MATHEMATICS, PHYSICS, CHEMISTRY

Corporate Office: CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-2434159 || Mock Test || Page # 1

Space for rough work

MOCK TEST FOR IIT-JEE 2010

MOCK TEST # 2 PAPER - II

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. Reason and Assertion type questions with only one correct answer in each. 3 marks will be awarded for each correct answer and –1 mark for each wrong answer.

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.

Corporate Office, CP Tower, Road No.1, IPIA, Kota (Rajasthan ) - 324 005 Phone (0744) -3040000, 2430505; Fax (0744) 2434159

email : [email protected] ; Website : www.careerpointgroup.com

Now, Schedule practice questions are available on internet also, Visit www.examtayari.com

Time : 3 : 00 Hrs. MAX MARKS: 243

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MATHEMATICS, PHYSICS, CHEMISTRY

Code-6



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)



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 rFkk 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. dFku rFkk dkj.k izdkj ds iz'u] ftuesa ls izR;sd esa dsoy ,d mÙkj lgh gSA izR;sd lgh mÙkj ds fy, 3 vad fn;s

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

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

[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

Time : 3 : 00 Hrs. MAX MARKS: 243

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



MATHEMATICS

Section – I Questions 1 to 9 are multiple choice questions. Eachquestion has four choices (A), (B), (C) and (D), out ofwhich ONLY ONE is correct. Mark your response inOMR sheet against the question number of thatquestion. + 3 marks will be given for each correct answerand – 1 mark for each wrong answer.

Q.1 The area enclosed by | y | – | x | = 1 andx2 + y2 =1 is

(A) 2 units2 (B) zero units2 (C) infinite units2 (D) none of these Q. 2 Least value of the expression 9sec2θ + 4cosec2θ, is- (A) 6 (B) 1 (C) 36 (D) 25

Q. 3 Let

f(x) = ∫ −−+−−x

1

223 dx))2x()1x(3)2x)(1x(2( ,

then- (A) f has exactly 4 critical points

(B) f has maximum at x = 2

(C) x = 57 is minima & x = 1 is maxima

(D) none of these

[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 oØ | y | – | x | = 1 ,oa x2 + y2 =1 ls ifjc) kS=kQy gS (A) 2 oxZ bdkbZ (B) 'kwU; oxZ bdkbZ (C) vuUr oxZ bdkbZ (D) buesa ls dksbZ ugh

Q. 2 O;atd 9sec2θ + 4cosec2θ dk U;wure eku gS-

(A) 6 (B) 1 (C) 36 (D) 25

Q. 3 ekukfd

f(x) = ∫ −−+−−x

1

223 dx))2x()1x(3)2x)(1x(2( , rks-

(A) f ds Bhd 4 ØkfUrd fcUnq gS (B) f dk mfPp"B eku x = 2 ij gS

(C) x = 57 fufEu"B o x = 1 mfPp"B gS

(D) buesa ls dksbZ ugh



Q. 4 The locus of the middle points of chords of a parabola which subtend a right angle at the vertex of the parabola is-

(A) Circle (B) Parabola (C) Ellipse (D) Straight line

Q. 5 The probability that a particular day in the month of

july is a rainy day is 43 . Two person whose

credibility are 54 and

32 respectively claim that

15th july was a rainy day. The probability that it was real a rainy day.

(A) 43 (B)

2524

(C) 98 (D) none of these

Q.6 Domain of f(x) =

−−

]x[]x[x2sin 1 , where [.]

denotes the greatest integer function, is (A) (–∞, 1) – 0

(B)

− 0,

34 ∪ 0

(C) (–∞, 0) ∪ I+ (D) (–∞, ∞) – [0, 1)

Q. 4 ,d ijoy; dh mu thokvksa ds e/; fcUnqvksa dk fcUnqiFk tks ml ijoy; ds 'kh"kZ ij ledks.k cukrh gS] gksxk -

(A) oÙk (B) ijoy; (C) nh?kZoÙk (D) ljy js[kk

Q. 5 tqykbZ ekg ds ,d fo'ks"k fnu ds o"kkZ okyk fnu gksus

dh izkf;drk 43

gS nks O;fDr ftuds lgh vuqeku

yxkus dh izkf;drk Øe'k% 54 o

32 gS] bu nksuksa ds

vuqlkj 15 tqykbZ dk fnu o"kkZ okyk fnu Fkk] rks ml fnu ds okLro esa o"kkZ okyk fnu gksus dh izkf;drk gS

(A) 43 (B)

2524

(C) 98 (D) buesa ls dksbZ ugh

Q.6 f(x) =

−−

]x[]x[x2sin 1 dk izkUr, tgk¡ [.] egÙke iw.kkZad

Qyu iznf'kZr djrk gS] gksxk- (A) (–∞, 1) – 0

(B)

− 0,

34 ∪ 0

(C) (–∞, 0) ∪ I+ (D) (–∞, ∞) – [0, 1)



Q.7 The number of different words of three letterswhich can be formed from the word "PROPOSAL",if a vowel is always in the middle are-

(A) 53 (B) 52 (C) 63 (D) 32

Q.8 Let a1, a2, a3, ...... be terms of an A.P. If

q21

p21

a.....aaa....aa

+++

+++ =

2

2

qp , p ≠ q, then

21

6

aa equals-

(A) 41/11 (B) 7/2

(C) 2/7 (D) 11/41

Q.9 The curve y = ax3 + bx2 + cx is inclined by 45º to

x-axis at origin and it touches x-axis at (1,0). Then-

(A) a = –2, b = 1, c = 1 (B) a = 1, b = 1, c = –2

(C) a = 1, b = –2, c = 1 (D) a = –1, b = 2, c = 1

This section contains 4 questions numbered 10 to 13,(Reason and Assertion type question). Each questioncontains Assertion (A) and Reason (R). Each questionhas 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Mark your response in OMR sheetagainst the question number of that question. +3 markswill be given for each correct answer and – 1 mark foreach wrong answer.

Q.7 rhu vkjksa ds fofHkUu 'kCnksa dh la[;k] tks fd 'kCn "PROPOSAL" ls cuk, tkrs gS ;fn ,d Loj lnSo e/; esa vk,] gksxh -

(A) 53 (B) 52 (C) 63 (D) 32

Q.8 ekuk a1, a2, a3, ...... ,d l-Js- ds in gS ;fn

q21

p21

a.....aaa....aa

+++

+++ =

2

2

qp , p ≠ q, rc

21

6

aa cjkcj gS-

(A) 41/11 (B) 7/2 (C) 2/7 (D) 11/41

Q.9 oØ y = ax3 + bx2 + cx ewy fcUnq ij x-vk ls 45º ds

dks.k ij >qdk gS rFkk ;g x-vk dks (1,0) ij Li'kZ djrk gS] rc -

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

bl [k.M esa 10 ls 13 rd 4 iz'u gSa, (dFku rFkk dkj.k izdkj ds iz'u)A izR;sd iz'u esa dFku (A) rFkk dkj.k (R) fn;s x;s gSaA izR;sd iz'u ds 4 fodYi (A), (B), (C) rFkk (D) fn;s x;s gSa] ftuesa ls dsoy ,d lgh gSA OMR 'khV esa iz'u dh iz'u la[;k ds lek viuk mÙkj vafdr dhft;sA izR;sd lgh mÙkj ds fy;s + 3 vad fn;s tk;asxs rFkk izR;sd xyr mÙkj ds fy, 1 vad ?kVk;k tk;sxkA



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

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

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

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

Q.10 Assertion : Let z be a complex number satisfying

|z – 3| ≤ |z – 1|, |z – 3| ≤ |z – 5|, |z – i| ≤ |z + i|

and |z – i| ≤ |z – 5i|. Then the area of region in

which z lies is 12 sq. units.

