04 03 2012 xiii vxy paper ii final test code a sol

7/30/2019 04 03 2012 XIII VXY Paper II Final Test Code a Sol

1/18

13th VXY (Date: 04-03-2012) Final Test

PAPER-2

Code-A

ANSWER KEY

PHYSICS

SECTION-2PART-A

Q.1 D

Q.2 B

Q.3 A

Q.4 C

Q.5 A

Q.6 D

Q.7 A

Q.8 B

Q.9 D

Q.10 B

Q.11 A

PART-B

Q.1 (A) P,R,T

(B) Q,T

(C) P,S

(D) Q,R,S,T

PART-C

Q.1 1250

Q.2 0270

Q.3 0005

Q.4 0002

Q.5 0002

Q.6 0009

MATHS

SECTION-1PART-A

Q.1 C

Q.2 C

Q.3 B

Q.4 A

Q.5 B

Q.6 C

Q.7 B

Q.8 D

Q.9 C

Q.10 D

Q.11 A

PART-B

Q.1 (A) P

(B) S

(C) R

(D) T

PART-C

Q.1 0005

Q.2 0006

Q.3 0009

Q.4 0016

Q.5 0106

Q.6 0006

CHEMISTRY

SECTION-3PART-A

Q.1 A

Q.2 A

Q.3 A

Q.4 A

Q.5 B

Q.6 C

Q.7 A

Q.8 B

Q.9 C

Q.10 B

Q.11 A

PART-B

Q.1 (A) S

(B) P

(C) T

(D) Q,R

PART-C

Instruction for Q.1 & Q.2

Partial Marks will be awarded (+2) for correct

value of 'ab' (+2) for correct value of 'cd' and

(+1) as bonus if both are correct.

Q.1 ab = 05 ; cd = 05

Q.2 ab = 20; cd = 80Q.3 0004

Q.4 0002

Q.5 0001

Q.6 0006

Q.6 0006


2/18

MATHEMATICS

Code-A Page # 1

PART-A

Q.1

[Sol. Equation of required plane is (x + y1) + (2yz) = 0

or x + (2 + 1) yz = 0 ........(1)

If plane (1) is perpendicular to plane x + y + 4z = 9, so

1(1) + 1(2 + 1) + 4 () = 0 22 = 0 = 1

On putting = 1 in equation (1), we getx + 3yz = 1. Ans.

So, on comparing we get k = 1. ]

Q.2

[Sol. x3dx

dyyx dxx2dyy

cx2

y 22

Now (1, 2) satisfy it, 2

1 = c = 1, so y2 = 2x2 + 2or 2x2y2 + 2 = 0. Ans.]

Q.3

[Sol. Given,

x

5

22dt)t('f3t

5

28)x(fx

Differentiating both sides with respect to x, we get

x2 f '(x) + 2x f(x) =

5

28x2

3f '(x)

Put x = 5 and using f(5) = 0, we get

25 f '(5) + 0 = 28 53f '(5) 28f '(5) = 28 5 f '(5) = 5. Ans.]

Q.4

[Sol. Any tangent to given hyperbola, is

11

tany

4

sec2x

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

Let (x1, y

1) be the middle point of the chord of ellipse.

Its equation is1

y

4

x

1

yy

4

xx 212111 ........(2)

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

1

y

4

x

1

y

tan

x

sec221

2111

sec =21

21

1

y4x

x2

and tan = 2

121

1

y4x

y4


3/18

MATHEMATICS

Code-A Page # 2

As, sec2tan2 = 1, so

1y4x

y16

y4x

x422

121

21

221

21

21

Locus is (x2 + 4y2)2 = 4(x24y2). Ans.]

Q.5

[Sol. A A A A1

A2

, M M M1

M2

, C1

C2

C3 12 fruits

A A1

A2

M1

M2

C1

C2

C3 AA MM

4 4

70!4!4

!8!2

!2!4!4

!8

If child C1

receives A in the distribution of 8 fruits then he can not get AA he has to take MM

Distribution of AA & MM can be done by only 1 ways.Hence, required number of ways = 70 1 = 70.]

Paragraph for Question no. 6 to 8

[Sol. Put z = x + iy in 0zz8zz 2 y2 = 4x ......(1)

Also, 41z1z

(6,0)x

y

3

22,

3

2P

3

22,

3

2Q

SO(0,0)

0,

3

2R

0,

3

2M

13

y

4

x22

or 3x2 + 4y2 = 12 ......(2)

Note that L (x1, y

1) =

2

3,1 .

(i) PQRArea

PQSArea

=

RMPQ2

1

SMPQ2

1

=

3

43

16

3

2

3

23

26

= 4. Ans.(C) is correct.

