iit maths

-1

MATHEMATICS PAPER : CT-1 TARGET DATE : 10-04-2014 PAPER LEVEL : MODERATE TO TOUGH SYLLABUS : Fundamentals of Mathematics-I, Quadratic Equation

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. Let n be an integer greater than 1 and let an = n

1log 1001

. If b = a3 + a4 + a5 + a6 and [BALG]

c = a11 + a12 + a13 + a14 + a15. Then value of (b ñ c) is equal to

ekuk n, 1 ls cM+h iw.kkZad la[;k gS rFkk ekuk an = n

1log 1001

;fn b = a3 + a4 + a5 + a6 vkSj c = a11 + a12 + a13 +

a14 + a15 rc (b ñ c) dk eku cjkcj gS - (A) 1001 (B) 1002 (C) ñ 2 (D*) ñ1

Sol. an = log1001n b = log10013 ◊ 4 ◊ 5 ◊ 6 and c = log1001(11◊12◊13◊14◊15)

(b ñ c) = log1001 3 4 5 6

11 12 13 14 15

= log1001(1001)ñ1 = ñ1 2. The number of integral values of 'a' for which both roots of the equation x2 ñ 2x ñ a2 = 0 lie between the

roots of equation x2 ñ 2x + a2 ñ11a + 12 = 0, is [QELR] 'a' ds iw.kkZad ekuksa dh la[;k ftlds fy, lehdj.k x2 ñ 2x ñ a2 = 0 ds nksuksa ewy lehdj.k x2 ñ 2x + a2 ñ11a + 12 = 0 es

ewyksa ds e/; fLFkr gS - (A) 0 (B) 1 (C*) 2 (D) 3 Sol. Let be the root of x2 ñ 2x ñ a2 = 0 ekuk lehdj.k x2 ñ 2x ñ a2 = 0 ds ewy gS 2 ñ 2 = a2 and roots of x2 ñ 2x ñ a2 = 0 lie between roots of x2 ñ 2x + a2 ñ 11a + 12 = 0 rFkk x2 ñ 2x ñ a2 = 0 ds ewy lehdj.k x2 ñ 2x + a2 ñ 11a + 12 = 0 ds ewyks ds e/; fLFkr gS 2 ñ 2 + a2 ñ 11a + 12 < 0 2 ñ 2 = a2 a2 + a2 ñ 11a + 12 < 0 2a2

ñ 11a + 12 < 0 (2a ñ 3)(a ñ 4) < 0

a 3 , 42

number of integral values of 'a' is = 2 'a' ds iw.kkZad ekuksa dh la[;k = 2 gS

-2

3. If x = 2 + 3i, then value of x4 ñ x3 + 10x2 + 3x ñ 5 is equal to [BAGQ] ;fn x = 2 + 3i rc x4 ñ x3 + 10x2 + 3x ñ 5 dk eku cjkcj gS - (A) 127 (B) 122 (C) 120 (D*) ñ 122 Sol. x = 2 + 3i (x ñ 2)2 = ñ 9 x2 ñ 4x + 13 = 0 x4 ñ x3 + 10x2 + 3x ñ 5 = (x2 ñ 4x + 13)(x2 + 3x + 9) ñ 122 = ñ 122

4. The complete solution set of inequation x 2

2(e ñ 2)(x ñ 5x 4)

(x ñ 5x 6)

0, is [BAIR]

vlfedk x 2

2(e ñ 2)(x ñ 5x 4)

(x ñ 5x 6)

0 dk lEiw.kZ gy leqPp; gS&

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

Sol. x(e ñ 2)(x ñ 1)(x ñ 4)

(x ñ 2)(x ñ 3)

0

x (ñ,n 2] [1, 2) (3, 4]

5. Product of the roots of the equation 22

12 xx

ñ 9 1xx

+ 14 = 0, is

lehdj.k 22

12 xx

ñ 9 1xx

+ 14 = 0 ds ewyksa dk xq.kuQy gS -

(A) 5 (B) 2 (C) 10 (D*) 1 Sol. Given equation become nh xbZ lehdj.k 2x4 ñ 9x3 + 14x2

ñ 9x + 2 = 0 product of roots ewyksa dk xq.kuQy = 1

Alter

Let ekuk x + 1x

= t

22 t ñ 2 ñ 9t + 14 = 0 2t2 ñ 9t + 10 = 0

t = 2, 52

x + 1x

= 2 or x + 1x

= 52

x = 1, x = 1 or x = 2, 12

product of roots = 1 ewyksa dk xq.kuQy = 1 6. The number of integral values of x satisfying

2 2x x x x 14 3.2 4

is [QEMS]

2 2x x x x 14 3.2 4

dks larq"V djus okys x ds iw.kkZad ekuksa dh la[;k gS - (A) 5 (B) 3 (C*) 4 (D) 2 Sol.

2 2x x x x 14 3.2 4

2 2x x x x 14 ñ 3.2 ñ 4 0

2 22x x x 2x 22 ñ 3.2 ñ 2 0

2 2 22x x x x x 2x 22 ñ 4.2 2 ñ 2 0

2 2 2x x x x x x 22 (2 ñ 4.2 ) 2 (2 ñ 2 ) 0

-3

2 2x x x x(2 ñ 4.2 )(2 2 ) 0

2x x(2 ñ 4.2 ) 0

2x x2 4.2 x2 x + 2 x2 ñ x ñ 2 0 (x ñ 2)(x + 1) 0 x [ñ1, 2] 4 integers iw.kk±d 7. Complete set of real values of k for which kx ñ x2 + 9 ñ x < 0 x R is [QEGR] (A) (0, ) (B) k (ñ, ) (C) k (ñ, ñ1] (D*) no such real k exists k ds okLrfod ekuksa dk lEiw.kZ leqPp; ftlds fy, kx ñ x2 + 9 ñ x < 0 x R gS - (A) (0, ) (B) k (ñ, )

(C) k (ñ, ñ1] (D*) k dk dksbZ okLrfod eku fo|eku ugh Sol. Roots are real as product of roots negative ewy okLrfod gS rFkk ewyksa dk xq.ku_.kkRed gSA

Alter ñ x2 + x(k ñ 1) + 9 < 0

D < 0 (k ñ 1)2 ñ 4 ◊ (ñ1)(+9) < 0

(k ñ 1)2 + 36 < 0 No real k exists k dk dksbZ okLrfod eku fo|eku ugha

8. If the equation (a2 + a ñ 30)x2 + (b ñ 1)(a2 ñ2a ñ 15)x + (b2 ñ 6b + 5) = 0 has more than two different solutions for x, then number of possible ordered pairs (a, b) is [QEGR]

;fn lehdj.k (a2 + a ñ 30)x2 + (b ñ 1)(a2 ñ2a ñ 15)x + (b2 ñ 6b + 5) = 0 ds x ds fy, nks ls vf/kd gy j[krs gS rc (a, b) ds laHkkfor Øfer ;qXeksa dh la[;k gS -

