13 11 2011 xiii vxy paper ii code at

7/30/2019 13 11 2011 Xiii Vxy Paper II Code At

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13th VXY (Date: 13-11-2011) Review Test-4

PAPER-2

Code-A

ANSWER KEY

PHYSICS

SECTION-2

PART-A

Q.1 D

Q.2 B

Q.3 D

Q.4 A

Q.5 C

Q.6 A

Q.7 B,C,D

Q.8 C,D

Q.9 B,C

Q.10 A,B,C,D

Q.11 A,C

Q.12 A,B

Q.13 B,D

PART-B

Q.1 (A) Q

(B) P,R

(C) S

PART-C

Q.1 0020

Q.2 0020

Q.3 0012

Q.4 0050

Q.5 0018

CHEMISTRY

SECTION-1

PART-A

Q.1 A

Q.2 C

Only XY Batch

Q.3 D

Only V-Group Batch

Q.3 A

Q.4 B

Q.5 D

Q.6 C

Q.7 A

Q.8 D

Q.9 B

Only For V & Y Batches

Q.10 A,B,C

Only For X Batches

Q.10 A,B,C

Q.11 A,B,C

Q.12 A,DQ.13 A,C

PART-B

Q.1 (A) Q,R

(B) P,Q,R,S

(C) P,Q,R,S

PART-C

Q.1 0087

Q.2 3100Q.3 0005

Q.4 2005

Q.5 6080

MATHS

SECTION-3

PART-A

Q.1 A

Q.2 C

Q.3 C

Q.4 D

Q.5 D

Q.6 D

Q.7 A,C,D

Q.8 A,C

Q.9 A

Q.10 B,D

Q.11 B,C,D

Q.12 A,C

Q.13 C,D

PART-B

Q.1 (A) S

(B) P

(C) P

PART-C

Q.1 0217

Q.2 0001

Q.3 0004

Q.4 0064

Q.5 0020


2/16

Code-A Page # 1

CHEMISTRY

PART-A

Q.1

[Sol. Weight of 1 proton =A

N

1

Weight of 1 mole proton = 1

Number of Moles of proton =

1

1000= 103 Ans. ]

Q.2

[Sol.

COOH

NaOH

COONa

CaO + CO

2]

Only XY Batch

Q.3

[Sol. [M(gly)(py)(OCN) (PPh3)N3]+2

(A) Calculation for oxidation state of central atom

x1 + 01 + 01 = +2

x = + 5 Oxidation of central atom

(B) [M(gly)(py)(OCN) (PPh3)N

3]+2

[M(AB)cdef]2+ Coordination number = 6

(C) [M(AB)cdef] 12 Geometrical isomers are possible(D) [M(gly)(py)(OCN) (PPh

3)N

3]+2

and Both complexes are linkage isomers of each other.[M(gly)(py)(NCO) (PPh

3

)N3

]+2 ]

Only V-Group Batch

Q.3

[Sol. Hg, Cu and Pb metals can be extracted by self reduction from their respective sulphide ores.]

Q.4

[Sol.dV

dP=

V

P

increases slope decreases]


3/16

Code-A Page # 2

CHEMISTRY

Q.5

[Sol. Rate of nucleophilic attack Electrophilic nature of carbonyl ]

Q.6

[Sol. (A) B

F

F F F

perfectly sp3 (B) Si

F

F F F

perfectly sp3

(C)C

F

F

F

F

%s

%s

%s

%simperfectly sp3 (D)

C

ClCl

Cl

Cl

perfectly sp3 ]

Paragraph for question nos. 7 to 9

[Sol.

KOH

MnO

KOH

H O3+

Red

Blue

litmu

s

H

C

O

CH=CH2 H

C

OH

HCH

O O

O

+ CH = CHOH CH CHO

2 3

(B)

(A)

(C)

CH CH=CHCHO3

CH OH+HCOONa3

(Aldol condensation)

(Cannizaro reaction)

]

Only For X Batches

Q.10

[Sol. CA

+ CB

+ CC

=0A

C

0dt

dC

dt

dC

dt

dCCBA

2

1

C

A

k

k

C

C

)kk(k

CCC

21

1

AA

B

0

]


4/16

Code-A Page # 3

CHEMISTRY

Q.11

[Sol. (A)

OH

|

CHCHCH 33 2INaOH

O||

CHIONaCCH 33

(B) CCH3

O

2INaOH

COONa

+ CHI3

(C) CH3CH

2Cl

2INaOHCHI

3+ HCOONa

(D)

O

||OHCMe 2INaOH MeCOONa ]

Q.12

[Sol. +2 +2

[PtCl(NH3)3][Cu(NH

3)Cl

3]

Triammine chlorido platinum (II)ammine trichlorido cuprate (II)

or[PtCl(NH3)3]+[Cu(NH

3)Cl

3]

Triammine chlorido platinum (+1)ammine trichlorido cuprate (1) ]

Q.13

[Sol. (A) H = H1

+ H3H

2

=(12830 + 1227013680)

=11420 Cal.

