2. ct-1_17 (paper-1) aug 2014xii

Page # 1

MATHEMATICS

TEST PATTERN BATCH : CT-1 TEST DATE : 17-08-2014 ELPD

SYLLABUS : Fundamentals of Mathematics, Quadratic Equation, Straight Line

S.No. Subject Nature of Questions No. of Questions Marks Negative Total

1 to 10 SCQ 10 3 ñ1 30

11 to 15 MCQ 5 4 0 20

16 to 20 Integer (double digits) 5 4 0 20

21 to 30 SCQ 10 3 ñ1 30

31 to 35 MCQ 5 4 0 20


41 to 50 SCQ 10 3 ñ1 30

51 to 55 MCQ 5 4 0 20


60 210

Paper-1 CT-1

Total Total

Maths

Physics

Chemistry

SCQ

1. Equations of the sides of a quadrilateral are 3 x y 0 , 3 y x 0 , 3x y 1 and

3y x 1 . The quadrilateral [SLPL] (A) is a parallelogram which is neither a rectangle nor a rhombus.

(B) is a rectangle which is not a square. (C*) is a rhombus which is not a square.

(D) has the area equals to 32

sq.units.

prqHkqZt dh Hkqtkvksa ds lehdj.k 3 x y 0 , 3 y x 0 , 3x y 1 rFkk 3y x 1 gS] rks

prqHkqZt&

(A) ,d lekUrj prqHkqZt tks u rks vk;r gS vkSj u gh leprqHkqZt gSA

(B) ,d vk;r tks oxZ ugha gSA

(C*) ,d leprqHkqZt tks oxZ ugha gSA

(D) dk {ks=kQy 3

2 oxZ bdkbZ ds cjkcj gSA

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Sol.

Here ;gk¡ p1 =12

and vkSj p2 =12

p1 = p2 and also none of the angles A, B, C or D is 90∫ p1 = p2 vkSj rFkk A, B, C ;k D dks.kksa esa ls dksbZ Hkh 90∫ ugha gSA parallelogram ABCD is a rhombus, but not a square. lekUrj prqHkqZt ABCD ,d leprqHkqZt gS tks oxZ ugha gSA

tan =

ñ1 331 1

= 13

= 30∫

cosec = 2

Area {ks=kQy = 12

◊

12

◊ 2 =

12

oxZ bdkbZ

2. Consider the following statements : S1: If the coordinates of vertices of a triangle are integers, then the triangle cannot be

equilateral. [SLDF] S2: If a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 are the equations of two straight lines where

a1 a2 + b1b2 < 0, then origin always lies in the acute angle. S3: If the points D, E and F divides the sides AB, BC and CA respectively in the same ratio

internally, then centroids of triangles ABC and DEF coincide. State, in order, whether S1, S2, S3 are true (T) or false (F) (A) F T F (B) T F F (C) T T T (D*) T F T fuEu dFkuksa ij /;ku nhft,&

S1: ;fn f=kHkqt ds 'kh"kksZ ds funsZ'kkad iw.kk±d gS] rc f=kHkqt] leckgq f=kHkqt ugha gks ldrk gSA

S2: ;fn ljy js[kkvksa ds lehdj.k a1x + b1y + c1 = 0 vkSj a2x + b2y + c2 = 0 gS rFkk a1 a2 + b1b2 < 0, rc

ewy fcUnq lnSo U;wudks.k esa fLFkr gSA

S3: ;fn fcUnq D, E vkSj F Hkqtkvksa AB, BC vkSj CA dks Øe'k% leku vuqikr esa vUr% foHkkftr djrk gS]

rks f=kHkqt ABC vkSj DEF ds dsUnzd lEikrh gSA

Øe esa crk,¡ fd S1, S2, S3 lR; (T) ;k vlR; (F) gSA (A) F T F (B) T F F (C) T T T (D*) T F T Sol. S1 and S3 are always true (standard results) S2 is true when c1 and c2 are positive. Here it is not mentioned. S2 is false Hindi S1 vkSj S3 lnSo lR; gS (ekud ifj.kke)

S2 lR; gS tc c1 vkSj c2 /kukRed gSA ;g fn;k x;k ugha gSA

S2 vlR; gSA

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3. The equations of the sides AB, BC, CA of a triangle ABC are 2x ñ y ñ 3 = 0, 6x + y ñ 21 = 0 and

2x + y ñ 5 = 0 respectively. The external bisector of angle A passes through the point [SLAB] f=kHkqt ABC dh Hkqtkvksa AB, BC, CA ds lehdj.k Øe'k% 2x ñ y ñ 3 = 0, 6x + y ñ 21 = 0 vkSj

2x + y ñ 5 = 0 gSA dks.k A ds cká v)Zd fdl fcUnq ls xqtjrk gS& (A) (3, 1) (B) (4, 2) (C*) (2, ñ 7) (D) (ñ 7, 1)

Sol. Equation of AD is y = 1 AD dk lehdj.k y = 1

external bisector of A is x = 2 A dk cká v)Zd x = 2

It passes through (2, ñ7) ;g (2, ñ7) ls xqtjrk gSA

4. If the product of roots of the equation x2 ñ 5x + 2log4 = 0 is 8, then is [QERC]

;fn lehdj.k x2 ñ 5x + 2log4 = 0 ds ewyksa dk xq.kuQy 8 gS] rks gS &

(A) ± 2 2 (B*) 2 2 (C) 3 (D) 5

Sol. = 2log4 =2

2log2 = 2 = 8 = ± 2 2

But > 0 = 2 2 only

ysfdu > 0 = 2 2 dsoy 5. If and ( < ) are the roots of the equation x2 ñ bx + c = 0, where c < 0 < b, then ;fn vkSj ( < ) lehdj.k x2 ñ bx + c = 0 ds ewy gS tgk¡ c < 0 < b, rc [QELR] (A) 0 < < (B) < 0 < < || (C) < < 0 (D*) < 0 < || < Sol. x2 ñ bx + c = 0 Here, + = b = c

c < 0 < 0 roots are of opposite sign ewy foijhr fpUg ds gSA b > 0 +> 0 positive root has the greatest magnitude

/kukRed ewy vf/kdre iw.kk±d j[krk gSA a < 0 < || < 6. The equation k sin + cos2 = 2k ñ 7 possess a solution if [BAMS] lehdj.k k sin + cos2 = 2k ñ 7 gy j[krh gS ;fn (A) k > 2 (B*) 2 k 6 (C) k > 6 (D) k < 2

Page # 4

Sol. k sin + 1 ñ 2sin2 = 2k ñ 7

2sin2 ñ ksin + (2k ñ 8) = 0

sin = 2k k ñ 8(2k ñ 8)

4

= 2k k ñ 16k 64

4

= k (k ñ 8)

4 = 2, k ñ 4

2

sin 2

sin = k ñ 4

2 ñ1 k ñ 4

2 1

ñ2 k ñ 4 2 2 k 6 7. If , R and the quadratic equations x2 + 2x + 7 = 0 and 4x2 + x + = 0 have atleast one

common roots, then the value of + is [QECR] ;fn , R rFkk f}?kkr lehdj.kksa x2 + 2x + 7 = 0 rFkk 4x2 + x + = 0 dk de ls de ,d ewy

mHk;fu"B gks] rks + dk eku gS & (A) ñ 4 (B*) 36 (C) 20 (D) 24 Sol. x2 + 2x + 7 = 0 has imaginary roots. x2 + 2x + 7 = 0 ds ewy dkYifud gSA Both roots of 4x2 + x + = 0 and x2 + 2x + 7 = 0 are common. 4x2 + x + = 0 rFkk x2 + 2x + 7 = 0 ds nksuksa ewy mHk;fu"B gSA

41 2 7

= 8 ; = 28

8. If one root of the quadratic equation x2 ñ bx ñ c = 0 (b, c Q) is tan8 , then the ordered pair

(b, c) is [QERC] (A) (2 2, ñ 1) (B*) (ñ2, 1) (C) (2, 2 2) (D) none of these

;fn f}?kkr lehdj.k x2 ñ bx ñ c = 0 (b, c Q) dk ,d ewy tan8 gS] rks Øfer ;qXe (b, c) gS&

(A) (2 2, ñ 1) (B*) (ñ2, 1) (C) (2, 2 2) (D) buesa ls dksbZ ugha

Sol. tan 8 = 2 ñ 1

Letekuk = ñ1 + 2

thenrks = ñ 1 ñ 2 + = ñ 2 = ñ 1 Quadratic equation is f}?kkr lehdj.k x2 + 2x ñ 1 = 0 b = ñ2, c = 1

Page # 5

9. Set of all the values of x satisfying 2

3(x 1) (x 2) log(1 x)

x (x 3)(x 4)

0 is [BAR ]

2

3(x 1) (x 2) log(1 x)

x (x 3)(x 4)

0 dks larq"V djus okys x ds lHkh ekuksa dk leqPp; gS&

(A*) (ñ1, 0) (0, 2] (3, 4) (B) (0, 2] (3, 4) (C) [1, 2] (3, 4) (D) (ñ1, 0) [1, 2] (3, 4)

Sol.