(A) 1 (B) 2 (C*) 3 (D) 4 Sol. for more than two different solution, the equation must be an identity and for that nks ls vf/kd gy ds fy, lehdj.k ,d loZlfedk gksxh a2 + a ñ 30 = 0 and vkSj (b ñ 1)(a2 ñ 2a ñ 15) = 0 and vkSj (b2 ñ 6b + 5) = 0 a = ñ6, a = 5 and vkSj b = 1, a = 5, a = ñ3 and vkSj b = 1, b = 5 possible ordered pairs (a, b) are lefor Øfer ;qXe (a, b) gS (5, 1), (5, 5), (ñ6, 1) three possible ordered pairs (a, b) are there. rhu laHkkfor Øfer ;qXe (a, b) gS 9. If , , are the roots of the equation 2x3 ñ 7x2 + 3x ñ 1 = 0, then the value of (1 ñ 2)(1 ñ 2)(1 ñ 2) is ;fn lehdj.k 2x3 ñ 7x2 + 3x ñ 1 = 0 ds ewy , , gS rks (1 ñ 2)(1 ñ 2)(1 ñ 2) dk eku gS - [QETE]

(A) 39 (B) ñ 39 (C) 394

(D*) ñ 394

Sol. 2x3 ñ 7x2 + 3x ñ 1 = 0 b

g

2x3 ñ 7x2 + 3x ñ 1= 2(x ñ )(x ñ )(x ñ ) ÖÖ..(i)

(i) put x = 1 j[kus ij, we get 2(1 ñ )(1 ñ )(1 ñ ) = ñ 3 ÖÖÖ.(ii) (ii) put x = ñ 1 in (1) esa j[kus ij, we get 2(ñ1 ñ )(ñ1 ñ )(ñ1 ñ ) = ñ13 2(1 + )(1 + )(1 + ) = 13 ÖÖÖ(iii) multiply (2) and (3), we get 4(1 ñ 2)(1 ñ 2)(1 ñ 2) = ñ39

10. The complete solution set of the inequation 1/ 2

1log | x |

> 1 is [BAMS]

-4

vlfedk 1/ 2

1log | x |

> 1 dk lEiw.kZ gy leqPp; gS -

(A) 1ñ2, ñ

2

1 , 12

(1, 2) (B) 1ñ2, ñ

2

1 , 22

(C*) (ñ2, ñ 1) 1ñ1, ñ

2

1 , 12

(1, 2) (D)

Sol. 1/ 2

1log | x |

> 1

1/ 2log | x | < 1 but ijUrq log1/2|x| 0

ñ1 < log1/2|x| < 1 but ijUrq |x| 1 x 1, ñ 1

2 > |x| > 12

12

< |x| < 2

x (ñ2, ñ 1) 1ñ1, ñ

2

1 , 12

(1, 2)

MCQ

11. If range of expression 2x ñ 12

2x ñ 7

(x R) is (ñ, a] [b, ) and let the solution of the equation

a alog b log x 3a.x b b is x = c, then [QEGR] (A) (a + b) and c are both prime (B*) (a + b) and c are coprime number (C*) (a + b + c) is a perfect square (D) a + b = c

;fn O;atd 2x ñ12

2x ñ 7

(x R) dk ifjlj (ñ, a] [b, ) gS rFkk lehdj.k a alog b log x 3a.x b b dk gy x = c gS rc -

(A) (a + b) vkSj c nksuksa vHkkT; gSA (B*) (a + b) rFkk c lgvHkkT; la[;k gSA (C*) (a + b + c) ,d iw.kZ oxZ gSA (D) a + b = c

Sol. Let ekuk y = 2x ñ 12

2x ñ 7

x2 ñ 2xy + 7y ñ 12 = 0

D 0 4y2 ñ 4 ◊ 1 ◊ (7y ñ 12) 0 y2 ñ 7y + 12 0 y (ñ, 3] [4, ) a = 3 and vkSj b = 4 equation becomes 3 3log 4 log x3.x 4 64 lehdj.k 3 3log 4 log x3.x 4 64 ls 3 3log x log x3(4 ) 4 64

3log x4 16 3log x = 2 x = 9 = c

12. Let a and b are the solutions of the equation 4 1/ 41 log x ñ1 log x265 55

such that a > b, then the value of ab

is

(A*) an even number (B*) a rational number [BALG] (C*) a composite number (D) a prime number

ekuk a vkSj b lehdj.k 4 1/ 41 log x ñ1 log x265 55

ds gy bl izdkj gS fd a > b rc ab

dk eku gS ñ

-5

(A*) le la[;k (B*) ifjes; la[;k (C*) la;qDr la[;k (D) vHkkT; la[;k

Sol. 4 4log x ñ log xñ1265.5 5 .55

Let ekuk 4log x5 = t

5t + 15t

= 265

25t2 ñ 26t + 1 = 0

t = 1 or t = 125

log4x = 0 or ;k log4x = ñ 2

x = 1 or ;k x = 116

ab

= 16

13. If ax2 + bx + c = 0 has imaginary roots and a,b,c R, then which of the following options are CORRECT? ;fn ax2 + bx + c = 0 ds dkYifud ewy gS rFkk a,b,c R, rc fuEu esa ls dkSuls fodYi lgh gS ? [QEGR] (A) ax2 + bx + c > 0 x R (B*) a(a ñ b + c) > 0 (C*) a(ax2 + bx + c) > 0 x R (D*) a2 + c2 + 2ac > b2 Sol. D < 0 (ax2 + bx + c) will be of same sign as that of 'a' (ax2 + bx + c) 'a' ds leku fpUg dk gksxk C is correct lgh gS (B) Let ekuk f(x) = ax2 + bx + c D < 0 Two graphs are possible nks vkj[ks laHko gS

ñ1

a > 0 and rFkk f(ñ1) > 0 a ñ b + c > 0 a(a ñ b + c) > 0

ñ1

a < 0 and rFkk f(ñ1) < 0 a ñ b + c < 0 a(a ñ b + c) > 0 (D) a2 + c2 + 2ac ñ b2 > 0 (a + c)2 ñ b2 > 0 (a ñ b + c)(a + b + c) > 0 f(ñ1) f(1) > 0 true lR; see the above two graphs nks vkjs[k mij fn[kk;s vuqlkj

14. The value of 2 2nb

a b( na)( nb)

6a log b log ae

is [BALG]

(A*) independent of a (B*) independent of b (C) dependent on a (D) dependent of b

2 2nb

a b( na)( nb)

6a log b log ae

dk eku gS -

(A*) a ls Lora=k (B*) b ls Lora=k (C) a ij fuHkZj (D) b ij fuHkZj

-6

Sol. 2 2nb

a bna nb

6a log b log ae

= nb

a b

nb

1 16.a . log b. log a2 2

a

= 64

= 32

15. Identify which of the following statement(s) are 'CORRECT' ? [QEGR]

(A*) For ax2 + bx + c = 0, (a 0) if 4a + 2b + c = 0, then roots are 2 and c2a

.

(B) If is repeated root of ax2 + bx + c = 0, (a 0) then ax2 + bx + c = (x ñ )2. (C*) For ax2 + bx + c = 0, (a 0, a,b,c Q) imaginary roots occur in conjugate pair only. (D*) If f(x) = ax2 + bx + c, (a 0) has finite maximum value and both roots of f(x) = 0 are of opposite sign,

then f(0) > 0. fuEu esa ls dkSulk dFku lgh gS ?

