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Space for rough work

TARGET COURSE FOR IIT-JEE 2010 PHASE- ALL (REVISION TEST-AT CENTRE)

TEST # 3 (R-1(1)) PAPER - 1

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. Write your Name & Roll No. in the space provided on this cover page of question paper. 3. The Question paper contains blank space for your rough work. No additional sheet will be provided for rough work. 4. The answer sheet, a machine readable Optical Mark Recognition (OMR) is provided separately. 5. Do not break the seal of the question-paper booklet before being instructed to do so by the invigilators. 6. Blank papers, Clipboards, Log Tables, Slide rule, Calculators, Cellular phones, Pagers and Electronic gadgets in any form are Not allowed to be carried inside the examination hall. B. FILLING THE OMR :

7. Fill your Name, Roll No., Batch, Course and Centre of Examination in the blocks of OMR sheet and darken circle properly. 8. DO NOT TAMPER WITH/ MUTILATE THE OMR.

C. MARKING SCHEME :

Each subject in this paper consists of following types of questions:- Section - I

9. Multiple choice questions with only one correct answer. 3 marks will be awarded for each correct answer and –1 mark for each wrong answer.

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

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

Section –II

12. Column matching type questions. 6 marks will be awarded for the complete correctly matched answer and No Negative marking for wrong answer. However, 1 mark will be given for a correctly marked answer in any row.

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Date : 14/02/2010 Time : 3 : 00 Hrs. MAX MARKS: 249

MATHEMATICS, PHYSICS, CHEMISTRY

SEA

L



Important Data (egRoiw.kZ vk¡dM+s)

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

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

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



MATHEMATICS

Section – I Questions 1 to 9 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 If ∫3

1

t dte2 = P then ∫

9e

e

dttln is equal to -

(A) 3e9 – e (B) 3e9 – p

(C) e9 – p (D) 3e9 – e – p

Q. 2 If f "(x) + f '(x) + f

2 (x) = x2 be the differential

equation of a curve and let P be the point of

maxima then number of tangents which can be

drawn from point P to x2 – y2 = a2 is -

(A) 2 (B) 1

(C) 0 (D) either 1 or 2

[k.M - I

iz'u 1 ls 9 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

1

t dte2 = P rc ∫

9e

e

dttln cjkcj gS -

(A) 3e9 – e (B) 3e9 – p

(C) e9 – p (D) 3e9 – e – p

Q. 2 ;fn f "(x) + f '(x) + f

2 (x) = x2 ,d oØ dh vody

lehdj.k gSA rFkk ekuk P mfPp"B fcUnq gSA rc mu

Li'kZ js[kkvksa dh la[;k tks fcUnq P ls x2 – y2 = a2 ij

[khaph tk lds, gksxh -

(A) 2 (B) 1

(C) 0 (D) ;k rks 1 ;k rks 2



Q. 3 A point P moves inside a triangle formed by

A (0, 0) B (2, 2 3 ). C (4, 0) such that

min. (PA, PB, PC) = 2, then the area bounded by

the curve traced by P is -

(A) 3 3 – 2

3π (B) 4 3 – 2π

(C) 3 – 2π (D) 2π

Q.4 The point in which the line x = 2 + k, y = 1 – k,

z = 2 – k, meets the coordinate plane xz is -

(A) (2, 0, 2) (B) (1, 0, 3)

(C) (3, 0, 1) (D) (2, 0, 0)

Q.5 A variable plane is at a constant distance 2p from

the origin. O meets the coordinate axes in A, B, C.

If locus of centroid of tetrahedron OABC is

2x1 +

2y1 +

2z1 =

2pk , then the value of k is -

(A) 1 (B) 2

(C) 3 (D) 4

Q. 3 ,d fcUnq P, ,d f=kHkqt tks fd A (0, 0) B (2, 2 3 )

C (4, 0) kjk cuk gS] ds vUnj bl izkj xeu djrk

gSA fd min. (PA, PB, PC) = 2 rc P kjk cuk, x,

oØ ls ifjc) ks=kQy gksxk -

(A) 3 3 – 2

3π (B) 4 3 – 2π

(C) 3 – 2π (D) 2π

Q.4 og fcUnq ftlesa js[kk x = 2 + k, y = 1 – k,

z = 2 – k funsZ'kh lery xz dks feyrh gS] gksxk -

(A) (2, 0, 2) (B) (1, 0, 3)

(C) (3, 0, 1) (D) (2, 0, 0)

Q.5 ,d pj lery ewy fcUnq O ls vpj nwjh 2p ij

funsZ'kh vkks dks Øe'k% A, B, C ij feyrk gSA ;fn

prq"Qyd OABC ds dsUnzd dk fcUnqiFk

2x1 +

2y1 +

2z1 =

2pk gSA rc k dk eku gS -

(A) 1 (B) 2

(C) 3 (D) 4



Q.6 In a certain country, the numerals in car registration

marks range from 1 to 999. Then the probability

that the first local car which you see while visiting

that country have at least two digits the same in its

registration mark is -

(A) 11130 (B)

11129

(C) 11128 (D)

11127

Q.7 If in a group of five married couples each male

shakes hand with exactly one female similarly each

female shakes hand with exactly one male, then the

probability that no husband shake hands with his

own wife, is -

(A) 2423 (B)

2511

(C) 120119 (D)

3011

Q.6 ,d fuf'pr ns'k esa dkj jftLVªs'ku] vadks 1 ls 999

rd gSA rc ml ns'k esa Hkze.k djrs le; ns[kus ij

og izkf;drk] ftlesa igyh LFkkuh; dkj fd ftlesa

jftLVªas'ku la[;k esa de ls de nks vad leku gks,

gksxh -

(A) 11130 (B)

11129

(C) 11128 (D)

11127

Q.7 ;fn ik¡p fookfgr tksM+ks ds lewg esa izR;sd iq:"k

dsoy ,d efgyk ls gkFk feykrk gSA mlh izdkj

izR;sd efgyk Hkh Bhd ,d iq:"k ls gkFk feykrh gSA

rc og izkf;drk, ftlesa dksbZ Hkh iq:"k Loa; dh iRuh

ls gkFk uk feyk,, gksxh -

(A) 2423 (B)

2511

(C) 120119 (D)

3011



Q. 8 The image of the point having the position vector

i + j3 + k4 in the plane mirror →r .( i2 – j + k )

+ 3 = 0 is -

(A) i3 + j2 + k5 (B) i3 + j5 + k2

(C) – i3 + j5 + k2 (D) i3 + j2 – k5

Q. 9 The members of a chess club took part in a competition in which each plays every one else once. All members scored the same number of points, except four juniors whose total score were 17.5. Assume that for each win a player scores 1 point, for draw 0.5 point and zero for losing. The total number of members in the club are -

(A) 26 (B) 27

(C) 28 (D) 29

Questions 10 to 14 are multiple choice questions. Each


which MULTIPLE (ONE OR MORE THAN 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. 8 ml fcUnq dk izfrfcEc ftldk fLFkfr lfn'k lery

niZ.k →r .( i2 – j + k ) + 3 = 0 esa i + j3 + k4 gks

gksxk -

(A) i3 + j2 + k5 (B) i3 + j5 + k2

(C) – i3 + j5 + k2 (D) i3 + j2 – k5

Q. 9 ,d 'krjat Dyc ds lnL;ksa us ,d izfr;ksfxrk esa Hkkx

fy;kftlesa izR;sd f[kykM+h ,d f[kykM+h ds lkFk [ksyk

dsoy pkj dfu"Bks (juniors) dks NksM+dj lHkh lnL;ksa

us leku vad izkIr fd, ftudk Ldksj 17.5 FkkA ekuk fd izR;sd thrus ds fy, 1 vad, cjkcj gksus ij 0.5 vad rFkk gkjus ij 'kwU; vad izkIr djrs gSA rc Dyc

esa lnL;ksa dh la[;k gS-

(A) 26 (B) 27 (C) 28 (D) 29

iz'u 10 ls 14 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. 10 ,d Fksys esa 3 yky rFkk 3 gjh xsans gSA rFkk ,d O;fDr 3

xsans ;knfPNd :i eas fudkyrk gSA rc og rhu uhyh

xsnksa dks okil Fksys esa, Mkyrk gSA rFkk iqu% 3 xsanks dks

;knfPNd :i ls fudkyrk gSA rc og izkf;drk ftlesa

ckn esa fudkyh xbZ rhuksa xsan fHkUu jaxksa okyh gks gksxh -

(A) 5027 (B)

10027

(C) 20 3

36

12

13

CC.C

(D)

20027

Q. 11 nks ?kVukvksa A o B ij fopkj djrs gSA fd

P(A) = 41 , P

AB =

21 , P

BA =

41 rc fuEu esa

ls lR; dFku gS?