Reason : Area of trapezium = 21 (sum of parallel

sides) (Distance between parallel sides)

Q. 11 Let f(x) = | 1 – x | and g(x) = sin–1(f | x |) Assertion (A) : Number of values of x, where g(x)

is non differentiable is 3. Reason (R) : Domain of g(x) is [–1, 1]

uhps fn;s x;s fuEufyf[kr iz'u "dFku" (A) rFkk

"dkj.k" (R) izdkj ds iz'u gSaA vr% mfpr mÙkj dk

p;u djus ds fy;s fuEu rkfydk dk mi;ksx dhft;sA

(A) ;fn (A) rFkk (R) nksuksa lR; gSa rFkk (R), (A) dk lgh

Li"Vhdj.k gSA

(B) ;fn (A) rFkk (R) nksuksa lR; gSa rFkk (R), (A) dk lgh

Li"Vhdj.k ugha gSA

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

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

Q.10 dFku: ekuk z ,d lfEeJ la[;k gS tks

|z – 3| ≤ |z – 1|, |z – 3| ≤ |z – 5|, |z – i| ≤ |z + i| rFkk

|z – i| ≤ |z – 5i| dks lUrq"V djrh gS rc og ks=kQy

ftlesa z fLFkr gS] 12 oxZ bdkbZ gS

dkj.k : leyEc prqHkqZt dk ks=kQy = 21 (lekUrj

Hkqtkvksa dk ;ksx) (lekUrj Hkqtkvksa ds e/; nwjh)

Q. 11 ekuk f(x) = | 1 – x | rFkk g(x) = sin–1(f | x |) dFku (A) : x ds ekuksa dh la[;k tgk¡ g(x) vodyuh;

ugha gS] 3 gS dkj.k (R) : g(x) dk izkUr [–1, 1] gS



Q. 12 dFku (A) : ;fn nks nh?kZoÙkksa dh mRdsUnzrk;sa leku gks rks mudk ks=kQy Hkh leku gksxk

dkj.k (R) : nh?kZoÙk 1by

ax

2

2

2

2=+ (a<b, a > 0, b > 0)

dk ks=kQy π ab oxZ bdkbZ gS

Q. 13 ekuk ,d oÙk S : (x – 2)2 + (y – 3)2 = 13 rFkk ,d js[kk L : y = x – 12 gS

dFku (A): L = 0 ds fdlh Hkh fcUnq ls S = 0 ij [khaph xbZ Li'kZ js[kkvksa dh Li'kZ thok P(3, 2) ls xqtjrh gS

dkj.k (R) : /kqzoh L = 0 dk S = 0 ds lkisk /kqzo P(3, 2) gS

bl [k.M esa 2 vuqPNsn fn;s x;s gSa] izR;sd esa 3 cgqfodYih iz'u gSaA (iz'u 14 ls 19) 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 vafdrdhft;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- 14 ls 16) ;fn f(xy) = f(x) . f(y) rFkk x = 1 ij f bl izdkj

vodyuh; gS fd f '(1) = 1 ,oa f(1) ≠ 0, rc Q.14 f(x) gS - (A) lHkh x ∈ R ds fy, larr (B) x = –1, 0, 1 ij vlarr (C) lHkh x ≠ 0 ds fy, vodyuh; (D) buesa ls dksbZ ugh

Q. 12 1Assertion (A) : If eccentricities of two ellipse are

same then their areas are also same.

Reason (R) : Area of the ellipse 1by

ax

2

2

2

2=+

(a < b, a > 0, b > 0) is π ab square units.

Q. 13 Consider a circle S : (x – 2)2 + (y – 3)2 = 13 and a line L : y = x – 12.

Assertion (A): Chord of contact of pair of tangentsdrawn from every point on L = 0 to S = 0 passesthrough P(3, 2)

Reason (R) : Pole of polar L = 0 with respect to S = 0 is P(3, 2)

This section contains 2 paragraphs; each has 3 multiplechoice questions. (Questions 14 to 19) Each question has4 choices (A), (B), (C) and (D) out of which ONLY ONEis correct. Mark your response in OMR sheet againstthe question number of that question. +4 marks will begiven for each correct answer and – 1 mark for eachwrong answer.

Passage # 1 (Ques. 14 to 16) If f(xy) = f(x) . f(y) and f is differentiable at x = 1

such that f '(1) = 1 also f(1) ≠ 0, then

Q.14 f(x) is - (A) continuous for all x ∈ R (B) discontinuous at x = –1, 0, 1 (C) differentiable for all x ≠ 0 (D) None of these



Q.15 f '(7) equals- (A) 7 (B) 14 (C) 1 (D) None of these Q.16 Area bounded by curve f(x), x axis and ordinate

x = 4, is- (A) 64/3 (B) 8 (C) 16 (D) None of these

Passage # 2 (Ques. 17 to 19)

There exists a G.P. with first term A1 and common

ratio A (A > 1). If we add 21 in the sum of first n

terms of the sequence, it equals to the sum of the coefficients of even power of x in the expansion of (1 + x)n. If we interchange the first term & common ratio of given G.P., the sum of new infinitely decreasing G.P. is equal to B, where A, B and n are related by the relation

∫−

−

+2B

2A

n dx)x1( = 3

364

Q.17 The value of Ax

Bn)x1(limA

Ax −−−+

→ is-

(A) 3 (B) 6 (C) e (D) 8

Q.18 Area bounded by f(x) = xA and g(x) = xB is-

(A) n

BA + (B) n

AB −

(C) nB2A

A2++

(D) BAn

B++

Q.15 f '(7) cjkcj gS- (A) 7 (B) 14 (C) 1 (D) buesa ls dksbZ ugh Q.16 oØ f(x), x – vk rFkk dksfV x = 4 kjk ifjc) kS=kQy gS - (A) 64/3 (B) 8 (C) 16 (D) buesa ls dksbZ ugh

x|ka'k # 2 (iz- 17 ls 19)

,d xq.kksÙkj Js.kh dk izFke in A1 rFkk lkoZvuqikr

A (A > 1) gSA ;fn Js.kh ds izFke n inksa ds ;ksx esa ge

21 tksM+ nsa rks ;g (1 + x)n ds izlkj esa x dh le?kkrksa

ds xq.kkadksa ds ;ksx ds rqY; gks tkrk gSA ;fn ge nh xbZ xq-Js- ds izFke in o lkoZ vuqikr dks ijLij cny nsa rks ubZ Ðkleku vuUr xq.kksÙkj Js.kh dk ;ksx B ds rqY; gS, tgk¡ A, B o n fuEu lEcU/k ls lacaf/kr gS

∫−

−

+2B

2A

n dx)x1( = 3

364

Q.17 Ax

Bn)x1(limA

Ax −−−+

→ dk eku gS-

(A) 3 (B) 6 (C) e (D) 8

Q.18 oØ f(x) = xA ,oa g(x) = xB ls ifjc) kS=kQy gS-

(A) n

BA + (B) n

AB −

(C) nB2A

A2++

(D) BAn

B++



Q.19 Number of real roots of the equation (xB – nxA)1/A = 6 are (A) 2 (B) 4 (C) 1 (D) 0

