rt solutions-04!03!2012 xiii vxy paper i final test code a sol

7/30/2019 Rt Solutions-04!03!2012 XIII VXY Paper I Final Test Code a Sol

1/19

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

PAPER-1

Code-A

ANSWER KEY

PHYSICS

SECTION-2PART-A

Q.1 B

Q.2 B

Q.3 A

Q.4 B

Q.5 C

Q.6 C

Q.7 D

Q.8 A

Q.9 D

Q.10 C

Q.11 A,C

Q.12 C

Q.13 B,C

Q.14 B,D

Q.15 C,D

PART-C

Q.1 0050

Q.2 0006

Q.3 0006

Q.4 0400

Q.5 1000

Q.6 0003

MATHS

SECTION-1PART-A

Q.1 C

Q.2 C

Q.3 D

Q.4 C

Q.5 A

Q.6 B

Q.7 B

Q.8 C

Q.9 A

Q.10 C

Q.11 B,C,D

Q.12 A,C,D

Q.13 B,C

Q.14 A,B,C

Q.15 B,C

PART-C

Q.1 0004

Q.2 0055

Q.3 0025

Q.4 0019

Q.5 0641

Q.6 0020

CHEMISTRY

SECTION-3PART-A

Q.1 B

Q.2 C

Q.3 B

Q.4 B

Q.5 C

Q.6 A

Q.7 CQ.8 B

Q.9 A

Q.10 C

Q.11 A,B,C

Q.12 A,B,C

Q.13 B

Q.14 A,B,C,D

Q.15 A,C

PART-C

Q.1 0050

Q.2 0009

Q.3 Bonus

Instruction for Q.4

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.4 ab = 20 and cd = 28

Q.5 0006

Q.6 0094


2/19

MATHEMATICS

Code-A Page # 1

PART-A

Q.1

[Sol. I =

1

1

xxxx dx6321nx l

Using King

I =

1

1

xxxxdx6321nx l

2I =

1

1

xxx

xxx

dx6321

6321nx l =

1

1xxx

xxxx

dx1236

63216nx l

(Multiplying Nr and Dr by 6x)

=

1

1

2 dx6nx l

2I = 2 ln 6 1

0

2 dxx =3

2ln 6 I = 3

1ln 6. Ans.]

Q.2

[Sol.

Let g(x) = ex f(x)

g'(x) = ex (f '(x) f(x)) > 0 Rxxx0

gI

g(x) is increasing on R

g(x) > g(x0) 0xx e

x f(x) >

0

0

x)x(fe 0

f(x) > 0 0xx [given f (x0) = 0]

Q.3[Sol. Since f(x) is continuous at x = 0, so at x = 0, both left and right limits must exist and both must be

equal to 3.

Now,2x

5xcos)xsinx1( =

2

2

x

.....x2

)5(

(Using series expansion of sin x and cos x)

If )x(fLim0x

exist, then + + 5 = 0 and )x(fLim0x

= 3 2

= 3

= 1, =

4.

Since, 3x

xx1Lim

x

1

2

3

0x

0

x

xxLim

2

3

0x

Now, )1(x1Lim x1

0x

= e = 3 e2 = 9

Hence, 2 + 2 + 2 + e2 = ( 1)2 + ( 4)2 + (0)2 + (3)2 = 1 + 16 + 0 + 9 = 26. Ans.]


3/19

MATHEMATICS

Code-A Page # 2

Q.4

Sol. Domain is x [1, 1]Given, sin1x = 2tan1x

2

1

x1

xtan =

2

1

x1

x2tan

2x1

x2

= 2

x1

x2

x = 0 or (1 x2)2 = 4 (1 x2) (1 x2) (3 + x2) = 0 x = 1, 0, 1 .......(1)

tan1 )1x(x + cosec1 2xx1 = 2

x(x 1) 0 x x2 0 x(x 1) = 0 x = 0, 1 .......(2)Now, (1) (2) gives x = 0, 1Hence, number of common solution are 2. Ans.]

Q.5

[Sol. Equation of tangent at M

2

1,3 is 3 x + 2y = 4

Also, distance of PQ from (0, 0) =7

4O

x

y

P

QM

x + y = 52 211

y

4

x 22

2

1,3

Now, PQ = 27

165 = 2

7

19

Area of (OPQ) = 719

27

4

2

1

= 7194

. Ans.]