ñ ñ ñ + ñ +

ñ1 0 1 2 3 4E E

not defined

10. The number of solutions of the equation [x] + 2{ñx} = 3x, is (where [ ] represents the greatest

integer function and {x} denotes the fractional part of x): [BAG ] lehdj.k [x] + 2{ñx} = 3x ds gyksa dh la[;k gSA (tgk¡ [ ] egÙke iw.kk±d Qyu dks iznf'kZr djrk gS rFkk {x}, x

ds fHkUukRed Hkkx dks fu:fir djrk gS½ (A) 1 (B) 2 (C*) 3 (D) 0 Sol. [x] + 2 {ñ x} = 3x Case- when x I. x = 3x ie x = 0 Case- when x , the equation becomes [x] + 2 (1 ñ {x}) = 3 ([x] + {x}) 2 ñ 2 [x] = 5 {x} 0 2 ñ 2 [x] < 5 ñ 2 ñ 2 [x] < 3

ñ 32

< [x] 1

[x] = 0 or 1 or ñ 1

if [x] = 0, then {x} = 25

x = 25

if [x] = 1, then {x} = 0 which is not possible

if [x] = ñ1, then {x} = 45

x = ñ 15

Hence there are 3 solutions Hindi. [x] + 2 {ñ x} = 3x Case- tc x I. x = 3x ie x = 0 Case- tc x ,

lehdj.k izkIr gksxk [x] + 2 (1 ñ {x}) = 3 ([x] + {x}) 2 ñ 2 [x] = 5 {x} 0 2 ñ 2 [x] < 5 ñ 2 ñ 2 [x] < 3

Page # 6

ñ 32

< [x] 1

[x] = 0 or 1 or ñ 1

;fn [x] = 0 gS] rks {x} = 25

x = 25

;fn [x] = 1 gS] rks {x} = 0 tks fd laHko ugha gSA

;fn [x] = ñ1 gS] rks {x} = 45

x = ñ 15

blfy, 3 gy gSaA

MCQ 11. If the coordinates of the extermities of diagonal of a square are (2, ñ1) and (6, 2), then the

coordinates of extremities of other diagonal are [SLPL]

;fn oxZ ds fod.kZ ds fljksa ds funsZ'kkad (2, ñ1) vkSj (6, 2) gS] rc nwljs fod.kZ ds fljksa ds funsZ'kkad gS&

(A*) 5 5,2 2

(B) 11 3,2 2

(C*) 11 3,ñ2 2

(D) 5 5,ñ2 2

Sol. AC = BD = 16 9 = 5

MD = MB = 52

PointfcUnq M 14,2

AC BD

Slope of AC dh izo.krk = 34

Slope of BD dh izo.krk = 4ñ

3

If is the inclination of BD then ;fn , BD dk >qdko gks] rks

tan = 4ñ

3

sin = 45

, cos = 3ñ

5

coordinates of extremities of diagonal BD are fod.kZ BD ds fljksa ds funsZ'kkad gS

5 1 54 cos , sin2 2 2

5 3 1 5 44 ñ ,

2 5 2 2 5

5 5,2 2

and vkSj 11 3,ñ2 2

12. If one vertex of an equilateral triangle of side a is origin and the other lies on the line x ñ 3 y = 0 then the coordinates of third vertex are [SLPL]

Page # 7

;fn a Hkqtk ds leckgq f=kHkqt dk ,d 'kh"kZ ewy fcUnq vkSj nwljk 'kh"kZ js[kk x ñ 3 y = 0 ij fLFkr gS] rc

rhljk 'kh"kZ gS&

(A*) (0, a) (B*) (0, ñ a) (C*) 3 a a,ñ2 2

(D*) 3 a añ ,

2 2

Sol.

13. If the equation ax2 + bx + c = 0, a, b, c R and a 0, has imaginary roots, then [QENR]

;fn lehdj.k ax2 + bx + c = 0, a, b, c R rFkk a 0, ds ewy dkYifud gks] rks & (A*) (a + b + c) (a ñ b + c) > 0 (B*) (a + b + c) (a ñ 2b + 4c) > 0

(C*) (a ñ b + c) (4a ñ 2b + c) > 0 (D*) (9a ñ 3b + c) (a ñ 3b + 9c) > 0 Sol. Let ¼ekukfd½ f(x) = ax2 + bx + c

f(1) f(ñ1) > 0 (a + b + c) (a ñ b + c) > 0

f(1) f 1ñ

2

> 0 (a + b + c) (a ñ 2b + 4c) > 0

f(ñ1) f(ñ2) > 0 (a ñ b + c) (4a ñ 2b + c) > 0

f(ñ3) f 1ñ

3

> 0 (9a ñ 3b + c) (a ñ 3b + 9c) > 0

14. The solution set of the system of inequations [BAMS] 2 sin2x ñ 3 sinx + 1 0 and x2 + x ñ 12 0 has (A*) three integers (B*) one prime numbers (C*) two natural numbers (D*) no composite number vlfedkvksa 2 sin2x ñ 3 sinx + 1 0 vkSj x2 + x ñ 12 0 ds fudk; dk gy leqPp; j[krk gS&

(A*) rhu iw.kk±d (B*) ,d vHkkT; la[;k

(C*) nks izkdr̀ la[;k,¡ (D*) la;qDr la[;k ugha

Page # 8

Sol. 2 sin2x ñ 3 sinx + 1 0 (2sinx ñ 1)(sinx ñ 1) 0

12

sinx 1

2n + 6 x 2n + 5

6 , n ....(1)

x2 + x ñ 12 0 (x + 4)(x ñ 3) 0 x [ñ 4, 3] .........(2) (1) and (2)

x ñ7

ñ4,

6

5,6 6

Integers are ñ 4, 1, 2

15. The equation 2 2x x

x 1 x ñ 1

= a (a ñ 1) has [QEPQ]

(A*) four real roots, if a > 2 (B*) two real roots, if 1 < a < 2 (C) no real roots, if a < ñ 1 (D*) four real roots, if a < ñ 1

lehdj.k 2 2x x

x 1 x ñ1

= a (a ñ 1) j[krk gS

(A*) pkj okLrfod ewy ;fn a > 2 (B*) nks okLrfod ewy ;fn 1 < a < 2 (C) dksbZ okLrfod ewy ugh ;fn a < ñ 1 (D*) pkj okLrfod ewy ;fn a < ñ 1

Sol. 2x x

x 1 x ñ 1

ñ 2

22x

x ñ 1

= a(a ñ 1)

22

22x

x ñ1

ñ 2

22x

(x ñ 1)

= a(a ñ 1)

Let ekuk 2

22x

x ñ 1

= t

t2 ñ t ñ a(a ñ1) = 0 t = a or t = 1 ñ a

2

22x

x ñ 1

= a x = ±

aa ñ 2

Page # 9

or ;k 2

22x

x ñ 1

= 1 ñ a x = ±

a ñ 1

a 1

a < ñ 1 all four roots are real lHkh pkj ewy okLrfod gS 1 < a < 2 two real roots nks okLrfod ewy gS a > 2 all four real roots lHkh pkj okLrfod ewy gS Integer 16. The combined equation of bisector of angles between the lines L1 and L2 is [SLAB]

2x2 ñ 3xy ñ 2y2 ñ x + 7y ñ 3 = 0. P(4, ñ 3) is a point on L1. If the equation of obtuse angle bisector is ax + by ñ 3 = 0, then find the value of |a ñ b|. js[kkvksa L1 rFkk L2 ds e/; ds dks.k ds v)Zdksa dk la;qDr lehdj.k 2x2 ñ 3xy ñ 2y2 ñ x + 7y ñ 3 = 0 gSA

L1 ij fcUnq P(4, ñ 3) gSA ;fn vf/kddks.k v)Zd dks.k dk lehdj.k ax + by ñ 3 = 0 gS] rc |a ñ b| dk eku

Kkr dhft,A Ans. 09 Sol. Solving the combined equation, we get the separate equations of angle bisectors as x ñ 2y + 1 = 0 ...........(B1) and 2x + y ñ 3 = 0 ...........(B2)

perpendicular distance of P(4, ñ3) from B1 , is p = 4 6 15

= 115

and perpendicular distance of P(4, ñ3) from B2 is q = 8 ñ 3 ñ 3

5 = 2

5

Since p > q B1 i.e. x ñ 2y + 1 = 0 is the obtuse angle bisector. or B1 is ñ 3x + 6y ñ 3 = 0 a = ñ 3, b = 6 or |a ñ b| = 9 Hindi la;qDr lehdj.k dks gy djus ij dks.k v)Zdksa dk lehdj.k x ñ 2y + 1 = 0 ...........(B1) rFkk 2x + y ñ 3 = 0 ...........(B2)

fcUnq P(4, ñ3) ls B1 ij yEcor~ nwjh p = 4 6 15

= 115

rFkk fcUnq P(4, ñ3) ls B2 ij yEcor~ nwjh q = 8 ñ 3 ñ 3

5 = 2

5

pawfd p > q

B1 i.e. x ñ 2y + 1 = 0 vf/kd dks.k v)Zd dks.k dk lehdj.k

;k B1 dk lehdj.k ñ 3x + 6y ñ 3 = 0

a = ñ 3, b = 6 ;k |a ñ b| = 9

Page # 10

17. O(0, 0) is a fixed point and P any point on the straight line x ñ 2y + 3 = 0. OP is joined and on it is

taken a point Q such that OP. OQ = 2. Locus of the point Q is a circle whose radius is k45

.