Section – II

This section contains 3 questions (Questions 1 to 3). 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.19 lehdj.k (xB – nxA)1/A = 6 ds okLrfod gyksa dh la[;k gS- (A) 2 (B) 4

(C) 1 (D) 0

[k.M - II

bl [k.M esa 3 iz'u (iz'u 1 ls 3) 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 vuqlkjmfpr 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



Q.1 Match the following:

Column-I Column-II

(A) The reflection of the point (P) 5

(t – 1, 2t + 2) in a line is

(2t + 1, t) then the line has slope

equal to

(B) If θ be the angle between two (Q) 6

tangents which are drawn to the

circle x2 + y2 – 6 3 x – 6y + 27 = 0

from the origin, then 2 3 tanθ equals to

(C) The shortest distance between (R) 72

parabolas y2 = 4x and y2 = 2x – 6

is d then d2 =

(D) Distance between foci of the (S) 1

curve represented by the equation

x = 1 + 4cosθ, y = 2 + 3sinθ is

Q.1 fuEu LrEHkksa dk feyku dhft, -

LrEHk -I LrEHk -II

(A) fcUnq (t – 1, 2t + 2) dk ,d js[kk (P) 5

esa izfrfcEc (2t + 1, t) gks rks ml

js[kk dh izo.krk gksxh

(B) ;fn θ ewyfcUnq ls oÙk (Q) 6

x2 + y2 – 6 3 x – 6y + 27 = 0 ij

[khaph xbZ Li'kZ js[kkvksa ds e/; dks.k

gks rks 2 3 tanθ cjkcj gS

(C) ijoy; y2 = 4x o y2 = 2x – 6 (R) 72

ds e/; dh U;wure nwjh d

gks rks d2 =

(D) oØ x = 1 + 4cosθ, y = 2 + 3sinθ (S) 1

dh ukfHk;ksa ds e/; dh nwjh gS



Q.2 Column-I Column-II

(A) If y = 2[x] + 9 = 3[x + 2], where (P) –1

[.] denotes greatest integer function,

then 61 [x + y] is equal to

(B) If x

x x1cos

x1sinlim

+

∞→ = ek/2 then (Q) 0

k is equal to

(C) If three successive terms of a G.P. (R) 2

with common ratio r, (r > 1) forms

the sides of a triangle then [r] + [–r]

is equal to (where [.] denotes

greatest integer function)

(D) Let f(x) = (x2 – 3x + 2)(x2+3x+2) (S) 3

and α, β, γ are the roots of

f '(x) = 0, then [α]+[β] + [γ] is

equal to (where [.] denotes greatest

integer function)

Q.2 LrEHk-I LrEHk -II

(A) ;fn y = 2[x] + 9 = 3[x + 2], tgk¡ (P) –1

[.] egÙke iw.kkZd Qyu gS,

rks 61 [x + y] dk eku gS

(B) ;fn x

x x1cos

x1sinlim

+

∞→ = ek/2 rks (Q) 0

k dk eku gS

(C) ;fn ,d xq-Js- ftldk lkoZvuqikr (R) 2

r, (r > 1) gS] ds rhu Øekxr in

,d f=kHkqt dh Hkqtk,sa gksa rks [r] + [–r]

dk eku gS (tgk¡ [.] egÙke iw.kkZd Qyu gS)

(D) ekukfd f(x)=(x2–3x+2)(x2+3x+2) (S) 3

tgk¡ α, β, γ lehdj.k f '(x) = 0 ds

ewy gSa, rks [α]+[β] + [γ] dk eku

gksxk (tgk¡ [.] egÙke iw.kkZd

Qyu dks n'kkZrk gS)



Q.3 Column-I Column-II

(A) The order and degree of the (P) 13 differential equation

0x7dx

yd4dxdy

2

23 =−− are

a and b then a + b is

(B) If k3j2ia ++=r , kji2b +−=

r (Q) 102

and kj2i3c ++=r and )cb(a

rrr××

is equal to czbyaxrrr

++ , then

x + y + z is equal to (C) The number of 4 digit numbers (R) 5 that can be made with the digits 1,2,3,4,3,2

(D) If ∫ ++ )4x)(1x(dx

22 = (S) 7

kdxtan

caxtan

ba 11 +

− −−

where k is constant of integration, then 2a + b + c + d is (where a & b and a & c are co-prime numbers)

Q.3 LrEHk-I LrEHk -II

(A) vody lehdj.k (P) 13

0x7dx

yd4dxdy

2

23 =−−

dh dksfV ,oa ?kkr Øe'k% a o b gksa rks a + b dk eku gS

(B) ;fn k3j2ia ++=r , kji2b +−=

r (Q) 102

,oa kj2i3c ++=r rFkk )cb(a

rrr××

O;atd czbyaxrrr

++ ds rqY; gks, rks x + y + z dk eku gS

(C) 4 vadks dh cuus okyh mu la[;kvksa (R) 5 dh la[;k] tks vadksa 1,2,3,4,3,2 ds iz;ksx ls cukbZ tkrh gS] gksxh

(D) ;fn ∫ ++ )4x)(1x(dx

22 = (S) 7

kdxtan

caxtan

ba 11 +

− −−

tgk¡ k lekdyu fu;rkad gS] rks

2a + b + c + d dk eku gS

(tgk¡ a o b rFkk a o c lg&vHkkT;

la[;k,sa gS)



PHYSICS

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 In the circuit shown, the cell is ideal, with

e.m.f. = 15 V. Each resistance is of 3Ω. The

potential difference across the capacitors in steady

state -

R=3Ω C=3µF

R R

R R

15V

+ –

(A) 0 (B) 9 V

(C) 12 V (D) 15 V

[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 n'kkZ;s ifjiFk esa lsy vkn'kZ gS rFkk mldk fo-ok- cy

= 15 V gSA izR;sd izfrjks/k 3Ω dk gSA larIr voLFkk

esa la/kkfj=k ds fljksa ij foHkokUrj gS -

R=3Ω C=3µF

R R

R R

15V

+ –

(A) 0 (B) 9 V

(C) 12 V (D) 15 V



Q.2 In a young double slit apparatus the screen is rotated

by 60º about an axis parallel to the slits. The slits

separation is 3mm, slits to screen distance (i.e AB) is

4 m & wavelength of light is 450 nm. The separation

between the 3rd dark fringe on the either side of B.