(B) H2H

1= 1368012830

= 850 Cal.

(C) H2H

3= 1368012270

= 1410 Cal.(D) H

2O H+ + OH

= + 13680 K ]

PART-B

Q.1

[Sol. (A) [Co(NH3)5Cl]2+ [Ma

5b]

No. G.I. C.No. = 6 EAN = 273 + 12 = 36

No. Linkage isomersim (Due to absence of ambidentate ligand.

(B) [Co(NH3)4(CN)

2]+ [Ma

4b

2]

Two GI's are possible C.No. = 6 EAN = 273 + 12 = 36 Shows linkage isomersm

(Because ambidentate ligand (CN) is present)

(C) [Co(NH3)2(NO

2)4]1 [Ma

2b

4]


5/16

Code-A Page # 4

CHEMISTRY

Two GI's are possible C.No. = 6 EAN = 273 + 12 = 36 Shows linkage isomersm

(Because ambidentate ligand (NO2) is present) ]

PART-C

Q.1[Sol. B

2(H

2PtCl

6)3 Pt

By POAC of Pt

PtofmassAtomic

Ptof.wt

saltofmassMole

slatof.wt3

195

1

saltofmassMole

4.23

Mole mass of salt = 3 2.4 195 = 1404

Mass of base =2

4103Msalt =2

41031404 = 87 Ans. ]

Q.3

[Sol. BeO , ZnO, PbO, Al2O

3, SnO

are amphoteric oxides ]

Q.4

[Sol. Decomposition of H2O

2is first order

and its t1/2

can be calculated as

2t1/2

= 40 sec

t1/2

= 20 sec

t1/2

= 60 sec 3t1/2

volume required =2

10= 5 ml

abcd = 2005 Ans. ]

Q.5

[Sol. 0Hf

for Br2(l)

(1) + (2) + (3) + (4) + (5)

0 + 2057 + 3015 + 0 + 1008 = 6080 Ans. ]


6/16

PHYSICS

Code-A Page # 1

PART-A

Q.1

[Sol. Q = ne ]

Q.2

[Sol.R

kQ= V

Q =k

VR

= 2R4

Q

= 2

kR4

VR

R

1]

Q.3

[Sol. Ui+ K

i= U

f+ K

f

0 + 200 = 100 + 1.5 10 h

h = 15

100

= 3

20

m ]

Q.4

[Sol. kx = m2 (l0

+ x)

5 100

5= 1 1 (l

0+ 0.05)

100

25= l

0+

100

5l

0= 20 cm ]

Q.5

[Sol.dt

dT=

ms

CA(TT

0) =

sR3

4

R4C

3

2

(TT

0)

R

1

2R

R

dt

dTdt

dT

P

Q

Q

P

]

Q.6

[Sol. Q

Q E 0 ; V = 0 ]

Q.8

[Sol. v cos = v0

v sin=m

qEt

t =0

v

a


7/16

PHYSICS

Code-A Page # 2

v sin =0

a

mv

qE

L

d= tan =

0

0

a

v

mv

qE= 2

0

a

mv

qE d = 2

0

a

mv

LqE]

Q.10

[Sol. collision is elastic '

2p = p1 = 3kg m/s

collision is inelastic '1

'2

pp = 1.5 kg m/s for other cases p is in between. ]

Q.11

[Sol. F = 4a

37

FN

1g

F sin 37 + N cos 37 = 10

F cos 37N sin 37 = 1a

3 3F + 4N = 50

4 4F3 N = a

25F = 150 + 4a

100 a4a = 150

a =96

150 F

8=

96

150 4 = 6.25 N ]

Q.13

[Sol. v =0

2

(Oy)

O

y

x

z= 02

y

= 2

y

]

PART-C

Q.1

[Sol. F = mg sin ~ mg tan

mg

tan =dx

dy= 40 x

m2x =mg 40 x

=400

= 20 rad/s ]