Find the value of k. [SLLO] O(0, 0) ,d fLFkj fcUnq gS rFkk ljy js[kk x ñ 2y + 3 = 0 ij dksbZ fcUnq P gSA OP dks feyk;k x;k gS rFkk

bl ij fcUnq Q bl izdkj fy;k x;k gS fd OP. OQ = 2 . fcUnq Q dk fcUnqiFk ,d oÙ̀k gS ftldh f=kT;k

k45

gS] rks k dk eku Kkr dhft,A

Ans. 05 Sol.

h = r1 cos, k = r1sin also r1=2 2h k

x1 = r2 cos, y1 = r2sin (x1 y1) lies on the line x ñ 2y + 3 = 0 r2cosñ 2r2 sin + 3 = 0

r2(cosñ 2sin) + 3 = 0 r2 = ñ3

cos ñ 2sin

r1r2 = 2 (Given)

r1 ñ3

cos ñ 2sin

= 2

or ñ 3r1 = 2cos ñ 4sin or ñ 3r1

2 = 2(r1cos) ñ 4(r1sin) ñ 3(h2 + k2) = 2h ñ 4k 3x2 + 3y2 + 2x ñ 4y = 0 which is a circle tks ,d oÙ̀k gSA

whose centre is ftldk dsUnz 1 2

ñ ,

3 3

and radius f=kT;k = 1 4ñ 0

9 9 = 5

3

k45

= 53

or ;k k3 5

= 53

or ;k k = 5

18. Find the greatest integral value of for which (2 ñ 1)x2 ñ 4x + (2 ñ 1) = 0, R has real roots.

dk iw.kk±d eku Kkr dhft, ftlds fy, lehdj.k (2 ñ 1)x2 ñ 4x + (2 ñ 1) = 0 , R ds okLrfod

ewy gSA [QEGR] Ans. 01

Page # 11

Sol. CasefLFkfr ñ I 12

16 ñ 4(2 ñ 1)2 0 (2 ñ 2 + 1)(2 + 2 ñ 1) 0

(2 ñ 3)(2 + 1) 0 1 3ñ ,

2 2

1 1ñ ,

2 2

1 3,2 2

integral value is, = 0, 1 iw.kk±d eku = 0, 1 gSA

CasefLFkfr ñ II 2 ñ 1 = 0 Rejected is not integral

2 ñ 1 = 0 vLohdk;Z gS D;ksafd iw.kk±d ugha gSA hence greatest integral value of = 1

vr% dk vf/kdre eku = 1. 19. Let f(x) = x3 + x + 1 suppose p(x) is a cubic polynomial such that p(0) = ñ1 and the roots of p(x) =

0 are the square of the roots of f(x) = 0. Find the value of p(1). [QETE] ekuk f(x) = x3 + x + 1 rFkk p(x) ,d ?kuh; cgqin bl izdkj gS fd p(0) = ñ1 vkSj p(x) = 0 ds ewy] f(x) = 0

ds ewyksa ds oxZ gSA p(1) dk eku Kkr dhft,A Ans. 03 Sol. f(x) = (x ñ a) (x ñ b) (x ñ c) a, b, c are roots of f(x) = 0 f(x) ds ewy a, b, c gSA

roots of p(x) = 0 ds ewy are a2, b2, c2

soblfy, p(x) = k(x ñ a2) (x ñ b2) (x ñ c2)

put x = 0 j[kus ij ñ k a2b2c2 = ñ1

Since pawfdabc = ñ1 so blfy, k = 1 p(x2) = (x2 ñ a2) (x2 ñ b2) (x2 ñ c2) = (x ñ a) (x ñ b) (x ñ c) (x + a) (x + b) (x + c) = ñf(x) f(ñx) p(1) = ñf(1) f(ñ1) = ñ3 ◊ (ñ1) = 3 20. Let y = [2x + 4] = 4[x ñ 3]. (where [ . ] denotes the greatest integer function). If is the sum of

possible values of [3x + y], then 89 is equal to [BAGI]

ekuk y = [2x + 4] = 4[x ñ 3]. (tgk¡ [ . ] egÙke iw.kk±d Qyu dks O;Dr djrk gS). ;fn [3x + y] ds lHkh

laHkkfor ekuksa dk ;ksxQy gS] rks 89 cjkcj gS&

Ans. 01 Sol. y = [2x + 4] = 4[x ñ 3] ..........(i) y = 4[x] ñ 12 ..................(ii) [2x] + 4 = 4[x] ñ12 [2x] = 4[x] ñ 16 x = + f 2I + [2f] = 4I ñ 16

Page # 12

[2f] = 2I ñ 16 ..................(iii) 0 f < 1 0 2f < 2 [2f] = 0, 1 Case- I : [2f] = 0 2I ñ 16 = 0 I = 8 [using (iii)] y = 4 ◊ 8 ñ 12 = 20 [using (ii)] [3x + y] = [3x] + 20 = [3I + 3f] + 20 = [24 + 3f] + 20 = 44 + [3f] ..........(iv) Since [2f] = 0 0 2f < 1

0 f < 12

0 3f < 32

[3f] = 0, 1 [3x + y] = 44 + 0 = 44 [using (iv)] or [3x + y] = 44 + 1 = 45 Case-II : [2f] = 1 2I ñ 16 = 1 2I = 17

I = 172

not possible

= 44 + 45 = 89

89 = 1

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Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555

P1-JRCT1170814C0-1

BATCH : JR- CT-1 TEST DATE : 17-08-2014

SYLLABUS : Geometrical Optics complete (excluding optical instruments), Rectilinear Motion, Projectile Motion, Relative Motion, NLM Photo electric effect. Submit_(E) :- 15-07-2014 (AJP Sir) Final_(E) :- 31-07-14_(AJP Sir) Submit_(H) :- 28-07-2014 (DKY Sir) Final_(H) :- 31-07-2014 (DKY Sir)


1 to 10 SCQ 10 3 ñ1 30

11 to 15 MCQ 5 4 0 20


21 to 30 SCQ 10 3 ñ1 30

31 to 35 MCQ 5 4 0 20


41 to 50 SCQ 10 3 ñ1 30

51 to 55 MCQ 5 4 0 20


60 210

Paper-1 CT-1

Total Total

Maths

Physics

Chemistry

PAPER-1

SECTION-1 : (Only One option correct type) [k.Mñ1 : (dsoy ,d lgh fodYi çdkj)

This section contains 10 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

bl [k.M esa 10 cgqfodYi ç'u gSA çR;sd ç'u esa pkj fodYi (A), (B), (C) vkSj (D) gS] ftuesa ls dsoy ,d lgh gSA

SCQ (10) 21. The figure below shows an object O placed at a distance R to the left of a convex spherical mirror that

has radius of curvature R. Point C is the center of curvature of the mirror. Size of the object is much smaller than the radius of curvature of the mirror. The image formed by the mirror is at [GO_SM] fp=k esa mÙky xksyh; niZ.k ds ck;h vksj R nwjh ij ,d oLrq O j[kh gqbZ gSA bl xksyh; niZ.k dh oØrk f=kT;k R gSA fcUnq C niZ.k dk oØrk dsUnz gSA oLrq dk vkdkj niZ.k dh oØrk f=kT;k dh rqyuk esa cgqr NksVk gSA niZ.k }kjk cuk;k x;k izfrfcEc fdruh nwjh ij gksxk &

(A) a distance R to the left of the mirror and inverted (B) a distance R to the right of the mirror and upright

(C) a distance R3

to the left of the mirror and inverted

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PHYS

ICS



P1-JRCT1170814C0-2

(D*) a distance R3

to the right of the mirror and upright

(A) niZ.k ds cka;h vksj R nwjh ij rFkk mYVk (B) niZ.k ds nka;h vksj R nwjh ij rFkk lh/kk

(C) niZ.k ds cka;h vksj R3

nwjh ij rFkk mYVk

(D*) niZ.k ds nka;h vksj R3

nwjh ij rFkk lh/kk

Sol. u = ñR ; f = R2

+ ; v = uf Ru f 3

Image is at a distance R3

to the right of the mirror. Its virtual and upright.

izfrfcEc niZ.k ds nka;h vksj R3

nwjh ij gksxkA ;g vkHkklh rFkk lha/kk gksxkA

22. Each of the figures below shows 3 blocks of masses 3m, 2m and m acted on by an external horizontal

force F. For each figure, which of the following statements about the magnitude of the force that block B exerts on block C is correct ? (Assume that the surface on which the blocks move is frictionless.)

[NL_SY] iznf'kZr izR;sd fp=k esa 3m, 2m rFkk m nzO;eku ds 3 CykWdksa ij ckg~; {kSfrt cy F dk;Zjr~ gSA izR;sd fp=k ds fy, cy ds ifjek.k ds ckjs esa fuEu esa ls dkSulk dFku lgh gS tks CykWd B, CykWd C ij vkjksfir djrk gS ? (;g ekfu, fd og lrg ftl ij CykWd xfr djrk gS] ?k"kZ.kjfgr gS)

Figure fp=k-1 Figure fp=k-2

(A*) FBC = F6

FBC = 5F6

(B) FBC = F6

FBC = F6

(C) FBC = 5F6

FBC = 5F6

(D) FBC = F FBC = F

Sol. In figure-1 FBC = m F F6m 6

In figure-2 FBC = F 5F5m6m 6

23. A small plane can fly at a speed of 200 km/h in still air. A 30 km/h wind is blowing from west to east.

How much time is required for the plane to fly 500 km due north ? [Made by AJP Sir 2014-2015]

[PL_TQ] ,d NksVk gokbZ tgkt gok esa 200 km/h dh pky ls mM+ ldrk gSA gok 30 km/h dh pky ls if'pe ls iwoZ dh vksj py jgh gSA gokbZ tgkt ds mÙkj fn'kk esa 500 km nwjh rd mM+us esa yxk vko';d le; fdruk gksxk ?