(A) P

BA =

43 (B) P

BA + P

BA = 1

(C) A rFkk B LorU=k ?kVuk,¡ gS

(D) A rFkk B ijLij viothZ ?kVuk,¡ gS

Q. 10 A bag contains 3red and 3 green balls and person

draws out 3 at random. He then drops 3 blue balls

into the bag and again draws out 3 at random. The

probability that the 3 later balls are all of different

colours is equal to -

(A) 5027 (B)

10027

(C) 20 3

36

12

13

CC.C

(D)

20027

Q. 11 Consider two events A and B such that P(A) = 41 ,

P

AB =

21 , P

BA =

41 . Then which of the

following statements is/are true?

(A) P

BA =

43 (B) P

BA + P

BA = 1

(C) A and B are independent events

(D) A and B are mutually exclusive events.



Q.12 If f(x) = ∫ +

x

3–2 dt

1tt ¸x ∈ [–1, 3], then correct

statement(s) is/are -

(A) minimum value of f(x) is 21 ln

101

(B) minimum value of f(x) is – 21 ln5

(C) maximum value of f(x) is 21 ln

51

(D) maximum value of f(x) is 0

Q. 13 Let y = (A + Bx) e3x be a solution of the differential

equation 2

2

dxyd + m

dxdy + ny = 0, m, n ∈ I then -

(A) m + n = 3 (B) n2 – m2 = 64

(C) m = – 6 (D) n = 9

Q.12 ;fn f(x) = ∫ +

x

3–2 dt

1tt ¸x ∈ [–1, 3], rc fuEu esa ls

lR; dFku gS -

(A) f(x) dk ;ure eku 21 ln

101 gS

(B) f(x) dk U;wure eku – 21 ln5 gS

(C) f(x) dk vf/kdre eku 21 ln

51 gS

(D) f(x) dk vf/kdre eku 0 gS

Q. 13 ekuk y = (A + Bx) e3x vody lehdj.k

2

2

dxyd + m

dxdy + ny = 0, m, n ∈ I dk gy gSA rc -

(A) m + n = 3 (B) n2 – m2 = 64

(C) m = – 6 (D) n = 9



Q. 14 If In = ∫π 4/

0

n ])x[–x((tan + tann–2 (x – [x]) dx and

n ∈ N, n ≥ 2, where [.] denotes greatest integer

function, then -

(A) 100I1 –

99I1 = 1

(B) I2, I3, I4 ……… are in H.P.

(C) I2, I3, I4………… are in A.P.

(D) ∑=

=n

2i i 2)1–n(n

I1

This section contains 2 paragraphs; each has

3 multiple choice questions. (Questions 15 to 20) 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. 14 ;fn In = ∫π 4/

0

n ])x[–x((tan + tann–2 (x – [x]) dx rFkk

n ∈ N, n ≥ 2, tgk¡ [.] egÙke iw.kkZad Qyu dks iznf'kZr

djrk gSA rc -

(A) 100I1 –

99I1 = 1

(B) I2, I3, I4 ……… g. Js. esa gS

(C) I2, I3, I4………… l. Js. esa gS

(D) ∑=

=n

2i i 2)1–n(n

I1

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

gSaA (iz'u 15 ls 20) 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 (Q. 15 to 17)

Given the curves y = f(x) passing through the

point (0, 1) and y = ∫∞

x

–

dt)t(f passing through the

point

21,0 .

The tangents drawn to both the curves at the points with equal abscissa intersects on x-axis.

Q.15 The equation of the curve y = f(x) is -

(A) log (x + 2) (B) e2x

(C) 2/3

32

xx–x3 (D) log

2/1

32

xx–x

Q.16 The area bounded by y = f(x), y =x, x = 0 and

x = 1 is -

(A) 21 (e2 – 1) (B)

21 (e2 – 2)

(C) (e2 – 1) (D) (e2 – 2)

x|ka'k # 1 (iz- 15 ls 17)

fn;k gS fd oØ y = f(x) fcUnq (0, 1) ls rFkk oØ

y = ∫∞

x

–

dt)t(f fcUnq

21,0 ls xqtjrk gSA

nksuksa oØks ds mu fcUnqvksa ftuds Hkqt leku gS] ij

[khaph xbZ Li'kZ js[kk,¡ x- vk ij izfrPNsn djrh gSA

Q.15 oØ y = f(x) dk lehdj.k gS -

(A) log (x + 2) (B) e2x

(C) 2/3

32

xx–x3 (D) log

2/1

32

xx–x

Q.16 y = f(x), y = x, x = 0 rFkk x = 1 kjk ifjc)

ks=kQy gS -

(A) 21 (e2 – 1) (B)

21 (e2 – 2)

(C) (e2 – 1) (D) (e2 – 2)



Q.17 The value of 0x

lim→ x

1–))x(f( 3is -

(A) 3 (B) 6 (C) 4 (D) 2 Passage # 2 (Q. 18 to 20) Consider the planes P1 ≡ 3x – 2y + 6z + 8 = 0 &

P2 ≡ 2x – y + 2z + 3 = 0 and a point P ≡ (– 1, – 1, – 1).

Q.18 The equation of obtuse angle bisector of the

planes P1 and P2 is -

(A) 5x – y – 4z + 3 = 0 (B) 23x – 13y + 32z + 45 = 0 (C) 5x – y – 4z – 3 = 0 (D) 23x + 13y + 32z + 45 = 0 Q.19 Assuming the plane P1 to be horizontal, then

equation of the line of greatest slope through the point P in the plane P2 is -

(A) 13

1x + = 2–1y + =

14–1z +

(B) 13

1x + = 12

1y + = 14

1z +

(C) 13

1x + = 2

1y + = 14

1z +

(D) 13

1x + = 12

1y + = 14–

1z +

Q.17 0x

lim→ x

1–))x(f( 3 dk eku gS -

(A) 3 (B) 6 (C) 4 (D) 2

x|ka'k # 2 (iz- 18 ls 20)

leryks P1 ≡ 3x – 2y + 6z + 8 = 0 rFkk P2 ≡ 2x – y + 2z + 3 = 0 ij fopkj dhft, rFkk fcUnq P ≡ (– 1, – 1, – 1) gSA

Q.18 leryks P1 rFkk P2 ds vf/kd dks.k v/kZd dk

lehdj.k gS -

(A) 5x – y – 4z + 3 = 0 (B) 23x – 13y + 32z + 45 = 0 (C) 5x – y – 4z – 3 = 0 (D) 23x + 13y + 32z + 45 = 0

Q.19 ekuk fd lery P1 kSfrt fLFkfr esa gS rc lery P2 esa egÙke <ky okyh js[kk dk lehdj.k tks fcUnq P ls xqtjrh gS gksxk -

(A) 13

1x + = 2–1y + =

14–1z +

(B) 13

1x + = 12

1y + = 14

1z +

(C) 13

1x + = 2

1y + = 14

1z +

(D) 13

1x + = 12

1y + = 14–

1z +



Q.20 The image of the point P in the plane P1 is -

(A)

4961–,

4945–,

4955–

(B)

49110–,

4994–,

49104–

(C)

49110–,

4994,

49104

(D)

4961–,

4945,

4955

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) 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, B-Q, B-R, C-P, C-Q and D-S, then the correctly bubbled 4 × 4 matrix should be as follows :

A B C D

P Q R S

S

P

P P Q R

R R

Q Q

S S

P Q R S

Q.20 lery P1 esa fcUnq P dk izfrfcEc gS -

(A)

4961–,

4945–,

4955–

(B)

49110–,

4994–,

49104–

(C)

49110–,

4994,

49104

(D)

4961–,

4945,

4955

[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) 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, B-Q, B-R, C-P, C-Q rFkk D-S gS, rks lgh fof/k ls dkys fd;s x;s xksyksa dk 4 × 4 eSfVªDl uhps n'kkZ;s vuqlkj gksxk :

ABCD

P Q R S

S

P

P P Q R

R R

Q Q

S S

P Q R S



Mark your response in OMR sheet against the question

number of that question in section-II. + 6 marks will be

given for complete correct answer and No Negative

marks for wrong answer. However, 1 mark will be

given for a correctly marked answer in any row.

