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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
36 to 40 Integer (double digits) 5 4 0 20
41 to 50 SCQ 10 3 ñ1 30
51 to 55 MCQ 5 4 0 20
56 to 60 Integer (double digits) 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
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
36 to 40 Integer (double digits) 5 4 0 20
41 to 50 SCQ 10 3 ñ1 30
51 to 55 MCQ 5 4 0 20
56 to 60 Integer (double digits) 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)
WEIGHTAGE IN PAPER-2
(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)
WEIGHTAGE IN PAPER-1
(BY FACULTY)
WEIGHTAGE IN PAPER-2
(BY FACULTY)
1. IUPAC nomenclature and Structural isomerism complete
Test Pattern :
Page #
2
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
36 to 40 Integer (double digits) 5 4 0 20
41 to 50 SCQ 10 3 ñ1 30
51 to 55 MCQ 5 4 0 20
56 to 60 Integer (double digits) 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
24 to 29 C o m pre he n s io n (3 C o m p. x 2 Q . ) 6 3 ñ1 18
30 to 32 M TC 3 8 0 24
33 to 39 M C Q 7 4 0 28
40 to 45 C o m pre he n s io n (3 C o m p. x 2 Q . ) 6 3 ñ1 18
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
57 Double Integer Type
(P) Atomic structure Photo electric effect E
58 Double Integer Type
(P) Mole concept Avg. atomic mass M
59 Double Integer Type
(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))
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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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(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