60º

A B

screen

(A) 6 mm (B) 8 mm

(C) 4 3 mm (D) 2 3 mm

Q.3 A black body emits radiation at the rate P when its

temperature is T. At this temperature the

wavelength at which the radiation has maximum

intensity is λ0. If at another temperature T' the

power radiated is P' & wavelength at maximum

intensity is 20λ then -

(A) P'T' = 32 PT (B) P'T' = 16 PT

(C) P'T' = 8 PT (D) P'T' = 4 PT

Q.2 ;ax ds f-fLyV midj.k esa insZ dks fLyVksa ds lekUrj

vk ds lkisk 60º kjk ?kqek;k tkrk gSA fLyVksa ds e/;

nwjh 3mm gS] fLyV rFkk insZ ds e/; nwjh (vFkkZr~ AB) 4

m o izdk'k dh rjaxnS/;Z 450 nm gSA B ds nksuksa vksj

3rd vnhIr fÝUt ds e/; nwjh Kkr djs

60º

A B

screen

(A) 6 mm (B) 8 mm

(C) 4 3 mm (D) 2 3 mm

Q.3 ,d df".kdk P nj ls fofdj.k mRlftZr djrh gS tc

mldk rki T gSA bl rki ij rjaxnS/;Z ftl ij

fofdj.k vf/kdre rhozrk j[krh gS λ0 gSA ;fn vU; rki

T' ij fofdfjr 'kfDr P' rFkk vf/kdre rhozrk ij

rjaxnS/;Z 20λ gS rc -

(A) P'T' = 32 PT (B) P'T' = 16 PT

(C) P'T' = 8 PT (D) P'T' = 4 PT



Q.4 If two identical string are stretched such that there is fractional increase in their length, the fractional increase in length of first string is f and second string is 2f. Then the ratio of their fundamental frequency is. (Assume both obey the Hook Law i.e. tension ∝ elongation in string) -

(A) 2

1f1f21

++ (B)

f1f212

++

(C) f21

f12

1++ (D)

f21f1

++

Q.5 A infinite line charge of charge density λ lies along the x axis and let the surface of zero potential passes through (0, 5, 12) m. The potential at point (2, 3, –4) is -

z

V = 0(0,5,12)

y

(A) 02πε

λ ln 5

13 (B) 0

2επλ

ln3

13

(C)04 επ

λ ln5

13 (D) 0επ

λln

313

Q.4 ;fn nks leku Mksfj;ksa dks bl izdkj [khapk tkrk gS fd mudh yEckbZ esa vkaf'kd of) gksrh gS] igyh Mksjh esa vkafa'kd of) f rFkk nwljh esa 2f gSA rc mudh ewy vkofÙk dk vuqikr gS (;g ekfu;s fd nksuksa gqd ds fu;e dk ikyu djrh gS vFkkZr~ ruko ∝ Mksjh esa foLrkj) -

(A) 2

1f1f21

++ (B)

f1f212

++

(C) f21

f12

1++ (D)

f21f1

++

Q.5 vkos'k ?kuRo λ dk ,d vuUr js[kh; vkos'k x- vk ds

vuqfn'k fLFkr gS rFkk ekuk 'kwU; foHko dh lrg (0, 5,

12) m ls xqtjrh gS A fcUnq (2, 3, –4) ij foHko gS - z

V = 0(0,5,12)

y

(A) 02πε

λ ln 5

13 (B) 0

2επλ

ln3

13

(C)04 επ

λ ln5

13 (D) 0επ

λln

313



Q.6 A satellite is put in an orbit just above the earth atmosphere with a velocity 5.1 times the velocity for a circular orbit at that height. The initial velocity imparted is horizontal. What would be the maximum distance of satellite from earth when it is in the orbit -

(A) 3R (B) 4R (C) 2 R (D) 5 R Q.7 The density of a uniform rod with cross section

area A is ρ, its specific heat capacity is C and the coefficient of its linear expansion is α. Calculate the amount of heat that should be added in order to increase the length of the rod by ∆l.

(A) α

∆ρ lCA2 (B) l∆

αρCA

(C) α

∆ρ lCA (D) α∆ρ

A2C l

Q.8 For the system shown each spring has a stiffness of 175 N/m. The mass of the pulleys may be neglected. The period of vertical oscillation of block – (mass of block is 28 kg)

k k

Block fricationless surface

28 kg

(A) 2 s (B) π 2 s (C) 5π s (D)

5π 2 s

Q.6 iFoh ds ok;qe.My ds Åij ,d dkk esa ,d mixzg] ml Å¡pkbZ ij ,d oÙkkdkj dkk ds fy;s osx dk

5.1 xquk osx ls LFkkfir fd;k x;k gSA fn;k x;k izkjfEHkd osx kSfrt gSaA iFoh ls mixzg dh vf/kdre nwjh D;k gksxh tc og dkk esa gS -

(A) 3R (B) 4R (C) 2 R (D) 5 R

Q.7 A vuqizLFk dkV ks=kQy dh ,dleku NM+ dk ?kuRo ρ gS] mldh fof'k"V Å"ek /kkfjrk C gS rFkk mldk js[kh; izlkj xq.kkad α gSA Å"ek dh og ek=kk Kkr dhft;s tks NM+ dh yEckbZ ∆l c<+kus ds Øe esa vo'; feykbZ tkuh pkfg;s

(A) α

∆ρ lCA2 (B) l∆

αρCA

(C) α

∆ρ lCA (D) α∆ρ

A2C l

Q.8 n'kkZ;s fudk; ds fy;s izR;sd fLizax dk cy fu;rkad 175 N/m gSA f?kjfu;ksa dk nzO;eku ux.; ekuk tk ldrk gSA CykWd ds Å/okZ/kj nksyuksa dk vkorZdky gS –(CykWd dk nzO;eku 28 kg gSA)

k k

Blockfricationless surface

28 kg

(A) 2 s (B) π 2 s (C) 5π s (D)

5π 2 s



Q.9 A vuqizLFk dkV ks=kQy dh ,dleku izR;kLFk NM+ dh izkjfEHkd yEckbZ L rFkk ;ax xq.kkad Y gS rFkk og ,d fpdus kSfrt ry ij fLFkr gSA vc nks kSfrt cy (F rFkk 3F ifjek.k ds) NM+ dh yEckbZ ds vuqfn'k funsZf'kr gS rFkk NM+ ds nks fljks ij foijhr fn'kk esa n'kkZ;s vuqlkj dk;Zjr gSA NM+ ds fu;r voLFkk izkIr djus ds ckn]¼ ftruk foLrkj gksuk Fkk gks pqdk½ NM+ dk foLrkj gksxk -

F 3F

Elastic rod

(A) YA

F2 L (B) YA

F4 L

(C) YAF L (D)

YA2F3 L

bl [k.M esa 10 ls 13 rd 4 iz'u gSa, (dFku rFkk dkj.k izdkj ds iz'u)A izR;sd iz'u esa dFku (A) rFkk dkj.k (R) fn;s x;s gSaA izR;sd iz'u ds 4 fodYi (A), (B), (C) rFkk (D) fn;s x;s gSa] ftuesa ls dsoy ,d lgh gSA OMR 'khV esa iz'u dh iz'u la[;k ds lek viuk mÙkj vafdr dhft;sA izR;sd lgh mÙkj ds fy;s + 3 vad fn;s tk;asxs rFkk izR;sd xyr mÙkj ds fy, 1 vad ?kVk;k tk;sxkA

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

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

Q.9 A uniform elastic rod of cross section area A, natural length L and young modulus Y is placed on a smooth horizontal surface. Now two horizontal force (of magnitude F and 3F) directed along the length of rod and in opposite direction act at two of its ends as shown. After the rod has acquired steady state (i.e. no further extension take place), the extension of the rod will be -

F 3F

Elastic rod

(A) YA

F2 L (B) YA

F4 L

(C) YAF L (D)

YA2F3 L

This section contains 4 questions numbered 10 to 13, (Reason and Assertion type question). Each question contains Assertion (A) and Reason (R). Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. +3 marks will be given for each correct answer and – 1 mark for each wrong answer. The following questions given below consist of

an "Assertion" (A) and "Reason" (R) Type questions. Use the following Key to choose the appropriate answer.

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



(B) If both (A) and (R) are true but (R) is not the

correct explanation of (A).

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

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

Q. 10 Assertion (A) :As the earth revolves around the

sun it has an acceleration which is directed towards

centre of the sun.