(A) 50 h23

(B) 50 h409

(C) 50 h20

(D*) 50 h391

Sol.

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P1-JRCT1170814C0-3

W

N

E

S

PWV

WV

PV

VP = 2 2

PW WV V = 10 391 km/hr

T = P

500kmV

= 50391

hr

24. In an experimental observation of the photoelectric effect, the stopping potential was plotted versus the

light frequency, as shown in the figure below. The best straight line was fitted to the experimental points. Which of the following gives the slope of the line ? (Work function of the metal is and symbols have their usual meaning.)

[MP_PE] izdk'k fo|qr izHkko ds izk;ksfxd isz{k.k esa fujks/kh foHko rFkk izdk'k dh vkof̀Ùk ds e/; xzkQ fp=kkuqlkj [khapk x;k gSA izk;ksfxd fcUnqvksa ds laxr mi;qDr ljy js[kk [khaph xbZ gSA fuEu esa ls dkSulk js[kk dh <ky dks n'kkZrk gS ? (/kkrq dk dk;ZQyu gS rFkk ladsrksa ds vius lkekU; vFkZ gSA)

vkof̀Ùk

fujks/kh foHko

(A) h (B*) he

(C) eh

(D) e

Sol. eV = h ñ hVe e

Slope of the line js[kk dk <+ky he

25. A particle is projected with speed 30m/s at angle 22.5∞ with horizontal from ground as shown. AB and

CD are parallel to y-axis and B is highest point of trajectory of particle. CD/AB is [PM_PH] ,d d.k dks /kjkry ls {kSfrt ls 22.5∞ dks.k ij 30m/s dh pky ls fp=kkuqlkj iz{ksfir fd;k tkrk gSA AB rFkk

CD, y-v{k ds lekUrj gS rFkk B d.k ds iFk dk mPpre fcUnq gSA CD/AB gksxkA

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P1-JRCT1170814C0-4

(A) 3 (B) 3/2 (C) 2 (D*) 4 Sol.

AB = 1/2 g(T/2)2 = 1/8 gT2 CD = 1/2 gT2 CD/AB = 4 26. A convex lens forms inverted image of a real object on a fixed screen. The size of image is 9 cm. When

lens is displaced 40 cm along principle axis it again forms a real image of size 4 cm on the screen. Focal length of the lens is [GO_LE]

,d mÙky ySUl fLFkj insZ ij okLrfod oLrq dk mYVk çfrfcEc cukrk gSA çfrfcEc dk vkdkj 9 cm gSA tc ySUl dks eq[; v{k ds vuqfn'k 40 cm foLFkkfir fd;k tkrk gS rks ;g iqu% insZ ij 4 cm vkdkj dk okLrfod çfrfcEc cukrk gSA ySUl dh Qksdl nwjh Kkr dhft,A

(A*) 48 cm (B) 100 cm (C) 30 cm (D) 10 cm

Sol. since m1 m2 = 1 1 2i i

o o

h h1

h h

h0 = 1 2i ih h = 6cm

for first position of the lens

m = 32

= 40xx

3x + 2x = 80 x = 80 u = ñ x , v = (40 + x) Substituting these values in lens formula we get f = 48 cm. Alternate From displacement method

Using f =2 2D d

4D =

2 2200 404 200

= 48 cm

pwafd m1 m2 = 1 1 2i i

o o

h h1

h h

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P1-JRCT1170814C0-5

h0 = 1 2i ih h = 6cm

ySal dh izFke fLFkfr ds fy,

m = 32

= 40xx

3x + 2x = 80 x = 80 u = ñ x , v = (40 + x) bu ekuksa dks ySal lw=k esa j[kus ij f = 48 cm. oSdfYid foLFkkiu fof/k ls

f =2 2D d

4D =

2 2200 404 200

= 48 cm

27. A point moves with uniform acceleration and v1 , v2 , v3 denote the average velocities in the three

successive intervals of time t1 , t2 , t3 respectively. Which of the following relation is correct ,d fcUnq ,d leku Roj.k ls xfr djrk gS rFkk v1 , v2 , v3 rhu Øekxr le;kUrjky Øe'k% t1 , t2 , t3 esa vkSlr osx

dks iznf'kZr djrs gS rc fuEu esa ls dkSulk lEcU/k lgh gSA [RM_AA] (A) (v1 ñv2) : (v2 ñ v3) = (t1ñt2): (t2 + t3) (B*) (v1 ñv2) : (v2 ñ v3) = (t1+ t2): (t2 + t3) (C) (v1 ñv2) : (v2 ñ v3) = (t1ñt2): (t1 ñ t3) (D) (v1 ñ v2) : (v2 ñ v3) = (t1ñt2): (t2 ñ t3)

Sol. v1 = u + 1at2

v2 = u + at1 + 2at2

v3 = u + at1 + at2 + 3at2

1 2

2 3

v vv v

= 1 2

2 3

t tt t

28. In the given figure a block of mass m is tied on a wedge by an ideal string as shown in figure. String is

parallel to the inclined surface of wedge. All the surfaces involved are smooth. Wedge is being moved towards right with a time varying velocity v = t2 (m/s) where t is in sec. At what time block will just break the contact with wedge. (use g = 10 m/s2) [NL_CM] fn;s x;s fp=k esa m nzO;eku dk ,d CykWd vkn'kZ Mksjh dh lgk;rk ls ost ij fp=kkuqlkj ca/kk gqvk gSA Mksjh ost dh ur lrg ds lekUrj gSA lHkh lrg ?k"kZ.kjfgr gSA ost nka;h vksj le; ifjorhZ osx v = t2 (m/s) ds lkFk xfr djrk gS tgk¡ t lsd.M esa gSA fdl le; ij CykWd ost ds lkFk Bhd lEidZ NksM+ nsxkA (g = 10 m/s2 ysa)

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P1-JRCT1170814C0-6

(A) 10 sec. (B*) 5 sec. (C) 2 sec. (D) 4 sec. Sol. v = t2 a = 2t

At the instant when the normal is just zero ml {k.k ij tc vfHkyEc Bhd 'kwU; gS mg cos = ma sin

a = g cot since = 45∞

a = 2t = 10 t = 5 sec. Alternative solutions oSdfYid gy v = t2 a = 2t Normal will be zero at the instant ma sin = mg cos vfHkyEc ml {k.k ij 'kwU; gksxk tc ma sin = mg cos

ma

T

m

a = g cot

since = 45∞ a = 2t = 10 t = 5 sec. 29. A particle is projected with speed 25 m/s at angle 53∞ from horizontal in front of an inclined plane mirror

as shown in figure. After how long speed of image w.r.t. object will be minimum (g = 10 m/s2) [PM_PH] fp=k esa n'kkZ;s vuqlkj >qds gq, lery niZ.k ds lkeus ,d d.k dks {kSfrt ls 53∞ dks.k cukrs gq,] 25 m/s dh pky

ls iz{ksfir fd;k tkrk gSA fdrus le; i'pkr~ izfrfcEc dh pky oLrq ds lkis{k U;wure gksxhA (g = 10 m/s2)

(A*) 7 s8

(B) 5 s3

(C) 4 s5

(D) 1s

Sol. Relative speed will be minimum (zero) when velocity of object will be parallel to plane mirror. lkis{k pky U;wure ¼'kwU;½ gksxh] tc oLrq dk osx lery niZ.k ds lekUrj gSA

tan = usin gtucos

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P1-JRCT1170814C0-7

34

=

425 10t5

3255

3 20 10t4 15

45 = 80 ñ 40t

40t = 35

t = 7 s8

30. In a photoelectric experiment, with light of wavelength , the fastest electron has speed v. If the exciting

wavelength is changed to 34

, the speed of the fastest emitted electron will become

izdk'k fo|qr iz;ksx esa iz;qDr rjaxnS/;Z ds izdk'k ls mRlftZr lcls rst bysDVªku dh pky v gSA ;fn iz;qDr izdk'k

dh rjaxnS/;Z 34

dj nh tk,s] rks mRlftZr lcls rst bysDVªku dh pky gksxh&

(A) v 34

(B) v 43

[MP_PE]

(C) less than v 34

ls de (D*) greater than v 43

ls vf/kd

Sol.(D) 21 hcmv2

21 mv '2

= hc(3 / 4)

ñ = 4hc3

Clearly Li"Vr% 4v ' v3

SECTION-2 : (One or more option correct type) [k.Mñ2 : (,d ;k vf/kd lgh fodYi çdkj)

This section contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE or MORE are correct.

bl [k.M esa 5 cgqfodYi ç'u gSA çR;sd ç'u esa pkj fodYi (A), (B), (C) vkSj (D) gS] ftuesa ls dsoy ,d ;k vf/kd lgh gSA