(ii) centroid (G) ofLRS =

3

0023

,3

6321

G

L

2

3,1

S(6, 0)R

0,

3

2

=

2

1,

9

19Ans. (B) is correct.


4/18

MATHEMATICS

Code-A Page # 3

(iii) Equation of normal to y2 = 4x, is y = mx2mm3. As it passes through D(3, 0) so

m = 0, 1,1.

A(0, 0), B(1,2), C(1, 2)

So, area of triangle ABC = 2|

121

121

100

|2

1

.

Hence, R = 24

455

4

abc

= 2

5

. Ans. (D) is correct.]

Paragraph for question nos. 9 to 11

[Sol. Let g"(x) = Ax, A 0

g'(x) =2

Ax2

+ B

As,

22

1'g = 0 0 =

8

1

2

A+ B B =

16

A

g'(x) =16

A

2

Ax2

So, g(x) =16

Ax

6

Ax3

+ C

As, g(0) = 0 C = 0

g(x) =16

Ax

6

Ax3 .

Also, g(1) = 5 5 =16A

6A A = 48

Hence, g(x) = 8x33x,

(i) g(2) = 646 = 58. Ans.

(ii) As, g(x) = 8x33x

So, )x(gf = sgn (8x33x) =

,3

8x;1

3

8,0x;1

0,3

8x;0

0,38x;1

3

8,x;1

x

y

0,

3

0,

3

O

Graph of y = g(x)


5/18

MATHEMATICS

Code-A Page # 4

So, area enclosed between ordinates x = to x = , is 2

0

dx = 2.

O

3

8

3

8

x

y

y = 1(0, 1)y = 1

y =1y =1 (0,1)(iii)dx

dy+ g'(x) y = 1)x(g g'(x)

Integrating factor = eg(x).

Also, general solution is y eg(x) = dx)x('g1)x(ge)x(g

Put g(x) = t g'(x) dx = dt

= dt)1t(et

= t et + c = g(x) eg(x) + C

h(x) = y = g(x) + C eg(x)

As, h(0) = 1 1 = 0 + C C = 1 h(x) = g(x) + eg(x)

Hence. h(1) = 5 + e5. Ans.]

PART-B

Q.1Sol.

(A) 2 + sin = x2 + x

a. x R{0}

1 2 + sin 3

1 x2 + x

a 3 ......(1)

Let f(x) = x2 + x

a f(x) = a2

x

ax

2

.min)x(f = 2 a , at x =

x

a x2 = a

But 2 a 1 a 2

1 a

4

1......(2)

Now, maximum of

2

2

x

ax = 3.

A.M. G.M. 2

x

ax

2

2

2x

ax

2

3 a a

4

9

a

4

9,

4

1. Ans.]


6/18

MATHEMATICS

Code-A Page # 5

(B) Given,dx

dy= 12x

On integrating, we get

y = xx2 + c

As (2,2) lies on it, so c = 0

y = xx2

For point of intersection, x = xx2 x = 0, 1m

Now,

area = 3

32dxmxxx

m1

0

2

m1

0

32

3

x

2

xm1

=

3

32

3

32

3

m1

2

m133

(1m)3 = 64 = ( 4)3 m =3, 5. Ans.

(C) On squaring, we get

(1sin x) = (1sin x)2 (1sin x) (1(1sin x)) = 0

Either sin x = 0 or sin x = 1

x = n or x = (4m + 1)2

, where n, m I.

But only x =2

, , 2,

2

5satisfy the condition | x |

2

3]

(D)We have h(x) = )x(fgf

h'(x) = )x(fg'f g'(f (x)) f '(x)

Now, h'(2) = f '(g(f (2))) g'(f (2) f '(2) = f '(g(1)) g'(1) f '(2) = 2)2('f g'(1) = (4)2 4 = 64

Hence )2('h = 64 = 8. Ans.]

PART-C

Q.1

[Sol. ABC = 1531

b

2

x02

5x1

203

2

= 153

4x

3bx

26

2

=

4x..

.15bx5.

..24

2

Tr(ABC) =24 + 5bx15x24 =x2 + 5bx43

Tr(ABC) 18 x Rx2 + 5bx43 18 x Rx2 + 5bx25 10 x RD 0 25b24 (1) (25) 0

b24 0b [2, 2]

Number of integral values of b are 5.Ans.]


7/18

MATHEMATICS

Code-A Page # 6

Q.2

Sol. I =

3

1

2

2dx)x1(n

x

1l

I = 3

1

2

2dxx1n

x

1

2

1l

2I =

3

12

3

1

2

dxx1

x1x2

xx1n

l

2I = 311 xtan2)2(n

3

4n ll

2I = ln 263

21

I = ln 2123

1

2

1

. Ans.]

