Rt Solutions-06!11!2011 XII ABCD Paper II Code A

7/30/2019 Rt Solutions-06!11!2011 XII ABCD Paper II Code A

1/19

12th ABCD (Date: 06-11-2011) Review Test-6

PAPER-2

Code-A

ANSWER KEY

CHEMISTRY

SECTION-2

PART-A

Q.1 C

Q.2 A

Q.3 B

Q.4 C

Q.5 C

Q.6 C

Q.7 D

Q.8 C

Q.9 B

Q.10 A,B,C,D

Q.11 A,B,C,D

Q.12 C,D

Q.13 A,B,C,D

PART-B

Q.1 (A) P,

(B) R,S

(C) Q,R,S

PART-C

Q.1 0007

Q.2 0005

Q.3 0040

Q.4 0004

Q.5 9260

MATHS

SECTION-3

PART-A

Q.1 A

Q.2 Bonus

Q.3 C

Q.4 C

Q.5 D

Q.6 D

Q.7 B

Q.8 A

Q.9 C

Q.10 A,C,D

Q.11 A,B,D

Q.12 A,B,C

Q.13 B,C,D

PART-B

Q.1 (A) Q

(B) P

(C) S

PART-C

Q.1 0003

Q.2 0001 or 0006

Q.3 0086

Q.4 0040

Q.5 0003

PHYSICS

SECTION-1

PART-A

Q.1 D

Q.2 D

Q.3 A

Q.4 B

Q.5 D

Q.6 C

Q.7 A,C,D

Q.8 A,B

Q.9 A,B

Q.10 B,D

Q.11 A,C,D

Q.12 A,C,D

Q.13 A,B

PART-B

Q.1 (A) R

(B) P,R

(C) R

PART-C

Q.1 0004

Q.2 0021

Q.3 0002

Q.4 0001

Q.5 5880


2/19

PHYSICS

Code-A Page # 1

PART-A

Q.1

[Sol. La

= Lb

mvar = 3mv

br

b

a

v

v= 3 : 1 ]

Q.2

[Sol. =E

hc=

1.1

1240nm ~ 1100 nm infra red. ]

Q.3

[Sol. AB = CD = T

T

= v

wave]

Q.4

[Sol. Fbottom

= L2 g h

hF

side= g

2

h hL

2

Lgh2

=6

1 ghL2

h =3

L= 20 cm ]

Q.6

[Sol.c

QiR = 0

i

Q= RC = constant. ]

Q.8

[Sol. (A) no net change in p. (B) no net change in p

(C) p along x-axis. (D) p along x-axis. ]


3/19

PHYSICS

Code-A Page # 2

Q.9

[Sol. = ( 1)A = (1.5 1) 180

9.0 =

400

rad.

dt

dN=

hc

= 834

9

1031063.6

102212.7

= 8 1018 /sec.

py = dtdN

h

sin

= 8 1018 9

34

10221

1063.6

400

= 6 1011 N ]

Q.10

[Sol. =f

v=

850

340=

5

2= 0.4 m

L =4

,4

3.......... = 10 cm, 30 cm, 50 cm etc.

l + e =4

3l = 2 cm ]

Q.12

[Sol. (A) EKE

L+ E

LE

N= E

KE

W= E

K. K

rray

= EKEM + EMEN ]

Q.13

[Sol.1

2

3

l v =2

4

5

l v

1

2

l

l

=6

5]

PART-B

Q.1

[Sol. Central maxima is at the point wherex = 0.