Reason (R) :Angular momentum of the earth about

the sun remains constant.

Q. 11 Assertion (A) :In electric circuits , wires carrying

currents in opposite direction are often twisted

together.

Reason (R) : The magnetic field in the surrounding

space of a twisted wire system in not precisely zero.

Q.12 Assertion (A) : A metal ball is floating in mercury.

Coefficient of volume expansion of metal is less

than that of mercury. If temperature is increased,

fraction of volume immersed of metal will increase.

Reason (R) : Effect of heating on density of

mercury will be more compared to that of metal.

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

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

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

Q. 10 dFku (A) : iFoh lw;Z ds pkjksa vksj pDdj yxkrh gS

blfy, og ,d Roj.k j[krh gS tks lw;Z ds dsUnz dh

vksj funsZf'kr gksrk gSA

dFku (R) : lw;Z ds lkisk iFoh dk dks.kh; laosx

fu;r jgrk gSA Q. 11 dFku (A) : fo|qr ifjiFkksa esa] foijhr fn'kkvksa esa

/kkjkokgh rkj izk;% ,d lkFk fyiVs gksrs gS

dFku (R) : ,d fyiVs gq, rkjksa ds fudk; kjk f?kjh

txg esa pqEcdh; ks=k 'kwU; ugha gksrk gSA

Q.12 dFku (A) : /kkrq dh ,d xsan ikjs esa rSj jgh gSA /kkrq

dk vk;ru izlkj xq.kkad ikjs ls de gSA ;fn rki c<+k;k tk;s] rks /kkrq dk Mwck gqvk vkaf'kd vk;ru c<s+xk

dkj.k (R) : rkieku esa of) dk ikjs ds ?kuRo ij

izHkko /kkrq ds ?kuRo dh viskk vf/kd gksrk gSA



Q.13 Assertion (A) : Work done by static friction is always zero.

Reason (R) : Static friction acts when there is no relative motion between two bodies in contact.

This section contains 2 paragraphs; each has 3 multiple choice questions. (Questions 14 to 19) 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. 14 to 16)

A conducting rod PQ of mass M rotates without friction on a horizontal plane about Ο on circular rails of diameter 'l'. The centre O and the periphery are connected by resistance R. The system is located in a uniform magnetic field perpendicular to the plane of the loop. At t = 0, PQ starts rotating clockwise with angular velocity ω0. Neglect the resistance of the rails and rod, as well as self inductance.

B

O ω0

Q

R P

⊗

Q.13 dFku (A) : LFkSfrd ?k"kZ.k kjk fd;k x;k dk;Z ges'kk 'kwU; gksrk gS

dkj.k (R) : LFkSfrd ?k"kZ.k rc dk;Zjr gksrk gS tc lEidZ esa fLFkr nks fi.Mksa ds e/; lkisfkd xfr ugha gksrh gSA

bl [k.M esa 2 vuqPNsn fn;s x;s gSa] izR;sd esa 3 cgqfodYih iz'u gSaA (iz'u 14 ls 19) 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- 14 ls 16)

M nzO;eku okyh ,d pkyd NM+ PQ kSfrt ry esa Ο ds lkisk 'l' O;kl okyh oÙkkdkj iVfj;ksa ij fcuk fdlh ?k"kZ.k ds ?kwe jgh gSA dsUnz O rFkk ifjf/k ,d izfrjks/k R kjk tqMs+ gq, gaSA fudk; ywi ds ry ds yEcor~ ,dleku pqEcdh; ks=k esa fLFkr gSA t = 0 ij] PQ dks.kh; osx ω0 ls nfk.kkorZ ?kweuk izkjEHk djrh gSA iVfj;ksa rFkk NM+ ds izfrjks/k rFkk lkFk gh LoizsjdRo dks ux.; ekusaA

B

O ω0

Q

R P

⊗



Q.14 Magnitude of current as a function of time

(A) t2

0 eR2

B α−ω l (B) t22

0 eR16

B α−ω l

(C) t2

0 eR8

B α−ω l (D) t22

0 eR8

B α−ω l

Where α = RM8B3 22l

Q.15 Total charge flow through resistance till rod PQ

stop rotating .

(A) B8M0ω (B)

B3M0ω (C)

B6M0ω (D)

B9M0ω

Q.16 Heat generated in the circuit by t = ∞

(A) 24

M 20

2ωl (B) 8

M 20

2ωl

(C) 3

M 20

2ωl (D) 32

M 20

2ωl

Passage # 2 (Ques. 17 to 19)

A cylindrical spool of radius R is rigidly attached

to a pulley of radius 2R. The mass of the

combination is m and the radius of gyration about

the centroidal axis G is R. A constant horizontal

force F is applied at one end of the tape. Assume

rolling motion between the pulley and the ground.

I is the moment of inertia about centroidal axis.

Q.14 le; ds Qyu ds #i esa /kkjk dk ifjek.k gksxk&

(A) t2

0 eR2

B α−ω l (B) t22

0 eR16

B α−ω l

(C) t2

0 eR8

B α−ω l (D) t22

0 eR8

B α−ω l

tgk¡ α = RM8B3 22l

Q.15 NM+ PQ dk ?kweuk cUn gksus ls igys izfrjks/k ls gksdj xqtjus okyk dqy vkos'k gksxk&

(A) B8M0ω (B)

B3M0ω (C)

B6M0ω (D)

B9M0ω

Q.16 t = ∞ rd ifjiFk esa mRiUu ÅtkZ gksxh&

(A) 24

M 20

2ωl (B) 8

M 20

2ωl

(C) 3

M 20

2ωl (D) 32

M 20

2ωl

x|ka'k # 2 (iz- 17 ls 19)

R f=kT;k dh ,d csyukdkj pj[kh (spool) dks 2R

f=kT;k dh ,d f?kjuh ls n<+rk ls tksM+k x;k gSA

la;kstu dk nzO;eku m rFkk dsUnzh; vk G ds lkisk

?kw.kZu f=kT;k R gSA ,d fu;r kSfrt cy Vsi ds ,d

fljs ij yxk;k tkrk gSA ;g ekfu;s fd f?kjuh rFkk

/kjkry ds e/; yksVuh xfr gksrh gSA dsUnzh; vk ds

lkisk tM+Ro vk?kw.kZ I gSA



2R

G R

PF

Q.17 The acceleration of point P on the tape relative to

the ground is -

(A) m2F (B)

m5F

(C) m3F (D)

m4F

Q.18 The length of the tape wound/unwound on the

spool in time t is -

(A) m10

Ft2 (B)

m5Ft2

(C) m4

Ft2 (D)

m2Ft2

Q.19 The linear acceleration of centre G is -

(A) I

FR2 2 (B) 2

2

mRIFR2

+

(C) 2

2

mR2IFR2

+ (D) 2

2

mR4IFR2

+

2R

GR

PF

Q.17 /kjkry ds lkisk Vsi ij fLFkr fcUnq P dk Roj.k gS -

(A) m2F (B)

m5F

(C) m3F (D)

m4F

Q.18 t le; esa pj[kh ij yisV@mrkjs x;s Vsi dh

yEckbZ gS-

(A) m10

Ft2 (B)

m5Ft2

(C) m4

Ft2 (D)

m2Ft2

Q.19 dsUnz G dk js[kh; Roj.k gS -

(A) I

FR2 2 (B) 2

2

mRIFR2

+

(C) 2

2

mR2IFR2

+ (D) 2

2

mR4IFR2

+



Section – II

This section contains 3 questions (Questions 1 to 3).