Q.3

[Sol. Given that p1, q

1, p

2q

2are in A.P.

(p2p

1)2 = (q

2q

1)2

(p2

+ p1)24p

1p

2= (q

2+ q

1)24q

1q

2

9

c4

9

b2

=

c

94

c

b2

81

9c4

81

b2 =

c

c36

c

b2

2

81

c36b2

=

c

c36b2

c2 = 81

c = 9 (As c =9 reject) think!. ]

Q.4

[Sol. We have f(x) = e + (1x)

x

1

dt)t(fe

xnl .......(1)

Differentiate both sides with respect to x, we get

f '(x) = (1

x)

e

x

nx

1l

(

1) + f(x)

f '(x)

f(x) =

e

x

n1x

1l

Multiplying both sides with ex, we get

ex f '(x)ex f(x) = ex

e

xn1

x

1l

)x(fedx

d x= ex

e

xn1

x

1l


8/18

MATHEMATICS

Code-A Page # 7

On integrating both sides with respect to x, we get ex f(x) = ex

dxx

1

e

xne x l

= ex + ex

e

xnl + f(x) = 1 +

e

xnl + ex .......(2)

Now, putting x = 1 in equation (1), we get f(1) = e .......(3)

Using (3) and (2), we get

= 1.

Thus, f(x) = 1 + ex +

e

xnl or f(x) = ex + ln x.

1Ox

y

Hence g(x) = x ln x ; Now A = 1

0

dxxnx l =4

1

A2 =

2

4

1

= 16. Ans.]

Q.5

[Sol. Two balls transferred from I to II can be 1 W and 1 B or both W or both B.

B1

: 1 W and 1 B From I II

25IIBag

34IBag

BWBalls

P(B1) =

27

13

14

C

CC=

21

12=

7

4

B2

: 2W from I II

P(B2) = 272

4

C

C

= 21

6

= 7

2

B3

: 2 B from I II

P(B3) =

27

23

C

C=

21

3=

7

1

Now, P (ball drawn from II is white) =

9

6

7=

63

24

63

43

+9

7

7

2

=

63

14

+9

5

7

1

=

63

5. ]


9/18

MATHEMATICS

Code-A Page # 8

Q.6

[Sol. C1

: 14

)2y(

9

)3x(22

A1 (6, 2) and A

2 (0, 2)

A (0,2)2 A (6,2)1(3,2)

23PAPA 21 .

Clearly, locus of P is a hyperbola whose A1A

2= 2ae = 6 foci are A

1and A

2and 2a = 23 .

e = 2 Locus of P is a rectangular hyperbola

a = b =2

3.

Equation of conic C2

is (x3)2(y2)2 =2

9.

D1 : Distance between foci of the conic is A1A2 = 6.

D2

: A1, A2 are the foci of the conic C2

Product of the perpendicular from A1 and A2 upon its any tangent is equal to b

2

i.e., 2

9

.

D3

: L =2

99 =

2

3.

3 2 L

3

Hence

2

3

21

D

DD= 6. Ans.]


10/18

PHYSICS

Code-A Page # 1

PART-A

Q.1

[Sol. mv2

L=

12

1ML22 .......(i)

mv = MV0

.......(ii)

V0

02/Lv0

=1 .......(iii) ]

Q.2

[Sol. t =80

Tt =

80

27.1cm = 0.015875 cm = 0.15875 mm

t = 0.159 mm ]

Q.3

[Sol. W = pdV = dVVC 3/2 P = CV2/3

=5C3 V5/3 =

53 (P

2V

2P1V1)

=5

3nR (T

2T

1) =

5

3 1

3

25 30 = 150 J ]

Q.4

[Sol. E goes from higher potential to lower potential

E =x

V

=

05.0

VV03

V

3= 400 0.05 = 20V ]

Q.5

[Sol.2

x=

l

l

1

x

2=

)2.0(1

2.0

l

l=

l

l

8.0

2.0

)8.0)(1(

)2.0(

ll

ll

= 1

l2 + 0.2 l = 0.8 + l20.8 ll

2l = 0.8l = 0.4 m

2

x=

6.0

4.0 x =

3

4

x

2=

)2.0(1

2.0

l

l

=l

l

2.1

2.0


11/18

PHYSICS

Code-A Page # 2

1 =)2.1)(1(

)2.0(

ll

ll

l20.2 l = 1.2 + l22.2 l

2l = 1.2

l = 0.6

x =4.0

6.02= 3 ]