MCQ (5) 31. Which of the following can never be the seperation between a point real object lying on the optical axis

of a converging lens of focal length 30 cm and its real image. 30 cm Qksdy nwjh ds ,d vfHklkjh ysal ds izdkf'kd v{k ij fLFkr ,d okLrfod fcUnq oLrq rFkk blds okLrfod izfrfcEc ds e/; nwjh fuEu esa ls dkSulh dHkh Hkh ugha gks ldrh &

(A*) 60 cm (B*) 90 cm (C*) 45 cm (D) 150 cm [GO_LE] Sol. Least seperation between a point real object lying on the optical axis of a converging lens of focal

length of f and its real image is 4f. f Qksdl nwjh ds vfHklkjh ySal dh izdkf'kd v{k ij fLFkr ,d okLrfod fcUnq oLrq rFkk blds okLrfod izfrfcEc ds

e/; U;wure nwjh 4f gksrh gSA 32. Particle thrown from O, parallel to the incline as shown hits the incline perpendicular to it. Choose the

correct options. [PM_PH] urry ds lekUrj fcUnq O ls, QSadk x;k d.k urry ij fp=kkuqlkj yEcor~ Vdjkrk gSA lgh fodYiksa dk p;u dhft,A

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P1-JRCT1170814C0-8

(A*) d is equal to 9375 m16

(B) Particle hits the incline after the time 10 sec

(C) d is equal to 7225 m16

(D*) Particle hits the incline after time 12.5 sec

(A*) d, 9375 m16

ds cjkcj gS (B) 10 sec i'pkr~ d.k urry ij Vdjk;sxk

(C) d, 7225 m16

ds cjkcj gS (D*) 12.5 sec i'pkr~ d.k urry ij Vdjk;sxk

Sol. yV60

= tan 37 = 3/4 vy = 45 m/s

ñ 45 = 80 ñ 10t 10t = 125

t = 252

sec

d + x0 = 25602

= 750 m.

(45)2 = (80)2 + 2(ñ10)y

y = 8754

m.

= tan 53 = 4/3 x0 = 3/4 y = 2625

16m.

d = 750 ñ 262516

= 937516

m

33. Out of the following, select the correct statement(s) : fuEu esa ls lgh dFku pqfu, & [GO_LE] (A*) Sun glasses which have curved surfaces (same radius of curvature at both surface) have no

optical power /kwi ds p'esa ftudh lrg oØh; gksrh gSa] dksbZ izdkf'k; 'kfDr ugha j[krs gSA ¼p'es dh nksuksa lrg dh oØrk f=kT;k

leku gSA½ (B*) When thin lenses are in contact, the magnifications get multiplied to get overall magnification tc irys ySUl lEidZ esa gksrs gS] rks dqy vko/kZu izR;sd ySal ds vko/kZu ds xq.kuQy ds cjkcj gks tkrk gSA (C*) The focal length of a lens (the surroundings on both sides of the lens have the same refractive

index) does not depend upon the direction from which light is incident ySUl dh Qksdl nwjh izdk'k ds vkiru dh fn'kk ij fuHkZj ugha djrh gS] ;fn ySUl ds nksuksa vksj okrkoj.k dk

viorZukad leku gSA (D) Decreasing the radii of the two surfaces of a double convex or double concave lens (keeping the

refractive index of lens and surrounding medium unchanged) increases its magnitude of focal length f}mÙky ;k f}vory ySUl dh nksuksa lrgksa dh f=kT;k ?kVkus ij ¼ySal rFkk vkl&ikl ds ek/;e dk viorZukad ugha

cnyrs gq,½ bldh Qksdl nwjh dk ifjek.k c<+rk gSA

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P1-JRCT1170814C0-9

Sol. Sunglasses are used to save the eyes from the heat radiations of sun. But the use of sun glasses should not affect the working of the eye. For that reason the sun glasses have zero optical power. To achieve this the radius of curvature of the two surfaces are same and their centres are also on the same side. Now mathematically

P = 1f

= (n ñ 1) 1 1R R

= (n ñ 1) (0) = 0

(B) m1 = 1hh

, m2 = 2

1

hh

....... ,

overall m = fhh

= m1 ◊ m2 ◊ m3 ◊ .......

Where hf = height of the final image.

(C) 1f

= (nrel ñ 1) 1 2

1 1R R

.... (1)

1f '

= (nrel ñ 1) 2 1

1 1R R

.... (2)

(1) and (2) gives f = f'

(D) 1 2

1 1 1(n 1)f R R

If the radius of curvature of both surfaces are 'a' and 'b' then 1f

= (n ñ 1) 1 1a b

If 'a' and 'b' are then f Sol. /kwi ds p'esa vk¡[kksa dks lw;Z dh Å"eh; fofdj.k ls cpkus ds fy, mi;ksx fd;s tkrs gSA ijUrq /kwi ds p'esa dk mi;ksx

vk¡[k dh dk;Z iz.kkyh ij izHkko ugha Mkyuk pkfg,A bl dkj.k ls /kwi ds p'eksa dh izdk'kh; 'kfDr 'kwU; gSA bldks izkIr djus ds fy, nksuksa lrgksa dh f=kT;k leku gksuh pkfg, o muds dsUnz Hkh leku fn'kk esa gksus pkfg,A vc xf.krh; :i ls

P = 1f

= (n ñ 1) 1 1R R

= (n ñ 1) (0) = 0

(B) m1 = 1hh

, m2 = 2

1

hh

....... ,

dqy feykdj m = fhh

= m1 ◊ m2 ◊ m3 ◊ .......

tgka hf = vfUre izfrfcEc dh ÅapkbZ

(C) 1f

= (nrel ñ 1) 1 2

1 1R R

.... (1)

1f '

= (nrel ñ 1) 2 1

1 1R R

.... (2)

(1) o (2) ls f = f'

(D) 1 2

1 1 1(n 1)f R R

;fn nksuks lrgksa dh oØrk f=kT;k 'a' o 'b' gS rks 1f

= (n ñ 1) 1 1a b

;fn 'a' o 'b' rks f 34. Both the blocks shown in figure have same mass 'm'. All the pulleys and strings are ideal. Choose the

correct options : [NL_TF]

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P1-JRCT1170814C0-10

n'kkZ;s x;s nksuks CykWdksa dk nzO;eku leku 'm' gSA lHkh f?kjuh o jLlh vkn'kZ gSA lgh fodYiksa dk p;u dhft,A

(A*) Acceleration of block A is 2g5

(B) Acceleration of block A is g5

(C*) Acceleration of block B is g5

(D*) Tension in the string attached with A is 3mg5

(A*) CykWd A dk Roj.k 2g5

gSA (B) CykWd A dk Roj.k g5 gSA

(C*) CykWd B dk Roj.k g5 gSA (D*) A ls tqM+h jLlh esa ruko

3mg5

gSA

Sol.

mg ñ T = 2ma ...........(i) 2T ñ mg = ma ...........(ii) Solving, gy djus ij mg = 5 ma

a = g5

T = mg ñ 2ma

= mg ñ 2m g5

= 3mg5

.

35. A beam of light having frequency is incident on an initially neutral metal of work function

(h > ). Choose the correct options : [GO_PE]

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P1-JRCT1170814C0-11

(A) all emitted photoelectrons have same kinetic energy equal to (h ñ ). (B*) all free electrons in the metal, that absorbs photon of energy h , may not be ejected out of the

metal. (C*) after being emitted out of the metal, the kinetic energy of photoelectrons decreases continuously

as long as they are moving away from the metal in free space. (D) the emitted photo electrons move with constant speed after escaping from the field of metal. vkof̀Ùk dk izdk'k iqUt dk;ZQyu (h > ) okys /kkrq ¼izkjEHk esa mnklhu½ ij vkifrr gksrk gS rks lgh dFkuksa dk

p;u dhft;s & (A) lHkh mRlftZr QksVks bysDVªkWuksa dh xfrt ÅtkZ (h ñ ) ds cjkcj gksxhA (B*) /kkrq esa mifLFkr lHkh eqä bysDVªkWu tks h ÅtkZ ds QksVkWuksa dks vko'kksf"kr djrs gS] /kkrq ds ckgj mRlftZr ugh

gks ldrs gSA (C*) /kkrq ls ckgj mRlftZr gksus ds i'pkr~ QksVks bysDVªkWu dh xfrt ÅtkZ fujUrj rc rd ?kVrh gS tc rd fd os

eqDr vkdk'k esa /kkrq ls nwj tkrs jgrs gSA (D) /kkrq ds {ks=k ls ckgj fudyus ds i'pkr~ mRlftZr QksVks bysDVªkWu fu;r pky ls xfr djrs gSA Sol. (A) All emitted photo electrons have kinetic energies ranging from 0 to h ñ 0. Hence false. (B) Some free electron may lose energy within metal due to collision and cannot escape the metal.