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 :

P Q R S

S

P

P R R

Q Q

S S

R Q P

S RQ P A BC D

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.

[k.M - II

bl [k.M esa 3 iz'u (iz'u 1 ls 3) 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 :

P Q R S

S

P

P R R

Q Q

S S

R Q P

S R Q PABCD

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 A charged particle passes through a region that

could have electric field only or magnetic field

only or both electric and magnetic field or none of

the fields. Match the possibilities

Column -I Column-II

(A) Kinetic energy of the

particle remain

constant.

(P) Under special

condition this is

possible when

both electric and

magnetic fields

are present

(B) Acceleration of the

particle is zero

(Q) The region has

electric field

(C) Kinetic energy of the

particle changes and it

also suffers deflection

(R) The region has

magnetic field

only

(D) Kinetic energy of the

particle changes but it

suffers no deflection

(S) The region

contains no field

Q.1 ,d vkosf'kr d.k ml ks=k ls xqtjrk gS tks dsoy

fo|qr ks=k ;k dsoy pqEcdh; ks=k ;k nksuksa fo|qr o

pqEcdh; ks=k ;k dksbZ Hkh ks=k ugha j[k ldrk gSA

laHkkoukvksa dk feyku dhft,A

LrEHk -I LrEHk -II (A) d.k dh xfrt ÅtkZ

fu;r jgrh gS

(P) fo'ks"k 'krZ ds v/khu

;g lEHko gS fd tc

fo|qr rFkk pqEcdh;

nksuksa ks=k mifLFkr gS

(B) d.k dk Roj.k 'kwU;

gksrk gS

(Q) ks=k fo|qr ks=k

j[krk gS

(C) d.k dh xfrt ÅtkZ

ifjofrZr gksrh gS rFkk

og fopyu Hkh n'kkZrh gS

(R) ks=k dsoy pqEcdh;

ks=k j[krk gS

(D) d.k dh xfrt ÅtkZ

ifjofrZr gksrh gS ysfdu

og fopyu ugha n'kkZrh gS

(S) ks=k dksbZ ks=k ugha

j[krk gS



Q.2 Match the column

Column -I Column-II Phenomena on which machine work

Machine or instrument

(A) Electromagnetic induction (P) photocell (B) Light of suitable frequency

falling on a material result in emissions of electrons from the material

(Q) DC motor

(C) change of orientation of a coil in a magnetic field results in e.m.f. across the coil

(R) AC generator

(D) Mutual induction (S) Transformer Q.3 Match the column

Column -I Column-II (A) A photon stimulates the

emission of another photon of

(P) Same direction

(B) Photons of electromagnetic wave of different wavelengths may have

(Q) Same energy

(C) Two points on a wavefront must have

(R) Same phase

(D) For constructive interference the waves must have

(S) Same frequency

Q.2 LrEHk dk feyku dhft;s LrEHk -I LrEHk -II

dk;Ziz.kkyh ftl ij e'khu dk;Z djrh gS

e'khu ;k midj.k

(A) fo|qr pqEcdh; izsj.k (P) izdk'k lsy (B) ,d inkFkZ ij mi;qDr vkofÙk

ds izdk'k ds fxjus ds ifj.kke:o:i inkFkZ ls bysDVªkWu mRlftZr gksrs gS

(Q) DC eksVj

(C) ,d fo|qr ks=k esa dq.Myh ds vfHkfoU;kl ds ifjorZu ds ifj.kke Lo:i dq.Myh ds fljksa ij fo-ok-cy izsfjr gksrk gS

(R) AC tfu=k

(D) vU;ksU; izsj.k (S) VªkalQkWeZj

Q.3 LrEHk feyku dhft;s LrEHk -I LrEHk -II

(A) ,d QksVksu ftldh nwljs QksVksu dk m)hfir mRltZu djrk gS ftldh---------gksrh gS

(P) leku fn'kk

(B) fHkUu-2 rjaxnS/;Z dh fo|qr pqEcdh; rjaxksa ds QksVkWu --------j[k ldrs gS

(Q) leku ÅtkZ

(C) ,d rjaxkxz ij nks fcUnq vo'; -------- j[krs gS

(R) leku dyk

(D) laiks"kh O;frdj.k ds fy;s rjaxsa vo'; -------- j[krh gS

(S) leku vkofÙk



CHEMISTRY

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 A reaction takes place in three steps; the rate constant are k1, k2 and k3. The overall rate constant

k = 2

31

kkk . If energies of activation are 40, 30 and

20 kJ, the overall energy of activation is (assuming 'A' to be constant for all)

(A) 10 (B) 15 (C) 30 (D) 60

Q.2 The number of moles of oxalate ions oxidized by

one mole of –4MnO ion in acidic medium -

(A) 25 (B)

52

(C) 53 (D)

35

[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 ,d vfHkfØ;k rhu inksa esa gksrh gSA osx fLFkjkad k1, k2

o k3 gSA dqy osx fu;rkad k = 2

31

kkk . ;fn lafØ;.k

ÅtkZ,a Øe'k% 40, 30 o 20 kJ gS rks dqy laafØ;.k ÅtkZ gS (lHkh ds fy, 'A' dks fLFkj ekuus ij)

(A) 10 (B) 15 (C) 30 (D) 60

Q.2 vEyh; ek/;e esa] ,d eksy –4MnO kjk vkWDlhdr

vkWDtsysV vk;uksa ds eksykas dh la[;k gS–

(A) 25 (B)

52

(C) 53 (D)

35



Q.3 A liquid which is confined inside an adiabatic container

and piston suddenly taken from state 1 to state 2 by a

single stage process. If the piston comes to rest at point

2 as shown. Then the enthalpy change for the process

will be –

V0 4V0 V

2

P2P0

P0

1

(A) ∆H = 1–VP2 00

γγ

(B) ∆H = 1–VP3 00

γγ

(C) ∆H = – P0V0 (D) None of these

Q.4 Analysis show that nickel oxide consist of nickel

ion with 96% ions having d8 configuration and 4%

having d7 configuration. Which amongst the

following best represents the formula of the oxide - (A) Ni1.02O1.00 (B) Ni0.96O1.00

(C) Ni0.98O0.98 (D) Ni0.98O1.00

Q.3 ,d nzo tks :)ks"e ik=k ds vUnj fLFkr gS rFkk ,d

inh; izØe kjk fiLVu vpkud fLFkfr 1 ls fLFkfr 2

esa fy;k tkrk gSA ;fn n'kkZ;sa vuqlkj fiLVu fLFkfr 2

ij fojke esa vk tkrk gS rks izØe ds fy, ,UFkSYih

ifjorZu gksxk–

V0 4V0 V

2

P2P0

P0

1

(A) ∆H = 1–VP2 00

γγ

(B) ∆H = 1–VP3 00

γγ

(C) ∆H = – P0V0 (D) buesa ls dksbZ ugha

Q.4 fo'ys"k.k tks n'kkZrk gS fd fufdy vkWDlkbM esa d8

foU;kl ds lkFk 96% vk;u rFkk d7 foU;kl ;qDr 4%

vk;u gksrs gSaA fuEu esa dkSulk vkWDlkbM ds lq=k dk

loZJs"B fu:i.k gSaA

(A) Ni1.02O1.00 (B) Ni0.96O1.00

(C) Ni0.98O0.98 (D) Ni0.98O1.00



Q.5 In a basic aqueous solution chloromethane undergoes a substitution reaction in which Cl– is replaced by OH– as CH3Cl(aq) + OH– CH3OH(aq) + Cl– (aq) The equilibrium constant of above reaction Kc = 1 × 1016. If a solution is prepared by mixing equal volumes of 0.1 M CH3Cl and 0.2 M NaOH (100% dissociated) then [OH–] concentration at equilibrium in mixture will be -