Q.6

[Sol. f(x) = sin1x + cos1x + tan1

x

1+ tan1(x)

x [ 1, 0) f(x) =22

= 0

y

(1, )

(1,0)(1,0)x

(0, /2)

x (0, 1] f(x) =

22Required area (shaded region)

= 2 2

122

1

sq. units. Ans.]


4/19


5/19

MATHEMATICS

Code-A Page # 4

(ii) Acute angle between L1

= 0 and P1

= 0 = P2

(x-axis)

cos =001109

0)1(0013

00,1,:axis-xofsDr'

10,3,:0LofsDr' 1

cos =10

3 cot = 3 = cot1 3 = cot1

= 3 Ans.(iii) Image of the point (5, 0, 3) in the plane P

2(x-y plane) is (5, 0, 3)

Point of intersection of the line L1 = 0 with x-y plane is -(3l + 2, 0 l + 4) l + 4 = 0 l = 4 (14, 0, 0)Now line passing through (5, 0, 3) & (14, 0, 0) is

3

z

0

y

9

14x

1

z

0

y

3

14x

Ans. ]

Q.11

[Sol. As, | cc1 cc

2| = | (r + r

1) (r + r

2) | = constant

where | r1 r

2| < c

1c

2

locus of C is a hyperbola with foci c1 and c2 i.e., (4, 0) and (4, 0).

Also, 2a = | r1 r

2| = 2 a = 1

Now, e =a2

ae2=

2

8= 4

C2C1

(4, 0)(4, 0)

C

So, b2 = 12 (42 1) = 15

Hence, locus of centre of circle is hyperbola, whose equation is15

y

1

x22

= 1.

Now, verify the options. Ans.]

Q.12Sol. Coordinates of B (p, q p2)

Coordinates of C

a

b1,

a

b2

Since mid point of B and C is origin. Hence2

a

bp

= 0 ap + b = 0 (D)

Similarly q p2 +a

b1

2

= 0 (C)

Since a > 0 and abscissa of point C is positive. Hencea

b> 0 b is negative

Hence ab < 0 (Option B is false).

Sum of the roots of equation (x2 + 2px + q) (ax2 + 2bx + 1) = 0 is

a

b2p2 = 2

a

bp =

a

2(ap + b) = 0 (A) is true. ]


6/19

MATHEMATICS

Code-A Page # 5

Q.13

[Sol.514/bin

10

2

2 2x

1x

=20

x

1x

[11th, 02-01-2011,P-2, PQ]

Tr + 1

= 20Crx20 r(1)r

rx

1= 20C

rx20 2r(1)r

20 2r = 0 r = 10So, term independent of x = 20C

10. Now verify alternatives. Ans.]

Q.14

[Sol. f(x) = x2 1

0

1

0

1

0

22dtxdt)t(fx2dt)t(f

f(x) = x2 + A + 2Bx x2 = A + 2Bx

A = 1

0

1

0

22 dtBt2Adt)t(f

B = 1

0

1

0

dtBt2Adt)t(f

f(x) = 3x 3Now verify all the option. Ans.]

Q.15

[Sol.

(A) disjoint mutually exclusiveTwo mutually exclusive events E and F with P(E) and P(F) > 0 cannot be independent

(B) P(E) + )FE(P + )FE(P = P(E) + P(F) P(E F) + )FE(P = 1 Exhaustive

(C) FEP > P(E) or)F(P

)FE(P > P(E) ......(1)

Now,)E(P

)EF(P >

)E(P

)F(P)E(P> P(F) (C) is correct.

(D) Every element of set of positive integer can be of the form 6n, 6n + 1, ......., 6n + 5.

But favourable are 6n + 1 or 6n + 5.

So, required probability =6

2=

3

1. Ans.]


7/19

MATHEMATICS

Code-A Page # 6

PART-C

Q.1

[Sol. Applying cosine rule, 2ab cosC = a2 + b2 c2 22 22 cos 8

= 2 + 2 + 2 c2

using cos8

=

2

22

2

2222

= 4 + 2 c2 c2 = 2 2 c = 22

Hence abc=

22222 = 2 (abc)2

= 4 Ans.]

Q.2

[Sol. A =

ww0b11

ca1

| A | = w(1 + b a c) 0 (for non-singular matrices) 1 + b a + cNow total matrices = 4 4 4 = 64 and 1 + b = a + c

2

2

4

1

i

i

1i

1

i

i

1i

i

i

1

1

i

i

1

1

1

111

cab

Total possibility = 9Total possibility of 1 + b = a + c are 9Hence 64 9 = 55 Ans.]