(A*) ax2 + bx + c = 0, (a 0) ds fy, ;fn 4a + 2b + c = 0 rc ewy 2 vkSj c2a

gSA

(B) ;fn , ax2 + bx + c = 0, (a 0) dk iqujkofÙk ewy gS] rc ax2 + bx + c = (x ñ )2. (C*) ax2 + bx + c = 0, (a 0, a,b,c Q) ds fy, dkYifud ewy dsoy la;qXeh gksrs gSA (D*) ;fn f(x) = ax2 + bx + c, (a 0) fu;r vf/kdre eku j[krk gS rFkk f(x) = 0 ds nksuksa ewy foijhr fpUg ds fpUg gS]

rc f(0) > 0. Sol. (A) 4a + 2b + c = 0 ax2 + bx + c = 0 has one root as 2

other root will be c2a

(B) ax2 + bx + c = a(x ñ )2 (C) coefficients are rational coefficients are real. imaginary roots occur in conjugate pair

(D)

f(0) > 0

Integer Type 16. If x R then absolute difference between the maximum and minimum values of the expression

2

2

x 14x 9x 2x 3

is [QEGR]

;fn x R rc O;atd 2

2

x 14x 9x 2x 3

ds vf/kdre vkSj U;wure ekuksa dk fujis{k vUrj gS&

Ans. 09 Sol. Let ekuk y =

2

2

x 14x 9x 2x 3

x2(y ñ 1)+ 2x(y ñ 7) + 3(y ñ 3) = 0 ÖÖÖÖ.(i) Case-1 : If y = 1, then equation (i) becomes

ñ12x ñ 6 = 0 x = 12

which is real tks fd okLrfod gSA

y = 1 is possible laHko gSA Case-2 : If y 1, then D 0 4(y ñ 7)2 ñ 4(y ñ 1).3(y ñ 3) 0 y2 ñ 14y + 49 ñ 3(y2 ñ 4y + 3) 0 ñ2y2 ñ 2y + 40 0 y2 + y ñ 20 0 ñ5 y 4 Absolute difference = 9 fujis{k vUrj = 9

-7

17. If a,b R, then the smallest natural number 'b' for which the equation x2 + 2(a + b)x + (a ñ b + 8) = 0 has unequal real roots for all a R, is [QENR]

;fn a,b R, rc U;wure izkdr la[;k 'b' ftlds fy, lehdj.k x2 + 2(a + b)x + (a ñ b + 8) = 0 ds lHkh a R ds fy, vleku okLrfod ewy gS&

Ans. 05 Sol. D = 4(a + b)2 ñ 4 ◊ 1 ◊ (a ñ b + 8) = 4[a2 + 2ab + b2 ñ a + b ñ 8] = 4[a2 + a(2b ñ 1) + (b2 + bñ 8)] for unequal real root for all a R lHkh a R ds fy, lHkh vleku okLrfod ewy gS D > 0 a R (2b ñ 1)2 ñ 4 ◊ 1 ◊ (b

2 + b ñ 8) < 0 4b2 ñ 4b + 1ñ 4b2 ñ 4b + 32 < 0 ñ 8b + 33 < 0 8b > 33

b > 338

smallest natural value of b is = 5 b dk U;wure izkdr eku = 5

18. Let the product of all the solutions of the equation 33 33 x 3(log 3x log 3x )log x = 2 be k, then find the

value of 18k. [BALG]

ekuk lehdj.k 33 33 x 3(log 3x log 3x )log x = 2 ds lHkh gyksa dk xq.kuQy k gS rc 18k dk eku Kkr dhft,

Ans. 02 Sol. squaring we get oxZ djus ij

1/3

1/ 3 33

3

log (3x)log (3x)

log x

3log3x = 4

Let ekuk log3x = t

13

(t 1)(t 1)t

.3t = 4

t + 1 = ±2

x = 3 or x = 127

product of roots xq.kuQy ds ewy = 19

= k

18k = 2 19. If a, b and c are real numbers such that a2 + 2b = 7, b2 + 4c = ñ7 and c2 + 6a = ñ14, then find the value of

2 2 2a b c2

. [BAMS]

;fn a, b vkSj c okLrfod la[;k bl izdkj gS fd a2 + 2b = 7, b2 + 4c = ñ7 vkSj c2 + 6a = ñ14 gS] rks 2 2 2a b c

2

dk eku Kkr dhft,A Ans. 07 Sol. a2 + b2 + c2 + 6a + 4c + 2b = ñ 14 (a + 3)2 + (b + 1)2 + (c + 2)2 = 0 a = ñ 3, b = ñ 1 and vkSj c = ñ 2

so blfy, 2 2 2a b c

2 = 9 1 4

2 = 7

-8

20. Find the number of integral values of 'm' less than 50, so that the roots of the quadratic equation mx2 + (2m ñ 1)x + (m ñ 2) = 0 are rational. [BAMS]

50 ls NksVs m ds iw.kk±d ekuksa dh la[;k Kkr dhft, tcfd f}?kkr lehdj.k mx2 + (2m ñ 1)x + (m ñ 2) = 0 ds ewy ifjes; gSA

Ans.06 Sol. D = 4m + 1 now for roots to be rational must be perfect square of a rational number but since m is an integer it will be perfect square of an integer. so let blfy, ekuk 4m + 1 = k2, k

m = 2k ñ 1

4 for m to be an integer k must be odd

k = ±1, ±3, ±5, ±7, ±9, ±11, ±13 but ijUrq m 0 There are 6 possible integral values of m.

JEE ñ ADVANCED (CT ñ 1) Date : - 10/8/2014

Test Syllabus : Mathematical Tools, Rectilinear Motion, Projectile Motion & Relative Motion complete


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. If the tangent on the curve 2y kx (where k is a constant) at x = 1makes an angle 45o with +x-axis,

then the value of k is ;fn oØ 2y kx (;gkW k ,d fu;rkad gS) ij x = 1 ij [khaph xbZ Li'kZ js[kk +x v{k ls 45∫ dks.k cukrh gks rks k dk

eku gksxkA

(A*) 12

(B) 14

(C) 2 (D) 4

Soln. (A) 2 45ody k x tandx

At x = 1 ij 2k(1) = 1 12

k

22. It is given that A R B and R A . Then the angle between A & B is

A R B rFkk R A fn;k gqvk gS] rks A & B ds e/; dks.k gksxkA

(A) 1 AcosR

(B) 1 BcosA

(C*) 1 RtanA

(D) 1 RsinA

Soln. (C)

B

aÆA

ÆR

tan = RA

= tanñ1 RA

23. If there are two vectors A and B such that 2à à àA B i j k and à àA B (i k) , then the angle between

A and B is .

;fn nks lfn'k A rFkk B bl izdkj gS fd 2à à àA B i j k rFkk à àA B ( i k) gS, rks A rFkk B ds e/; dks.k

gksxk

(A) 30o (B) 45o (C*) 60o (D) 120o

Soln. (C) After solving gy djus ij :-

à à2A = 2i +2j

à à2B = 2j+2k

à àA i j , à àB j k

A.BcosAB

12

cos 60o

24. A particle moves in a straight line. Its speed (v) increases linearly with time. If the initial speed of the

particle is vo and its speed at t = 4 sec is 4vo, then 4

0vdt is equal to

,d d.k ljy js[kk ij xfr'khy gSA bldh pky (v) le; ds lkFk js[kh; :i ls c<+rh gSA ;fn d.k dh izkjfEHkd

pky vo rFkk t = 4 lSd.M ij pky 4vo gks rks 4

0vdt dk eku gksxkA

(A) 4 ov (B) 6 ov (C) 8 ov (D*) 10 ov

Soln. (D) acceleration Roj.k = a

4v0 = v0 + a (4)

a = 4

03v

4 4

0 00 0

3vdt v v t dt

4 2

0

314 42 4

oo

VVdt S V ( ) ( )

= 10 ov

25. A ball is thrown vertically upwards. The velocity at one fourth of the maximum height is 10 3 m/s. then the velocity with which the ball was thrown is .