(A) 0.1 M (B) 0.5 M (C) 0.2 M (D) 0.05 M

Q.6 The product formed in the reaction :

CH≡C–COOH 2

2

Hg

H/OH+

+

→ ? ; is -

(A)

CH3 C COOH

O

(B)

CH C COOH

OH

(C) CHOCH2COOH

(D) CH2 C COOH

OH

Q.5 kkjh; tyh; foy;u esa] DyksjksesFksu izfrLFkkiu vfHkfØ;k

nsrh gS ftlesa Cl– fuEu izdkj OH– ls izfrLFkkfir gksrss gSa

CH3Cl(aq) + OH– CH3OH(aq) + Cl– (aq)

mijksDr vfHkfØ;k dk lkE; fLFkjkad Kc = 1 × 1016 gSA

;fn foy;u dks 0.1 M CH3Cl ,oa 0.2 M NaOH

(100% fo;ksftr) ds cjkcj vk;ruksa dks fefJr djds

cuk;k x;k gks rks feJ.k esa lkE; ij [OH–] lkUnzrk

gksxh - (A) 0.1 M (B) 0.5 M

(C) 0.2 M (D) 0.05 M

Q.6 fuEu vfHkfØ;k esa cuk mRikn gS :

CH≡C–COOH 2

2

Hg

H/OH+

+

→ ?

(A)

CH3 C COOH

O

(B)

CH C COOH

OH

(C) CHOCH2COOH

(D) CH2 C COOH

OH



Q.7 fuEu vfHkfØ;k esa mRikn (b) igpkfu;s %

O

CHO+

HO

HO a

(i) CH3MgBr

(ii) H3O+b

(A)

O

CH CH3

OH

(B) OH

CHO

CH3

(C) A o B nksuksa curs gS

(D)

CH CH3

OH

OHCH3

Q.8 fuEu vfHkfØ;k dk mRikn gS :

OHH+

∆ X

(A)

(B)

(C) CH3

(D)

Q.7 Identify the product (b)

O

CHO +

HO

HO a

(i) CH3MgBr

(ii) H3O+b

(A)

O

CH CH3

OH

(B) OH

CHO

CH3

(C) Both A and B formed

(D)

CH CH3

OH

OH CH3

Q.8 The product of the reaction :

OH H+

∆ X ; is

(A)

(B)

(C) CH3

(D)



Q.9 The end product in the reaction :

OH

CHCl3/KOH X

50%KOH Y ; is/are

(A)

OH OH

(B)

OH

OH and

OH COOK

(C) COOK CH2OH

+

(D)

OH O COOK

This section contains 4 questions numbered 10 to 13, (Reason and Assertion type question). Each question contains Assertion (A) and Reason (R). Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. +3 marks will be given for each correct answer and – 1 mark for each wrong answer.

Q.9 fuEu vfHkfØ;k dk vfUre mRikn gS@gSa :

OH

CHCl3/KOHX

50%KOH Y

(A)

OH OH

(B)

OH

OH and

OH COOK

(C) COOK CH2OH

+

(D)

OH OCOOK

bl [k.M esa 10 ls 13 rd 4 iz'u gSa, (dFku rFkk dkj.k izdkj ds iz'u)A izR;sd iz'u esa dFku rFkk dkj.k fn;s x;s gSaA izR;sd iz'u ds 4 fodYi (A), (B), (C) rFkk (D) fn;s x;s gSa] ftuesa ls dsoy ,d lgh gSA OMR 'khV esa iz'u dh iz'u la[;k ds lek viuk mÙkj vafdr dhft;sA izR;sd lgh mÙkj ds fy;s + 3 vad fn;s tk;asxs rFkk izR;sd xyr mÙkj ds fy, 1 vad ?kVk;k tk;sxkA



The following questions given below consist of an "Assertion" (A) and "Reason" (R) Type questions. Use the following Key to choose the appropriate answer.

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

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

(C) If (A) is true but (R) is false. (D) If (A) is false but (R) is true. Q. 10 Assertion (A) : N2F3

+ is planar at each nitrogen atom.

Reason (R) : In N3H, the bond angle H–N–N is 120° and both the N–N bond lengths are not equal.

Q. 11 Assertion (A) : A solution of [Ni(H2O)6]2+ is green but a solution of [Ni(CN)4]2– is colourless

Reason (R) : Energy difference between d levels (i.e. ∆) for H2O complex (paramagnetic) is in the visible region and that for the cyano complex (diamagnetic) is in the UV region

Q.12 Assertion (A) :Alkyl halides form cyanides with KCN but isocyanide with AgCN.

Reason (R) : KCN is an ionic compound and C–C bond is stronger than C–N bond. AgCN is a covalent compound and lone pair is present only on nitrogen.

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

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

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

(C) ;fn (A) lR; gS ysfdu (R) vlR; gSA (D) ;fn (A) vlR; gS ysfdu (R) lR; gSA Q. 10 dFku (A) : N2F3

+ izR;sd ukbMªkstu ijek.kq ij leryh; gSA

dkj.k (R) : N3H esa, ca/k dks.k H–N–N, 120° rFkk nksuksa N–N ca/k yEckbZ;kW cjkcj ugh gSA

Q. 11 dFku (A) : [Ni(H2O)6]2+ dk foy;u gjk gS ijUrq

[Ni(CN)4]2– dk foy;u jaxghu gSA

dkj.k (R) : H2O ladqy ¼vuqpwEcdh;½ ds fy, d ryksa ds

e/; ÅtkZ vUrj (∆) n'; ks=k esa gksrk gS rFkk lk;uks ladqy ¼izfrpqEcdh;½ ijkcsaxuh ks=k esa gksrk gSA

Q.12 dFku (A) : ,fYdy gsykbM KCN ds lkFk lk;ukbM ijUrq AgCN ds lkFk vk;lkslk;ukbM cukrk gSA

dkj.k (R) : KCN vk;fud ;kSfxd gS o C–C c/k C–N ca/k ls vf/kd izcy gSA AgCN lgla;kstd ;kSfxd gS o ,adkdh ;qXe dsoy ukbVªkstu ij mifLFkr gSA



Q.13 Assertion (A) : But-2-yne does not give a red precipitate with ammonical Cu2Cl2.

Reason (R) : But-2-yne is a non-terminal alkyne.

This section contains 2 paragraphs; each has 3 multiple choice questions. (Questions 14 to 19) 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. 14 to 16) For the purpose of systematic qualitative analysis,

cations are classified into various groups on the basis of their behaviour against some reagents. The group reagents used for the classification of most common cations are hydrochloric acid, hydrogen sulphide, ammonium hydroxide, and ammonium carbonate. Classification is based on whether a cation reacts with these reagents by the formation of precipitates or not.

Q.14 Which one among the following pairs of ions

cannot be separated by H2S in presence of dilute

hydrochloric acid ?