C

I

Here x = ( 1)t d sin = 0

sin= 1d

t)1( (Not possible)

So central maxima does not exist in part A & C. ]


4/19

PHYSICS

Code-A Page # 3

PART-C

Q.1

[Sol.X

4

4

9V

1

x = P.d. across 4 = 4V ]

Q.2[Sol. Q = U + W

Q W = UmL PV =U1 540 4200 105 (VS 10

3) = U

VS

=P

nRT=

M

m RT

= 31018

1

3

25

105

373=

216

373m3

= 2.268000 105

216

373= U

= 105

216

37368.22 ~ 21 105 J ]

Q.3

[Sol. k BjBiBB321

4 109 v1 kBjBiBk45sinj45cos 321 = iF1

321BBB2

1

2

1

0

kji

=

2

B2

B 23 i + j2B1

k2

B1

B1

= 0 kBjB32

2F

= q 2 104 kBjBi32

= q 2 104 jBkB32

B2

= 0

F2

= 4 105 = 4 109 2 104 B3

B3

= 2T ]

Q.4

[Sol. Initially, only m2oscillates

m1 m2 t =2

T=

k

m2 =

25

next both m1

and m2

oscillate about their CM.

t' =2

'T=

k

=

k)mm(

mm

21

21

= 210


5/19

PHYSICS

Code-A Page # 4

m2

=50

k= 3kg

)m3(

3m

1

1

150 =

200

1

4m1

+ 3 + m1 m

1= 1 kg ]

Q.5

[Sol. B = 0ni B =

0

Ni N = i

B

0

L = N 2r = ir2B

0

=

2104

1001.024.1168.07

= 5880 m ]


6/19

CHEMISTRY

Code - AA Page # 1

PART-A

Q.1

[Sol. Dissociation of water will be maximum when [H+] = [OH] at 25C

[H+] = [OH] = 107 pH = 7 ]

Q.2

[Sol. Compouds which produce H+ ion in solution is called Bronsted Acid

(A) H[Co(CO)4

] H+ + [Co(CO)4

]

(B) LiAlH4 Li+ [AlH4]

It provides hydrides [H]

in the solution. ]

Q.3

[Sol. (I)

MeS Br

NGPS

KOHaq

N

MeS OH

It is formed by SNNGP path

(II)

MeS Br

2S

KOHaq

N

MeS OHS

N2 path ]

Q.4

[Sol. Mass of neutron is greater than that of proton. ]

Q.5

[Sol. M(s) + O2

(g) I MO

2(s) S = a

M(l) + O2

(g) II MO

2(s) S = b

M(g) + O2(g)

III MO2(s) S = c

|a| < |b| < |c|

i.e.

M(g) + O2(g) MO

2(s) minimum entropy change

M(g) + O2(g) MO

2(s) maximum entropy change

We know that

G = H + TS

G = (S)T + H

y = mx + C

where y = G, m =S, x = T, C = H

M(s) + O2

(g) I MO

2(s) m

I=S =(a) = a

M(l) + O2

(g) II MO

2(s) m

II=S =(b) = b

M(g) + O2(g)

III MO2(s) m

III=S =(c) = c

|c| > |b| > |a| mIII

> mII

> mI


7/19

CHEMISTRY

Code - AA Page # 2

(C) III

III

G

(KJmol )1

T(C)

]

Q.6

[Sol. (a)

D H

H

HH

Cl

KOH.alc

)major(CHCHD 2 + CH2 = CH2

(b)

H

D

HH

Cl

H

KOH.alc

)major(CHCHD 2 + CH2 = CH2

There is free rotation along CC bond & CH bond fission take place more easily in comparison of

CD bond fission. ]

Paragraph for question nos. 7 to 9

[Sol.

CH3 CH3

O3Zu/H O2

CH3CH CH CH2 2

CH3CH CH CH = CCH2 2

(A) (B)

CH+

H O/H2 NaOH,

NaOH, CH2CH3

CH OH2

COONa

+

Aldol reaction

CannizarosReaction

Electrophilic Acidreaction

(C)

(D)

(E)

CH3CH CH CHCH2 2 2

CH3CH CH = CHCH2 2

CH3CH CH2

CH3CH CH = CCH2

OHCH3

CH3

CH3

CH3

Acid d-Base reactionEllimination reaction

H /

O /Zn, H O3 2

NaOH, NaOH,

Aldol reaction

+ CHCH2 (F)

