dk O;kl gksxk (Q) x = 0

(C) ‘A’ esa izkIr Li'kZ js[kk ds yEcor~ (R) y = 0

js[kk gksxh

(D) S dh ,d thok] tks fd O;kl (S) x – y – 2 = 0

ugha gS] gksxh (T) 2x – 5 = 0



Q.2 Column-I Column-II (A) If f(x) = sin–1x and (P) 2

+

→

21x

lim f(3x – 4x3) = a – 3, +

→

21x

lim f(x),

then [a] = (B) If f(x) = tan–1g(x) where (Q) 3

g(x)=2

3

x31xx3

−− and

0lim→h h3

)a(f)h3a(f −+

=2a1

3+

, when 31− < a <

31 ,

then find

−

+

→ h621fh6

21f

lim0h

=

(C) If cos–1(4x3 – 3x) = (R) 4

a + b cos–1x for –1 < x < 21− ,

then [a + b + 2] = (D) If f(x) = cos–1(4x3–3x) and (S) –2

+

→

21x

lim f ′(x) = a and –

21x

lim

→

f ′(x) = b,

then a + b + 3 = (T) 1

Q.2 LrEHk-I LrEHk-II

(A) ;fn f(x) = sin–1x rFkk (P) 2

+

→

21x

lim f(3x – 4x3) = a – 3, +

→

21x

lim f(x) gS,

rc [a] = (B) ;fn f(x) = tan–1g(x) tgk¡ (Q) 3

g(x)=2

3

x31xx3

−− rFkk

0lim→h h3

)a(f)h3a(f −+

=2a1

3+

gS, tc 31− < a <

31 ,

rc

−

+

→ h621fh6

21f

lim0h

=

(C) ;fn cos–1(4x3 – 3x) = (R) 4

a + b cos–1x, –1 < x <21− ds fy, gS,

rc [a + b + 2] =

(D) ;fn f(x) = cos–1(4x3–3x) rFkk (S) –2

+

→

21x

lim f ′(x) = a rFkk –

21x

lim

→

f ′(x) = b,

rc a + b + 3 = (T) 1



PHYSICS

Section – I

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

Q.1 A rod of mass m is lying on a smooth horizontal

surface. Two particles of mass m and 2m moving with a velocity v0 and –2v0 strike the rod at ends A and B and sticks to rod. The velocity of CM of the system will be –

B

A

x

y 2v0 2m

v0 m

(A) ^0 i4

v (B) )i(

4v ^0 −

(C) )i(4v3 ^0− (D) ^0 i

4v3

[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 ?kVk;k tk;sxkA

Q.1 m nzO;eku dh NM+ ,d ?k"kZ.kghu kSfrt lrg ij fLFkr gSA m o 2m nzO;eku ds nks d.k Øe'k% v0 rFkk –2v0 osx ds lkFk xfr djrs gq, Øe'k% NM+ ds fljs A o B ij Vdjkrs gS o NM+ ls fpid tkrs gSaA fudk; ds nzO;eku dsUnz dk osx gksxk –

B

A

x

y 2v0 2m

v0m

(A) ^0 i4

v (B) )i(

4v ^0 −

(C) )i(4v3 ^0− (D) ^0 i

4v3



Q.2 A ring of mass m is suspended to a spring of force constant k. Consider mass of the spokes to be negligible. If the ring is slightly pulled down and released, time period of oscillations will be

(A) km

π (B) km2π

(C) km

2π (D)

km22π

Q.3 A ring of mass m is in pure rolling. CM of ring has a linear velocity v0. KE of mass m attached to the ring is K1 and KE of system is K2 at the instant shown in the figure, K1/K2 is

m

ω0

v0

(A) 1/2 (B) 2/3 (C) 3/2 (D) 3/4

Q.2 m nzO;eku dh ,d oy; k cy fu;rkad ds ,d fLi z ax ls yVdh g SA Lik sDl dk n zO;eku ux.; ekfu;sA ;fn oy; dks gYdk lk uhps [khapdj NksM+ fn; k t k; s r k s n k syu k s a d k v kor Zd ky g k s x k