,d xsan Å/okZ/kj Åij dh rjQ iz{ksfir dh tkrh gSA vf/kdre~ ÅWpkbZ dh ,d pkSFkkbZ ÅWpkbZ ij bldk osx 10 3 m/s gks rks xsan dk iz{ksi.k osx gksxkA

(A) 5 m/s (B) 10 m/s (C) 15 m/s (D*) 20 m/s

Soln. (D) Let u be velocity of throwing and h be maximum height.

ekuk u iz{ksi.k osx rFkk h vf/kdre~ ÅWpkbZ gSA

Then rc 2uh

2g

2 2V u 2as

u

h 10 3 m/s

h4

2 2 h(10 3 ) u 2( g)

4

22 u300 u 2g.

2g(4)

2u 400 u 20m/s .

26. A particle is moving on a curve given by y = 2 sin2x( x& y are in meters). The x-component of velocity is always 2 m/s. If the particle starts from origin at t = 0, then displacement (in meters) of the particle from

t = o to 2

t sec is

,d d.k oØ y = 2 sin2x( x rFkk y ehVj esa gSA) ij xfr'khy gSA osx dk x ?kVd ges'kk 2 m/s jgrk gSA ;fn t = 0

ij d.k ewy fcUnq ls xfr izkjEHk djrk gks rks t = o ls 2

t lSd.M rd d.k dk foLFkkiu (ehVj esa) gksxkA

(A*) m (B) 2 m (C) 0 m (D) 2

m

Soln. (A) 22

x V x t

2 2 0y sin( )

Displacement foLFkkiu =

27. A particle is moving along a straight line with uniform acceleration 25 m / s and initial velocity 12.5 m/s. then distance travelled by it in 3rd second of motion is

,d d.k lh/kh js[kk ij le:i Roj.k 25 m / s rFkk izkjfEHkd osx 12.5 m/s ds }kjk xfr'khy gSA 3rd (rhljs) lSd.M esa xfr ds nkSjku blds }kjk r; dh xbZ nwjh Kkr djksA

(A) 1 m (B) 0 (C*) 1.25 m (D) 2 m

Soln. (C) V = u + at = 0 12 5 5 0. t

t = 2.5 sec

V = 0

t = 2 sec 2.5 m/s

t = 3 sec 2 5t . sec (i-e particle turns back) 2 5t . sec (vFkkZr~ d.k okil ykSVsxkA) Distance travelled in 3rd sec

3rd lSd.M esa r; nwjh

2.5 10 12

2 2 = 1.25 m

28. A particle is projected with speed u m/s from origin O along y-axis as shown in figure. If the acceleration of the particle is à à(a i aj) where a is constant, then the equation of trajectory of the particle is

fp=kkuqlkj ,d d.k dks ewy fcUnq O ls y v{k ds vuqfn'k pky u m/s ds }kjk iz{ksfir fd;k tkrk gSA ;fn d.k dk

Roj.k à à(a i aj) (tgkW a fu;r gS) gks rks d.k ds iFk dk lehdj.k gksxkA

X

Y

u

(A) 2

212

axyu

(B*) 2

2 2xu(y x)

a (C)

22 2yu

(x y)a

(D) 2

2 2 2xux y

a

Soln. (B) 212

x at Ö Ö Ö .. (i)

212

y ut at Ö Ö Ö .. (ii)

y xt

u, putting in equation (i)

y xt

u, lehdj.k (i) esa j[kus ij

2

2 2xu(y x)a

29. Two particles are projected simultaneously with the same speed u m/s in the same vertical plane with angles of projection 30∫ and 60∫ . At what time (after start) their velocities will become parallel ?

nks d.kksa dks leku Å/okZ/kj ry esa ,d lkFk leku pky u m/s }kjk Øe'k% 30∫ rFkk 60∫ ds iz{ksi.k dks.k ij iz{ksfir fd;k tkrk gSA iz{ksi.k ds fdrus le; i'pkr~ buds osx ,d&nwljs ds lekUrj gks tk;sxsa ?

(A*) 3 1

u

g sec (B)

3 1

u

g sec (C)

3u

g sec (D) 3

2u

g sec

Soln. (A)

30 60

30 60

o o

o ou sin gt u sin gt

u cos u cos

3 1

utg

30. A particle is thrown vertically upwards with initial speed u m/s (with respect to lift) inside a lift moving downwards with constant velocity. Its time of flight is T sec. Again the particle is thrown vertically upwards with same initial speed u m/s. (with respect to lift) inside a lift moving upwards with constant acceleration g/2, then new time of flight is

uhps dh rjQ fu;r osx ls xfr'khy fy¶V ds vUnj ,d d.k dks fy¶V ds lkis{k Å/okZ/kj Åij dh rjQ izkjfEHkd

pky u m/s ls iz{ksfir fd;k tkrk gSA bl le; bldk mMM~;u dky T lSd.M gSA vc nwckjk Åij dh rjQ fu;r Roj.k g/2 ls xfr'khy fy¶V ds vUnj d.k dks fy¶V ds lkis{k Å/okZ/kj Åij dh rjQ leku izkjfEHkd pky u m/s ls iz{ksfir fd;k tk;s rks bl d.k dk u;k mM~M;u dky gksxkA

(A) 2T

sec (B) 2T sec (C) 3

2T sec (D*)

23T

sec

Soln. (D) 2u

Tg

2

2' uT

g g /2 23

ug

23T

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. A particle is projected at an angle with the horizontal from a point O on a plane which is inclined at an angle to the horizontal. The particle is moving horizontally when it strikes the plane at a point A.

,d d.k dks fcUnq O ls {kSfrt ls dks.k ij urry ij iz{ksfir fd;k tkrk gSA urry dk {kSfrt ls urdks.k gSA ;fn urry dh fcUnq A ij d.k ds Vdjkrs le; ;g {kSfrt fn'kk esa xfr'khy gks rks

Fixed

A

BO

u

(A) Time of flight 2usinTg

sec

mMM~;u dky 2usinT

g lSd.M gSA

(B*) Maximum height of the particle 2 2

2u sin

g m

d.k dh vf/kdre~~ ÅWpkbZ 2 2

2u sin

g ehVj gSA

(C*) Horizontal range OB = 2u sin cos

gm

{kSfrt ijkl OB = 2u sin cos

gehVj gSA

(D*) 1 2tan ( tan )

Soln. (B, C, D)

usinTg

2u sin ( )gcos

tan = 2 tan , 1 2tan ( tan )

2 2

2maxu sinH

g

OB (u cos ) (T) = 2u sin cos

g

32. The vectors A is given by 2à à àA t i (sin t) j t k where t is time. Then which of the following is (are) correct ?

lfn'k A fuEu }kjk fn;k tkrk gSA

2à à àA ti (sin t) j t k tgkW t le; gS] rks fuEu ls dkSuls fodYi lgh gS ?

