12 02 2012 xiii vxy paper ii code a sol

7/30/2019 12 02 2012 Xiii Vxy Paper II Code a Sol

1/18

13th VXY (Date: 12-02-2012) Review Test-7

PAPER-2

Code-A

ANSWER KEY

MATHS

SECTION-2PART-A

Q.1 B

Q.2 C

Q.3 A

Q.4 A

Q.5 D

Q.6 B

Q.7 B

Q.8 C

Q.9 D

Q.10 B

Q.11 C

Q.12 B,C,D

Q.13 C,D

Q.14 A,B

PART-B

Q.1 (A) R

(B) P

(C) Q

(D) T

PART-C

Q.1 0007

Q.2 0007

Q.3 0009

Q.4 0001

CHEMISTRY

SECTION-1PART-A

Q.1 B

Q.2 D

[Only V-Group]

Q.3 A

[Only XY-Batch]

Q.3 B

[Only V-Group]

Q.4 C

[Only XY-Batch]Q.4 C

Q.5 A

Q.6 B

[Only V-Group]

Q.7 A

[Only XY-Batch]

Q.7 D

Q.8 D

Q.9 B

Q.10 BQ.11 A

Q.12 A

Q.13 B

Q.14 B

PART-A

[Only V-Group]

Q.1 (A) S,T (B) P,Q,R

(C) Q,R,S (D) P,R[Only XY-Batch]

Q.1 (A) PR (B) PQS

(C) T (D) S

PART-A

Q.1 0040

Q.2 0003

Q.3 0006

Q.4 3424

PHYSICS

SECTION-3PART-A

Q.1 B

Q.2 C

Q.3 A

Q.4 B

Q.5 A

Q.6 B

Q.7 B

Q.8 C

Q.9 B

Q.10 A

Q.11 A

Q.12 B,C

Q.13 A,C

Q.14 A,D

PART-B

Q.1 (A) P,R,S,T

(B) P,Q,T

(C) R,S

(D) Q

PART-C

Q.1 0003

Q.2 1200

Q.3 0001

Q.4 0017


2/18

CHEMISTRY

Code-A Page # 1

PART-A

Q.1

[Sol. V = 0

W = PextV = 0 ]

Q.2

[Sol. Oxidation state of metal in the bead can be determined from bead colour but oxidation state of the metal

in the given salt can not be predicted. ]

[Only V-Group]

Q.3

[Sol. Alanine zwitter ion is

Thus Ans is A]

[Only XY-Batch]

Q.3

[Sol.

ClH

OH2 H O2

OH OH

+

'R' 'S'

98% racemisation means 49% R and 49% Sand 2% inversion (S) take place

Total % S = 49 + 2 = 51 % ]

[Only V-Group]

Q.4

[Sol. KOCPh||

O

KOCPh||

O

OCPh||

O

2CO

Ph PhPh

PhI

Cu PhPh

PhClAlCl3

poor yield

PhClTHF

Na PhPh ]


3/18

CHEMISTRY

Code-A Page # 2

[Only XY-Batch]

Q.4

[Sol.

14

OMe

Br

14

OMe

Br

14 14

OMe OMe

14

OMe

Br

14

14

14

14 14

14

14

14

OMe

OMe

OMe

OMe OMe

OMe

OMe

OMe

3

2

NH.liq

NaNH

NH2

NH2

NH2

+

+

+

NH2

NH2

NH2

NH2

NH2

NH2

NH2

NH2

Solvent

Solvent

Solvent

Solvent

]

Q.5

[Sol. In 100 gm of dried sample 10 gm H

2O , 50 gm silica and (1005010) = 40 gm nonvolatile impurities

If Wgm of original sample contained 50 gm silica and 40 gm nonvolatile impurities then

W = 0.28 W + 40 + 50

W = 125 gm

% of Silica = 100125

50 = 40 %]


4/18

CHEMISTRY

Code-A Page # 3

Q.6

[Sol. O=P P=OO

O O

P

P

O

O

O

OO

OH2 2

O=POH HOP=O

OO

P

P

O

O

O

O

OH

OH

Tetrametaphosphoric acid

H O2

OHOHOHOH||||

OHPOPOPOPHO||||||||

OOOO

(Tetra polyphosphoric acid)

H O2

acidoricPyrophosphOHOH||

OHPOPHO||||

OO

2

2H O2

acidhoricOrthophospOH|

OHPHO||O

4

]

[Only V-Group]

Q.7

[Sol. PhCHMe2

H)ii(

h/O)i( 2

OHO|CMePh 2

H PhOH + Me2C = O

PhCl KOH.aq (Poor yield)

OH

; PhMgBr OH2 PhN + MgBr (OH)

PhNH2

C50

HClNaNO2

NClNPh

]


5/18

CHEMISTRY

Code-A Page # 4

[Only XY-Batch]

Q.7

[Sol.