(A) km

π (B) km2π

(C) km

2π (D)

km22π

Q.3 m nzO;eku dh ,d oy; 'kq) yksVuh xfr dj jgh gSA oy; dk nzO;eku dsUnz v0 js[kh; osx j[krk gSA fp=k esa n'kkZ;s k.k ij oy; ls layXu m nzO;eku dh xfrt ÅtkZ K1 gS rFkk fudk; dh xfrt ÅtkZ K2 gS rks K1/K2 gS&

m

ω0

v0

(A) 1/2 (B) 2/3 (C) 3/2 (D) 3/4



Q.4 Two capacitors C1 and C2, can be charged to a potential V/2 each by having -

S2S1

R R 0

C2C1

V

(A) S1 closed and S2 open (B) S1open and S2 closed (C) S1 and S2 both closed (D) cannot be charged at V/2

Q.5 Relative accuracy of a screw gauge can be increased by

(A) By increasing the size of pitch (B) By increasing the number of division on the

circular scale (C) By taking large number of observation (D) By having a device free from zero error Q.6 The least count of a stop watch is 0.1 sec. The time

of 20 oscillations of pendulum is 20 sec. The percentage error in time period is –

(A) 0.25 % (B) 0.5 % (C) 0.75 % (D) 1 %

Q.4 uhps n'kkZbZ ifjiFk O;oLFkk esas nks la/kkfj=kksa C1 o C2 esa ls izR;sd dks ,d V/2 foHko rd vkosf'kr fd;k tk ldrk gS] blds fy, ;g djuk gksxk -

S2S1

R R 0

C2C1

V

(A) S1 dks can o S2 dks [kqyk j[ks (B) S1 dks [kqyk o S2 can j[kdj (C) S1 o S2 nksuks dks can djs (D) V/2 ij vkosf'kr ugha fd;k tk ldrk gS

Q.5 ,d LØwxst dh vkisfkd ;FkkFkZrk c<+kbZ tk ldrh gS]

ds kjk

(A) fip (pwM+h vUrjky) dk vkdkj c<+kdj

(B) ofÙk; iSekus ij foHkktuksa dh la[;k c<+kdj

(C) cM+h la[;k esa izsk.k ysdj

(D) 'kwU; =kqfV ls eqDr midj.k ysdj

Q.6 ,d LVkWi okWp (stop watch) dk vYirekad 0.1 sec gSA

,d yksyd ds 20 nksyuksa dk le; 20 sec gSA

vkorZdky esas izfr'kr =kqfV gS –

(A) 0.25 % (B) 0.5 % (C) 0.75 % (D) 1 %



Q.7 1000small drops of radius r are combined to form a

single large drop. If T is the surface tension then

energy released will be

(A) 3600 πr2T (B) 7200 πr2T

(C) 1800 πr2T (D) 5400 πr2T

Q.8 A person speaking normally produces a sound of intensity 40 dB at a distance of 1 meter. If the threshold intensity for reasonable audibility is 20 dB. The maximum distance at which he can hear clearly

(A) 4 meter (B) 5 meter (C) 10 meter (D) 20 meter

Questions 9 to 12 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct. Mark your response in OMR sheet against the question number of that question. + 4 marks will be given for each correct answer and – 1 mark for each wrong answer.

Q.9 When a wave is reflected from a mirror, there is no

change in its

(A) amplitude (B) frequency

(C) wavelength (D) velocity

Q.7 r f=kT;k dh 1000 y?kq cwanksa dks ,d cM+h cwan fufeZr

djus ds fy, la;ksftr fd;k x;k gSA ;fn T i"B ruko gS rks eqDr ÅtkZ gksxh

(A) 3600 πr2T (B) 7200 πr2T

(C) 1800 πr2T (D) 5400 πr2T

Q.8 ,d O;fDr kjk lkekU; cksyus ls 1 ehVj nwjh ij 40 dB rhozrk dh /ofu mRiUu gksrh gSA ;fn mi;qDr JO;rk ds fy, nSgyh rhozrk 20 dB gSA og vf/kdre nwjh tgk¡ rd og Li"V :i ls lqu ldrk gS] gksxh&