CH3CH2CH = C

CH3O = CH

O O

O

O O

O

CH CH3 2 CH CH = CH2

Sol.7 In complete paragraph hetrolytic fission takes place so no any free radical reaction


8/19

CHEMISTRY

Code - AA Page # 3

Sol.8

OOO

O

CH CH3 2 CH CH = CH2

CH CH3 2 CH CHCH2

CH CH3 2 CH CH2 CH CH3 2 CH CH2

CH CH CH CH3 2 2

CH CH3 2 CH CH2

CH3

CH3

O3

O O

O

(A) (B) (C) (D)

O O

+

+

+CH CHCH3

CH3

CH3CH3

CH3

O O O

O OO OO O

CH CH CH CH CH CH2 2 3CH

Combination of A&B or C&D Combination of B&C Combination of A&D

Sol.9 So three ozonides

E & F both give aldol condensation. ]

Q.10

[Sol. Using phenolphthalein,

1 32CONa

M 10 = 2.5 5

1= 0.5

32CONaM = 0.05 M

Using methyl orange

2 32CONa

M 10 + 1 3NaHCO

M 10 = 7.5

5

1

2 0.05 10 + 103NaHCO

M = 1.5

3NaHCOM = 0.05 M

Mass of32CONa

M in 10 ml solution = 10 0.05 106 gm 103 = 0.053 gm

Mass of NaHCO3

in 10 ml solution = 10 0.05 84 103 gm = 0.042 gm ]

Q.12

[Sol. A2(g) 1

K2A (g)

B3(g) 2

K3 B (g)

r1

= k1[A

2]

r2

= k2[B

3] r

1= r

2 k

1= k

2

10RT

200005

RT

14000

T = K314.8

1200


9/19

CHEMISTRY

Code - AA Page # 4

Initial moles of A2

and B3

are same so

22 BAPP and Rate of disappearance of AA

2and B

3is also same

so1

P

P

timeanyatB

A

3

2

Pi= atm237.0

314.8

12000821.0

100

2

so P can't be less than 0.237 Total moles are increasing.

1P

P

B

A

Ans. (C) and (D) ]

PART-B

Q.1

[Sol. (A) 64OP

3

+ 6H2O 33POH43

(B) 4P0

+ 3NaOH (aq.) + 3H2O

3

PH3

+22

PONaH31

(Disproportionation reaction)

(C)33

POH43

3PH

3+ 33POH3

5

(Disproportionation reaction) ]

PART-C

Q.1

[Sol. AgBr Ag+ + Br

G = 80104(100)

= 76

G =2.303 RT logKP

1000 76 =2.303 PKlog4.1303.2314.8

7600314.8

log KP

=7600

4.1761000

log KSP =1.4 10logK

sp=14 or K

sp= 1014

Solubility = 107

Ans. = 7 ]


10/19

CHEMISTRY

Code - AA Page # 5

Q.2

[Sol. [Zn(gly)2] Td [M(AB)

2]

M

A

BAB

[Optically Active]

[Mn BrCl (PH3) (H

2O)] Td [Mabcd]

M

a

bcd

[Optically Active]

[Pt(NH3) (H

2O) FCl] Square planar [Mabcd]

M

a b

cd

[Optically inactive]

[RhCl(CO)(PPh3) (NH

3)] Square planar [Mabcd]

M

a b

cd[Optically inactive]

[Be4O(CH

3COO)

6] Symmetrical

or

[Be4O(OAC)

6] Be

OAC

Be

OAC

Be

OAC

OA

C

Be

OAC

OAC

O

[Fe(EDTA)] Oh Optically active

[Cr(NH3)3FCl Br] Oh [Ma

3bcd]


11/19

CHEMISTRY

Code - AA Page # 6

M

ac

d

[Optically active]

a

b

a

[Co(en)3]3+ Oh [M(AA)

3]3+

M

AA

A

[Optically active]

A

A

A

5 Molecules show optical isomerism. ]

Q.3

[Sol. (i) In presence of NaCl, Ist reaction shifts backward and IInd reaction shifts forwards. So solubility of

AgCl may increase or decrease.