Q.3

[Sol.22

221

1

at

at2

at

at2 = 1

t1t2

= 4

M

T

90

Q(t )2

P(t )1

at t , a(t + t )1 2 1 2

O

N

x

y

(h,k)

Now T a t1t2, a(t

1+ t

2)

and N a )2tttt( 212221 , a t1t2(t1 + t2)Hence 2h = a((t

1+ t

2)2 + 2)

& 2k = a(t1

+ t2) (1 t

1t2)

2k = 5a(t1

+ t2)

t1

+ t2

=a5

k2

2h =

2

a5

k22

2h =

2

22

a25a50k4

50ah = 4k2 + 50a2

25ah = 2k2 + 25a2

25 2(h 2) = 2k2 or y2 =2

25(x 1)

Hence latus rectum =2

25 2L = 2

2

25= 25. Ans.]


8/19

MATHEMATICS

Code-A Page # 7

Q.4

[Sol. Bowl A

6R

4B

5 ballsPut inBowl B

Let A : One ball drawn from the bowl B found to be blue.

B1

: 1 R + 4 B from bowl A to B.

B2


B3


B4 : 4 R + 1 B from bowl A to B.B

5: 5 R + 0 B from bowl A to B.

P (B1) =

510

44

16

C

CC=

510C

6; P (B

2) =

510

34

26

C

CC=

510 C

60; P (B

3) =

510

24

36

C

CC=

510 C

120

P (B4) =

510

14

46

C

CC=

510C

60; P (B

5) =

510

04

56

C

CC=

510 C

6

1BAP = 54

; 2BAP = 53

; 3BAP = 52

; 4BAP = 51

; 5BAP = 0

ABP 2 =

5

1i

ii

22

BAPBP

BAPBP

ABP 2 =124836

5

24

36

=

1435

2

3

=

5

402

3

= 4215

=14

5

n

m

Hence (m + n) = 19. Ans.]

Q.5[Sol. log

2| z 3 | > log

2| z 3 | > 0

log2| z 3i | > log

2| z 3 | (z 3, 3i)

put z = x + iy

| z 3i | > | z 3 |

| x + i (y 3) | > | x 3 + iy |

x2 + (y 3)2 > (x 3)2 + y2

x > y

| amp ( z ( 1 + i)) | 4

(3, 0)

(5, 0)

(0, 2)

(1, 1)

A

| z | = 5

O

B

(0, 5)

x

y

x = y

y > x

x > y

4 amp (z (1 + i))

4

Region enclosed by z satisfying all the three given inequalities is sector AOB in which point (3, 0) and the

points on the line segment OB are not included.

Area of the sector AOB = Area of quater circle =4

25=

16

625

a + b = 625 + 16 = 641. Ans.]


9/19

MATHEMATICS

Code-A Page # 8

Q.6

[Sol. If and are the roots of x2 + t2x 2t = 0, then we have + = t2 and = 2t

so that 2

2

2

22

)(

2)(

)(

=

t

1

4

t2 and

t2

1

t4

11

)(

122

.

Now, I = dx11

x1

x

2

1

22

= dxt2

1

t4

1x

t

1

4

tx

2

1

2

22

= 3

t4

3

8

t32

2

.

Now 2

2

t4

3

8

t3

32

6=

4

23(Using A.M. G.M.)

Hence, I 8

9+ 3

Hence a = 3, b = 9 and c = 8. a + b + c = 20. Ans.]


10/19

PHYSICS

Code-A Page # 1

PART-A

Q.4

[Sol. P1A = P

2A + mg

P1

> P2

P1 P2 ]

Q.5

[Sol. F = qE = 5 106d

9= 1.5 104

d = 4

6

105.1

1045

= 0.30 m ]

Q.6

[Sol.

y

x+ve x-direction. ]

Q.7

[Sol. IA

= I + 4I + 2I42 cos 2

= 5I

IB

= I + 4I + 2I42 cos = I

IA

IB = 4I ]

Q.8

[Sol. n = 4 3 E = 3eV

=3

1242= 414 nm ]

Q.9

[Sol. E = exactly the energy difference ]

Q.10[Sol. E

4 2 = 8eV

E2 1 = 10 eV

< 8eVE

3 2 = 5eV

> 5eV ]


11/19

PHYSICS

Code-A Page # 2

Q.11

[Sol. Current leads the voltagei VR

V0

V VC L

V0

= 2R

2CL

V)VV(

= 25100 = 55 V

P.F. =0

R

V

V=

55

5=

5

1]