Hence true. (C) Since metal is positively charged it will attracted emitted photoelectrons. Hence true. (A) lHkh mRlftZr QksVks bysDVªkWuksa dh xfrt ÅtkZ 0 ls h ñ 0 ijkl esa gksrh gSA vr% oDrO; xyr gSA (B) /kkrq esa mifLFkr eqä bysDVªkWuksa dh VDdj ds i'pkr~ ÅtkZ esa gkfu gks ldrh gS rFkk /kkrq ls ckgj ugh fudy

ldrs gSA vr% oDrO; lR; gSA (C) pwafd /kkrq /kukRed :i ls vkosf'kr gS ;g mRlftZr QksVksa bysDVªkWuksa dks vkdf"kZr djsxk vr% oDrO; lR; gSA

SECTION-3 : (Integer value correct Type) [k.M ñ 3 : (iw.kk±d eku lgh çdkj)

This section contains 5 questions. The answer to each question is a Two digit integer, ranging from 00 to 99 (both inclusive).

bl [k.M esa 5 ç'u gSaA çR;sd ç'u dk mÙkj 00 ls 99 rd ¼nksuksa 'kkfey½ ds chp dk nks vadksa okyk iw.kk±d gSA Integer (Double Digits) (5) 36. In the figure velocity of bodies A, B and C are shown with directions. Values b and c are w.r.t. ground.

Whereas a is velocity of block A w.r.t. wedge C. velocity of block A w.r.t. ground is 3 m/s. Where a is

a pure number. Find (use b = 5 m/s, c = 5 m/s, = 60∞) [NL_CM] fp=k esa oLrq A, B rFkk C ds osx dh fn'kk,sa n'kkZ;h xbZ gSA osx b rFkk c tehu ds lkis{k gS tcfd a ost C ds

lkis{k CykWd A dk osx gSA tehu ds lkis{k CykWd A dk osx 3 m/s gSA tgk¡ a ,d 'kq) la[;k gSA Kkr dhft,A (b = 5 m/s, c = 5 m/s, = 60∞ ysa)

A

B

C c

b

a

Ans. 25 Sol. a = b + c Net acceleration of A = 2 2a c 2ac cos( )

A dk dqy Roj.k = 2 2a c 2ac cos( )

2 2(b c) c 2(b c)c cos .

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P1-JRCT1170814C0-12

37. A thin cylindrical parallel light beam of beam diameter d and intensity 0 falls on a convex lens parallel to

the principal axis symmetrically. Light refracted from lens is obtained on screen, which is perpendicular to the principal axis. Intensity obtained on screen is 16 0 for two positions of screen, separated by a displacement of 10cm. Focal length of lens in cm is : (Assume all the light to be transmitted from the lens)

d O;kl rFkk 0 rhozrk dk ,d iryk csyukdkj lekUrj izdk'k iaqt mÙky ySal ij eq[; v{k ds lekUrj lefer :i ls fxjk;k tkrk gSA ySal ls viofrZr izdk'k insZ ij izkIr gksrk gS tks eq[; v{k ds yEcor~ gSA insZ ij 10 cm dh nwjh ij fLFkr nks fLFkfr;ksa ij izkIr rhozrk 16 0 gSA ySal dh Qksdl nwjh cm eas gksxh : (;g ekfu, fd lEiw.kZ izdk'k ySal

ls ikjxfer gksrk gS) [GO_LE] Ans. 20 Sol.

f

2= 10

f = 20 cm. 38. In the following arrangement the system is initially at rest. The 5 kg block is now released. Assuming

the pulleys and string to be massless and smooth. If the acceleration of block 'C' is x10

m/s2, then find

value of x. Take g = 9.8 m/s2. uhps fn;s x;s la;kstu esa fudk; izkjEHk esa fLFkj voLFkk esa gSA vc 5kg nzO;eku dks eqDr djrs gSA f?kjfu;ksa ,oa jLlh

dks nzO;eku jfgr ,oa fpduk ekusA ;fn xqVds 'C' dk Roj.k x

10m/s2 gS rc x dk eku Kkr dhft,A

g = 9.8 m/s2 ekus [NL_CM] ]

Ans. 7

Sol. Block B will not move. CykWd B xfr ugha djsxkA 5g ñ T = 5a .............(1)

2T ñ 8g = 8 a2

..............(2)

10g ñ 2T = 10a ..............from (1) ls

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P1-JRCT1170814C0-13

2g = 14a

a = g7

a2

= g14

= 9.814

= 710

m/s2 = 0.7

x = 7 39. A particle is projected with speed v = 150 m/s from the horizontal surface such that its range on the

horizontal plane is twice the greatest height attained by it. The range of the projectile in metre is : (use g = 10 m/s2 ) [PM_PH]

,d d.k dks {kSfrt lrg ls v = 150 osx ls bl izdkj iz{ksfir fd;k tkrk gS fd {kSfrt ry ij bldh ijkl blds }kjk izkIr vf/kdre~ ÅWpkbZ dh nqxquh gksrh gSA iz{ksI; dh ijkl ehVj esa gksxh (tgkW g xq:Roh; Roj.k gSA)

Ans. 12

Sol. 22v sin cos

g=

2 22v sin2g

tan =2, sin = 25

, cos = 15

R = 24v

5g

40. When a metallic surface is illuminated with monochromatic light of wavelength , the stopping potential

is 5 V0. When the same surface is illuminated with light of wavelength 3 , the stopping potential is V0. If

work function of the metallic surface is hcx

then 'x' is :

tc ,d /kkfRod lrg dks , rjaxnS/;Z ds ,do.khZ; çdk'k ls çdkf'kr fd;k tkrk gS rks fujks/kh foHko 5 V0 gSA tc bl lrg dks 3 rjaxnS/;Z ds çdk'k ls çdkf'kr fd;k tkrk gS rks fujks/kh foHko V0 gSA ;fn /kkfRod lrg dk

dk;ZQyu hc

x gS rks 'x' gksxkA [MP_PE]

Ans : 6

Sol. hc = 5 eV0 +

hc3

= eV0 + 2hc3

= 4eV0 = hc6

Ans.

Page # 1

Course : JR CT-1(ELPD) Test Date : 17.08.2014 Test Type : CT-1 Paper-1 Time Duration : 3 Hrs.

Paper Level - Moderate to Tough

SYLLABUS : Mole Concept, Quantum Number, Periodic Table, Basic Inorganic

Nomenclature, Gaseous State

ORGANIC

REVISE FROM ASW SIR SYLLABUS :

IUPAC Nomenclature, Structure isomers, Structure identification, POC, Electronic Effect and Aromaticity & Carbocation

SYLLABUS SCHEDULED

SYLLABUS SCHEDULED WEIGHTAGE

(BY FC)

SR. NO.

TOPIC NAME

(I) (II)

WEIGHTAGE IN PAPER-1

(BY FACULTY)

WEIGHTAGE IN PAPER-2

(BY FACULTY)

1. IUPAC Nomenclature SCQ(1) + Double

integer(1)

2. Structure isomers SCQ(1) 3. Structure

identification SCQ(1) MTC(1)

4. POC SCQ(1) MCQ(1)

MCQ(1) + Comp. (mixed)

5. Electronic Effect and Aromaticity

MCQ(1) MCQ(1)

6. Carbocation Double integer(1)

MCQ(1)

Test Pattern :

Page # 2


1 to 10 SCQ 10 3 ñ1 30

11 to 15 MCQ 5 4 0 20


21 to 30 SCQ 10 3 ñ1 30

31 to 35 MCQ 5 4 0 20


41 to 50 SCQ 10 3 ñ1 30

51 to 55 MCQ 5 4 0 20


60 210

Paper-1 CT-1

Total Total

Maths

Physics

Chemistry


1 to 7 MCQ 7 4 0 28

8 to 13 Comprehension (3 Comp. x 2 Q.) 6 3 ñ1 18

14 to 16 MTC 3 8 0 24

17 to 23 MCQ 7 4 0 28


30 to 32 MTC 3 8 0 24

33 to 39 MCQ 7 4 0 28


46 to 48 MTC 3 8 0 24

48 210

Paper-2 CT-1

Total Total

Maths

Physics

Chemistry

Physical paper1 Organic paper 1 SCQ(6) SCQ(4) MCQ(3) MCQ(2) Integer(Double)(3) Integer(Double (2) Physical paper 2 Organic paper 2 MCQ (4) MCQ(3) Comp.(3 x 2Q)(2) Comp. (3 x 2Q) (1) MTC (4 vs 4) (2) MTC((4 vs 4)(1)

Page # 3

JEE (ADVANCED) CHEMISTRY PAPER SKELETON Faculty Name : Test Name : JR (CT-1)

PAPER - 1 S.

No. TYPE (P) (I) (O) TOPIC(S) SUBTOPIC(S) DIFFICULTY LEVEL : Easy (E), Moderate (M), Tough (T)

41 SCQ (P) Gas Avg. GMM M

42 SCQ (P) Gas Laws T / M

43 SCQ (P) Gas V. W. Eq. E

44 SCQ (P) Mol Stoichiometry M

45 SCQ (P) Atomic Quantum M

46 SCQ (P) Mol M. Avg. M

47 SCQ (O) Electronic effect Inductive/Resonance E

48 SCQ (O) POC No. of action E

49 SCQ (O) POC Lab test M

50 SCQ (O) E.E. Resonance M

51 MCQ (P) Atomic Quantum M

52 MCQ (I) PT Mix E

53 MCQ (P) Gas Tc M

54 MCQ (O) POC Lab test M

55 MCQ (O) E.E. Order of I-effect M

56 Double Integer Type

(I) BIN M


(P) Mol Conc. Term M


(P) Mol Basic T / M

Page # 4


(O) Str. Iso Isomer connting M

60. Double Integer Type

(O) POC Lab test M

Faculty preparing the TEST PAPER should fill it according to paper pattern and submit it with finalisaion of paper at SMD.