Q.1 Match the following :

Column -I Column-II

(A) The vector )b3a(→→

+ is

perpendicular to )b5–a7(→→

and

vector )b4–a(→→

is perpendicular

to )b2–a7(→→

. Then angle

between →a &

→b is

(P) 30º

(B) If |b.a|→→

= |ba|→→

× then angle

between →→b&a is

(Q) 45º

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 + 6 vad fn;s tk;saxs rFkk xyr mÙkj ds fy;s

dksbZ +_.kkRed vadu ugha gS (vFkkZr~ dksbZ vad ugha

?kVk;k tk;sxk)A fdUrq] fdlh iafDr esa lgh :i ls

fpfUgr mÙkj ds fy, 1 vad fn;k tk;sxkA

Q.1 LrEHk lqesy dhft;s :

LrEHk-I LrEHk -II

(A) lfn'k )b3a(→→

+ lfn'k

)b5–a7(→→

ds yEcor~ rFkk

lfn'k )b4–a(→→

lfn'k

)b2–a7(→→

ds yEcor~ gS rc →a

o →b ds e/; dks.k gS

(P) 30º

(B) ;fn |b.a|→→

= |ba|→→

× rc →→ba rFkk ds e/; dks.k gS

(Q) 45º



(C) Angle between the line

43x + =

2–4–y =

1–5z + and plane

4x – 2y – z = 1 is

(R) 60º

(D) If a, b, c are different real numbers and kcjbia ++ ,

kajcib ++ & kbjaic ++ are position vectors of three non collinear points A, B & C. Then

angle between →

AB and →

AC is

(S) 90º


Column -I Column-II

(A) The number of non congruent

rectangles that can be found on a

chess board are

(P) 31

(B) In an examination minimum is

to be scored in each of 5 subject to

pass. The number of ways in which

student can fail is

(Q) 32

(C) js[kk 4

3x + = 2–4–y =

1–5z +rFkk

lery 4x – 2y – z = 1 ds e/; dks.k gS

(R) 60º

(D) ;fn a, b, c fHkUu okLrfod

la[;k,s gSA rFkk kcjbia ++ ,

kajcib ++ o kbjaic ++ rhu vlajs[kh; fcUnqvksa A, B o C ds fLFkfr

lfn'k gS rc →

AB rFkk→

AC ds e/; dks.k gS

(S) 90º


LrEHk-I LrEHk -II

(A) 'krjat cksMZ ij cuus okys

vlokZaxle vk;rks dh la[;k gksxh

(P) 31

(B) ,d ijhkk esa ikl gksus ds fy, 5 fo"k;ksa esa izR;sd esa U;wure vad ykuk

vko';d gSA rc mu rjhdks dh

la[;k ftuesa fo|kFkkhZ Qsy gks lds, gksxh

(Q) 32



(C) The number of divisors of

157500 which are divisible by 30

are

(R) 33

(D) The remainder obtained when

1 + 2 + 3 + ……… + 2009

is divided by 35 is

(S) 36

(C) 157500 ds Hkktdksa dh la[;k tks

30 ls foHkkftr gks gksxh

(R) 33

(D) 1 + 2 + 3 + ……… +

2009 dks 35 ls foHkkftr djus ij

'ks"kQy izkIr gksxk

(S) 36



PHYSICS

[k.M - I iz'u 1 ls 9 rd cgqfodYih iz'u gSaA izR;sd iz'u ds pkj fodYi





Q.1 ydM+h dk ,d CykWd 'kq) ty esa rSj jgk gS ftlds vk;ru dk nks frgkbZ Hkkx ty esa Mwck gqvk gSA rsy esa tc ;g CykWd rSjrk gS] rks bldk ,d pkSFkkbZ Hkkx Mwck jgrk gSA rsy dk ?kuRo gksxk& (A) 2666.7 kg/m3 (B) 5333.3 kg/m3

(C) 1333.3 kg/m3 (D) 3333.3 kg/m3

Q.2 ty ls Hkjk gqvk ,d chdj ,d fLizax rjktw ds IysVQkWeZ ij j[kk tkrk gSA rjktw 1.5 kg dk ikB~;kad n'kkZrk gSA 0.5 kg nzO;eku rFkk 500 kg/m3 ?kuRo ds ,d iRFkj dks chdj dh nhokjksa dks Li'kZ fd, fcuk ty esa iw.kZr;k Mwcks;k tkrk gSA vc rjktw dk ikB~;kad gks tk,xk& (A) 2 kg (B) 1 kg

(C) 2.5 kg (D) 3 kg





question. + 3 marks will be given for each correct

answer and – 1 mark for each wrong answer.

Q.1 A block of wood floats in fresh water with two third of its volume submerged. In oil the block floats with one fourth of its volume submerged. The density of oil is - (A) 2666.7 kg/m3 (B) 5333.3 kg/m3

(C) 1333.3 kg/m3 (D) 3333.3 kg/m3

Q.2 A beaker containing water is placed on the platform of a spring balance. The balance reads 1.5 kg. A stone of mass 0.5 kg and density 500 kg/m3 is completely immersed in water without touching the walls of beaker. Now the balance reading will be - (A) 2 kg (B) 1 kg

(C) 2.5 kg (D) 3 kg



Q.3 /kukRed x-vk dh fn'kk esa lapfjr rjax fokksHk dk

vk;ke t = 2 lsd.M ij y = 1x2–x

12 +

kjk rFkk t = 6

lsd.M ij y = 5x2x

12 ++

kjk fn;k tkrk gS] tgk¡

x rFkk y ehVj esa gSA foLian dk osx gksxk& (A) /kukRed x-fn'kk esa 1 m/s (B) _.kkRed x-fn'kk esa + 2 m/s (C) _.kkRed x-fn'kk esa 0.5 m/s

(D) _.kkRed x-fn'kk esa 1 m/s

Q.4 nks vkWxZu ikbi tks fd ,d fljs ij ls cUn gS] 5 foLian izfr lsd.M nsrk gS] tc ;s ewy Loj ÅRlftZr djrk gSA ;fn budh yEckbZ;k¡ 50 : 51 ds vuqikr esa gks] rks budh ewy vkofÙk;k¡ t (Hz esa) gksaxh -

(A) 255, 250 (B) 255, 260 (C) 260, 265 (D) 265, 270

Q.5 ,d eksVj lkbfdy fojkekoLFkk ls izkjEHk gksrh gS rFkk ljy js[kh; iFk ds vuqfn'k 2 m/s2 ls Rofjr gksrh gSA eksVj lkbfdy ds izkjfEHkd fcUnq ij ,d fLFkj fo|qr lk;ju yxk gqvk gSA eksVj lkbfdy fdruh nwj tk pqdh gksrh gS tc Mªkboj lk;ju dh vkofÙk] tc eksVj lkbfdy fojkekoLFkk esa gksrh gS] ij blds eku dk 94% Hkkx lqurk gS ? (/ofu dh pky = 330 ms–1)

(A) 49 m (B) 98 m (C) 147 m (D) 196 m

Q.3 The amplitude of wave disturbance propagating in

the positive x-axis is given by y = 1x2–x

12 +

at

t = 2 sec and y = 5x2x

12 ++

at t = 6 sec, where

x and y are in meters. Velocity of the pulse is - (A) 1 m/s in positive x-direction (B) + 2 m/s in negative x-direction (C) 0.5 m/s in negative x-direction

(D) 1 m/s in negative x-direction Q.4 Two organ pipes, each closed at one end, gives

5 beats per sec, when emitting their fundamental notes. If their lengths are in the ratio of 50 : 51, their fundamental frequencies (in Hz) are -

(A) 255, 250 (B) 255, 260 (C) 260, 265 (D) 265, 270

Q.5 A motor cycle starts from rest and accelerates along a straight path at 2 m/s2. At the starting point of the motor cycle there is a stationary electric siren. How far has the motor cycle gone when the driver hears the frequency of the siren at 94% of its value when the motor cycle was at rest ? (Speed of sound = 330 ms–1)