(A) 4 ehVj (B) 5 ehVj

(C) 10 ehVj (D) 20 ehVj

iz'u 9 ls 12 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, + 4 vad fn;s tk;saxs rFkk izR;sd xyr mÙkj ds fy, 1 vad ?kVk;k tk;sxkA

Q.9 tc ,d rjax ,d niZ.k ls ifjofrZr gksrh gS] blds

dkSuls xq.k esa ifjorZu ugha gksrk

(A) vk;ke (B) vkofÙk

(C) rjaxnS/;Z (D) osx



Q.10 uhps n'kkZ;k fp=k m nzO;eku dk ,d CykWd ,d kSfrt

?k"kZ.kghu lrg ij n<+ nhokj ls ,d l nwjh ij fojke

esa fLFkr gSA CykWd dks 3 l /2 nwjh rd nkbZa vksj

/kdsyk tkrk gS o NksM+ fn;k tkrk gSA tc CykWd

bldh e/;koLFkk ls xqtjrk gS m1 nzO;eku dk ,d

vU; CykWd bl ij fLFkr fd;k tkrk gS tksfd bl ij

?k"kZ.k ds dkj.k fpid tkrk gS bl izdkj la;qDr

CykWd ckbZa nhokj ls Bhd Vdjkrk gSA mDr ds lEcU/k

es lgh dFku pqfu,s &

P m

k

l

(A) m1 = 8m5

(B) m1 = 4m5

(C) ek/; fLFkfr ij CykWd m dk osx mK

23l gS

(D) ek/; fLFkfr ij CykWd m dk osx mK

43l

gS

Q.10 Figure shows a block P of mass m resting on a

horizontal smooth floor at a distance l from a rigid

wall. Block is pushed toward right by a distant 3 l /2

and released. When block passes from its mean

position another block of mass m1 is placed on it

which sticks to it due to friction so that the

combined block just collides with the left wall.

Now choose the correct option (s)

P m

k

l

(A) m1 = 8m5

(B) m1 = 4m5

(C) Velocity of block m at mean position is mK

23l

(D) Velocity of block m at mean position is mK

43l



Q.11 X-ray from a tube with a target A of atomic number Z show strong K lines for target A and weak K lines for impurities. The wavelength of Kα lines is λz for target A and λ1 and λ2 for two impurities.

λz/λ1 = 4 and λz/λ2 =1/4 Screening constant of Kα lines to be unity. Select

the correct statement(s) (A) The atomic number of first impurity is 2z – 1 (B) The atomic number of first impurity is 2z + 1

(C) the atomic number of second impurity is 2

)1z( +

(D) The atomic number of second impurity is 12z

+

Q.12 An ideal coil of 20 H is joined in series with a resistance 10Ω and ideal battery of 10V. After two seconds the current flowing (in ampere) in the circuit will be

(A) e (B) e–1 (C) (1 – e–1) (D) (1 – e)

This section contains 2 paragraphs; each has 3 multiple choice questions. (Questions 13 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. + 4 marks will be given for each correct answer and – 1 mark for each wrong answer.

Q.11 ,d X-fdj.k ufydk ftleas y; A dk ijek.kq Øekad Z gS] ls X-fdj.k y; A ds fy, izcy K js[kk o v'kqf) ds fy, nqcZy K js[kk n'kkZrh gSA y; A ds fy, Kα js[kk dh rjaxnS/;Z λz gS o nks v'kqf);ksa ds fy, rjaxnS/;Z λ1 rFkk λ2 gS

λz/λ1 = 4 o λz/λ2 =1/4 Kα js[kk dk LØhfuax fu;rkad bdkbZ gSA fuEu esa ls

lgh fodYi pqfu,& (A) izFke v'kqf) dk ijek.kq Øekad 2z – 1 gS (B) izFke v'kqf) dk ijek.kq Øekad 2z + 1 gS