Q.2

[Sol. =60

260 rad/s = 2 rad/s

0 = 0t

=40

2=

20


8/16

PHYSICS

Code-A Page # 3

= I = 4020

= 22 ~ 20 Nm ]

Q.3

[Sol. v = 0 at a point between both charges and to left of 2q.4 x

2q 3q

9x

q2k

x5

kq3

= 0

102x = 3x

x = 2to left of 2q,

y

2q 3q

y

q2k

y5

kq3

= 0

10 + 2y = 3y

y = 10 distance = 12 m ]

Q.4

[Sol. 11.25 = x tan 2

1 2

2

20

x10(1 + tan2)

11.25 = x tan 80

x2

80

x2

tan2

80

x2

tan2 + x tan 80

x2

+ 11.25 = 0

b24 ac 0

80

xx4x

22 025.11

80

x2

1 +20

1

80

x25.11

2

0

20

25.31

1600

x2

x2 312.5 8 = 625 4

x 50 m ]

Q.5

[Sol. 5.15

1= (1.504

1)

21 R1

R

1

f

1= (1.4341)

21R

1

R

1

5.15

f=

434.0

504.0

f = 15.5 217

252= 18 cm ]


9/16

MATHEMATICS

Code-A Page # 1

PART-A

Q.1

[Sol. I = e

1)xn1(nxx

dx

llPut 1 + ln x = t2

I =

2

12

1tt

dtt2= 2 12n l = 223n l . Ans.]

Q.2

[Sol.MB

f(x) = min. 1x,1x = x + 1 xR

x

yy=x+1 y = x + 1

O(0,0)

]

Q.3

[Sol. If f (x) < 0 then | f (x) | =

f (x)Hence, f (x) > f (x) is not possible.

If f (x) > 0 then | f (x) | = f (x)

Hence, f (x) > f (x)

2 f (x) > 0 f (x) > 0 x24x + 3 > 0 x (, 1) (3, ). Ans.]

Q.4

[Sol. As, AA1 = I =

1001

22

2

sinsin

0sin2

21

01=

1001

2

2

sin20

0sin2=

10

01

2 sin2 = 1 sin2 =2

1= sin2

4

= n

4

, n I

Hence, number of values of in [, ] is 4 i.e.4

3,

4

Ans. ]

Q.5

[Sol. We have

x2 + 2cx + b = 0

and x + 2bx + c = 02

So, 2 + 2c + b = 0 .........(1)

and 2 + 2b + c = 0 ........(2)

(Subtracting)

2(cb) = (cb)


10/16

MATHEMATICS

Code-A Page # 2

As, b c so =2

1

Now, putting =2

1in equation (1), we get

0b

2

1c2

2

12

4

1+ c + b = 0

Hence, (b + c) =4

1. Ans.]

Q.6

Sol. Let ABD = and B = 45.

2

1BD BA sin = 3

2

1 BD BC sin (45 )

BC

BAsin=

2

3(cossin)

75

45

60C

D A

B

75sin

60sinsin=

2

3cos

2

3sin

31 3 sin = 31 cos

tan

= 3

1

= 30

4,

8 ]

Paragraph for question nos. 7 to 9

[Sol. )x(ff = a3

a

2b2x)b2(x2

a

bb2x)b2(x2 , a 0

Now, we shall consider 3 cases:

Case I: When x (2 + b)x +2

5 5

a

2b2 = 0

2 + b = 10 b = 8

O(0,0) X

Y

(2,0) x = 5 (8,0)

Graph of f(x) = (x2)(x8)9

2

(x = 5, y = 2)

9

32,0

Also, 2ba

2= 25 1625 =

a

2 a =

9

2

Let, D = discriminant of

a

bb2x)b2(x2


11/16

MATHEMATICS

Code-A Page # 3

= (2 + b)24

a

bb2 = (2 + 8)28 8 +

9/2

84

= 10064144 < 0

So, f(x) =9

2(x2) (x8) =

9

2(x5)2 + 2

Case II: When x (2 + b)x +2

5 5

a

bb2 = 0

2 + b = 10 b = 8

1625 =a

8 a =

9

8

Let D' = discriminant of

a

2b2x)b2(x

2= (2 + b)24

a

2b2

= (2 + 8)28 8 +9/8

8

= 100649 > 0. So, above case is rejected.

