MT-CET 2013 PCM Solution - 15.04.2013

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.


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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


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EDUCATIONAL TRUSTS



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


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EDUCATIONAL TRUSTS



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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MT-CET 2013 PCM Solution - 15.04.2013