(C) frh; v'kqf) dk ijek.kq Øekad 2

)1z( +gS

(D) frh; v'kqf) dk ijek.kq Øekad 12z

+ gS

Q.12 20 H dh ,d vkn'kZ dq.Myh] ,d 10Ω izfrjks/k o ,d

10V vkn'kZ cSVjh ds lkFk Js.khØe esa la;ksftr gSA nks

lSd.M i'pkr~ ifjiFk esa izokfgr /kkjk (,fEi;j esa) gksxh& (A) e (B) e–1

(C) (1 – e–1) (D) (1 – e)

bl [k.M esa 2 vuqPNsn fn;s x;s gSa] izR;sd esa 3 cgqfodYih iz'u gSaA (iz'u 13 ls 18) izR;sd iz'u ds pkj fodYi (A), (B), (C) rFkk (D) gSa, ftuesa ls dsoy ,d fodYi lgh gSA OMR 'khV esa iz'u dh iz'u la[;k ds lek viuk mÙkj vafdr dhft;sA izR;sd lgh mÙkj ds fy;s + 4 vad fn;s tk,saxs rFkk izR;sd xyr mÙkj ds fy, 1 vad ?kVk;k tk;sxkA



Passage # 1 (Ques. 13 to 15)

The conservation of angular momentum, which

corresponds to the conservation of linear

momentum states that the angular momentum about

an axis of a given rotating body or system of bodies

is constant, if no external torque acts about that

axis.

The earth rotates about an axis passing through its

geographic north and south poles with a period of

one day. If its is struck by meteorities, then since

action and reaction are equal, no external couple

acts on the earth and meteorites. Their total angular

momentum is thus conserved. Neglecting the

angular momentum of the meteorites about the

earth's axis before collision compared with that of

the earth, then angular momentum of (earth +

meteorites) after collision = angular momentum of

earth before collision. Since the effective mass of

the earth has increased after collision, the moment

of inertia has increased. So the earth will slow

down slightly.

x|ka'k # 1 (iz- 13 ls 15)

dks.kh; laosx lajk.k dk fu;e tks fd js[kh; laosx

lajk.k ls lEcfU/kr gS tks fd crkrk gS fd fdlh

?kw.kZu djrh fdlh oLrq ;k fudk; dk fdlh nh gqbZ

vk ds ifjr% dks.kh; laosx lajfkr jgrk gS] ;fn bl

vk ds ifjr% dksbZ ckg~; cy vk?kw.kZ dk;Zjr u gksA

iFoh HkkSxksfyd mÙkjh o nfk.kh /kqzoksa ls xqtjrh ,d

vk ds ifjr% ,d fnu vkorZdky ds ds lkFk ?kw.kZu

dj jgh gSA ;fn ;g mYdkfi.M (meteorities) ls

Vdjk;s] D;kasfd fØ;k o izfrfØ;k cjkcj gS] iFoh o

mYdkfi.M ij dksbZ ckg~; cy ;qXe dk;Z ugh djrkA

bl izdkj budk dqy dks.kh; laosx lajfkr jgrk gSA

VDdj ls iwoZ] iFoh vk ds lkisk mYdkfi.M ds

dks.kh; laosxksa dks ux.; ekurs gq,

VDdj ds i'pkr (iFoh + mYdkfi.M) dk dks.kh;