(A*) 1 2A (at t ) (B*) 1 2dA à à à(at t ) i j kdt

(C*) 21 2 1dAA (at t )dt

(D*) 1 3dAA (at t )dt

(A) 1 2A (t )ij (B) 1 2

dA à à à(t ) i j kdt

ij

(C) 21 2 1dAA (t )dt

ij (D) 1 3dAA (t )dt

ij

Soln. (A, B, C, D)

2à à àA t i sin t j t k

1 à à àA(t ) i sin t j k

2A

2dA à à ài cos t j t kdt

1 2dA à à à(t ) ( i j k)dt

2dA à à à à àA (i k) ( i j ) k)dt

à à à( i j k)

22 1dAAdt

1 2 3dAA.dt

33. A car is travelling north along a straight road at 50 km hr ñ1 . An instrument in the car indicates that the

wind is directed towards east. If carís speed is 80 km hrñ1, then instrument indicates that the wind is directed towards south ñ east. Then angle made by windís direction is given by. [RL - TD]

,d dkj lh/kh lM+d ij mÙkj fn'kk ds vuqfn'k 50 km hr ñ1 dh pky ls xfr'khy gSA dkj esa yxk gqvk ;a=k iznf'kZr

djrk gS fd gok iwoZ fn'kk dh rjQ izokfgr gSA ;fn dkj dh pky 80 km hrñ1 gks rks ;a=k gok dh fn'kk nf{k.k&iwoZ

dh rjQ iznf'kZr djrk gS rks gok dh fn'kk }kjk cuk;k x;k dks.k gksxkA

(A) 1 3tan Nof E5

(B*) 1 5tan Nof E3

(C) 1 1tan Nof W2

(D) = tanñ1(5) N of E

(A) 1 3tan5

iwoZ ls mÙkj dh rjQ (B*) 1 5tan3

iwoZ ls mÙkj dh rjQ

(C) 1 1tan2

if'pe ls mÙkj dh rjQ (D) = tanñ1(5) iwoZ ls mÙkj dh rjQ

Sol. (B)

WGV = x yà àV i V j

1WCV is towards east

1WCV iwoZ dh rjQ

Vx

Vy

50 Vy = 50 m/s ; Vx > 0

2WCV is towards south-east

2WCV nf{k.k iwoZ dh rjQ

Vx

V y

80

45∞

Vx = 30 m/s

50

30

1 5tan Nof E3

34. A particle moving in a straight line with constant acceleration has speeds 7 m/s and 17 m/s at A

and B points respectively. If M is the mid-point of AB, then which of the following is (are) correct?

ljy js[kk ij fu;r Roj.k ls xfr'khy d.k dh fcUnq A rFkk B ij pky Øe'k% 7 m/s rFkk 17 m/s gSA ;fn AB js[kk dk e/; fcUnq M gks rks fuEu es ls dkSulk@dkSuls fodYi lgh gSA

(A*) the speed of particle at M is 13 m/s

M ij d.k dh pky 13 m/s gSA

(B*) the average speed between A and M is 10 m/s

A rFkk M ds e/; vkSlr pky 10 m/s gSA

(C) the average speed between M and B is 14 m/s

M rFkk B ds e/; vkSlr pky 14 m/s gSA

(D*) the ratio of the time to go from A to M and that from M to B is 3 : 2.

A ls M rd rFkk M ls B rd tkus esa fy;s x;s le; dk vuqikr 3 : 2 gSA.

Soln: (A,B,D)

M

x xA

V=7 V=17B

ax27V 22M .......(i)

ax2V17 2M

2 .......(ii)

s/m13VM

s/m102137AMVavg

s/m152

1713MBVavg

1ta713

2at1317 2:3t:t 21

35. A man swims at a speed hr/kmjà6ià3V1 relative to water. If the water flows with a speed

hr/kmià5V2 . If the width of the river is jàm500d .Then (A*) path of man is straight line. (B*) time of crossing the river is 5 min.

(C) velocity of man is 10 m/s. (D) drift of man in the direction of flow is 600 m.

,d O;fDr ikuh ds lkis{k pky hr/kmjà6ià3V1 ls rSj ldrk gSA ;fn ikuh hr/kmià5V2 pky ls izokfgr

gSA ;fn unh dh pkSM+kbZ jàm500d gks rks (A*) O;fDr dk iFk ljy js[kh; gSA (B*) unh dks ikj djus esa yxk le; 5 min gSA

(C) O;fDr dk osx 10 m/s gSA (D) izokg dh fn'kk esa O;fDr dk foLFkkiu 600 m gSA

Soln: (A, B)

Resultant path of man is straight line.

O;fDr dk ifj.kkeh iFk ljy js[kh; gSA

hr/kmjà6ià8VVV RMRM

s/m925hr/km10VM

min51000

606

500t

Drift fopyu m3

200010006058 .

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_(5)_(Double Digit)

36. A ball is projected with speed 10 m/s from ground at angle 300 with the vertical. After some time it again fall on the ground, then the magnitude of average velocity of the ball in this interval (in m/s) is

,d xsan dks tehu ls 10 m/s dh pky ls Å/okZ/kj ls 30∞ dks.k ij iz{ksfir fd;k tkrk gSA dqN le; i'pkr~ ;g okil tehu ij fxjrh gS] rks bl le;kUrjky esa xsan dk vkSlr osx dk ifjek.k (m/s esa) Kkr dhft,A

Ans. 05 Soln : (5)

oavgV ucos 10cos60 5m / s

37. If c,b,a are three vectors having magnitudes 1, 2, 3 respectively such that a b c 0 then value of

a.cc.bb.a is :

;fn c,b,a rhu lfn'kksa dk ifjek.k Øe'k% 1, 2, 3 bl izdkj gS fd a b c 0 gks rks a.cc.bb.a dk eku

Kkr dhft, : Ans. 07 Soln : (7)

0cba

0a.cc.bb.a2cba 222

2

cbaa.cc.bb.a222

72

9412

cbaa.cc.bb.a222

38. A particle moves along the curve x2y2 where 2tx2

(x & y are in meters), (t is time in sec). Then

the magnitude of the acceleration of the particle at sec2t (in m/s2) is

,d d.k oØ x2y2 ds vuqfn'k xfr'khy gS ;gk¡ 2tx2

gSA (x o y ehVj esa gS), (t le; lsd.M esa gS) A rks le;

sec2t ij d.k ds Roj.k dk ifjek.k (m/s2 esa) Kkr dhft,A Ans. 01

Soln: (1) 2tx2

1atdtdxV xx

0a1Vtytyx2y yy222

2xa a 1 m /s

39. The velocity of a particle is given by V = (2 + 3x) m/s(x is position in meters). Then the acceleration of

the particle at m32x is (in m/s2).

,d d.k dk osx V = (2 + 3x) m/s (fLFkfr x ehVj esa gS) }kjk fn;k tkrk gSA rks m32x ij d.k dk Roj.k (m/s2

esa) Kkr dhft,A Ans. 12 Soln : (12)

x96dxdvVa

2s/m1232xata

40. The velocity vector of a particle moving in xy plane is given by jàxiàtV where t is time and x is position. If initially the particle was at origin, and equation of trajectory (path) of the particle is

23 byax , then the value of (a + b) is

xy ry esa xfr'khy d.k dk osx lfn'k jàxiàtV }kjk fn;k tkrk gS ;gk¡ t le; rFkk x fLFkfr gSA ;fn izkjEHk

esa d.k ewy fcUnq ij gks rFkk d.k ds iFk dh lehdj.k 23 byax }kjk nh tk;s rks (a + b) dk eku Kkr dhft,A Ans. 11 Soln: (11)

At t = 0, x = 0, y = 0 ij

jàxiàtV

2txt

dtdx 2

2t

xdtdy 2

6ty3

Eliminating t dks izfrLFkkfir djus ij

23 y9x2 (i-e. equation of path iFk dh lehdj.k)

a = 2, b = 9 11ba

Page #

1

Course : (ELPD ) Test Date : 10.08.2014 Test Type : CT-1 Paper-1 Time Duration : 3 Hrs. Paper-2 Time Duration : 3 Hrs.