Physical paper-1

SCQ(6) 41. A mixture of two gases A and B in the mole ratio 2 : 3 is kept in a 2 litre vessel. A second 3 litre vessel

has the same two gases in the mole ratio 3 : 5. Both gas mixtures have the same temperature and same pressure. They are allowed to intermix and the final temperature and pressure are the same as the initial values, the final volume being 5 litres. Given that the molar masses are MA and MB, what is the mean molar mass of the final mixture ? (GST)(P)

,d nks yhVj ds ¶ykLd esa nks xSlksa A o B ds feJ.k dks eksy vuqikr 2 : 3 esa j[kk tkrk gSA ,d vU; 3 yhVj ds ¶ykLd esa leku nksuks xSlksa dks eksy vuqikr 3 : 5 esa fy;k tkrk gSA nksuks xSl feJ.kksa dk rki o nkc leku gSA nksuksa xSlksa dks feyk;k tkrk gs rFkk vfUre rki o nkc izkjfEHkd eku ds cjkcj gS rFkk vfUre vk;ru 5 yhVj gks tkrk gSA fn;k x;k gS fd eksyj nzO;eku MA rFkk MB gS] rks vfUre feJ.k dk ek/; eksyj nzO;eku D;k gSA

(A*) 77 123200

A BM M (B) 123 77200A BM M (C) 77 123

250A BM M (D) 123 77

250A BM M

Sol.

2 3

a mol b mol 5 +

nA = 0.4a nA = 38

b nA = 0.4a + 38

b

nB = 0.6a nB = 58

b nB = 0.6a + 58

b

T, P T,P T,P

2P = aRT 3P = bRT 5P = (a + b) RT

a b2 3

or b = 1.5a

Mav. = A B

3 50.4a b M 0.6a b M8 8

a b =

A B3 50.4a 1.5a M 0.6a 1.5a M8 8

a 1.5a

= A B

7.7a 12.3aM M8 8

2.5a= A B77M 123M

200

Page # 5

42.

50 g H2

N2 Balloon(10ml)

Glass Bulb(110ml)

Calculate the volume of balloon if 25g H2 is further added in glass bulb at constant temperature. ;fn fu;r rki ij Xykl cYc esa vksj vkxs 25g H2 dks feyk;k tkrk gS rks xqCckjs dk vk;ru ifjdfyr dhft, : (GST)(P) (A) 7.45 ml (B) 8.125 ml (C) 4.2 ml (D*) 6.875 ml

Sol. Apply iz;qDr djus ij 2H

Vn

= 2N

Vn

10025

= 10n

, n = 2.5

then rc 100 x 37.5

= 102.5

x

43. If in the given apparatus a third sphere ëCí of radius r/2 is placed inside the sphere A as shown, then the

free volume Vi for the sphere C will be (GST)(P) ;fn fn;s x;s midj.k esa n'kkZ;s vuqlkj xksyh; A ds vUnj f=kT;k r/2 dk ,d rr̀h; xksyk ëCí j[kk tkrk gS] rks xksys C

ds fy, eqDr vk;ru Vi fuEu gksxk :

(A*) 43

(r/2)3 (B) 23

r3 (C) 43

(2r)3 (D) 43

(r/2)3 ◊ 4

Sol.

excluded volume is fudkyk x;k vk;ru = 43

3

2rr = 4

3

3

2r

44. a A(g) + b B(g) c C(g) + d D(g) (Mol)(P) Reaction is taking place at constant Temperature, Pressure & Volume, then correct statement is: fu;r rki] nkc o vk;ru ij mijksDr vfHkfØ;k gksrh gS] rks lgh dFku fuEu gS% (A*) a + b = c + d (B) Mavg. may increase or decrease depending upon limiting reagent. M

vkSlr esa of̀) ;k deh gksrh gS ;g lhekUr vfHkdeZd ij fuHkZj djrk gSA

(C) Vapour density of mixture will not remain same throughout the course of reaction. lEiw.kZ vfHkfØ;k ds nkSjku feJ.k dk ok"i ?kuRo leku ugha cuk jgrk gSA (D) Total moles will change with progress of reaction. vfHkfØ;k dh izxfr ds lkFk dqy eksy ifjofrZr gkssaxsA 45. Number of electrons having + m value equal to zero in 26Fe may be (ATS)(P) 26Fe esa mu bysDVªksuks dh la[;k D;k gks ldrh gSA ftuds fy;s + m dk eku 'kwU; ds cjkcj gS%

50 g H 2

N2 xqCckjk(10ml)

Xykl cYc(110ml)

Page # 6

(A*) 13 (B) 15 (C) 7 (D) 12 Sol. 26Fe ñ 1s2 , 2s2 2p6 , 3s2 3p6, 3d6, 4s2 + m = 0 = 0, m = 0 i.e. s-subshell = 1, m = ñ1 i.e. one orbital of p. = 2, m = 2 i.e. one of d-orbitals hence there are 13 or 14 electron as in d-orbital it may be one or two electron having m = ñ2. gy- 26Fe ñ 1s2 , 2s2 2p6 , 3s2 3p6, 3d6, 4s2 + m = 0 = 0, m = 0 i.e. s- midks'k = 1, m = ñ1 i.e. p. dk ,d d{kd = 2, m = 2 i.e. d- dk ,d d{kd bl izdkj ;gk¡ 13 ;k 14 bysDVªkWu gS pqafd d- d{kd esa ,d ;k nks bysDVªkWu m = ñ2. ds j[ksxsaA 46. Calculate percentage change in Mavg. of mixture, if PCl 5 undergoes 50% decomposition.

PCl5 PCl3 + Cl2 (MOl)(P) feJ.k ds M

vkSlr esa izfr'kr ifjorZu ifjdfyr dhft;s ;fn PCl5 dk 50% fo;kstu gksrk gS :

PCl5 PCl3 + Cl2 (A) 50% (B) 66.66% (C*) 33.33 % (D) Zero Sol. PCl5 PCl3 + Cl2

1ñx x x x = 0.5 1 ◊ 208.5 = (1+0.5) ◊ Mav. vkSlr

Mav.vkSlr = 139, % change (izfr'kr ifjorZu) = 69.5208.5

◊ 100 = 33.33%

SCQ(4) 47. Which of the following statement is correct ? (A) The complete transfer of electron takes place in the inductive effect. (B) Inductive effect increases with increase in distance. (C*) The resonance structures are hypothetical structure and they do not represent any real molecule. (D) The energy of resonance hybrid is always more than that of any resonating structure. fuEu esa dkSulk dFku lgh gS \ (A) izsjf.kd izHkko esa bysDVªkWuksa dk iw.kZr;k LFkkukUrj.k gksrk gSA (B) izsjf.kd izHkko nwjh c<+us ds lkFk c<+rk gSA (C*) vuquknh lajpuk;sa dkYifud lajpuk;sa gS rFkk ;s okLrfod lajpuk dks iznf'kZr ugha djrh gSA (D) fdlh Hkh vuquknh lajpuk dh rqyuk esa vuquknh ladj dh ÅtkZ vf/kd gksrh gSA 48. How many number of active hydrogen atoms are present in Harmone with mol.mass 10,000, 50 g of which

when treated with Na metal liberates 112 ml of the H2 gas at STP. gkeksZu (vkf.od nzO;eku = 10,000) esa mifLFkr lfØ; gkbMªkstu ijek.kqvksa dh la[;k D;k gksxh] ;fn blds 50 g STP ij Na /kkrq ls vfHkfØ;k dj H2 xSl ds 112 ml eqDr djs %&

(A) 1 (B*) 2 (C) 3 (D) 4 Sol. 112 ml of H2 is obtained from 50 g

22400 ml of H2 is obtained from 50 22400112

= 10,000 g

10,000 g compound give one mole H2 gas i.e. 2H obtained from 1 mole of compound. Ans. No. of active H = 2 gy% 50 xzke ls H2 ds 112 ml izkIr gksrs gSA

22400 ml H2 izkIr gksxh 50 22400

112 = 10000 xzke

10,000 xzke ;kSfxd ,d eksy H2 xSl nsxkA vFkkZr~] 1 eksy ;kSfxd ls 2H izkIr gksxhA mÙkj% lfØ; H dh la[;k = 2

Page # 7

49. PhCONH2 and PhCOOH can be distinguished by (POC(O)) NaOH, NaHCO3 2, 4-DNP C2H5OH + H2SO4 (I) (II) (III) (IV) (A) II and III (B) I and IV (C*) I, II and IV (D) I, II, III and IV PhCONH2 rFkk PhCOOH dks foHksfnr fd;k tk ldrk gS % NaOH, NaHCO3 2, 4-DNP C2H5OH + H2SO4 (I) (II) (III) (IV) (A) II rFkk III (B) I rFkk IV (C*) I, II rFkk IV (D) I, II, III rFkk IV

Sol. PhCONH2

PhCOOH

Sol. PhCONH2

PhCOOH

50. In which of the following delocalisation of anion is not possible ? fuEu esa ls fdl ;kSfxd esa _.kk;u dk foLFkkuhdj.k lEHko ugha gS\