(A) 49 m (B) 98 m

(C) 147 m (D) 196 m



Q. 6 R f=kT;k rFkk l yEckbZ ds ,d yEcs dqpkyd flys.Mj dh lrg ij Q vkos'k ,dleku #i ls forfjr gSA flys.Mj ds pkjkas vkjs ,d jLlh dks yisVk tkrk gS ftlls mnzO;eku dk ,d CykWd yVdk gqvk gSA ;g nzO;eku uhps dh vksj xfr djus ds fy, Lora=k gS rFkk flys.Mj dks ?kqek ldrk gSA flys.Mj ds tM++Ro vk?kw.kZ dks ux.; ekudj] CykWd ds Roj.k dh x.kuk dhft,&

(A)

lm4Q1

g22

0

πµ

+

(B)

lm4Q1

g2

0

πµ

+

(C)

lm4Q1

g42

0

πµ

+

(D)

lm4Q1

g62

0

πµ

+

Q.7 +q rFkk –q vkos'k okys nks le#i pkyd xksyksa dks

ok;q esa ,d nwljs ls 'd' nwjh ij j[kk tkrk gSA izR;sd

xksys dh f=kT;k r rFkk buds dsUnzksa ds chp dh nwjh

d (d >> r) gSA fudk; dh /kkfjrk dh x.kuk dhft, –

++ ++ +

++ +

+ +

+q –qrr

d (A) 4πε0r (B) 2πε0r

(C) 4πloged

r0ε (D) 4πlogedr

Q. 6 A long insulating cylinder of radius R and length l carries a uniformly distributed surface charge Q. A string is coiled around the cylinder from which a block of mass m hangs. The mass is free to move downwards and can rotate the cylinder. Neglecting the moment of inertia of the cylinder, calculate the acceleration of the block –

(A)

lm4Q1

g22

0

πµ

+

(B)

lm4Q1

g2

0

πµ

+

(C)

lm4Q1

g42

0

πµ

+

(D)

lm4Q1

g62

0

πµ

+

Q.7 Two similar conductor spheres having charges +q and –q are placed at a separation 'd' from each other in air. The radius of each sphere is r and the separation between their centres is d (d >> r). Calculate the capacitance of the system –

+ + + + +

+ + +

+ +

+q –qr r

d (A) 4πε0r (B) 2πε0r

(C) 4πloged

r0ε (D) 4πlogedr



Q.8 ,d dkcZu izfrjks/kd ij Hkwjs] dkys] gjs rFkk lqugjs jax dh oy;sa fpfUgr dh tkrh gSA bldk izfrjks/k (vkse esa) gksxk&

(A) 3.2 × 105 ± 5 % (B) 1 × 106 ± 10 % (C) 1 × 107 ± 5 % (D) 1 × 106 ± 5 % Q.9 n'kkZ, x, ifjiFk ds fy,] og izfrjks/k ftlesa vf/kdre

Å"ek dk k; gksrk gS] gksxk&

4Ω

12Ω6Ω

2Ω

(A) 4Ω (B) 2Ω (C) 6Ω (D) 12Ω

iz'u 10 ls 14 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


Q.8 A carbon resistor is marked with the rings coloured brown, black, green and gold. The resistance (in ohm) is -

(A) 3.2 × 105 ± 5 % (B) 1 × 106 ± 10 % (C) 1 × 107 ± 5 % (D) 1 × 106 ± 5 %

Q.9 For the circuit shown, the resistance in which the maximum heat is dissipated is -

4Ω

12Ω6Ω

2Ω

(A) 4Ω (B) 2Ω (C) 6Ω (D) 12Ω Questions 10 to 14 are multiple choice questions.

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

out of which MULTIPLE (ONE OR MORE THAN

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.10 ,d yEch u uyh esa ‘ρ’ ?kuRo dk nzo bl izdkj Hkjk tkrk gS fd nksuksa Hkqtkvksa esa nzo ds Åij uyh dh yEckbZ ‘a’ gSA uyh dh ,d Hkqtk (nk;ha Hkqtk) dks lhy cUn dj fn;k tk, rks ifjofrZr uyh eas (‘P0’ ok;qe.Myh; nkc)

a

(A) nzo ck;ha uyh ls ckgj fudysxk ;fn a < g4

P0ρ

(B) ‘a’ ds fdlh Hkh eku ds fy, nzo ck;ha uyh ls ckgj fudysxk

(C) ck;ha Hkqtk eas nzo lrg xfr ugha djsxh ;fn a = g2

P0ρ

gks

(D) nk;ha Hkqtk esa nzo lrg ds uhps vk tk,xh

Q.11 ,d ik=k esa ty dkss α m3/s dh fu;r nj ls Mkyk tkrk

gSA VSad dh ryh eas a ks=kQy dk ,d y?kq fNnz gSA ik=k esa

ty dk vf/kdre Lrj fuEu ds lekuqikrh gksxk&

(A) α (B) α2 (C) a–1 (D) a–2

Q.10 A long u tube is filled with a liquid of density ‘ρ’ such that length of tube above liquid is ‘a’ in both arm. One side of tube (right arm) is sealed the tube is inverted (‘P0’ atmospheric pressure)

a

(A) Liquid will spill out the left tube if a < g4

P0ρ

(B) Liquid will not spill out the left tube for any value of ‘a’

(C) Liquid surface in left arm will not move if

a = g2

P0ρ

(D) Liquid surface in right arm will come down

Q.11 Water is being poured in a vessel at a constant

rate α m3/s. There is a small hole of area a at

the bottom of the tank. The maximum level of

water in the vessel is proportional to :

(A) α (B) α2 (C) a–1 (D) a–2



Q.12 tc ,d rjax lapfjr gksrh gS] rks& (A) ,d lery rjax ds fy, rjax rhozrk fu;r jgrh gS (B) ,d xksyh; rjax ds fy, rjax rhozrk L=kksr ls nwjh ds O;qRØe ds vuqlkj ?kVrh gSA (C) ,d xksyh; rjax ds fy, rjax rhozrk L=kksr ls nwjh ds oxZ ds O;qRØe ds fy, rjax rhozrk nwjh ds O;qRØe ds vuqlkj ?kVrh gS

(D) ,d js[kh; L=kksr ds fy, rjax rhozrk nwjh ds O;qRØe ds vuqlkj ?kVrh gS

Q.13 gekjs ikl ux.; eksVkbZ dh vuUr vkdkj dh ,d vpkyd ijr gS] ftl ij ,dleku i"Bh; vkos'k ?kuRo–σ gS rFkk blds lekUrj D eksVkbZ dh ,d vuUr vkdkj dh ifêdk j[kh gqbZ gS ftl ij ,dleku vk;ru vkos'k ?kuRo +ρgSA lHkh vkos'k fLFkj gSaA

+ρ D

– σ

(A) _.kkRed #i ls vkosf'kr ijr ds uhps h nwjh ij fo|qr

ks=k dk ifjek.k 02

Dε

σ−ρ gS

(B) _.kkRed #i ls vkosf'kr ijr ds uhps h nwjh (h < D) ij ifêdk ds vUnj fo|qr ks=k dk ifjek.k

02)h2D(

ε−ρ+σ gS

(C) ifêdk ds ry ls h nwjh uhps fo|qr ks=k dk

ifjek.k 04

Dε

σ−ρ gS

(D) ifêdk ds ry ls h nwjh uhps fo|qr ks=k dk ifjek.k

02D

εσ−ρ gS

Q.12 As a wave propagates –(A) The wave intensity remains constant for a plane wave (B) The wave intensity decreases as the inverse of the distance from the source for a spherical wave (C) The wave intensity decreases as the inverse square of the distance from the source for a spherical wave (D) The wave intensity decreases as the inverse of the distance for a line source

Q.13 We have an infinite non-conducting sheet of

negligible thickness carrying a uniform surface charge density –σ and next to it , an infinite parallel slab of thickness D with uniform volume charge density +ρ. All charges are fixed.

+ρD

– σ

(A) Magnitude of electric field at a distance h above

the negatively charged sheet is 02

Dε

σ−ρ

(B) Magnitude of electric field inside the slab at a distance h below the negatively charged

sheet (h < D) is 02

)h2D(ε

−ρ+σ

(C) Magnitude of electric field at a distance h

below the bottom of the slab is 04

Dε

σ−ρ

(D) Magnitude of electric field at a distance h

below the bottom of the slab is 02

Dε

σ−ρ



Q.14 R = 10Ω o E = 13 V gks rFkk oksYVehVj ;k vehVj vkn'kZ

gks] rks

V

A

c 6V

b 3Ω

8V

a

R

E

(A) vehVj dk ikB~;kad 2.4 A gksxk

(B) vehVj dk ikB~;kad 8.4 A gksxk

(C) oksYVehVj dk ikB~;kad 8.4 V gksxk

(D) oksYVehVj dk ikB~;kad 27 V gksxk

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

gSaA (iz'u 15 ls 20) 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

Q.14 R = 10Ω & E = 13 V and voltmeter & ammeter

are ideal then -

V

A

c 6V

b 3Ω

8V

a

R

E

(A) Reading of Ammeter is 2.4 A

(B) Reading of Ammeter is 8.4 A

(C) Reading of voltmeter is 8.4 V

(D) Reading of voltmeter is 27 V This section contains 2 paragraphs; each has

3 multiple choice questions. (Questions 15 to 20) Each

question has 4 choices (A), (B), (C) and (D) out of



question. + 4 marks will be given for each correct

answer and –1 mark for each wrong answer.