Me

Me

Me

OH

Cl

Cl

D

D

D

H

H

H

SOCl2

pyridine

PCl3

(S 2)N

S 2N

]

Q.8

[Sol. The metals which forms volatile iodides are purified by Van-Arkel method.]

Paragraph for question nos. 9 to 11Q.9

[Sol.(i) Oxidation half : Ag(s) + Br (aq) AgBr(s) + e

Reduction half : e +1/2 Hg

2Cl

2(s) Hg(l) + Cl(aq)

(ii)sp

o

/Ag(s)Ag

o

g(s)/AgBr(s)/ABr K

1log

1

06.0EE

= 0.80.06 12

= 0.08

(iii) Ecell =]Br[]Cl[log

106.0EE o /HgClCl/Hgo (s)/BrAg(s)/AgBr 22

=0.08 + 0.280.06 log8

1

= 0.2 + 0.054 = 0.254 V Ans. ]


6/18

CHEMISTRY

Code-A Page # 5

Paragraph for question nos. 12 to 14

[Sol. (NH ) Cr O74 2 2

CrCl3CoCl 6H O2 2.

Mg N3 2

NH3

CoCl2

H2O

Conc. HCl

Cr O32 2 2(s) + N + H O (g)(A) (B) (C)

Pink(E)

(I)

(J)

(K)

(F)

(D)Orange solid Green

Mg

NiCl2sol.

[Ni(NH ) ]3 62+

Deep blue

Sol.12 (NH4)2

Cr2O

7

Cationic NH4+ sp3

Anionic Cr2O72 d3s

Sol.13 Ni2+ in [Ni (NH3)6]2+

3d 4s 4p 4d

Hyb.:sp d3 2

eff=2.8 B.M.

Sol.14 NH3(J) Bond angle 107.5

N2(C) Bond angle not defined

Difference of Bond angle between (J) and (C) is 107.5]


7/18

CHEMISTRY

Code-A Page # 6

PART-B

[Only XY-Batch]

Q.1

[Sol. 1 / 2 / 3 alcohol are differentiate by Lucas Reagent, Victor Mayer test group having

lg|

RCHCH3 or

OH|

RCHCH3 shows haolform test 1,2 alcohol do not show test with aq. AgNO3. 1 amine and 2

amine can be differentiated by isocyanide test. ]

PART-C

Q.1

[Sol. KP

=2

OH2p OH2

p = 102 atm

V.P. of H2O = 19 torr =

760

19atm

% relative humidity =19

10076001.0 = 40 %]

Q.2

[Sol. Paramagnetic unpaired electrons

KO2 K+ O

2 according to MOT paramagnetic

Na2O

2 2Na+ O

22 diamagnetic

CaO Ca+2

O2

diamagneticKF K+ F diamagnetic

NO2

N

OO

odd electron species Paramagnetic

NO N = O odd electron species Paramagnetic ]


8/18

CHEMISTRY

Code-A Page # 7

Q.3

[Sol. Following sets of aqueous solutions can form a buffer solution.

Set (1): CH3COOH + NaOH

acidweak3COOHCH +

basestrongwithsalt3COONaCH

Set (2): HCN + KCN acidweakHCN +

basestrongwithsaltKCN

Set (4): CH3COOH + NH

3

baseweak&acidweakofsalt43COONHCH

Set (6): CH3COONa + HCl

acidweak3

basestrongwithsalt3

COOHCHCOONaCH

Set (7) : NH4Cl + NaOH

acidstrongwithSalt4

baseweak4

ClNHOHNH

Set (8) : NH3

+ HCl acidstrongwithSalt

4baseweak

4ClNHOHNH ]

Q.4

[Sol. (i)

NMe3

OH

(major)

(3-H)

(ii)H

O

O

Cl MeMgBr

Me

Me

OMgBr

MeMgBr

+

-

OMgBr

+ Mg

Br

Cl

(4-MeMgBr)

(iii)

D

O

HCN

D

OHCN

+

D

OH CN

(2-products)

(iv) 2-Chiral centre in part (iii) so all possible stereoisomers = 22 = 4.