Paragraph for question nos. 6 to 8

[Sol. Case-I : TFRest

OSC

Rest

Case-II : TF OSCRest

v

Case-III : TF OSCRest

u

(6) For PA

& PC K

A= K

C

belong to case-I or case-IIIFor case-III, frequency is more. P

Abelongs to case-III

PC

belongs to case-I

PB

belongs to case-II

I-C, II-B, III-A(7) C = 320 m/s f = 248 Hz

f ' =VC

C

f

256 =

V320

320

248 V = 10 m/s

(8) f'' =C

uCf

2

527=

320

u320 248 u = 20 m/s ]

Q.10

[Sol.2

1 v

r

2E

2

1

2

m(v0)2 0.5

2

1mv2 1 MeV ]


12/18

PHYSICS

Code-A Page # 3

Q.11

[Sol. Q = (mT

+ mH2m

D)C2

4 = (mT

+ 1.00782 2.0141) 931.5

4.294 103 = mT

+ 1.00784.0282

4.02821.00780.0043 = mT

= 3.0161 amu ]

PART-C

Q.1

[Sol. = B r2 =R2

I0

r2 =R2

0

R2

r2 =

2

20 rR4

ind

=2

20

R4

r

dt

d

iind

=)r2(

ind

=

20

8

r

dt

d= 1.25 AA ]

Q.2

[Sol.30

4+

60

1

=

30

13

4

=90

1

60

30Rv3

4=

90

1

60

1

v3

4=

60

32

v = 240 cm ]

Q.3

[Sol. r =21

22

21

vv

EE

=

sind

kq

cosd

kq

)sind(

kq

)cosd(

kq2

2

2

2

=

cossin

cossind

cossin

cossin44

44

=

cossin]cos[sindcossin

4

= 5 ]


13/18

PHYSICS

Code-A Page # 4

Q.4

[Sol. P = P0

+A

mg

mg

P A0

PAQ = ncpT =

VP1

Power =1

PAvv

i2R =1

(P

0A + mg)v

v =7

2

1010101010

350)2.0(45

2

m/s = 2 cm/s ]

Q.5

[Sol. T = 2g

nT

Tn=

l

l

2

T100

T=

l

l

2

l

l=

100

200= 2 ]

Q.6

[Sol. E =2

1

5.01

5.01

(1(0.5))2

4

11

=3

5.0 1.52 1.5

4

3=

16

5.4J

1 1000 T =216

5.4

T = 64

9

103

C ]


14/18

CHEMISTRY

Code-A Page # 1

PART-A

Q.1

[Sol. Given EnE

1= 108.9 eV

eV9.1081

Z6.13

n

Z6.13

2

2

2

2

13.6 Z2

2n

11 = 108.9 eV ....(i)

Given n

1= 2r

1 2

11n r22

1

r2

n

r2

(by 2r

n= n)

11

21 r4r2n

nr2

n1 = 2

n = 3Putting n back in eq. (i)

Z = 3 ]

Q.2

[Sol. Co(CO)4

EAN = 27 +2(4) = 35

(i) Co(CO)4

ductionRe 4)CO(Co EAN = 27(1) + 2(4) = 28 + 8 = 36

(ii) Co(CO)4

Dimerises Co2(CO)

8EAN =

2

)bondtodue(2)8(2227 = 36

(iii) Co(CO)4

Oxidation 4)CO(Co EAN = 271 + 2(4) = 34Not gain extra stability by oxidation ]

Q.3

[Sol. (A) N2is evolved on heating (B) gives red colour as it is having N as well as S (D) Inorganic compound

do not give Lassaigne's test for nitrogen ]

Q.4

[Sol. [Cr(H2O)

5NO

2] Br

2

Isomer

linkage [Cr(H2O)5ONO] Br2

Pentaaquanitrito-'O' chromium (III) bromide.]

Q.5

[Sol. Na 4[Fe(CN)5NOS]

Fe+2d6 sys. Hybridisation of Fe in complex is d2sp3 ]


15/18

CHEMISTRY

Code-A Page # 2

Q.6

[Sol. Cr2O

72 + Fe2+ Fe3+ + Cr+3 E = +ve

Cr2O

72 + C

2O

42 Cr3+ + CO

2E = +ve

MnO4 + Fe+2 Mn+2 + Fe+3 E = +ve

MnO4+ C

2O

42 Mn+2 + CO

2E = +ve

Cr2O

72+ Cl Cr3+ + Cl

2E =ve

MnO4 + Cl Mn+2 + Cl

2E = +ve

MnO4 can't be used to estimate FeC2O4 in presence of dil.HCl as it will also oxidise Cl to Cl2.]