Passage # 1 : (Q. 15 to 17) A cylindrical container of length L is full to the

brim with a liquid which has mass density ρ. It

is placed on a weight-scale; the scale reading is

w. A light ball which would float on the liquid

if allowed to do so, of volume V and mass m is

pushed gently down and held beneath the

surface of the liquid with a rigid rod of

negligible volume as shown on the left.

rigid rod

L

Fig.

Q.15 What is the mass M of liquid which overflowed while the ball was being pushed into the liquid?

(A) ρV (B) m (C) m – ρV (D) None of these

x|ka'k # 1 (iz- 15 ls 17) L yEckbZ dk ,d csyukdkj ik=k ,d nzo ls Åij rd Hkjk gqvk gS ftldk nzO;eku ?kuRo ρ gSA bls ,d Hkkj ekih Ldsy ij j[kk tkrk gS( Ldsy dk ikB~;kad w gSA V vk;ru rFkk m nzO;eku dh ,d gYdh xsan nzo ij rSjsxh ;fn bls ,slk djus fn;k tk,A vc bls lko/kkuh ls uhps /kdsyk tkrk gS rFkk ux.; vk;ru dh ,d n<+ NM+ ds lkFk nzo dh lrg ds uhps j[kk tkrk gS] tSlk fd ck;ha vksj ds fp=k esa n'kkZ;k x;k gSA

rigid rod

L

Fig.

Q.15 tc nzo esa xsan dks /kdsyk tkrk gS] rks cg fudyus

okys nzo dk nzO;eku M D;k gksxk ?

(A) ρV (B) m

(C) m – ρV (D) buesa ls dksbZ ugha



Q.16 What is the reading of the scale when the ball is

fully immersed –

(A) w – ρVg (B) w

(C) w + mg – ρVg (D) None of these

Q.17 If instead of being pushed down by a rod, the

ball is held in place by a thin string attached to

the bottom of the container as shown on the

right. What is the tension T in the string?

(A) (ρV – m)g (B) ρVg

(C) mg (D) None of these

Passage # 2 (Ques. 18 to 20) A ball of radius R carries a positive charge whose

volume charge density depends only on the

distance r form the ball's centre as :

ρ = ρ0

Rr–1

where ρ0 is a constant. Assume ε as the

permittivity of the ball.

Q.16 tc xsan dks iw.kZr;k nzo esa Mwcks;k tkrk gS rks

Ldsy dk ikB~;kad D;k gksxk ?

(A) w – ρVg (B) w

(C) w + mg – ρVg (D) buesa ls dksbZ ugha

Q.17 ;fn ,d NM+ kjk xasn dks uhps /kdsyus ds vfrfjDr bls ik=k dh ryh ls ,d iryh jLlh kjk tksM+dj j[kk tkrk gS tSlk fd nk;ha vksj ds fp=k esa n'kkZ;k x;k gS] rks jLlh esa ruko T D;k gksxk?

(A) (ρV – m)g (B) ρVg (C) mg (D) buesa ls dksbZ ugha

x|ka'k # 2 (iz- 18 ls 20)

R f=kT;k dh ,d xsan ftl ij /kukRed vkos'k gS]

dk vk;ru vkos'k ?kuRo xsan ds dsUnz ls r nwjh ij

fuEukuqlkj fuHkZj djrk gS&

ρ = ρ0

Rr–1

tgk¡ ρ0 ,d fu;rkad gSA ekuk ε xsan dh fo|qr'khyrk gSA



Q.18 The magnitude of the electric field as a function of the distance r outside the ball is given by :

(A) E = 2

30

r8R

ερ (B) E =

2

30

r12Rε

ρ

(C) E = 3

20

r8R

ερ (D) E =

3

20

r12Rε

ρ

Q.19 The value of distance rm, at which electric field

intensity is maximum, is given by :

(A) rm = 3R (B) rm =

2R3

(C) rm = 3R2 (D) rm =

3R4

Q.20 The maximum electric field intensity is :

(A) Em = ε

ρ9

R0 (B) Em = R90ερ

(C) Em = ε

ρ3

R0 (D) Em = ε

ρ6

R0

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) 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, B-Q, B-R, C-P, C-Q and D-S, then the correctly bubbled 4 × 4 matrix should be as follows :

Q.18 fo|qr ks=k dk ifjek.k xsan ds ckgj r nwjh ds Qyu ds #i esa fuEu kjk fn;k tkrk gS&

(A) E = 2

30

r8R

ερ (B) E =

2

30

r12Rε

ρ

(C) E = 3

20

r8R

ερ (D) E =

3

20

r12Rε

ρ

Q.19 nwjh rm dk eku] ftl ij fo|qr ks=k dh rhozrk vf/kdre gksrh gS] fn;k tkrk gS&

(A) rm = 3R (B) rm =

2R3

(C) rm = 3R2 (D) rm =

3R4

Q.20 fo|qr ks=k dh vf/kdre rhozrk gksxh&

(A) Em = ε

ρ9

R0 (B) Em = R90ερ

(C) Em = ε

ρ3

R0 (D) Em = ε

ρ6

R0

[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) 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, B-Q, B-R, C-P, C-Q rFkk D-S gS, rks lgh fof/k ls dkys fd;s x;s xksyksa dk 4 × 4 eSfVªDl uhps n'kkZ;s vuqlkj gksxk :



A B C D

P Q R S

S

P

P P Q R

R R

Q Q

S S

P Q R S

Mark your response in OMR sheet against the question number of that question in section-II. + 6 marks will be given for complete correct answer and No Negative marks for wrong answer. However, 1 mark will be given for a correctly marked answer in any row.

Q.1 In each situation of column-I, some charge distributions are given with all details explained. The electrostatic potential energy and its nature is given situation in column-II. Then match situation in column-I with the corresponding results in column-II -

Column-I Column-II

(A) A thin shell of radius (P) a

Q8

1 2

0πεin

a and having magnitude a charge –Q uniformly distributed over its surface as shown

– Q

a

ABCD

P Q R S

S

P

P P Q R

R R

Q Q

S S

P Q R S

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 + 6 vad fn;s tk;saxs rFkk xyr mÙkj ds fy;s dksbZ +_.kkRed vadu ugha gS (vFkkZr~ dksbZ vad ugha ?kVk;k tk;sxk)A fdUrq] fdlh iafDr esa lgh :i ls fpfUgr mÙkj ds fy, 1 vad fn;k tk;sxkA

Q.1 LrEHk-I dh izR;sd fLFkfr esa] dqN vkos'k forj.k lEiw.kZ foLrr tkudkjh ds lkFk fn;s x, gSaA bl fLFkfr esa fLFkjoS|qr fLFkfrt ÅtkZ rFkk bldh izdfr LrEHk-II esa nh xbZ gSA rc LrEHk-I esa nh xbZ fLFkfr dk LrEHk-II esa laxr ifj.kkeksa ls feyku dhft, -

LrEHk-I LrEHk-II (A) a f=kT;k dh ,d iryh ijr (P) ifjek.k esa

a

Q8

1 2

0πε

ftldh lrg ij –Q vkos'k ,dleku #i ls forfjr gS] n'kkZ, vuqlkj gS

– Q

a



(B) A thin shell of radius 2a5 (Q)

aQ

203 2

0πε

and having a charge –Q in magnitude uniformly distributed over its surface and a point charge –Q placed at its centre as shown

– Q

2a5

(C) A solid sphere of radius a (R) a

Q5

2 2

0πε

and having a charge –Q in magnitude uniformly distributed throughout its volume

as shown

– Q

a

(D) A solid sphere of radius a (S) positive in and having a charge –Q sign uniformly distributed throughout its volume. The solid sphere is surrounded by a concentric thin uniformly charged spherical shell of radius 20 and carrying charge –Q as shown