(A) (B)

(C*) (D)

Sol. Octet of nitrogen is complete. ukbVªkstu dk v"Vd iw.kZ gSA MCQ(3) 51. Which is/are correct about the Clñ ion (AST)(P) Clñ vk;u ds ckjs esa lgh gS@gSa % (A*) For the last electron n = 3, = 1 (B) Eight electrons have m = 0 (C*) Eight electron have (n + ) = 3 (D*) It's magnetic moment is Zero. (A*) vfUre bysDVªkWu ds fy;s n = 3, = 1 (B) vkB bysDVªkWu m = 0 j[krs gSa (C*) vkB bysDVªkWu (n + ) = 3 j[krs gSa (D*) bldk pqEcdh; vk?kw.kZ 'kwU; gSA

Page # 8

52. In which of the following arrangements, the order is correct according to the property indicated against it:

fuEufyf[kr O;oLFkkvksa esa ls] Øe ds vkxs fy[ks x;s xq.kksa ds vuqlkj lgh Øe esa gS@gSa% (PTB)(InO) (A*) increasing size : Cu2+ < Cu+ < Cu (B) increasing EA1 (magnitude): S > O > Se > Te (C*) increasing EA1 (magnitude) : < Br < F < Cl (D) increasing E1 : Li < Na < K < Rb

(A*) c<+rk gqvk vkdkj : Cu2+ < Cu+ < Cu (B) c<+rh gqbZ EA1 (ifjek.k) : S > O > Se > Te (C*) c<+rh gqbZ EA1 (ifjek.k) : < Br < F < Cl (D) c<+rh gqbZ E1 : Li < Na < K < Rb 53. Incorrect statement about critical temperature is/are : (GST)(P) (A*) It is minimum temperature at which liquid and vapour can coexist. (B) Surface tension vanishes at critical temperature (C*) Gas and liquid phase has different critical density at critical temperature. (D*) Critical temperature of hydrogen and helium is very high. ØkfUrd rki ds ckjs esa xyr dFku fuEu gS / gSa : (GST)(P) (A*) ;g og U;wure rkieku gS ftl ij nzo o ok"i ,d lkFk izkIr gks ldrs gSA (B) ØkfUrd rkieku ij i"̀Bh; ruko foyqIr gks tkrk gSA (C*) ØkfUrd rkieku ij xSl rFkk nzo izkoLFkk fHkUu ØkfUrd ?kuRo j[krs gSA (D*) gkbMªkstu o ghfy;e dk ØkfUrd rkieku cgqr mPp gksrk gSA MCQ(2) 54. Which reagent will give positive response with following compound : fuEu ;kSfxd ds lkFk dkSulk vfHkdeZd /kukRed ijh{k.k nsrk gS \

(A*) Nañmetal (/kkrq½ (B*) NaHCO3 (C) 2,4-DNP (D) AgNO3 + NH4OH 55. Which of the following shows ñ effect stronger than ñF group ? fuEu esa ls dkSulk ;kSfxd ñF lewg dh rqyuk esa izcyre ñ izHkko n'kkZrk gS\

(A*) (B*) ...... ñCHO

(C*) ...... NO2 (D) ñOCH3 Integer(Double)(3) 56. How many of the following have -ate suffix. (BIN) (InO) fuEu esa ls fdrus izR;; ¼vuqyXu½ -ate (,V) j[krs gSA Na2PbO3, K2WO4, Ca(HPO3)2, Mg(H2PO2)2, Na4 P2 O7, Na2 SO5, K2SnO2, Ca(OCl)2, K2CrO4, Na2Cr2O7,

KMnO4, Na2S4O6, Pb(CH3COO)2, KCNO, Na2S2O8, K2H2P2O5. Ans. 11.

Page # 9

57. 150 g of a HNO3 solution (d = 1.2 g/mL) contains 63 g of HNO3. What volume of this HNO3 (in mL) solution will be required to react with a NaOH solution, containing 24 g NaOH : (Topic-Mole concept-I)

Reaction : HNO3 + NaOH NaNO3 + H2O ,d 150 g HNO3 foy;u (d = 1.2 g/mL), tksfd] 63 g HNO3 j[krk gSA NaOH foy;u] tks 24 g NaOH ;qDr gS] ds

lkFk iw.kZ fØ;k djus ds fy, bl HNO3 foy;u dk fdruk vk;ru (mL esa) vko';d gksxk %

vfHkfØ;k : HNO3 + NaOH NaNO3 + H2O Ans. 75 mL

Sol. Mole of HNO3 = 6363

= 1 ; Volume of HNO3 solution = 125 mL

Molarity of HNO3 = 8 M ; Letís assume that volume of HNO3 solution required for NaOH = V L Mole of HNO3 required for NaOH = 8 V.

Mole of NaOH = 2440

= 610

; 8 V = 610

V = 0.075 L = 75 mL.

gy HNO3 ds fy, eksy = 6363

= 1 ; HNO3 foy;u dk vk;ru = 125 mL

HNO3 dh eksyjrk = 8 M ; ekuk fd NaOH ds fy, vko';d HNO3 foy;u dk vk;ru = V L NaOH ds fy, vko';d HNO3 ds eksy = 8 V.

NaOH ds eksy = 2440

= 610

; 8 V = 610

V = 0.075 L = 75 mL.

58. Two liquids 'A' (molecular mass = 40) and 'B' (Molecular mass = 20) are partially miscible. When 1 mol of A and 3 mol of B are shaken together and allowed to settle, two layer L1 and L2 are formed as shown in diagram. (Mol)(P) [T]

Layer 'L1' contains 0.1 mole fraction of 'A' and layer 'L2' contains 0.4 mole fraction of A calculate simple

ratio of masses of layer L1 to layer L2. If your answer is xy

then report as (x + y).

nks nzo 'A' (v.kqHkkj = 40) o 'B' (v.kqHkkj = 20) vkaf'kd :i ls feJ.kh; gSaA tc 1 eksy A o 3 eksy B dks ,d lkFk fgykdj j[k fn;k tkrk gS] rks js[kkfp=k esa n'kkZ;s vuqlkj nks ijr L1 o L2 curh gSA (Mol)(P) [T]

ijr 'L1' 'A' dh 0.1 eksy fHkUu o ijr 'L2' 'A' dh 0.4 eksy fHkUu j[krh gSA ijr L1 o ijr L2 ds nzO;eku dk ljy

vuqikr ifjdfyr dhft;sA ;fn vkidk mÙkj xy

gS rks bls (x + y) ds :i esa nhft;sA

Ans. 25.

Page # 10

Sol. Let. moles of A in L1 = x Let. moles of B in L1 = y

x 0.1x y

so 9x = y .........(i)

In L2 mole of A = 1 ñ x In L2 mole of B = 3 ñ y

1 x 0.41 x 3 y

So 1 ñx = 1.6 ñ0.4 x ñ0.4 y 0.4y ñ0.6 x = 0.6 .........(ii) by eqn. (i) and (ii) 3x = 0.6 and x = 0.2 & y = 1.8 Mass of L1 = 0.2◊40+1.8◊20 = 44 g Mass of L2 = 0.8◊40+1.2◊20 = 56 g

Mass Ratio L1 to L2 = 44 1156 14

gy. ekuk L1 esa A ds eksy = x ekuk L1 esa B ds eksy = y

x 0.1x y

blfy, 9x = y .........(i)

L2 esa A ds eksy = 1 ñ x L2 esa B ds eksy = 3 ñ y

1 x 0.41 x 3 y

blfy, 1 ñx = 1.6 ñ0.4 x ñ0.4 y 0.4y ñ0.6 x = 0.6 .........(ii) lehdj.k (i) ,oa (ii) }kjk 3x = 0.6 rFkk x = 0.2 & y = 1.8 L1 dk nzO;eku = 0.2◊40+1.8◊20 = 44 g L2 dk nzO;eku = 0.8◊40+1.2◊20 = 56 g

L1 o L2 dk nzO;eku vuqikr = 44 1156 14

Integer(Double)(2) 59. How many structurally isomeric carbonyl compounds with molecular formula C5H10O are possible? v.kqlw=k C5H10O ds fdrus lajpukRed leko;oh dkcksZfuy ;kSfxd lEHko gS\

Ans. 07

Sol. (i) CH3ñCH2ñCH2ñCH2ñCHO (ii) (iii) (iv)

Page # 11

60. Identify the number of reagents which can distinguish between the following compounds ? I. AgNO3 + NH4OH II. HCl + ZnCl2 (anhydrous) III. Neutral FeCl3 IV. I2/NaOH V. 2, 4-DNP VI. aq. NaHCO3 VII. NH4OH + Cu2Cl2 VIII. Na-Metal

uhps fn;s x;s ;kSfxdksa ds e/; fuEu esa ls fdrus vfHkdeZdksa }kjk foHksnu fd;k tk ldrk gS \ I. AgNO3 + NH4OH II. HCl + ZnCl2 (futZy) III. mnklhu FeCl3 IV. I2/NaOH V. 2, 4-DNP VI. tyh; NaHCO3 VII. NH4OH + Cu2Cl2 VIII. Na-/kkrq

Ans. 05

2. ct-1_17 (paper-1) aug 2014xii

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