Paper Level - Moderate to Tough

SYLLABUS :

Introduction to chemistry , Atomic structure ( Upto Heisenberg uncertainity principle)

SYLLABUS SCHEDULED SYLLABUS

SCHEDULED WEIGHTAGE

(BY FC)

SR. NO.

TOPIC NAME

(I) (II)

WEIGHTAGE IN PAPER-1

(BY FACULTY)


(BY FACULTY)

1. Introduction to chemistry

30%

30%

2. Atomic structure ( Upto Heisenberg uncertainity principle)

70%

70%

Organic chemistry

SYLLABUS : IUPAC nomenclature and Structural isomerism complete.

SYLLABUS SCHEDULED

SYLLABUS SCHEDULED WEIGHTAGE

(BY FC)

SR. NO.

TOPIC NAME

(I) (II)


(BY FACULTY)


(BY FACULTY)

1. IUPAC nomenclature and Structural isomerism complete

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

S .N o . S u b je c t N a tu re o f Q u e stio n s N o . o f Q u e stio n s M a rk s N e g a tive T o ta l

1 to 7 M C Q 7 4 0 28

8 to 13 C o m pre he n s io n (3 C o m p. x 2 Q . ) 6 3 ñ1 18

14 to 16 M TC 3 8 0 24

17 to 23 M C Q 7 4 0 28


30 to 32 M TC 3 8 0 24

33 to 39 M C Q 7 4 0 28


46 to 48 M TC 3 8 0 24

48 2 10

P a p e r-2 C T -1

T o ta l T o ta l

M a th s

P h ysics

C h e m istry

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 : HM SIR Test Name : JB&JB* (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) Atomic structure Bohr model M

42 SCQ (P) Atomic structure Photo electric effect M

43 SCQ (P) Atomic structure Bohr model E

44 SCQ (P) Mole concept M

45 SCQ (P) Atomic structure de-Broglie T

46 SCQ (P) Mole concept M 47 SCQ (O) IUPAC M 48 SCQ (O) IUPAC M

49 SCQ (O) IUPAC M

50 SCQ (O) Structural isomerism M

51 MCQ (P) Mole concept E 52 MCQ (P) Atomic structure Bohar model M

53 MCQ (P) Atomic structure H-spectrum de-Broglie T

54 MCQ (O) IUPAC M

55 MCQ (O) IUPAC M

56 Double Integer Type

(P) Atomic structure Quantum theory M


(P) Atomic structure Photo electric effect E


(P) Mole concept Avg. atomic mass M


(O) IUPAC M

60. Double Integer Type

(O) Isomerism counting M

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

Page #

4

Physical paper-1

SCQ(6) 41. The ratio of area of 4th circular orbit of He+

ion to that of 3rd orbit of Li2+ ion is : (ATS(P)) He+ vk;u dh 4th o Li2+ vk;u dh 3rd o Ùkkdkj d{kk ds {k s=kQy dk vuqikr fuEu gS& (A*) 64 : 9 (B) 4 : 3 (C) 8 : 3 (D) 3 : 4

Sol. 4 2

A B4 2B A

n Zn Z

= 4 2

4 24 33 2

= 649

2

2 narea r , rZ

22 nr , r

Z{ks=kQy

42. Light of wavelength 2 falls on a metal having work function hc/ 0. Photoelectric effect will take

place only if: (ATS(P)) dk;Z Qyu hc/ 0 ;qDr ,d /kkrq ij 2 rjax}S/;Z dk i zdk'k vkifrr gk srk gSA izdk'k fo|qr

i zH kko dsoy rHkh mRiUu gk sxk tc% (A) 0 (B) 2 0 (C*) 2 0 (D) 0 / 2 Sol. 43. An electron in a hydrogen like species jumps from an energy level to another energy level in such

a way that its kinetic energy changes from 'a' to a4

. The change in total energy of electron will be :

,d gkbMªkstu leku Lih'kht esa ,d bysDVªkWu ,d ÅtkZ Lrj ls vU; ÅtkZ Lrj esa bl izdkj LFkkukUrfjr

gksrk gS] fd bldh xfrt ÅtkZ 'a' ls a4 rd ifjofrZr gksrh gS] rks bldh dqy ÅtkZ esa ifjorZu gksxk :

(ATS(P))

(A*) + 34

a (B) ñ 38

a (C) + 32

a (D) ñ 34

a

Sol. Change in total energy = añ ñ ( a)4

dqy ÅtkZ esa ifjorZu = añ ñ ( a)4

= 3 a4

44. LPG contains n-butane and isobutane. Mass of carbon in 14.5 Kg of LPG. LPG n-C;wVsu o vkblksC;wVsu j[krk gSA LPG ds 14.5 Kg esa dkcZu dk nzO;eku fuEu gS & (Mole-1(P)) (A) 3 Kg (B*) 12 Kg (C) 14.5 Kg (D) 10 Kg

Sol. mass of Carbon dkcZu dk nzO;eku = 14.5 4858

= 12 Kg 45. Calculate ratio of de-Broglie wavelength of O2 molecule to He atom if ratio of their kinetic energy

is 1 : 18. O2 v.kq o He ijek.kq dh Mh&czksXyh rjax}S/;Z dk vuqikr ifjdfyr dhft, ;fn budh xfrt ÅtkZ dk

vuqikr 1 : 18 gksa&

(A) 1:23 (B*) 3 : 2 (C) 2 : 3 (D) 23:1

Page #

5

Sol. 2O = 1KE322

h & He =

2KE42h

He

O2=

1

2

KE32KE4

= 132184

= 23 .

46. Which of the following has maximum volume at STP - (Mole-1(P)) STP ij fuEu esa ls fdldk vk;ru vf/kdre gksrk gS& (A) 2 gm molecules of CH4 (B) 28 gm of CO (C*) 5 gm H2 (D) 200 gm of H2O STP ij fuEu esa ls fdldk vk;ru vf/kdre gksrk gS& (A) 2 gm CH4 v.kq (B) 28 gm CO v.kq (C*) 5 gm H2 v.kq (D) 200 gm H2O v.kq Sol. (A) volume of CH4 = 2 22.4 = 44.8 liter. (B) volume of CO = 1 22.4 = 22.4 liter. (C) volume of H2 = 2.5 22.4 = 56 liter. (D) volume of H2O = 200 ml. Sol. (A) CH4 dk vk;ru = 2 22.4 = 44.8 yhVj. (B) CO dk vk;ru 1 22.4 = 22.4 yhVj. (C) H2 dk vk;ru = 2.5 22.4 = 56 yhVj. (D) H2O dk vk;ru = 200 ml. SCQ(4)

47. IUPAC name of compound

Cl

Br is (IUPAC(O))

(A) 7-Bromo-3-chloro-5-(1,1-dimethylethyl)-3-ethyl-7-methyl-5-(2-methylpropyl)nonane

(B*) 3-Bromo-7-chloro-5-(1,1-dimethylethyl)-7-ethyl-3-methyl-5-(2-methylpropyl)nonane

(C) 3-Bromo-7-chloro-7-ethyl-3-methyl-5-(1,1-dimethylethyl)-5-(2-methylpropyl)nonane

(D) 3-Bromo-5-(1,1-dimethylethyl)-5-(2-methylethyl)-7-chloro-7-ethyl-3-methylnonane

Cl

Br ;kSfxd dk IUPAC uke gS

(A) 7-czkseks-3-Dyksjks-5-(1,1-MkbZesfFky,fFky)-3-,fFky-7-esfFky-5-(2-esfFkyizksfiy) uksusu

(B*) 3-czkseks-7-Dyksjks-5-(1,1-MkbZesfFky,fFky)-7-,fFky-3-esfFky-5-(2-esfFkyizksfiy) uksusu

(C) 3-czkseks-7-Dyksjks-7-,fFky-3-esfFky-5-(1,1-MkbesfFky,fFky)-5-(2-esfFkyizksfiy)uksusu

(D) 3-czkseks-5-(1,1-MkbesfFky,fFky)-5-(2-esfFky,fFky)-7-Dyksjks-7-,fFky-3-esfFkyuksusu

Page #

6

Sol.