Q.12

[Sol. T2 =GM

42

R3

T =GM

2R3/2 ]

Q.13

[Sol. r = 90I

r

II sin I = 1 sin r = cos I = cot I

sin QC

= 1 sin 90

sin QC

= 1

sin QC

=

1= tan I

QC

= sin1 (tan I) = sin1 (cot r) ]

Q.14

[Sol. v = 22

2

AA

=

2

3 v

max

a = 2x =2

A2

=2

amax

]

PART-C

Q.1

[Sol. 6rv = B = 3

4

r3

PL g

=9

2r2

v

gPL

=9

2

7.0

100075.1)9.0(2

= 50 poise ]


12/19

PHYSICS

Code-A Page # 3

Q.2

[Sol. f = mg 8

R3cos 60

30

60

mg

3R/8

30

N1

N2

60

N2

= f

N1

= mg

f =16

mg3

fN1

16

3 32 = 6 ]

Q.3

[Sol. q = q1

11CR

t

e

= q2

22CR

t

e

2

1

q

q=

2211 CR

1

CR

1t

e

R = 6 ]

Q.4

Sol. PL 6 102 g = 600 g

mg + 600 g = PL 1000 g

m = 1000600 = 400 gm

Q.5

[Sol.dt

dE=ATT4

T4

= )1067.5()05.0(4

)1067.5(

82

2

= 1012

[A = 4r2

]

T = 1000 K ]

Q.6

[Sol. L1

= L2

=2

L= 9 105 H

R1

= R2

=2

R= 3

eqL

1

=1L

1

+2L

1

Leq = 2

L1

= 2

9

10

5 H

eqR

1=

1R

1+

2R

1 R

eq=

2

R1

= 1.5

=R

L=

2

3

102

9 5= 3 105 sec. ]


13/19

CHEMISTRY

Code-A Page # 1

PART-A

Q.1

[Sol. (B) Trigonal also called Rhombohedral

= = 90 and a = b = c ]

Q.2

[Sol. (A) A

Hybridisation : sp3d

Number of angles at 90 = 12

2

x x

x x

x

x

90

90

(B) A

Hybridisation : sp3

d2

Number of angles at 90 = 12

x x

x x

x

x

(C) A

Hybridisation : sp2

dNumber of angles at 90 = 4

x x

x x90

(D) A

x

x

xx

Hybridisation : sp3d

Number of angles at 90=6

x 12090

]

Q.3

[Sol. Cl C CH HOCCH

Na OC CONa+ +

Na OCCH OH+

2

Cl CCH

O O

OO

O O

O

Omorereactive

OH

O

OH

NaOH Cannizaro

+

]

Q.4

[Sol. (B) 0.3 M K2Fe[Fe(CN)

6] will give maximum number of ions in the solution. ]


14/19

CHEMISTRY

Code-A Page # 2

Q.5

[Sol.

423 SOHHNO

ArN

S,

ONaCH3

]

Q.6[Sol. (A)

7N15 n/p = 8/7

n/p > 1 so7N15 will be -emitter ]

Q.7

[Sol.

OH OH

H HH

OH H

CHOH2

O

OOH

H OH

H OH

HO

H

CHOH2

Glycosidiclinkage

Due to this it will be reducing sugar the ring can open into aldehydic

form and it can undergo mutarotation.]

Q.8

[Sol.

ClBeCl ClBeOHH O2sp

+slow HCl

HCl

FastBe

B

Cl

Cl

Cl

OH

OH2

OH2

+

+

H O2

Be(OH)2[Be(OH) ]2 n or

white ppt

OH[Be(OH) ]4

2

Beryllate ion is formed in basic medium

]

Q.9

[Sol. BCl3

+ 3H2O H

3BO

3+ 3HCl

H3BO3 + 2H2O H3O+ + 4)OH(B

During the hydrolysis of BCl3the medium becomes acidic due to formation of HCl hence in the presence

HCl feasibility of second reaction ie formation of 4)OH(B is least.]


15/19

CHEMISTRY

Code-A Page # 3

Q.10

[Sol. (i) Under ordinary condition, H2O molecule can not approach to the vaccant antibonding M.O. of

ClCl bond (yet having low energy) due to steric crowding over central carbon atom.