= 3424 ]


9/18

MATHEMATICS

C d A P # 1

PART-A

Q.1

[Sol. We have 2xy xcosye 2 = 5 ........(1)Put x = 0, we get

1 + y = 5 y = 4 (0, 4) lies on the given curve.

Now, differentiating (1) with respect to x, we get

2xy ydx

dyy2xe

2

y 2x sin (x2) + cos (x2)dx

dy= 0

As (0, 4) satisfy it, we get

16 +)4,0(dx

dy

= 0

)4,0(dx

dy

=16. Ans.]

Q.2

[Sol. As, AD =

cb

bc2 cos2A

AD2

Acos2

=c1

b1

A

B C

bc

a

EF

D

A2

A2

B

2

B

2

C2

C2

AD2

Acos

=c

1

b

1

a

1 =

7

1

5

1

3

1 =

105

71Ans.]

Q.3

[Sol. For continuity, a = 8 ; b = 2.

Director circle is x2 + y2 = 84 = 4 (A) is correct ]

Q.4

[Sol. Case-I: When '0' is not at terminal position T T3C

2 2! 5C

4 4! = 6 120 = 720

Case-II: When 0 is at last place3C

1 1 5C

4 4! = 3 120 = 360

Total number of six digit numbers = 720 + 360 = 1080. Ans. ]

Q.5

[Sol. A =

0

0

0

Number of skew symmetric matrices = 3! 8 = 48. Ans.

[ As, diagonal element must be 0 and conjugate pair elements are additive inverse of each other

in skew-symmetric matrix. ]

Aliter: 1 can be put by 6 ways

1 can be put by 1 way

2 can be put by 4 ways


3 can be put by 2 ways


Number of skew symmetric matrices = 6 1 4 1 2 1 = 48. Ans.]


10/18

MATHEMATICS

C d A P # 2

Q.6

[Sol. The rough graph of f (x) is shown below and from the graph, f (x) is an odd function.

x=1 x= 1

y=x(1,1)

(1,1)

Ox

y

]

Q.7

[Sol. Given, x2 + ax4 = 0

+ =a, = 4

Now, 2(2 + 2) = ( + ) + (3 + 3) 2[( + )22] = ( + ) + [( + )33( + )] 2(a2 + 8) =a + (a312a) a3 + 2a2 + 13a + 16 = 0Let F (a) = a3 + 2a2 + 13a + 16

F ' (a) = 3a2 + 4a + 13 > 0 a R F (a) is increasing function on a (, ).So, F (a) = 0 has exactly one real root.

Also, f (1) = (1)3 + 2(1)2 + 13(1) + 16 = 1 + 213 + 16 = 4

and f (2) = (2)3 + 2(2)2 + 13(2) + 16 = 8 + 826 + 16 = 16

f (1) f (2) < 0

F (a) = 0 has a root in (

2,

1). (B) is correct. ]

Q.8

[Sol. Given, 2 66 sincos = (1 + cos 2) .......(1)Now squaring on both sides, we get

4

2sin4

31

2= (1 + 2cos 2 + cos22)

43 (1cos22) = 1 + 2cos 2 + cos22 2 cos 2 (cos 21) = 0

Either cos 2 = 0 or cos 2 = 1Now, cos 2 = 1 2 = 2m, m I = m, m I

Also, cos 2 = 0 2 = (2n + 1)2

, n I = (2n + 1)

4

, n I

But, for = (2n + 1)4

, n I

L.H.S. of equation (1) equals 0 and R.H.S. of equation (2) equals 1

So, L.H.S. R.H.S.Hence, = m, m I

So, possible solutions are =, 0, .Hence, number of solution are three. Ans.]


11/18

MATHEMATICS

C d A P # 3


[Sol. Equation of plane passing through the intersection of planes

P1

: 2x5y + z3 = 0 and P2

: x + y + 4z5 = 0 is P1

+ P2

= 0.

i.e., (2x5y + z3) + (x + y + 4z5) = 0or (2 + ) x + (5 + ) y + (1 + 4) z(3 + 5) = 0, ........(1)

which is parallel to plane P3 : x + 3y + 6z1 = 0.

So,6

41

3

5

1

2

=

2

11

From equation (1), we get

2

49z21

2

y21

2

x7

= 0 or P

4: x + 3y + 6z = 7

(i) As, equation of plane P4

is x + 3y + 6z = 7.

So on comparing with x + 3y + 6z = k, we get k = 7 (D) is correct.

(ii) We have P3

: x + 3y + 6z 1 = 0 and P4

: x + 3y + 6z 7 = 0

So, the least distance between planes P3

and P4

= 222 )6()3()1(

|)7(1|

=46

6 (B) is correct.