3-Bromo-7-chloro-5-(1,1-dimethylethyl)-7-ethyl-3-methyl-5-(2-methylpropyl)nonane 3-czkseks-7-Dyksjks-5-(1,1-MkbZesfFky,fFky)-7-,fFky-3-esfFky-5-(2-esfFkyizksfiy) uksusu

48. The IUPAC name of

HOñCñCOOH |

|

CH2ñCOOH

CH2ñCOOH

is (IUPAC(O))

;kSfxd

HOñCñCOOH |

|

CH2ñCOOH

CH2ñCOOH

dk lgh IUPAC uke gS %&

(A) 3-Carboxy-3-hydroxypentanedicarboxylic acid (B) 2-Hydroxypropane-1,2,3-trioic acid (C) 3-Hydroxypropane-1,2,3-tricarboxylic acid (D*) 2-Hydroxypropane-1,2,3-tricarboxylic acid (A) 3-dkcksZDlh -3-gkbMªksDlhisUVsuMkbdkcksZfDlfyd vEy (B) 2-gkbMªksDlhizksisu -1,2,3-VªkbZvksbd vEy (C) 3- gkbMªksDlhizksisu -1,2,3- VªkbZdkcksZfDlfyd vEy (D*) 2- gkbMªksDlhizksisu -1,2,3- VªkbZdkcksZfDlfyd vEy

Sol.

HOñCñCOOH |

|

CH2ñCOOH

CH2ñCOOH

1

2

3

2-Hydroxypropane-1,2,3-tricarboxylic acid

gy%

HOñCñCOOH |

|

CH2ñCOOH

CH2ñCOOH

1

2

3

2- gkbMªksDlhizksisu -1,2,3- VªkbZdkcksZfDlfyd vEy

49. The IUPAC name of

CHñCCl3

Cl

Cl

is (IUPAC(O))

Page #

7

CHñCCl3

Cl

Cl

;kSfxd dk lgh IUPAC uke gS %

(A) Dichlorodiphenyltrichloroethane (B) Trichloromethylbis-(4 chlorophenyl) methane (C*) 1,1,1-Trichloro-2,2-bis(4-chlorophenyl) ethane (D) 2,2,2-Trichloro-1,1-bis(4-chlorophenyl) ethane (A) MkbZDyksjksMkbQSfuy MªkbZDyksjks,sFksu (B) VªkbZDyksjksesfFkyfcl-(4 DyksjksQsfuy) esFksu (C*) 1,1,1-VªkbZDyksjks -2,2-fcl (4-DyksjksQsfuy) ,sFksu (D) 2,2,2-VªkbZDyksjks -1,1-fcl (4-DyksjksQsfuy) ,sFksu

Sol.

CHñCCl3

Cl

Cl

12

1,1,1-Trichloro-2,2-bis(4-chlorophenyl) ethane 1,1,1- VªbZDyksjks -2,2-fcl (4-DyksjksQsfuy) ,sFksu

50.

CHO

COOCH3

and CHOCOOH

are :

(Isomerism(O)) (A) Identical compounds (B) Positional isomer (C*) Functional isomer (D) Chain isomer

CHO

COOCH3

rFkk CHOCOOH

gS :

(A) le:ih ;kSfxd (B) fLFkfr leko;oh (C*) fØ;kRed leko;oh (D) Ja[kyk leko;oh Sol. Both compounds having same molecular formula but different functional group.

nksuksa ;kSfxdksa ds v.kqlw=k leku gS ysfdu fØ;kRed lewg fHkUu&fHkUu gSA MCQ(3) 51. Which of the following samples contain 5 NA atoms : (Mole-1(P)) (A*) 22.4 L of CH4 at STP (B*) 1 gram molecules of N2O3 (C*) 1.25 moles of P4 (D) 0.5 NA molecules of Ethene (C2H4) fuEu esa ls dkSuls uewus 5 NA ijek.kq j[krs gSa % (A*) STP ij 22.4 L CH4 (B*) N2O3 ds 1 xzke v.kq (C*) 1.25 eksy P4 (D) ,Fkhu (C2H4) ds 0.5 NA v.kq

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8

Sol. (A) mole of CH4 = 1 number of atoms = 5NA (B) mole of N2O3 = 1 number of atoms = 5NA (C) mole of P4 = 1.25 number of atoms = 5NA (D) number of atoms in C2H4 = 3NA

Sol. (A) CH4 ds eksy = 1 ijek.kqvksa dh la[;k = 5NA (B) N2O3 ds eksy = 1 ijek.kqvksa dh la[;k = 5NA (C) P4 ds eksy = 1.25 ijek.kqvksa dh la[;k = 5NA

(D) C2H4 esa ijek.kqvksa dh la[;k = 3NA 52. Which of the following options is/are independent of both n and Z ? (ATS(P)) Un = Potential energy of electron in nth orbit KEn = Kinetic energy of electron in nth orbit n = angular momentam of electron in nth orbit vn = Speed of electron in nth orbit fn = Frequency of revolution of electron in nth orbit Tn = Time period of revolution of electron in nth orbit fuEu esa ls dkSuls fodYi n rFkk Z nksuksa ij fuHkZj ugh djrk gS@gSa\ Un = nth d{kk esa bysDVªkWu dh fLFkfrt ÅtkZ KEn = nth d{kk esa bysDVªkWu dh xfrt ÅtkZ n = nth d{kk esa bysDVªkWu dk dks.kh; laosx vn = nth d{kk esa bysDVªkWu ds ?kw.kZu dh xfr fn = nth d{kk esa bysDVªkWu ds ?kw.kZu dh vkofÙk Tn = nth d{kk esa bySDVªkWu ds ?kw.kZu dk vkorZdky

(A*) n

n

UKE

(B*) n

nn vr (C*) nn FT (D*) 2

n

nn

vf

Sol. (A) n

n

KEU

= ñ 12

(B) n n

n

r v =

2nz

zn

1n

= 1

(C) Tn fn = 3

2nz

2

3zn

= 1

(D) n n2n

fv

= 2

3n z

n

2

2nz

= 1

53. Electron in a sample of H atoms are returned to ground state from an excited state so that change

in de-Broglie wavelength of electron corresponding to the transition of maximum energy = ( 8 0.529) ≈ . Select correct options (ATS(P))

(A*) Number of orbit of original excited state is 5. (B) Total number of different spectral lines in visible region is 4. (C*) Total number of different spectral lines in infrared region is 3. (D*) Change in angular momentum of electron corresponding to the transition of minimum energy

=

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9

,d izkn'kZ esa bySDVªkWu ,d mÙksftr voLFkk ls vk| voLFkk esa ykSVdj vkrs gSa rkfd vf/kdre ÅtkZ dh laØe.k ds laxr bySDVªkWu dh Mh&czksXyh rjax}S/;Z esa ifjorZu = ( 8 0.529) ≈ . gks tk,A lgh fodYi dk p;u dhft,A