laosx = VDdj ls iwoZ iFoh dk dks.kh; laosx] D;ksafd

VDdj ds i'pkr~ iFoh dk izHkkoh nzO;eku c<+ tkrk

gS] tM+Ro vk?kw.kZ c<+ tkrk gSA vr% iFoh dh ?kw.kZu

nj FkksM+h lh de gks tk;sxhA



Q.13 A bullet of mass m moving horizontally with

velocity u sticks to the top of a solid cylinder

of mass M and radius R resting on a rough

horizontal surface as shown. (Assume

cylinder rolls without slipping). Angular

velocity of cylinder is

m u

(A) R)M3m8(

mu4+

(B) R)M3m8(

mu+

(C) R)M3m(

mu4+

(D) R)Mm8(

mu+

Q.14 In the previous problem, velocity of the

bullet just after the collision is

(A) u (B) )Mm8(

mu8+

(C) M3m8

mu8+

(D) None of these

Q.13 m nzO;eku dh ,d xksyh u os ls kSfrt :i ls

xfr djrs gq, M nzO;eku o R f=kT;k ds ,d

Bksl csyu tks fd ,d kSfrt ?k"kZ.kghu lrg ij

fojke esa gS ds 'kh"kZ (Top) ij fpid tkrh gS]

n'kkZ;s fp=kkuqlkjA (;g ekfu;s fd csyu fQlys

fcuk yq<+d jgk gS), csyu dk dks.kh; osx gS&

m u

(A) R)M3m8(

mu4+

(B) R)M3m8(

mu+

(C) R)M3m(

mu4+

(D) R)Mm8(

mu+

Q.14 mijksDr iz'u esas VDdj ds rRdky i'pkr~

xksyh dk osx gS

(A) u (B) )Mm8(

mu8+

(C) M3m8

mu8+

(D) buesa ls dksbZ ugh



Q.15 A uniform heavy disc is rotating at constant

angular velocity about a vertical axis through its

centre, some sand was dropped gently on the

disc, the angular velocity of the disc.

(A) does not change (B) increases

(C) decreases (D) becomes zero

Passage # 2 (Ques. 16 to 18)

A solid, insulating ball of radius 'a' is

surrounded by a conducting spherical shell of

inner radius 'b' and outer radius 'c' as shown in

the figure. The inner ball has a charge Q which

is uniformly distribute throughout its volume.

The conducting spherical shell has a charge –Q.

Answer the following questions.

–Qc

b

Q a

Q.15 ,d le:i Hkkjh pdrh viuh dsUnz ls xqtjrh

,d Å/okZ/kj vk ds ikfjr% ,d fu;r dks.kh; osx

ls ?kw.kZu dj jgh gS bl pdrh ij /khjs ls dqN

jsr fxjk nh tkrh gS] pdrh dk dks.kh; osx&

(A) ifjofrZr ugha gksrk (B) c<+rk gS

(C) ?kVrk gS (D) 'kwU; gks tkrk gS

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

,d Bksl vpkyd xsan ftldh f=kT;k 'a' gS ,d pkyd

xksyh; dks'k ftldh vkUrfjd f=kT;k 'b' o ckg~;

f=kT;k 'c' gS ls f?kjh gSA vkUrfjd xsan ,d vkos'k Q

j[krh gS tks fd blds iwjs vk;ru esa ,d leku :i

ls forfjr gSA xksyh; pkyd dks'k–Q vkos'k j[krk gSA

fuEu ds mÙkj nhft,&

–Qc

b

Q a



Q.16 Assuming the potential at infinity to be zero, the

potential at a point located at a distance a/2

from the centre of the sphere will be -

(A)

−

πε b1

a2

4Q

0 (B)

−

πε b1

a811

4Q

0

(C)

−

πε b1

a1

4Q

0 (D) None of these

Q.17 Work done by external agent in taking a charge

q slowly from inner surface of the shell to

surface of the spherical ball will be–

(A)

−

c1

a1kQq (B)

−

a1

b1kQq

(C)

−

b1

a1kQq (D)

−

a1

c1kQq

Q.18 Now the outer shell is grounded, i.e. potential of

the outer surface is fixed to be zero. Now the

charge on the inner ball will be

(A) zero (B) Q

(C)

−+

b1

c1

a1

CQ (D)

−+

b1

c1

a1

bQ

Q.16 ;g ekurs gq, fd vuUr ij foHko 'kwU; gS] xksys ds

dsUnz ls a/2 nwjh ij fLFkr ,d fcUnq ij foHko gksxk -

(A)

−

πε b1

a2

4Q

0 (B)

−

πε b1

a811

4Q

0

(C)