(ii) CCl bond dissociation energy is not very high as its magnitude is less than that of CF. ]

Q.11

[Sol. (A) K4[Fe(CN)

6] 4K+ [Fe(CN0

6]4

Na2HPO

3 2Na+ + HPO

32

higher negative charge over [Fe(CN)6]4so it can cause more coagulation in a basic dye.

(B) Starch aqua-sol is a lyophilic sol so it can be used as protective colloid.

(C) Freundlich adsorption isotherm fails at high pressure ]

Q.12

[Sol. (A) O O O

O O

O O

Si Si(B) O O O

O O

O O

Cr Cr

(C) O O O

O O

O O

S S

(D) or

Si

SiSiO

O

OO

O

OO

O O

]

Q.13

[Sol. H

H

Me

Me

It can show GI It do not posses COSIt posses POS It can show OI (it is meso isomer) ]


16/19

CHEMISTRY

Code-A Page # 4

Q.14

[Sol. (A) Z =C

CC

RT

VP=

8

3

Z < 1

attractive forces are dominating

(B)TC

T1

T2

(L+G)Gas

Liquid

P

V

Liquid and gas can not be distinguished above critical temperature.

(C) At very high pressure , Z > 1

1)ideal(V

)real(V

m

m

Vm

(real) > Vm

(ideal)

(D) For y-Intercept, X = 0

PVmT

P

When PO

Gas shows ideal behaviour

PVm

= RT

RT

PVm

y = R ]


17/19

CHEMISTRY

Code-A Page # 5

Q.15

[Sol. (A) & (C) Mozingo & WolfKishner is applicable

(B) N2H

2donot effect C

||O

(D) Ring expension take place ]

PART-C

Q.1

[Sol. Mole of H2O2 at t = 0 1

Kt

693.0

2/1

t1/2

= 10 sec.

Number of half lives in 20 sec. =10

202

Mole of H2O

2remaining after 20 sec. =

4

1

H2O

2+ 2H+ + 2I I2 + 2H2O

(Reactant)H(Product)HH ofofoReaction= [60 +2 (290)][200 + 2 (60)]

=520 + 320 200 kJ

Energy released when4

1mole H

2O

2reacts

200 4

1

50 kJ Ans. ]

Q.3

[Sol.O

H

O

O

O

87

6

5

4

3

2

1

+HCHO

O

O

O

O

O

H

H

HCHO

+ CH CHO3

8

7

6

5

4

3 21

]


18/19

CHEMISTRY

Code-A Page # 6

Q.4

[Sol. Ag2CrO

4+ AgNO

3

0.5 litre 0.5 litre

saturated solution 2 106 M

Ag2CrO

4 2Ag+ + CrO

42

2S S

When mixed with 0.5 litre AgNO3, total volume1 litre

AgNO3Ag+ + NO

3

2 106 2 106

In the solution:

[Ag+] =2

S21026

= [S + 106]

[CrO42] =

2

S

[NO3] =

2

1026

= 106

Given :2

10

2

S 6

S = 106

KSP

[Ag2CrO

4] = [Ag+]2 [CrO

42]

= [106 + 106]22

106

= 4 10122

106

= 2 1018

ab = 19

18

10

102

= 20

In the solution we have

Ag+, CrO42, NO

3, H+, OH

M

1000K]Ag[

Agm

;

6

Ag

102

1000K50

Ag

K = 107 ;M

1000K]CrO[

24CrO2

4m

2/10

1000K100

6

CrO24

; 0.5 107 = 24CrO

K

6

NO

3m10

1000K]NO[ 3

; 3NO

K = 0.75 107


19/19

CHEMISTRY

7H

m10

1000K]H[

; 350 = 7H

10

1000K

H

K = 0.35 107

7

OHm

10

1000K]OH[

; 7OH

10

1000K200

OHK = 0.2 107

K = 107 + 0.5 107 + 0.75 107 + 0.35 107 + 0.2 107

= 2.8 107

= 28 108

cd = 28 ]

Q.5

[Sol. Complex Number of geometrical isomers [CoCl2Br

2]2

0

[Rh(en)3]3+ 0

[Cr(en)2Br

2]+ 2

[Pt(en)Cl2] 2

[Co(NH3)3(NO

2)3] 2

[Pt(NH3)2(SCN)

2] 2

[Cr(NH3)2

Br2Cl

2] 5 ]

Q.6

[Sol. Isoelectric pH of a aminoacid =2

21

aa pKpK

8.2 =2

7 2apK 2

apK = 16.47 = 9.4 ]

rt solutions-04!03!2012 xiii vxy paper i final test code a sol

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