Case III : When, x (2 + b)x +2

5 5

a

2b2 = 0 and x (2 + b)x +2

5 5

a

bb2 = 0

But, it is not possible for any real value of a and b. So, this possibility is also rejected.

We conclude that f(x) =9

2(x2) (x8)

(i) From above graph of f(x), we can say that the maximum value of f(x) is 2 and attained at x = 5.

(ii) Let the line passing through O(0, 0) be y = mx.

Solving y = mx and y =9

2(x5)2 + 2 ; we get

mx =9

2(x5)2 + 2 9mx =2(x210x + 25) + 18 2x2 + (9m20)x + 32 = 0

As, y = mx is tangent line, so put D = 0

(9m20)2 = 4 2 16 (9m20) = 16 9m = 20 16 m =9

36,

9

4

m = 4,9

4 Ans.

(iii) 5

2

dx)x(f =

5

2

2dx2)5x(

9

2=

5

2

3

3

)5x(

9

2

+ 6 =

27

2(0 + 27) + 6 = 2 + 6 = 4 Ans.]


12/16

MATHEMATICS

Code-A Page # 4

Q.10

[Sol. P = sin 25 sin 35 sin 60 sin 85

= sin 25 sin (6025) sin 60 sin (60 + 25)

= sin 60 sin 25 sin (6025) sin (60 + 25)

P = sin 60 4

1 sin 75 .......(1)

Q = sin 20 sin 40 sin 75 sin 80 = sin 20 sin (60 20) sin 75 sin (60 + 20)

= sin 75 sin 20 sin (60 20) sin (60 + 20)

Q = sin 75 4

1 sin 60 .......(2)

Hence P = Q (B) and (D) ]

Q.11

[Sol. P1: x + 3y + 2z = 6

P2: x + ay + 2z = 7

P3: x + 3y + 2z = bNote that the planes P

1and P

3are parallel hence system of equation will never have unique solution.

If b = 6 and a = 4 then P1

and P3

are identical and P2

will intersect P1

and P3

at infinite number of

points because they have common line of intersection.

If b 6, then P1

and P3

are parallel and system will have no solution for every aR.

Alternatively:

Given, x + 3y + 2z = 6 ........(1)

x + ay + 2z = 7 ........(2)

x + 3y + 2z = b ........(3)

(A) If a = 2, then= 0 so unique solution is not possible. (A) is incorrect.(B) If a = 4, b = 2; consider the equation (1) and (2), we get

x + 3y = 62z

x + 4y = 72z

y = 1 and x = 32z

Now, on substituting in equation (3), we get

32z + 3 + 2z = 6, (True)

infinite solution exists. (B) is correct.

(C) If a = 5, b = 7 ; consider the equation (2) and (3), we get

x + 5y = 7

2xx + 3y = 72x

y = 0 and x = 72x

Now, on substituting in equation (1), we get

72z + 2z = 6 (False)

No solution exists. (C) is correct.

(D) If a = 3, b = 5 then equation (1) and (2) have no solution No solution exists. (D) is correct.]


13/16

MATHEMATICS

Code-A Page # 5

Q.12

[Sol.mb

We have f(x)2

)say(A

4

0

5

2

dt)t(ftcosxcos

xsin

=xcos

xsin5

2

f (x)2Axcos

xsin5

2=

xcos

xsin5

2 f (x) = (2A + 1)

xcos

xsin5

2......(1)

Now, A =

4

0

5

2

dttcos

tsin)1A2(tcos =

4

0

4

2

dttcos

tsin)1A2( =

4

0

22 dttsecttan)1A2(

Put tan t = y sec2t dt = dy

we get A = (2A + 1)

1

0

2

dyy = (2A + 1) 3

1

3A = 2A + 1

A = 1

Hence from equation (1), we get

f(x) =xcos

xsin35

2

(A) Clearly,xcos

xsin3Lim)x(fLim

5

2

3x

3x

= 5

2

2

1

2

33

= 72 Ans.

(B) As, f(x) =xcos

xsin35

2

So, f (x) is periodic with period 2.

(C) f(x) =xcosxsin3

5

2

f ' (x) =

xcos

)xsinxcos5(xsin)xcosxsin2(xcos 10

425

f ' () = 0

M (x = , y = 0)

So, equation of normal to the graph of f (x) at point M whose abscissa is , is given by x= 0

(D) As, f (x) = 0 xcos

xsin35

2

= 0 sin x = 0


14/16

MATHEMATICS

Code-A Page # 6

x = n, n I

So, the equation f (x) = 0 has no root in (0, 3). ]

Q.13

[Sol.