(ii) SAgCl

= [Ag+] + [AgCl2]

= ]Cl[k]Cl[

k2

1

(iii) For minimum solubility

]Cl[ddS = 22

1 k]Cl[d

k = 0

[Cl] =08.0

102

k

k 10

2

1

= 5 105

31

2

2

]Cl[

k2

]Cl[d

Sd

> 0

(iv) If complex formation is not occurring, then only reaction 1 will occur and solubility will decrease as

[Cl] increases.

13 + 27 = 0040 Ans. ]


12/19

CHEMISTRY

Code - AA Page # 7

Q.4

[Sol.

Na CO

(aqueous)2 3

SrCl2

CaCl2

BaCl2

Hg(NO)2 3 2

HgCl2CuCl2

Pb(CHCOO)3 2

NHCl4

BaCO3(white)

HgCO3(Yellow)

HgCO 3HgO3(Reddish Brown)

Cu(OH) 2CuCO

(Blue)2 3

Pb(CH) 2PbCO

(White)

2 3

(NH ) CO

(Soluble)4 2 3

SrCO

(White)3

CaCO

(White)3

]

Q.5

[Sol. (a) C C = C C

C

(b)

(c) (d) CH2

= CH2

]


13/19

MATHEMATICS

Code-A Page # 1

PART-A

Q.1

[Sol. I = 2

2

12

1

dx1xx

xtan.......(1) put x =

t

1

I =

2

2

12

1

dt1tt

t

1

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

Now, (1) + (2) gives

2I =

2

2

122

2

3

2

1x

dx

2

I =

2

2

1

1

3

1x2tan3

24

=

0332

=36

2 A. Ans. ]

Q.2

[Sol. P(E) = P( R R W W B or R R R W W or W W W R R)

= !2!2

!5

53

1

+ !2!3

!5

53

1

+ 53!2!3

!5

Simplifying P(E) = 53

50 k = 50 Ans.]

Q.3

[Sol. )x(fLim

2x

=h

(sinh)sinLim

0h = 1 and )x(fLim

2x

= 1h

)1(sinLim

0h

.

So, )x(fLim2

x does not exist. Ans.]

Q.4

[Sol. 1, x, y G.P. Let x = k, y = k 2

and x, y, 3 A.P. 2y = x + 3 2k2 = k + 3

2k2 k 3 = 0 k = 1,2

3.


14/19

MATHEMATICS

Code-A Page # 2

x + y = k2 + k =4

1

2

1k

2

.

x + y|max

=4

15

4

14 at k =

2

3. Ans.]

Q.5

[Sol. As, ar.(ABC) = ar. (ACD) + ar. (BCD)

2

1(3) (4) =

2

1(4) (CD) sin 60 + 30sinCD3

2

1

A

CB

D

60

30

b=4

a=3

c=5

CD =334

24

=

39

)334(24 = )334(

13

8 . Ans.]

Q.6

[Sol. Possible cases :

(i) 0 0 0 0 0 3 !5!6

(when digit 0 comes at first place then number of arrangement =!4

!5)

Number of six digit numbers with 0 0 0 0 0 3 =!5

!6

!4

!5= 1

(ii) Similarly for 0 0 0 2 2 1

!2!2

!5

!2!3

!6= 30

(iii) 1 1 1 1 1 2 !5

!6= 6

Total = 1 + 30 + 6 = 37. Ans.]

Paragraph for question nos. 7 to 9

Sol.

(i) No American together

A1A

2A

3A

4B, C, D, E

For A1

we have four (B, C, D, E) favourable cases out of total cases 7.

Hence, probability =7

4(say) A

1and B are paired and the remaining

are, A2A

3A

4C D E

for A2, favourable cases 3 out of total cases 5

Hence, probability =5

3say AA

2and C are paired and the remaining are

Now, A3

A4

D, E


15/19

MATHEMATICS

Code-A Page # 3

for A3, favourable cases 2 out of total 3.