(A*) ewy mÙksftr voLFkk ds fy, d{kk dh la[;k 5 gSA (B) n'; {ks=k esa fofHkUu LiSDVªy js[kkvksa dh dqy la[;k 4 gSA (C*) vojDr {ks=k esa fofHkUu LiSDVªy js[kkvksa dh dqy la[;k 3 gSA (D*) U;wure ÅtkZ ds laØe.k ds laxr bySDVªkWu ds dks.kh; laosx esa ifjorZu = gSA

Sol.

n

1

= 02 rz

(nñ1)

n ñ 1 = 4 n = 5. Total number of lines in IR region = 3. IR {ks=k esa js[kkvksa dh dqy la[;k = 3

Change in angular momentam (5 4) = h2

=

dks.kh; laosx esa ifjorZu (5 4) = h2

=

MCQ(2) 54. Which of the following is/are represent correct IUPAC name? (IUPAC(O))

(A*) NC

C=C CN CN

NC Ethenetetracarbonitrile

(B*)

CHO| CHO

Ethanedial

(C*)

CHO

CHO

CHOOHC Methanetetracarbaldehyde

(D*)

O

O

O

O 3,3-Di(1-Oxoethyl)pentane-2,4-dione

fuEu esa ls dkSulk@dkSuls ;kSfxdksa ds le{k fn;s x;s IUPAC uke lgh gS\

(A*) NC

C=C CN CN

NC ,sFkhuVsVªkdkcksZukbVªkby

(B*)

CHO| CHO

,FksuMkb,sy

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10

(C*)

CHO

CHO

CHOOHC esFksuVsVªkdkcsZfYMgkbM

(D*)

O

O

O

O 3,3-MkbZ (1-vkWDlk,fFky) isUVsu -2,4-MkbZvksu

Sol. (A) NC

C=C CN CN

NC

1 2 Ethenetetracarbonitrile

(B)

CHO| CHO1

2 ,Fksu Mkb,sy

(C)

CHO

CHO

CHOOHC 1 Methanetetracarbaldehyde

(D)

O

O

O

O 1 2 34

5 3,3-Di(1-Oxoethyl)pentane-2,4-dione

gy% (A) NC

C=C CN CN

NC

1 2 ,sFkhuVsVªkdkcksZukbVªkby

(B)

CHO| CHO1

2 ,FksuMkbZ,sy

(C)

CHO

CHO

CHOOHC 1 esFksuVsVªkdkcsZfYMgkbM

(D)

O

O

O

O 1 2 34

5 3,3-MkbZ (1-vkWDlk,fFky) isUVsu -2,4-MkbZvksu

55. and

Which is/are true about above two structures. (IUPAC & Structural Isomers(O)) (A*) Index of hydrogen deficiency of each is 3 (B*) Both are metamers.

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11

(C) Both are chain isomers (D*) Both have same general formula.

rFkk

mijksDr nks lajpukvksa ds fy, dkSulk@dkSuls dFku lR; gS %

(A*) izR;sd esa gkbMªkstu U;wurk dk lwpdkad rhu gSA

(B*) nksuksa e/;ko;oh gSaA

(C) nksuksa Ja[kyk leko;oh gSaA

(D*) nksuksa leku lkekU; lw=k j[krs gSaA Integer(Double)(3) 56. An LED of powers X watt emits twice as many photons at 1000 nm as another LED of power 5

watt at 400 nm in one second. Find X. (ATS(P))

1 lSd.M esa X okWV 'kfDr dh ,d LED }kjk mRlftZr 1000 nm ds QksVksuksa dh la[;k] 5 okWV 'kfDr dh vU; LED }kjk mRlftZr 400 nm ds QksVksuksa dh la[;k dh rqyuk esa nqxquh gksrh gS] rc X Kkr dhft,A

Ans. 04

Sol. Time = 1 second LED 1 ; 152LED

Energy = X ◊ 1 = XJ ; 5J = 1000 nm ; = 400 nm no. of photons = 2n ; no. of photon = n

1000

hcn2X ; 400hcn5

1000

54002X = 4.

Sol. le; = 1 lSd.M LED 1 ; 152LED

ÅtkZ = X ◊ 1 = XJ ; 5J = 1000 nm ; = 400 nm QksVkWuksa dh la[;k = 2n ; QksVkWuksa dh la[;k = n

1000

hcn2X ; 400hcn5

1000

54002X = 4.

57. Threshold frequency of a metal is 0. When light of frequency = 3 0 is incident on the metal

plate, maximum kinetic energy of emitted photoelectron is x. When frequency of incident radiation is 5 0, kinetic energy of emitted photoelectron is y. If threshold energy of metal is z. Find value of

2

zyx

: (ATS(P))

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12

,d /kkrq dh nsgyh vkofÙk 0 gSA tc /kkrq IysV ij vkofÙk = 3 0 dk izdk'k vkifrr gksrk gS] rks mRlftZr izdk'k] bySDVªkWu dh vf/kdre xfrt ÅtkZ x gSA tc vkifrr fofdj.k dh vkofÙk 5 0 gS] rks mRlftZr izdk'k

bySDVªkWu dh xfrt ÅtkZ y gSA ;fn /kkrq dh nsgyh ÅtkZ z gS] rks 2

zyx

dk eku Kkr dhft,A

Ans. 36 Sol. 0 0 0x 3h h 2h 0 0 0y 5h h 4h 0z h

2x y

z = 36.

58. An element ex ist in three isotopic form : 60A, 62A and 64A. Re lative abundance of 60A = 30% by mole. I f average atomic mass of ëA' is 62.6 u, f ind out the sum of % abundance (by mole)

of 60A and 64A. (Mol-1)(P)) ;fn ,d rRo] rhu leLFkkfud :i 60A, 62A rFkk 64A esa ik;k tkrk gSA 60A dh vkisf{kd ckgqY;rk = 30% ;fn ëA' dk vkSlr ijek.oh; nzO;eku 62.6 u gS] rks 60A o 64A dh izfr'kr ckgqY;rk ¼eksy@eksy esa½ dk

;ksx Kkr dhft,& Ans. 90

Sol. Average atomic mass = (30 60) (70 x)62 (x 64)100

x = 60% mole% of 60A + mole% of 64A = 90

Sol. vkSlr ijek.oh; nzO;eku = (30 60) (70 x)62 (x 64)100

x = 60% 60A dk eksy % + 64A dk eksy % = 90 Integer(Double)(2) 59. The sum of number of functional group and index of hydrogen deficiency in the following

compound is - fuEu ;kSfxd eas fØ;kRed lewg o gkbMªkstu U;wurk dk lwpdkad dh la[;k dk ;ksx gSA

(IUPAC(O))

OH

O

N NH2 COOH

Ans. 19 Sol. Index of hydrogen deficiency = 15 Functional group = 4 gkbMªkstu U;wurk dk lwpdkad = 15 fØ;kRed lewg = 4 60. How many isomers are possible of molecular formula C7H16 having word root pent. (Isomerism(O)) C7H16 v.kqlw=k j[kus okys ;kSfxd ftldk ewy 'kCn isUV gS] ds fdrus leko;oh lEHko gS\ Ans. 05

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13

Sol.

CñCñCñCñC |

|

C

C

CñCñCñCñC |C

|C

CñCñCñCñC |C

| C

CñCñCñCñC|

|

C

C

CñCñCñCñC | C | C

iit maths

Documents

x x x x x

x x x2

ekuk x

bagq fn x

bair vlfedk x

number of integral values

dks larqv djus okys

a2 a2 a2