(A) The above statement is false because A1 exist only when det. (A) 0.

(B) The above statement is false. As, det. (A B

1) = det. (A) det. (B

1) = )B(.det

)A(.det

= 2

6=

3.

(C) Given, A =

111111111

AA2 =

333333333

= 3A.

A3 = 3A2 A3 = 3(3A) A3 = 9A. So, this statement is correct.

(D) Given, A2 = A and B = IA

Now, AB + BA + I( IA)2 = AB + BA + I(I + A22A) = AB + BA + A (As, A2 = A)

= A(IA) + (IA) A + A = AA + AA + A = A (As, A2 = A and B = IA)

So, this statement is correct. Ans.]

PART-B

Q.1

Sol.(A) Given, 2a sin xa sin 3x 6 xRor, a [2sin x(3sin x4sin3x)] 6or, a sin x (4sin2x1) 6 xRAs, maximum value of a sin x (4sin2x1) is 3 | a |.

So, we must have 3 | a | 6 | a | 2

So, possible integral values of a are

2,

1, 0, 1, 2.Hence, number of integral values of a are 5.

(B) x2 + 2x sin (xy) + 1 = 0 (As, x 0)

Dividing by x, we get )xysin(2x

1x

which is possible when x = 1 and y =2

or x =1 and y =

2

.

Hence total number of ordered pairs (x, y) is 2. Ans.

(C) L = 3231

x

0

2

0x x21nxsin

dttcosLim

l

=

32

32

31

31

x

0

2

0x

x2

x21n

x

xsin

dttcosLim

l

=x211

dttcosLim

x

0

2

0x

=2

1xcosLim

2

0x=

2

1 L1 = 2 ]


15/16

MATHEMATICS

Code-A Page # 7

PART-C

Q.1

[Sol. Sn

=2

13n

; Sn + 1

=2

131n

Tn

= 5 5n 1

Tn + 2

= 5n + 2

1n 2n

1n

T

S=

1nn

n

2552

133=

1nn

1n

n

5

1

50

1

5

3

50

3

=

5

11

5

1

50

1

5

31

5

3

50

3

=

4

1

50

1

2

3

50

3 =

200

1

200

18 =

200

17Ans.]

Q.2

[Sol. As, )C.(detB.adj.det

= )A5.(detB.adj.adj.det

=

)A.(det5

A.adj3

13 2

=125

A3

.

As, det. (A) = 5

So,)C(.det

)B.adj.(det=

125

53

= 1. Ans.]

Q.3

Sol. Let I =

4

65

4

xsinxcos )21)(21(

dx=

416

4

xsinxcos )21)(21(

dx

=

16

0

2periodwithperiodic

xsinxcos )21)(21(

dx

=

2

0xsinxcos)21)(21(

dx8 = 8I

1

where I1 =

2

0xsinxcos )21)(21(

dx

Using King and add

2I1

=

2

0

xcos)21(

dx


16/16

MATHEMATICS

Code-A Page # 8

Again using King and add

4I1

= 2

0

dx = 2

Hence I1

=2

I = 8I1 = 8 2

= 4 k. Hence k = 4. Ans.]

Q.4

[Sol. In

= 1

0

n

I

2

IIdxx11 =

1

0

21n21

0

n2 dxxx1n2xx1

= 2n2n

1

0

21n2dx)1x1(x1

In = 2n

2n

1

0

1n21

0

n2

dxx1dxx1

In

= 2n2n In

+ 2n In1

(2n + 1) In

= 2n + 2n In1

or (2n + 1) In2n I

n1= 2n

Put n = 6,

13 I612 I

5= 26 = 64. Ans.]

Q.5

[Sol. We have g(x) =

2

1

x1

x2

sin2 = 2 +

2

1

x1

x2sin

As, sin1 2x1

x2

2

,2

2

1

x1

x2sin =2,1, 0, 1.

Range of g(x) = {0, 1, 2, 3} for )x(gf < 0 x R f(0) < 0 and f(3) < 0Now, f(0) < 0 a2 < 0 a < 2

and f(3) < 0 96a + a2 < 0

a >5

7

0 1 2 3

x-axis

f(x) = x 2axa22

a

2,

5

7.

Hence, k1

=5

7, k

2= 2

(10k1

+ 3k2) = 14 + 6 = 20. Ans.]

13 11 2011 xiii vxy paper ii code at

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