−

πε b1

a1

4Q

0 (D) buesa ls dksbZ ugh

Q.17 ,d ckg~; dkjd kjk ,d q vkos'k dks dks'k dhs

vkUrfjd lrg ls xksyh; xsan dh lrg rd /khjs

ls ys tkus esa fy;k x;k dk;Z gksxk–

(A)

−

c1

a1kQq (B)

−

a1

b1kQq

(C)

−

b1

a1kQq (D)

−

a1

c1kQq

Q.18 vc cká dks'k dks Hkw lEifdZr dj fn;k x;k gS] vFkkZr~ cká lrg ds foHko dks 'kwU; ij fQDl fd;k gSA vc vkUrfjd xsan ij vkos'k gksxk –

(A) 'kwU; (B) Q

(C)

−+

b1

c1

a1

CQ (D)

−+

b1

c1

a1

bQ



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

marks for wrong answer.

[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 (vFkkZr~ izR;sd lgh iafDr feyku

ds fy, +2 vad fn, tk,asxs) rFkk xyr mÙkj ds fy;s dksbZ

_.kkRed vadu ugha gS (vFkkZr~ dksbZ vad ugha ?kVk;k

tk;sxk)A



Q.1 Match the column I with column II (Each

resistor has resistance R)

Column-I Column-II

(A)

BA (P) R

158

(B)

B A

C

O

D

Tetrahedron

(Q) R127

(C)

G

H

C

F

E

A

D

B

(R) 54 R

(D)

A B

(S) R78

(T) R

Q.1 LrEHk I dks LrEHk II ls lqesfyr dhft, (izR;sd izfrjks/kd] izfrjks/k R j[krk gS)

LrEHk-I LrEHk-II

(A)

BA (P) R158

(B)

B A

C

O

D

prq"Qydh;

(Q) R127

(C)

G

H

C

F

E

A

D

B

(R) 54 R

(D)

A B

(S) R78

(T) R



Q.2 Match the column I with column II

Column-I

Column-II

(A) Rainbow (P) Refraction

(B) Mirage (Q) Dispersion

(C) Twinkling of stars (R) Scattering

(D)

Blue sky

(S)

Total internal

reflection

(T) Reflection only

Q.2 LrEHk I dks LrEHk II ls lqesfyr dhft,

LrEHk-I LrEHk-II

(A) bUnz /kuq"k (P) viorZu

(B) ejhfpdk (Q) foiFku

(C) Rkjksa dk fVefVekuk (R) izdh.kZu

(D)

uhyk vkdk'k (S)

iw.kZ vkUrfjd

ijkorZu

(T) Dsoy ijkorZu



Time : 3 : 00 Hrs. MAX MARKS: 240

INSTRUCTIONS TO CANDIDATE

A. lkekU; :

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

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

B. vadu i)fr: bl iz'ui=k esa izR;sd fo"k; esa fuEu izdkj ds iz'u gSa:-

[k.M – I

4. cgqfodYih izdkj ds iz'u ftuesa ls dsoy ,d fodYi lgh gSA izR;sd lgh mÙkj ds fy, 3 vad fn;s tk;saxs o izR;sd xyr mÙkj ds

fy, 1 vad ?kVk;k tk,xkA

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

xyr mÙkj ds fy, 1 vad ?kVk;k tk,xkA

6. x|ka'k ij vk/kkfjr cgqfodYih izdkj ds iz'u ftuesa ls dsoy ,d fodYi lgh gSA izR;sd lgh mÙkj ds fy, 4 vad fn;s tk;saxs rFkk

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

[k.M – II

7. LrEHkksa dks lqesfyr djus okys iz'u (4 × 5 izdkj) gSaA iw.kZ :Ik ls lgh lqesfyr mÙkj ds fy, 8 vad fn;s tk;saxs ¼vr% lgh lqesfyr izR;sd iafDr ds fy, +2 vad fn, tk,saxs½ rFkk xyr mÙkj ds fy, dksbZ _.kkRed vadu ugha gSA

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

012

3

4

56

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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