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7/30/2019 MT-CET 2013 PCM Solution - 15.04.2013
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EDUCATIONAL TRUSTS
Junior Science CollegeANDHERI / BORIVALI / DADAR / NERUL / POWAI / THANE
STD: XII MT-CET EXAMINATION 2013
DATE:15/04/2013 PHYSICS SOLUTION
Topics : . Wave Theory of Light, Interference & Diffraction, Semiconductors, Communication systems
2. (a)
88
8
13 10
sinsin 3 10 sin 30 2, 2.119 10 / sec1sin sin sin 45
2
a am
m
C C riC m
r C i
3. (c) 8 82 10 / sec, 2.25 10 / secg wC m C m
8
8
2 10 0.88882.25 10
g
g w
w
C
C
4. (d) .a a a aww w w w
d C t t
d C t t
64 3 10 18008 sec
3 900w a w
w
a
t dt
d
5. (d) 1 1 0tan tan (1.5) 57p g
6. (c)
2 21 2max 1
2 2min 1 2 2
(5 1) 36 9 5,
( ) (5 1) 16 4 1
a aI a
I a a a
7. (b)7
4 4
4 4 5 10 22
2 10
DX cm
d
8. (b) 1 2X X 1 1 2 2
1 2
,D D
d d
1 2 1
2 1 2
2 2 4
1 1 1
D d
D d
9. (b)1.22
d
and
x
a where d is diameter of objective, X is distance between stars, a is
distance of stars from the earth.1.22 d
Xa
7 15
1
1.22 6 10 10 9.5 10
5 10
=
111.391 10 m
10. (a) 671. 1.952 101.22 1.22 4.2 10dR P
11. (b) Increase with increases in its temperature.
7/30/2019 MT-CET 2013 PCM Solution - 15.04.2013
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16. (b)100
0.991 101
23. (d) ( )V R VIBGYOR
min max
33. (b)
75.8 102.4162 . . 2 0.12
Xd mN A
36. (b) E = VKnee + IR35.7 0.7 10 , 5R R K
44. (a)0.9
91 1 0.9
49. (a)1
' 3 , , '3
Xd d X X
d
50. (c) 2 2
1 1
,X
XX
2
60001 1.2
5000X mm
7/30/2019 MT-CET 2013 PCM Solution - 15.04.2013
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EDUCATIONAL TRUSTS
Junior Science CollegeANDHERI / BORIVALI / DADAR / NERUL / POWAI / THANE
STD: XII MT-CET EXAMINATION 2013
DATE: 15/04/2013 CHEMISTRY SOLUTION
Topics : Halogen derivative of alkanes, Alcohols, phenols, ether, Aldehydes, Ketones and Carboxylic Acids
2. (b) 3 2 %CH CH Br
. 80% 100 100 73.39 74 75
109
Wt of BrBr
Total wt of compound
4. (b) 3 3CH CHBrCH and KCN
3 3
3 3
| |
CH CH Br KCN CH CH CN KBr
CH CH
6. (a) Only this contains -hydrogen atom.
7. (b)
23. (b) second carbon is asymmetric hence optically active.
24. (a) Mercaptans are Thio alcohols.R S H are mercaptans.
27. (a) In ether oxygen atom undergoes 3sp hybridization.
28. (b) Oxygen of alcohols contains lone pair of electrons. Hence Lewis base.
32. (a) Alkaline solution of phenols reacts with electrophile 2CO to give salicylic acid it is called
Kolbes Schmidt reaction.
33. (b) Cresol has 3 6 4CH C H OH cresol phenolic OH
34. (b) Phenols contain polar OH group. Hence shows hydrogen bonding.
3 2 3
|
CH C H CH CH
Cl
3 2 3
|
CH CH CH CH
OH
O H
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35. (d) 6 5 2C H CH OH . Among primary alcohols only ethanol show iodoform test a, b contain
3CH CO group. C contains
3
|
CH CH
OH
group. They show iodoform test.
36. (b) Due to 2sp hybridization of carbonyl carbon carbonyl compounds are planner in nature.
37. (a) 2 2 72 4[ ]
2K Cr O O
dil H SOR CH OH R CHO R COOH
39. (a) CO andHCl {gattermann Koch reaction}
40. (a) Phenyl ethanalThe Etard reaction alkyl side chain linked to aromatic ring is oxidized in aldehyde group.
41. (d) Butanone is least reactive due to more +I and steric hinderance of alkyl group.
44. (d)
49. Soda ash 2 3( )Na CO is decomposed by formic acid where as phenol can not.
CH - CH - CH3
OH
K2
Cr2O
7
H+
CH2 - C - CH3
O
NH2
- NH2
KOH /
CH2
- CH2
- CH3
7/30/2019 MT-CET 2013 PCM Solution - 15.04.2013
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EDUCATIONAL TRUSTS
Junior Science CollegeANDHERI / BORIVALI / DADAR / NERUL / POWAI / THANE
STD: XII MT-CET EXAMINATION 2013
DATE:15/04/2013 MATHEMATICS & STATISTICS SOLUTION
Topics : Matrices, Mathematical Logic, Binomial distribution, Statistics - Bivariate Frequency
Distribution, Probability distribution.