(A) Bi3+, Cd2+ (B) Al3+, Hg2+

(C) Zn2+, Cu2+ (D) Ni2+, Cu2+

Q.13 dFku (A) : C;wV-2-vkbu veksfudr Cu2Cl2 ds lkFk yky voksi ugh nsrkA

dkj.k (R) : C;wV-2-vkbu fcuk fljs okyh (non-terminal) ,Ydkbu gSA

bl [k.M esa 2 vuqPNsn fn;s x;s gSa] izR;sd esa 3 cgqfodYih iz'u gSaA (iz'u 14 ls 19) 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- 14 ls 16) O;ofLFkr xq.kkRed fo'ys"k.k ds mn~ns'; ls] dqN

vfHkdkjdksa ds lkisk muds O;ogkj ds vk/kkj ij /kuk;uksa dks fofHkUu oxksaZ esa foHkDr fd;k x;k gSA lewg vfHkdeZd dks lokZf/kd lkekU; /kuk;uksa ds oxhZdj.k ds fy, iz;qDr fd;k tkrk gS tks gkbMªksDyksfjd] vEy] gkbMªkstu lYQkbM] veksfu;e gkbMªkWDlkbM o veksfu;e dkcksZusV gSaA oxhZdj.k dk bl rF; ij vk/kkfjr gS fd /kuk;u bu vfHkdeZdks ds lkFk fØ;k dj voksi cukrk gS ;k ughaA

Q.14 fuEu esa ls vk;uksa dk dkSulk ;qXe ruq gkbMªksDyksfjd

vEy dh mifLFkfr esa H2S kjk iFkd ugha fd;k tk

ldrk ? (A) Bi3+, Cd2+ (B) Al3+, Hg2+ (C) Zn2+, Cu2+ (D) Ni2+, Cu2+



Q.15 An aqueous solution contains Hg2+, +22Hg , Pb2+

and Cd2+. The addition of 2 M HCl will precipitate - (A) HgCl2 only (B) PbCl2 only (C) PbCl2 and Hg2Cl2 (D) PbCl2 and CdCl2

Q.16 An aqueous solution which is slightly acidic contains cations Fe3+, Zn2+ and Cu2+. The reagent that when added in excess to this solution would identify the separate Fe3+ ion in one step is -

(A) 2M HCl (B) 6 M NH3 (C) 6M NaOH (D) H2S gas

Passage # 2 (Ques. 17 to 19) Consider the following two compounds A and B.

1

CNH-NH

2

O

3

(A)

NH

4

O

5

(B)

6

Q.17 In compound A correct order of electron density

inside the benzene ring will be - (A) 1 > 2 > 3 (B) 2 > 3 > 1 (C) 2 > 1 > 3 (D) 3 > 2 > 1

Q.15 Hg2+, +22Hg , Pb2+ o Cd2+ ;qDr tyh; foy;u gSA

2 M HCl ds feykus ij fdldk voks.k gksxk -

(A) dsoy HgCl2 (B) dsoy PbCl2

(C) PbCl2 ,oa Hg2Cl2 (D) PbCl2 ,oa CdCl2

Q.16 tyh; foy;u tks FkksM+k vEyh; gS] esa Fe3+, Zn2+ ,oa Cu2+ /kuk;u gSaA vfHkdeZd] tks bl foy;u ls

vf/kdrk esa feyk;k tkrk gS tks ,d in esa Fe3+ vk;u

dks iFkd ifjfkr djsxk] gS - (A) 2M HCl (B) 6 M NH3

(C) 6M NaOH (D) H2S xSl

x|ka'k # 2 (iz- 17 ls 19)

fuEu ;kSfxdksa A o B ij fopkj djus ij

1

CNH-NH

2

O

3

(A)

NH

4

O

5

(B)

6

Q.17 ;kSfxd A esa cSUthu oy; ds vUnj bysDVªkWu ?kuRo

dk lgh Øe gksxk - (A) 1 > 2 > 3 (B) 2 > 3 > 1 (C) 2 > 1 > 3 (D) 3 > 2 > 1



Q.18 In compound B correct order of electron density inside the benzene ring will be -

(A) 4 > 5 > 6 (B) 6 > 4 > 5 (C) 5 > 6 > 4 (D) 5 > 4 > 6 Q.19 In compound B correct order of electron density

inside the benzene ring when N is replaced by S will be -

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

Section – II This section contains 3 questions (Questions 1 to 3). 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.18 ;kSfxd B esa cSUthu oy; ds vUnj bysDVªkWu ?kuRo

dk lgh Øe gksxk -

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

Q.19 ;kSfxd B esa cSUthu oy; ds vUnj] tc N, S ls

izfrLFkkfir gksrk gS] bysDVªkWu ?kuRo dk lgh Øe

gksxk -

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

[k.M - II

bl [k.M esa 3 iz'u (iz'u 1 ls 3) 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 followings :

Column –I (Arrangement of the atoms/ions)

Column –II (Planes in fcc lattice)

(A)

(P)

(B)

(Q)

(C) (R)

(D)

(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 lgh mÙkj ds fy;s 6 vad fn;s tk;saxs rFkk xyr mÙkj ds fy;s 'kwU; vad fn;s tk;saxs (vFkkZr~ dksbZ vad ugha ?kVk;k tk;sxk)A

Q.1 LrEHk lwesfyr dhft, :

LrEHk –I (ijek.kq /vk;uksa dh O;oLFkk)

LrEHk –II (fcc tkyd esa ry)

(A)

(P)

(B)

(Q)

(C) (R)

(D)

(S)



Q.2 Match the reactions in Column I with nature of the reactions/type of the products in Column II.

Column -I Column-II

(A) +24MnO + H+

(P) One of the products of transition element is paramagnetic

(B) Cu+ (aq) → (Q) Disproportation reaction

(C) Cr2O72–(s) + H+(conc.) +

Cl–(s) →

(R) One of the products is liberated as coloured vapours

(D) Fe2(SO4)3 + I–

(S) In one of the products central atom exhibits its highest oxidation state

Q.3 Column I Column II (Compounds) (Reagent for distinction) (A) Phenol and Picric acid (P) NaHCO3 solution) (B) Isopropyl alcohol and (Q) NaOI tert. butyl alcohol (C) Methanol and Ethanol (R) Sodium (D) Ethanol and Diethyl ether (S) HCl conc. + anhyd. ZnCl2

Q.2 LrEHk I esa vfHkfØ;k dk LrEHk II esa mRikn dh izdfr@vfHkdeZd dh izdfr ds lkFk lwesfyr dhft,A

LrEHk -I LrEHk-II

(A) +24MnO + H+ (P) laØe.k rRo ,d

mRikn vuqpqEcdh; gS

(B) Cu+ (aq) → (Q) forrkuqikfrr vfHkfØ;k

(C) Cr2O72–(s) + H+(conc.)+

Cl–(s) →

(R) ,d mRikn jaxhu ok"i ds lkFk eqDr gksrk gS

(D) Fe2(SO4)3 + I–

(S) ,d mRikn esa dsUnzh; ijek.kq vf/kdre vkWDlhdj.k voLFkk n'kkZrk

Q.3 LrEHk I LrEHk II (;kSfxd) (foHksnrk ds fy, vfHkdeZd) (A) fQukWy ,oa fidfjd vEy (P) NaHCO3 foy;u) (B) vkblksizksikby ,YdksgkWy ,oa (Q) NaOI rrh;d C;wfVy ,YdksgkWy (C) esFksukWy o ,FksukWy (R) lksfM;e (D) ,FksukWy o MkbZ,fFky bZFkj (S) HCl conc. + anhyd. ZnCl2

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