– Q

a 20

– Q

(T) negative in Sign

(B) 2a5 f=kT;k dh ,d iryh ijr (Q) ifjek.k esa

ftldh lrg ij –Q a

Q20

3 2

0πε vkos'k ,dleku #i ls forfjr

gS rFkk blds dsUnz ij ,d fcUnq vkos'k –Q n'kkZ, vuqlkj j[kk gqvk gS

– Q

2a5

(C) a f=kT;k dk ,d Bksl xksyk (R) ifjek.k esa

ftlds lEiw.kZ vk;ru ij –Q a

Q5

2 2

0πε

vkos'k ,dleku #i ls forfjr gS]

n'kkZ, vuqlkj gS

– Q

a

(D) a f=kT;k ds ,d Bksl xksys ds (S) fpUg esa /kukRed lEiw.kZ vk;ru ij –Q vkos'k ,dleku #i ls forfjr gSA bl Bksl xksys dks 20 f=kT;k rFkk –Q vkos'k okys ,d ladsUnzh; irys ,dleku vkosf'kr xksyh; dks'k kjk <dk tkrk gS tSlk fd n'kkZ;k x;k gS

(T) fpUg esa

_.kkRed

– Q

a20

– Q



Q.2 In the circuits drawn in column I of the following table, all the bulbs are identical. Match the entries of column I with the entries of column II

Column I Column II

(A)

E

(P) Current drawn

from the battery is maximum

(B)

E EE

(Q) Current drawn

from the battery is the least

(C) E (R)Bulbs will lit the

brightest

(D)

EE

(S) Bulbs will lit

with brightness lying between maximum and minimum value

Q.2 fuEu lkj.kh ds LrEHk I esa cuk, x, ifjiFkksa esa lHkh cYc ,dleku gSaA LrEHk –I dh izfof"V;ksa dk LrEHk&II dh izfof"V;ksa ls feyku dhft,

LrEHk I LrEHk II

(A)

E

(P)cSVjh ls fudyus

okyh /kkjk vf/kdre gS

(B)

E EE

(Q) cSVjh ls fudyus

okyh /kkjk U;wure gS

(C) E (R) cYo vf/kdre iznhfIr

ls izdkf'kr gksaxs

(D)

EE

(S) cYc vf/kdre rFkk

U;wure eku ds e/; dh iznhfIr ls izdkf'kr gksaxs



CHEMISTRY





question. + 3 marks will be given for each correct answer

and – 1 mark for each wrong answer.

Q.1 The reduction potential of hydrogen half cell will be negative if (T = 298K) -

(A) 2HP = 1atm and [H+] = 1.0 M

(B) 2HP = 2atm and [H+] = 2.0 M

(C) 2HP = 2atm and [H+] = 1.0 M

(D) 2HP = 1atm and [H+] = 2.0 M

Q. 2 Given (i) MnO4

– + 8H+ + 5e– → Mn2+ + 4H2O E° = x1V (ii) MnO2 + 4H+ + 2e– → Mn2+ + 2H2O E° = x2V Find Eº for the following reaction −+− ++ e3H4MnO4 → MnO2 + 2H2O

(A) x2 – x1 (B) x1–x2

(C) 3

x2x5 21 − (D) 3

x5x2 21 −

[k.M - I iz'u 1 ls 9 rd cgqfodYih iz'u gSaA izR;sd iz'u ds pkj fodYi





Q.1 gkbMªkstu v)Zlsy dk vip;u foHko _.kkRed gksxk] ;fn (T = 298K) -

(A) 2HP = 1atm rFkk [H+] = 1.0 M

(B) 2HP = 2atm rFkk [H+] = 2.0 M

(C) 2HP = 2atm rFkk [H+] = 1.0 M

(D) 2HP = 1atm rFkk [H+] = 2.0 M

Q. 2 fn;k gS (i) MnO4

– + 8H+ + 5e– → Mn2+ + 4H2O E° = x1V (ii) MnO2 + 4H+ + 2e– → Mn2+ + 2H2O E° = x2V fuEu vfHkfØ;k ds fy, Eº Kkr dhft,A −+− ++ e3H4MnO4 → MnO2 + 2H2O (A) x2 – x1 (B) x1–x2

(C) 3

x2x5 21 − (D) 3

x5x2 21 −



Q. 3 The specific conductivity of an aqueous solution of a weak monoprotic acid is 0.00033 ohm–1 cm–1 at a concentration, 0.02 M. If at this concentration the degree of dissociation is 0.043, calculate the value of 0Λ (ohm–1 cm2/eqt).

(A) 483 (B) 438 (C) 348 (D) 384

Q.4 Which of the following complexes produces three moles of silver chloride when its one mole treated with excess of silver nitrate ?

(A) [Cr(H2O)3Cl3] (B) [Cr(H2O)4Cl2]Cl (C) [Cr(H2O)5Cl]Cl2 (D) [Cr(H2O)6]Cl3 Q.5 Which metal is extracted using a hydrometallurgical

process involving complexation ? (A)Mg (B) Ag (C) Cu (D) Zn

Q.6 A liquid which is confined inside an adiabatic container and piston is suddenly taken from state 1 to state 2 by single stage process. If the piston comes to rest at point 2 as shown. Then the enthalpy change for the process will be.

P

2P0

P0

V0 4V0 V

1

2

(A) 1VP2

H 00

−γγ

=∆ (B) 1VP3

H 00

−γγ

=∆

(C) 00VPH −=∆ (D) None of these

Q. 3 ,d nqcZy eksuksizksfVd vEy ds tyh; foy;u dh fof'k"V pkydrk 0.02 M lkUnzrk ij 0.00033 ohm–1 cm–1 gSA ;fn bl lkUnzrk ij fo;kstu dh ek=kk 0.043 gks rks

0Λ dk eku Kkr dhft,A (ohm–1 cm2/eqt)

(A) 483 (B) 438 (C) 348 (D) 384 Q.4 fuEu esa ls dkSulk ladqy flYoj DyksjkbM ds rhu eksy

cukrk gSA tc blds ,d eksy dks flYoj ukbVªsV ds vkf/kD; ds lkFk mipkfjr fd;k tkrk gS ?

(A) [Cr(H2O)3Cl3] (B) [Cr(H2O)4Cl2]Cl (C) [Cr(H2O)5Cl]Cl2 (D) [Cr(H2O)6]Cl3

Q.5 dkSulh /kkrq dks tfVyrk (complexation) ds lkFk tyh; /kkrq izØe dks iz;qDr djds fu"d£"kr fd;k tkrk gS ?

(A)Mg (B) Ag (C) Cu (D) Zn

Q.6 nzo tks :)ks"e ik=k esa gSA fiLVu dks vpkud fLFkfr 1 ls fLFkfr 2 eas ,dy izØe kjk yk;k tkrk gSA ;fn fiLVu dks n'kkZ;s vuqlkj fcUnq 2 ls fojke ij yk;k tkrk gS] rks izØe ds fy, ,UFkSfYi ifjorZu gksxk

P

2P0

P0

V0 4V0 V

1

2

(A) 1VP2

H 00

−γγ

=∆ (B) 1VP3

H 00

−γγ

=∆

(C) 00VPH −=∆ (D) buesa ls dksbZ ugha



Q.7 10 lt box contain O3 and O2 at equilibrium at 2000 K. The ∆G° = – 534.52 kJ at 8 atm equilibrium pressure. The following equilibrium is present in the container

2O3(g) 3O2(g) The partial pressure of O3 will be (In 10 = 2.3, R =

8.3 Jmole–1K–1)

(A) 8 × 10–6 (B) 22.62 × 10–7 (C) 9.71 × 10–6 (D) 9.71 × 10–2

Q.8 The average person can see the red colour imparted by the complex [Fe(SCN)]2+ to an aqueous solution if the concentration of the complex is 6 × 10–6 M or greater. What minimum concentration of KSCN would be require to make it possible to detect 1 ppm (part per million) of Fe (III) in a natural water sample ? The instability constant for

Fe(SCN)2+ Fe3+ + SCN– is 7.142 × 10–3. (A) 0.0036 M (B) 0.0037 M (C) 0.0035 M (D) None of these