1. In sentence q, is not defined so it is an open sentence not a statement.
2. Given statement is ( ) ( )p p q
contra positive is ( ) ( )p q p p q p
22. 2 3y x
where 2 0
The two variable are in perfect indirect correlation.1
xyr
23.16
804(0.5)
y
25. 5u x 5v y 2
u 2
v uv -1 0 1 0 02 3 4 9 63 1 9 1 3-2 -2 4 4 4-1 0 1 0 0
1 2 19 14 13
2 2
1 213
5
1 219 . 14
5 5
xy uvr r
=
63
94 . 66
26. 1 1r
27. Two variables are connected by the relation7x y
( 1) 7y x x andy are in perfect indirect correlation
1xy
r
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29.cov( , ) 8 8 2
3 4 3var . var 9 16
X Yr
X Y
30. Covariance is independent of origin.
31.5
1
( ) 1x
P X x
( 1) ( 2) ( 3) ( 4) ( 5) 1P X P X P X P X P X
1( ) 2( ) 3( ) 4( ) 5( ) 1k k k k k
15 1k
1
15k
32. ( )f x is a p.d.f.
( ) 1f x dx
5
0
sin 15
xk dx
5
0
5. cos 1
5
xk
5
. cos cos 0 1k
5(2). 1k
10k
33. P(X is odd) = P(X = 1 or 3) = 0.1 + 0.5 = 0.6P(X is even) = P(X = 2 or 4) = 0.2 + 0.2 = 0.4
35. (1,2) (4,5)A B
( ) ( ) ( )P A B P A P B
So,2 5
1 4
( ) ( ) ( )P A B f x dx f x dx
2 5
2 2
1 4
1 1dx dxx x
2 5
1 1
1 1
x x
1 1 1
12 5 4
110.55
20
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36.2
1 1
( ) ( )( 1)
n n
x x
xE X x P x
n n
2
1
1
( 1)
n
x
xn n
( 1) (2 1).
( 1) 6
n n n
n n
2 1 16 3 6
n n
37. 2( )P X K CK
( 0) 0, ( 1)P X P X C
( 2) 4 , ( 3) 9 , ( 4) 16P X C P X C P X C 4
0
1i
i
P
0 4 9 16 1C C C C
30 1C
1
30C
38. HHT, HTT, HHH, TTT(2, 2, -1) (2, -1, -1) (2, 2, 2) (-1, -1, -1)
3, 0, 6, 3 is range of X.
39. (2,2),(2,3),(3,2),(3,3)S
Mean = i ix P =1 1 1
4 5 6 54 2 4
40. 1, 2, 3, ...... .X k
1 1 1 1( ) , , ,....P X
k k k k
1( )
2
kE X
22( ) ( )
i iVar X X P E X
221 4 9 1.....
2
k k
k k k k
2
2 2 21 11 2 ....2
kk
k
21 ( 1) (2 1) 1
6 2
k k k k
k
2 1
12
k
X 4 5 6P(X) 1
4
1
2
1
4
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41.1
5,2
X B n p
( )V X npq
1 1 55
2 2 4
42. 8n 1
2p 1
2q
(1 41 2) (2 6)P x P x
1 ( 0,1,7or8)P X
8
8 8 8 80 1 7 8
11
2C C C C
81
1 (1 8 8 1)2
18 1191
256 128
43. 15np npq and 2 2 2 2 2 117n p n p q 2
2
1 117
(1 ) 225
q
q
26 13 6 0q q
2
3q ;
1
3p
15np npq
1 1 2 153 3 3
n
27n
9np
44.90 9
;100 10
p 1
10q
5 0 5
55
9 1 9
10 10 10C
45. 3,n (3or6)P = 1 1 2 16 6 6 3
1 21 1
33q p q
Required probability= P(exactly two) + P(exactly three)
2 3 0
3 32 3
1 2 1 2
3 3 3 3C C
2 1 7
9 27 27
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46. n = 8 p =5
100=
1
20, q =
19
20
0 8 1 7
8 80 1
1 19 1 19( 2)
20 20 20 20P X C C
727 19
20 20
48.1
2p
1
2q
( 1) ( 3) ..... ( 99)P X P X P X 100
100 100 1001 3 99
1....
2C C C
100
100 11 122 2
49.21 ( 3)
8x
Ke
=2
2
1( 3)
2(2)x
Ke
4
50. ( ) 5E X np
22( ) ( )E X E X npq
= 44
5q
1
5p 25n
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