Hence probability =3

2.

P (No two Americans together) =35

8

3

2

5

3

7

4 (B)

(ii) P (delegates of the same country form both pairs)

A1A

2A

3A

4E, B, C, D

= 1151

73

[For A1

we have 3 favourable and of 7 and two A2

only 1 favourable out of 5 are for the remaining, no

constraints.]

EBCD

=35

3(B)

(iii) P (delegates of the same country not forming any pair

+ forming both pair + forming exactly one pair) = 1

P (forming exactly one pair) = 1 353

358 =

3524 (C). Ans.

Alternatively: A1, A

2, A

3, A

4, B, C, D, E

n(S) =!4)!2(

!84 = 105

(i) The probability that no two delegates of the same country are paired =105

CCC1

2

1

3

1

4

=

105

234=

35

8

(ii) The probability that delegates of the same country form two pairs =

105

33=

35

3

(As A1, A

2, A

3, A

4can be paired in 3 ways and B, C, D, E can be paired in 3 ways.)

(iii) The probability that exactly two delegates of the same country are paired together

=105

34C2

4 =

35

24Ans.]

Q.10

[Sol. We have

g(x) =

1,0xif,0

1,0x,1x

sin1xx

sinx22

Clearly, g(x) is differentiable x R.(As sum and product of differentiable functions is also differentiable function.)

Now, g '(x) =

1,0x,0

1,0x,1x

sin)1x(21x

cosx

sinx2x

cos


16/19

MATHEMATICS

Code-A Page # 4

Clearly, g'(x) is discontinuous at both x = 0 and x = 1.

Also, g(0) = 0 = g(1) (Given)

So, Rolle's theorem is applicable for g(x) in [0, 1].

Q.11

[Sol.

(A) P(A/B) = P(B/A) P(A) = P(B)Now, P(A B) = P(A) + P(B) P(A B) P(A B) = P(A) + P(B) P(A B)

P(A B) = 2P(A) 1 > 0; P(A) >21 True

(B) P(B) =4

3; P(A/ B) =

2

1

)B(P

)BA(P =

2

1 P(A B) =

4

3

2

1=

8

3

Now, P(A) + P(B) P(A B)= P(A B) 1

P(A) +4

3

8

3 1

P(A) 8

5

P(A)]max.

=8

5 True

(C) P(AC BC)C = C4443)aa()aa( = (a4)

C = a1

+ a2

+ a3

= P(A + B)

P(AC BC) = (a1

+ a3

+ a4)C = a

2= P(AB)

Hence P(AC BC) + P(AC BC) = P(A) + P(B)] =6

23

3

1

2

1 =

6

5.

(D) Given, A is subset of B.

Now, P(B/A) =)A(P

)AB(P =

)A(P

)A(P= 1.

S

A

B

Q.12

[Sol. A vector coplanar with kji2a

, kjib

and perpendicular to k6j2i5c

will be

along cba

= acbbca

= kji29kji18 = kj39k9j27

Vector will be along k

j

3 .This vector will lie in the plane which will be parallel to it.

Normal of plane will be perpendicular to vector. Ans.]

Q.13

[Sol. Equating the coefficient of x20, we get

220 a20 = 1 20 20 12a (C)


17/19

MATHEMATICS

Code-A Page # 5

put, x =2

1, we get

0 1020

q2

p

4

1b

2

a

0q2

p

4

1b

2

a1020

b2

a and 0q

2

p

4

1

a + 2b = 0 (B)

b =2

1220 20

.

put, x = 0 we get

1 b20 = q10

1

2020 20

2

12

= q

10

10

20

20

q2

121

10

20q

2

1 q =

4

1

Using, 0q2

p

4

1

04

1

2

p

4

1

p =

1

Hence B, C, D. Ans.]

PART-B

Q.1

[Sol.