Q. 9 During poling which of the following reaction occurs in extraction of Cu.

(A) 4CuO + CH4 → 4Cu↓ + CO2 + 2H2O (B) Cu2S + 3/2O2 → Cu2O + SO2 (C) Cu2S + 3/2O2→ CuO + SO2 (D) Cu2S + 2Cu2O → 6Cu + SO2

Q.7 10 lt ds ckWDl esa 2000 K ij lkE; ij O3 o O2 gSa 8

atm lkE; nkc ij ∆G° = – 534.52 kJ gS ik=k esa fuEu

lkE; LFkkfir gksrk gS

2O3(g) 3O2(g); O3 dk vkaf'kd nkc gksxk (In 10

= 2.3, R = 8.3 mole–1K–1)

(A) 8 × 10–6 (B) 22.62 × 10–7

(C) 9.71 × 10–6 (D) 9.71 × 10–2

Q.8 ,d vkSlr O;fDr fn;s x;s tyh; foy;u esa [Fe(SCN)]2+ kjk fn;s x;s yky jax dks ns[k ldrk gS] ;fn ladqy dh lkUnzrk 6 × 10–6 M ;k blls vf/kd gks rks izkÑfrd ty ds uewus esa Fe(III) ds 1 ppm dks tk¡pus ds fy, lEHko vko';d KSCN dh U;wure lkanzrk D;k gksxh] fuEu vfHkfØ;k ds fy, LFkkf;Ro xq.kkad 7.142 × 10–3 fn;k x;k gSA

Fe(SCN)2+ Fe3+ + SCN–

(A) 0.0036 M (B) 0.0037 M

(C) 0.0035 M (D) buesa ls dksbZ ugha

Q. 9 iksfyax ds nkSjku fuEu esa ls dkSulh vfHkfØ;k Cu ds fu"d"kZ.k esa iz;qDr gksrh gS&

(A) 4CuO + CH4 → 4Cu↓ + CO2 + 2H2O (B) Cu2S + 3/2O2 → Cu2O + SO2 (C) Cu2S + 3/2O2→ CuO + SO2 (D) Cu2S + 2Cu2O → 6Cu + SO2



Questions 10 to 14 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE THAN 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. 10 The following reaction attains equilibrium at high temperature.

N2(g) + 2H2O(g) + heat 2NO(g) + 2H2(g) The yield of NO is affected by (A) increasing the nitrogen concentration (B) decreasing the hydrogen concentration (C) compressing the reaction mixture (D) none of these

Q. 11 Solid ammonium carbamate, NH4CO2NH2(s), dissociates into ammonia and carbon dioxide when it evaporates as shown by

NH4CO2NH2(s) 2NH3(g)+ CO2(g) At 25°C, the total pressure of the gasses in

equilibrium with the solid is 0.116 atm. If 0.1 atm of CO2 is introduced after equilibrium is reached then–

(A) final pressure of CO2 will be less than 0.1 atm (B) final pressure of CO2 will be more than 0.1 atm (C) pressure of NH3 will decrease due to addition

of CO2 (D) pressure of NH3 will increase due to addition of

CO2

iz'u 10 ls 14 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. 10 fuEu vfHkfØ;k mPp rki ij lkE; LFkkfir djrh gS N2(g) + 2H2O(g) + Å"ek 2NO(g) + 2H2(g)

NO dh yfC/k fdlls izHkkfor gksrh gS (A) ukbVªkstu dh lkUnzrk c<+kus ls (B) gkbMªkstu dh lkUnzrk ?kVkus ls (C) vfHkfØ;k feJ.k ds laihMu ls (D) buesa ls dksbZ ugha Q. 11 Bksl veksfu;e dkcZsesV NH4CO2NH2(s) veksfu;k o

dkcZu MkbZ&vkWDlkbM esa fo;ksftr gks tkrk gS tc

bls uhps n'kkZ;s vuqlkj okf"ir fd;k tkrk gS NH4CO2NH2(s) 2NH3(g)+ CO2(g) 25°C ij xSalksa dk Bksl ds lkFk lkE; ij dqy nkc

0.116 atm gS ;fn lkE;koLFkk ig¡qpus ds i'pkr~

0.1 atm CO2 dks feyk;k tk;s rks–

(A) CO2 dk vfUre nkc 0.1 atm ls de gksrk gS

(B) CO2 dk vfUre nkc 0.1 atm ls vf/kd gksrk gS

(C) CO2 feykus ds dkj.k NH3 dk nkc ?kVrk gSA

(D) CO2 feykus ds dkj.k NH3 dk nkc c<+rk gSA



Q. 12 fuEu esa ls fdl ;qXe esa [kfut Lor% vip;u fof/k

kjk /kkrq esa ifjofrZr gks tkrk gS?

(A) Cu2,S,PbS (B) PbS, HgS

(C) PbS, ZnS (D) Ag2S, Cu2S

Q. 13 fuEu esa ls dkSulk lgh lqesfyr gS (A) dkWij-cslsej ifjorZd (B) vk;ju-okR;kHkV~Vh (C) Øksfe;e-,Y;qfeuksrkih fof/k (D) fVu-oS|qr vi?kVuh; vip;u

Q. 14 PbS ls Pb ds fu"d"kZ.k esa CaO feyk;k tkrk gS] ;g:

(A) PbSO4 ds fuekZ.k dks jksdrk gS

(B) PbSiO3 ds fuekZ.k dks jksdrk gS

(C) /kkrqey ds fuekZ.k esa iz;qDr gksrk gS

(D) mijksDr esa ls dksbZ Hkh lgh ugha gS

bl [k.M esa 2 vuqPNsn fn;s x;s gSa] izR;sd esa 3 cgqfodYih iz'u gSaA (iz'u 15 ls 20) 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

Q. 12 In which of the following pair (s), the minerals are

converted in to metal by self-reduction process?

(A) Cu2,S,PbS (B) PbS, HgS

(C) PbS, ZnS (D) Ag2S, Cu2S

Q.13 Which of the following is/are correctly matched?

(A) Copper - Bessemer converter

(B) Iron – Blast furnance

(C) Chromium – Aluminothermic process (D) Tin – Electrolytic reduction

Q. 14 In the extraction of Pb from PbS, CaO is added, it :

(A) prevents formation of PbSO4

(B) prevents formation of PbSiO3

(C) is used as a slag (D) none of the above is correct

This section contains 2 paragraphs; each has 3 multiple choice questions. (Questions 15 to 20) 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. 15 to 17) One method for the manufacture of synthesis gas

(primarily a mixture of CO & H2) is the catalytic

reforming of CH4 with steam at high temperature

and atmospheric pressure.

CH4(g) + H2O(g) → CO(g) + 3H2(g) ∆H298 = 206 KJ

The only other reaction which occurs to an

appreciable extent is the water gas shift reaction

CO(g) + H2O(g) → CO2(g) + H2(g) ∆H298 = – 41.2 KJ

If the reactants are supplied in the ratio, 2 mole

steam to 1 mole CH4, and CH4 is completely

converted and product stream contains 17.4 mole %

CO.

Given CPm of

CH4 = 5.3 R

H2O = 4.2 R (In the temperature range of 298 to 598 K)

CO = 3.8 R

H2 = 3.6 R

CO2 = 6 R

H2O = 4.7 R (In the temperature range of 598 to 1298 K)

x|ka'k # 1 (iz- 15 ls 17)

mPp rki ok;qe.Myh; nkc ij CH4 dk Hkki (steam) ds lkFk mRizsjdh ty vi?kVu] la'kys"k.k xSl (synthesis gas) (CO rFkk H2 izkFkfed feJ.k) fuekZ.k ds fy, ,d fof/k gSA CH4(g) + H2O(g) → CO(g) + 3H2(g) ∆H298 = 206 KJ

,d vU; vfHkfØ;k ty xSl foLFkkiu vfHkfØ;k gksrh gS

CO(g) + H2O(g) → CO2(g) + H2(g) ∆H298 = – 41.2 KJ

;fn fØ;kdkjd 2 mole Hkki (steam) rFkk 1 mole CH4,

ds vuqikr esa izokfgr fd;s tkrs gS rFkk CH4 iw.kZr

ifjofrZr gks tkrh gS rFkk mRikn 17.4 mole % CO

j[krk gSA

CPm ds eku fn;s x;s gSa

CH4 = 5.3 R

H2O = 4.2 R (298 ls 598 K rkiØe rd)

CO = 3.8 R

H2 = 3.6 R

CO2 = 6 R H2O = 4.7 R (598 ls 1298 K rkiØe rd)