(iii) The equation of plane P4

in intercept form is67

z

37

y

7

x = 1.

So, area of triangle ABC =222222 accbba

2

1

= 49

36

49

36

49

9

49

9

4949

2

1

y

x

z

A(7, 0, 0)

0,

3

7,0 B

6

7,0,0

C

=36

1

369

1

9

1

2

49

=

63

46

2

49

=

36

4649. (C) is correctect. Ans.]


Sol. Any tangent to given parabola is

x secy tan = 1 ... (1)Let P(h, k) be the point of intersection of tangents to the parabola y2 = 4x,

so equation of chord of contact of (h, k) is ky = 2(x + h) ... (2)

As, (1) and (2) are identical, so

2

sec=

k

tan =

h2

1(on comparing)

sec =h

1and tan =

h2

k

As, sec2tan2 = 1

2h

1 2

2

h4

k= 1 4k2 = 4h2

x=-1

y

x=1

(0, 2)

(0, -2)

(-1, 0) (1, 0)Ox


12/18

MATHEMATICS

C d A P # 4

Locus of P(h, k) is

E : 4x2 + y2 = 4, where is an ellipse or1

x2+

4

y2= 1 ... (3)

(i) As, e2 = 14

1=

4

3 e =

2

3= sin 60 (A) is incorrect.

The foci of ellipse are (0, be) 3i.e., 3,0 (B) is correct.

Also, l (L.R) =b

a22

=2

)1(22

= 1 (C) is correct.

Note, that distance between vertices of ellipse E

= distance between (0,2) and (0, 2) = 4 (D) is correct.

(ii) The circle described on vertices of ellipse E as diameter is x2 + y2 = 4 .......(4)

As, circle in equation (4) intersects orthogonally the circle x2 + y2 4x 2y + k2 = 0,

so using condition of orthogonality, we get 0 = k24 k = 2 (C) and (D) are correct.

(iii) From above figure, the equation of common tangent to hyperbola x2 y2 =1 and

ellipse E can be x =1 or x = 1. (A) and (B) are correct. ]

PART-B

Q.1

[Sol.

(A) For,

2

x

logsin

2

2

1

to be defined,

1

2

x

log

2

2 1

2

x

2

1

2

1 x2 4 x [2,1] [1, 2] .......(2)

Also, for

x|x|1

log2 to be defined, | x |x > 0 x < 0 .......(2)

From (1) and (2), we get domain of f(x) = [2,1].

So, number of integers in domain of f(x) is two i.e.,

2,

1. Ans.

(B) Let I =

4

11

4

3

dx)x1(2

cos)1x(

Put (x1) = t dx = dt

So, I =

47

4

7

even

odd

dtt2

cost

= 0 Ans.


13/18

MATHEMATICS

C d A P # 5

Aliter : Let I = dx2

xsin1x

4

11

4

3

)P.B.I()II(

)I(

=

4

11

4

3

4

11

4

3dx

2

xcos

2

2

xcosx1

2

=4

11

4

32 2

xsin

4

8

3cos

4

7

8

11cos

4

72

=

83

cos4

7

8

3cos

4

72+

83

sin8

11sin

42

=

8

3sin

8

3sin

42 = 0. Ans.

(C) Focus of given parabola is (5, 2).

Now, any line through (5, 2) is (y2) = m (x5)

This will be a tangent to the circle (x7)2 + (y2)2 = 2,

if 2m1

m202

4m2 = 2 + 2m2 m = 1. Ans.

(D) As, | w | = 2 x2 + y2 = 4 .......(1),where w = x + iy

Let z = h + ik, then

z =w

1w (h + ik) = (x + iy)

iyx

1

h + ik = (x + iy)22 yx

)iyx(

= (x + iy)

4

)iyx( (As, x2 + y2 = 4)

h =3

h4x

4

x3 and k =

5

k4yy

4

5

As, x2 + y2 = 4 425

k16

9

h16 22 1

4

25

k

4

9

h22

.

Locus of P (z) is 1

4

25

y

4

9

x 22 ,

which is an ellipse whose diameter of auxiliary circle = 22

5= 5. Ans.]