(A) We know that | adj A | = | A |2 for a 3 3 matrix

Given adj A = KAT |adj A| = |KAT| = K3 | A | (|AT| = | A | ) K3 | A | = | A |2

K3 = | A | ; Now det A = 1 (1 4) 2(2 4) + 2 (4 + 2) = 27 k3 = 27K = 3.

(B) Given, f (x) = (2x + 1)50 (3x 4)60

f ' (x) = 220(2x + 1)48 (3x 4)58 (2x + 1)(3x 4)(3x 1)

1

2

1

3

4

3

+ +

Sign scheme of f '(x)

Least positive integer is k = 2


18/19

MATHEMATICS

Code-A Page # 6

(C) By applying continuity and differentiability at x = 1, we get a = 3, b =1.

Hence, (2a + b) = 2 (3) 1 = 5. Ans.]

PART-C

Q.1

[Sol. Line L1

is parallel to vector jin1

and line L2

is parallel to vector k2j2in2

. As, normal of

plane is perpendicular to both lines L1

and L2

, so n

= normal vector of plane

=211

011

kji

nn21

= j2i2 .P(x,y,z)

N(1,1,5)

n = 2i 2j ^ ^

So, equation of plane is 2(x 1) 2(y + 1) + 0(z 5) = 0 x + y = 0 ........(1)

So, distance of plane from 0,8,2M =2

82 = 3. Ans.]

Q.2 Let f(x) be a differentiable function satisfying

x

0

x

0

dtxttandtttan)t(f)x(f = 0

2

x .

Find the number of solutions of the equation f(x) = 0.

[Ans. 1]

[Sol. x

0

x

0 KingUsing

dtxttandtttan)t(f)x(f

= 0

x

0

x

0

dtttandtttan)t(f)x(f = 0

Differentiate w.r.t. x

f '(x) + 1)x(f tan x = 0

1)x(f

)x('f

+ tan x = 0

integrate w.r.t. x

ln 1)x(f + ln (sec x) = C

f(0) = 0 C = 0.

Hence

1xcos

1)x(f

f(x) = cos x 1 f(x) = 0 cos x = 1

Hence only solution in

2

,2

is x = 0.

Hence number of solution is 1.

Note that domain of f(x) is

2

,2

Ans.]


19/19

MATHEMATICS

C d A P # 7

Q.3

[Sol. We have, f '' (x) = 12x 4

f ' (x) = 6x24x + cAs, f ' (1) = 0 c = 2 f ' (x) = 6x24x 2 f (x) = 2x32x22x + As, f (1) = 0 = 2Hence, f (x) = 2(x3x2x + 1)

or f (x) = 2(x

1)2

(x + 1).Now, M (x = 2, y = 6) and f ' (2) = 14.

So, the equation of normal at M is (y 6) =14

1(x 2)

For x-intercept, put y = 0

we get x = 86 Ans.]

Q.4

[Sol. Let O be the centre of polygon

Area of rectangle = 4 OA1A

2= 6 .........(1)

and Area OA1A2 = n1 area of polygon .........(2) A

1

A2

Ak + 1

Ak

O

(1) and (2)4

6= 60

n

1

n = 40. Ans.]

Q.5

[Sol. Using tan1 + tan1 + tan1 = tan1

1

In L.H.S. we get

tan1 2)c)(bx()bx(ax)ax(c1

bxaxcbxaxc

where c =

8

x

x

1.

Now, 1 = ax

8

x

x

1+ abx2 + bx

8

x

x

1 1 = 222 x

8

bbabxx

8

aa x R

1 = (a + b) + x2

8

ba

ab

On comparing, we geta + b = 1 ..........(1)

and ab =8

ba =

8

1.......(2) (Using (1))

Now, a2 + b2 + 2ab = 1 a2 + b2 +4

1= 1

Hence, 4(a2 + b2) = 3. Ans.]

Rt Solutions-06!11!2011 XII ABCD Paper II Code A

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