Q.15 25°C ij izfr eksy CH4 ds fy, ,UFkSYih ifjorZu

(HkV~Vh (reactor) ds fy, vko';d Å"ek) dh x.kuk

dhft,

(A) 280. 6 KJ (B) 200.6 KJ

(C) 196 KJ (D) 176 KJ

Q.16 CH4 ds izfreksy [kpZ gksus ij vfHkfØ;k ds iw.kZ gksus ds

i'pkr~ CO ds eksyksa dh x.kuk dhft, –

(A) 0.13 (B) 0.87

(C) 3.17 (D) buesa ls dksbZ ugh

Q.17 ekuk fd fØ;kdkjd igys 598 K ij xeZ fd;s tkrs gS

rFkk HkV~Vh (reactor) dks Å"ek bl izdkj nh tkrh gS

fd mRikn dk rki 1298 K rd igqap tk,A mijksDr

ifjorZu ds fy, HkV~Vh esa izfreksy esFksu dh [kir ds

fy, vko';d Å"ek dh x.kuk djsa

(A) – (15.33 R + 200.6) KJ (B) (15.33 R + 200.6) KJ (C) – (8.33 R + 200.6) KJ

(D) buesa ls dksbZ ugh

Q.15 Calculate enthalpy change for above changes/reactions

(heat requirement for the reactor) at 25°C per mole of

CH4 consumed -

(A) 280. 6 KJ (B) 200.6 KJ

(C) 196 KJ (D) 176 KJ

Q.16 Calculate moles of CO after completion of reaction

per mole of CH4 consumed –

(A) 0.13 (B) 0.87

(C) 3.17 (D) none of these

Q.17 Assume the reactants to be preheated to 598 K.

Finally the products are at 1298 K. Calculate

enthalpy changes for above changes heat

requirement for the reactor per mole of the CH4

consumed-

(A) – (15.33 R + 200.6) KJ

(B) (15.33 R + 200.6) KJ

(C) – (8.33 R + 200.6) KJ

D) none of these



x|ka'k # 2 (iz- 18 ls 20)

fuEu x|ka'k dks lko/kkuhiwoZd if<+;s o iz'uksa ds mÙkj

nhft,

2A2 A4 …..(i) 11p atm

812K −=

A2 + 2C A2C2 …..(ii)

A2 C2 2AC …..(iii)

A2 o C dks fuf'pr rki ij fuf'pr vk;ru ds cUn ik=k esa 3 : 1 eksy vuqikr esa fy;k x;k gSA o mijksDr rhuksa lkE; ,d lkFk LFkkfir gksrs gSaS izFke vfHkfØ;k ds fy,

1pK , 2/81 atm–1 gSA lkE; ij A4(g) o AC(g)

izR;sd dk vkaf'kd nkc 1/2 atm ik;k tkrk gS rFkk lkE;

ij dqy nkc 4

27 atm ik;k tkrk gS rc –

Q.18 lkE; ij A2C2 dk vkaf'kd nkc gS –

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

Q.19 lkE; ij A2 o AC xSlksa dk eksy vuqikr gS –

(A) 9/2 (B) 7/2 (C) 8 (D) 9 Q.20 fuEu vfHkfØ;k ds fy, lkE; fLFkjkad gS

(A) (B) (C) (D)

Passage # 2 (Ques. 18 to 20) Read the following passage carefully and answer

the questions. 2A2 A4 …..(i)

11p atm

812K −=

A2 + 2C A2C2 …..(ii)

A2 C2 2AC …..(iii)

A2 and C are taken in 3 : 1 mole ratio in a closed container of a certain volume at a fixed temperature and above three equilibriums are established simultaneously.

1pK for the first

reaction is 2/81 atm–1. At equilibrium partial pressure of A4(g) and AC(g) are found to be 1/2 atm each and the total pressure at equilibrium is

found to be 4

27 atm. Then

Q.18 The partial pressure of A2C2 at equilibrium is

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

Q.19 The mole ratio of gases A2 and AC at equilibrium is

(A) 9/2 (B) 7/2 (C) 8 (D) 9 Q.20

(A) (B) (C) (D)



[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) 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, B-Q, B-R, C-P, C-Q rFkk D-S gS, rks lgh fof/k ls dkys

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

ABCD

P Q R S

S

P

P P Q R

R R

Q Q

S S

P Q R S

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 + 6 vad fn;s tk;saxs rFkk xyr mÙkj ds fy;s dksbZ +_.kkRed vadu ugha gS (vFkkZr~ dksbZ vad ugha ?kVk;k tk;sxk)A fdUrq] fdlh iafDr esa lgh :i ls fpfUgr mÙkj ds fy, 1 vad fn;k tk;sxkA

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) 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, B-Q, B-R,

C-P, C-Q and D-S, then the correctly bubbled 4 × 4

matrix should be as follows :

A B C D

P Q R S

S

P

P P Q R

R R

Q Q

S S

P Q R S

Mark your response in OMR sheet against the question

number of that question in section-II. + 6 marks will be

given for complete correct answer and No Negative

marks for wrong answer. However, 1 mark will be given

for a correctly marked answer in any row.



Q.1 LrEHk feyku dhft,

LrEHk -I LrEHk-II

(A) ckWDlkbV (P) fyfpax

(B) vtZs.VkbV (Q) oS|qr vi?kVuh vip;u

(C) xsysuk (R) >kx Iyou

(D) dkusZykbV (S) lYQkbM v;Ld


LrEHk-I LrEHk -II

(A) mRØe.kh; :nks"e izØe (P) ∆s = 0

(B) vuqRØe.kh; :nks"e izØe (Q) Q = 0

(C) lerkih; izØe (R) ∆s = ∫ volt

(D) ,lsUVªkWfid (Isentropic) izØe

(S) ∆H = ∆ U


Column -I Column-II

(A) Bauxite (P) Leaching

(B) Argentite (Q)Electrolytic reduction

(C) Galena (R) Froth floatation

(D) Cornallite (S) Suphide ore


Column -I Column-II

(A) Reversible adiabatic process (P) ∆s = 0

(B) Irreversible adiabatic process (Q) Q = 0

(C) Isothermal process (R) ∆s = ∫ volt

(D) Isentropic process

(S) ∆H = ∆ U



Name : _________________________________________________________ Roll No. : __________________________

ijhkkfFkZ;ksa ds fy, funsZ'k %

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. iz'u&i=k ds bl eq[k i"B ij fn, x;s [kkyh LFkku esa viuk uke] vuqØek¡d fyf[k;sA

3. iz'ui=k esa jQ dk;Z gsrq [kkyh LFkku fn;s x, gSaA jQ dk;Z gsrq dksbZ vfrfjDr iqfLrdk ugha nh tk,sxhA

4. mRrj ds fy,] OMR vyx ls nh tk jgh gSA

5. ifjohkdksa kjk funsZ'k fn;s tkus ls iwoZ iz'u&i=k iqfLrdk dh lhy dks ugha rksMsa+A

6. [kkyh dkxt+] fDyicksMZ] y?kqx.kd lkj.kh] LykbM :y] dsYdqysVj] lsY;qyj Qksu] ist+j rFkk bysDVªkWfud midj.kksa dh

fdlh Hkh voLFkk esa ijhkk dk esa vanj ys tkus dh vuqefr ugha nh tk,sxhA

B. OMR dh iwfrZ :

7. OMR 'khV ds CykWdksa esa viuk uke] vuqØek¡d] cSp] dkslZ rFkk ijhkk dk dsUnz Hkjsa o xksyksa dks mi;qDr :i ls dkyk djsaA

8. OMR 'khV dks xUnk@eksMsa+ ughaA

C. vadu i)fr:

bl iz'ui=k esa izR;sd fo"k; esa fuEu izdkj ds iz'u gSa:-

[k.M – I

9. 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

10. 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;saxas rFkk izR;sd xyr mÙkj ds fy, 1 vad ?kVk;k tk,xkA

11. 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 xyr mÙkj ds fy, 1 vad ?kVk;k tk,xkA

[k.M –II

12. LrEHkksa dks lqesfyr djus okys iz'u gSaA iw.kZ :Ik ls lgh lqesfyr mÙkj ds fy, 6 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, 1 vad fn;k tk;sxkA

Date : 14/02/2010 Time : 3 : 00 Hrs. MAX MARKS: 249

SEA

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

phase- all (revision test-at centre) mathematics, …...corporate office: cp tower, road no.1, ipia,...

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