Q.7

[Sol. (nf= 6) (n

f= 3)

Cr2O

72 + FeC

2O

4Cr3+ + Fe3+ + CO

2

50 ml

0.1 M

5 millimoles

5 6 Milliequivalents

Meq of Cr2O72

= Meq of feC2O4 = 30

Millimoles of FeC2O

4=

3

30= 10 ]

Q.8

[Sol. Millimoles of KMnO4

used = 50 0.1 = 5

Meq

of KMnO4

used = 5 5 = 25 = Meq

of Cl + Meq

of FeC2O

4

(nf= 5)

Meq of Cl = Meq of Cl2 = 3.5 2 = 7

(nf= 2)

Meq

of FeC2O

4= 257 = 18

(nf= 3)

Millimoles of FeC2O

4=

3

18= 6 ]

Q.9

[Sol. Stability of radical = c > a > b

Bond energy = b > a > c]

Q.10

[Sol. Stability h > g > dNspofCspof 3

ENEN

Acidic strength = h > g > d ]


16/18

CHEMISTRY

Code-A Page # 3

Q.11

[Sol. Kb

f > e > d ]

PART-C

Q.1

[Sol. after adding of NaOH to weak monoprotic acid, pH is in the acidic medium i.e. NaOH added is

quite lesser

After addition of 50 ml NaoH 0.1 MHA + NaOH NaA + H

2O

(50 ml, xM) (50 ml, 0.1 M)

50 x millimoles 5 millimoles

Left mmoles in solution (50x5) 0 5

pH = pKa

+ log]HA[

]A[

4.699 = pKa

+ log

5x50

5....(i)

After adding of 75 ml NaOH 0.1 M

Left millimoles of HA = 50 x7.5

Millimoles of NaA formed = 7.5

pH = pKa

+ log]HA[

]A[

5 = pKa

+ log

5.7x50

5.7....(ii)

On subtracting (ii)(i)

0.3010 = log

5

)5x50(

5.7x50

5.7= log 2

on solving x = 0.3

pKa

= 5 ab = 05

On adding 7.5 millimoles of HCl

NaA + HCl NaCl + HA

15 Millimioles 7.5 Millimoles

7.5 Millimoles 0 7.5 Millimoles

pH = pKa

+ log]HA[

]A[ = pKa

+ log5.75.7

pH = pKa

= 5 cd = 05 ]


17/18

CHEMISTRY

Code-A Page # 4

Q.2

[Sol. A2B

3(aq) 2A+3 (aq) + 3B2 (aq)

Initial conc. a 0 0

at time t = 10 min ax 2x 3x

a 2

1 + 4x 6

3

a

x4a

a = 2x x =2

a

t1/2

= 10 min. t3/4

= 2 t1/2

= 20 min.

ab = 20

at t = 20 min

A2B

3(aq) 2A+3 (aq) + 3B2 (aq)

a

t = 20 min4

a

2

a3

4

a9

= 2 t1/2

a 2

8a44

a9

2

a3

4

a

at t = 20 min osomotic rise = 8 mm of Solution

= h g = 8 103 1 103 10

= 80 Pascal

cd = 80 ]

Q.3[Sol. [Cr(NH

3)6]Cl

3 [Cr(NH

3)6]3+ + 3Cl

One complex cation + 3 simple anion

Total = 4 ]

Q.4

[Sol. ideal

=3

A

0

a

N

MZ

24

24

310A

10125

1067.125.314

)10500(

N25.314

= 1.67 g/ml

actual

= 1.6075 g/ml

Differenve due to Schottkey effect = 1.671.6075 0.0625 g/ml

= 0.0625 103 g/L

= 62.5 g/L


18/18

CHEMISTRY

C d A P # 5

= L/Mole25.31

5.62

= 2 mole/L ]

Q.5

[Sol. [Cr(H2O)

6]3+

Cr3+ 4s0 3d3

Hund's rule is not followed then unpaired electron is one ]

Q.6

[Sol. [M(AB)a2bc] n

cbyreplaceb [M(AB)a2c

2]n

ma

b

aA

Bc

m

a

ba

A

Bc

ma

c

a

A

Bc

m

c

a

aA

Bc

mb

c

aA

Ba

m

b

c

a

A

Ba

mc

a

A

Ba

c

ma

c

A

Bc

a

ma

c

a

A

Bb

m

c

a

aA

Bb

ma

b

A

Bc

aG. I. 4 S.I. = 6

7 G.I. Total change in

12 S.I. Reduce

dno.of

G.I.isome

rsbythese

S.I. = 126 = 0006 ]

04 03 2012 xiii vxy paper ii final test code a sol

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