14/18

MATHEMATICS

C d A P # 6

PART-C

Q.1

[Sol. Given, f (x) = x

0

ttx

0

tt dt)43(t3dt)43(3x

For maxima / minima, we have

f ' (x) = 0

)43(x3)43(3xdt)43(3xxxx

x

0

tt

f ' (x) = x

0

ttdt)43(3

f ' (x) = 3n2

1

l (32x

8 3x

+ 7)

f ' (x) =3n2

1

l(3x1) (3x7)

f ' (x) = 0 x = 0, log37

0 log 73

+ve veve + ve

sign scheme of f '(x)

x-axis x = log

37 is the point of minima.

Hence, 3a =7log33 = 7 Ans.]

Q.2

[Sol. Let z = r ei ; | z | = r ; arg z =

Now,

2

z

1z2 = 1 (squaring the given relation)

z

1z2

z

1z2 = 1

zz

1

z

z

z

z2zz4

= 1

4r2 + 2 (ei2 + ei2) + 2r

1= 1 4r2 + 2

r

1+ 4cos 2 = 1

2

r

1r2

+ 4 + 4 (1 2 sin2) = 1

2

r

1r2

+ 3 + 4 = 8 sin2

8 sin2 = 7 +2

r

1r2

Hence, 8 sin2 |min.

= 7 when 2r =r

i.e. r2 =

2

1. Ans. ]


15/18

MATHEMATICS

C d A P # 7

Q.3

[Sol. Given, wuvu

.........(1)

and vuw

.........(2)

Putting w

from equation (1) in equation (2), we get

vuuvu

vuvu

vuvuvu 2

0vu

)given(1u,As

vu

.

Now, wvuwvu

= )uvu(vu = uv)vu(vu

= uv0u|v|u 2 = 22 vu 3vand1u,As

= (1)2 (3)2 = 9. Ans.

Aliter vuw

taking dot product with v

on both sides

v

uw

= 2v

wvu

= 9. Ans.]

Q.4

[Sol. We have f '(x) = 3x2 + 6x + 4 + b cos xc sin x

Now, for f(x) to be one-one, only possibility is f '(x) 0 x R.i.e., 3x2 + 6x + 4 + b cos x c sin x 0 x Ri.e., 3x2 + 6x + 4 c sin xb cos x x R

i.e., 3x2 + 6x + 4 22 cb x R

i.e., 1x2x3cb 222 + 1 x Ri.e., 11x3cb 222 x R

Rx1cb 22 b2 + c2 1 x R. Ans.]


16/18

PHYSICS

Code-A Page # 1

SOLUTION

Q.1

[Sol. The second charge is kept just outside surface.

Flux will not change but another carge influences the shape. ]

Q.2

[Sol. mg(Rr) =10

7mv20

V = )rR(g7

10 ]

Q.3

[Sol. t =g

U2= 10 s ]

Q.5

Sol. x0

xI

=f2

xI=

0

2

x

f]

Q.6

[Sol. g = 3R

GMx]

Q.7

[Sol.21

100 42 =21

400 x02

x0

= 2 cm ]

Q.8

[Sol. Limiting frictionfmax

=Mgcos= 0.5 80 = 40 NMgsinT = 6050 = 10 Newtons


17/18

PHYSICS

Code-A Page # 2

Step-2 :Tl

2

Tl1

T /3.5l1

Tl3

T /3.5l3

Tl2

Fnet l2

= 0 ]

Q.Q.12 & 13

[Sol.

B 0~f = 6 rv

n 1

mgv

1

qE

6 rv2mg

v2

3

4r3 g = 6r v

1........ (1) qE =

3

4r3g + 6rv

2........ (2)

q =E

v

v1gr

3

4

1

23

]

Q.14

[Sol. q1

=

01.0

05.01 = 6

q2

=

01.0

04.01 = 5 ]

Q.16

[Sol. A B

t = 0 N0

N0

t0

= 3 days 2N N

2N = N0

(0.5) t0/

1

N = N0

(0.5)t0/

2

2 = (0.5) t0

21

11

0.51 = (0.5)

2

3

1

1 =1

3

2

1= 3 days ]


18/18

PHYSICS

C d A P # 3

Q.17

Sol. = Blv = 36

v =5.006.0

36

= 1200 m/s ]

Q.18

[Sol.m2

p2

=hc

= ~

hc = 2

2

m2h

=mc2

h= 831

34

103101.92

1060.6

=91

11 1011 m = 1.2 pm ]

Q.19

[Sol. P =dt

dQ=

dt

dmL ........ (i)

PV = nRT

dt

dV=

M

dt/dm

0P

RT......... (ii)

av0

=M

L/P

0P

RT

v0

= aMLP

PRT

0= 17 m/s ]

12 02 2012 xiii vxy paper ii code a sol

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